What is the discrepancy between time measurements in different frames?

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SUMMARY

The forum discussion revolves around the discrepancy in time measurements between two reference frames in special relativity, specifically involving a person traveling at 0.8c between two stars, A and B, which are 12 light-years apart. The person calculates that light from star A takes 18 years to reach him at the midpoint, while using Lorentz transformation, it is determined that star A would claim the light pulse took only 6 years to travel. The calculations reveal that the time elapsed in the star's frame differs significantly from the observer's frame due to relativistic effects, leading to confusion about the correct interpretation of time intervals in different frames.

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Homework Statement



There are two stars A and B relatively at rest in the universe with the proper length of 12 ly.
A person is traveling at the speed 0.8c relative to the stars along the extended line of stars A & B (towards A).
When the person gets to the midpoint between A & B, he sees a light pulse from stars A and B respectively.


Homework Equations



How many years ago does he claim the Star A emits the light pulse? (18 years)

Using Lorenz's transformation equation, t = G [ t' + vx'/c2 ]
What was the time difference for A when the person's time has elapsed these 18 years?


The Attempt at a Solution



After length contraction, 12 ly => 7.2 ly for the person.
3.6 ly / (c-0.8c) = 18 years.

By using the Lorenz's transformation equation from the person's R.F. to A's R.F., the result is not 6 ly / c = 6 years as I expected.

Primed variables are with respect to the person, while the unprimed ones are to the Star A.

t = G [ t' - vx/c2 ]

G=0.6
tf = 0.6 [ t'f - 0.8c*3.6ly/c2 ] = 0.6t'f - 1.728y

ti = 0.6 [ t'i - 0.8c*( 3.6ly + 0.8c*18y)/c2 ] = 0.6t'i - 8.64y

tf - ti = 0.6( t'f - t'i ) - 1.728 + 8.64
= 0.6*18y + 6.912y = 17.712y (Relative to Star A.)

Star A would claim that the light pulse only travels 6 years from himself to the person who's in the midpoint.
Why is it not 6ly/c= 6y?? (Relative to Star A.)

This is the discrepancy I'm asking about.


What's wrong??
 
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Could you show some work for your first answer? You just said the person sees the distance from A to B as 7.2 ly, so how come it takes 18 years for light to travel to him when he is at the midpoint?
 
Sure!
What I did was that,
when the person sees the light, it is 3.6 ly apart from the Star A.
And, light is traveling at c whereas the Star travels at 0.8c relative to the person.
Therefore, it takes ""3.6 ly / (c-0.8c) = 18 years "" for the ligth pulse and A to be 3.6 ly apart from each other relative to the person.
As a result, 18 years is the duration the person would claim before the ligth pulse is emitted.

That's how I got 18 years. Hopefully it's clear enough.
I would later on show the time difference when using the Lorenz's Trasformation.
 
Ah, I see how you did it. Well, since the star frame is not the inertial frame, its change in time should be longer. For example using your lorentz transformation, I get:

t = \frac{1}{\sqrt{1-0.8^2}}\left(18+0.8*0\right)=30\textrm{ yrs}

Although check me, since I am prone to mistakes :)

By the way it took me awhile to realize he didn't start from B, but passed B along his way to A. That is why I got confused with the 18 years. Which means in his frame, the light was sent a distance 18 lyrs before reaching him. That also transforms as...

18 lyrs => to planet frame: G*18lyrs = 30 lyrs

So the planet A sent the light beam 30 years ahead of time.
 
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Sorry but I don't really agree with your calculation since it should not be 0 for the second term. Let me show you my calculation.

Primed variables are with respect to the person, while the unprimed ones are to the Star A.

t = G [ t' - vx/c2 ]

G=0.6
tf = 0.6 [ t'f - 0.8c*3.6ly/c2 ] = 0.6t'f - 1.728y

ti = 0.6 [ t'i - 0.8c*( 3.6ly + 0.8c*18y)/c2 ] = 0.6t'i - 8.64y

tf - ti = 0.6( t'f - t'i ) - 1.728 + 8.64
= 0.6*18y + 6.912y = 17.712y (Relative to Star A.)

