Speed of Light Paradox: What Did I Miss?

In summary, the conversation discusses the concept of measuring the speed of light in a moving train and how it differs from the measurement of a stationary observer. It is explained that the theory of special relativity takes into account the relativity of simultaneity and the contraction of the tube, but also mentions that the motion of the tube affects the time it takes for the light to travel. The conversation concludes with a discussion on the distance traveled by the light as measured by the stationary observer.
  • #1
bgq
162
0
Hi, there is something I can't understand:

Consider a stationary observer at A. Now consider an observer B in a train that moves with constant velocity v with respect to A. In the train, B tries to measure the speed of light using an empty tube of length L0 (proper length as measured by B). He sends a light signal at extremity E, the signal reaches extremity F (where a mirror exists) and return back to the extremity E. B measure this duration T0 (proper period as measured by B).
Now B measure the speed of light as:

CB = 2L0/T0

Now according to A the length of the tube is contracted and the time is dilated, so he measure the speed of light as:

CA = 2L/T = (2L0/γ)/(γT0) = (L0/T0)/(γ^2) = CB/(γ^2)

which is different from the value measure by B (divided by Gamma squared)

What did I miss in this analysis?

thanks.
 

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  • #2
bgq said:
What did I miss in this analysis?

Probably relativity of simultaneity.
 
  • #3
jtbell said:
Probably relativity of simultaneity.

I didn't use anything related to simultaneity. The emitting and reception of light events are not simultaneous to both observers.
 
  • #4
bgq said:
What did I miss in this analysis?
From A's frame the tube is moving. The light going from E to F does not merely travel a distance L0/γ since F moves while the light is in transit.
 
  • #5
Doc Al said:
From A's frame the tube is moving. The light going from E to F does not merely travel a distance L0/γ since F moves while the light is in transit.
According to the theory of special relativity, the propagation of the light is not affected by the motion of the tube.
 
  • #6
bgq said:
According to the theory of special relativity, the propagation of the light is not affected by the motion of the tube.
Right! So?

Each frame sees the light travel at speed c with respect to their own frame.
 
  • #7
The speed of light is the same (side note, there are things that are changed about the observation of light, but they have nothing to do with this question), but the position of the tube's extremities are changing with time. You have taken into account the contraction of the tube, but not the fact that the "stationary" observer also sees it moving. The normal derivation of SR uses light moving perpendicular to the motion of the reference frame for this reason.

Essentially what you missed is that the light is moving at c for the "stationary" observer, but distance the light has to travel on the way back will be shorter than the way to the mirror since the "extremity E" has moved toward "extremity F" as time progresses.
 
  • #8
DrewD said:
Essentially what you missed is that the light is moving at c for the "stationary" observer, but distance the light has to travel on the way back will be shorter than the way to the mirror since the "extremity E" has moved toward "extremity F" as time progresses.

The light moves towards the extremity E with a speed c (not c+v), so the time will be the same (dilated of course). The issue here is that light does not behave in the same classical way. The motion of the tube has nothing to do with the propagation of the light (According to the postulates of SR).
 
  • #9
bgq said:
The light moves towards the extremity E with a speed c (not c+v),
Yes, both frames will see the light travel at speed c with respect to themselves.
so the time will be the same (dilated of course).
The same as what?
The issue here is that light does not behave in the same classical way. The motion of the tube has nothing to do with the propagation of the light (According to the postulates of SR).
The motion of the tube certainly affects the time it takes for the light to go from E to F (and from F to E) as seen by A. It doesn't affect the speed, of course, but it does affect the distance and thus the travel time.
 
  • #10
The time it takes for the light to go from E to F and back is T0 in frame B and γT0 in frame A. That part is correct.

But the distance traveled by the light (according to A) is not simply 2L0/γ.
 
  • #11
It has nothing to do with the propagation of the light, but it does close the distance that the light has to move before it reaches the extremity. You are correct that the tube motion has nothing to do with the light's motion, but the distance that the light travels before hitting "extremity E" will be less than the length contracted distance because the tube is moving toward the light. The light's velocity is not changing, but the point where you will stop measuring the time IS changing. The "stationary" observer will measure the light moving at c in one direction and the tube moving v in the other direction. That means that the distance between the light and the end of the tube is changing at -c-v as measured by the stationary observer. Nothing is actually moving at that speed so there is no paradox.
 
