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For the electromagnetic field, we can define the EM energy:

$$\mathcal{E}_{EM}= \frac{1}{2} \int_V \left(\varepsilon_0\boldsymbol{E}^2+\frac{\boldsymbol{B}^2}{\mu_0}\right)dV$$

And because charged particles experience a change in kinetic energy given by:

$$\frac{\text{d}\mathcal{E}_K}{\text{d}t} = \int_V \boldsymbol{j}_q\cdot \boldsymbol{E}\ dV$$

where $\boldsymbol{j}_q$ is the density current, then Maxwell's equations imply that:

$$\frac{\text{d}\mathcal{E}_K}{\text{d}t} + \frac{\text{d}\mathcal{E}_{EM}}{\text{d}t} = 0$$

In the case of the gravitational field it is still true that:

$$\frac{\text{d}\mathcal{E}_K}{\text{d}t} = \int_V \boldsymbol{j}_m\cdot \boldsymbol{g}\ dV$$

where $\boldsymbol{j}_m$ is the mass flux. Thus, one would expect in the gravitational case to have a conservation law of the type:

$$\frac{\text{d}\mathcal{E}_K}{\text{d}t} + \frac{\text{d}\mathcal{E}_{G}}{\text{d}t} = 0$$

However, it is not clear that even in the Newtonian context this is true (of course, in general relativity in the case of non-static space-time it is not possible to define global energy that is conserved).

My attempt was to define Newtonian gravitational energy as:

$$\mathcal{E}_{G}= \frac{1}{2} \int_V \frac{\boldsymbol{g}^2}{4\pi G}dV$$

but this does not seem to work unless:

$$ {4\pi G}\boldsymbol{j}_M + \frac{\text{d}\boldsymbol{g}}{\text{d}t} = 0$$

But it is not clear that this can be true in general.

Davius
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1 Answers1

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It seems that the specific condition $\boldsymbol{j} + 4\pi G \dot{\boldsymbol{g}} =0$ is satisfied in a general way. To see this we use Gauss's theorem of divergence:

$$ \nabla\cdot \mathbf{g} = 4\pi G \rho_m$$

deriving with respect to time (assuming sufficient continuity in $\mathbf{g}$) one obtains:

$$ \frac{\partial(\nabla\cdot \mathbf{g})}{\partial t} = \nabla\cdot \frac{\partial\mathbf{g}}{\partial t} = 4\pi G \frac{\partial\rho_m}{\partial t}$$

Where we can transform the right-hand side of the equation using the continuity equation we have:

$$ \frac{\partial\rho_m}{\partial t} = -\nabla\cdot\boldsymbol{j}, \qquad \Rightarrow \nabla\cdot \frac{\partial\mathbf{g}}{\partial t} = 4\pi G \frac{\partial\rho_m}{\partial t} = -4\pi G\nabla\cdot \boldsymbol{j}$$

Thus we have in general:

$$\nabla\cdot\left(\frac{\partial\mathbf{g}}{\partial t} + 4\pi G\boldsymbol{j} \right) = 0$$

Thus, in general, we have that:

$$\frac{\partial\mathbf{g}}{\partial t} + 4\pi G\boldsymbol{j} = c_1 \nabla\times \boldsymbol{b}$$

for some vector field $\boldsymbol{b}$ and since nothing prevents choosing $c_1 = 0$ we have that the condition would be fulfilled.

Davius
  • 1,700