# Bending of Light

1. Dec 13, 2014

### David Diwik

Okay so according to this link (https://briankoberlein.com/2014/08/01/bend-like-newton/):

"The catch is that the amount of bending predicted by Newton’s model is half of what Einstein’s model predicted."
(This is right after the changing diagram with light going past the sun in different theories and is describing the bending of light around the sun).

The author used the relativistic mass of light with Newton's formula. Although this may be self contradictory because special relativity may lead into general relativity and therefore away from Newton's gravitational formula (haven't gotten there yet), I will consider it as an assumption that one might be able to do this. My problem is that I would think the curving between Einstein and this breed of Newton wouldn't differ by such a high degree.

This is because when I consider the deviation between Einstein and Newton's formulas in terms of raw prediction, I usually think that Einstein's predictions really start to vary when either both masses are comparable to each other (in mass) or when they are a sizable radius away from each other. In the first case of variance I would suggest a case where there are two identical earths being attracted to each other rather than a satellite and the earth; the other case I think is pretty clear.
In the case cited by the article of the photon (light?), the photon's relativistic mass is relatively insignificant compared to the sun's, and the radius between the photon passing by the sun and the sun isn't really that large either (comparatively to something like Mercury, which I believe gets a sizable perihelion shift due to the distance between, with Mercury's mass as contributing factor) . Therefore I would predict that if one were to take the relativistic mass of a photon and plug it into Newton's formula the deviation of that from Einstein's formula would be negligible; certainly they wouldn't predict a curvature that is half of what Einstein would.

I know I probably shouldn't look at random websites for information, but I get the sense that the author is actually legitimate. I haven't actually taken general relativity yet and only base my analysis on my philosophy class on space and time and some geometric thinking. Also, I'm not even sure how you get relativistic mass for a photon, since it has zero rest mass so I can't use the relativistic formula for mass increase; I also am not sure I can use E=mc^2 because that may rely on relativistic mass for derivation (but this is really a side question).

I hope this question is not a burden at all; such that I can ask freely without worrying about causing anyone an unfortunate amount of extra trouble.

Diwik

By the way I did look through the forum and while there were similar questions, I don't think they really answered what I was looking at. I hope that's right.

Last edited: Dec 13, 2014
2. Dec 13, 2014

### Jonathan Scott

To avoid problems thinking about mass, just remember that relative to local space everything accelerates with the same acceleration, regardless of mass.

The double bending of light is due to the fact that in GR gravity causes the shape of local space near a gravitational source to be curved relative to distant space. The curvature for Newtonian acceleration $g$ has a radius of curvature $r = c^2/g$. When something is moving in a curve with radius $r$ with tangential speed $v$ its acceleration is $v^2/r$. In this case, light is moving at speed $c$ so the additional acceleration relative to distant space caused by the curvature of local space is $v^2/r = c^2/(c^2/g) = g$.

3. Dec 13, 2014

### David Diwik

So maybe to best hit my question I'm going to try and break down what you said.

"The double bending of light is due to the fact that in GR gravity causes the shape of local space near a gravitational source to be curved relative to distant space."

So a gravitational source is mass right? So general relativity makes mass curve space. I follow so far (although I haven't yet gotten to where the double bending of light is actually justified). (Although I am trying not to ignore your warning about thinking about mass, I will translate it that way for now).

"The curvature for Newtonian acceleration g has a radius of curvature r=c2/g."

Ok so with this curvature my final resulting Einstein prediction will equal my Newton prediction for acceleration due to gravity, which is g. I'm ok with this in principle.

"When something is moving in a curve with radius r with tangential speed v its acceleration is v2/r. In this case, light is moving at speed c so the additional acceleration relative to distant space caused by the curvature of local space is v2/r=c2/(c2/g)=g."

Ok so these look like calculations using that curvature to actually get you the g as an acceleration value.
__________
I will try to interpret now. I think in your calculations the acceleration v^2/r is meant to be derived from the curvature that actually occurs due to the sun, which I will assume is right. This will get you an acceleration of g; this acceleration will lead to light bending a certain amount. Now if just plug in Newton's equations I will get an acceleration of that same g (by definition based on how I think you defined it), so I should get the same bending amount; so why are the bending amounts supposed to be different?

Also even if I am formulaically supposed to get a different acceleration, is there a qualitative reason for the deviation that I can look at using the geometric model of the curving of spacetime (I'm used to the bowling ball on the trampoline)?

Thank-you for the answer by the way; I'm trying to make sure I'm getting this not just sort of believing you. I do appreciate the reply. That's the real reason why I'm punching this through.

Last edited: Dec 13, 2014
4. Dec 13, 2014

### Jonathan Scott

In general relativity, the field equations determine the curvature of space-time paths both with respect to space and with respect to time. The curvature of paths with respect to time affects everything equally, relative to local space, and is just like the acceleration for Newtonian gravity. The effect of local space curvature on moving objects, relative to distant space, adds a further acceleration which depends on the speed. In the weak field approximation to GR (which applies fairly accurately almost everywhere except near very dense objects such as neutron stars and black holes) the curvature of space in this sense is equal to the Newtonian acceleration in appropriate units.

5. Dec 13, 2014

### Bandersnatch

Would this be a correct way of putting it as succinctly as possible?

"Newtonian gravity and space-time cruvature are equivalent, but in GR there is an additional curvature of space itself unaccounted for in Newtonian description."

