Canonical invariance vs. Lorentz invariance

[gasar8]

Homework Statement

I have an assignment to prove that specific intensity over frequency cubed is Lorentz invariant. One of the main tasks there is to prove the invariance of phase space $d^3q \ d^3p$ and I am trying to prove it with symplectic geometry. I am following Jorge V. Jose and Eugene J. Salentan: Classical dynamics.

Homework Equations

$$\omega = dq^{\alpha} \wedge dp_{\alpha}\\ v = d^3q \ d^3p = \frac{1}{n!} \omega^{\wedge n} = \frac{1}{n!} \omega \wedge \omega \wedge \cdots \wedge \omega = dq^1 \wedge dp_1 \wedge \cdots \wedge dq^n \wedge dp_n,$$
of course in my case $n=3$.

The Attempt at a Solution

The book says that this $v$ is invariant under canonical transformations, because $\omega$ is. I am now wondering if this is enough also for the Lorentz invariance?

 

[jambaugh]

It would be sufficient if Lorentz transforms manifest as canonical transformations in phase space. Do they? If not then you'll need to show it directly. To answer the question I think you cannot avoid explicitly dipping into the details of how the Lorentz group acts on phase space. So get too it.

 

[gasar8]

I tried now in another way. By the definition of forms and change of coordinates (the Jacobi matrix) which is for $p$ done also in Goodman. I write Lorentz matrix for boost in $x$ direction, which is$$\Lambda_{\mu}^{\nu} = \begin{pmatrix} \gamma & \gamma \beta & 0 & 0\\ \gamma \beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix},$$

so that the transformation is

$$x'_0 = \gamma(x_0 + \beta x_1),\\ x'_1 = \gamma(x_1 + \beta x_0),\\ x'_2 = x_2,\\ x'_3 = x_3.$$

I know that $v = p/E = p^1/p^0$. Following Goodman, I can get the $d^3p/p^0$ invariance, but when I try to do Jacobain determinant for $d^3x$ in the same way, I get

$$\begin{vmatrix} \frac{\partial x'_1}{\partial x_1} & \frac{\partial x'_1}{\partial x_2} & \frac{\partial x'_1}{\partial x_3} \\ \frac{\partial x'_2}{\partial x_1} & \frac{\partial x'_2}{\partial x_2} & \frac{\partial x'_2}{\partial x_3} \\ \frac{\partial x'_3}{\partial x_1} & \frac{\partial x'_3}{\partial x_2} & \frac{\partial x'_3}{\partial x_3} \end{vmatrix} = \begin{vmatrix} \gamma (1 + \beta \frac{\partial x_0}{\partial x_1}) & \gamma \beta \frac{\partial x_0}{\partial x_2} &\gamma \beta \frac{\partial x_0}{\partial x_3}\\ 0&1&0\\ 0 & & 1 \end{vmatrix} = \gamma(1+\beta \frac{\partial x_0}{\partial x_1}).$$

I then assume that $\frac{\partial x_0}{\partial x_1} = 1/v$, since $x_0 := t$ and also $v = \frac{p_1}{p_0}$, so the final determinant is equal to $\gamma(1+\beta \frac{p_0}{p_1})$, which can be rewritten as
$$\frac{p'_1}{p_1},$$
which is clearly not what I want. Should be $\frac{p_0}{p'_0}$. Where am I mistaken?