📂Electrodynamics

Conservation of Momentum in Electrodynamics

Overview¹

In electrodynamics, the law of conservation of momentum is as follows.

$$ \dfrac{d \mathbf{p}}{dt} =-\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau + \oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} $$

Explanation

According to Newton’s second law, the force acting on an object and the change in the object’s momentum are equal.

$$ \mathbf{F} = \dfrac{d \mathbf{p}}{dt} $$

$\mathbf{p}$ is the total mechanical momentum of particles within volume $\mathcal{V}$. To distinguish it from the momentum stored in the electromagnetic field, we’ll refer to $\mathbf{p}$ as ‘mechanical’ momentum. The electromagnetic force acting on the charge within the volume is as follows.

$$ \mathbf{F} =\oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} -\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau $$

Therefore,

$$ \dfrac{d \mathbf{p}}{dt} =-\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau + \oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} $$

This equation is the law of conservation of momentum in electrodynamics. Since its form is similar to the Poynting’s theorem, it can be understood in a similar way.

The first integral on the right-hand side represents the momentum stored in the electromagnetic field within volume $\mathcal{V}$. In other words, $\epsilon_{0} \mu_{0} \mathbf{S}$ represents the momentum density of the electromagnetic field in unit volume space; simply put, it’s the momentum density of the field. This is expressed as follows.

$$ \mathbf{g} =\epsilon_{0} \mu_{0} \mathbf{S} = \epsilon_{0} \mathbf{E}\times\mathbf{B} $$

The second integral on the right-hand side represents the momentum flowing into the surface (boundary) $\mathcal{S}$ enveloping volume $\mathcal{V}$ per unit time. Therefore, if the mechanical momentum $\mathbf{p}$ increases, it means either the momentum stored in the field is decreasing, or momentum is being carried into the field through the boundary surface. When mechanical momentum within volume $\mathcal{V}$ does not change over time, as in a vacuum,

$$ \begin{align*} && 0 &= - \int_{\mathcal{V}} \dfrac{\partial \mathbf{g}}{\partial t} d\tau + \oint_{\mathcal{S}} \mathbf{T}\cdot d\mathbf{a} \\ \implies && \quad \int_{\mathcal{V}} \dfrac{\partial \mathbf{g}}{\partial t} d\tau &= \oint_{\mathcal{S}} \mathbf{T}\cdot d\mathbf{a}=\int_{\mathcal{V}} \nabla \cdot \mathbf{T} d\tau \end{align*} $$

The second bracket in the second line is valid due to the divergence theorem. From the above result, the following equation is valid.

$$ \dfrac{\partial \mathbf{g}}{\partial t} =\nabla \cdot \mathbf{T} $$

The above equation is the continuity equation for electromagnetic momentum. $\mathbf{g}$ and $\rho$ serve similar roles, as do $-\mathbf{T}$ and $\mathbf{J}$. This implies that electromagnetic momentum is locally conserved, but not generally. Since charges and the electromagnetic field exchange momentum, the total momentum of both is conserved; in other words, the combined momentum of matter and the electromagnetic field is preserved.