Canonical invariance vs. Lorentz invariance

gasar8

1. 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.

2. 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 $n=3$.

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?

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jambaugh

Gold Member
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?