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?)

  • 1
    $\begingroup$ Hint: What is the determinant of a Lorentz matrix? $\endgroup$
    – Qmechanic
    Commented Jun 16, 2019 at 14:28
  • $\begingroup$ Could the issue be that Lorentz boosting won't leave rectangles in the $x-t$ plane as rectangles? The intuition is that the volume increases by $\gamma$ due to time dilation and decreases by $\gamma$ due to length contraction, but eg length contraction refers to events that are simultaneous. $\endgroup$
    – jacob1729
    Commented Jun 16, 2019 at 15:01
  • $\begingroup$ The point is that you have an overall wrong idea. It is false that you can interpret the equation you wrote before the Lorentz matrix ($d^4k'=$...) as an identity connecting the measure of a volume in different coordinate frames. $\endgroup$ Commented Jun 16, 2019 at 15:58
  • $\begingroup$ Thanks all. But now @Valter Moretti, why is Eq.~(1) incorrect? Suppose we have an integral of a function f(x,y) expressed in terms of Cartesian coordinates in 2D (x,y) as the independent variables. When we go to polar coordinates (r, $\theta$) we have $dx \, dy \rightarrow dr \, r \, d\theta$. What elementary step am I missing here? Is my Eq.~(1) lacking some scale factors? $\endgroup$
    – JayDee.UU
    Commented Jun 16, 2019 at 16:50
  • $\begingroup$ The point is that the numbers $dx^k$ are not the $n$ lengths of the edges of an elementary $n$-cube of $\mathbb{R}^n$, but they are the components of a given vector represented along the natural basis associated to the used coordinates. Also dealing with polar coordinates you obtain the wrong result if following your idea since terms proportional to $d\theta^2$ and $dr^2$ appear. $\endgroup$ Commented Jun 16, 2019 at 17:35

1 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 $1$.

  • $\begingroup$ Thanks for your detailed post. Much appreciated! I'll need time to absorb this. $\endgroup$
    – JayDee.UU
    Commented Jun 16, 2019 at 18:54
  • $\begingroup$ I'm not sure that it's critical to the argument, but I am uncertain (and not for the first time) about the definition of a unit vector $\mathbf{e}_k$ for the $k$-th coordinate direction as $\partial / \partial x^k$. Defined this way it is in effect an operator, and carries units. $\endgroup$
    – JayDee.UU
    Commented Jun 17, 2019 at 0:47
  • $\begingroup$ No, it has nothing to to with the issue. If you want, simply ignore that notation (which is however the standard notation in differential geometry for the basis in the tangent space associated to a coordinate system). $\endgroup$ Commented Jun 17, 2019 at 3:32
  • $\begingroup$ Am I correct in assuming that in the notation $\delta x^j_{(k)}$ for the components of $d_{(k)}$ the role of the brackets $(k)$ around $k$ is to discourage the confusion of (potentially) thinking $k$ is an index, as opposed to merely a label for one of the "side-vectors" of the parallelpiped "$Q$"? $\endgroup$
    – JayDee.UU
    Commented Jun 17, 2019 at 14:32
  • $\begingroup$ Sure! You are right! $\endgroup$ Commented Jun 17, 2019 at 15:53

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