Special Relativity problem

  • Thread starter m00nd0g68
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  • #1
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I need help with this one...
Observer S notes that two colored lights, separated by 2400m, occur at positions on the +x-axis of his reference frame. A blue flash occurs first, followed after 5.00μs by a red flash. The red flash is farther away from his origin than the blue. Another observer, R, obtains exactly the same differences in time and position, but declares that the red flash occurred FIRST.

a) What is the relative speed of R, with respect to S, and in what direction
(+x or –x)

b) Calculate which flash occurred closer to the origin of R.

If anyone could get me started I would appreciate it.
moondog
 
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Answers and Replies

  • #2
Doc Al
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Start by writing what you know. I'll start you off: What does each observer measure as the difference in time and position between these two events? (Red Flash - Blue Flash):

S frame: [itex]\Delta x = 2400[/itex]m; [itex]\Delta t = 5[/itex]μs
R frame: [itex]\Delta x' = 2400[/itex]m; [itex]\Delta t' = -5[/itex]μs

Now apply the Lorentz transformations to solve for the relative velocity between the frames. Once you've solved part a, a little thinking will solve part b.
 
  • #3
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Thank you

I appreciate the hint. I will post what I have later.
moondog
 
  • #4
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What I have so far

For S frame:
x'a = (0 -u*a)γ=0 x'b = (2400-u*5E-6)γ
y'a = 0 y'b = 0
z'a = 0 z'b = 0
t'a =(0-u/c^2*0)γ=0 t'b = (5E-6-u/c^2*5E-6)γ

Is this a good start or have I missed some fundamentat concept?

moondog
 
  • #5
Doc Al
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Lorentz transformations

Start with the Lorentz transformations in this form:
[tex]\Delta x' = \gamma(\Delta x - v\Delta t)[/tex]
[tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]
Pick either one and you can solve for v, which is the speed of the primed frame with respect to the unprimed frame. Let S be the unprimed frame.
 
  • #6
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gamma

I have solved the Δx' equation and have:
v=(x'-Δx*γ)/(Δtγ)
Now how do I find gamma? This is where I am stuck...
moondog
 
  • #7
Romperstomper
I haven't done the math, but as far as where they are, could observer R be on the opposite side of the lights going the same speed as observer S, but in the opposite direction of observer S?
 
  • #8
Doc Al
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m00nd0g68 said:
I have solved the Δx' equation and have:
v=(x'-Δx*γ)/(Δtγ)
Now how do I find gamma? This is where I am stuck...
You can't solve for v without first writing [itex]\gamma[/itex] in terms of v:
[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
Substitute that into the Δx' equation and then solve for v.
 
  • #9
Doc Al
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Romperstomper said:
I haven't done the math, but as far as where they are, could observer R be on the opposite side of the lights going the same speed as observer S, but in the opposite direction of observer S?
Not sure what you are saying here. Clearly, each observer sees himself as at rest and the other as moving with the same speed (v) in the opposite direction. (If the velocity of R with respect to S is +v, then the velocity of S with respect to R is -v.)
 

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