When light is deflected by gravity, (as when passing near the Sun) does frequency influence the value of the angle of deflection ?
Any link is welcome.
Thanks

When light is deflected by gravity, (as when passing near the Sun) does frequency influence the value of the angle of deflection ?
Any link is welcome.
Thanks
...and yet I think it should because of the einstein frequency shift.
Gravitational redshift  Wikipedia, the free encyclopedia
https://www.google.ca/url?sa=t&rct=j...71198958,d.aWw
Since the field strength of a gravitional lense changes as the distance from the gravitational source changes there should be some chromatic distortion.
This should be observable as the lense moves in relation to the source and observer or as the distance between the observer and the lense changes.
However in most cases the distances involved and the rate of motion would tend to minimize chromatic aberations so you would need strong gravitional fields at a fairly close distance to observe the effect.
So far most attempts to observe gravitional lensing have relied on narrow frequency sensors (usually gamma ray) to record the transit so they are unable to observe any frquency shift since they simply see only a narrow frequency band.
You are mixing up two DIFFERENT effects.
Nothing to do with the light deflection.
I read the original paper by Soldner : On the Deflection of a Light Ray from its Rectilinear Motion  Wikisource, the free online library
He talks of a hyperbola, shouldn't the angle be calculated by an integration of the normal force?
Last edited by whizkid; July 20th, 2014 at 02:21 AM.
I have some years ago done the calculation for gravitational light deflection from start to finish  here it is :
https://www.dropbox.com/s/50rswg7ft7...ction%20SM.pdf
Gravitational red shift is an effect along the radial direction, whereas gravitational light deflection is an effect along the azimuthal angle. They are not the same things.
Last edited by Markus Hanke; July 19th, 2014 at 01:27 AM.
No. A consequence of the equivalence principle is that the effect of gravity on light is the same for all frequencies of the light.
Beautiful!
You can add to your paper the calculation of the Shapiro delay, this would be, in my opinion the most elegant calculation.
Start with the fact that the trajectory is a parabola of equation that extends from to , makes an angle with the radial direction (you have already calculated the deflection angle in your paper) and has the value (the "closest impat parameter).
Use the conditions to get the parameters:
Now, calculate the length of the arc of the parabola .
Divide by c in order to get the Shapiro delay.
Last edited by Howard Roark; July 18th, 2014 at 05:32 PM.
So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
The Shapiro delay is a GR effect based the same principle as gravitational lensing and frame dragging. It is essentially frame dragging in reverse. With frame dragging, a rapidly rotating massive body will accelerate the motion of a smaller body within its gravitational field and the acceleration is in the same direction as the rotation of the more massive body. The gravitational field of the massive body 'drags' the smaller body along with it. The orbit of Mercury about the sun is a familiar example of frame dragging. The Shapiro delay works in the opposite direction where a rapidly moving body enters the gravitational field of a slow or nonrotating body and the motion of the fast moving object is slowed as it passes through a static gravitational field. A nonrotating body is one that is not rotating relative to the distant galaxies.
There is also a cosmological Shapiro delay where any rapidly moving object is slowed by its passage through the static gravitational field of the universe as a whole. This is a bit contrary to Newton's law where moving objects remain in motion. The cosmological Shapiro delay has yet to be observed. Possibly because it is too slight to be observed by present day technology but some think it may be a contributing factor to the redshifting of distant galaxies.
That is interesting. Can you tell me anything about how it works with moving lenses?
I also wonder if the cosmological Shapiro delay would not just average out? Since the Einstein shift has to be observed from different gravitational wells with the difference between the deppth of the wells determining the degree and colour of the frequency shift shouldn't they simply cancel out on average?
Thanks Markus, the discussion is very interesting, but I wanted to discuss the classical method.
If someone can understand the rationale of the cited paper, please explain it to me, and tell me if with the current parameters 0.84 is still valid ,following that method.
