# Thread: Photon decay when bending in a gravitation field

1. A photon is influenced of all gravitation gtot from every direction and the proportion of time * acceleration =v bend in the direction of strongest sum of field. This mean the deviation is much bigger than a hypothetic materia particle, even tough the magnitude is same. Since all colors bend equally (no dispersion), it follows that bending use a proportion of the photons energy (color). When passing a celestial body (ies) a perpendicular speed v=(sigma)g*t iis added depending on a body of gravitational constant µ and closest fly by distance r as follows v=2*µ/(c*r) where c is speed of light. The angle of curving (from - infinity to +infinity) a=arcsin(v/(c-v)). Electromagnetic energy transformed to kinetic momentum E*c is (delta)E/E=2*v/c where E=c*hi/(lambda). hi is my (indifferent) Planck constant 6,22*10^-34 Js and (lambda) the wavelength (color). The "lost" energy(decay) (delta)E=4*µ*hi/(c*r*(lambda)). The energy is "lost" quant by quant n*hi=2*v/(lambda), and the bend is stepped n steps, for weak gravity/long wavelength even distinguishable. Total redshift z is the sum of 4*µ/(c^2*r) regardless of direction.
Calculation ex. The volume of a r=13,5*10^9 ly universe can fit 2,2*10^12 galaxies 3,3*10^6 ly apart. Looking trough a r=13*10^9 ly and 13*10^9 ly long cylinder , it is having 1,06*10^12 galaxies of 1,2*10^31 m3/s2 (90 billion sun masses ). Randomly dispersed average distance from centerline is 8,2*10^25m giving average of 500 simulated z=8,6*10¨-12 times 1,06*10^12 give redshift about 9. Reminding me of something I've read. The distances r,l,d are assumed but their ratio hold.
For shorter stints or distances s=c/t the redshift is z=gtot*t/c where gtot is sum of all gravitation from all directions. Using 21.century celestial masses Mi that are 1,13 times mainstream masses and mn mass of nucleon 1,671*10^-27kg
Formulas can be written (delta)E/E=Mi*mn*c*t/(2*r¨2) and (delta)E/t=Mi*mn*c^2/(2*r¨2*(lambda)) =gtot*hi/(lambda). Close to planets gtot values are very different from classic g (mainstream). Here a few picked values from earth altitudes for using in calculating photon decay onboard satellites.
Altitude over sea level 0km 15,92m/s2 10km 15,82m/s2 20km15,73m/s2 50km15,48m/s2 100km15,11m/s2 200km14,45m/s2 300km13,86m/s2 400km13,32m/s2 500km12,82m/s2 1000km10,77m/s2 2000km8,02m/s2 5000km4,12m/s2 10000km1,936m/s2 20000km0,736m/s2 30000km0,385m/s2. For higher altitudes gtot=1,27*classic g +0,0075m/s2 for sun.
For the moon I have not calculated, but rough comparisons surprised me by suggesting proportionally same size heavy core as earths. This would mean that moon is a miniature of earth with high altitude g=1,27*classic.

2.

3. Light also bends in an accelerated frame of reference.

4. Originally Posted by KJW
Light also bends in an accelerated frame of reference.
I have left that out (ignored) and only calculated mass induced bending and the redshift resulting from that, via the "spent" energy (decay). This approach is by my account not regarded earlier. And its magnitude is in the same ballpark as redshift due to expansion of universe. (Or even alternative to that)

5. Originally Posted by Timo Moilanen
Originally Posted by KJW
Light also bends in an accelerated frame of reference.
I have left that out (ignored) and only calculated mass induced bending and the redshift resulting from that, via the "spent" energy (decay). This approach is by my account not regarded earlier. And its magnitude is in the same ballpark as redshift due to expansion of universe. (Or even alternative to that)
It should be noted that with light bending in an accelerated frame of reference, we have light bending without any photon "decay". That is, bending light does not necessitate photon "decay", so that the bending of light in a gravitational field does not necessitate photon "decay" either. In an accelerated frame of reference, we still have redshift and blueshift just as we have in the case of gravity, and the redshift and blueshift determine the acceleration of the frame of reference. Thus, redshift and blueshift in the case of gravity determine the acceleration of the gravitational frame of reference. That being in a gravitational field is the same as being in an accelerated frame of reference is the equivalence principle.

6. Originally Posted by KJW
Originally Posted by Timo Moilanen
Originally Posted by KJW
Light also bends in an accelerated frame of reference.
I have left that out (ignored) and only calculated mass induced bending and the redshift resulting from that, via the "spent" energy (decay). This approach is by my account not regarded earlier. And its magnitude is in the same ballpark as redshift due to expansion of universe. (Or even alternative to that)
It should be noted that with light bending in an accelerated frame of reference, we have light bending without any photon "decay". That is, bending light does not necessitate photon "decay", so that the bending of light in a gravitational field does not necessitate photon "decay" either. In an accelerated frame of reference, we still have redshift and blueshift just as we have in the case of gravity, and the redshift and blueshift determine the acceleration of the frame of reference. Thus, redshift and blueshift in the case of gravity determine the acceleration of the gravitational frame of reference. That being in a gravitational field is the same as being in an accelerated frame of reference is the equivalence principle.
I say that gravity cause only and always redshift (blueshift due to gravity is impossible ), and photons react to all gravity from every direction at the same time. This total gravity (mainstream (sigma)M*G)*4/pi (my physics) all ad to both bending and redshift. My formulas build on these "facts" and the numbers ad up, which also means the universe do NOT expand
Yes I have read the text books and understand what they try to teach. I just don't believe it, and even less since I found the missing gravity. Black holes is as far as mass can "hide", rest is coming to take it's place in history as fiction.

7. Originally Posted by Timo Moilanen
I say that gravity cause only and always redshift (blueshift due to gravity is impossible), and photons react to all gravity from every direction at the same time.
This is false. Otherwise, we would've detected this with our interferometers.

Note that in general relativity, in a stationary spacetime (a three-dimensional space that does not change with time), light that reflects off a mirror and returns to the source does not undergo an overall change in frequency. Even if the light is redshifted or blueshifted at the mirror, this redshift or blueshift will be undone upon return to the source. Thus, in this case, interferometry, which depends on mixing light from different paths at the same location, will not detect a change in frequency.

Originally Posted by Timo Moilanen
I have read the textbooks and understand what they try to teach.
Do you? In order for you to have any chance in convincing us, you need to demonstrate your understanding of general relativity. You can do this by telling us what specifically is wrong with general relativity. Since your theory conflicts with the equivalence principle, you could probably start with that.

8. This is false. Otherwise, we would've detected this with our interferometers.
First all electromagnetic waves are not photons ( just oscillating fields perpendicular to original "signal"). The Changes are small, but this is exactly what Pound and Rebka is considered to have proven. Except they measured z=2,5*10^-15 and my theory give gtot/c^2*22,5m=3,98*10^-15. This for 22,5 m "flight" and is hardly measured (very little) by accident in other experiments.
I wonder who you are demanding me to convince you. It may just be a waste of time to entertain your whatever vagary.
As for the equivalence principle, the weak feather hammer is obvious, but there is a (in large scale) geometrical aspect to it. Remember I have said the shell theory don't work, and the math to correct those flaws are the basis of my grav. theory.
The strong equivalence principle is of course problematic to measure since the total gravitation (essential to at least photons) has not "been found" till i did. Compare the gravitation on sea level g=9,81m/s2 vs. gtot=15,9m/s2. That is why I picked a few examples at altitude, and I must stress that all fields beyond near surface of stellar bodies are 4/pi times stronger than mainstream.
About (Einstein's) equiv.pr. I have no sentiment to say intact atoms should react to grav. (proton/electron mass, photon original color and some other constants and ratios )
I have different masses different gravitation constant so I pretty much just have to ignore the gravitation part of the theory of relativity.

9. Originally Posted by Timo Moilanen
First all electromagnetic waves are not photons
A decrease in the energy of photons is equivalent to a decrease in the frequency of electromagnetic waves. So, if your theory is claiming that gravity causes a decrease in the energy of photons, then it would be detected by interferometry as a decrease in frequency. But you are claiming that gravity causes a redshift which is a decrease in frequency anyway, so why are you saying that electromagnetic waves are not photons?

Originally Posted by Timo Moilanen
The Changes are small, but this is exactly what Pound and Rebka is considered to have proven. Except they measured z=2,5*10^-15 and my theory give gtot/c^2*22,5m=3,98*10^-15. This for 22,5 m "flight" and is hardly measured (very little) by accident in other experiments.
The Pound-Rebka experiment was not measured using interferometry. It was measured using the principles of Mössbauer spectroscopy. As I said in my previous post, interferometry would not detect the redshift expected in the Pound-Rebka experiment, but it would detect the redshift that you are claiming to be caused by gravity. It's not a question of magnitude, interferometry and Mössbauer spectroscopy are entirely different, so the Pound-Rebka experiment is irrelevant to the notion that interferometry would invalidate your theory.

Originally Posted by Timo Moilanen
I wonder who you are demanding me to convince you.
I am the one who is replying to your posts. I didn't say convince "me", I said convince "us", referring to all the people looking at your posts on this forum. If it is not your purpose to convince us that what you are saying is worthwhile, then why are you posting here? And if you are not interested in demonstrating the value of your theory, then the moderators will place it in the Trash Can.

Originally Posted by Timo Moilanen
Remember I have said the shell theory don't work, and the math to correct those flaws are the basis of my grav. theory.
In general relativity, one has Birkhoff's theorem. I personally have not seen the proof of Birkhoff's theorem, but I have seen the derivation of the Schwarzschild metric, which I believe implies Birkhoff's theorem. This is because obtaining the Schwarzschild metric involves solving a first-order ordinary differential equation in the radial variable. This differential equation, like all differential equations, describes local behaviour, and the solution over a region is obtained by specifying the solution on a spherical surface (a given value of the radial variable). Thus, the solution outside this spherical surface is determined entirely by the solution on the spherical surface, which in turn is determined by what is inside the spherical surface as a single-value parameter. If we have a matter distribution consisting of concentric spherical shells, then the regions between the shells are also described by Schwarzschild metrics, with each region described by a different mass parameter.

Originally Posted by Timo Moilanen
I have different masses different gravitation constant so I pretty much just have to ignore the gravitation part of the theory of relativity.
Based on your theory, it appears to me that you don't really understand general relativity. For example, can you provide a general relativistic explanation of why objects fall when dropped? Whether you agree with general relativity or not, having an understanding of general relativity is important because it goes toward credibility. You're trying to provide an alternative theory of gravity, so you need to understand the theory it is alternative to. And if you don't demonstrate your understanding of general relativity, then your theory itself will be taken as a lack of understanding of general relativity.

10. so why are you saying that electromagnetic waves are not photons?
I'm saying not all are, most probably are when counting how may. Then there are those propagating 90 degrees compared to photons.
I am sure metrological interferometers would be able in principle to do Pound Rebkas measurement. About the astronomical interferometers I cant guess what kind you might refer to, and they are not my trade anyhow. So I could need some good faith specifier to maybe find out what sort of interferometer at what kind of experiment invalidate what exactly?

I try to answer to all comments and have done till now, and if I'm not agreeing with every mainstream detail, this would not be a very new idea. Threatening do not help and this time do absolutely not change my theory. I'm not into every trade so "questions" could be like everyone can understand them.
I'm modifying Newtons gravity, and a object drops for the same reason or reasons if the theories give same result. I am not going to copy textbooks here to prove NOTING. And it should be taken as relative theory do not work and can in future not be coupled to proper gravitational theories.
Just come to my mind we are here to talk about new (own) theories, that's why I'm not writing this in Physics Forum.

11. Originally Posted by Timo Moilanen
I am sure metrological interferometers would be able in principle to do Pound Rebkas measurement. About the astronomical interferometers I cant guess what kind you might refer to, and they are not my trade anyhow. So I could need some good faith specifier to maybe find out what sort of interferometer at what kind of experiment invalidate what exactly?
I didn't have a particular interferometer or even a particular experiment in mind when I suggested that they would invalidate your theory. You seem to be saying that light decreases in frequency as it spends time in a gravitational field. If that were true, then any experiment in which a laser beam is split into two paths of different length then recombined would detect the change in frequency. But for a stationary spacetime in general relativity, the gravitational potential is a spacetime function and therefore the two paths of the laser beam will recombine at the same frequency. Because such experiments would be very sensitive to changes in frequency, it is reasonable to assume that changes in frequency have not been detected even if the experiments were not designed to test for changes in frequency.

Originally Posted by Timo Moilanen
I'm not agreeing with every mainstream detail, this would not be a very new idea.
No one is expecting you to agree with every mainstream detail. I also disagree with the mainstream regarding particular details. For example, I disagree with the mainstream regarding the emission of gravitational radiation from a rotating arbitrarily shaped rigid object.

Originally Posted by Timo Moilanen
I'm modifying Newtons gravity
I thought that to be the case because it seemed to me that you have a somewhat Newtonian view of gravity. However, you are modifying Newtonian gravity in a way that disagrees with general relativity. In fact, your theory disagrees with general relativity worse than Newtonian theory.

Originally Posted by Timo Moilanen
an object drops for the same reason or reasons if the theories give same result.
That's not true. Newtonian theory and general relativity have a very different reason for why a dropped object falls. But this difference does not itself lead to different results. But it does turn out that Newtonian theory leads to a contradiction when applied to the Pound-Rebka experiment, even though it does give the same result.

Originally Posted by Timo Moilanen
I am not going to copy textbooks here
Nobody wants you to copy textbooks here. You were being asked to provide your understanding of general relativity.

Originally Posted by Timo Moilanen
And it should be taken as relative theory do not work and can in future not be coupled to proper gravitational theories.
You need to justify this statement.

Originally Posted by Timo Moilanen
Just come to my mind we are here to talk about new (own) theories, that's why I'm not writing this in Physics Forum.
Even though you are posting in the Personal Theories & Alternative Ideas forum, you still have to provide justification. Also, in discussing your theory with you, I am under no obligation to agree with it, and if I find that your theory is somehow flawed, I am within my right to say so. Yes, the Personal Theories & Alternative Ideas forum does allow for less rigor on your part, but it doesn't protect what you post from scrutiny from other forum members.

