In general relativity 4-volume element $\mathrm{d}^4 x = \mathrm{d} x^0\mathrm{d} x^1 \mathrm{d} x^2\mathrm{d} x^3$ is clearly a pseudoscalar (or scalar density) of weight 1 since it transforms as $\mathrm{d}^4 x \to \mathrm{d}^4 x' = J \,\mathrm{d}^4 x$, with $J$ Jacobian determinant of the coordinates transformation.

In special relativity we consider only Lorentz transformations, which have $J = \pm 1$, so shouldn't $\mathrm{d}^4 x$ be a pseudoscalar also in special relativity? Landau states ("The Classical Theory of Fields", § 6) that «the element is a scalar: it is obvious that the volume of a portion of four-space is unchanged by a rotation of the coordinate system», but this doesn't proof that $\mathrm{d}^4 x$ is a scalar rather than a pseudoscalar since a pseudoscalar looks like a scalar under a proper rotation of the coordinate system ($J = 1$), differences arise only for an axis inversion ($J = -1$).


2 Answers 2


I) Consider an arbitrary coordinate transformation

$$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$


$$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$

denote the corresponding Jacobian.

Traditionally in physics,

  1. a scalar $\sigma$ transforms as $$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma, $$

  2. a pseudo-scalar $\sigma$ transforms as $$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~{\rm sgn}(J) \sigma,$$

  3. a density $\rho$ transforms as $$ \rho ~\longrightarrow~ \rho^{\prime}~=~\frac{\rho}{J}, $$

  4. a pseudo-density $\psi$ transforms as $$ \psi ~\longrightarrow~ \psi^{\prime}~=~{\rm sgn}(J)\frac{\psi}{J}, $$

  5. a density $\rho$ of (integer) weight $w$ transforms as $$ \rho ~\longrightarrow~ \rho^{\prime}~=~\frac{\rho}{J^w}, $$

  6. a pseudo-density $\psi$ of (integer) weight $w$ transforms as $$ \psi ~\longrightarrow~ \psi^{\prime}~=~{\rm sgn}(J)\frac{\psi}{J^w}. $$

For tensor, pseudo-tensor, tensor-density, pseudo-tensor-density, etc, see the linked Wikipedia page.


  1. On a Lorentzian manifold $(M,g)$ of signature $(-,+,\ldots,+)$, the square root $\sqrt{-\det(g_{\mu\nu})}$ is a density.

  2. In General Relativity (GR), the four-form $\mathrm{d} x^0\wedge \mathrm{d} x^1\wedge \mathrm{d} x^2\wedge\mathrm{d} x^3$ transforms as an inverse density, i.e. a density of weight $w=-1$.

II) Within Special Relativity (SR), the Jacobian $J=\pm1 $ is plus/minus one, as OP correctly notes, so that $\mathrm{d} x^0\wedge \mathrm{d} x^1\wedge \mathrm{d} x^2\wedge\mathrm{d} x^3$ transforms as a pseudo-scalar.

III) Landau and Lifshitz (L&L), The Classical Theory of Fields, $\S 6$ p. 21 around eq. (6.13), indeed states that the element of integration

$$dx^0 dx^1 dx^2 dx^3 $$

is a scalar. Here are some suggestions:

  1. Perhaps L&L are only considering proper Lorentz transformations $\Lambda \in SO(3,1)$ where $J=1$ by definition?

  2. Perhaps L&L are viewing $dx^0 dx^1 dx^2 dx^3$ not as a four-form but as a manifestly positive infinitesimal volume element, which by definition transforms with the absolute value $|J|$ of the Jacobian $J$?

However, neither of the two above interpretations (1 and 2) seem to fit particularly well with what is said in the rest of $\S 6$, in particular the footnote on p.21.


Regarding Landau's book, I am reading it in the Russian edition, and in that paragraph when he talks about elements of length/area/volume he neglects the pseudo- prefix. The explanation that the 4-volume element is a scalar he explicitly only considers transformations with unit determinant, and it can be easily deduced from what is written that in general, it is a pseudoscalar.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.