Thread: The great error of Einstein (the proof)

Equivalence principle of the General Theory of Relativity is absolutely wrong!

This is the proof.

http://www.tsolkas.gr/english/docume...uivalence.html


tsolkas

Did you understand any of this?
I didnt

Elaugree, dont bother, he is a crackpot.
wroomfondel went throu his mathematic and saw many many errors if im not misstagen, i know atleast he found some. shouldnt this be in pseudoscience

Originally Posted by Zelos

shouldnt this be in pseudoscience

Yes.

Hmmm... is this trying to say that objects of different mass will hit the ground at a different time?






J0N

yes, he i using fancy math to make normal people confused and trust him becuase of his fancy math

Originally Posted by Jon

Hmmm... is this trying to say that objects of different mass will hit the ground at a different time?






J0N

Galileo Galilei will be turning in his grave

they went to his grave, apperently his tombestone says now "I wasted my life going against the church just to some know-nothing smart boy called tsolkas-1 can threw it away?"

Originally Posted by Zelos

they went to his grave, apperently his tombestone says now "I wasted my life going against the church just to some know-nothing smart boy called tsolkas-1 can threw it away?"

i think this is dumb

this topic? sure is

Since you both think so, why not form a joint team to dissect the paper by Tsolkas, in detail. From this dissection prepare a formal, mathematical proof demonstrating the error(s) in Tsolkas's thinking.

Clearly this should not be difficult for either of you, since you have been able to determine, almost instantly, and beyond any personal doubt, that Tsolkas is talking rubbish. For those of us lest gifted in mathematics and less knowledgeable in the physical sciences such a demonstration would be both educational and reassuring.

I look forward to your presentation. It will be much more edifying than the one line put downs you have been delivering up until now. Even I can do those.

Ophiolite, dont u remember wroomfondel did that a while ago?

Hi,
Just a quick thought...

Is "R" in equation (1) constant or not? You canceled them in your equation for "A" in your proof thus implying R is constant.

If R doesn't change, you have disproven yourself! (Think about it...)



I sort of stopped there, but as a footnote, I think you are dividing by zero in your equation (2) if v1 and v2 are indeed equal. You can prove 1=2 if you divide by zero....

william

Having read this thread I wish it to be known that I, the author of the hammer and feather thread have no association with tsolkas? whatever.

Having briefly read his paper I do not agree with his findings, I do NOT have the time at present to dissect it. My experience is that if something looks 'rong' it probably is. :wink:

Billco,
Are you sure you're not collaborating with tsolkas-1 working on a paper to be published in Nature???

His drawings are looking more and more like that hammer and feather you were talking about.... :-D :wink:

I want in!
william

Originally Posted by william

Billco,
Are you sure you're not collaborating with tsolkas-1 working on a paper to be published in Nature???

His drawings are looking more and more like that hammer and feather you were talking about.... :-D :wink:

I want in!
william

As an engineer(retd) and tutor(private) I would not have any truck with anybody who tried to 'blind' others with mathematics - if his argument is aimed at lessor mathematicians THAT is the way he should have written it. I state that the object of 'Hammer and feather' is purely to get you guys to question science - good science will always answer with replies you can understand. Since others have confirmed my findings (at a Newtonian, Pythagorian) level there is no need for me to publish my proof. If you are smart you can do it yourself. I have posted the figures used, and detailed the formula and conditions. If I told you a pure capacitance of 100picofarads and a pure inducatance of 3.2 micro-henrys has a resonant frequency (in perfect conditions) of >8Mhz using the standard formula for resonance in a tuned circuit - that should be enough I may add that Fo= 8.897031Mhz +/- 1Hz - using f=1/(2pi root(lc)) I do not need to show the working UNLESS I am teaching. - My original post said "Does anybody agree" - IF I had shown my working everybody would have agreed - I invited others to come to the same conclusion or otherwise from their own working. What happened was a barrage of 'everybody knows you're wrong' - hardly a scientific rebuttle.

Now young William ,what is an accurate formula for determining how far away the horizon is? to an accuracy of <1%

You may ignore, Atmospheric abberations, effects of curved space time,
you may assume the body you are standing on is a perfect sphere, of uniform density, and the horizon is clearly visible.