6. Dec 13, 2014

### Jonathan Scott

No, I don't think that's quite right. "Space-time" covers both space and time so the additional effect is already included as part of space-time curvature.

You could say that free-fall paths in space-time are curved with respect to time and with respect to distant space (by approximately the same amount) in the GR description, so the curvature with respect to time affects everything (causing velocity vectors to curve towards the direction of the Newtonian gravitational field) but the curvature with respect to space only affects things moving with respect to the source. (It actually affects things moving radially as well as tangentially; the effect on the momentum is exactly the same for a given speed, but this is not due to a change of direction but rather due to a decrease in the coordinate speed of light caused by the same curvature).

You also have to be careful about the use of the word "curvature", as GR people also use it to describe the intrinsic curvature caused by the presence of matter as well as the ordinary curvature of a path in curved space-time relative to some coordinate system.

7. Dec 13, 2014

### A.T.

That analogy shows the spatial geometry (Flamm's paraboloid), which is responsible for the "additional" bend, that isn't there in Newtons flat space, and affects only moving objects. But light is not moving on geodesics on that spatial paraboloid. It also has radial coordinate acceleration due to the gravitational time dilation gradient (lower part of the beam has less coordinate speed). That radial coordinate acceleration corresponds to Newtonian gravity, and affects everything equally regardless of its movement.

See:
http://www.mathpages.com/rr/s8-09/8-09.htm

And:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html
Note that Fig. 2.6 her is not
what actually happens, because it ignores the light bend due to gravitational time dilation, The whole section 2 pretends that only a spatial distortion exists, in order to explain it better on its own.

8. Dec 13, 2014

### Staff: Mentor

Do you have a reference for this?

9. Dec 14, 2014

### Jonathan Scott

I don't know of anything at a suitable level for the original poster. The fact that the curvature of a tangential path in the Schwarzschild solution is $g/c^2$ is of course just the gradient of the $dr$ term in the metric, and the rest is just algebra to show the result in a more familiar form for Newtonian mechanics.

10. Dec 14, 2014

### Staff: Mentor

I was asking for me. Do you have a reference?

11. Dec 14, 2014

### Jonathan Scott

What are you looking for? I thought this was very well known. You can get the curvature of local rulers relative to the coordinate system from the spatial factor in the Schwarzschild metric in isotropic coordinates just using simple geometry. That metric can be found in just about any GR text book. The factor is of course exactly $(1+Gm/2rc^2)^2$ using isotropic $r$ and can be approximated by the usual Newtonian $1/(1-Gm/rc^2)$ or $1+Gm/rc^2$ in the weak field approximation. In linearized GR the approximation $1+\sum(Gm/rc^2)$ holds even for multiple sources.

I don't have the tools and patience to draw a picture, but just imagine a horizontal ruler with proper size $a$. The coordinate width is then $a' = a (1-Gm/rc^2)$, so the rate of change of coordinate width with respect to radial position is approximately $a (Gm/r^2 c^2) = a g / c^2$ where $g$ is the Newtonian acceleration. That is the fraction by which the coordinate width of a box of proper width $a$ increases for a small increase in height so it is the angle between the sides, and hence the angle through which space has curved in distance $a$, and the curvature is therefore $g / c^2$.

12. Dec 14, 2014

### Staff: Mentor

I am looking for a reference that makes the computation in post 2, particularly the part relating to Newtonian g. If it is very well known then you should find it easy to point me to a reference.

As a matter of policy on this forum, all posts must be in agreement with the professional scientific literature. When anyone asks for a reference on any topic then one should always be provided without this hesitance, and regardless of how well known the topic is.

Last edited: Dec 14, 2014
13. Dec 14, 2014

### David Diwik

I have to spend more time analyzing this. I'm trying look for deviation factors and the one that seems to be coming up has something to do with motion of the object that's passing the sun (for example). This has to have something to do with speed otherwise any object passing by the sun (not just a photon) would deviate from Newton's approximation heavily (not true), and that's the only sizable change in variable that I can think of that is particular to a photon. I have looked through the links (etc.), but I have to fit them into a structure of analysis so I can't post right away.

It's seems somewhat clear that anything moving tangentially to a curvature will deviate from Newton (similar to the arguments from the second link)(dependent on the degree of the curvature), but in that case speed doesn't seem to be a factor. Perhaps it is because the second link didn't play into time dilation and speed leads to time dilation. I will eventually analyze the second link further because something doesn't seem right about drawing a elliptical and then bending space around it as a method of seeing the motion according to Einstein converted from that of Newton (this will take a little bit of thought, which I will do).

I think I pulled this from, "The curvature of paths with respect to time affects everything equally, relative to local space, and is just like the acceleration for Newtonian gravity. The effect of local space curvature on moving objects, relative to distant space, adds a further acceleration which depends on the speed."
Later note:
Ok I think I've grasped the Mercury method w/o time dilation.

Last edited: Dec 15, 2014
14. Dec 15, 2014

### A.T.

That is correct. The amount of bending that comes from curved space alone (without time) doesn't depend on the speed though space. If you had a rocket flying past the Sun, with it's engine controlled to offset just the Newtonian gravity (or gravity due to gravitational time dilation gradient, or the local g value measured by a stationary accelerometer), it's path would be bent only due to spatial curvature. And the amount of bend would not depend on how fast the rocket is.