12. Since you refer to "if made" experiments I am sure that it would be found that light ,sea level earth, is losing energy E*1/1,79*10^-16 for every traveled meter in any direction if measured .The experiment would maybe need to be out of the box like Pound and Rebka:s experiment. That by the way will work horizontally in my favor.
Yes I disagree more with general relativity than Newton, even if Einstein was aware of shortcoming with dimensional masses. Which always pointed out by mentioning the validity is for point like masses.
I meant that there can be several correct and valid theories on the same phenomenon. In this case we have two (Newton,Einstein)
that give somewhat useful calculations to a part of necessary implementations.
About your "obsession" with my understanding of a falling ball. As I understand it the ball is following a geodesic line, induced by earths mass (a very hypothetic line from an equivalently so equation) and do not do any work, do not accelerate (at least not in that frame) and would not "feel" anything if undisturbed by air resistance or such. An besides a still observer would have to work while observing P=0,5W*9,81m/s2^2*m(weight of observer). The other half watt is done by the floor.
This statement is my personal knowledge (opinion) and both living and dead, scientists or not, have to "live" with that. This is a discussion forum (by my understanding) and personal floccinaucinihilipificaution is plainly uncivilized even on this informal forum.
And criticism is kind of the idea, but I hope for more discussion that just "no", since the math isn't complex this time.

13. Originally Posted by Timo Moilanen
Since you refer to "if made" experiments I am sure that it would be found that light ,sea level earth, is losing energy E*1/1,79*10^-16 for every traveled meter in any direction if measured .The experiment would maybe need to be out of the box like Pound and Rebka:s experiment. That by the way will work horizontally in my favor.
Interferometry is extremely sensitive to changes in relative phase between the light paths. A difference in frequency results in a constantly changing relative phase and I think if such a frequency difference existed in interferometry experiments, that it would already have been detected, even if the experiments weren't designed to measure the difference in frequency.

Originally Posted by Timo Moilanen
I meant that there can be several correct and valid theories on the same phenomenon. In this case we have two (Newton,Einstein) that give somewhat useful calculations to a part of necessary implementations.
That depends on what you mean by "correct and valid". For example, Newton's theory predicts results that are quite close to observation. Because Newton's theory was based on observation in the first place, it is only natural that it would be quite close to observation. Also, the difference between observation and the simplicity of Newton's theory is fortuitously quite small, so Newton's theory remains quite useful where its simplicity is a real advantage, and the accuracy of Einstein's theory is not needed.

On the other hand, from an ontological and epistemological perspective, Newton's theory is wrong. Newton's theory does not describe what gravity actually is. Newton's theory describes gravity as a force, whereas gravity is not a force but an accelerated frame of reference along with the associated time dilation, as described by Einstein's theory. Because Newton's theory is based on observation, it lacks the insight required to make predictions that go beyond observation. By contrast, Einstein's theory has a logical and mathematical basis, and therefore its scope is not limited to mere observation.

Originally Posted by Timo Moilanen
As I understand it the ball is following a geodesic line, induced by earths mass (a very hypothetic line from an equivalently so equation) and do not do any work, do not accelerate (at least not in that frame) and would not "feel" anything if undisturbed by air resistance or such. An besides a still observer would have to work while observing P=0,5W*9,81m/s2^2*m(weight of observer). The other half watt is done by the floor.
Are you trying to fake an understanding of general relativity? For example, you said that falling objects:

do not accelerate (at least not in that frame)

whereas an object that is freefalling in a vacuum is not accelerating in every coordinate system. This is such an important point that failing to understand it can be considered a failure to understand general relativity. Do you know what acceleration is in general relativity?

And this:

An besides a still observer would have to work while observing P=0,5W*9,81m/s2^2*m(weight of observer). The other half watt is done by the floor.

is just plain wrong and suggests you are mixing Newtionian notions with general relativity.

Originally Posted by Timo Moilanen
This is a discussion forum (by my understanding) and personal floccinaucinihilipificaution is plainly uncivilized even on this informal forum.
Sorry, but you are making claims that conflict with a theory that (1) has passed every test that has been put to it, and (2) has a sound logical foundation. It is only natural that I start from a position of assuming you are wrong. And because you made a prediction that I believe would have already failed to be observed, my initial assumption seems to be validated. Also, you haven't really provided any basis for your theory, or any basis for the belief that general relativity is wrong. You said that you have read the textbooks and understand what they try to teach, which I regard as an attempt to elevate your level of authority concerning accepted physics. Anybody can say they are knowledgeable in an effort to convince others of the value of what they say. That is why I felt it necessary to put your knowledge of general relativity to the test.

Originally Posted by Timo Moilanen
And criticism is kind of the idea, but I hope for more discussion that just "no"
But you have gotten more discussion than just "no". I don't know what you're expecting, but I have mentioned how interferometry may have already invalidated your theory. And I have asked you to justify your claims, in particular your claim that general relativity is wrong. Personally, I see no value in asking you to justify your theory itself, so I focus on where your theory conflicts with accepted physics. It seems to be a failure of many people putting forward alternative theories that they expect physicists to directly test their theories without realising that their theories may have already been indirectly tested by the tests of the accepted theories they seek to replace.

Originally Posted by Timo Moilanen
the math isn't complex this time.
The mathematics is not well presented and is difficult to read. Each formula should be on a line on its own and have some form of explanation of why the formula is what it is. If a formula uses superscripts or subscripts, then you should use the corresponding BB codes. Unfortunately, this forum has stopped supporting LaTeX.

14. The best (most suitable) interferometers I "found" are LIGO (4km) and Virgo that should be run at 1,5THz /n where n<10 for this magnitude of (delta)f.
I meant that for example same constants can be determined by different paths, probably wishful thinking of me. I cant remind myself of any correct.
I have not invented any "out of this world" theory and it is based on done measurements (wrongly interpreted are the cornerstone). I only use classic time as unit mostly. Neither do I go into quantum physics, smallest is the nucleon that I use instead of mass (1neutron+1proton+1electron=2) and gravity comes from n1(nucleons)*n2(nucleons)*constant/geometry. Gravity for orbits is not linear ever, for light it is always linear.
Since I don't understand gravity or was it equivalence principle, could you please explain it to me coz I'm obviously a bad reader (that is actually a problem for many years now). Then I can pretend smart next time.
I can prove G is wrong and not constant. So masses are also substantially of. But by measuring for instance the Standard Gravitational Parameter are put "right". The "proper formula for µ =MG is µ=mn*pi/8*c^2*Mi (indexed M=correct mass) So you will no have a facile task to "defend mankind" from me. One useful thing that I have calculated that is unmeasured is earths and Mars:s gravitation profiles per altitude. ( This where I'm called flat earther ) Well they have smashed landers all over Mars, learned the hard way. It is good NASA is cautious about the moon they do not yet have enough practice to replace physics as on earth that is much simpler (The moon is messing the orbit the "right" way and gravity is compensated with fuel)
I have not written "almost any" math about gravitation this thread I planned to discuss light.
I still would be thankful for some short clarification on the falling ball, I'm probably unable to handle relativity theory mainstream unviolated.

15. Originally Posted by Timo Moilanen
I have not invented any "out of this world" theory
You're right, "tired light" explanations of cosmological redshift are not new. However, they have been disproven by observation. For example, the Tolman surface brightness test rules out "tired light" explanations. Note that time dilation not only decreases the frequency of the light, but also decreases the rate of photon emission, unlike "tired light". One thing that I realised is that redshifts caused by "tired light" cannot have the change in frequency directly proportional to the frequency. That is because a change in frequency directly proportional to the frequency proves that the redshift is a time dilation. That is, if the change in frequency is directly proportional to the frequency, then the rate at which mechanical clocks are observed to tick will change proportionally by the same amount.

Originally Posted by Timo Moilanen
Since I don't understand gravity or was it equivalence principle, could you please explain it to me
Consider the following diagram:

The rocket is being observed from an inertial frame of reference, and in this frame of reference, it is seen to be accelerating upward. When the ball is dropped, it remains motionless relative to the inertial frame of reference. But relative to an observer in the rocket (not shown), the ball appears to accelerate downward in the same way as it does on Earth. In the rocket, the observer observes a spacetime metric corresponding to observing a Minkowskian spacetime from an accelerated frame of reference. On Earth, the spacetime curvature associated with the energy-momentum density of the earth induces a curvature of the spacetime surrounding the earth. Locally, the spacetime metric observed by an observer on Earth is more or less the same as the spacetime metric observed by an observer in the rocket. Thus, just like the observer in the rocket, the observer on Earth is accelerating upward. However, over an extended region of spacetime around the earth, the spacetime metric does differ from that observed in the rocket, and this difference is due to the spacetime curvature around the earth, but zero in the rocket. The spacetime curvature around the earth produces the tidal effect, which does not exist in the rocket. Thus, the equivalence between gravitation and acceleration applies only over distances small enough that the tidal effect can be ignored.

16. "Tired light" have no solid (or other ) basis whereas I have an exact physical connection to mass and distance, but is not time dilation, and I avoid time as parameter (especially relativistic ). Since you explained ( copy/pasted ) the "tidal effect exempt" falling ball better than I could have written, even in Swedish, could you help me with the relativistic bending of light at distance. I compare whit all available but here I can find no verbal reference. The formulas in this case are not very good for the why and the result for suns surface grazing are exactly the same. That can not easily be an accident. Please you have the experience ( I admit no more for mow ).
For the Tolman surface brightness test they got factors 2,6 and 3,4 that I think is much of 4 for the (1+z)^4 that brightness must decrease ( and newer come back ). OK they introduced a brightness "resurrecting" theory ( Bruzal et Charlot ) to save the day, yet a patch plus the earlier exception. My theory on the other hand is readily fit. Since I do not consider time dilation but a phenomenon that has a greater effect (small that it is) on a ( particle ) photon than equivalent gravity have on a mass particle, a mechanical clock has nothing to do with it.
Simply put, photon decay in a gravitational field is not a relativistic occurrence.

17. Originally Posted by Timo Moilanen
"Tired light" have no solid (or other ) basis whereas I have an exact physical connection to mass and distance, but is not time dilation
As I see it, any explanation of cosmological redshift that isn't time dilation is "tired light". I didn't mention any particular mechanism for the "tired light", and the tests that I see as ruling out "tired light" are tests that rule out any explanation that isn't time dilation. For example, the Tolman surface brightness test distinguishes time dilation from other explanations of cosmological redshift, whereas I didn't mention tests of "tired light" explanations based on scattering because I assumed that your explanation isn't based on scattering.

Originally Posted by Timo Moilanen
For the Tolman surface brightness test they got factors 2,6 and 3,4 that I think is much of 4 for the (1+z)^4 that brightness must decrease ( and newer come back ). OK they introduced a brightness "resurrecting" theory ( Bruzal et Charlot ) to save the day, yet a patch plus the earlier exception. My theory on the other hand is readily fit.
I should point out that the (1+z)^4 is for a simple expanding universe. The observed exponent values of 2.6 or 3.4 are still for an expanding universe, just not a simple expanding universe. It still rules out explanations of cosmological redshift that aren't time dilation, for which the exponent value is zero. Perhaps you can explain how your theory satisfies the Tolman surface brightness test without time dilation.

Originally Posted by Timo Moilanen
Since I do not consider time dilation but a phenomenon that has a greater effect (small that it is) on a ( particle ) photon than equivalent gravity have on a mass particle, a mechanical clock has nothing to do with it.
This seems to be a misunderstanding of the point I was making. What I am saying is that if the decrease in frequency is directly proportional to the frequency, then the redshift must be a time dilation. The observed slowing of the ticking of a mechanical clock that occurs if the decrease in frequency is directly proportional to the frequency is indicative that we are dealing with time and not simply the energy of photons. I chose a mechanical clock because that unequivocally represents the measurement of time. You're saying that the redshift is a loss of photon energy but not a time dilation. In that case, the mechanical clock would appear redder, but not tick any slower, and therefore the decrease in frequency cannot be directly proportional to the frequency.

Originally Posted by Timo Moilanen
Since you explained ( copy/pasted ) the "tidal effect exempt" falling ball better than I could have written, even in Swedish, could you help me with the relativistic bending of light at distance.
My explanation of the falling ball was not copy/pasted. The image came from Google but the text is all my own. As for helping you with the relativistic bending of light, that is too complicated for me to do in a post on this forum. However, general relativity predicted the correct value of 1.75 arcseconds for light that grazes the Sun, which is double the value predicted by Newton's theory.

18. There actually are a small scattering (indirectly caused by gravity and its strength) that makes it "useful", but even those "stray photons" are red shifted.
"Favoring" the relativity theory do not rule out gravitational redshift, to which I have added it do not go back blue again.The Tolman test have redshift as base line and (1+z) is base factor value=1, it is just assumed to be "doppler" expansion redshift.
My theory do not satisfy the Tolman test since the redshift depends on the gravity fields passed arriving to observation. The redshifts are redder from far away (more passed masses) but no distance measure.
I'm lazy to be exact, but this time dE/E=df/f, and the photon lose a proportion of its energy ( "same percentage for all colors" ). And no tired quants self destruct. The clock would look red only "with a good telescope" as it is beside the light source. I do not consider "time dilation" speed red shift, there is not the speed. The universe is not expanding, remember?
The bend at distance with Einstein's formula should be simply (delta)(phi)=4M/(b/r0*rx). Where r0 is the r used to calculate b and rx the new wished distance. As for mine I left r "open"(delta)(phi)=arcsin(4µ/(c^2*r). Both have ignored that r should be a little longer to counter the deflection in the "rounding", or as in wiki " r when deflection absent ". It can be necessary to consider near dense bodies ("small" black holes).
I used Newtonian mechanics except how the photon reacts.

19. Originally Posted by Timo Moilanen
The Tolman test have redshift as base line and (1+z) is base factor value=1, it is just assumed to be "doppler" expansion redshift.
It actually occurred to me some time after I had posted that I may have made an error with regards to the zero exponent. I was thinking of "brightness" in terms of the number of photons whereas in the Tolman test, it is in terms of energy, which decreases due to the redshift. Thus, the exponent for "tired light" would be 1, not 0.