Enough of the smugness billco - if criticisms are invalid explain why, don't just cop out with "if you're smart you can do it yourself". My main criticism, especially without seeing your "proof", is that there are probably a multitude of factors you did not and probably cannot consider because the magnitude of the difference in when the feather and hammer hit is so small. Rotation of said hammer and feather, for example. And that being the case, it's hard to have confidence in the conclusion. Simply repeating that you are right isn't very convincing.

Originally Posted by Neutrino

Enough of the smugness billco - if criticisms are invalid explain why, don't just cop out with "if you're smart you can do it yourself". My main criticism, especially without seeing your "proof", is that there are probably a multitude of factors you did not and probably cannot consider because the magnitude of the difference in when the feather and hammer hit is so small. Rotation of said hammer and feather, for example. And that being the case, it's hard to have confidence in the conclusion. Simply repeating that you are right isn't very convincing.

Rotation? caused by what?


I am dropping a hammer and feather not an olympic freestyle high diver!

Did I miss something in the Formula a=G(m1*m2)/r^2

Or was it in F=ma

Or perhaps a^2=b^2+c^2

Whoah...
I was only teasing you billco. That's all.
I said I agree with your hammer/feather assessment. I did the math that I suggested in that thread (see the "hammer/feather" thread in the physics section for those unfamiliar with this discussion). Neglecting constants (other than G of course), I get that the collision time between two point masses goes like

t ~ r^(3/2)/sqrt(G(m1+m2)).

You'll notice both masses are in the answer. Since the masses are arbitrary, one of them can be your moon or what-have-you and the other can be either the feather or hammer. Either way, I'm sure you see that the time (t) depends on both masses => the greater the masses, the shorter the time => hammer hits first. If there is any critique of using point masses instead of extended masses... my reply would be that the same reasoning applies, except the math gets uglier.

Also, billco and neutrino, I think there is a misunderstanding between you two. Billco is not the author of this particular thread.

Take it easy billco. I'm on your side!

billco wrote:

Now young William ,what is an accurate formula for determining how far away the horizon is? to an accuracy of <1%

You may ignore, Atmospheric abberations, effects of curved space time,
you may assume the body you are standing on is a perfect sphere, of uniform density, and the horizon is clearly visible.

Firstly, thanks for calling me young.

Okay...
For the answer,

s=sqrt(2hR)

where R is the radius of the object you are on (earth, moon, sun (ouch!), whatever)
h is the height above the surface (specifically, the height of your eyes I guess)
and s is the distance measured along the surface (that is, not the direct line of sight).

This is accurate to within 1% for h<153.71 km on earth (unless I goofed punching in the numbers...). For a person ~ 183cm tall, on earth, this is accurate to within 0.0013% (again, unless I goofed punching in the numbers...). I used R(earth) = 6.378 * 10^8 cm.

I neglected all the things you said I could, and you did say I could neglect curved space-time, and photons are massless....

I am guessing that you somehow use the mass (or density) of the earth in your formula. As you can see, I did not....

I look forward to your derivation.

Cheers,
william

P.S. billco, I'm starting to like you. So don't take what I say in a negative way. Okay?

Ok,

Well done for NOT using the ugly formula. Now just type in the following numbers and tell be what answer you get! My eye height above ground is 2 Metres, the Radius of the planet I am standing on is ONE METRE.
Off the top of my head the distance should be around 2.828M - what does your Formula say ??? It's 2.449 ?? close but not perfect.

Now draw this out: A sphere centre point (origin = 'O') = Standing on the surface of the sphere is little old me, 2.0 Metres from planet surface to eye height, I am standing upright, I can see the horizon,my eye is at position 'E' and the horizon is at point 'H' points O,H,E form a right angle triangle with the right angle being angle 'OHE'. Side OE is therefore the Hypotenuse. Now thanks to the tin man in 'The Wizard of OZ' we know that: "The Square of the Hypotenuse is equal to the sum of the squares on the other two sides" - (And you thought it was Pythagorus!)

NOW - we know the length of the Hypotenuse 'OE' it is the radius of the planet + my eyeheight. We also know the opposite 'OH' and this is simply the Radius of the planet. The distance to the horizon is then

Let r = OH and h = OE-OH ie my height.
simply Sqrt( (r+h)^2 - r^2) or (multiply out) = Sqrt(h^2+2rH)

I think somewhere you lost h^2 So go back over your work, and find it!

