2D Lorentz-transform question

[kleinwolf]

If we consider 2 Lorentz transformation along x and y :

 

Lx

 

gammax -betax*gammax 0

 

-betax*gammax gammax 0

0 0 1

 

Ly

 

gammay 0 -betay*gammay

 

0 1 0

 

-betay*gammay 0 gammay

 

 

 

Why is L_x*L_y not equal to L_y*L_x ?

 

---------------------

BTW is it to confond with Lorenz, with his Goose (?)

[ydoaPs]

If we consider 2 Lorentz transformation along x and y :

 

Lx

 

gammax -betax*gammax 0

 

-betax*gammax gammax 0

0 0 1

 

Ly

 

gammay 0 -betay*gammay

 

0 1 0

 

-betay*gammay 0 gammay

 

 

 

Why is L_x*L_y not equal to L_y*L_x ?

 

---------------------

BTW is it to confond with Lorenz, with his Goose (?)

 

Try again with LaTeX. It's incomprehensible as is.

[D H]

The general form for relativistic velocity addition is

 

[math]

\mathbf u + \mathbf v =

\frac

{\mathbf u + \mathbf v_{\parallel} + \sqrt{1-u^2/c^2}\,\mathbf v_{\perp}}

{1+\mathbf u\cdot\mathbf v/c^2}

[/math]

 

where [math]\mathbf v_{\parallel}[/math] and [math]\mathbf v_{\perp}[/math] are the components of [math]\mathbf v[/math] parallel and perpendicular to [math]\mathbf u[/math].

 

This obviously commutes only when the two velocities are parallel. When they aren't parallel you get Thomas precession.

[kleinwolf]

Yes, I'm sorry, I couldn't find out how to make matrices with the tex commands.

 

I had several question from this out :

 

-It does not commute, so what corresponds physically to the order of addition of speed ?

 

-In 2D, the boost makes that Oxy is no more orthogonal (decomposition of L : [math]R^{-1}(\theta)L R(\theta),\theta=atg(v_y/v_x)[/math], where R is a rotation based on x and y component of v, and L a 1D boost with total v speed)

 

-In the Minkowski diagram, the boosts correspond to "lean" the t-axis towards x-axis. In 2D shouldn't then t-axis be leaned over every direction, such that the space x-y should be conical ?

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Merged post follows:

Consecutive posts merged

(Add.: Sorry, not conical, but curved, by considering that the Minkowski-angle theta between the phi-direction in the xy-plane and x'y'-frame:

 

[math] \theta=atan(vx/c*cos(\phi)+vy/c*sin(\phi)) [/math].

 

whereas with linear transformation of boost with rotation (see above) one gets :

 

[math]\theta=atan(vx/c)*cos(\phi)+atan(vy/c)*sin(\phi)[/math]

)