I would like to consider light as any other orbiting body, and the rationale should be:
R = 7.5 *10^10 = 2.5 C, C/R = .4
In order to bend the path by 1 radian F_r must be 0.4 C = 1.2*10^10 by 1'' .4 C/206265 F_1'' = 58 177
the pull at the tangent (cos alpha = 1) is 23600
the correct way to find the angle is deflection is to integrate the force normal to the direction of the light.
Is this correct?
Is this what Soldner did ? I saw no integration
I do not know calculus, let alone differentials, is it the integration of normal force?
What I asked is if the rationale of Soldner is the one I described and if with current parameters we get same value of .84''
Integrating force the way I showed, if the logic is correct, we get a much greater value , greater than GR
P.S. It would be advisable that you stopped insulting or showing off, it doesn't do justice to your intelligence.
Yes, that much is pretty obvious. You should try learning them. Much better use of your time than spamming threads on subjects that are way out of your level of understanding.
There is no "integrating of force" in the method. You do not understand the physics because you cannot follow the simple calculus in the paper.Integrating force the way I showed, if the logic is correct, we get a much greater value , greater than GR
Is this approach accepatable in classical physics? What is wrong in it?
In order to get the correct GR value we must get F = 102 000, is this right?
I do not understand why Soldner considers both sin /cos phi
What is relevant here is F, which is cos phi or , better, CA/CM which multiplied by F_x (CM being the xaxis) gives the the value of the force at each point x
If this approach is correct we need only one simple formula instead of the log derivation by Soldner
Last edited by whizkid; July 20th, 2014 at 03:03 AM.
If you do not know calculus then you will find it very difficult to discuss this subject. I would advise you to first get a very solid grounding in multivariable calculus and linear algebra before getting involved with differential geometry and GR.I do not know calculus, let alone differentials
I am not sure what you are trying to say, but there are no forces in GR. The derivation of gravitational light deflection is done by calculating the null geodesic along which light propagates, and then find the angle between their asymptotes. It's purely geometrical.Integrating force the way I showed, if the logic is correct, we get a much greater value , greater than GR
Thanks Markus, but I would not dare to attempt to meddle with GR.
I said that I want to discuss if the Soldner method is correct, I 'd like to understand what he did and if that value is correct with current parameters.
Besides that, if nobody can tell me that, I'd like to know if my approach is valid in classical physics, applying to light the same laws we use for massive bodies : F = v^2/r
The GREinstein value of 1.75'' corresponds in classical physics to F = 102000,
I think we can get a similar value by classical physics, I do not know where Soldner went wrong, but only in the 5 seconds light takes to cover two R (1.5*10^11cm) while at the tangent it gets a force in the excess of 83000
If you or anyone else is prepared to do that, I 'll explain my approach again, but I suppose it is standard physics.
I do integrations at Wolfram, and so make up for my ignorance.
Any idea is welcome
Last edited by whizkid; July 20th, 2014 at 03:44 AM.
This is incorrect. The calculation of the perihelion precession of Mercury's orbit does not use the rotation of the sun as a parameter. Indeed, the perihelion precession of Mercury's orbit is the result of the general relativistic deviation from the perfect inverse square law, this deviation causing nonconservation of the LaplaceRungeLenz vector. Frame dragging is an extremely tiny effect that has only recently been tested by Gravity Probe B.
The Shapiro delay only applies to slow or nonmoving gravitational fields but even in this case it would have a negligible effect on gravitational lensing.
As you mentioned, gravitational effects should cancel as objects or even light move in and out of a gravitational field but the cosmological Shapiro delay is a different animal because it involves the entire gravitational field of the universe where there is only an in but no out. We can't escape from the gravity of some 200 billion galaxies. There may appear to be zero gravity in deep space because gravity is equal in all directions but there is no place in space where there is no gravity. Gravity is curved spacetime and the entire universe is a spacetime curved upon itself. It is this static gravitational field that is suspected to have a Shapiro delaying effect on all moving bodies.
