Break the upper limit: 2x speed of light?

[matttan]

I have a very important qs to ask on relativity.

Picture this:
2 person "A" and "B" : A is stationary and B is moving in a rocket (lets assume a half transparent and half white painted "rocket") at the speed of light TOWARDS A.
But B is holding 2 touch light one shining straight at the transparent part towards A and the 2nd touch light at the white painted "wall" inside the rocket.

So my qs is from A point of view, does the light from B travel 2 times the speed of light towards A ?

If so, then from B point of view, when he shines the 2nd touch light on the white wall inside the rocket, it will appear at the 1 time speed of light.

Otherwise if the upper limit of speed is observe, then from A point of view, light from B touch light is only 1 times the speed of light.
Then from B point of view, shouldn't the light from B to the wall appear 0.5 the speed of light or it will be 1 times the speed of light?


Can anyone advise

Thanks a lot

 

[JesseM]

I have a very important qs to ask on relativity.

Picture this:
2 person "A" and "B" : A is stationary and B is moving in a rocket (lets assume a half transparent and half white painted "rocket") at the speed of light TOWARDS A.

It's impossible for an object with mass, like a rocket, to reach the speed of light--it would take infinite energy. Let's say the rocket is moving at 0.6c in the frame of A instead (also note that all motion is relative, so A cannot be 'stationary' in any absolute sense; in the frame of B, B is stationary and A is moving at 0.6c, and this frame is just as valid as A's).

But B is holding 2 touch light one shining straight at the transparent part towards A and the 2nd touch light at the white painted "wall" inside the rocket.

So my qs is from A point of view, does the light from B travel 2 times the speed of light towards A ?

Nope, you might think that if B is moving at 0.6c in A's frame, then the light would have to be moving at 1.6c in A's frame, but velocities don't add the same way in relativity as they do in Newtonian mechanics. Each observer defines speed in his frame in terms of distance/time as measured by rulers and clocks at rest in his own frame, and remember that in A's frame, B's rulers are all shrunk due to Lorentz contraction, and B's clocks are slowed down due to time dilation (and there is also something called the 'relativity of simultaneity' that says that if two clocks at different locations are synchronized in B's frame, then they will be out-of-sync in A's frame). So, velocities add according to the relativistic velocity addition formula--if an object is moving at velocity v in B's frame, and B is moving at velocity u in the same direction in A's frame, then the object will not be moving at (v + u) in A's frame as you'd expect in Newtonian mechanics, but will instead be moving at (v + u)/(1 + uv/c^2). You can see that in the special case where v=c, the velocity in A's frame will be:

(c + u)/(1 + uc/c^2) = (c + u)/(1 + u/c) = (c + u)/[(1/c)*(c + u)] = c*(c + u)/(c + u) = c.

To understand how this relates to the fact that A sees B's rulers shrunk and his clocks slowed-down and out-of-sync, you have to know that in A's frame, the ruler that B used to measure the distance was shrunk by a factor of \sqrt{1 - v^2/c^2}, the time between ticks on the moving clocks is expanded by 1 / \sqrt{1 - v^2/c^2}, and two of B's clocks which are synchronized in B's frame will be out-of-sync in A's frame by vx/c^2 (where x is the distance between the clocks in B's rest frame). On another thread I gave an example of how these factors come together to ensure that both observers measure the light to be moving at c:

Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor (which determines the amount of length contraction and time dilation) is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.

Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.

If you want to also consider what happens if, after reaching the front end of the moving ruler at 100 seconds in my frame, the light then bounces back towards the back in the opposite direction towards the back end, then at 125 seconds in my frame the light will be at a position of 75 light-seconds on my ruler, and the back end of the moving ruler will be at that position as well. Since 125 seconds have passed in my frame, 125/1.25 = 100 seconds will have passed on the clock at the back of the moving ruler. Now remember that on the clock at the front read 50 seconds when the light reached it, and the ruler is 50 light-seconds long in its own rest frame, so an observer on the moving ruler will have measured the light to take an additional 50 seconds to travel the 50 light-seconds from front end to back end.