Originally Posted by Timo Moilanen
There actually are a small scattering (indirectly caused by gravity and its strength) that makes it "useful", but even those "stray photons" are red shifted.
Scattering leads to the blurring of images, which is not observed, and rules out "tired light" mechanisms based on scattering.

Originally Posted by Timo Moilanen
the photon lose a proportion of its energy ( "same percentage for all colors" )
In that case, the redshift must be a time dilation.

Originally Posted by Timo Moilanen
And no tired quants self destruct
This was assumed to be the case. It means that in the absence of time dilation, the photons are received at the same rate as they were emitted at the source. This enables a redshift based on time dilation to be distinguished from a redshift not based on time dilation.

Originally Posted by Timo Moilanen
The clock would look red only "with a good telescope" as it is beside the light source.
Well, obviously one would need a good telescope to see all those mechanical clocks across the universe. But even though a mechanical clock is not emitting light, the light that it reflects and which we see is still subject to the redshift whatever its cause. A mechanical clock behaves as an amplitude modulator of the light it is reflecting, and even though the source of that light may be monochromatic, the reflected light itself contains multiple frequency components due to the modulation, based on the trigonometric identities such as:

cos(A) cos(B) = ½ cos(A–B) + ½ cos(A+B)

Here, A is assumed to be the frequency of the source light and B is assumed to be the amplitude modulation frequency that is the rate at which the mechanical clock ticks. A–B and A+B are the frequencies subjected to the redshift. It should be noted that there are no photons corresponding to the modulation frequency, and therefore there is no energy loss at the modulation frequency. But if the energy loss of all the photons is directly proportional to the energy of the photons, then both A–B and A+B will be proportionally decreased by the same amount. This results in both A and B being proportionally decreased by the same amount. That is, the modulation frequency, the tick rate of the clock, has proportionally decreased by the same amount. This is time dilation.

Originally Posted by Timo Moilanen
I do not consider "time dilation" speed red shift, there is not the speed. The universe is not expanding, remember?
All the observational evidence indicates that the universe is expanding, and that your theory cannot correctly say otherwise.

20. It actually occurred to me some time after I had posted that I may have made an error with regards to the zero exponent. I was thinking of "brightness" in terms of the number of photons whereas in the Tolman test, it is in terms of energy, which decreases due to the redshift. Thus, the exponent for "tired light" would be 1, not 0.

A absolute minimum factor (besides the "whished 4) for a assumed redshift by speed, is 2 (redshift +incoming photon count), that make a number 2,6 seem like "lame".
Since I have coupled irreversible redshift to gravitation and will ad scattering, I'm going to insist my theory to be called
"Permanent Gravitational Redshift"

Scattering leads to the blurring of images, which is not observed, and rules out "tired light" mechanisms based on scattering.
Scattering occur always, that is why they targets with good enough "angular resolution"

In that case, the redshift must be a time dilation.

There are "doppler induced" red- and even blueshift, but they are random and overlayed with permanent gravitational redshift, that mathematically can redshift only in energy proportion, otherwise the light would refract (into colors).

This was assumed to be the case. It means that in the absence of time dilation, the photons are received at the same rate as they were emitted at the source. This enables a redshift based on time dilation to be distinguished from a redshift not based on time dilation.

The geometrics make the bundle of light (the image) stretch in direction of the gravity source, to become slightly egg formed, the bigger angular size per distance to the "impacting" mass the the more stretched. Long distances make objects only slightly redder, shorter distances stretches them and dilute the brightness and very close make overly red and stray "pixels" from the higher gravity edge. A few million treatments of this make a range of images of originally (even similar) objects. One thing is for sure redshifted objects are always more or less diluted (and equally magnified). So there can not be objects of close to Tolman factor 1,00 of any significance of "red". And the more blurred a light emitter the redder, especially those seeming to be "out of context" close could be telling my "story". B but it is probable that these low value (hard to analyze) are less investigated. Assuming redshifted correlates to distance and blur not speed could give strong correlation.

21. Originally Posted by Timo Moilanen
Originally Posted by KJW
In that case, the redshift must be a time dilation.
There are "doppler induced" red- and even blueshift, but they are random and overlayed with permanent gravitational redshift, that mathematically can redshift only in energy proportion, otherwise the light would refract (into colors).
No. If the change in frequency is directly proportional to the frequency, the redshift is entirely a time dilation. I have already proven this. It is quite simply impossible for such a redshift to be anything other than a time dilation. If your theory is saying that the frequency is directly proportional to the frequency, and not entirely a time dilation, then it is wrong. In my proof:

A mechanical clock behaves as an amplitude modulator of the light it is reflecting, and even though the source of that light may be monochromatic, the reflected light itself contains multiple frequency components due to the modulation, based on the trigonometric identities such as:

cos(A) cos(B) = ½ cos(A–B) + ½ cos(A+B)

Here, A is assumed to be the frequency of the source light and B is assumed to be the amplitude modulation frequency that is the rate at which the mechanical clock ticks. A–B and A+B are the frequencies subjected to the redshift. It should be noted that there are no photons corresponding to the modulation frequency, and therefore there is no energy loss at the modulation frequency. But if the energy loss of all the photons is directly proportional to the energy of the photons, then both A–B and A+B will be proportionally decreased by the same amount. This results in both A and B being proportionally decreased by the same amount. That is, the modulation frequency, the tick rate of the clock, has proportionally decreased by the same amount. This is time dilation.

I made no mention of any mechanism of the redshift, nor of the time dilation. It is thus universally applicable to every mechanism for which the change in frequency is directly proportional to the frequency.

Your theory has failed an important hurdle... logical validity.

22. What a coincidence just what I thought of The Theory of Relativity 40 years ago.

Your theory has failed an important hurdle... logical validity.[/QUOTE]

So here I write about redshift as I understand the relativistic interpretation of a photon from "start" till "arrival" to observation.
A photon is emitting with quite exact energy from a otherwise intact atom. Pretty much regardless of physical circumstances the photon get a "signature" frequency fs from the potential energy a electron when "jumping" to a lower (higher potent) shell in a atom. At the same time a planet ,some lengthy distance away where the photon is going to end up, is moving away from the star (and/or vice versa). This reciprocal speed difference v is determining the Doppler redshift when the photon is measured at the planet years away. But it turns out that the speed between the sun and planet accelerate so the already redshifted photon will have a timetable adjustment. The path to the planet will take even longer, and the photon shortly experience a further redshift due to this acceleration, called time dilation. Notice that this redshift is not a feature of the photon but change depending on the planets acceleration. Here I leave out the gravitations of the sun an planet. We get a photon "stretched out" signature frequency fa that at arrival exactly matches the speed of the sun relative to the planet. From this we get the speed between the sun and planet at the "arrival" moment(time) v(sun-pla)=(fs-fa)*c/fs. To "pinpoint" time/distance where the photon "was born" we must assume a speed of the sun at that time, from which we calculate the frequency drop df=v/c*fs and beginning speed for acceleration fv=fs-df. From these can be calculated speeds an distances unravel. Ex. the sun has accelerated (delta) vs =((fv-fa)/fv*c since the photon left sun, and so on.
For me it is problematic to accept that the photons fate is determined quite randomly depending on where and when it lands and it have no "hard" values, like a set energy, before measured in maybe in a few billion years.

In my theory, with no universe expansion, speeds are small (little fraction of light) and their Doppler shifts omittable. Now ignoring suns planets and their "neighbors" mases. So on its way the photon passes suns, galaxies and so on, at perpendicular distances r. The masses I enter as Standard Gravitational Parameters µ, and for every pass the photon lose some frequency, proportional to the actual frequency at that time, and df=z*f=4*µ/(r*c^2)*f. For every pass, of the millions, the trajectory will bend in same proportion z (redshift). This mean the spent energy to the bend is proportional to photon energy at the time (color) dE=df*h(Planck constant). The more "energetic" the photon at the time the more energy to bend it. Otherwise colors would bend differently (refract). The photon carry its "history" with it and when arriving to any observation anywhere will have frequency fa=fs*(1-SUM z) The photon will at all times have an energy E=f*h that of course can not change retroactively. The photon is a "sum" of the random gravitational environment it have passed. All energy is accounted for and the redshift is proportional both to frequency and energy. How could it not be (both).

23. Originally Posted by Timo Moilanen
All energy is accounted for and the redshift is proportional both to frequency and energy. How could it not be (both).
This statement took me by surprise, so I shall address it first. I believe that we are in agreement that the energy of a photon is directly proportional to its frequency. In fact, the direct proportionality between the energy and frequency of light can be derived from general relativity. Consider the thought experiment of a person on earth shining a beam of light upward into space. Ignoring the atmosphere, the energy density of the light beam decreases with elevation for two reasons: (1) divergence of the beam, and (2) gravitational time dilation. Removing the divergence of the beam, the correlation between the decrease in energy density and the concomitant decrease in frequency leads to the direct proportionality between energy and frequency (it is quantum mechanics that produces the concept of the photon and the specific value of the Planck constant). Thus, we both agree that a redshift is also a decrease in energy. We differ in whether redshift is a time dilation, and also the cause of the energy decrease. In your theory (and any "tired light" theory), the decrease in energy is some form of degradation of the light where part of the energy of the light is transferred to something else. In general relativity, the light does not actually lose any energy. The decrease in energy has two causes: (1) the light is being observed in a frame of reference that is different to that in which it was emitted, and (2) the conservation of energy within the spacetime geometry of time dilation. The thing about general relativity, the thing that makes general relativity essentially correct, is that it expresses the laws of physics the same way in every coordinate system. A consequence of this is that energy might not appear to be conserved even though it is.

Originally Posted by Timo Moilanen
What a coincidence just what I thought of The Theory of Relativity 40 years ago.

Originally Posted by KJW
Your theory has failed an important hurdle... logical validity.
So here I write about redshift as I understand the relativistic interpretation of a photon from "start" till "arrival" to observation.
A photon is emitting with quite exact energy from a otherwise intact atom. Pretty much regardless of physical circumstances the photon get a "signature" frequency fs from the potential energy a electron when "jumping" to a lower (higher potent) shell in a atom. At the same time a planet ,some lengthy distance away where the photon is going to end up, is moving away from the star (and/or vice versa). This reciprocal speed difference v is determining the Doppler redshift when the photon is measured at the planet years away. But it turns out that the speed between the sun and planet accelerate so the already redshifted photon will have a timetable adjustment. The path to the planet will take even longer, and the photon shortly experience a further redshift due to this acceleration, called time dilation. Notice that this redshift is not a feature of the photon but change depending on the planets acceleration. Here I leave out the gravitations of the sun an planet. We get a photon "stretched out" signature frequency fa that at arrival exactly matches the speed of the sun relative to the planet. From this we get the speed between the sun and planet at the "arrival" moment(time) v(sun-pla)=(fs-fa)*c/fs. To "pinpoint" time/distance where the photon "was born" we must assume a speed of the sun at that time, from which we calculate the frequency drop df=v/c*fs and beginning speed for acceleration fv=fs-df. From these can be calculated speeds an distances unravel. Ex. the sun has accelerated (delta) vs =((fv-fa)/fv*c since the photon left sun, and so on.
For me it is problematic to accept that the photons fate is determined quite randomly depending on where and when it lands and it have no "hard" values, like a set energy, before measured in maybe in a few billion years.
What you see as logical invalidity about general relativity is really a disagreement between general relativity and how you think reality ought to behave (by contrast, your theory is disagreeing with trigonometry). To be honest, I can't see what it is that you think is a fault with general relativity. Consider two points in spacetime that are separated by a small interval of time such as a single cycle of an electromagnetic wave. From each of these two points, consider null geodesic spacetime trajectories that both end at the location of the observer. If the time interval at the start of the trajectories is a single cycle of an electromagnetic wave, then the time interval at the end of the trajectories is also a single cycle of an electromagnetic wave because the phase of an electromagnetic wave does not change over a null geodesic spacetime trajectory. Then in general, the time interval between these two trajectories will differ between the start location and the end location. This is time dilation. Even in an accelerated frame of reference in flat spacetime, there is time dilation, but in the case of gravitation, the time dilation is spacetime curvature. It should be noted that the concept of spacetime curvature emerges as a consequence of the laws of physics being expressed the same way in all coordinate systems.

24. Ignoring the atmosphere, the energy density of the light beam decreases with elevation for two reasons: (1) divergence of the beam, and (2) gravitational time dilation.

Since the photon passes (huge) masses in real world or the beam is pointed more horizontally, the beams photons closer to earth will bend more than the upper ones ( in the not diverged part of the beam ). I called it scatter since it is not color selective. Its presence is undisputed, and Eddington used this exact phenomenon when measuring how much the stars close to sun "moved" compared to those little farther away. And in my theory photons bend and lose a proportional piece of energy in any direction.
If light don't lose energy on the path, and all "amplitudes" is still there (only longer) then they would just arrive "later" but the caloric energy sum would be the same on a stretched time. But I have newer seen or heard anyone saying that just the power has decreased.

Consider two points in spacetime that are separated by a small interval of time such as a single cycle of an electromagnetic wave.

I do not consider a by light speed progressing and alternating electric (magnetic) field the same as a photon, the photon particle is elemental a requirement. This is about light (photons), not a long wave radio that can be heard beyond the horizon. If a electrically induced magnet field needed photon particles a supra conductor would be impossible.
On big flaw vit relativity is that it do not account for all mass. I get the rel. phenomenon with double the magnitude.

25. Originally Posted by Timo Moilanen
Originally Posted by KJW
Ignoring the atmosphere, the energy density of the light beam decreases with elevation for two reasons: (1) divergence of the beam, and (2) gravitational time dilation.
Since the photon passes (huge) masses in real world or the beam is pointed more horizontally, the beams photons closer to earth will bend more than the upper ones ( in the not diverged part of the beam ). I called it scatter since it is not color selective. Its presence is undisputed, and Eddington used this exact phenomenon when measuring how much the stars close to sun "moved" compared to those little farther away. And in my theory photons bend and lose a proportional piece of energy in any direction.
If light don't lose energy on the path, and all "amplitudes" is still there (only longer) then they would just arrive "later" but the caloric energy sum would be the same on a stretched time. But I have newer seen or heard anyone saying that just the power has decreased.
I don't know what your point is here. I was describing a thought experiment of a person directing a beam of light upwards. It provides an understanding of how to derive the proportionality between energy and frequency in general relativity. I could've just provided the maths. I find it interesting because we normally think of the formula E=hf as being from quantum mechanics. But the proportionality between energy and frequency (where h is unspecified) comes from general relativity.