In some of the stuff I was reading there seemed to be a suggestion the Shapiro delay could account for the Hubble redshift.
(Some of the claims seemed to focus on the idea that there is about six times as much dark matter as visible matter in the universe.)
Is this possible?
Also, if the universe is spherical (or nearly spherical) shouldn't most of the gravitational field inside the universe cancel naturally just as it would inside a thin shell?
Soldner method, using Newtonian mechanics produces an incorrect answer.
GR predicts TWICE the angular deviation compared to Soldner method. Experiment confirms GR and contradicts Soldner.
Even if you want to reproduce Soldner method you need to LEARN calculus, you cannot do it by using 9th grade algebra as you keep trying. You cannot reduce a problem to your level of understanding.
Last edited by Howard Roark; July 20th, 2014 at 09:35 AM.
If you were in a capsule at the center of the Earth you could float weightlessly because gravity is acting equally in all directions but this does not mean that there is no gravity at the center of the Earth. Gravitational fields add rather than cancel in much the same way that air or water pressures increase with depth. We don't notice the air pressure around us because it is equal in all directions but the air pressure is an essential part of our environment and gravitational effects exists even when they are acting in all directions.
Clocks run slower in a strong gravitational field than in one that is weak so a clock should run slower at the center of the Earth than at the surface and a clock at the surface should run slower than a clock in space. Likewise, the Shapiro delay should increase with the total strength of the gravitational field even when it is equal in all directions.
Some speculation has come from neoMachian physics that the Shapiro delay may be a part of the Hubble redshift. Amitabha Ghosh has even suggested that the universe is not expanding and the Hubble redshift is entirely Shapiro effect even without considering the possibility of exotic dark matter. The presence of exotic dark matter would amplify the effect but all this is hypothetical at this time. The part of this that I find incredible is the idea that there is six times as much dark matter as visible matter.
I also find it interesting that the ratio of (DM + VM)/VM seems to be converging on 2 * Pi as more accurate results come in.
It's like what you would expect to see if rest mass and relative mass were being conflated on a universal scale (i.e. 2 * Pi is the difference between the reduced Compton wavelength and the reduced Planck constant and their standard counterparts).
Actually I do not think this part of what you are saying is quite right.
Shell theorem  Wikipedia, the free encyclopedia
I'll start a new thread. please reply if you answer directly the questions
Last edited by whizkid; July 21st, 2014 at 03:07 AM.
I don't see how the Shell theorem applies. The gravitational vectors may all go to zero at the center of a spherical body but this does not mean that gravity does not exist at that point. Where did it go? And when the spherical body is the universe itself, the universe is a 4D hypersphere and every point within that sphere is essentially "at the center" where the gravitational forces are equal in all directions so we don't notice their effect. We only only notice the local effects of gravity from objects like the Earth, sun, and moon. Mach demonstrated the overwhelming effect of the total gravitational mass of the universe with a gyroscope. A freely rotating gyroscope orients itself with what Mach called the "combined masses of the distant stars" and a gyroscope's orientation rotates in sidereal time as the Earth rotates beneath it. In other words, a gyroscope always points to the same position in the heavens and it is not influenced by the rotation of the Earth or by the rotation of its surroundings if it is on a ship or an airplane. The local gravitational effects of the Earth and sun are not strong enough to effect the orientation of a gyroscope because a gyroscope aligns with the total mass of the universe which is far, far greater.
A series of hollow shells is a solid. The instant you pass through the first shell, gravity does not go to zero because you are still on the outside of a myriad of other shells. In a solid sphere, gravity is not equal in all directions until you reach the center of the center shell.
Depends what you mean by "gravity". If you mean "gravitational force", that is zero anywhere inside the shell, not only at the center. See here for exact proof.
If, by "gravity" you mean "gravitational potential", that is not zero, indeed.
So, what do you mean?