Rgds, BillCo.

Real quick billco (before you retire to bed...)
Do realize my formula gives the distance along the surface, not the line-of-sight distance. Did you take that into account?
I'll now give your post some more thought. But... my solution is correct within the given errors I stated....

Regards,
william

Hi Billco,
There is some confusion here about "distance to the horizon." The confusion is "line-of-sight" distance versus distance "along the surface."


Billco wrote:

Well done for NOT using the ugly formula. Now just type in the following numbers and tell be what answer you get! My eye height above ground is 2 Metres, the Radius of the planet I am standing on is ONE METRE.
Off the top of my head the distance should be around 2.828M - what does your Formula say ??? It's 2.449 ?? close but not perfect.

Okay, you have moved the goal posts...
Using your new numbers, my formula gives the distance as 2 meters (along the surface) - not 2.449 as you stated. Note that 'h' is the distance above the surface, not from center.

On the 1-meter radius planet, my formula is only accurate to within 62.5%. This is because 'h' in my formula should be quite a bit smaller than 'R' to maintain an inkling of accuracy. If this were earth, you'd be 12,756 km tall! That is, you'd be as tall as the earth is in diameter! In that case, I'd have to tighten up the assumptions I used when deriving my formula.


Yes, yes... you are using the tin-man (Pythagorean) theorem. That gives the line-of-sight distance. My formula gives the distance along the surface. I suspected this was the source of the confusion.

I want to make sure you understand what I mean by "distance along the surface." Let's say you are on a tower of height h and you see the horizon. Now, you want to go to that spot (the horizon). The distance along the surface is the distance you would walk or drive to that spot once you climed down from the tower.

Allright, if I had known that all you wanted was the Pythagorean theorem, it would have saved me some work. So... to make it up to me, I'll ask you to give me a formula for the distance to the horizon along the surface. That is, I'll ask you to derive the formula I gave (or another one - as long as it gives the distance along the surface). And no cheating. I'm actually asking you to derive it, not just restate it.

Are you up to the challenge?

Cheers,
william

The answer is infinity - you cannot walk to the horizon!

otherwise 2*pi*r = circumference

now just find our angle as a portion of 360 degree using C A H rule.

G'night!

Billco wrote:

The answer is infinity - you cannot walk to the horizon!

Well... you know what I meant. Distance to THAT spot. This is not one of those 'end of the rainbow' thingys....

G'night,
william

you guys can get off of topic ANYWHERE. :P

Okay, I realize that this is not the proper thread for this, and is way off topic, but if anyone wants to brush up on some math, I recommend deriving the formula I gave above

s=sqrt(2hR)

where R is the radius of the object you are on (earth, moon, sun (ouch!), whatever)
h is the height above the surface (specifically, the height of your eyes I guess)
and s is the distance measured along the surface (that is, not the direct line of sight).

for the horizon problem.

It involves a Taylor expansion and a binomial expansion and you start this task with good-old trig. It might be good practice for any future physicists (vroom, zelos, leap). If anyone cares to do this, and wants more hints, feel free to PM me.

For further practice, see if you get the same numbers as I for the uncertainty

This is accurate to within 1% for h<153.71 km on earth (unless I goofed punching in the numbers...). For a person ~ 183cm tall, on earth, this is accurate to within 0.0013% (again, unless I goofed punching in the numbers...). I used R(earth) = 6.378 * 10^8 cm.

This too requires a bit of thought.

Cheers,
william

something seems wrong here. Is that just the first term in the taylor expansion? because, according to that, as h -> infinity, so will s. But that isnt true, it will approach a quarter of the circumference of the circle.

Originally Posted by Vroomfondel

something seems wrong here. Is that just the first term in the taylor expansion? because, according to that, as h -> infinity, so will s. But that isnt true, it will approach a quarter of the circumference of the circle.

You are quite correct,you cannot see further than pi*r/2

See my Post: Wed Aug 30, 2006 3:44 pm

Hi Vroom,
I'll have to see your work. PM it to me if you want.

