# Behaviour of light

1. Apr 18, 2007

### Nuclear on the Rocks

Hello! I am somewhat a layman, but i have a question.When we talk of bodies with large gravities bending light, are they accelerating it?

2. Apr 19, 2007

### Danger

No, if I understand the situation properly. The light is still following what to it is a straight line, so there is no acceleration imparted to it. Gravity bends the space itself, but the light has no way of 'knowing' that.

3. Apr 19, 2007

### Integral

Staff Emeritus
The Speed remains constant, but the apparent direction changes, therefore this is a change in velocity which implies an acceleration.

4. Apr 19, 2007

### pervect

Staff Emeritus
I think much of the question is a matter of semantics - what does one mean by "accelerating" the light?

I think that I generally prefer the explanation that light travels along a geodesic, though, for the following reason. If we say that light "accelerates", how do we assign a number to this acceleration to quanfity it?

We can talk about the proper acceleration of physical objects by looking at the 4-acceleration, but I don't see how to make this approach work for light. (For instance, if we try to take the limit of a 4-velocity of a massive object as it approaches c - it doesn't have a limit, so we don't even have a 4-velocity for light, much less a 4-acceleration).

Maybe there is some way around this difficulty, but I don't know what it is, and I don't recall reading anything on the topic. It might be possible to replace proper time with some affine parameter, for instance. So on the whole, I'd rather stick with saying that light follows a geodesic, it seems the simplest approach.

5. Apr 19, 2007

### MeJennifer

What some call "bending of light" is simply an effect of the curvature of spacetime (e.g geodesic convergence and divergence). In general relativity a photon in a gravitational field obviously does not accelerate for it would completely invalidate the foundations of the theory.

Of course that is possible. But then one ought to explain the physical meaning of that parameterization as well, otherwise it is justs playing mathematics with GR.

Last edited: Apr 19, 2007
6. Apr 19, 2007

### Mentz114

Isn't the gravitational red/blue shift rather like the light having been 'accelerated' or 'decelerated' ?

Question - if light reaches me from a source accelerating away from me, is there a frequency shift component for instantaneous velocity, and a separate contribution from the acceleration ?

I suppose I should go and look again at the Rindler space-time.

7. Apr 19, 2007

### MeJennifer

What is called gravitational red- and blueshift of photons are the effects of the curvature of spacetime. The photon itself undergoes no changes whatsoever.

The curvature of spacetime determines the ratio between the emitter's and absorber's measurement of the frequency of a photon.

The Ridler chart is very interesting but notice that Minkowski spacetime only deals with the kinematics of acceleration. A more complete understanding of acceleration can only be obtained in general relativity, where the kinematics, if we even can speak of kinematics here, is simply a reflection of the dynamics.

Afterall: In the general theory of relativity the doctrine of space and time, or kinematics, no longer figures as a fundamental independent of the rest of physics. The geometrical behaviour of bodies and the motion of clocks rather depend on gravitational fields which in their turn are produced by matter.

Last edited: Apr 19, 2007
8. Apr 19, 2007

### Coto

Looking at it from a semi-classical standpoint. Could you think of it like this?

E(photon) = p*c .. So gravity imparts a force based off of this photon's relativistic momentum. The gravitation force of course being something along the lines of centripetal force (i.e acceleration perpindicular to the direction of motion of the photon).

9. Apr 19, 2007

### Mentz114

MJ, this puzzles me. In the Harvard tower experiment, light was beamed downwards i.e. it 'fell', and the blue shift was very precisely measured.
The gain in the light energy due to the blue shift was calculated to be exactly ( within the exp. error) the amount that a massive body would have gained by falling. They did it the other way around also and got the same agreement.

The experiment was by Pound and Snider in 1965.

A description is here -
http://www.mth.uct.ac.za/omei/gr/chap5/node2.html

Last edited: Apr 20, 2007
10. Apr 20, 2007

### Gib Z

Well if you say Massive Body, I would think GR.

11. Apr 20, 2007

### pmb_phy

Actually 1/2 the value of the deflection is due to coordinate acceleration while the other 1/2 is due to spacetime curvature.

Pete

12. Apr 20, 2007

### MeJennifer

Well that is true, that 1/2 is actually due to the principle of equivalence alone while the other 1/2 is due to the curvature of spacetime.
Note that the first 1/2 applies to all metric theories.

Einstein was initially wrong in his calculations, he only took the equivalence principle in consideration, and so he calculated the first half only. Later he corrected it and included the curvature part as well.

Last edited: Apr 20, 2007