  • #12
Doc Al said:
But the distance traveled by the light (according to A) is not simply 2L0/γ.

Why not? If it is not the distance, then what is the distance as measured by A?
 
  • #13
bgq said:
Why not? If it is not the distance, then what is the distance as measured by A?
Let the time it takes for the light to go from E to F according to A be called T1. Then the distance traveled by the light must be (again, according to A):
cT1 = L0/γ + vT1

That's the length of the tube (in A's frame) plus the distance the tube has moved (in A's frame).

Do a similar analysis for the return trip.
 
  • #14
Doc Al said:
Let the time it takes for the light to go from E to F according to A be called T1. Then the distance traveled by the light must be (again, according to A):
cT1 = L0/γ + vT1

That's the length of the tube (in A's frame) plus the distance the tube has moved (in A's frame).

Do a similar analysis for the return trip.

This leads to a contradiction:

cT1 = L0/γ + vT1

cT1 - vT1 = L0

T1(c-v) = L0

This shows that the speed of light becomes c-v (not c); I believe that the distance is L0/γ. It arises directly from lorentz transformations, so we don't have too much to deal with this.
 
  • #15
bgq said:
This leads to a contradiction:

cT1 = L0/γ + vT1

cT1 - vT1 = L0

T1(c-v) = L0
No contradiction at all.

This shows that the speed of light becomes c-v (not c);
Clearly not. Note that I made use of the speed of light as c when I wrote that the distance = speed*time = cT1.

You are confusing the speed of light measured by A, which is of course simply c, with the rate at which the light and the end of the tube move towards each other as seen by A. That is c-v, but as already noted that is not the speed of the light.
I believe that the distance is L0/γ.
That's the length of the tube as seen by A. But as already explained, that's not the distance that light travels according to A.
 
  • #16
Doc Al said:
That's the length of the tube as seen by A. But as already explained, that's not the distance that light travels according to A.

The light travels from F to E, but covers a distance not equal to FE. How is this possible?
 
  • #17
bgq said:
The light travels from F to E, but covers a distance not equal to FE. How is this possible?
Forget relativity for the moment. Imagine you are on a train moving at some speed. You walk from the rear of the car to the front of the car. As seen from the train tracks, would you not agree that you traveled a greater distance than just the length of the car? Same idea here.
 
  • #18
The light 4-vector in the rest frame of the emitter is a null vector (ω,k,0,0), with ω/k=c=1 (the last equality because I'm using gemetric units). If this vector is boosted by β , the result is another null vector ( because the proper length is invariant) (√(1+β)(1-β), √(1+β)(1-β), 0, 0). The velocity of propagation is obviously still 1 ( ratio of first and second components).

So the Lorentz transformation ensures that the speed of propagation is unchanged, but the frequency and wave-number are identically Doppler shifted.
 
  • #19
Doc Al said:
Forget relativity for the moment. Imagine you are on a train moving at some speed. You walk from the rear of the car to the front of the car. As seen from the train tracks, would you not agree that you traveled a greater distance than just the length of the car? Same idea here.

This is OK according to Galilean relativity, where the speed is changed which ensures the change in the distance in the same invariant time. In SR there is a critical point is that the speed of light is not affected by either the observer or source or both. Anyway, even in Galilean relativity the distance is not changed between E and F, the speed does.
 
  • #20
bgq said:
This is OK according to Galilean relativity, where the speed is changed which ensures the change in the distance in the same invariant time. In SR there is a critical point is that the speed of light is not affected by either the observer or source or both. Anyway, even in Galilean relativity the distance is not changed between E and F, the speed does.
You seem unable to grasp the fact that the distance traveled is not equal to the length of the tube. I suggest that you really think about the scenario that Doc Al posted and try to understand the difference between length and distance traveled. If you are unwilling to do that then it might help to write down the definitions of distance traveled and length.
 
