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}\\<br /> 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} =<br /> <br /> \begin{pmatrix}<br /> <br /> \gamma &amp; \gamma \beta &amp; 0 &amp; 0\\<br /> <br /> \gamma \beta &amp; \gamma &amp; 0 &amp; 0\\<br /> <br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> <br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> <br /> \end{pmatrix},

so that the transformation is

<br /> <br /> x&#039;_0 = \gamma(x_0 + \beta x_1),\\<br /> <br /> x&#039;_1 = \gamma(x_1 + \beta x_0),\\<br /> <br /> x&#039;_2 = x_2,\\<br /> <br /> x&#039;_3 = x_3.<br /> <br />

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}<br /> <br /> \frac{\partial x&#039;_1}{\partial x_1} &amp; \frac{\partial x&#039;_1}{\partial x_2} &amp; \frac{\partial x&#039;_1}{\partial x_3} \\<br /> <br /> \frac{\partial x&#039;_2}{\partial x_1} &amp; \frac{\partial x&#039;_2}{\partial x_2} &amp; \frac{\partial x&#039;_2}{\partial x_3} \\<br /> <br /> \frac{\partial x&#039;_3}{\partial x_1} &amp; \frac{\partial x&#039;_3}{\partial x_2} &amp; \frac{\partial x&#039;_3}{\partial x_3}<br /> <br /> \end{vmatrix} = \begin{vmatrix} \gamma (1 + \beta \frac{\partial x_0}{\partial x_1}) &amp; \gamma \beta \frac{\partial x_0}{\partial x_2} &amp;\gamma \beta \frac{\partial x_0}{\partial x_3}\\<br /> 0&amp;1&amp;0\\<br /> 0 &amp; &amp; 1<br /> \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&#039;_1}{p_1},
which is clearly not what I want. Should be \frac{p_0}{p&#039;_0}. Where am I mistaken?