# Question about how the speed of light is constant

1. Oct 1, 2009

### endusto

i already posted a homework question in the homework forum. i'm asking this question so i can finish my homework but its not just for one particular question so i'm just gonna ask it here.

i'm not 100% sure about how the speed of light stays constant in all reference frames. say i'm driving on the surface of earth and there is a beam of light at the very back end of my car traveling towards the front of my car. so in my reference frame, relative to me, the light beam is moving at the speed of light right? and relative to say a tree on earths surface, it is still moving at the speed of light.

so say i want to know how long the beam of light will take to go from the back of my car to the front of it, relative to my reference frame. the beam of light will take t' = d'/c where d is the proper length of my car relative to my reference frame, to travel all across it. so say my car is 10m long, then it will take the light 10m/c to travel from the back to the front of my car?

i think that is correct. so if i want to know how long it will take the light to travel from the back to the front of my car relative to the tree's reference frame it it will be t = y*t = sqrt(1 - v^2/c^2) * 10m/c where v is the velocity of the car? which will end up being slightly smaller than t' am i right?

so basically in the tree's reference frame it will measure fewer seconds for the light to go from the back to the front of the car, and in the car's reference frame it will measure more seconds for the light to go from the back to the front? can someone please just tell me if i'm correct about this. i'm in my first modern physics course and learning about relativity for my first time.

edit:
so lets say light is moving in the opposite direction of the car. lets say the car is going 10m/s in the +x direction and the light is going at speed c in the -x direction. does that mean the light relative to the car is still going at c? And not (c + 10m/s)?

would it take light the same amount of time to travel from the front to the back of my car while im driving straight as it would to travel from the back to the front if im moving straight?

Last edited: Oct 1, 2009
2. Oct 1, 2009

### Bob_for_short

First of all, it is an experimental fact. It was a "coup de pouce" in formulating relativistic equations and variable transformations (academician H. Poincaré, A. Einstein, H. Minkowski, ...). It contradicts the common sense based on wave in air or on water propagation. In order to get used to it you may think that the light medium (an ether) is connected in a complicated way with any observer.

3. Oct 1, 2009

### endusto

thanks for responding but honestly that doesn't answer any of my questions. i guess i did ask a lot in an unorganized fashion. i basically want to know if i was right about how i went about solving that little example i made up.

and the question i really want to know most is:
would it take light the same amount of time to travel from the front to the back of my car while im driving straight as it would to travel from the back to the front if im moving straight?

4. Oct 1, 2009

### DaveC426913

The Earth and the car are in two different reference frames. The rate at which you experience time in the car is different from the rate at which an observer on the roadside experiences time - your frame of reference is slightly dilated compared to the roadside.

When you measure the distance tavelled over the time taken, it will balance out perfectly that you both see the light beam travel at c.

5. Oct 1, 2009

### DaveC426913

From the frame of reference of the car, the light beam will travel from front to back or back to front in the same length of time. (exactly as you would expect it to if you couldn't see out your windows to tell if you were moving relative to the ground).
From the frame of reference of the roadside, the light beam will take less time to travel backwards and forwards (exactly as one would expect when watching a car and a light beam race).

6. Oct 3, 2009

### yuiop

Endusto, you seem to have missed a couple of important points in your analysis. You have taken time dilation into account, but not length contraction of the car nor "the relativity of simultaneity".

Not calculated it, but from memory the observer in the tree frame will measure more than the (10m/c) seconds than the observer onboard the car measures.

The car observer measures (10m/c) seconds for the light to go from the back to the front of the car, whether the car is moving or not and this is less time the the tree observer measures.

Yes, if you are talking about measurements made by the car observer and no if you are talking about measurments made by the tree observer. The car observer measures the same time interval (10m/c) in either direction, whether the car is moving or not. The tree observer measures less than (10m/c) seconds for the light to go from the front to the back and more than (10m/c) seconds to go the other way as already stated.

Bear in mind that the car observer has to use two clocks (one at the front and one at the back) to make his measurement. When he synchronises his clocks, using the standard procedure, they will not be synchronised from the point of view of the tree observer. The tree observer can understand why the car observer makes measurements that are different to his own, because he can see (measure) that the clocks onboard the car are running slow, are out of sync and the car is length contracted. Look up "relativity of simultaneity". It is key to understanding these sort of problems.