3
$\begingroup$

What does it mean, from the geometrical point of view, use (in General Relativity) of the constraints on the metric tensor's coefficients such that $\Delta g_{ij}=0$? (where $\Delta$ is the Beltrami-Laplace Operator, $g_{ij}$ the metric tensor).
With $\Delta g_{ij}=0$, I mean the Laplace-Beltrami operator, applied componentwise to the components of the metric tensor.

$\endgroup$
  • 1
    $\begingroup$ Which book are you getting this from? As far as I remember, $\Delta g_{ij} \neq 0$. So this gets me confused too. $\endgroup$ – Zhengyan Shi Jul 4 '16 at 12:41
  • $\begingroup$ Hi Zhengyan, it's just my hypothesis...I have not found in any book. $\endgroup$ – Alexander Pigazzini Jul 4 '16 at 13:12
  • $\begingroup$ I think that $\Delta g_{ij} \neq 0$ in general curved spacetime, because you don't have enough freedom to use coordinate transformations and make all the second derivatives vanish. However, you can do it for first derivatives. In fact, physically speaking, that means the spacetime is locally flat and Minkowski-like, which is basically the strong equivalence principle. $\endgroup$ – Zhengyan Shi Jul 4 '16 at 13:23
  • $\begingroup$ So for example, if in (1+1) dimension General Relativity, I have this metric: $g=h(t,x)(dt^2+dx^2)$ where $h>0$ satisfies $h_{xx}+h_{yy}=0$ would it serve not purpose? $\endgroup$ – Alexander Pigazzini Jul 4 '16 at 14:04
  • $\begingroup$ That metric is conformally flat and therefore a very special metric. But in general it doesn't work. Also it should be $g = h(t,x) (-dt^2+dx^2)$ to preserve the Minkowski signature. $\endgroup$ – Zhengyan Shi Jul 4 '16 at 14:10
2
$\begingroup$

That $\Delta g_{ij} = 0$ as you define it is equivalent to saying that the gradient of all metric components have vanishing divergence $$ g_{ij;k}{}^k \equiv g^{k\ell}g_{ij;k\ell} = 0. $$ Here it is important to remember that the indices $i,j$ denote functions. To clarify this we will let $g_k$ represent the gradient of an arbitrary component function $g_{ij}$. Thus $$ g_k := g_{ij;k}. $$

Then consider the Hodge star operator $*: \Omega^k(M) \to \Omega^{n-k}(M)$ which we can take to be defined by $$ *\alpha = \sqrt{|\det[g_{ij}]|}\alpha^{\vec{J}}\epsilon_{\vec{J}\vec{I}}\omega^{\vec{I}}, $$ where $\omega^i$ is the local frame field on $T^*M$, $\epsilon_{i_1\ldots i_n}$ is the $n$-dimensional Levi-Civita symbol, and $\vec{I},\vec{J},\ldots$ denote strictly increasing multi indices of appropriate lengths (above $\vec{J}$ is of length $k$ and $\vec{I}$ is of length $(n-k)$). To simplify notation we take $\varepsilon_{I} := \sqrt{|\det[g_{ij}]|}\epsilon_{I}$ to be the Levi-Civita tensor.

In the case of a 1-form the Hodge dual is essentially a $(n-1)$-form completely orthogonal to the original 1-form. For let $\alpha_i$ be the 1-form, then contraction on some index with $*\alpha$ yields $$ \alpha^i\alpha^j\varepsilon_{ik_1\ldots k_{r-1} j k_{r+1} \ldots k_n} = 0. $$

It therefore should come as no surprise that the Hodge star operator can be used when considering flows across a surface. To make matters more precise consider a hypersurface, $\Sigma$, defined by the normal $\eta_i$, which we take at first to be non-null and normalized. Then the invariant surface element of $\Sigma$ is given by $*\eta$, and by abuse of notation we may designate the directed surface element by $d\Sigma_i := \eta_i{*\eta}$. However, we can also write $$ d\Sigma_i = \varepsilon_{i\vec{J}}\omega^{\vec{J}}|_\Sigma, $$ where $|_\Sigma$ denotes projection onto the surface $\Sigma$. To see this note that contraction with $\eta_i$ produces the same result for both expressions of $d\Sigma_i$, as does contraction with any 1-form or vector orthogonal to $\eta_i$. The latter expression is however also well defined for null hypersurfaces, and by continuity also gives the directed surface element in these cases.

This can easily be seen to imply that the component functions have no local maxima or minima, by Stoke's theorem on smooth manifolds. For suppose there is a local minima at $p$, then there is a region with boundary, $V$, containing $p$ such that $$ \int_{\partial V} g^{k}d\Sigma_k \neq 0, $$ But by Stoke's theorem on smooth manifolds $$ \int_{\partial V}g^{k}d\Sigma_k = \int_Vd\left(g^{k}d\Sigma_k\right) = \int_V g^k{}_{;k}dV. $$

Depending on your preferences, the final expression can be confirmed in a coordinate frame from the formula $$ X^j{}_{;j} = \frac{1}{\sqrt{|\det g|}}\left(\sqrt{|\det g|}X^j\right)_{,j}, $$ which is valid in coordinate frames and follows from the Jacobi formula for the derivative of a determinant; in a rigid frame from the first Cartan equation $$ d\omega^i = \omega^j\wedge\gamma^i{}_j, $$ where $\gamma^i{}_j$ are the connection forms; and in a mixed frame from both.

EDIT: I had previously concluded in error that the component functions must be constants, using local compactness. However, the above argument only works in the interior of the region. We can therefore only say that the function is such that any local minimum or maximum must be at the boundary, and in particular there must be no local minima nor maxima on open regions. Of course constant components functions still satisfy the condition, but there is a wider class of functions that do.

$\endgroup$
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Jul 7 '16 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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