Laplace 1749-1827 |

\begin{align}

\nabla^2\Phi=4\pi G\rho&\phantom {10000}(1)\nonumber

\end{align}What does ##\nabla^2## mean? A quick look on the internet reveals that it is the Laplace operator or Laplacian sometimes written ##\Delta## (capital delta) or possibly ##∆## (which Microsoft describes as increment). They come out different in latex \Delta and ∆ respectively. I should use the former.

I have found various formulations for the Laplacian and I want to check that they are all really the same. Two are from Wikipedia and the third is from Carroll. They are:

A Wikipedia formula in ##n## dimensions:

\begin{align}

\nabla^2=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^i}\left(\sqrt{\left|g\right|}g^{ij}\frac{\partial}{\partial x^j}\right)&\phantom {10000}(2)\nonumber

\end{align}A Wikipedia formula in "in 3 general curvilinear coordinates ##(x^1,x^2,x^3)##":

\begin{align}

\nabla^2=g^{\mu\nu}\left(\frac{\partial^2}{\partial x^\mu\partial x^\nu}-\Gamma_{\mu\nu}^\lambda\frac{\partial}{\partial x^\lambda}\right)&\phantom {10000}(3)\nonumber

\end{align}And Carroll's formula (from exercise 3.4) which I did not understand at the time because I had not spotted the second step:

\begin{align}

\nabla^2=\nabla_\mu\nabla^\mu=g^{\mu\nu}\nabla_\mu\nabla_\nu&\phantom {10000}(4)\nonumber

\end{align}The Wikipedia also gives a formula for the Laplacian in spherical polar coordinates:

\begin{align}

\nabla^2f&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(5)\nonumber\\

&=\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(rf\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(6)\nonumber

\end{align}where ##\phi## represents the azimuthal angle and ##\theta## the zenith angle or co-latitude. So the metric will be

\begin{align}

g_{\mu\nu}=\left(\begin{matrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2{\theta}\\\end{matrix}\right)&\phantom {10000}(7)\nonumber

\end{align}I assumed that the coordinates are ordered ##r,\theta,\phi## although Wikipedia does not say that.

I want to prove that

A) (2), (3) and (4) both give (5) or (6) the Laplacian in spherical polar coordinates.

B) (4) is equivalent to (3) the general 3-dimensional expression.

C) (4) is equivalent to (2) the general 𝑛-dimensional expression.

A was quite easy. B follows immediately from the formula for the covariant derivative. But in 2020 I could not prove C. In January 2021 I found and proved the Voss-Weyl formula and C becomes easy.

The proofs are in Commentary 4.1 Laplacian.pdf on the first three pages. The other four record my 2020 struggles.

I spent far too long on this but

1) Revisited using the Levi-Civita symbol for expanding determinant (47)

2) Met the log, exponent, trace of a matrix (63) and for diagonal matrices (67)

3) Used diagonalizing matrices (65)

4) Used the product operator ##\prod\ ##for the first time at (27) and again at (66)

5) Proved that ##\ln{\left(\det{\left(A\right)}\right)}=tr{\left(\ln{\left(A\right)}\right)}## (66)

6) Found a formula for the log of a derivative (73)

7) Found a formula for the log of a matrix (76)

Thanks to Physics forums with whose I proved C for a diagonal metric but they did not know about Voss-Weyl.

This comment has been removed by the author.

ReplyDeleteLogs of matrices only work as log of ordinary numbers in certain cases. You have to notice that e^(A + B), in general, is not equal to e^A e^B, unless A and B commute.

ReplyDeleteIn fact, what to need to prove is that the logarithm and the trace operations commute. If you can express the log of A as a power series, then, log(PDP') = P log(D) P' (for instance (PDP')^2 = P D^2 P', and the result follows because the trace of ABC is equal to the trace of all even permutations of ABC. Then Tr (log(PDP')= Tr (P log(D) P') = Tr (log(D)) = log ( det (D)) = log (det(PDP')).

DeleteThanks vvv. Your comment brought me back to this problem and I have the complete solution!

Delete