Deriving relativistic mass from inelastic collision in different frames

[mainguy]

Hi guys, thanks for helping with this! I'm a little stuck with this question about the derivation for relativistic mass.

1. Homework Statement
By considering the inelastic collision of two balls as perceived in different reference frames show that the relativistic mass is equal to the rest mass multiplied by the gamma factor (sqrt(1-u^2/c^2).

Homework Equations

So I know the factor is 1/(1-u²/c²) but proving it is tough.
I've considered a reference frame moving at u in a direction perpendicular to the collision, so basically this:

The Attempt at a Solution

The vertical velocity u₀ is transformed by a gamma factor, u = u^'*gamma
So it slows down slightly as expected in the moving frame

It seems to me that the ball moving sidelong, say B in the first image, will have it's velocity altered in two parts. The vertical component will be multiplied by a gamma factor, and the horizontal component will transform as a lorentz:
u^'= (u -v)/(1-uv/c²)

It seems clear to me that the vertical velocities of A and B are identical, and that they transform in an identical manner.

From class I know this isn't true, apparently they are identical velocities but they transform in a different manner. But I don't see how B could be transformed via the Lorentz formula if only a compnent of it's velocity is along the line parallel to the motion of the moving frame. Help would be much appreciated![/B]

 

[Ray Vickson]

Hi guys, thanks for helping with this! I'm a little stuck with this question about the derivation for relativistic mass.

1. Homework Statement
By considering the inelastic collision of two balls as perceived in different reference frames show that the relativistic mass is equal to the rest mass multiplied by the gamma factor (sqrt(1-u^2/c^2).

Homework Equations

So I know the factor is 1/(1-u²/c²) but proving it is tough.
I've considered a reference frame moving at u in a direction perpendicular to the collision, so basically this:

View attachment 221980

The Attempt at a Solution

The vertical velocity u₀ is transformed by a gamma factor, u = u^'*gamma
So it slows down slightly as expected in the moving frame

It seems to me that the ball moving sidelong, say B in the first image, will have it's velocity altered in two parts. The vertical component will be multiplied by a gamma factor, and the horizontal component will transform as a lorentz:
u^'= (u -v)/(1-uv/c²)

It seems clear to me that the vertical velocities of A and B are identical, and that they transform in an identical manner.

From class I know this isn't true, apparently they are identical velocities but they transform in a different manner. But I don't see how B could be transformed via the Lorentz formula if only a compnent of it's velocity is along the line parallel to the motion of the moving frame. Help would be much appreciated![/B]

This is a problem that was treated by Planck in a 1906 paper, but the arguments there are not particularly enlightening. A much nicer and more convincing argument was advanced in a 1909 paper by Lewis and Tolman. You can find this argument nicely laid out more-or-less completely on pages 48--50 of "A History of the Theories of Aether and Electricity, Volume II", by Sir Edmund Whittaker, Harper (1953).

 

[tms]

According tot he diagram, frame $S$ is moving to the right at speed $V$, but according to what you wrote, you have it moving vertically at speed $u$. You also use $u$ for the speed of one of the particles in the collision.