Stationary solutions to Klein Gordon equation in spherical symmetry

[Clvrhammer]

TL;DR Summary

The question is concerned about the correctness of a proof on the non-existence of nontrivial time-independent solutions to the Klein-Gordon equation with zero mass. Specifically, the author asks whether the application of Gauß' divergence theorem is valid in the context of his proof as well as whether the proof is valid overall.

I am trying to prove that in spherically symmetric spacetimes there are no nontrivial time-independent solutions to the Klein-Gordon equation (with mass $= 0$) (**is this even true?**). My Ansatz is as follows:

A spherically symmetric spacetime has metric

$$g = g_{tt} \, dt^2 + g_{tr} \, dt \, dr + g_{rr} \, dr^2 + r^2 \, d\Omega^2$$

The Klein Gordon equation for a time-independent and spherically symmetric field $\phi \equiv \phi(r)$ is

$$g^{rr} \phi_{,rr} + \frac{1}{\sqrt{|g|}} \partial_r \big( \sqrt{|g|} g^{rr} \big) \phi_{,r} = 0 \quad \Longrightarrow \quad \sqrt{|g|} g^{rr} \phi_{,r} = \text{const}$$

Now, if I integrate over a hypersurface $A$, defined as $A = \{ t = \text{const}, r \leq r_0\}$, then by the divergence theorem I have

$$0 = \int_A dV \, \frac{1}{\sqrt{|g|}} \partial_r ( \sqrt{|g|} g^{rr} \phi_{,r} ) = \int_A dV \, \operatorname{div} (\nabla \phi) = \oint_{\partial A} dS \, n^r \phi_{,r}$$

Since spacetime and field are both spherically symmetric, the integrand of the integral over $\partial A$ is either consistently positive or consistently negative, so that the integral only vanishes if $n^r \phi_r = 0$. But then $\phi_r = 0$ and so $\phi$ is constant along $r$.

---
I am not sure about the mathematics in this, especially if the divergence theorem may be applied on this submanifold the way I did.

 

[renormalize]

TL;DR Summary: The question is concerned about the correctness of a proof on the non-existence of nontrivial time-independent solutions to the Klein-Gordon equation with zero mass.

Aren't you just asking about solutions to Gauss' Law?
A massless Klein-Gordon field $\phi$ in vacuum satisfies the 4D wave equation $0=\square\phi\left(t,x\right)\equiv\frac{\partial^{2}\phi\left(t,x\right)}{\partial t^{2}}-\nabla^{2}\phi\left(t,x\right)\Rightarrow-\nabla^{2}\phi\left(x\right)=0$ in the time-independent case. This is just Gauss' law written in terms of the scalar potential $\phi$ with the well-known spherically-symmetric solution $\phi\left(r\right)=\frac{k}{r}+\phi\left(\infty\right)$; i.e., it arises from a point-charge at the origin. Do you consider this to be a "nontrivial time-independent solution to the Klein-Gordon equation with zero mass"?

 

[Clvrhammer]

Aren't you just asking about solutions to Gauss' Law?
A massless Klein-Gordon field $\phi$ in vacuum satisfies the 4D wave equation $0=\square\phi\left(t,x\right)\equiv\frac{\partial^{2}\phi\left(t,x\right)}{\partial t^{2}}-\nabla^{2}\phi\left(t,x\right)\Rightarrow-\nabla^{2}\phi\left(x\right)=0$ in the time-independent case. This is just Gauss' law written in terms of the scalar potential $\phi$ with the well-known spherically-symmetric solution $\phi\left(r\right)=\frac{k}{r}+\phi\left(\infty\right)$; i.e., it arises from a point-charge at the origin. Do you consider this to be a "nontrivial time-independent solution to the Klein-Gordon equation with zero mass"?

Thank you for your answer. Yes, I am looking for spherically symmetric solutions to the time-independent and massless Klein-Gordon equation on spherically symmetric and static curved spacetimes, i.e. solutions to $\nabla^{\mu} \nabla_{\mu} \phi = 0$. What I failed to mention (sorry for that), however, is that I am interested in regular solutions. Your example is not differentiable at the origin. Are there even such solutions or do only the trivial ones ($\phi = \text{const}$) exist?

 

[Clvrhammer]

This is just Gauss' law written in terms of the scalar potential $\phi$ with the well-known spherically-symmetric solution $\phi\left(r\right)=\frac{k}{r}+\phi\left(\infty\right)$; i.e., it arises from a point-charge at the origin.

Moreover, doesn't
$$\nabla_{\mu} \nabla^{\mu} \Big( \frac{1}{r} \Big) \propto \delta(r)$$
hold instead, and so the solution you provided is in fact mo solution to the equation?

 

[renormalize]

Yes, I am looking for spherically symmetric solutions to the time-independent and massless Klein-Gordon equation on spherically symmetric and static curved spacetimes, i.e. solutions to $\nabla^{\mu} \nabla_{\mu} \phi = 0$. What I failed to mention (sorry for that), however, is that I am interested in regular solutions. Your example is not differentiable at the origin. Are there even such solutions or do only the trivial ones ($\phi = \text{const}$) exist?