Originally Posted by Timo Moilanen
Originally Posted by KJW
Consider two points in spacetime that are separated by a small interval of time such as a single cycle of an electromagnetic wave.
I do not consider a by light speed progressing and alternating electric (magnetic) field the same as a photon, the photon particle is elemental a requirement. This is about light (photons), not a long wave radio that can be heard beyond the horizon. If a electrically induced magnet field needed photon particles a supra conductor would be impossible. On big flaw vit relativity is that it do not account for all mass. I get the rel. phenomenon with double the magnitude.
What about laser light? But even if we only consider long wave radio, the change in the interval of time between the two null geodesics is still a time dilation.

26. Sorry I'm still hang up on the shattering of light via "uneven" bending, was thinking you forgot. It should interest you too because it is missing from the Tolman test. And Planck constant of course is the beginning of quantum mechanics. I just use my own nonrelativistic h=12*k*e/c^3 that fit with classic Joule (not electric).

Laser light is photons bend in gravity radio waves conductors, isolators, other el.fields and so on but not mass permanent redshift.

27. Originally Posted by Timo Moilanen
Sorry I'm still hang up on the shattering of light via "uneven" bending, was thinking you forgot. It should interest you too because it is missing from the Tolman test.
The Tolman test is not the only evidence of an expanding universe. For one thing, there's the cosmic microwave background radiation. There is also that if the change in frequency due to redshift is directly proportional to the frequency, then the redshift is a time dilation.

Originally Posted by Timo Moilanen
And Planck constant of course is the beginning of quantum mechanics.
But the proportionality between energy and frequency can be derived from general relativity.

Originally Posted by Timo Moilanen
I just use my own nonrelativistic h=12*k*e/c^3 that fit with classic Joule (not electric).
Dimensional analysis:

h = kg⋅m2⋅s–1
k = kg⋅m3⋅s–4⋅A–2
e = A⋅s
c–3 = m–3⋅s3

On the left is kg⋅m2⋅s–1. On the right is kg⋅A–1. The dimensions don't match... FAIL.

Originally Posted by Timo Moilanen
Laser light is photons
Laser light is coherent monochromatic light. As such, it behaves as a wave. Thus, we can consider two points in spacetime that are separated by a small interval of time such as a single cycle of the laser light wave, and then from each of these two points, consider null geodesic spacetime trajectories that both end at the location of the observer.

28. Dimensional analysis:
The kg can be tricky to transform, but the formula can be written with (epsylon nough) E0. So h=e/(E0*c^3) that give easier to transform units. Mark that I have E0=1/(12*k) , since it turned out that the 4pi assumption wasn't quite perfect. Well 12 would have been too odd even for a wild guess. This brought the big change to the whole net of formulas and constants all way up to the Rydberg constant. Geometrically this come from that permittivity is trough a plane surface between two spherical charges (as in definition of Coulomb) rather than a curved plane from where the pi was "taken" to explain the some 12 to fill the gap. Now all formulas "line up"
But this is in an earlier of my threads.

29. Originally Posted by Timo Moilanen
the formula can be written with (epsylon nough) E0. So h=e/(E0*c^3)
No, this formula has exactly the same problem as the previous formula... the dimensions as represented by their SI units are not the same on each side of the equation.

Originally Posted by Timo Moilanen
Mark that I have E0=1/(12*k) , since it turned out that the 4pi assumption wasn't quite perfect. Well 12 would have been too odd even for a wild guess. This brought the big change to the whole net of formulas and constants all way up to the Rydberg constant. Geometrically this come from that permittivity is trough a plane surface between two spherical charges (as in definition of Coulomb) rather than a curved plane from where the pi was "taken" to explain the some 12 to fill the gap.
The replacement of 4pi with 12 is not the issue. The issue is that the dimensions as represented by their SI units are not the same on each side of the equation. This means that the formula cannot be correct. But it should be noted that 4pi arises because it is the surface area of a unit sphere.

If you really want a formula that expresses the Planck constant in terms of electrical stuff, then there is this formula:

h = e2/2ε0αc

where α is the fine-structure constant, a dimensionless constant whose value is approximately 0.00729735. The dimensions of this formula:

h = kg⋅m2⋅s–1
e2 = A2⋅s2
ε0–1 = kg⋅m3⋅s–4⋅A–2
c–1 = m–1⋅s

e2ε0–1c–1 = A2⋅s2 × kg⋅m3⋅s–4⋅A–2 × m–1⋅s
= kg⋅m2⋅s–1

Thus, the dimensions of h and e2/2ε0αc are the same.

30. Thus, the dimensions of h and e2/2ε0αc are the same.
For readers that use less math, I need to point out that units can be expanded and reduced exactly like parameters. For example vacuum permittivity have SI units [E0]=F/m=C/(V m)=C^2/(N m^2)=C^2*s^2/(kg m^3)=A^2s^4/(kg m^3). So when formulas are right the units will fit (usually easily ) with a few "moves". Whit formulas even more "flexible" that makes physics interesting.

31. Originally Posted by Timo Moilanen
[E0]=F/m=C/(V m)=C^2/(N m^2)=C^2*s^2/(kg m^3)=A^2s^4/(kg m^3).
If you examine my examples, you should take note that all the quantities are expressed in terms of the SI base units. The SI standard uses the following base dimensions: time (T), length (L), mass (M), electric current (I), absolute temperature (Θ), amount of substance (N) and luminous intensity (J). Up to ordering, the dimension of a quantity in terms of the base dimensions is unique. In the case of vacuum permittivity, that would be kg–1⋅m–3⋅s4⋅A2. None of the other units you mention are entirely in terms of the SI standard base units. Therefore, it is clear that the left and right sides of your formula for h are incommensurable.

32. What is the fault with for example 1V*1C=1J ?

33. Originally Posted by Timo Moilanen
What is the fault with for example 1V*1C=1J ?
Let's have a look at the base dimensions (expressed as SI base units):

V = kg⋅m2⋅s–3⋅A–1
C = A⋅s
J = kg⋅m2⋅s–2

VC = kg⋅m2⋅s–3⋅A–1 × A⋅s = kg⋅m2⋅s–2

Therefore, VC and J are commensurable.

Being commensurable doesn't mean that the formula is correct, but at least it won't be dismissed immediately as invalid. Of course, we know that the formula is a standard formula from physics, but I thought it would be helpful to see the dimensional analysis of a formula that we know is correct.

34. Being commensurable doesn't mean that the formula is correct, but at least it won't be dismissed immediately as invalid.
My units at simplest goes like ks=4/(3(mpi+mei)*c2) and the existing units "crosses" [ks]=kg m2/s2 and the rest must be added. When multiplication/dividing a more complicated "editing" is needed since so many arithmetically prefect and math. correct answers must be same units (in the end),I do not do numerology.
I"index" all calculated constant, i=indifferent . A few more examples of Planck.

hi0i*ei /c = mei*1000/( E0i*c2*sqrt(3) = 16*ei /((mpi+mei)*c5 = 16*1000*mei /((mpi+mei)*sqrt(3)*c4 =
16000/((mpi/mei +1)*sqrt(3)*c4 =6,22488445*10-34 Jis for mpi/mei =1836,1526734311 and hi=mei*ei2*c5 /(8*Rinf) .

All my electrodynamics is derived from proton/electron mass ratio an c (light speed). I have put in 4/3 to Planck const. from a simple geometric calculation and changed E0i=1/(k*4*pi) to E0i=1/(12*ks) , it did not fit anyway and is now a constant worthy parameter. The "web" of constants and units are kg compatible and give right return only with my indexed constants and all mechanic constants (traditional non electric). The kg comparable current Ai=0,971135456 A(SI) and voltage Vi=1,04615695 V(SI). Here used constants:
ks=8,8762526*109 Nm2/Ci2
ei=1,57464444*10-19 Ci
E0i=9,388346308*10-12 Fi/m .
mpi=1,6704412965*10-27kg .
mei=9,0975076347*10-31kg . .
µ0i=1,1851395545*10-6 N/Ai2
Rinf=10969066,527 m-1
The times 1000 that I use here and there when the parameter at right is one atom ,molecule ,ion (one of NA) compared to the units of the "measure" Ex. ei2=4*10002*(mpi+mei)/(3*c2) This same phenomenon one would guess is common , but I never have heard.

Now that units are done "planned" could abuse relativity theory again. . .. .

35. Originally Posted by Timo Moilanen
My units at simplest goes like ks=4/(3(mpi+mei)*c2) and the existing units "crosses" [ks]=kg m2/s2 and the rest must be added. When multiplication/dividing a more complicated "editing" is needed since so many arithmetically prefect and math. correct answers must be same units (in the end),I do not do numerology.
I"index" all calculated constant, i=indifferent . A few more examples of Planck.

hi0i*ei /c = mei*1000/( E0i*c2*sqrt(3) = 16*ei /((mpi+mei)*c5 = 16*1000*mei /((mpi+mei)*sqrt(3)*c4 =
16000/((mpi/mei +1)*sqrt(3)*c4 =6,22488445*10-34 Jis for mpi/mei =1836,1526734311 and hi=mei*ei2*c5 /(8*Rinf) .

All my electrodynamics is derived from proton/electron mass ratio an c (light speed). I have put in 4/3 to Planck const. from a simple geometric calculation and changed E0i=1/(k*4*pi) to E0i=1/(12*ks) , it did not fit anyway and is now a constant worthy parameter. The "web" of constants and units are kg compatible and give right return only with my indexed constants and all mechanic constants (traditional non electric). The kg comparable current Ai=0,971135456 A(SI) and voltage Vi=1,04615695 V(SI). Here used constants:
ks=8,8762526*109 Nm2/Ci2
ei=1,57464444*10-19 Ci
E0i=9,388346308*10-12 Fi/m .
mpi=1,6704412965*10-27kg .
mei=9,0975076347*10-31kg . .
µ0i=1,1851395545*10-6 N/Ai2
Rinf=10969066,527 m-1
The times 1000 that I use here and there when the parameter at right is one atom ,molecule ,ion (one of NA) compared to the units of the "measure" Ex. ei2=4*10002*(mpi+mei)/(3*c2) This same phenomenon one would guess is common , but I never have heard.

Now that units are done "planned" could abuse relativity theory again. . .. .
I don't intend to do a dimensional analysis of all the above. Nor am I going to check that the above is numerically correct. With regards to equating incommensurable quantities, is numerical equality maintained if you change units, for example from metres to feet, seconds to hours, or kilograms to ounces? I don't think so.

Anyway, it seems to me that you are applying your own rules to such things as units. That might be ok if you had told us what your rules are. But because you haven't told us, we have no other option than to assess your work on the basis of the standard rules. And according to the standard rules, your formulae are invalid. Are the standard rules correct? For any disagreement between your rules and the standard rules, the standard rules are assumed to be correct because they are the standard rules. Of course, the possibility does exist that the standard rules are not correct. But to determine this would require some independent assessment that could also be used to determine if your rules are correct. Ultimately, there has to be something that we both agree on. Then and only then can your theory be assessed in a way that you agree with regardless of the outcome of the assessment. It seems to me that you try to avoid admitting that your theory is incorrect by the claim that your theory differs from the standard theory in every way that could be used to invalidate your theory. By doing this, your theory becomes a kludge. The simple fact of the matter is that your theory can never be correct unless it can be shown to be incorrect. So far, it cannot be shown to be incorrect because you are not accepting the basis of any assessment. And without that, it will be declared incorrect on the basis of standard physics without your acceptance. Let's face it, it is you who has the burden of proof, not us.

36. I don't intend to do a dimensional analysis of all the above. Nor am I going to check that the above is numerically correct. With regards to equating incommensurable quantities, is numerical equality maintained if you change units, for example from metres to feet, seconds to hours, or kilograms to ounces? I don't think so.

The change I have done connect kg and C (Coulomb) units, and there is no need for the bla. bla. for this electrodynamic detail since it is not new physics, just an upgrade of the current (100 y old) one . My gravitation theory we have barely mentioned yet, and we were discussing a little detail of light decaying while bending, or not bending nor decaying when "crossing space".

Ones again who are "us" , on what "organizations" behalf are you "declaring" any kind of decision. This is the most absurd "discussion" I have ever had, or heard of. Are here more people than we two, us and I ?

37. Originally Posted by Timo Moilanen
The change I have done connect kg and C (Coulomb) units, and there is no need for the bla. bla. for this electrodynamic detail since it is not new physics, just an upgrade of the current (100 y old) one.
This doesn't remove the necessity of only equating quantities that are commensurable. I don't know why you are not accepting this. It is quite obvious that only commensurable quantities can be added, subtracted, equated, or compared.

Originally Posted by Timo Moilanen
My gravitation theory we have barely mentioned yet, and we were discussing a little detail of light decaying while bending, or not bending nor decaying when "crossing space".
I reject the notion that light decays while bending. You have not provided any support for this notion and have been unable to counter my argument that it has likely already failed to be detected. Also, you have been unable to counter my argument that a change in frequency proportional to the frequency proves that the change in frequency is a time dilation. I should remark that not only does gravity produce a time dilation, but the time dilation itself produces gravity. In other words, general relativity provides a genuine explanation of gravity that Newtonian theory cannot provide. As I see it, your theory not only changes how gravity behaves, but also what gravity is.

Originally Posted by Timo Moilanen
Ones again who are "us" , on what "organizations" behalf are you "declaring" any kind of decision. This is the most absurd "discussion" I have ever had, or heard of. Are here more people than we two, us and I?
What do you hope to achieve by posting your theory on this forum? I assume that you would like to convince the people visiting the forum that your theory has value. But so far, you haven't done anything to convince the only person replying to your posts that you have anything worthwhile to offer. If you expect your theory to be accepted without question, then that it not what science is about. Your theory is going to be scrutinised, and how you handle that scrutiny will determine your credibility and the value of your theory. Why should I continue to discuss your theory with you if I don't think you are willing to accept a reasoned argument? I should remark that if the moderators do not believe you are being genuine, they will place the thread in the Trash Can, where only logged-in forum members can access.