  • #21
From the stationary observer's perspective:

If T1 is the time of light travel in the same direction as the motion af the train, then

c*T1 = L/g +v*T1, which gives T1 = (L/g)/(c-v). Likewise, the time of light travel in the opposite direction is T2 = (L/g)/(c+v). Adding this gives, after a little algebra:
T1+T2 = 2gT, where T is the one way time measured by the moving observer, just as it should be. Thus, the assumption that the stationary observer measures the velocity to c leads to the right answer for the time.
 
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  • #22
DaleSpam said:
You seem unable to grasp the fact that the distance traveled is not equal to the length of the tube. I suggest that you really think about the scenario that Doc Al posted and try to understand the difference between length and distance traveled. If you are unwilling to do that then it might help to write down the definitions of distance traveled and length.

Hmm, I think I now understand the difference between length and the covered distance, but unfortunately this leads to the same result:

d = (L0/γ + vT1) + (L0/γ - vT1) = 2L0
 
  • #23
bgq said:
hmm, i think i now understand the difference between length and the covered distance, but unfortunately this leads to the same result:

D = (l0/γ + vt1) + (l0/γ - vt1) = 2l0
t1 ≠ t2
 
  • #24
bgq said:
Hmm, I think I now understand the difference between length and the covered distance, but unfortunately this leads to the same result:

d = (L0/γ + vT1) + (L0/γ - vT1) = 2L0
I cannot decipher that formula in this context. There should be two speeds, one for the speed of the object and one for the signal. Also, there should be two times, one for reaching the far end, and one for returning to the near end.
 
  • #25
The situation is as following:

A sees B moving at vrel, with a ruler/tube in front of him. At some point, an event happens. B shoots some photons towards the direction of the ruler/tube.
Both draw a coordinate system where they consider this event to happen at x=0/t=0.

As you correctly calculated, the time for the lightbeam to travel the ruler and back seen from B's point of view is 2Lo/cA will see the same ruler length contracted L = Lo*1/γ

So LOGICALLY, as light travels at c for both observers. If the ruler was not moving, it would be trivial 2L/c. Unfortunately the ruler end is moving away of the lightbeam, while the back is moving towards it on the turn around trip.
So we have L/(c-vrel) for the time it takes to move towards the end, while we have L/(c+v) for the time it takes to move back.
L/(c-vrel) + L/(c+vrel) is the time it takes seen from A's point of view.
Time-interval expansion(dilation) is measured LOCALLY in B's rest frame. Meaning that two events which happen at the SAME space but at different times and are measured to have a time interval of t seconds, are measured in A's rest frame to have a higher time-interval (dilation/expansion).
They do NOT happen at the same space-distance in A's frame.

In your example, the two local events seen from B's point of view are:
E1) B shoots lightbeam at x=0/t=0 towards the back of the rocket.
E2) Lightbeam returns to b after being reflected at x=0/t= 2Lo/c
This is NOT equivalent to length contraction, because while similarly here we measure the distance between two events at the SAME time seen from B's rest frame.
We however do NOT measure the distance between those two events in A's frame which of course happen at different times, just like in the case of time dilation the events A measures happen at a different space-distance.
INSTEAD we measure the distance between two events which happen at the SAME time on the back and front of an object which is moving at vrel.

I did not think this out, but i find it quite remarkable.
It seems to be a human centric view. Whatever seems to make more sense to us.
 
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  • #26
Where are E and F with respect to A at the different points when the light reflects? How far does the light have to travel in frame A. Draw where the points would be with respect to A if you need to.

I mean heck take a car moving at 60 mph. You have only a single point in the car let's call this point E. With respect to the car E never moves it's still in the same spot. But from the ground at the start there is spot E0, and spot E1, these after an hour are 60 miles away from each other.
 
  • #27
To actually answer the question.

The distance light will travel in A's frame is NOT 2L. You forgot that A sees the tube moving. This increases the distance on the trip towards the end, while shortening it on the round-trip.

L/(c-v) = t1 is the time the lightbeam takes to the end of the tube, seen from A's point of view
L/(c+v) = t2 the time it takes to the back.

Light travels at c. So we have a total length of c*t1 + c*t2 = Ltotal

So instead of just 2L/c you have,
Ltotal/c should be T as seen from A's point of view. I did not calculate it, so there might be some mistake if my mind tricked me.
 