Why is it not 6ly/c= 6y?? (Relative to Star A.)

This is the discrepancy I'm asking about.
 
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What was the time difference for A when the person's time has elapsed these 18 years?

It only asks for the time difference in A-frame compared to person-frame, and not when A sent the beam of light in the A-frame. If you wanted to calculate when A sent the beam of light then you would do...

t = \frac{1}{\sqrt{1-0.8^2}}\left(18-0.8*18\right)=6\textrm{ yrs}

Where the two events are separated by a time of 18 yrs in person-frame, and a distance of 18 lyrs in person-frame.

Also your G should be 1/sqrt(1-0.8^2) = 5/3.
 
Hmm.. Yeah, my calculation was wrong in the G.
I wonder how you get 0.8*18. Are you subtracting two equations and plug in only the difference?

I guess the equation will become:
\Deltat = G [ \Deltat' - v\Deltax/c2 ]

Therefore, I would get t = \frac{1}{0.6} [ 18 - 0.8c(0.8c*18yrs)/c2]
As a result, I will get the t to be 10.8 yrs instead.

Why don't you have another 0.8c*18yrs, aside from equation's v= 0.8c, in the second term for the difference in position of A relative to the person?
 
The part in question is vx/c^2. Well v=0.8c which you agree, and x=18lyrs (which you don't agree yet). We are in the person-frame (so his speed is 0 relative to himself and light travel towards him at v=c) so the initial event (a flash of light) ocurred 18 yrs away from him and traveled towards him at the speed of light. Therefore x = 18yrs * c = 18lyrs.
 
[Ahh.. I kind of get it. I understand what you are calculating now.] (Before modified.)
Wait.. No.. If you are calculating the time difference for the light, then I do not really agree with 0.8c since it is the speed of Star A relative to me. I would use c instead as it is the speed of the light relative to me.

Why can't I use the difference in position for the Star A instead?
I thought I am calculating the time difference relative to Star A, and therefore, vx/c^2 for x I should care about the positions of Star A rather than the light.
What I'm thinking is just that initially Star A is at some position relative to me and moving towards me, so is the final condition. Hence, I should be able to claim the initial time and final time (also time difference) for those positions where the Star A was and is located. That should be the time for A, shouldn't it?

Anyway, to put it simple, I guess my question comes back to "What is v and x in the equation?" if we are to transform the time in two reference frames.

Thanks a lot for you patience by the way!
 
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  • #10
v and x depends on what you are asking...

Are you asking "What was the time difference for A when the person's time has elapsed these 18 years?" (30 years)

Or are you asking "What was the time difference for A when the light was traveling to the ship?" (6 years)

Those both will get you different answers.
 
  • #11
Hmm.. Why will we have the discrepancy like this? I thought the person could claim that 18 years ago, Star A just emitted the light and according to the Lorentz transformation, the time for A should elapsed 6 years. What's wrong with this idea?

Would you please explain to me in more details about why you plugged all those corresponding numbers in the equations? I don't really get how you plugged in each datum since, to me, your R.F. keeps jumping back and forth.

I greatly appreciate your kindness and patience!
 
  • #12
I thought the person could claim that 18 years ago, Star A just emitted the light and according to the Lorentz transformation, the time for A should elapsed 6 years.

The differences are the locations of the events in the person's-frame along his worldline.

A clock on the spaceship will measure 18 years but not move in the spaceship... That means x'=0 and t'=18yrs. Because the first event is the clock starting at 0, and the 2nd event is the clock saying 18 yrs, but the clock never moves in the spaceship frame. (x'=0, t'=18)

Now for the light event. In the person frame, first event occurs at x'_0=18 lyrs away, and t'_0 = 0 yrs. The 2nd event occurs at x'_1 = 0 lyrs away (light hits spaceship) and t'_1 = 18 yrs (light traveled at speed c). So now (x' = 18, t' = 18).

So both situations have different events and therefore different times in the A-frame.