Cheers,
william

Okey lets make 2 things clear here
1: im perfect so never argue with me( )
2: Resistence is futile( )
adding a third
3: you are going way of topic as i see it
adding forth
4: Negotiation is irrelevant( )

Originally Posted by Zelos

Okey lets make 2 things clear here
1: im perfect so never argue with me( )
2: Resistence is futile( )
adding a third
3: you are going way of topic as i see it
adding forth
4: Negotiation is irrelevant( )

1) We don't argue with you, we stand B4U in Awe.
2) Resistance is voltage divided by current.
3) (still trying to work out what it means).
4) to what?

1) We don't argue with you, we stand B4U in Awe.
2) Resistance is voltage divided by current.
3) (still trying to work out what it means).
4) to what?

1: HAIL
2: Yet its futile
3: Me to
4: negotiation is irrelevant

Originally Posted by Borg

We are the borg.
You will be assimilated.
Existens as you know it is over
Your biological and technological distinctivness will be added to our own
You culture will adept to serv us
Resistence is futile
Negotiation is irrelevant. You will be assimilated

Originally Posted by Zelos

1) We don't argue with you, we stand B4U in Awe.
2) Resistance is voltage divided by current.
3) (still trying to work out what it means).
4) to what?

1: HAIL
2: Yet its futile
3: Me to
4: negotiation is irrelevant

Originally Posted by Borg

We are the borg.
You will be assimilated.
Existens as you know it is over
Your biological and technological distinctivness will be added to our own
You culture will adept to serv us
Resistence is futile
Negotiation is irrelevant. You will be assimilated

And he dares to call me childish! :wink:

im young, i still have that right. Your if i understand you correcly is by old egyptian standards dead 3 times

You have yet to get to my age, you have much to learn.
Am I supposed to put up my feet, curl up and die?
Oh no, I have only one life, I have much more to learn before I am dragged kicking and screaming away, I intend to live long enough to see man land on the moon AND Mars, I can wait. - I'll be 103years old But I'll see it!

moon is already done.

I'll be 103years old But I'll see it!

i bet my money on you won´t

Originally Posted by Zelos

moon is already done.

I'll be 103years old But I'll see it!

i bet my money on you won´t

That's a bet I'll take on ANY DAY how much? - just remember I can't lose that bet!!

bieng over 100 years is less than 1% so im the one sure on that. you are going down "grandpa"

Even if you win he'll be dead so it doesnt matter anyway.

Vroomfondel wrote:

something seems wrong here. Is that just the first term in the taylor expansion? because, according to that, as h -> infinity, so will s. But that isnt true, it will approach a quarter of the circumference of the circle.

I see now what you are asking. At first I thought you were asking about how I derived it.

Okay, yes, as h --> infinity, so does s. In that case, the formula is no longer accurate (AT ALL!!). And even for small values of h, the formula is only an approximation (albeit a fairly accurate one).

You'll notice in a previous post, billco had me use it on a planet of 1-meter radius and a 2-meter tall billco (h). In that case, I mentioned that I only had something like 65% accuracy.

You might wonder, what good is this formula then? It's only an approximation! Well, it works very well on earth for modest h, and it's simple.

Plus, it was fun to derive.

I'll give y'all a hint:
You should draw the earth, a tower on earth of height 'h', and connect a line from the tower to the horizon. Call the angle of the triangle near the center of earth 'theta' and the radius of earth 'R'. Oh yeah, and the portion of the circle (earth) from the base of the tower to the horizon we'll call s (the distance along the surface of earth).

Okay, from that, you should be able to get to the formula

h/R = sec(theta) - 1

.
This is where you'll have to figure out how to apply the Taylor series and then a binomial series and use s = R*theta somewhere too.

Have fun!
Cheers,
william

P.S., Once you get that, see if you can get a formula for the error. Afterall, we'll need to know the error if we use an approximation....
More fun!

Originally Posted by billco

You have yet to get to my age, you have much to learn.
Am I supposed to put up my feet, curl up and die?
Oh no, I have only one life, I have much more to learn before I am dragged kicking and screaming away, I intend to live long enough to see man land on the moon AND Mars, I can wait. - I'll be 103years old But I'll see it!