The gravity I have in mind is both the net gravitational force of the total mass of the universe with its two billion some galaxies and the total gravitational potential of all the masses in the universe. I am not including anything smaller than the Whole Shebang. That is the "Big M" in calculations where M represents the total mass of the universe.
The Shell theorem applies to smaller 3D spheres that have a measurable radius and center unlike the 4D universe that surrounds us. The Shell theorem can be used for calculating things like the weight of a miner deep inside a mine but it does not apply in the same way when the sphere under consideration is the entire universe.
The net gravitational attraction of all the distant galaxies is essentially zero because it is uniform in all directions. We are at the apparent center of the universe and inside all the shells where the net gravity goes to zero but a remote observer in a far corner corner of the universe could make exactly the same observation. We are not within a 3D sphere where one can calculate his distance from the center.
The net gravitational attraction from the distant galaxies may be zero but we are still under the influence of their total gravitational potential which is considerable. It influences the rate at which time passes and our measurements of distance and all the other GR effects that we associate with a gravitational potential including the Shapiro delay.
So? As you said yourself " And when the spherical body is the universe itself, the universe is a 4D hypersphere and every point within that sphere is essentially "at the center" where..." and since the universe is infinite we are at the center of an infinite series of shells no matter where we are so I think the point becomes a bit moot since the effect is the same.
Xyzt's point about gravitational potential is worth noting too because I have been thinking in terms of force which relates to acceleration (or curvature) instead of gravitational potential which relates to work. Gravitational potential is negative almost everywhere and increasingly negative the deeper down a gravity well you are.
Just to add to what has already been well explained by xyzt and others  the gravitational net force in the interior of a thin shell is everywhere zero, so a massive object placed anywhere inside the shell remains at rest. The gravitational potential is the same everywhere in the interior region of such a shell ( obviously, or else the net force couldn't vanish ). In terms of GR, spacetime in the interior of a thin shell is completely flat and Minkowskian ( which follows directly from Birkhoff's theorem ).
However, globally speaking, the uniform gravitational potential in the interior region of the shell has a value that is different from that of a reference observer at infinity; this means that a clock placed into the interior of a shell is gravitationally time dilated with respect to a faraway observer, even though spacetime in that interior region is completely flat. This can be the source of immense confusion if not properly understood, hence I thought I'd point it out here.
Thank you, Markus
Here is an excellent presentation that explains the variation of the gravitational potential on the Earth surface.
The physics of this appears clear as stated but could you clarify some of your terms? The term "thin shell" appears strange when speaking globally (universe wide) because the visible portion of our "thin shell" is the entire several billion ly of galaxies that surrounds us. Every observer at every point in the universe should find himself the apparent center of the universe so there is no part of our universe that is not a part of an observer's "thin shell." The term "thin" appears to be a bit of an understatement.
And can you call it a "shell" when it has no outer surface?
And, by "reference observer at infinity" would that be a god like observer ( outside ?) our universe?
No, there seems to be a misunderstanding here. By "thin shell" I mean a hollow and perfectly spherical shell of ordinary matter and finite size; for example what you would get if you were to hollow out the Earth and just leave an empty shell a few kilometres thick. I wasn't referring to the universe as a whole, since Birkhoff's theorem does not apply to FLRW spacetimes. By "global" I just meant the totality of the shell, its interior, and a faraway observer, not the universe as it is modelled in cosmology. What I have said does not apply to the universe as a whole.
That makes sense and thanks for the clarification. You were NOT discussing the universe as a whole but I WAS discussing the universe as a whole and the misunderstanding predates your post. I was discussing the cosmological Shapiro delay and its possible effect on the Hubble redshift when Don Hunter introduced the Thin Shell Theorem in Post #31. I tried to explain in Post #35 that the Shell Theorem does not apply to the universe as a whole but the confusion continued. I thought you were also trying to apply the Shell Theorem as a cosmological effect which is why your phrasing sounded so strange to me.
From Post #35
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