38. This doesn't remove the necessity of only equating quantities that are commensurable.
All the "traditional" equations work as earlier with same name an units on constants an parameters. The values are slightly different (adapted to match kg). Only change I have done is the 4pi to 12 for calculating vacuum permittivity, that now fit in like accordingly also the magnetic permeability. My adapting electro dynamics to kg comparable units work flawlessly, plus I fixed a flaw, The excess unit loosing equations are not necessary, they are just a idiosyncrasy showing you cant get a miscalculated numerical value while cutting corners. The "fitted" ampere and volt units integrate all units used in this sector.
It is quite obvious that only commensurable quantities can be added, subtracted, equated, or compared.
So now the constants and parameters are even more of all the above than before. Which was my initial idea.
As I see it, your theory not only changes how gravity behaves, but also what gravity is.
I can hardly adapt to all relativistic phenomenon while saying the theory do not work, so i can not see what a personal theory might mean at this cite. And you have not proven that light do not scatter, it did for Eddington to enable his measurement.
don't think you are willing to accept a reasoned argument
So you demand me to admit my theory is wrong to continue. That is impossible. The proportional to energy "demise" of a photo do not exclude all but relativity. My numbers hold for light bending and the redshift is there. I have no contradictions in my theory but I can of course not adapt all to relativity theory only explain the main phenomenon by my theory.

39. Originally Posted by Timo Moilanen
Originally Posted by KJW
This doesn't remove the necessity of only equating quantities that are commensurable.
All the "traditional" equations work as earlier with same name an units on constants an parameters. The values are slightly different (adapted to match kg). Only change I have done is the 4pi to 12 for calculating vacuum permittivity, that now fit in like accordingly also the magnetic permeability. My adapting electro dynamics to kg comparable units work flawlessly, plus I fixed a flaw, The excess unit loosing equations are not necessary, they are just a idiosyncrasy showing you cant get a miscalculated numerical value while cutting corners. The "fitted" ampere and volt units integrate all units used in this sector.

Originally Posted by KJW
It is quite obvious that only commensurable quantities can be added, subtracted, equated, or compared.
So now the constants and parameters are even more of all the above than before. Which was my initial idea.
For the original formula you presented that led to this discussion:

h = 12⋅k⋅e⋅c–3

The SI base units are:

h = kg⋅m2⋅s–1
k = kg⋅m3⋅s–4⋅A–2
e = A⋅s
c–3 = m–3⋅s3

On the left (h) is kg⋅m2⋅s–1.
On the right (12⋅k⋅e⋅c–3) is kg⋅A–1.

These two sides of the formula that you are equating have different SI base units. Thus, the formula is wrong, plain and simple. There is no getting around this. There is no "my way is better". Your attempts to justify the difference in units rather than acknowledging that you made an error damage your credibility in producing a physics theory.

Originally Posted by Timo Moilanen
Originally Posted by KJW
As I see it, your theory not only changes how gravity behaves, but also what gravity is.
I can hardly adapt to all relativistic phenomenon while saying the theory do not work, so i can not see what a personal theory might mean at this site.
There is no "adapting" to relativistic phenomena. If there is disagreement between your theory and relativity, it is up to you to show that (1) relativity is wrong, and (2) your theory is right. Being in the Personal Theories & Alternative Ideas forum allows your theory to be presented, but it will still be subjected to the scrutiny expected of any new scientific theory. That is, although you are permitted to present an alternative theory, you are still required to justify that theory. And if discussion of the theory leads to the conclusion that it is wrong, then so be it. Being in the Personal Theories & Alternative Ideas forum doesn't protect a theory from being assessed as worthless.

Originally Posted by Timo Moilanen
And you have not proven that light do not scatter, it did for Eddington to enable his measurement.
I don't have to prove that light doesn't scatter in a gravitational field. You have to prove that it does. You'll have to supply a reference that there was scattering in Eddington's measurements.

I have provided a reasonable argument that if there is redshift due to photon decay in a gravitational field, it would already have been detected.

Originally Posted by Timo Moilanen
Originally Posted by KJW
Why should I continue to discuss your theory with you if I don't think you are willing to accept a reasoned argument?
So you demand me to admit my theory is wrong to continue. That is impossible.
If your theory has been proven to be wrong, then admitting it is wrong is a way to show good faith. But what I meant was that I have the expectation that your disagreements with what I say be reasonable. It is unreasonable to adopt the position of "all science is wrong". Therefore, there has to be scientific principles that we can both agree on that form the basis of reasonable argument. If I am unable to prove your theory wrong, that does not make it right. And if I provide arguments indicating that your theory is wrong based on standard science, but you counter those arguments based on your own assumptions, then the weight of the arguments based on standard science will crush your theory whether you accept it or not.

Given that very few people get to change the way that physics view the world, it is delusional of you to expect that your theory is the theory that is going to replace the currently accepted theory. It is far more likely that your theory is wrong. A reasonable person would expect that their theory will be shown to be wrong and accept that when it happens. A person who considers it impossible to accept that their theory is wrong is not acting in good faith.

Originally Posted by Timo Moilanen
The proportional to energy "demise" of a photo do not exclude all but relativity.
I have said the proportionality between the change in frequency (change in photon energy) and frequency (photon energy) proves that the redshift is a time dilation. Therefore, if your theory is saying that the change in photon energy is proportional to photon energy, and that the redshift is not a time dilation, then that is an error in your theory. It is not merely a disagreement with relativity.

40. There is no "adapting" to relativistic phenomena. If there is disagreement between your theory and relativity, it is up to you to show that (1) relativity is wrong, and (2) your theory is right. Being in the Personal Theories & Alternative Ideas forum allows your theory to be presented, but it will still be subjected to the scrutiny expected of any new scientific theory. That is, although you are permitted to present an alternative theory, you are still required to justify that theory. And if discussion of the theory leads to the conclusion that it is wrong, then so be it. Being in the Personal Theories & Alternative Ideas forum doesn't protect a theory from being assessed as worthless.

That was an of guard admission, that I can precent my theory here. The insulting intimidating and ridiculing is just life, but I do not consider me having any obligation to do the same to the relativity theory. But if you insist I could give a few "constructive" opinions.

About the units you are right I need to make them univocal, by disposing (alpha) a. Now I have a=e*c^2/2 , e*c^2/(2*a)=1,and since E0=1/(12*ks) ks=Coulomb constant for two speres. giving h=e^2/(2*a*E0*c). Some two years ago I introduced a Coulomb constant ka for a sphere (nucleus, proton) and a point representing a distance at where a electron at average or a anion "stay". ka=6*ks/a =12*ks/(e*c^2) =7,526395*10^9 with a unit needing to be N* m^2/(As*n) where n is how many point charges (electrons) the particle have. For n >1 spells trouble since the ion is gaining directions. For single charges (of q=e ) ka is equal to the integral over a proton (with mass) and a massless electron adjusted for the largely unnoticed kmol/mol dilemma "corrected" with (alpha). ka~=6/7*1000*ks. Two point charges (electrons) will make the constant =1000^2/((mp+me)*c^2)=6,659*10^15 N m^2/(2 charges of e). So much for this "stoichio/mass" lesson.

I don't have to prove that light doesn't scatter in a gravitational field. You have to prove that it does. You'll have to supply a reference that there was scattering in Eddington's measurements.

Eddington's measurement could not have detected any other scattering than the atmospheric. When two "stars" distance is measured at the edge of the "solar disc", they move apart more the closer to the sun the beams have travelled. That is what Eddington measured , and scientist still today are. Saying that a beam from say a planet with a bigger angular diameter would stay circular by some magic is a "toll call". The planet disc will stretch (most at the sun side edge) and slightly " dilute" the brightness. As happens brightness measures have been done (Tolman experiment) and the brightness is measured to decrease, not as much as the relativity theory would predict ,but the brightness diminishes even in the middle of objects.

I have said the proportionality between the change in frequency (change in photon energy) and frequency (photon energy) proves that the redshift is a time dilation. Therefore, if your theory is saying that the change in photon energy is proportional to photon energy, and that the redshift is not a time dilation, then that is an error in your theory. It is not merely a disagreement with relativity.

The photon has constant speed c, but the energy f*h need to be "redirected" when it bend, the more energetic photon the "bigger" work to cause the bend, proportionally more. And besides the "proportionality" of time dilation redshift do not bother me.
Some write that the GPS "frequency correction" some 10,2MHz compared to the satellite signals 1,2 to 1,5 GHz is time dilation (gravitational at that). Telescopes would reach barely 1 ly (light year) into space. I no way buy any relativistic explanation.

41. Originally Posted by Timo Moilanen
About the units you are right
Although I insisted that both sides of your formulae have the same base units, I should point out that there is an exception... the use of natural units. There are a number of systems of natural units in use, created by setting fundamental constants to a value of 1. For example, geometrized units set c=1 and G=1. Thus, equations containing these constants are expressed in a simpler form without these constants in geometrized units. Geometrized units are often used in general relativity. For example, the Einstein equation of general relativity, which in ordinary units is:

Gij + Λ gij = 8πGc–4 Tij

becomes simplified to:

Gij + Λ gij = 8π Tij

in geometrized units.

Consider:

T = c–1⋅L

L⋅T–2 = G⋅M⋅L–2
M = G–1⋅L3⋅T–2 = G–1⋅c2⋅L

For c = 1 and G = 1:

T = L
M = L

That is, the use of geometrized units has eliminated the dimensions of time and mass, converting them both to the dimension of length. Expressed in terms of the dimension of length, all the terms of the simplified Einstein equation have the dimension of L–2, so these terms are still commensurable within the reduced number of dimensions. But the terms of the simplified Einstein equation are not commensurable when considered in terms of length, time, and mass without c and G.

However, the above doesn't contradict what I said to you about commensurable quantities because you were not attempting to use natural units. Indeed, you appear to place an undue significance to the values of the fundamental constants. The above shows that we can set their value to whatever we like (such as 1) with the corresponding change in the units. Thus, c = 1 gives time the same units as length, which makes these quantities comparable when considered from the perspective of four-dimensional spacetime.

Originally Posted by Timo Moilanen
And besides the "proportionality" of time dilation redshift do not bother me.
It should "bother" you because if gravity is caused by time dilation, it is not being caused by whatever mechanism you think causes gravity. Thus, if your theory is saying that the change in photon energy is proportional to photon energy then that is an error in your theory. Even in the case of Newtonian theory, the change in photon energy being proportional to photon energy leads to a contradiction.

42. Geometrized units do not help me. The 1/(mp+me)*c^2 is the coupling to mass, F is proportional how many pairs of that weight is dragging, times c^2 give energy to cross over to right, my integrals has length unit R for easy math distance p=r/R. So m^2 fit as choice, definition say the other is named Coulomb. I see no way of integrating units to all formulas. Shortcuts are handy but probably would be best to use only those formulas with functioning units, they cover the "structure"

Sorry I slip on those (charge, energy,quant) often , n*h=E and f*h=E (photon). For gravitation I have discarded the shell theorem and calculated new transformation parameter from the little measure results available, getting Ti=7,462*10^-11Nm^2/kg^2. I am now very confident it will be mn*c^2/2=7,507*10^-11 Nm^2/kg^2. All planets have an nonlinear gravity/alt. profile, a orbit need 2,55 times pot. energy compared to kinetic energy at that gravity (alt.) Galaxies have 25% stronger grav. field horizontally than vertically at even long distances (they never get round grav. field) Very thin discs collapse at their outer perimeter if they become wider than R*9/800 (think Saturn rings will heal). All "known" masses are about 13% bigger and assumed free fall gravity *4/pi=1,273 stronger. Moon and especially Mars have much stronger gravity at altitude, at surface geology diminishes it about 30%, earth about27%. There is only too many wise men for too many generations, that already know everything.

43. Originally Posted by Timo Moilanen
Geometrized units do not help me. The 1/(mp+me)*c^2 is the coupling to mass, F is proportional how many pairs of that weight is dragging, times c^2 give energy to cross over to right, my integrals has length unit R for easy math distance p=r/R. So m^2 fit as choice, definition say the other is named Coulomb. I see no way of integrating units to all formulas. Shortcuts are handy but probably would be best to use only those formulas with functioning units, they cover the "structure"
I didn't introduce geometrized units to "help" you. I introduced them to you to show you how the dimensions can be unified. For example, c unifies length and time so that one can speak of distances in spacetime even when the separation is in time. Also, the quantities of general relativity are based on the length dimension.

Originally Posted by Timo Moilanen
Sorry I slip on those (charge, energy,quant) often , n*h=E and f*h=E (photon). For gravitation I have discarded the shell theorem and calculated new transformation parameter from the little measure results available, getting Ti=7,462*10^-11Nm^2/kg^2. I am now very confident it will be mn*c^2/2=7,507*10^-11 Nm^2/kg^2. All planets have an nonlinear gravity/alt. profile, a orbit need 2,55 times pot. energy compared to kinetic energy at that gravity (alt.) Galaxies have 25% stronger grav. field horizontally than vertically at even long distances (they never get round grav. field) Very thin discs collapse at their outer perimeter if they become wider than R*9/800 (think Saturn rings will heal). All "known" masses are about 13% bigger and assumed free fall gravity *4/pi=1,273 stronger. Moon and especially Mars have much stronger gravity at altitude, at surface geology diminishes it about 30%, earth about27%. There is only too many wise men for too many generations, that already know everything.
It should be noted that the solar system provides evidence that the "force" of gravity very closely obeys the inverse square law, and that deviations from the inverse square law would have significant effects on the orbits of planets and moons. You should read about the Laplace-Runge-Lenz vector. This vector is a constant of motion when the force law is inverse square. But deviations from the inverse square law lead to a rotation of the vector which results in a precession of the elliptical orbits. It seems to me that if gravity was as different to the accepted values as you claim, we would see very different motion of the planets and moons than we actually see. Therefore, it seems to me that your theory is irreconcilable to observations of the solar system.

44. It should be noted that the solar system provides evidence that the "force" of gravity very closely obeys the inverse square law, and that deviations from the inverse square law would have significant effects on the orbits of planets and moons. You should read about the Laplace-Runge-Lenz vector. This vector is a constant of motion when the force law is inverse square. But deviations from the inverse square law lead to a rotation of the vector which results in a precession of the elliptical orbits. It seems to me that if gravity was as different to the accepted values as you claim, we would see very different motion of the planets and moons than we actually see. Therefore, it seems to me that your theory is irreconcilable to observations of the solar system.

