Questions regarding relativistic momentum

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SUMMARY

The discussion centers on the derivation of relativistic momentum, specifically the transition from the equation for relativistic mass involving x-velocity to the general relativistic mass equation. The user questions how the derivation ignores the relativistic effects of y-velocities when moving from the equation for the red ball's mass to the general relativistic mass equation. The hypothesis presented suggests that both y-velocities approach zero, allowing for the cancellation of relativistic effects. Additionally, the recommendation to learn about four-vectors is made as a simpler alternative for understanding relativistic momentum.

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Hi everyone,

I found a derivation for relativistic momentum on this page:
http://en.wikibooks.org/wiki/Special_Relativity/Dynamics

I understand it all the way through, except the point where they have come to this equation.

m[itex]_{red} =[/itex][itex]\frac{m_{blue}}{\sqrt{1-(\frac{u(x,r)}{c})^{2}}}[/itex]

Where(as I understand correctly) mblue is the mass according to the frame moving at the same x-velocity as the blue ball and where mred is the mass of the red ball according to the frame moving with the same x-velocity as the blue ball.

Then they go to:

M =[itex]\frac{m_{0}}{\sqrt{1-(\frac{u}{c})^{2}}}[/itex]

Where u is now the relative velocity of the red ball for the moving frame.
But that is what I don't understand. How can they suddenly ignore the relativistic effects of the y-velocities? The first equation gives the mass of red ball in terms of the relative x-velocity of the red ball. Then how do they get to relativistic mass in terms of the relative velocity u? Because ux,red is only the x component velocity and ured has both a x-component and y-component.

This is my hypothesis:

They let both uy,red and uy, blue approach zero. So that the mass of the blue ball will correspond to the rest mass and that the relativistic effects of the y-velocities will cancel.

Thank you.
 
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