• Support PF! Buy your school textbooks, materials and every day products Here!

Lorentz transformations - relative speeds of frames

  • #1
ElijahRockers
Gold Member
270
10

Homework Statement



Suzanne observes 2 light pulses to be emitted from the same location, but separated in time by 3μs. Mark sees the emission of the same two pulses separated in time by by 9μs.

a) How fast is Mark moving relative to Suzanne?

b) According to Mark, what is the separation in space of the two pulses?

Homework Equations



Time dilation, Lorentz transforms

The Attempt at a Solution



Part A

My first attempt had Mark as the 'primed' observer, since I am used to that frame moving, and they are asking about Mark's relative velocity to Suzanne, and I got an answer that didn't make any sense for v.

I realized that since Mark is observing a larger time interval than Suzy, she must be in the 'primed' frame for Δt = γΔt' to make any sense. So I did it that way, and found a v of .943c which I think is correct. (The answers to this question are not in the back of the book)

But as for the question 'How fast is Mark moving relative to Suzanne' I am not sure. Is the answer to this question -.943c, since Suzy's frame is moving away from Mark in the positive direction?

Part B

Suzy observes the pulses to be emitted from the same location, so Δx' = 0. Mark's observation however, Δx, is going to be non-zero, since the pulses are moving in his frame.
So in this case, would I just use (vt = d) .943c * 9μs ≈ 2.5km ? or is there something about the speed of light and the pulses moving away from him that would distort this answer? Or should I just assume he observes the pulses instantaneously? If the latter, then would 2.5km be the correct answer?

Thanks for the help in advance..
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
44,904
1,169
Part A

My first attempt had Mark as the 'primed' observer, since I am used to that frame moving, and they are asking about Mark's relative velocity to Suzanne, and I got an answer that didn't make any sense for v.
When using the full Lorentz transformations, the prime and unprimed designations are arbitrary.

I realized that since Mark is observing a larger time interval than Suzy, she must be in the 'primed' frame for Δt = γΔt' to make any sense. So I did it that way, and found a v of .943c which I think is correct. (The answers to this question are not in the back of the book)
When you use the special case formula for time dilation, Δt = γΔt', then you have to make sure that the primed frame is the one in which the events take place at the same location. (Like the ticks of a clock.)

But as for the question 'How fast is Mark moving relative to Suzanne' I am not sure. Is the answer to this question -.943c, since Suzy's frame is moving away from Mark in the positive direction?
You weren't given any information about direction, so just give the magnitude of the speed.

Part B

Suzy observes the pulses to be emitted from the same location, so Δx' = 0. Mark's observation however, Δx, is going to be non-zero, since the pulses are moving in his frame.
So in this case, would I just use (vt = d) .943c * 9μs ≈ 2.5km ? or is there something about the speed of light and the pulses moving away from him that would distort this answer? Or should I just assume he observes the pulses instantaneously? If the latter, then would 2.5km be the correct answer?
Yes, that's all there is to it. You can also use the full Lorentz transformations. That would a good exercise for you.
 
  • #3
ElijahRockers
Gold Member
270
10
Ok.... I couldn't use the special time dilation formula for Mark because his events didn't occur at the same location in his frame.

So when you say the full Lorentz Transformations, are you meaning

x' = γ(x - vt)

y' = y

z' = z

t' = γ(t- vx/c2)

Where the primed frame is moving along the positive x-axis with respect to the unprimed frame? If so, what is the main difference between these and their 'inverse' transformations:

x = γ(x' + vt')

t = γ(t' + vx'/c2)

Is this still describing a situation where the primed frame is moving along the positive x-axis with respect to the unprimed frame?

If so, then I think it's all starting to come together... thanks again for your help.
 
  • #4
20,111
4,191
The problem statement seems to imply that the source is at rest in Suzanne's frame of reference. There is no doppler correction for her frame, but there will be a doppler correction for Mark's frame, and it will depend on whether Mark is moving toward Suzanne's source or away from Suzanne's source.
 
  • #5
Doc Al
Mentor
44,904
1,169
Ok.... I couldn't use the special time dilation formula for Mark because his events didn't occur at the same location in his frame.
Right.
So when you say the full Lorentz Transformations, are you meaning

x' = γ(x - vt)

y' = y

z' = z

t' = γ(t- vx/c2)
I mean those and the inverse transforms.
Where the primed frame is moving along the positive x-axis with respect to the unprimed frame?
Definitely.
If so, what is the main difference between these and their 'inverse' transformations:

x = γ(x' + vt')

t = γ(t' + vx'/c2)

Is this still describing a situation where the primed frame is moving along the positive x-axis with respect to the unprimed frame?
Events happen for everyone. The 'regular' LT lets you transform measurements from the unprimed frame to the primed frame; the 'inverse' LT does the reverse. They are both saying the same thing.
 
  • #6
Doc Al
Mentor
44,904
1,169
The problem statement seems to imply that the source is at rest in Suzanne's frame of reference. There is no doppler correction for her frame, but there will be a doppler correction for Mark's frame, and it will depend on whether Mark is moving toward Suzanne's source or away from Suzanne's source.
Good point. The problem is ambiguous in its language where it says: "Mark sees the emission of the same two pulses separated in time by by 9μs."

It's not clear whether it's just sloppy language or if they really did mean the problem to be about the Doppler effect.

@ElijahRockers: Can you tell from context? Have you covered the Doppler effect yet?

The point is this: The Lorentz Transforms relate measurements made in the various frames after light travel time has been accounted for. So they don't necessarily describe what an observer literally sees--that depends on how close they are to the light source and how long it takes for the light to travel to their eyes.

If you've covered the Doppler effect for light, then you'll need to rework the problem. Try it.
 
  • #7
20,111
4,191
Mark's velocity relative to Suzanne also depends on whether Suzanne is located between the source and Mark, whether the source is located between Suzanne and Mark, or whether Mark is located between Suzanne and the source.
 
  • #8
Doc Al
Mentor
44,904
1,169
Mark's velocity relative to Suzanne also depends on whether Suzanne is located between the source and Mark, whether the source is located between Suzanne and Mark, or whether Mark is located between Suzanne and the source.
It's worse than that. Even if Suzanne is collocated with the source, what Mark sees will depend on how the source is moving with respect to him: Whether directly toward or away from him or transverse to him.

So I seriously doubt this was meant to be a Doppler problem.
 
  • #9
ElijahRockers
Gold Member
270
10
Mark's velocity relative to Suzanne also depends on whether Suzanne is located between the source and Mark, whether the source is located between Suzanne and Mark, or whether Mark is located between Suzanne and the source.
It's too late in the evening for this, lol
 

Related Threads on Lorentz transformations - relative speeds of frames

Replies
4
Views
1K
Replies
1
Views
490
  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
Replies
0
Views
1K
Replies
1
Views
5K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
2
Views
887
Replies
4
Views
1K
Top