6
$\begingroup$

What exactly is the relationship between the continuity equation and the wave equation?

Suppose $J^\mu$ is a contravariant vector that satisfies the continuity equation $\partial_\mu J^\mu=0$. Let $J^\mu$ be defined by $J^\mu=\partial^\mu\varphi$, where $\varphi$ is some Lorentz scalar. Performing a simple substitution yields $\partial_\mu\partial^\mu\varphi=0$, or $\square\varphi=0$, where $\square$ is the D'Alembert operator. This seems to be a manifestation of the wave equation. Is this a correct derivation? If so, what is the physical interpretation behind it?

$\endgroup$
3
  • 1
    $\begingroup$ Well, have you studied that free field theory on its own? Hint: If you define a trajectory by $x''=0$ in Newtons theory, then the kinetic energy is $\propto x'^2$ which you can also express via momenta $p\propto x'$ and by the way $p'=0$. If you're interested in $\partial^2\varphi=0$, the Lagrangian is $\propto (\partial \varphi)^2$ and now what you do is introducing the letter $J\propto\partial\varphi$. $\endgroup$
    – Nikolaj-K
    Jul 2, 2014 at 22:43
  • 4
    $\begingroup$ I'm confused by your question. If for a given $\varphi$ you define $J^\mu = \partial^\mu\varphi$, then it won't always be the case that $J^\mu$ is conserved. Could you clarify what you're trying to prove from what? $\endgroup$ Jul 2, 2014 at 22:46
  • $\begingroup$ I should have written my question more clearly. My question is: Under what conditions does the wave equation emerge from the continuity equation? As an example, this page (en.wikipedia.org/wiki/Incompressible_flow#Derivation) derives the divergence-free velocity field condition from the incompressibility condition and the continuity condition. In a similar way, under what conditions can one derive the wave equation from the continuity equation (plus other assumptions)? I suspect this involves the energy-momentum tensor of the fluid. $\endgroup$
    – user76284
    Jul 5, 2014 at 19:23

1 Answer 1

1
$\begingroup$

The wave equation can be written as $\nabla_i\nabla^i\varphi = 0$ where $\nabla$ is the Levi-Civita connection on Minkowski space, $\varphi$ does not need to be Lorentz invariant. $\partial_i\partial^i\varphi = 0$ is the Laplace equation on both Minkowski and Euclidean space since ordinary partial derivatives do not respect the metric.

In (pseudo)Riemannian geometry the covariant derivative $\nabla_i$ replaces partials $\partial_i$. The Laplace-Beltrami operator $$\Delta \triangleq \nabla^i\nabla_i$$ is a common generalization of both the D'Alembertian and the Laplacian, $$\Delta \varphi = 0$$ is the Laplace-Beltrami equation.

Your observation then generalizes to say that, for scalar field, a vanishing Laplace-Beltrami operator is the same as a divergence free contravariant gradient. This is universally valid since $$\Delta \varphi = \nabla^i\nabla_i\varphi = g^{ij}\nabla_i\nabla_j\varphi= \nabla_i\nabla^i\varphi$$ It does not depend on the metric or coordinate system. However, the precise geometric meaning of this depends on the metric.

In Euclidean spaces it means a function is harmonic iff it has a divergence free gradient since the Laplace-Beltrami operator is just the Laplacian.

In Minkowski space the Laplace-Beltrami operator is the D'Alembertian, and the Laplace-Beltrami equation becomes the wave equation. In Minkowski space covariant derivatives are just ordinary partial derivatives, as in Euclidean space, since there is no curvature, making Christoffel symbols vanish. Furthermore for scalars $$\nabla_i \varphi = (d\varphi)_i = \partial_i\varphi$$ is true in all metrics. However the contravariant derivative $\nabla^i\varphi$ is not the usual gradient vector since index raising depends on the the metric. The gradient is therefore more naturally thought of as a covector or a 1-form. Geometrically the Minkowski gradient vectors are the time reflections of what the Euclidean gradient vectors would be. I do not have a good intuitive picture of contravariant derivatives and their divergences in general non Euclidean spaces though.

Note that for $$\varphi \triangleq t^2 + x^2$$ in Euclidean space $$\nabla_i\nabla^i\varphi = 4$$ while in Minkowski space $$\nabla_i\nabla^i\varphi = 0$$ making it a solution to the equation even though it is not Lorentz invariant.


A general vector field $J^i$ may satisfy $\nabla_i J^i = 0$, but still have non zero curl or in more generally $\nabla_{[i} J_{j]} \neq 0$. Such a field is not a gradient of a scalar field. So at least $\nabla_{[i} J_{j]} = 0$ is required. The Poincare lemma says this is sufficient for fields on contractible subsets of Euclidean space.

$\endgroup$
2

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.