# Light velocity measurements

1. Oct 30, 2008

### brightonb

Will an accelarating observer obtain the same value for light velocity as one at rest or moving at constant velocity? Will the measurement be the same for linear and radial (circular motion) acceleration?

2. Oct 31, 2008

### George Jones

Staff Emeritus
An accelerating observer measures the local speed of light to be c in all directions.

3. Nov 2, 2008

### Naty1

Can someone reconcile the following from the post above:

"An accelerating observer measures the local speed of light to be c in all directions..."

with the following from another thread:

Is the difference that the first refers to local frames, while the second refers to extended (distant) frames as well?? Does the accelerating observer have accelerating clocks?

4. Nov 3, 2008

Yes.
No.

5. Nov 3, 2008

### Naty1

George posted:
which is correct. But I do not believe that has been experimentally verified; it is a theoretical construct .....for clarification to the question:

I would add: (This is my own interpretation, not sourced...so comments are welcome...)

While the local speeds of light will be measured as c by each observer, distant observations will yield different results for the speed of light between the two observers. The constant velocity inertial observer will see all light as c; for her, all spacetime is straight.
The accelerating observer will observe things as if immersed in a gravitational field where spacetime is curved; his observations change since curvature (gravitational strength) becomes apparent with increasing distance.

6. Nov 3, 2008

### Naty1

I forgot to address the rotational part of the question....acceleration is acceleration; so although I can't picture an analogy (any takers???????????) ..

linear and rotational acceleration are equivalent because each produces a force...the only difference is in the direction of the force relative to that of the velocity and the type of change in velocity. A steady linear force will eventually propel an object to near light speed; a rotational force of angular origin maintains a fixed speed while direction changes.

(sure seems like there should be some other effects...hmmmmmmm. well, one is a linear force doesn't make you dizzy...)

7. Nov 3, 2008

### DrGreg

All things being equal, yes. I.e. both observers must be at the same place at the same time, measuring light "locally" i.e. at their own location.

This is true by definition. Q: How does a non-inertial observer measure local distance and local time? A: by asking a co-moving (i.e. relatively stationary) inertial observer to make the measurement for him/her. This is how local distance and time are defined for a non-inertial observer.

Don't confuse space-time curvature with non-inertial ("accelerating") observers. The two often occur hand-in-hand but they have different effects.

Inertial observers (a.k.a. free-falling observers) in the presence of gravitation do not see "all spacetime" as straight -- it only looks straight locally. The curvature or flatness of spacetime is an intrinsic property of spacetime itself and does not depend on the motion of observers. An inertial observer can tell if spacetime is flat or curved by looking at other, non-local inertial observers and seeing if they all move at constant velocity or if some don't.

Curvature is not "gravitational strength"; it is (roughly speaking) the "rate-of-change of gravitational strength" over distance i.e. "gravitational gradient".

8. Nov 3, 2008

### Naty1

I have been puzzled by the boldface part for sometime....and I think it does depend on frame.....Let me disagree for discussion: A free falling inertial observer in the presence of gravitation sees local spacetime as flat; all other frames will observe the same local space as curved.

In fact, when light is observed from a distance approaching a black hole and slowing, blue shifting, as viewed say from earth, and never reaching the event horizon, (curved space) while a local free falling observer sees that same light proceed to the event horizon and disappear (flat space)....isn't that an example of different observers seeing different shape spacetime??

9. Nov 3, 2008

### Jonathan Scott

Note that there are two types of "curvature". The "curvature" caused by mass is like the curvature of the surface of a ball, and cannot be transformed away. The "curvature" of a gravitational field in a vacuum is more like the curvature of a cone made of flat paper, where it is possible to use a flat view to describe local events. (Don't stretch this analogy too far, though!)

10. Nov 4, 2008

### MeJennifer

No, the boldefaced statement is exactly right, the curvature of spacetime is observer independent.

Note that the (arguably useless and confusing) notion of the curvature of space (as opposed to spacetime), is neither observer nor coordinate independent.

11. Nov 4, 2008

### DrGreg

What is curvature?

What we call the "curvature of spacetime" has a technical meaning; the equations that describe it are very similar to the equations that describe, say, the curvature of the earth's surface in terms of latitude and longitude coordinates, or any other pair of coordinates you might choose. This "curvature" need not manifest itself as a physical curve "in space".

For the rest of this post let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. This is adequate for many scenarios, even for black holes provided we are interested only in particles moving radially.

In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame (but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarise, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer.

12. Nov 20, 2008

### Naty1

DrGreg.....very clear description (I Think)...Thanks!!!!
I missed this post until a day or so ago....

Do I understand what it means?

It seems the same boldface concept should also apply with gravity (right?) : an inertial object seems to follow a (slightly different) curved trajectory through space....one slightly different from the physical curvature... hence a value for lightspeed different from c?

Is this the 'conventional' explanation??

Someone already explained a different convention: an inertial observer coincident with the accelerating observer would measure "c".

13. Nov 20, 2008

### Naty1

My prior post appears inconsistent with George's post..and I'll bet he's right, my #12, not.

Also, another thread references this which appears to be consistent with George:

http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf

14. Nov 21, 2008

### DrGreg

A free-falling object (either a massive particle or a photon) has a worldline (trajectory in spacetime) that is "as straight as possible" constrained only by the curvature of "the paper on which it is drawn". Usually curved trajectories through space (rather than spacetime) are due to the proper-acceleration of the observer (remember someone stood on the earth's surface is properly-accelerating). The "as straight as possible" line seems to be curved but that's relative to the observer's curved gridlines (which the accelerating observer perceives to be straight). Relative to a free-falling observer's locally-inertial frame (=locally square grid), the worldline is locally as-good-as straight (but curves in the distance).

It's worth thinking what is the spacetime geometrical interpretation of the speed of light. In flat spacetime and an inertial frame, i.e. a square grid on flat graph paper, a photon's worldline runs diagonally across the squares of the grid (assuming 1 second x 1 light-second squares).

On curved graph paper you cannot draw an everywhere-square grid, so light that crosses the diagonals locally won't (in general) be crossing the diagonals further away. So the slope of the worldline measured on the grid won't be 1 in grid coordinates everywhere on the grid. But at any point you can always draw another grid where where the light really does cross the diagonal there (in fact you just stretch the axes to make it fit, which is why I've previously said the local speed of light is c by definition).

You might like to have a look at this thread and in particular the diagrams of mine in post #9 and of kev in post #13, which illustrate the idea of switching between a square grid ("Minkowski") and a curved grid ("Rindler") in flat spacetime i.e. on flat graph paper.

In kev's diagram (post #13) on the right, the coordinate speed of light is c at the red spot (because the blue lines are at 45 degrees) but not in most other places.

Last edited: Nov 21, 2008
15. Nov 21, 2008

### Naty1

drGreg ..again thanks for for patience....I finally made the critical connection I was missing:

George's post, # 2:

is by convention measured according to DrGregs definition, post #7

Well I'll be gosh derned...no wonder I did not get it.....!!!!!! Who knew???

Ok now I can return to the current piece of the discussion....how acclerating frame (non inertial) observers see things....via the equivalence principle, I hope....

I need to think about Dr Gregs post # 14....

Last edited: Nov 22, 2008