I am sure there are many formulas during hundreds of years for mainstream "assumptions". With my theory the planets fit in very well and even Mercury fall in naturally in the orbits. Satellites around earth fit in much better than the simple inverse square ( yes I have done a ton of calculations and tables). My theory can of course be proved while explaining quite a few , by mainstream so called anomalies, The "gravitation constant" Ti can be calculated from the thousands of experiments for G , and have a constant value unlike G. A simple variation of the "Cavendish experiment" will disapprove both G and shell theorem at the same time when giving correct value for Ti. Above all this theory diminish the "need" for dark mass.

A gravitation field strength at a distance r from a even density sphere is g=M*Ti/r^2/(3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1))) , where M is mass of sphere Ti=7,5*10^-11 Nm2/kg2 and p=r/R where R is radius of the sphere. This is measurable and no kind of complicated. The Shell Theorem "magic" is although gone, sorry for that.

45. Originally Posted by Timo Moilanen
I am sure there are many formulas during hundreds of years for mainstream "assumptions". With my theory the planets fit in very well and even Mercury fall in naturally in the orbits. Satellites around earth fit in much better than the simple inverse square ( yes I have done a ton of calculations and tables). My theory can of course be proved while explaining quite a few , by mainstream so called anomalies, The "gravitation constant" Ti can be calculated from the thousands of experiments for G , and have a constant value unlike G. A simple variation of the "Cavendish experiment" will disapprove both G and shell theorem at the same time when giving correct value for Ti. Above all this theory diminish the "need" for dark mass.

A gravitation field strength at a distance r from a even density sphere is g=M*Ti/r^2/(3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1))) , where M is mass of sphere Ti=7,5*10^-11 Nm2/kg2 and p=r/R where R is radius of the sphere. This is measurable and no kind of complicated. The Shell Theorem "magic" is although gone, sorry for that.
What is the source of your astronomical data?

What are the anomalies? Note that an anomaly is not just a deviation from the inverse square law. It is a deviation that cannot be explained by either Newtonian theory or general relativity. For example, an oblate spheroid deviates from the inverse square law, but this is fully accounted for in Newtonian theory. Thus, if you assume that the earth is a sphere, then it is only natural that satellites will deviate from the inverse square law because the earth is approximately an oblate spheroid rather than a sphere. Then there are the inhomogeneities within the earth that turn it from being a "spherical chicken in a vacuum" to a real-world object.

What is the simple variation of the Cavendish experiment to which you refer?

Disprove G? In geometrized units, G = 1, and therefore does not even appear in general relativity equations.

46. What is the source of your astronomical data?

Unfortunately I'm dependent on the Wikipedia side of paywall (free public data, in contrast to "costly" kind of "public" data). There are more detailed measurements and statics with random coverage, mostly not enough for any precise calculations.
As anomalies for example the "long distance" probes (Voyagers and others), sourced from here unspecified news and articles from last few decades. But most of all I consider dark mass an anomaly (for all new and old ,actual and outdated theories).
Ellipticity of a planet is for now waste of time pondering when all density layers all the way to the heave cores need to be accounted for. My models (earth Mars) have 15-20 layers and the ellipticity is in queue when and if the major "parameters" are sufficiently accounted for. So as I see it mainstream is tinkering with minor trifles while I'm working on the M*G +27% and especially what it bring in to galaxies "flat" type of mass distribution and the gravity field forms magnitude. Most worried I am now about moon an Mars landings and even more takeoffs, since the non-anomal way seem to be smashing the probes into the landscape.

The variant of measure at simplest would be making source masses (spheres) of two kinds (two pairs) of same weight. One pair with denser core material and lighter outer 20,6% of radius (50% volume). The second same diameter light core and denser outer layer. These speres will make no sense G values but fit perfectly in my formulas. A sphere have "today" 3 gravitation features : biggest total gravity "pressure" is true for photons and probably other particles with natural light speed, mid strength the mainstream g-values (gm) and G that are true as measured but useful only to calculating the real gravitation (g/m) and real masses. For all bodies a exact relation is g/m=(gm)^2/gtot where I use k=gtot/gm. Sigma (Ti*dM/s^2)dxdydz =gtot and Sigma (Ti*dM*cosb/s^2)dxdydz=gm

Since G is a (not constant) number measured at a distance about 1,47*radius of source mass it have no use for other distances and all assumed (calculated) masses are equally "valuable". How can a theory based on "fictive" masses and a assumed G=1 (maybe close enough) give useful calculations? My guess, the Standard Gravitational Parameter is close to reality, but by no means a constant measure , and enables realistic calculated values. (even with 27% of gravity missing for staters)
And of course I have a improved Keplerian elliptic orbit with no minus energies and more realistic shape.
So I think I have very rationalized and explored theory compared to mainstream "hotheads" (self declared experts).
KWJ on the other hand have knowledge, is extravagant only half the time and could have a scientific career when done "playing" with us.

47. Originally Posted by Timo Moilanen
Unfortunately I'm dependent on the Wikipedia side of paywall (free public data, in contrast to "costly" kind of "public" data). There are more detailed measurements and statics with random coverage, mostly not enough for any precise calculations.
But earlier, you claimed to have the data that shows that your theory is a better fit than the currently accepted theory.

Originally Posted by Timo Moilanen
As anomalies for example the "long distance" probes (Voyagers and others), sourced from here unspecified news and articles from last few decades.
How did you eliminate possible causes of this anomaly other than an error in the law of gravity? These probes are quite small and are therefore quite sensitive to things that might occur in space. Because you don't have control over the probes, I don't think you are in a position to say that any anomaly is evidence of an error in the law of gravity. You really need to be considering the motion of the planets and moons rather than small artificial probes.

Originally Posted by Timo Moilanen
But most of all I consider dark mass an anomaly (for all new and old ,actual and outdated theories).
As far as anomalies that could reveal an error in the law of gravity, I think you should stay within our solar system. I think there are too many uncertainties with regards to the composition of other galaxies to claim that dark matter is an anomaly. There are two basic possibilities of what dark matter is: (1), a material with properties quite unlike anything we are familiar with, (2), a modification of the law of gravity. You obviously believe in (2), whereas the evidence indicates (1).

Originally Posted by Timo Moilanen
The variant of measure at simplest would be making source masses (spheres) of two kinds (two pairs) of same weight. One pair with denser core material and lighter outer 20,6% of radius (50% volume). The second same diameter light core and denser outer layer.
The equivalence principle has been tested with different materials.

Originally Posted by Timo Moilanen
Since G is a (not constant) number measured at a distance about 1,47*radius of source mass it have no use for other distances and all assumed (calculated) masses are equally "valuable".
For an elliptical orbit, the distance between the masses do vary. Also, for different planets orbiting the sun at different distances, Kepler's third law holds:

Log-log plot of period T vs semi-major axis a (average of aphelion and perihelion) of some Solar System orbits (crosses denoting Kepler's values) showing that a³/T² is constant (green line)

Originally Posted by Timo Moilanen
How can a theory based on "fictive" masses and a assumed G=1 (maybe close enough) give useful calculations?
It depends on what units are being used for mass. If one is using kilograms, then the more familiar value of G will be needed. But if mass is specified in metres, then G=1 and c=1 is just as correct. Of course, what does it mean for mass to be a length, and how is it measured? Consider the standard gravitational parameter, μ, defined as:

μ = G⋅M

This has SI units m3⋅s–2. But:

μ/c2 = G⋅M/c2

has SI units of m, and is thus a mass in units of metres (but note that c indicates that all times are specified as the distance travelled by light). I think that you could benefit from considering the question: What is mass?

48. But earlier, you claimed to have the data that shows that your theory is a better fit than the currently accepted theory.
Available data do not show nor claim that (mainstream) theories match perfectly.
It was in an article written by Frank Wilczek where while talking about a big chunk of gravity missing mentioned a discussion with a college who had from Voyagers calculated that some 25% was un accounted for. Later has been new of probes running into a wall, and so on. I keep my eyes open. Mercury is "explained" many times, in my theories it fall in just like that and the other planets have no "privileges" for being further away. The fact that I have a more advanced orbit model do not matter. The fact that the orbits work in a times 4/pi stronger gravity field is the essential. This is seen best when trajectories deviate from orbits.
As far as anomalies that could reveal an error in the law of gravity, I think you should stay within our solar system. I think there are too many uncertainties with regards to the composition of other galaxies to claim that dark matter is an anomaly. There are two basic possibilities of what dark matter is: (1), a material with properties quite unlike anything we are familiar with, (2), a modification of the law of gravity. You obviously believe in (2), whereas the evidence indicates (1).
What I mean is that dark matter is actually treated as an anomaly, but I say it fi very neatly in because it is the gravity of "visible" matter and some black holes (black holes are natural matter)
The equivalence principle has been tested with different materials.
This has nothing to do with equivalence principle, it is geometry and the 1,47 R I mean the gravity measure experiments. My constant fall in with those at infinite distance.
μ = G⋅M The standard gravitational parameter is not meant to be G*M but µ=v^2*r because G is too "unsure" for astronomers.
What is mass?
We are all waiting. please tell us.

49. Originally Posted by Timo Moilanen
The standard gravitational parameter is not meant to be G*M but µ=v^2*r because G is too "unsure" for astronomers.
The standard gravitational parameter is defined as μ = GM. The formula µ=v²r is how this parameter can be measured, but only applies to a circular orbit. For an elliptical orbit, it is given by the formula:

µ=4π²a³/T²

where T is the orbital period, and a is the semi-major axis.

Using a pendulum, it is given by the formula:

µ=4π²r²L/T²

where r is the radius of the gravitating body, L is the length of the pendulum, and T is the period of the pendulum.

It is true that for astronomical objects, the product GM is far more accurately determined than either G or M separately. This is not a flaw in the theory, but a difficulty in accurately performing the Cavendish experiment.

Originally Posted by Timo Moilanen
What is mass? We are all waiting. please tell us.
I thought that you might like to tell us what you think mass is, given that you think that there is substantially more of it in the various astronomical objects than what the mainstream knows about.

50. I thought that you might like to tell us what you think mass is, given that you think that there is substantially more of it in the various astronomical objects than what the mainstream knows about.
Mass is just the "ordinary stuff" and much of it has been "missed" because there is obvious problems weighing planets , suns and so on. Main reason for the unaccounted mass is a "stubborn" mainstream hold on to the shell theorem. For a even density sphere the math ads up to F=const.*M/r^2 but this is only the first half of the "story" (very convenient though). The "assumption" that this simple formula would be good for varying densities is mathematical gimmick seen trough time and time again, but hold on by "mainstream" probably because the "inconvenience to all gravitational theories. I have just integrated an exact formula for calculating the gravity including all mass of a spere. That simply lead to making G a proper constant by getting its value at infinity. This in praxis give a 12,5% bigger G and also 13% bigger celestial masses. For "other shapes" as galaxies this have a even bigger impact, and I think that mainstream on hold to "satellite" orbits in galaxies is not stupidity but the exempt of alternatives, all of which change with both more gravity and more mass. And your ridiculing of my every attempt to close in on the "hated" math. is surely undermining everything for only one obvious reason......

A example with the horrid calculator. A 100km high beam standing upright is "drag" by 9,81m/s2 in its low end and 9,81*(6371000^2/6471000^2)=9,50914m/s2 on its upper end, average=9,65957m/s2. Mainstream (this little antic) calculate the total is 9,81*(6371000^2/6421000^2)=9,65781m/s2 The average value is obtained at6420415 from center of earth (the gravitational mid point ), a small 585m difference (1/100 of a%). This work god enough for anyone, but when "reversed" and calculated via all density layers the gravitational midpoint falls in at 5653348 meters depth. Quite close but the smallish difference mean that earth need to be 1,13 times heavier. This applied to the speres in measurements for G make a 12,5% error. The "sum" of these 1,125*1,13=4/pi ads up as the ratio in energy between a free falling small "stone" from infinity towards the center of body to a height of r above the center and a stone falling to a point r besides the body (point mass is good enough). It turn out that the straight at give the stone a kinetic energy equivalent to half the potential energy at r distance from the body center. The stone falling r besides get pi/4 of that energy in orbital speed.
A second notice: The work of a orbiting body(1kg) W=F*s =2pi*r*g for one lap and the body is also accelerated and braked 8 times to orbital speed and back to 0 giving v^2*8. as an example moon earth 2pi*385E6*0,0034m/s2=8,22MJ/ (1kg) and 1014m/s^2*8=8,22MJ/kg, MARK! This is true only for The new values for Me=6,77*10^24kg and G=Ti=7,506Nm2/kg2. Since the gravity midpoint is much "closer" at short distances all the mass and "G" is needed. This apply directly to the solar system, but planets satellites have yet a surprise or two, while the shape of galaxies need their own formulas (4 of them). From here the theory is plain "arithmetics" done by math. formulas, which if necessary since there are no point masses. Einstein liked to refer to point like masses, there are no such either.