Last edited:
  • #28
YES!
I think I got it!

Let t1 be the time from E to F, and t2 from F to E as measured by A.

According to Lorentz transformation:

t1 = γ(T0/2 + vL0/c2)

t2 = γ(T0/2 - vL0/c2)

d = (L0/γ + vt1) + (L0/γ - vt2)

= 2L0/γ + v(t2 - t2)

= 2L0/γ + v(2γvL0/c2)

= 2L0/γ + 2γL0v2/c2

= 2L0(1/γ + γv2/c2)

CA = d/t = 2L0(1/γ + γv2/c2)/γT0

= (2L0/T0)(1/γ2 + v2/c2)

= (2L0/T0)(1 - v2/c2 + v2/c2)

= (2L0/T0)(1)

= 2L0/T0

= CB

Thank you very much for your help, it was very difficult to me to solve it alone. Thank you very much.

I am happy :smile:
 
Last edited:
  • #29
bgq said:
The light moves towards the extremity E with a speed c (not c+v), so the time will be the same (dilated of course). The issue here is that light does not behave in the same classical way. The motion of the tube has nothing to do with the propagation of the light (According to the postulates of SR).
Quite the contrary: c+v is used (or more precisely c-v), and that light behaves in a classical way is expressed by saying that the motion of the tube is independent of the light propagation (=second postulate).
I recently explained the use of c-v here:
- https://www.physicsforums.com/showthread.php?p=3992825
- https://www.physicsforums.com/showthread.php?p=4082811

But I see now that you already solved it by a different route - glad to see that you are happy. :smile:
 
  • #30
  • #31
"bgq
Posts: 59
YES!
I think I got it!
Let t1 be the time from E to F, and t2 from F to E as measured by A.
According to Lorentz transformation:
t1 = γ(T0/2 + vL0/c2)
t2 = γ(T0/2 - vL0/c2)
d = (L0/γ + vt1) + (L0/γ - vt2)
= 2L0/γ + v(t2 - t2)
= 2L0/γ + v(2γvL0/c2)
= 2L0/γ + 2γL0v2/c2
= 2L0(1/γ + γv2/c2)

so then solving, d = gamma * 2Lo, not divided by gamma as orginally posted,
this confuses me, any explanations would be helpful, thanks
 
Last edited:
  • #32
randyu said:
[..]
Let t1 be the time from E to F, and t2 from F to E as measured by A.
According to Lorentz transformation:
t1 = γ(T0/2 + vL0/c2)
t2 = γ(T0/2 - vL0/c2)
d = (L0/γ + vt1) + (L0/γ - vt2)
= 2L0/γ + v(t2 - t2)
= 2L0/γ + v(2γvL0/c2)
= 2L0/γ + 2γL0v2/c2
= 2L0(1/γ + γv2/c2)

so then solving, d = gamma * 2Lo, not divided by gamma as orginally posted,
this confuses me, any explanations would be helpful, thanks
Welcome to physicsforums. :smile:

You referred to post #28.

How did you get from
d= 2L0(1/γ + γv2/c2)
to
d = γ * 2Lo ?

Perhaps you did not look at it carefully.
(1/γ + γv2/c2) = 1/γ(1 + γ2v2/c2) ≠ γ
[EDIT: I was wrong, see next]
 
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  • #33
hi harrlylin,

Y=1/sqt(1-v2/c2) > v2/c2=(Y2-1)/Y2
then,
d=2Lo * 1/Y+Y(Y2-1)/Y2= 2Lo*Y
is this not right, thanks
 
  • #34
randyu said:
hi harrlylin,

Y=1/sqt(1-v2/c2) > v2/c2=(Y2-1)/Y2
then,
d=2Lo * 1/Y+Y(Y2-1)/Y2= 2Lo*Y
is this not right, thanks
Oops - I must admit that I did not check bqq's derivation. :blushing:
And yes you are right. Regretfully I did not follow that conversation, [STRIKE]and I won't look into that now. Maybe someone else will, or I will later. [/STRIKE]
What I can say already, is that I see nowhere claimed that d= 2L0/Y :tongue2:

Oh OK I see it: the answer is in post #22. Did you read that?

d is the path length that the light ray travels, and as d>2Lo I suppose that it is the path length in the rest system, with respect to which the apparatus is moving. Do you follow that?

