Speed of a moving object, according to another moving object

Phantoful
Messages
30
Reaction score
3

Homework Statement



Special Relativity Question.

Consider objects 1 and 2 moving in the lab frame; they both start at the origin, and #1 moves with a speed u and #2 moves with a speed v. They both move in straight lines, with an angle θ between their trajectories (again in the lab frame). What is #1's speed as viewed by #2?

crFYNA6.png

Homework Equations



x' = γ(x-vt)
t' = γ(t-(vx/c2))
u' = (u±v)/(1±(uv/c2))
Length Contractions
Lorentz Transformations
[/B]
Where γ = 1/((1-(v2/c2))),
v is the velocity of the frame S' relative to S,
and u' is the velocity of some object in the new frame S'.
(These are general equations, not variables from the question).

The Attempt at a Solution


So I assumed the lab frame (I'll call it S) was at rest, and I had #2's (Flash) velocity along the x-axis, and #1's (Superman) velocity at some angle θ1 above that, both starting from the origin. To simplify things, I made another frame S' where the new Vframe=v, so in this new frame object 2's speed is v' = 0, and u' is what we're looking for I think.

In the j direction, nothing would change in S' for object 1, and the velocity in the i direction would be less so the angle would change, as well as the magnitude from that, so we now have θ2. However, I'm not really sure how I would use this new angle. From using the u' equation I have above, I got:

u'x = (u*cos(θ1) - v) / (1 - (u*cos(θ1)*v / (c2)))

(Would this be 1+ uc/c^2 or 1- uv/c^2 in the denominator? I'm not sure how to tell)

u'y = (u*sin(θ1) - 0) / (1 - 0)

I got 1 - 0 because the V of the frame in the j direction is 0

However none of that really took into account θ2, and I'm not sure how to get my final answer with the x and y components. With relativity can I just do the Pythagorean theorem as normal?
 

Attachments

  • crFYNA6.png
    crFYNA6.png
    10.6 KB · Views: 484
Physics news on Phys.org
Phantoful said:
In the j direction, nothing would change in S' for object 1,
This is not correct. You need to think more about this.

Phantoful said:
From using the u' equation I have above, I got:

u'x = (u*cos(θ1) - v) / (1 - (u*cos(θ1)*v / (c2)))
This is not correct either. You cannot treat the components separately. You need to Lorentz transform the worldline of 1 to S’ and identify tge corresponding velocity. Alternatively you can use 4-vectors if you are familiar with them.
 
  • Like
Likes   Reactions: Phantoful
Orodruin said:
Phantoful said:
u'x = (u*cos(θ1) - v) / (1 - (u*cos(θ1)*v / (c2)))
This is not correct either.
I’m hesitant to go against you, but I’m pretty sure that part is correct, isn’t it?

Anyway @Phantoful the y component is definitely not the same when changing frames. (A boost along the x-direction leave y-displacements unchanged, not y-velocities.)

Try considering two events located on superman that are Δt apart in the lab frame. What is Δx and Δy in the lab frame? What is Δt’, Δx’, Δy’ between these events in Flash’s frame?
 
  • Like
Likes   Reactions: Phantoful
Nathanael said:
I’m hesitant to go against you, but I’m pretty sure that part is correct, isn’t it?

Yes, you are correct. I blame lack of sleep last night. :rolleyes: Don't hesitate to point out when I screw up.
 
  • Like
Likes   Reactions: Phantoful

Similar threads

  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
8
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
4K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K