# Measuring the speed of light at different velocities

1. Jun 14, 2012

### coktail

Let's say there is a football field with a mirror set up in one of the end zones. If I am standing at the 50 yard line with a laser, and I point it at the mirror (facing me) in the end zone and shoot of a single blast of the laser and then record it's velocity as it bounces back and hits a sensor that I am holding, it will measure c (speed of light).

If I do the same thing but move away from the mirror at a certain velocity after shooting off the blast of the laser, I'd still get a measurement of c because my both myself and my sensor have time dilated, so the sensor measures more slowly, thus compensating for my movement and keeping c consistent.

If I do the same thing, but move towards the mirror at the same velocity that I moved away from it, c still remains consistent, but I don't understand why. i would think that me and my slowing down would actually anti-compensate for my movement, and that I would measure time as faster because I'm moving towards it AND my instrument is slowing down.

What am I missing here?

Thanks!

2. Jun 14, 2012

### TheEtherWind

If you walked away while sending a photon down the field you will not have changed the propagation speed of the photon but you will start and end at a different point when you sent it off and retrieve it again. It will take a slightly longer amount of time to get back to you if you're moving backwards simply because it has more distance to travel. But accordingly it will travel a greater distance as well. Distance/time elapsed will always be c.

3. Jun 14, 2012

### coktail

Thank you, TheEtherWind.

I suppose the essence of what I'm asking is, do we measure time as c regardless of our velocity towards or away from a light source because our instruments of measurement are subject to time dilation and length contraction, or is it because the light source "slows down" from our frame of reference, or both. Or neither, and I'm thinking about this incorrectly.

4. Jun 14, 2012

### yuiop

If you observe another observer moving towards a stationary light source, then you will conclude that the reason he does not measure a higher value for the speed of light is due to time dilation, length contraction and importantly, the relativity of simultaneity. If on the other hand you consider yourself to be stationary and you see a light source coming towards you, then there is no need for an explanation because the speed of light is independent of the speed of the source and that is a simple fact.

Also, you do not measure time as c. Time has units of seconds and c has units of metres per second. I think you meant to say we measure the speed of light as c regardless of our velocity towards or away from a light source.

5. Jun 14, 2012

### coktail

Thank you! That's quite helpful. And yes, I did mean "speed of light."

6. Jun 14, 2012

### ghwellsjr

You only get the same measurement of c when both you and the mirror are traveling at the same speed, which, of course means that you are at rest with respect to the mirror.

In your first measurement, where both you and the mirror are stationary with respect to the football field, you don't have to worry about when the light hit the mirror because you always know how far away it was whenever that happened and you calculate the speed of light by taking its total round trip distance divided by its total round trip time. You will always get the same answer even if you perform the measurement on a high speed space ship.

You can consider your second measurement as if you were stationary on the 50 yard line but the mirror is moving away from you. Then you have the problem of defining when and where the light hit the mirror. You cannot answer that question until after you have established a Frame of Reference. So even though you know how long it took for the light to make the round trip, you don't know what distance to use in the calculation.

If you make an assumption about the speed of light being the same as your stationary measurement, then you have already defined a Frame of Reference and you use that to determine where the mirror was when the light hit it. At this point, you can use the Lorentz Transformation to see what happens in a different Frame of Reference where the mirror (and the football field) is stationary and you are moving.

When you do that, you will see that the relative motion between you and the mirror is the same in both frames, so that cannot be the explanation of why the speed of light is the same in both frames. Also, since the mirror is stationary, time dilation doesn't apply for it. It's only your clock that is time dilated. And in the football field's FoR, although you are length contracted, other distances are not, so that cannot be part of the explanation.

The only thing left is what we started with: where was the mirror in relation to you when the light hit it? And, since we are defining instead of measuring the speed of light to be the same in every FoR, we can now calculate where the mirror was when the light hit it.

7. Jun 14, 2012

### yuiop

If you measure the velocity of the light by measuring the total distance it travels (from time it left you until the time it returns) then the time dilation of you clock due to motion relative to the football field would make the light appear fast, but the length contraction of the football field due to your motion relative to it exactly compensates so you still measure the speed of light to be c.

If you insist on just using your hand held light speed measuring device, then you need to know the exact details of how it is designed to be able to analyse it. It is not just about time dilation. There is much more to relativity.

8. Jun 14, 2012

### coktail

Ok, so moving towards the mirror, my clock time dilates, but that is compensated for by the length contraction of the distance between myself and the mirror from my FoR.

When I move away from the mirror (or it moves away from me), my clock time dilates, but does the distance between me and the mirror still contract? If so, what changes to compensate for the clock slowing and length contracting? I'd think it would need to expand in order to compensate, but I don't think it does.

9. Jun 14, 2012

### yuiop

The light has to travel from the 50 yard line to the end zone and back to the 50 yard line and then the extra distance that you travelled in that time. This extra distance obviously requires extra time for the light to travel it even from a purely Newtonian point of view. Again, when we take only time dilation onto account the light speed would seem faster and again, the length contraction compensates just as in the first case.

Last edited: Jun 14, 2012
10. Jun 14, 2012

### ghwellsjr

In the football field's FoR, which is the one in which you are moving, you experience time dilation and length contraction but nothing else does. The distance to the mirror is not contracted in that FoR because it is not moving. When you do your experiment, you will get some measurement for the time that it took for the light to make the round trip but how will you determine the distance?

11. Jun 14, 2012

### jartsa

Let's consider driving in a car at speed 0.86 c. A clock runs at half speed at this speed, and a meter stick is half meters long at this speed.

I calculated that the driver will see 516 000 kilometer post go by in one second.

The driver will see a passing light beam gaining distance to himself the same way as always: 300 000 kilometers in one second. That's equivalent to 600 000 kilometer posts along the length contracted road.

According to the driver, a light beam that is going in the same direction as him is passing 516 000 + 600 000 kilometer posts in one second.

A light beam going in the opposite direction is passing 600 000 - 516 000 kilometer posts in one second, according to the driver.