Is the variation of a metric with respect to a metric with a different signature, zero?

Question

I have a problem that involves calculating the variation of a metric $ \bar{g}_{\alpha\beta} $ with +3 signature with respect to a metric $ g_{\alpha\beta} $ with a signature of +1. Both metrics have the same spatial dimension of 3. The metrics are related by $ g_{\alpha\beta}=\bar{g}_{\alpha\beta}-2u_{\alpha}u_{\beta} $ where u is a unit vector in $ \bar{g}$, i.e. ($ u_{\alpha}u^{\alpha}=1 $), and $ u_{\alpha}u^{\alpha}=-1 $ in g. Since the signature of a metric is constant everywhere, it seems trivial that $\dfrac{\delta \bar{g}_{\alpha\beta}}{\delta g_{\alpha\beta}} =0$. Am I missing something?

Answer 1

If you treat $\bar g = g + 2u\otimes u$ as a function of $g$ and $u$ separately, the "partial variational derivatives" are $$\frac{\delta \bar g_{\alpha\beta} }{ \delta g }[h]= h_{\alpha\beta}$$ and $$\frac{\delta \bar g_{\alpha \beta} }{ \delta u }[v] = 2 u_{(\alpha} v_{\beta)},$$ so once you have chosen a functional relationship $g \mapsto u(g)$ the total derivative is $$\frac{\delta}{\delta g} \left(\bar g (g,u(g))\right)[h] = h + 2 u \odot \frac{\delta u}{\delta g}[h]$$ where $\odot$ denotes the symmetric tensor product.

Thus we have reduced your question to asking how $u$ varies when you vary $g$. The constraint $g(u,u)=-1$ is not enough to determine this, since this defines a hyperboloid on which we can freely slide $u$ around. It does somewhat constrain the variation, however: differentiating the constraint with respect to $g$ we find $$h(u,u) + 2g \left(u,\frac{\delta u}{\delta g}[h]\right) = 0.$$

As an example, we could choose $\delta u/\delta g[h]=0$ in directions $h$ satisfying $h(u,u) = 0$, in which case the total derivative in these directions would be $\delta \bar g/\delta g[h] = h \ne 0.$ If you have a more explicit description of $u(g)$ then perhaps I can say more, but I don't think there's any choice that makes the derivative vanish.

Edit: Here's an explicit example in which the derivative is nonzero. Suppose we are working on $\mathbb R^3$, and the point at which we are differentiating is where $g_0$ is the Minkowski metric, $\bar g_0$ is the Euclidean metric and $u_0 = \partial_t$ is the standard time translation field. (This is really just for convenience - all of this could be adapted locally to any example.) For metrics $g$ sufficiently close to $g_0$ in the sup norm, the field $$ u(g) =\frac{\partial_t}{\sqrt{-g(\partial_t,\partial_t)}}$$ is smooth (in both spacetime and with respect to $g$) and satisfies your requirements. Now, if we choose the variation $h = dx \otimes dx$, note that $u(g_0 + \epsilon h) = u(g_0)$ since $u$ depends only on the component $g_{tt}$. Thus we have $$\frac{\delta \bar g_{xx}}{\delta g}\Big|_{g_0} = \frac{\delta \bar g}{\delta g}\Big|_{g_0}[h] = \frac{d}{d \epsilon}\Big|_{\epsilon = 0}(g_0+\epsilon h) = h \ne 0.$$

Answer 2

The question is more complex than I thought. However, it seems that $\dfrac{\delta \bar{g}_{\alpha\beta}}{\delta g_{\mu\nu}} =0$ is the only condition that satisfies the two metrics defined by $g_{\alpha\beta}=\bar{g}_{\alpha\beta}-2u_{\alpha}u_{\beta} $ where u is a unit vector in $ \bar{g} $, i.e. $u_{\alpha}u^{\alpha}=1 $, and $ u_{\alpha}u^{\alpha}=-1 $ in g. We can check that $ g^{\alpha\beta}=\bar{g}^{\alpha\beta}-2u^{\alpha}u^{\beta} $ is the inverse of $ g_{\alpha\beta} $ by calculating $ g_{\alpha\beta}g^{\alpha\mu}=\delta_{\beta}^{\mu} $. Let's assume $\dfrac{\delta \bar{g}_{\alpha\beta}}{\delta g_{\mu\nu}} \neq 0$. With $ \delta $ standing for $\dfrac{\delta} {\delta g_{\mu\nu}} $, we have $ g^{\alpha\beta}\delta(u_{\alpha}u_{\beta})=-u_{\alpha}u_{\beta} \delta g^{\alpha\beta}$ from $ u_{\alpha}u^{\alpha}=-1 $ and $ \delta g_{\alpha\beta}=\delta \bar{g}_{\alpha\beta}-2\delta (u_{\alpha}u_{\beta}) $. Calculating $ g^{\alpha\beta}\delta g_{\alpha\beta} $ gives $\bar{g}^{\alpha\beta}\delta g_{\alpha\beta}=g^{\alpha\beta}\delta \bar{g}_{\alpha\beta} $. Setting $\alpha=\mu$ and $\beta=\nu$ gives $ \bar{g}^{\alpha\beta}{1}_{\alpha\beta}=(\bar{g}^{\alpha\beta}-2u^{\alpha}u^{\beta})(\dfrac{\delta\bar{g}}{\delta g}) _{\alpha\beta}$. This requires $ (\dfrac{\delta\bar{g}}{\delta g}) _{\alpha\beta}={1}_{\alpha\beta} $ and $u^{\alpha}u^{\beta}{1}_{\alpha\beta}=0 $ which is impossible. We assumed and used $\dfrac{\delta \bar{g}_{\alpha\beta}}{\delta g_{\mu\nu}} \neq 0$ so the assumption must be wrong. The only solution is then $\dfrac{\delta \bar{g}_{\alpha\beta}}{\delta g_{\mu\nu}} =0$ and $\dfrac{\delta (u_{\alpha}u_{\beta})}{\delta g_{\alpha\beta}} =-1/2{1}_{\alpha\beta}$.