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Homework Statement
Prove that the element [tex]dt\ dx\ dy\ dz[/tex] is invariant under Lorentz boost with velocity [tex]\beta[/tex] along [tex]z[/tex] axis.
Homework Equations
Convention [tex]c=1[/tex]
Lorentz boost in z direction:
[tex]L(z)=\left[ \begin{array}{cccc} \gamma & 0 & 0 & -\gamma\beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma\beta & 0 & 0 & \gamma \end{array} \right],\ \gamma=\frac{1}{\sqrt{1-\beta^2}}[/tex]
Definition of invariance:
[tex]dt'dx'dy'dz'=dt\ dx\ dy\ dz[/tex]
The Attempt at a Solution
Looks simple.
[tex]\left[ \begin{array}{c} dt' \\ dx' \\ dy' \\ dz' \end{array} \right]=\left[ \begin{array}{cccc} \gamma & 0 & 0 & -\gamma\beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma\beta & 0 & 0 & \gamma \end{array} \right]\left[ \begin{array}{c} dt \\ dx \\ dy \\ dz \end{array} \right][/tex]
[tex]dt'=\gamma dt-\gamma\beta dz,\ dx'=dx,\ dy'=dy,\ dz'=-\gamma\beta dt+\gamma dz[/tex]
[tex]dt'dx'dy'dz'=\gamma^2(dt-\beta dz)\ dx\ dy\ (dz-\beta dt)[/tex]
I got stuck, but then I noticed that [tex]\beta=dz/dt[/tex], so:
[tex]dt'dx'dy'dz'=\gamma^2 dt\left(1-\beta \frac{dz}{dt}\right)\ dx\ dy\ dz\left(1-\beta \frac{dt}{dz}\right)=\frac{1}{1-\beta^2} dt\left(1-\beta^2\right)dx\ dy\ dz\left(1-\frac{dz}{dt}\frac{dt}{dz}\right)=\left(1-1\right)\ dt\ dz\ dy\ dz=0[/tex]
It shouldn't be zero! Please help!
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