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Lorenz's Transformation

  • Thread starter raymondtco
  • Start date
  • #1
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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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Answers and Replies

  • #2
674
2
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?
 
  • #3
12
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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.
 
  • #4
674
2
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:

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

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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  • #5
12
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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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  • #6
674
2
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...

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

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.
 
  • #7
12
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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:
[tex]\Delta[/tex]t = G [ [tex]\Delta[/tex]t' - v[tex]\Delta[/tex]x/c2 ]

Therefore, I would get t = [tex]\frac{1}{0.6}[/tex] [ 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?
 
  • #8
674
2
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 travelled towards him at the speed of light. Therefore x = 18yrs * c = 18lyrs.
 
  • #9
12
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[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
674
2
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 travelling to the ship?" (6 years)

Those both will get you different answers.
 
  • #11
12
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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
674
2
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 travelled at speed c). So now (x' = 18, t' = 18).

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

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