Conservation of linear momentum at relativistic speeds

[carl fischbach]

The question I ask is linear momentum conserved in
in instance cited below?


You place a particle at the origin on a x-y axis
and accelerate it to 61% of c in the y direction.
Then you accelerate it to 61% of c in the x direction.
The net velocity of the particle will be
86% of c at 45 degrees.The key here is that it
takes approximately 3 times the energy to
accelerate the particle in the x direction than
the y direction, due to the fact that the net
velocity change in the y direction is 0%-61% of c
and in x direction the net velocity change is
61%-86% of c.If the rate of acceleration,distance
of acceleration and time of acceleration are the
same on the x and y axis, then force of acceleration on
the x-axis has to be greater than
on the y axis, since the energy of acceleration
on the x-axis is 3 times that of the y axis.
Therefore the momentum on the x-axis is greater
the y axis.

If the particle's final velocity is 86% of c at
45 degrees then the momentum of acceleration
should be equal on both the x and y axis.
Is there a discrepancy in momentum here?

 

[Labguy]

Originally posted by carl fischbach
The question I ask is linear momentum conserved in
in instance cited below?


You place a particle at the origin on a x-y axis
and accelerate it to 61% of c in the y direction.
Then you accelerate it to 61% of c in the x direction.
The net velocity of the particle will be
86% of c at 45 degrees.The key here is that it
takes approximately 3 times the energy to
accelerate the particle in the x direction than
the y direction, due to the fact that the net
velocity change in the y direction is 0%-61% of c
and in x direction the net velocity change is
61%-86% of c.If the rate of acceleration,distance
of acceleration and time of acceleration are the
same on the x and y axis, then force of acceleration on
the x-axis has to be greater than
on the y axis, since the energy of acceleration
on the x-axis is 3 times that of the y axis.
Therefore the momentum on the x-axis is greater
the y axis.

If the particle's final velocity is 86% of c at
45 degrees then the momentum of acceleration
should be equal on both the x and y axis.
Is there a discrepancy in momentum here?

I don't think so. No acceleration above 0.61c is used or provides momentum. The 0.86c is the apparent VECTORED V (in your example). But, I don't think that your example is correct since you seem to be using two-dimensional (graph-paper) trig here, while with excluding time. In V calculations, a Y velocity at 0.61c and an X velocity at 0.61c takes time to reach the "end-point" from where you measure the Hypotenuse. The "actual V" of the vectored triangle cannot exceed the V of the greater of the XV or the YV if time is included. The "apparent" V can though, as is seen in apparent superluminal expansion around some supernova remnants.

 

[R0man]

Yes, there is a discrepancy in momentum in this scenario. While linear momentum is conserved in all frames of reference, the conservation of momentum at relativistic speeds is not as straightforward as it is in classical mechanics. This is due to the fact that as an object approaches the speed of light, its mass increases and its momentum also increases. In this scenario, the particle's final velocity is 86% of c, which means its mass has increased and therefore its momentum has also increased. However, the discrepancy in momentum arises from the fact that it takes 3 times the energy to accelerate the particle in the x direction compared to the y direction. This means that the force of acceleration on the x-axis is greater, resulting in a greater momentum in that direction. Therefore, while the total momentum of the particle is conserved, the distribution of momentum between the x and y directions is not equal.