Special Relativity: 3 Objects, Momentum & Time Dilation

[learning]

TL;DR Summary

Let us have an observer at rest at (0,0) in a coordinate plane. Another at (1,0) moving away from the first observer at 5 m/s in the x direction as seen by observer 1 and another object at (1,1) moving away from observer 2 at 0.5C in the y direction as seen by observer 2. Should observer 1 see observer 2 moving in the x direction faster than observer 3.

Let us have an observer at rest at (0,0) in a coordinate plane. Another at (1,0) moving away from the first observer at 5 m/s in the x direction as seen by observer 1 and another object at (1,1) moving away from observer 2 at 0.5C in the y direction as seen by observer 2. Should observer 1 see observer 2 moving in the x direction faster than observer 3?

The confusion I have is because if an object has more mass at higher velocities would it slow down in directions where it is not moving relativistically because of conservation of momentum or time dilation or something?

 

[PeroK]

Summary:: Let us have an observer at rest at (0,0) in a coordinate plane. Another at (1,0) moving away from the first observer at 5 m/s in the x direction as seen by observer 1 and another object at (1,1) moving away from observer 2 at 0.5C in the y direction as seen by observer 2. Should observer 1 see observer 2 moving in the x direction faster than observer 3.

Let us have an observer at rest at (0,0) in a coordinate plane. Another at (1,0) moving away from the first observer at 5 m/s in the x direction as seen by observer 1 and another object at (1,1) moving away from observer 2 at 0.5C in the y direction as seen by observer 2. Should observer 1 see observer 2 moving in the x direction faster than observer 3?

They will have the same x-component of velocity, but different y-components. You can use the velocity transformation rule. Here we have the second frame moving to the right at some speed $v$. And in that frame, the velocity of the third object is $u'_x = 0, u'_y = 0.5c$.

When we transform the velocity of this object to the first frame we get: $$u_x = \frac{u'_x + v}{1 + vu'_x/c^2} = v$$ And both objects 2 and 3 have $v$ as the x-component of their velocity, as measured in the first frame.

The confusion I have is because if an object has more mass at higher velocities would it slow down in directions where it is not moving relativistically because of conservation of momentum or time dilation or something?

This makes no sense. Generally, relativistic mass is a red herring and not used any more.

Conservation of momentum means "the same momentum over time". Momentum varies between reference frames, as it must: you can be at rest in one frame but not in another.

 

[PeroK]

PS The y-component of the velocity transforms according to: $$u_y = \frac{u'_y}{\gamma(1 + vu'_x/c^2)} = \frac{u'_y}{\gamma} < u'_y$$

 

[anuttarasammyak]

@learning positions you said as (0,0)or (1,1) does not matter.
May I interpret your question as :

No. 2 observes No.1 is moving x direction with v1, you say -5 m/s.
No. 1 observes No. 2 is moving x direction with -v1, you say 5m/s.
No. 2 observes No.3 is moving y direction with v3, you say 0.5 c =1.5E8 m/s.
No. 3 observes No.2 is moving y direction with -v3, you say -0.5 c =-1.5E8 m/s.
Question: How fast does No.1 observe No.3 moves ?

 

[learning]

They will have the same x-component of velocity, but different y-components. You can use the velocity transformation rule. Here we have the second frame moving to the right at some speed $v$. And in that frame, the velocity of the third object is $u'_x = 0, u'_y = 0.5c$.

When we transform the velocity of this object to the first frame we get: $$u_x = \frac{u'_x + v}{1 + vu'_x/c^2} = v$$ And both objects 2 and 3 have $v$ as the x-component of their velocity, as measured in the first frame.This makes no sense. Generally, relativistic mass is a red herring and not used any more.

Conservation of momentum means "the same momentum over time". Momentum varies between reference frames, as it must: you can be at rest in one frame but not in another.

Thank you this clears things up.