51. µ=4π²a³/T²

Since this came up I need point out that that I of course do not just like that accept random tables as very "correct", even if the one you pasted is lax for anything. I have mass in abundance so I can freely assume there is all the energy for all orbits, no need for Keplers negative energy or any tricking with inertia. "Wrote" a new ellipse with one focal point and better fitting profile (a=-pi to+pi) r^2=(ap*sin(a/2))^2+(per*cos(a/2))^2 looks "common" but have for real a different profile from the traditional math. one. The perimeter is per=(a+p)^0,5*(2*a^2+2*p^2)^0,25*pi I have overdone the accuracy by removing the barycenter length so a=-2bary/(ap+per)*ap+ap and p eq. By doing the energy integral over the angle I was able to counter two small "inexacts". From the elliptical energy I removed the excess part since "I have energy to spare" and the orbit is also ment to cover longer orbits (comets). E=µ*((a^2+p^2)^0,5*(8a^2+8p^2)-2^(5/2)*p^3-3*2^(5/2)*a^2*p)/(3*2^0,5*(p^4-2+a*2*p^2+a^4)) +
µ*((a^2+p^2)^0,5*(8a^2+8p^2)-2^(5/2)*a^3-3*2^(5/2)*p^2*a)/(3*2^0,5*(p^4-2+a*2*p^2+a^4))
Now since the math is "mostly" denoted, I remove a few moons and add the wobble of the sun, that is 1,51*10^9m sum of the planets barycenters and 2,0*10^9m for covering Jupiter and Saturns tidal wobbling (the moons did not explain their high energies +0,4% and 0,13% while others were spot on <10^-4. The 2*10^9m is also for asteroids and such, but do not seem to imply that the sun could have any wobble separate from the "system" After lining up a reasonable average is 1,3285*10^20m3/s2 a 0,1% more than Wiki but I trust my own better which do not mean anything, not even to me, but the mat seem to work very exactly. Using the perimeter to calculate the average orbital speed and divided by 2pi for r give a very similar result v^2*r for the less elliptic orbits, and are several days faster
So there is the energy from orbiting the sun, this later plus the energy of the planet pendling back and forth "by its own" (the excess from the "longer calculation". Maybe Kepler was right after all, except for the minus sign (and quantity)

1. This is calculation from the orbit so its length need be exact. Eliminating sun movement.
a=ap-2*bary*ap/(ap+per) and p=per-2*bary*per/(ap+per)

2. True length of orbit. Perimeter =(a+p)^0,5*(2*a^2+2*p^2)^0,25*pi give a true orbital speed vo=perim./T T is total orbit time. A average radius rel=perim./(2pi)

3. The standard gravitational parameter µ=vo^2*rel This is also "mainstream" µ=G*M even for elliptic orbits because the ellipse fit rel and exact vo.

4. The integrated formula give total energy of the orbit per kg of planet(moon). The amount that is over µ/rel is possible because the center mass is enough to "carry" substantially more energetic orbits.

6. The wobbling of the sun btot (the sum of all barycenter) is accounted for by "adding" it to formula for ellip. radius. rel=(a+p)^0,5*(2*(a+btot)^2+2*(p-btot))^0,25/2 The orbit speed is still from original rel. btot for sun is 3,5*10^9m to 5*10^9m. The orbiting planets may "wobble" each other doubling their effect from 1,5 to 3*10^9m
This last one is for extracting the µ value and do not "depict" any real orbit. The influence of suns wobbling is of course in the measured aps and pers.

Mass of the sun is Ms=8*µ/(pi*mn*c^2) =2,2521*10^30kg for µ=1,3285*10^20m3/s2 and mn (nucleon mass) =1,671351047*10^-27kg

52. Originally Posted by Timo Moilanen
a "stubborn" mainstream hold on to the shell theorem
It's you alone against the entire physics community throughout history since Sir Isaac Newton. How likely is it that all the physicists have made an error in their proofs of the shell theorem. IIRC, I have also proven the shell theorem, but that was a long time ago.

Originally Posted by Timo Moilanen
For a even density sphere the math ads up to F=const.*M/r^2 but this is only the first half of the "story" (very convenient though). The "assumption" that this simple formula would be good for varying densities is mathematical gimmick seen trough time and time again, but hold on by "mainstream" probably because the "inconvenience to all gravitational theories.
It's called the "shell theorem" because it deals with the gravitation from a thin spherical shell of matter. By combining concentric spherical shells, one can obtain the gravitation from a spherically symmetric distribution of matter with an arbitrary radial density distribution. The Wikipedia article "Shell theorem" provides a number of proofs of the shell theorem. Perhaps you could show why these proofs are wrong. A straightforward proof, albeit using more advanced maths, is from the integral form of Gauss's law for gravity. The differential form of Gauss's law for gravity is:

This formula describes the source of the gravitational field at each point in terms of the density at that point. The reason why I chose the differential form is because the differential form is local. That is, at each point in space and time, it describes what happens at that point in terms of what is at that point. This is in contrast to Newton's law which specifies the acceleration of an object in terms of a mass that is some distance away. I see it as a fundamental principle that all physics is ultimately local. Presumably, you disagree with the inverse square law of gravity. Otherwise, you would be arguing against mathematics with regards to your rejection of the shell theorem applied to a non-uniform radial density distribution. Gauss's law for gravity produces the inverse square law, so you are also disagreeing with Gauss's law for gravity. If not Gauss's law for gravity, what local law satisfies your theory of gravity? I specify "local law" because the formula you provided in post #43 looks made up in that it doesn't look like a reasonable law of physics. It doesn't look to me like a law in which the gravitational field behaves at a point in a way that depends only on whatever is at that point. From the differential form of Gauss's law for gravity, the integral form of Gauss's law for gravity:

is obtained by the divergence theorem, which is a three-dimensional generalisation of the fundamental theorem of calculus. What this formula says is that the total inward flux¹ of the gravitational field over an arbitrary closed surface is directly proportional to the total mass inside the closed surface. This is a more general statement than the shell theorem. But the shell theorem can be obtained from it. For one thing, it is immediately obvious that the gravitational field inside a hollow spherical shell is zero. Secondly, for a spherically symmetric density distribution, the total flux over an enclosing spherical surface depends only on the total mass inside the surface and is independent of the radial density distribution of that mass. Therefore, the shell theorem is proven.

In general relativity, there is Birkhoff's theorem, which says that for any spherically symmetric spacetime, any vacuum region has Schwarzschild geometry. In post #8, I said:

In general relativity, one has Birkhoff's theorem. I personally have not seen the proof of Birkhoff's theorem, but I have seen the derivation of the Schwarzschild metric, which I believe implies Birkhoff's theorem. This is because obtaining the Schwarzschild metric involves solving a first-order ordinary differential equation in the radial variable. This differential equation, like all differential equations, describes local behaviour, and the solution over a region is obtained by specifying the solution on a spherical surface (a given value of the radial variable). Thus, the solution outside this spherical surface is determined entirely by the solution on the spherical surface, which in turn is determined by what is inside the spherical surface as a single-value parameter. If we have a matter distribution consisting of concentric spherical shells, then the regions between the shells are also described by Schwarzschild metrics, with each region described by a different mass parameter.

Thus, even in general relativity, a version of the shell theorem is true. Perhaps you should detail where the mainstream has made their error.

¹ The surface integral is defined as an outward flux, hence the minus sign for the mass.

53. because the formula you provided in post #43 looks made up in that it doesn't look like a reasonable law of physics.

I years ago very carefully examined Gauss's law and the gravity I think have gone wrong due to "resembling" too much the "electric" ones by logics. I could not agree with the gravity. About proving this or that about Gauss can be forgotten since I can not write down his logics in math and point to math. error. What comes to looking this or that, the formula is essential since it give the gravitational magnitude to what the photon is "experiencing" reacting to. It is originally the total gravity at a point outside a sphere g=M*Ti/r^2*(3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1))). So I would not call it a law of anything it is just the integral used for calculating (traditionally) the gravity for a sphere. g=M*G*(INT) dM cos(b)/s^2 dx,dy,dz= M*G/r^2 without the cos(b) gtot=M*G*(INT)dM/s^2dx,dy,dz =as above. The first integral I have read should be problematic some how. No problems for even density (did it years ago) for varying densities I think it is undoable at least with pencil. The gtot integral nothing special, and i have done two kinds of arithmetic check on both since for the first there was often it shall be and must be and no integral to be found.
As I see it weighing is to count the nucleons (protons+neutrons) on the scale, so gravity can be seen as N1kg*1000NA*N2kg*1000NA/r^2*const. NA=Avogadro's number and the gravity is how many combinations of nucleon pairs are there (n1*n2/2 combnations) and I put in mn*c^2 to "make" Newton F=N1*N2*1000000*NA^2/2*mn*c^2/r^2*X=M1*M2*Ti/(r^2*k1*k2). For long distances k1 and k2 is close to1. I call 1/k volumetric efficiency k= 3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1)) where p1=r/R1, r is distance between spheres and R1 radius of shere1. This give the opportunity to "test" the experiments "quality" c^2/(2000*G*k1*k2)=NA. G is the value calculated from the experiment, averages of course are not so exactly 5,98*10^23. I have not exactly got invitations to "validate" institutions "doings". This I think prove quite much of my theory, but I have no bunch of proven experiment results to calculate and show.
For a shell gtot=M*Ti*(ln(p+1)-ln(p-1))/(2*r*R)

So what comes to what I have against me,..........NOTHING CONVINCING

I think Einstein's 1,75arcsec got right because even if G was 1 it was 1,125 times to small in correlation to other parameters and the sun mass was 1,13 times too small

54. Originally Posted by Timo Moilanen
I years ago very carefully examined Gauss's law and the gravity I think have gone wrong due to "resembling" too much the "electric" ones by logics. I could not agree with the gravity.
Why is Gauss's law for gravity resembling Gauss's law for the electric field a problem? Actually, they are opposite in sign as a result of one being attractive and the other being repulsive. But the differential form of either law is just an expression of the source of the field being a density.

Originally Posted by Timo Moilanen
No problems for even density (did it years ago) for varying densities I think it is undoable at least with pencil.
If you can work out the gravity from a thin shell of arbitrary size and density, then for a ball of matter, it makes no difference whether the radial density distribution is uniform or varying. You still haven't justified your claim that the shell theorem is invalid.

You still haven't accepted my request to provide a local law for gravity. If you don't know what the source of the gravitational field is, then how can you calculate the total gravity at a point outside a sphere?

Also, you haven't addressed Birkhoff's theorem, the general relativistic form of the shell theorem. Indeed, you have not provided any justification for your opposition to general relativity.

55. This is supposed to be true for any volume. I have only vectors from point "masses" so this is not true for any volume

Putting M=4pi*R and integral (R=0 to1) of M*Ti*(ln(p+1)-ln(p-1))/(2R*r) is always smaller than 4pi/(3*r^2) so it is not true with even densities. I have nonlinear gravity "close" outside the sphere, g=m*Ti/r^2 is for point masses only.

The gravity g outside an even density sphere of radius R at distance r with mass M. g=gm^2/(gtot*r^2)=M*Ti/(k*r^2).
gm for a sphere is the familiar gm=M*Ti/r^2 where Ti is G at infinite distance, That is when measuring G in the "experiment" Ti=G*k1*k2.
gtot=m*Ti/r^2*(3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1)) where P=R/r. For spheres I use gtot/gm=k since the easy formula. For example on surface of a even density sphere k=3/2 and on sea level earth about 1,27 due to the different densities. These density "layers" project on the low orbits and less higher, Geostationary k=1,004 Moon k=1,00005. The very low orbits are not much "disturbed" of the almost nonexistent atmosphere but have a weaker gravity (higher k-value).

Also, you haven't addressed Birkhoff's theorem, the general relativistic form of the shell theorem. Indeed, you have not provided any justification for your opposition to general relativity.
I build on Newton remember, Birkhoff do not agree with me. From relativity theory I agree whit principles of equivalence till they are abused, from the rest I think you have tried to "sell" me most with little coot.

Here a quite careful calculation that I have had no luck sending to NASA. I would be forever indebt to anyone who can forward this table. And they will too the sooner the better.

56. Originally Posted by Timo Moilanen
I have nonlinear gravity
Gravitation in general relativity is non-linear. However, in the weak-field limit of Newtonian gravity, it is linear. But, the non-linearity of general relativity is a difficulty.

For a spherically symmetric ball of matter, the external gravitational field of the Schwarzschild metric:

g = GMr–2(1–2GMc–2r–1)–1

In the weak-field limit:

(1–2GMc–2r–1)–1 = 1

and therefore:

g = GMr–2

It is worth noting that r is the standard radial parameter of the Schwarzschild metric and is not exactly equal to the distance from the centre of the ball of matter. However, for C defined as the circumference of the circle centred at the centre of the ball of matter, and passing through the point at r:

r = C/2π

This is exact by definition of the r parameter of the Schwarzschild metric. The non-exactness of r as the distance from the centre of the ball of matter is due to the spacetime curvature.

Originally Posted by Timo Moilanen
I build on Newton remember
Yes, you've said that, but why build on Newton when general relativity is the more correct theory? Newton is still in use because it is relatively simple, and gives good enough results where the gravitation is weak. But, by building on Newton, you remove the main benefit of simplicity, in which case you may as well be using general relativity.

Originally Posted by Timo Moilanen
From relativity theory I agree with principles of equivalence till they are abused
I don't see any agreement with the principle of equivalence. This thread is about a notion that demands that gravitation is not equivalent to acceleration.

Originally Posted by Timo Moilanen
Here a quite careful calculation that I have had no luck sending to NASA. I would be forever indebt to anyone who can forward this table. And they will too the sooner the better.
I think NASA is in a much better position than either you or I in knowing the gravitation of astronomical objects such as Mars. You haven't convinced me that your theory has validity, so why would you think that NASA would be any more likely to be convinced? One reason why you haven't convinced me is that you haven't provided any principles upon which your theory is based. You seem to think that formulae and numerical values are enough. No, formulae and numerical values can be made up, and without the principles to back them up, I can only assume that the formulae and numerical values are made up.

In post #13, you said that NASA have smashed landers all over Mars. I do not believe that NASA have ever crashed a lander due to an error in gravity. Please supply a reference to support this statement.

57. I don't see any agreement with the principle of equivalence. This thread is about a notion that demands that gravitation is not equivalent to acceleration.

Yes I agree to equivalence, I have added that for different densities the shape (size) must be the same. For a hammer and feather this seem like trifle but for a planet moon pair the is often huge difference. For our moon it is about 1/1,00005. Well the moon don't fall down for any calculated "values" but there is abandoned orbits, especially around the moon that only need "adjustment".
The gravitation that we measure and calculate (even tough I get "mostly bigger values) is absolutely accelerating falling objects very classically but is needed 4/pi times more (than mainstream) to "make" a orbit. And no problem all the gravity is there (in abundance), we need to consider that "objects" are falling faster than orbits predict (mainstream). For light all gravity from all directions (gtot) must be accounted for.

Everything points to that NASA do not know accurately enough, and they can hardly blame a gravity they do not know existing.
My table I refer to seem to have disappeared, so much for this attempt.

As for principles " the shell theorem do not work " and everything must be calculated the hard way geometrically. After a few years grinding the math and confirming by very sparce data the details fall in. As for relativity it is interesting but do not work with my masses. I think, I have not tested very mush with masses times1,13 and G times1,125.