It's like the Michelson-Morley experiment. And it is very well explained in post #17.
So, to elaborate: You walk from the rear of the car to the front of the car and back. As seen from the train tracks, the distance is greater.
- http://en.wikipedia.org/wiki/Michelson–Morley_experiment
 
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  • #35
bgq said:
Hi, there is something I can't understand:

Consider a stationary observer at A. Now consider an observer B in a train that moves with constant velocity v with respect to A. In the train, B tries to measure the speed of light using an empty tube of length L0 (proper length as measured by B). He sends a light signal at extremity E, the signal reaches extremity F (where a mirror exists) and return back to the extremity E. B measure this duration T0 (proper period as measured by B).

Now B measure the speed of light as:

CB = 2L0/T0

Now according to A the length of the tube is contracted and the time is dilated, so he measure the speed of light as:

CA = 2L/T = (2L0/γ)/(γT0) = (L0/T0)/(γ^2) = CB/(γ^2)

which is different from the value measure by B (divided by Gamma squared)

thanks harrylin, what I was thinking is from post #1, d=2Lo/Y and t=YTo which gave c as "divided by gamma squared".
This was wrong from the beginning, the length in the stationary frame is longer than the moving length Lo as you state, confusing because each is moving relative to the other. And also the time is YTo in the "stationary" frame, longer.
I guess this all makes sense, I just get confused about time dilation/expansion using t in different ways it seems to me, sometime an interval, sometimes ticks. Oh well, will come together sooner or later.
Thanks.
 
<h2>1. What is the speed of light paradox?</h2><p>The speed of light paradox is a thought experiment that questions our understanding of the speed of light. It suggests that if an observer is traveling at the speed of light, time would stand still for them, and they would not experience the passage of time. This goes against our understanding of time and the laws of physics.</p><h2>2. How does the speed of light paradox challenge our understanding of time?</h2><p>The speed of light paradox challenges our understanding of time because it suggests that time can be relative and can be affected by the speed of an observer. This goes against the concept of time being constant and universal.</p><h2>3. What is the significance of the speed of light paradox in physics?</h2><p>The speed of light paradox is significant in physics because it highlights the limitations of our current understanding of the universe. It also challenges us to think outside the box and explore new theories and concepts to explain the behavior of light and time.</p><h2>4. How do scientists explain the speed of light paradox?</h2><p>Scientists explain the speed of light paradox using Einstein's theory of relativity. This theory states that the laws of physics are the same for all observers, regardless of their relative motion. It also suggests that time and space are relative and can be affected by the speed of an observer.</p><h2>5. Can the speed of light paradox be resolved?</h2><p>Currently, there is no definitive answer to the speed of light paradox. However, scientists continue to explore and research different theories and concepts to try to resolve this paradox. It is possible that with further advancements in technology and understanding, we may one day have a better understanding of the speed of light and its paradox.</p>

1. What is the speed of light paradox?

The speed of light paradox is a thought experiment that questions our understanding of the speed of light. It suggests that if an observer is traveling at the speed of light, time would stand still for them, and they would not experience the passage of time. This goes against our understanding of time and the laws of physics.

2. How does the speed of light paradox challenge our understanding of time?

The speed of light paradox challenges our understanding of time because it suggests that time can be relative and can be affected by the speed of an observer. This goes against the concept of time being constant and universal.

3. What is the significance of the speed of light paradox in physics?

The speed of light paradox is significant in physics because it highlights the limitations of our current understanding of the universe. It also challenges us to think outside the box and explore new theories and concepts to explain the behavior of light and time.

4. How do scientists explain the speed of light paradox?

Scientists explain the speed of light paradox using Einstein's theory of relativity. This theory states that the laws of physics are the same for all observers, regardless of their relative motion. It also suggests that time and space are relative and can be affected by the speed of an observer.

5. Can the speed of light paradox be resolved?

Currently, there is no definitive answer to the speed of light paradox. However, scientists continue to explore and research different theories and concepts to try to resolve this paradox. It is possible that with further advancements in technology and understanding, we may one day have a better understanding of the speed of light and its paradox.

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