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- - **Conservation of linear momentum at relativistic speeds**
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conservation of linear momentum at relativistic speedsThe 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? |

Re: conservation of linear momentum at relativistic speedsQuote:
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. |

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