Suppose we have an integral int d^4 k f(k) we want to evaluate and that we're in Minkowski space with some metric (+,−,−,−).

Victor Harris

Victor Harris

Open question

2022-08-20

Suppose we have an integral d 4 k   f ( k )
we want to evaluate and that we're in Minkowski space with some metric ( + , , , ).
Is it true that:
d 4 k = d k 0   d 3 k = d k 0 | k | 2 d | k | d ( cos θ ) d ϕ
just like in ordinary space?

Answer & Explanation

Jaydon Villanueva

Jaydon Villanueva

Beginner2022-08-21Added 7 answers

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 , , x n )
| det ( g μ ν ) | d x 1 d x n
where det ( g μ ν ) 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.
Bellenik3

Bellenik3

Beginner2022-08-22Added 2 answers

Note. The notational convention in physics is d n x = d x 1 d x n

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