To prove: the integration measure is Lorentz invariant (Schwartz's problem 2.6b)

Question

I am stuck on Schwartz's ("Quantum Field Theory and the Standard Model") problem 2.6b, and would be grateful for clarification. (I'm aware that this question has been asked and answered elsewhere (Stack Exchange), but I am seeking a less formal proof.)

Let $k'$ and $k$ be four-vectors in two inertial frames, implying that ${k'}^{\mu} = \Lambda_{\nu}^{\mu} \, k^{\nu}$ where $\Lambda_{\nu}^{\mu}$ is the Lorentz tensor $\Lambda_{\nu}^{\mu} = \partial{k'}^{\mu}/\partial k^{\nu}$. One is asked to prove that the "integration measure" is a Lorentz invariant, i.e. that $d^4k'$ equates to $d^4 k$.

We have ${dk'}^{\mu} = \Lambda_{\nu}^{\mu} \, dk^{\nu}$ and so \begin{equation}\label{eq:one} d^4 k' = \Lambda^0_{\nu} \Lambda^1_{\mu} \Lambda^2_{\rho} \Lambda^3_{\lambda} \; dk^{\nu} dk^{\mu} dk^{\rho} dk^{\lambda} \;\; \; \; \; (1) . \end{equation}

Now testing this for a special case, that of a constant relative motion (with speed $v$) of the two frames along their co-oriented $(x,x')$ axes, we have \begin{equation} \Lambda_{\nu}^{\mu} = \; \left[\begin{array}{rrrr} \gamma & - \gamma \beta & 0 & 0\\ -\gamma \beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right] \end{equation} with $\mu$ being the row index and $\nu$ the column index (as usual, $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$). It easily follows that \begin{eqnarray*} \Lambda^0_{\nu} dk^{\nu} &=& \gamma \; dk^0 - \gamma \beta \; dk^1 \, , \\ \Lambda^1_{\mu} dk^{\mu} &=& -\gamma \beta \; dk^0 + \gamma \; dk^1 \, , \\ \Lambda^2_{\rho} dk^{\rho} &=& dk^2 \, , \\ \Lambda^3_{\lambda} dk^{\lambda} &=& dk^3 \, . \end{eqnarray*} My difficulty is that I don't think the product \begin{equation*} (\gamma dk^0 - \gamma \beta dk^1) \, (-\gamma \beta dk^0 + \gamma dk^1 ) \end{equation*} evaluates to unity, as would be required to confirm that $d^4k'= d^4 k$. (Response to comment: I'm aware that the determinant of $\boldsymbol{\Lambda}$ has unit magnitude, but surely in any case I ought to be able to prove the assertion - or at least, prove it true in this special case - by an elementary argument along the lines shown?)

Answer 1

Let us consider $\mathbb{R}^n$ equipped with the standard metric. Now consider $n$ vectors $d_{(1)},\ldots, d_{(n)}$ viewed as the edges of a parallelepiped $Q(d_{(1)},\ldots, d_{(n)})$. As is known $$vol(Q(d_{(1)},\ldots, d_{(1)})) = \det(d_{(1)},\ldots, d_{(n)})\:.\tag{1}$$ Let us pass to describe these facts in Cartesian (orthonormal) coordinates. First of all, observe that $$d_{(k)} = \delta x^j_{(k)} {\bf e}_j$$ where I adopted Einstein's convention rule on repeated indices and I denoted by ${\bf e}_k = \frac{\partial}{\partial x^k}$ the unit vector associated to the $k$-th (orthonormal) Cartesian coordinate $x^k$. In particular $$vol(Q(d_{(1)},\ldots, d_{(1)})) = \det \left( [\delta x^{j}_{(k)}]_{j,k=1,\ldots, n}\right)\tag{2}$$ It is important to stress that the volume of $Q(d_{(1)},\ldots, d_{(n)})$ has therefore a complicated expression in function of the components of the $n$ vectors defining it.

In the special case where the vectors $d_k$ respectively parallel to the coordinates curves $x^k$, i.e., $$\delta x^j_{(k)} = \delta^j_k \delta x_k \quad \mbox{(no sum over repeated indices)}$$ we find $$vol(Q(d_{(1)},\ldots, d_{(1)})) = \delta x_1 \cdots \delta x_k\tag{3}\:.$$

Changing coordinates, passing to a coordinate patch $y^1,\ldots, y^n$ we have that $$d_k = \delta y^j_{(k)} \frac{\partial}{\partial y^k}$$ where $$\delta x^j_{(k)} = \frac{\partial x^j}{\partial y^i} \delta y^i_{(k)} \tag{3}$$ Here the right-hand side of (2) becomes $$vol(Q(d_{(1)},\ldots, d_{(1)})) = \det\left(\left[\frac{\partial x^{i}}{\partial y^l}\right]_{i,l=1,\ldots, n} \right)\det \left( [\delta y^{j}_{(k)}]_{j,k=1,\ldots, n}\right)\tag{5}$$ You see that the formula is exactly (2), but it is corrected with the Jacobian factor. It is clear that, even if $d_1, \ldots, d_n$ are parallel to the initial coordinate curves we have $$vol(Q(d_{(1)},\ldots, d_{(1)})) = \delta x_1 \cdots \delta x_n = \frac{\partial x^1}{\partial y^{i_1}} \delta y^{i_1}_{(1)}\cdots \frac{\partial x^n}{\partial y^{i_n}} \delta y^{i_n}_{(n)}\:.$$ You see that, in principle there could be many different factors, say, $\delta y^1_{j}$ with different $j$. This is because the vectors $d_j$ are not parallel to the corresponding coordinate line $y^j$ in general.

In your approach you instead considered all those factors as identical. When you wrote $$ \Lambda^0_{\nu} dk^{\nu} = \gamma \; dk^0 - \gamma \beta \; dk^1 \, , \\ \Lambda^1_{\mu} dk^{\mu} = -\gamma \beta \; dk^0 + \gamma \; dk^1 \, , $$ $dk^0$ on the first and that on the second line are actually two different numbers in general. The same holds for $dk^1$.

The correct approach is using (5) directly, computing the Jacobian determinant and proving that it is $1$.