OK, the most general static, spherically-symmetric line element can always be brought into the form:$$ds^{2}=A\left(r\right)dt^{2}-B\left(r\right)dr^{2}-r^{2}\left(d\theta^{2}+\sin^{2}\theta\,d\varphi^{2}\right)\tag{1}$$Given this, the massless wave-equation for the static, spherically-symmetric scalar field $\phi\left(r\right)$ is:$$0=\frac{1}{\sqrt{-g}}\partial_{\mu}\left(\sqrt{-g}\,g^{\mu\nu}\partial_{\nu}\phi\right)=\frac{1}{\sqrt{A\left(r\right)B\left(r\right)}\,r^{2}}\left(\sqrt{\frac{A\left(r\right)}{B\left(r\right)}}r^{2}\phi^{\prime}\left(r\right)\right)^{\prime}\tag{2}$$ with the first-integral:$$\phi^{\prime}\left(r\right)=\frac{k}{\sqrt{\frac{A\left(r\right)}{B\left(r\right)}}r^{2}}\tag{3}$$Now everything comes down to the behavior of $\sqrt{A\left(r\right)/B\left(r\right)}$ as $r\rightarrow 0$. If this quantity happens to approach $1/r^2$, then near the origin we have the non-trivial regular solution $\phi\left(r\right)\sim kr+\text{const}$. In contrast, for the Schwarzschild metric $A\left(r\right)=B^{-1}\left(r\right)=1-\frac{2M}{r}$ which gives:$$\phi^{\prime}\left(r\right)=\frac{k}{r^{2}\left(1-\frac{2M}{r}\right)}\Rightarrow\phi\left(r\right)=\frac{k}{2M}\log\left(1-\frac{2M}{r}\right)+\text{const}\tag{4}$$ This solution is singular at the Schwarzschild radius and at the origin unless $k=0$.

 

[Clvrhammer]

Thank you for your time and effort. This already helped me a lot. One last follow-up -- if you don't mind -- about another idea I had on this issue...

Assume as an additional condition the existence of some point $r_0$ whereat the field $\phi(r_0) = 0$ vanishes. The initial differential equation is a linear ODE. Could one therefore not simply apply a uniqueness theorem, such that $\phi(r) \equiv 0$ is a solution and therefore the ONLY solution? In order for this to be possible one must have that

$$\sqrt{-|g|} g^{rr} \phi_{,r} = k \quad \Longleftrightarrow \phi_{,r} = \frac{k}{\sqrt{-|g|} g^{rr}} \tag{#}$$

is differentiable, right? The metric coefficients must be at least twice-differentiable (because they are solutions to Einsteins Field equations, which is second order). Moreover, as $|g|$ cannot be zero due to non-degeneracy of the metric, the composition $\sqrt{-|g|}$ is differentiable again, right? And so as long as $g^{rr} \neq 0$ everywhere, this should prove the initial proposal (no regular and non-constant static solutions). Is there any flaw to this logic?

Note that this is not contradicting your post, as in the Schwarzschild case $g^{rr} = 0$ at some point and so by equation (#) one finds $\phi_{,r} \equiv 0$ anyway.

 

[renormalize]

Moreover, as $|g|$ cannot be zero due to non-degeneracy of the metric...

While true, you have to be careful using this fact to prove something in a given coordinate system. After all, even in flat spacetime, in spherical-coordinates $\sqrt{-g}=r^{2}\sin\theta$ vanishes at $r=0$, $\theta=0$ and $\theta=\pi$. (These are coordinate singularities.)
I think it would be helpful if you would clearly enumerate your assumptions, followed by an explicit statement of what you are trying to prove. Like:

1. Consider a 4D spacetime that is static and spherically-symmetric.
2. In this spacetime there exists a static scalar field $\phi(r)$ which satisfies the massless scalar wave equation.
3. $\phi(r)$ is everywhere regular. (?)
4. Etc.

From these, it follows that: ...

 

[PeterDonis]

in the Schwarzschild case $g^{rr} = 0$ at some point

Only in Schwarzschild coordinates, and that point (the horizon $r = 2M$) is a coordinate singularity, and as @renormalize has pointed out, you can't prove anything in that case.

 

[Clvrhammer]

Only in Schwarzschild coordinates, and that point (the horizon $r = 2M$) is a coordinate singularity, and as @renormalize has pointed out, you can't prove anything in that case.

I must disagree, $g^{rr} = 0$ also holds in e.g. Eddington-Finkelstein coordinates, which are perfectly regular at the event horizon.

 

[PeterDonis]

$g^{rr} = 0$ also holds in e.g. Eddington-Finkelstein coordinates

Yes, that's true.

which are perfectly regular at the event horizon.

At one horizon, yes. But in maximally extended Schwarzschild spacetime, there are two horizons, not one.

The only chart I'm aware of that is regular everywhere in maximally extended Schwarzschild spacetime is the Kruskal-Szekeres chart, for which the $t$ - $r$ portion of the metric is diagonal, non-singular, and non-degenerate everywhere.