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QUESTION: Hello,

Thank you for your time. I am trying to wrap my head around special relativity. I have read several thought experiments, and I think I understand most of them, but there is one scenario for which, I have not been able to find a good explanation.

...

Consider two light sources, A and B, which are co-moving (e.g. same inertial reference frame). A spaceship sits between A and B and is in the same reference frame. If the spaceship measures the speed of light from A or B it will, of course, be c (speed of light in vacuum).

Now the spaceship starts moving away from A and toward B, at a rate of 0.5 c. If light obeyed Galilean relativity, we would expect the ship to measure the speed of light from source A to be 0.5c and the light from source B to be 1.5c. However, in accordance with special relativity, the spaceship will still see the light from A and B as both being c.

What happens that allows the light from A to appear faster than Galilean relativity predicts, and also allows the light from B to appear slower than Galilean relativity predicts? Playing it out in my head, any manipulation of time or space that would explain A seems to directly contradict B, and vice versa.

ANSWER: Once you accelerate to that speed, you're in a different inertial reference frame where time is defined differently relative to points A and B as observed by your spaceship. That time shift accounts for the constant speed of light and the distances.

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QUESTION: I figured it was along those lines, but what I don't understand is what the time and space effects are. Can you explain from lets say the perspective of the inertial frame of A and B? Here's what I know (please correct me if I'm wrong):

As observed by someone in the inertial frame of A and B:

1. The spaceship will appear contracted due to Lorentz contraction.

2. Time on the spaceship will appear to move more slowly due to time dilation.

By my calculations (which I'm sure are missing something), the light from source A would move from the rear of the ship to the front of the ship in 2L/c seconds (where L is the length of the ship). Light from source B would move from the front of the ship to the rear in (sqrt(3) L)/(3 c) seconds. This would seem to imply that in the ships frame of reference the light from B would be moving faster. What am I missing?

ANSWER: Your first two effects are true, but you have to realize that when you accelerate into an inertial reference frame you essentially re-set your version of where t=0 was for your time calculations. You have to shift it to where time was zero before you started accelerating, in the direction you accelerated in. So you're essentially shifting your starting point or shifting your zero for time, depending on how you want to look at it in the final inertial reference frame. How quickly you accelerate matters, but the net change in velocity is what really shakes out in the end.

I'm also not sure where you got your calculations, but they don't seem right. But you also don't even attempt to account for acceleration, which makes the calculation fantastically more difficult.

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QUESTION: First of all, thank you for your continued help.

I don't want to get bogged down calculating acceleration, so let me write a simpler problem that gets at the heart of what I'm trying to understand. Consider there are two light sources A and B, and an observer O all in the same inertial frame of reference. There is a box (lets name it S) between A and B moving at 1/2 c toward B:

A S-> B

O

Let me describe my calculations so you can tell me where I am going wrong.

First, I will attempt to calculate the time it will take the light from A to pass through the box, from the perspective of the observer.

1. S will be Lorentz contracted so its length as seen by O will be:

L' = L sqrt(1 - (1/2 c)^2 / c^2), where L is the real length of the box, and L' is the contracted length

which reduces to

L' = L sqrt(3) / 2

2. For simplicity, consider the time the light first hits the left edge of S to be t = 0

let Prs be the position of right edge of S along the axis of motion as seen by O

let Pa be the position of A along the axis of motion as seen by O

Prs and Pa can be calculated by:

Prs = 1/2 ct + L'

Pa = ct

3. To determine the time when the light reaches the right edge of S, find the time where Prs = Pa

ct = 1/2 ct + L'

which reduces to

t = 2 L' / c

4. Substituting in the equation for L' gives us

t = 2 (L sqrt(3) / 2) / c

which reduces to

t = L sqrt(3) / c

5. From the observers perspective, time took 2 L'/c seconds to pass through the box, but to get the box's perspective we have to consider time dilation. Which can be calculated by:

t' = t / sqrt(1 - (1/2c)^2/c2) , where t' is time in the box's

subsituting in for t yeilds..

t' = (L sqrt(3) / c) / sqrt(1 - (1/2c)^2/c2)

which reduces to

t' = 2L / c

6. I believe this is wrong since t' should be L/c, but I am not sure where it went wrong.

Next I will attempt to calculate the time it will take the light from B to pass through the box, from the perspective of the observer.

1. Same Lorentz contraction as before,

L' = L sqrt(3) / 2

2. Consider the time the light first hits the right edge of S to be t = 0

let Pls be the position of left edge of S along the axis of motion as seen by O

let Pa be the position of A along the axis of motion as seen by O

Pls and Pa can be calculated by:

Pls = 1/2 ct - L'

Pa = -ct

3. To determine the time when the light reaches the left edge of S, find the time where Pls = Pa

-ct = 1/2 ct - L'

which reduces to

t = 2L' / (3c)

4. Substituting in the equation for L' gives us

t = 2(L sqrt(3) / 2) / (3c)

which reduces to

t = sqrt(3) L / (3c)

5. Calculate the time passed in S's perspective

t' = t / sqrt(1 - (1/2c)^2/c2) , where t' is time in the box's

subsituting in for t yeilds..

t' = (sqrt(3) L / (3c)) / sqrt(1 - (1/2c)^2/c2)

which reduces to

t' = 2L/(3c)

6. I believe this is also wrong since t' should be L/c.

What baffles me more is that the time it takes the light from A to cross the box seems to be greater than the time it takes the light from B to cross the box. This would violate speed of light invariance. What am I missing?

Woah, back up. You want to assume that point S is already moving? Then time is already re-defined. If S were stopped then A and B would emit light "at the same time" if it were halfway. But if S is moving as you have drawn, then S will see that B emitted its light before A did and there's no conflict. The speed of light is still c, the distances are still L/(2gamma), where gamma is the usual relativistic factor (1.51 in this case). From the perspective of both S and O, the light arrives at S from A in a time t = L/2c + t*v/c. Soving for t, we get time t=(L/2c)/(1-v/c). For light from B the sign is reversed and t=(L/2c)/(1+v/c). That's the answer, end of story.

Where you put the observer and S (moving perpendicular to O's sight line), there's no relativistic shift that the observer sees between O and S because it's not moving towards observer O. Common mistake, your math looks good but you were wrong in the physics from the get-go by trying to apply a contraction to what observer O sees.

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I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.**Education/Credentials**

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