# Observing the Speed of Light

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1. Jan 9, 2014

### AlfieD

Hi,

I've read that the speed of light is consistent for all observers, no matter where they are. So, for instance, a photon on a train is going at the same speed whether you are on the train or off the train. If it was a football, instead of a photon, it would be different speeds right (the speed of the ball for the observer on the train and the speed of the ball plus the train's speed for the observer not on the train)? When I've questioned this, I was told that time dilates and I think the word 'gamma' was mentioned. Could you please explain fully what actually happens for the light to be moving at the same speed for all observers?

AlfieD

2. Jan 9, 2014

### A.T.

That is the Galilean transformation, which fails for high speeds. The correct one is the Lorentz transformation. The difference is visualized here:

3. Jan 9, 2014

### AlfieD

So in order for the speed of light to be preserved for all observers, moving objects are shrunk and moving clocks are slowed down?

First, what does it exactly mean 'moving objects are shrunk'? Does it literally mean that one of the observers becomes physically smaller? Probably not but could this be explained further?

Second, when it says moving clocks are slowed down, for whom is time slowed down? And how can you have anything but a moving clock?

4. Jan 9, 2014

### Staff: Mentor

No. You're expecting that if the train is moving at speed $u$ relative to the ground and the football is moving at speed $v$ relative to the train, then the football will be moving at speed $u+v$ relative to the ground. It's not - it's actually $\frac{u+v}{1+\frac{uv}{c^2}}$, and this works for footballs, photons, and everything else.

It's almost impossible to tell the difference for any reasonable ball or train, but it starts to show up if one or both speeds is getting close to $c$. And if either speed is equal to $c$, the whole thing comes out to $c$ no matter what the other speed is.

5. Jan 9, 2014

### phinds

Time dilation and length contraction are artifacts of remote observers. That is, in your own frame of reference, you are always the same size and your clock always ticks at one second per second. Some other observer, from a different frame of reference, SEES you as being length contracted and with a slow moving clock.

You, right now as you read this, are moving at almost the speed of light from the reference frame of an accelerated particle at CERN. Do you feel any different?

6. Jan 9, 2014

### WannabeNewton

No you're thinking of this in a backwards manner. Time dilation, length contraction, and the Lorentz transformations in general are a consequence of the invariance of $c$ between inertial frames and the flatness of space-time (the latter of which amounts to a set of "symmetry" assumptions). Time dilation and length contraction don't "enforce" the invariance of $c$ but rather they are simply derivatives of this postulation.

Imagine we have a rod being carried by an observer $O$. Furthermore $O$ is moving with some velocity $v$ relative to another observer $O'$ along the line joining $O$ and $O'$. Now $O$ wants to know the length of the rod he's carrying. How can he determine its length?

Well here's one way: $O$ attaches a mirror to the front end of the rod (he holds the back end of the rod). He then emits a beam of light towards the mirror which bounces off instantaneously from the front mirror and arrives back to him. He uses a clock to record the time for the round-trip and says it is $\Delta t$. Well if the speed of light in the forward direction is the same as the speed of light in the backwards direction, like we assumed, then surely the light beam reached the mirror when $O$'s clock read $\frac{\Delta t}{2}$ and hence the distance traveled would obviously be $L_0 = c \frac{\Delta t}{2}$. This distance spans the length of the rod as measured by $O$.

Can you see why this "radar echo" method would yield a different length for the rod when a similar measurement is made by $O'$ because of the relative motion between $O'$ and $O$? $O'$ simply measures the rod to be a different (in fact shorter) length from its length as measured by $O$ (the "rest" length since $O$ is carrying the rod); however neither measurement is any more correct than the other i.e. they are both equally valid after being attributed to the respective observers.

Switching gears a bit, $O$ will record a time $\Delta t$ for one round-trip of the light-clock he carries with him. However $O'$ will measure a time $\Delta t' > \Delta t$ for one round-trip of the light-clock that O carries.

7. Jan 9, 2014

### AlfieD

Sorry but what does the bottom right of the equation actually say? Please forgive both my poor eyesight and my browser's zoom functionality. :)

I feel OK. :D Funny you should mention CERN because I just went there to visit ATLAS.

8. Jan 9, 2014

### DrGreg

Here it is a bit bigger:$$\frac{u+v}{1+uv/c^2}$$

9. Jan 9, 2014

### AlfieD

Haha thanks, looked like a 'w' and an 'o' to me. Very poor eyesight and guessing skills! :D

10. Jan 9, 2014

### AlfieD

Sorry, but I don't get how $O'$ is even measuring it? Could you just explain a bit further as to how they differ in measurements? Thanks.

I thought you said that $O'$ measures it shorter but $\Delta t' > \Delta t$ would suggest that $O'$ measures a time that is greater than $O$'s time.

11. Jan 9, 2014

### WannabeNewton

Read through this, it should be instructive to you especially since it uses space-time diagrams to explain everything: http://people.uncw.edu/hermanr/GR/Minkowski/Minkowski.pdf

No I said length intervals gets contracted, not temporal intervals. We're talking about two different sets of measurements here so don't conflate them otherwise you might end up confused (hence why I said "switching gears"). The link above will hopefully clarify things via the space-time diagrams.

12. Jan 9, 2014

### AlfieD

13. Jan 9, 2014

### AlfieD

Last edited by a moderator: May 6, 2017
14. Jan 9, 2014

### WannabeNewton

No problem! I hope it helps; space-time diagrams are unequivocally the best way to understand measurements in special relativity especially when the measurements involve radar. Also, given the title of your thread, keep in mind the important difference between the one-way speed of light and the two-way speed of light.

15. Jan 10, 2014

### A.T.

The terms "artifacts", "remote observer" and "seeing" are potentially misleading. The effects have nothing to do with observing from a distance, and are not visual artifacts due to signal travel time. The effects are still there after you already accounted for distances and signal travel time. The effects depend only on the relative movement, not on the remoteness.

16. Jan 10, 2014

### BruceW

yeah, I agree. The object really is physically shorter according to some other observer. Although, you could say that only the proper length of the object is important, and any other length is just an artifact. But I think this is quite a 'hardline philosophy'.

17. Jan 10, 2014

### A.T.

Yes, you could call all frame dependent quantities "artifacts". Then the velocity of the object would also be just an artifact. But I don't think this is the common usage of the word, hence misleading. Especially combined with talking about "remoteness" and "seeing".

18. Jan 10, 2014

### phinds

Yeah, by "remote" I meant "in another frame of reference", not "far away" so that was a very bad choice of words. Likewise, "seeing" is misleading, as you point out. My use of "artifact" is arguable, but this has been discussed on this forum before and I stand by my use, although I do understand your point. "Difference from proper length" or something like that would be a less arguable way of stating it.

19. Jan 10, 2014

### AlfieD

Ok, I'm getting confused with all of these terms now... I think I get it though thanks to the link by @WannabeNewton but let's not make it any more complicated than it needs to be haha. Thanks though for everyone who spent their own free time to answer this. Your help was appreciated. :)