# Volume element $\mathrm{d}^4k =\mathrm{d}k^0 \,|\mathbf{k}|^2\,\mathrm{d}|\mathbf{k}| \,\mathrm{d}(\cos\theta) \,\mathrm{d}\phi$ in Minkowski space?

Suppose we have an integral

$$\int \mathrm{d}^4k \,\ f(k)$$ we want to evaluate and that we're in Minkowski space with some metric $(+,-,-,-)$.

Is it true that: $$\mathrm{d}^4k = \mathrm{d}k^0\ \mathrm{d}^3\mathbf{k} = \mathrm{d}k^0 \,\,\,\,|\mathbf{k}|^2\,\mathrm{d}|\mathbf{k}| \,\,\mathrm{d}(\cos\theta) \,\mathrm{d}\phi$$

just like in ordinary space?

If not, what are the differences between this and an Euclidean integral in (say) 4 dimensions?

## 1 Answer

Yes it is.

The volume form on any (pseudo-)Riemannian manifold $(M,g)$ of dimension $n$, where $g$ is the metric, is given in local coordinates $(x^1, \dots, x^n)$ $$\sqrt{|\det (g_{\mu\nu})|}dx^1\wedge \cdots \wedge dx^n$$ where $\det(g_{\mu\nu})$ is the determinant of the metric in these coordinates. In cartesian coordinates, the determinant of the Euclidean metric is $+1$ why the determinant of the Minkowski metric is $-1$. However, the absolute value in the square root factor of the volume form kills the sign difference, so the volume forms are the same.

NOTE. The notational convention in physics is $$d^n x = dx^1\wedge \cdots \wedge dx^n$$ See, for example, Carroll's Spacetime and Geometry eq. 2.95. So, the answer to your question is really "yes" due to notational convention, but then this begs the question "Why would one use $d^nx$ as the volume form for both Minkowski space and Euclidean space?" whose answer is given abave.

• @PlaneWaves Sorry the answer is kind of abstract. You might find it helpful to read the link I included on the term "volume form" that talks about volume forms on manifolds. Also, I added a note that might clear things up a little bit. May 18, 2013 at 16:52