A second small adjustment is the elliptic orbit, that I cant say is perfect match to real orbits but it is better and enables the 21,5% lower "entropy" compared to the by now historical centripetal force = gravitational force. It is surprising that a labile system (orbit) that has a positive feedback can have been used for hundred of years. There must have been no alternatives since it is very clear that planets and satellites would fly away or drop all the way down, with zero energy marginals.

Probably I could have written to NASA the American way, "You guys are doing these rocket flying por attempt totally wrong, let me tell you how it really shall be done
. But the ting is now that there are still phenomenon that need to be taken into account, the more dynamic parts of gravity (orbits). For galaxies I have math but little inputs and not the computer power to simulate whit any flexibility.

58. Originally Posted by Timo Moilanen
Originally Posted by KJW
I don't see any agreement with the principle of equivalence. This thread is about a notion that demands that gravitation is not equivalent to acceleration.
Yes I agree to equivalence, I have added that for different densities the shape (size) must be the same. For a hammer and feather this seem like trifle but for a planet moon pair the is often huge difference. For our moon it is about 1/1,00005. Well the moon don't fall down for any calculated "values" but there is abandoned orbits, especially around the moon that only need "adjustment".
The gravitation that we measure and calculate (even tough I get "mostly bigger values) is absolutely accelerating falling objects very classically but is needed 4/pi times more (than mainstream) to "make" a orbit. And no problem all the gravity is there (in abundance), we need to consider that "objects" are falling faster than orbits predict (mainstream). For light all gravity from all directions (gtot) must be accounted for.
Hmmm, it seems that I need to explain why this thread is about a notion that demands that gravitation is not equivalent to acceleration. The title of this thread is about the notion that a beam of photons decay while bending in a gravitational field. Suppose one is an observer in an inertial frame of reference far away from any gravitation. A beam of photons travels in a straight line and does not decay according to this observer. Consider another observer in an accelerated frame of reference observing the same beam of photons. This observer sees the beam of photons bend, but it does not decay because it is the same beam of photons being observed by the observer in the inertial frame of reference. But in a gravitational field, you claim that a beam of photons bends and decays, and therefore behaves differently to being in an accelerated frame of reference. In other words, in your theory, gravitation is not equivalent to an accelerated frame of reference.

This non-equivalence arises because, in your theory, the photons physically interact with the gravitational field. In other words, the gravitational field is something which interacts with photons and presumably other things. This is different to gravitation in general relativity, which is not a thing that interacts with anything¹. In general relativity, objects behave in a gravitational field the same way they behave in an accelerated frame of reference, which is the same way they behave in an inertial frame of reference but viewed from an accelerated frame of reference. If the gravitational field is something which interacts with photons and presumably other things, then how can you guarantee that it doesn't interact differently with different things?

¹ Assuming the object is small enough that the tidal effect can be ignored. In the case that the tidal effect cannot be ignored, the behaviour of the object can be determined by considering the effect of different accelerations on different parts of the object.

Originally Posted by Timo Moilanen
As for principles " the shell theorem do not work " and everything must be calculated the hard way geometrically.
This is not a principle. It is a conclusion arising from a made-up law of gravitation. Because you haven't said anything to convince me that your law of gravitation isn't made-up, I can conclude that the shell theorem is still valid. And you still haven't properly addressed Birkhoff's theorem.

Originally Posted by Timo Moilanen
Everything points to that NASA do not know accurately enough
What is this "everything" that points to "that NASA do not know accurately enough"? Please supply a reference that supports your answer to this question.

59. In other words, in your theory, gravitation is not equivalent to an accelerated frame of reference.

Accelerating a beam (torch) with a rocket do not bend or degrade it, except for the rockets modest mass. When in a "capsule" in gravitation field the photons bend and degrade regardless of the capsule accelerating (or falling) or not moving in relation to the field. Speed induced red/blueshift do not decay light. And above all light interact with the total gravitation fields from every direction. On the surface of a even density sphere the weight of a particle is m*M*Ti/(k*r^2) a photon reacts to the field M*Ti*k/r^2, k on a sphere surface 1,5 . Light (and probably some subatomic particles) react to total gravity field while mass react to the vector sum of all fields. g on earth 9,81m/s2 and gtot=15,8m/s2 since k on earth surface is about 1,27

About Birkhoff theorem I agree that an other gravitational field do not influence the symmetrical spherical field of a body, but that it should vanish I can not agree on. The field remain inverse square for infinity.
My law of gravity is actually made by Newton I just have "finished" it. I suppose that male it a modified theory.

I put in a large table that vanished from my input that would have bee mush telling for people reding "numbers"

60. Originally Posted by Timo Moilanen
Accelerating a beam (torch) with a rocket do not bend or degrade it, except for the rockets modest mass.
I did not mention a rocket. I'm talking about a beam of light that is straight in an inertial frame of reference, but observed by an observer in an accelerated frame of reference to be bent. I did not say that the beam of light was being accelerated by a rocket. The accelerated observer can be in a rocket, but the beam of light itself is outside the rocket and not connected to the rocket in any way. It is important to understand that nothing is causing the beam of light to bend, it is merely the way the accelerated observer observes the beam of light. And that is a key point concerning how general relativity regards gravity.

Originally Posted by Timo Moilanen
And above all light interact with the total gravitation fields from every direction.
And that is a violation of the equivalence principle.

Originally Posted by Timo Moilanen
About Birkhoff theorem I agree that an other gravitational field do not influence the symmetrical spherical field of a body, but that it should vanish I can not agree on. The field remain inverse square for infinity.
What are you talking about? Your statement makes no sense with regards to Birkhoff's theorem.

Originally Posted by Timo Moilanen
I put in a large table that vanished from my input that would have bee mush telling for people reding "numbers"
No, I asked for a reference, not a large table.

61. And that is a violation of the equivalence principle.
I am sorry for that but for my theory it is essential that light "adsorb" all gravity and curve more than a mass particle would.
Strong- weak- Einstein's principle , how?

What are you talking about? Your statement makes no sense with regards to Birkhoff's theorem.
Did you mean that a symmetrically pulsating star can not emit gravitational waves.

No, I asked for a reference, not a large table.
The table is the reference. There are no site for what NASA don't know to refer to on Wikipedia, and that's why a good assumption is they know everything (at least next to you)?

There is very little to gain if you see everything and all only through relativistic "glasses" since I specifically said this is not a relativistic theory. And what comes to any constructive discussion you have already scared away decent people.
Thanks for nothing
And on Mars gravity has been the main challenge from day one, and still will be.

62. Originally Posted by Timo Moilanen
for my theory it is essential that light "adsorb" all gravity and curve more than a mass particle would.
And if it doesn't?

Originally Posted by Timo Moilanen
Did you mean that a symmetrically pulsating star cannot emit gravitational waves.
I wasn't specifically referring to that, but it is a consequence of Birkhoff's theorem. Birkhoff's theorem can be regarded as an extension of Newton's shell theorem to general relativity. It allows one to cookie-cut a spherically symmetric matter distribution into a Schwarzschild metric. However, one of the results to emerge from solving the Einstein field equations for a spherically symmetric vacuum is that the solution must be static. A static spacetime is a spacetime that does not change with time and is also irrotational (it is a special case of a stationary spacetime, which is a spacetime that does not change with time but can rotate). This implies that spherically symmetric pulsating matter cannot emit gravitational radiation. Indeed, spherically symmetric gravitational radiation cannot even exist.

Originally Posted by Timo Moilanen
The table is the reference.
No, if this were a court of law, the table would be considered as hearsay.

Originally Posted by Timo Moilanen
There are no site for what NASA don't know to refer to on Wikipedia
Then where are you getting your information from?

Originally Posted by Timo Moilanen
There is very little to gain if you see everything and all only through relativistic "glasses" since I specifically said this is not a relativistic theory.
Let's not forget that your theory has to agree with reality. And given that general relativity is the best theory about gravitation that we have, it is only natural that your theory be compared to general relativity. Even if your theory is not intended to be a relativistic theory, it is still required to be better than Newtonian theory, otherwise there is no point to the theory. But the weak-field limit of general relativity is Newtonian theory, so your theory is not even agreeing with general relativity as a non-relativistic approximation. I do recognise the difference between something that is an approximation, and something that is wrong, and your theory is not an approximation. The main reason why I refer to general relativity is because of its explanatory power. General relativity actually explains what gravitation is, and when I see you claim something that flies in the face of general relativity's explanation, I naturally see this as wrong even for a non-relativistic theory.

Originally Posted by Timo Moilanen
Thanks for nothing
Stop being petulant. Your theory is getting the discussion it deserves.

Originally Posted by Timo Moilanen
And on Mars gravity has been the main challenge from day one, and still will be.
You say this but you don't supply a reference to back it up. Why should I believe you?

63. And if it doesn't?
It is how much it is measured to bend and I suppose light don't stop bending due to gravity.
No, if this were a court of law, the table would be considered as hearsay.
Gravity is more than an assumption, but dark matter could qualify as hearsay your honor.
Then where are you getting your information from?
When there is no measurements calculations are better than just guesses.
I naturally see this as wrong even for a non-relativistic theory.
Happy hunting for dark matter you need tons of it (luck)
You say this but you don't supply a reference to back it up. Why should I believe you?
Well I can not read the news for you. Maybe digital news would go as evidence.

64. Originally Posted by Timo Moilanen
Originally Posted by KJW
And if it doesn't?
It is how much it is measured to bend and I suppose light don't stop bending due to gravity.
No, my question was about why light bends. General relativity also gives the correct value for the deflection, but an entirely different explanation. Actually, general relativity did something better... it predicted the correct value before the measured value became known.

Originally Posted by Timo Moilanen
Originally Posted by KJW
No, if this were a court of law, the table would be considered as hearsay.
Gravity is more than an assumption, but dark matter could qualify as hearsay your honor.
Non sequitur.

Originally Posted by Timo Moilanen
Originally Posted by KJW
Then where are you getting your information from?
When there is no measurements calculations are better than just guesses.
No, calculations are not a substitute for measurements. You have claimed that the gravitation of Mars is stronger than what NASA believes. This involves measured data, the measured data indicating that the gravitation of Mars is actually stronger than standard theory. Your theoretical calculation that the gravitation of Mars is stronger than standard theory is of no value to me. I want to see evidence that your theory matches reality. This involves getting data for reality, actual data from a reputable source, in other words, not you.

There is a broader issue here, the underlying reason for my question: You make the claim that your theory agrees with measurements of reality. But if you are unable to provide a source of the measured data, then this calls into question your ability to compare your theory with reality, and therefore the veracity of your claim that your theory agrees with reality.

Originally Posted by Timo Moilanen
Originally Posted by KJW
I naturally see this as wrong even for a non-relativistic theory.
Happy hunting for dark matter you need tons of it (luck)
The scientific community do not know what dark matter is, but neither do you, so this high and mighty attitude is unwarranted. But as I mentioned in post #46, the observational evidence indicates that dark matter is some form of matter and not a modified law of gravity.

Originally Posted by Timo Moilanen
There is very little to gain if you see everything and all only through relativistic "glasses" since I specifically said this is not a relativistic theory.
There is one other thing to say in reply to this: In saying that you specifically said this is not a relativistic theory, either you are providing a non-relativistic approximation to relativity, or you are disagreeing with relativity. The former case is that you accept relativity but are providing a simplification that is intended to work under similar conditions as Newtonian gravity, but not under much more extreme conditions. The latter seems to be your case. It seems to me that you believe your theory is THE theory that will eventually replace general relativity. If that's the case, then you don't get to tell me that there is very little to gain if I see everything through relativistic glasses. That's because you've set yourself in opposition to relativity, and therefore seeing everything through relativistic glasses is the only way to see everything.

Originally Posted by Timo Moilanen
Originally Posted by KJW
You say this but you don't supply a reference to back it up. Why should I believe you?
Well I can not read the news for you. Maybe digital news would go as evidence.
Well, I haven't seen those stories, so could you supply a link?

65. <span style="color: rgb(51, 51, 51); background-color: rgb(246, 246, 246);">And if it doesn't?<br>It is how much it is measured to bend and I suppose light don't stop bending due to gravity.<br></span><span style="color: rgb(51, 51, 51); background-color: rgb(246, 246, 246);">No, if this were a court of law, the table would be considered as hearsay.<br>Gravity is more than an assumption, but dark matter could qualify as hearsay your honor.&nbsp;<br></span><span style="color: rgb(51, 51, 51); background-color: rgb(246, 246, 246);">Then where are you getting your information from?<br>When there is no measurements calculations are better than just guesses.<br></span><span style="color: rgb(51, 51, 51); background-color: rgb(246, 246, 246);">&nbsp;I naturally see this as wrong even for a non-relativistic theory.<br>Happy hunting for dark matter you need tons of it (luck)<br></span><span style="color: rgb(51, 51, 51); background-color: rgb(246, 246, 246);">You say this but you don't supply a reference to back it up. Why should I believe you?<br>Well I can not read the news for you. Maybe digital news would go as evidence.</span>

66. Here an attempt for Mars gravitation altitude table

The same via gmail. The last column is "mainstream" orbital speed

67. Originally Posted by Timo Moilanen
Here an attempt for Mars gravitation altitude table
This wants me to sign in to Google. Anyway, I suspect that it doesn't satisfy my request for data that doesn't come from you.

Originally Posted by Timo Moilanen
[ IMG]https://mail.google.com/mail/u/0?ui=2&ik=4814565b7f&attid=0.1&permmsgid=msg-a:r4930753229906386414&th=185cee3c10de7243&view=fi mg&fur=ip&sz=s0-l75-ft&attbid=ANGjdJ-FYNbUXqcY_to74NceAmz2CkHHi2gWzF-Eo1PLhO6mABwsqhAjoBKO5sAK3L8FqRJWAEvHqH8uGailbvRR3 T_Xwvr6C5bonIotAl5hQ8uRpAeb7hdG46suan0&disp=emb&re alattid=ii_ld4fd7ss0[ /IMG]
The image (whose url I revealed by modifying the img tags) didn't display. Anyway, I suspect that it doesn't satisfy my request for data that doesn't come from you.

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