Only one life ey? Those young philosophers I have talked to either thought that being born was simply to become something that is "I" from not being "I". A carrot for instance, is not a carrot anymore if you have eaten it. So apperently the carrot is untrue in existence before it existed and after it existed. So are you, and yet you exist. Apperently you can come to exist from not existing. Elseways you always existed, cause you do exist. And isn't it very fortunate that you, out of all mass in the universe, just happened to become as rare as a person who lives. Perhaps it was unavoidable, or what do you think?

After all, what makes you exist is not there when you don't exist. Like position.

Myself I think we allways live, and not until we are 103

I'm a retired engineer, you know engineers, we make practical things. Philosophers? I have yet to see anything they have contributed to reality, except increasing headache tablet sales....

Originally Posted by billco

I'm a retired engineer, you know engineers, we make practical things. Philosophers? I have yet to see anything they have contributed to reality,

Philosophy, the contemplation of why and what and how things are, led to Natural Philosophy, the study of nature. Natural Philosophy was what we used to call physics. Physics is the basis of what engineers do. Claiming philosphers have produced nothing of value is like saying that grass has no relationship to the steak produced from the cow that ate the grass, philosophically speaking.

Well, there is wrong and there is right. Philosophy is not wrong if it helps you make good decisions. Knowledge is not right if you make bad decisions with it. So if you have the right philosophy and the right knowledge you can make the right decisions without feeling bad about it. But if you knew the right philosophy, like the one I have you will also know that there is nothing right or wrong, cause happiness only makes you sad and sadness only makes you happy. Yeh can't eat the cookie and still have it. Yeh can't become a singularity without exploding eventually. It ain't gonna be like this forever. Yeh can't make a mess without loosing energy, the energy will order mess again. But I know for certain, that it ain't gonna end. Cause it can't.

Originally Posted by Ophiolite

Originally Posted by billco

I'm a retired engineer, you know engineers, we make practical things. Philosophers? I have yet to see anything they have contributed to reality,

Philosophy, the contemplation of why and what and how things are, led to Natural Philosophy, the study of nature. Natural Philosophy was what we used to call physics. Physics is the basis of what engineers do. Claiming philosphers have produced nothing of value is like saying that grass has no relationship to the steak produced from the cow that ate the grass, philosophically speaking.

Looks like you popped in, pooped, and passed out.... :wink:

It's the joy that drives us and it's the pain that keep us. Anyway Einstein did the best he could and sometimes that leads us to somewhere. Like in his case alot of stuff. Everything you say is not right, you have to trial to succeed. Perhaps he did something wrong that he later published. For instance, I don't know where his trial and errors are. Perhaps that was an error in itself. That's real philosophy!

By the way, he did make some errors in school so perhaps there is where they are.

Originally Posted by william

Vroomfondel wrote:

something seems wrong here. Is that just the first term in the taylor expansion? because, according to that, as h -> infinity, so will s. But that isnt true, it will approach a quarter of the circumference of the circle.

I see now what you are asking. At first I thought you were asking about how I derived it.

Okay, yes, as h --> infinity, so does s. In that case, the formula is no longer accurate (AT ALL!!). And even for small values of h, the formula is only an approximation (albeit a fairly accurate one).

You'll notice in a previous post, billco had me use it on a planet of 1-meter radius and a 2-meter tall billco (h). In that case, I mentioned that I only had something like 65% accuracy.

You might wonder, what good is this formula then? It's only an approximation! Well, it works very well on earth for modest h, and it's simple.

Plus, it was fun to derive.

I'll give y'all a hint:
You should draw the earth, a tower on earth of height 'h', and connect a line from the tower to the horizon. Call the angle of the triangle near the center of earth 'theta' and the radius of earth 'R'. Oh yeah, and the portion of the circle (earth) from the base of the tower to the horizon we'll call s (the distance along the surface of earth).

Okay, from that, you should be able to get to the formula

h/R = sec(theta) - 1

.
This is where you'll have to figure out how to apply the Taylor series and then a binomial series and use s = R*theta somewhere too.

Have fun!
Cheers,
william

P.S., Once you get that, see if you can get a formula for the error. Afterall, we'll need to know the error if we use an approximation....
More fun!

we are just using s=rTheta in my pre-calc class! lol