Electromagnetic Force on a Charge inside a Volume
Overview1
The electromagnetic force experienced by all charges in volume $\mathcal{V}$ is as follows.
$$ \mathbf{F} =\oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} -\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau $$
$\mathcal{S}$ is the boundary surface of volume $\mathcal{V}$, $\mathbf{T}$ is the Maxwell stress tensor, $\mathbf{S}$ is the Poynting vector.
Derivation
- Part 1.
By Lorentz force law, the force experienced by a charge is
$$ \mathbf{F}=q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) $$
If we express the charge quantity in terms of charge density, it becomes $q=\int \rho d\tau$
$$ \mathbf{F}=\int_{\mathcal{V}} \rho (\mathbf{E} + \mathbf{v} \times \mathbf{B})d\tau $$
Volume current density is therefore $\rho\mathbf{v}=\mathbf{J}$
$$ \mathbf{F}=\int_{\mathcal{V}} (\rho\mathbf{E} + \mathbf{J} \times \mathbf{B})d\tau $$
Differentiating both sides to represent the force experienced by the charge per unit volume,
$$ \dfrac{d\mathbf{F}}{d \tau}=\mathbf{f}=\rho\mathbf{E} + \mathbf{J} \times \mathbf{B} $$
$\mathrm{(i)}\ \rho=\epsilon_{0} \nabla \cdot \mathbf{E}$, $\mathrm{(iv)}\ \mathbf{J}=\dfrac{1}{\mu_{0}}\nabla \times \mathbf{B}-\epsilon_{0}\dfrac{\partial \mathbf{E}}{\partial t}$
Using Maxwell’s equations to express $\mathbf{f}$ purely in terms of electromagnetic fields,
$$ \begin{equation} \mathbf{f} = \epsilon_{0}(\nabla \cdot \mathbf{E})\mathbf{E} + \dfrac{1}{\mu_{0}}(\nabla \times \mathbf{B})\times \mathbf{B} - \epsilon_{0} \dfrac{\partial \mathbf{E}}{\partial t} \times \mathbf{B} \end{equation} $$
- Part 2.
To adjust the last term on the right side of $(1)$, let’s examine the following formula.
$$ \dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) = \dfrac{\partial \mathbf{E}}{\partial t} \times \mathbf{B} + \mathbf{E} \times \dfrac{\partial \mathbf{B}}{\partial t} $$
$\frac{\partial \mathbf{B}}{\partial t}=-\nabla \times \mathbf{E}$
Reorganizing the above formula using Faraday’s Law,
$$ \dfrac{\partial \mathbf{E}}{\partial t} \times \mathbf{B} = \dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) + \mathbf{E} \times (\nabla \times \mathbf{E} ) $$
Substituting it into $(1)$ and rearranging,
$$ \begin{align} \mathbf{f} &= \epsilon_{0}(\nabla \cdot \mathbf{E})\mathbf{E} + \dfrac{1}{\mu_{0}}(\nabla \times \mathbf{B})\times \mathbf{B} - \epsilon_{0} \left[ \dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) + \mathbf{E} \times (\nabla \times \mathbf{E} ) \right] \nonumber \\ &= \epsilon_{0} \bigg[ (\nabla \cdot \mathbf{E})\mathbf{E} - \mathbf{E} \times (\nabla \times \mathbf{E}) \bigg] - \dfrac{1}{\mu_{0}} \bigg[ \mathbf{B} \times (\nabla \times \mathbf{B} ) \bigg] -\epsilon_{0}\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B} ) \end{align} $$
- Part 3.
To make the formula symmetric, in other words, to make it look nicer, add $(\nabla \cdot \mathbf{B})\mathbf{B}$ inside the second brackets of $(2)$. Since $\nabla \cdot \mathbf{B}=0$, there is no problem. Then,
$$ \begin{equation} \begin{aligned} \mathbf{f} &= \epsilon_{0} \bigg[ (\nabla \cdot \mathbf{E})\mathbf{E} - \mathbf{E} \times (\nabla \times \mathbf{E}) \bigg] + \dfrac{1}{\mu_{0}} \bigg[(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B} ) \bigg] \\ &\quad -\epsilon_{0}\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) \end{aligned} \end{equation} $$
$$ \nabla(\mathbf{A} \cdot \mathbf{B}) = \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})+(\mathbf{A} \cdot \nabla)\mathbf{B}+(\mathbf{B} \cdot \nabla) \mathbf{A} $$
Also, by using the multiplication rule,
$$ \nabla(\mathbf{A} \cdot \mathbf{A} ) =\nabla(A^2) = 2(\mathbf{A} \cdot \nabla) \mathbf{A} + 2\mathbf{A} \times (\nabla \times \mathbf{A} ) $$
Thus,
$$ \begin{cases} \mathbf{E} \times (\nabla \times \mathbf{E} ) = -(\mathbf{E}\cdot\nabla) \mathbf{E} + \dfrac{1}{2}\nabla(E^2) \\ \mathbf{B} \times (\nabla \times \mathbf{B} ) = -(\mathbf{B}\cdot\nabla) \mathbf{B} +\dfrac{1}{2}\nabla(B^2) \end{cases} $$ and substituting into $(3)$,
$$ \begin{align*} \mathbf{f} &= \epsilon_{0} \bigg[ (\nabla \cdot \mathbf{E})\mathbf{E} +(\mathbf{E}\cdot\nabla)\mathbf{E} - \dfrac{1}{2}\nabla(E^2) \bigg] + \dfrac{1}{\mu_{0}} \bigg[ \nabla \cdot \mathbf{B}) \mathbf{B} +(\mathbf{B}\cdot\nabla)\mathbf{B} -\dfrac{1}{2}\nabla(B^2) \bigg] -\epsilon_{0}\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B} ) \end{align*} $$
Because the equation is too complex, using the Maxwell stress tensor $\mathbf{T}$ to represent the first three terms as $\nabla \cdot \mathbf{T}$ and the last term with the Poynting vector $\mathbf{S}$, the form of the equation becomes
$$ \mathbf{f} = \nabla \cdot \mathbf{T} - \epsilon_{0}\mu_{0}\dfrac{\partial \mathbf{S}}{\partial t} $$
Integrating both sides over the volume and using the divergence theorem on the first term on the right-hand side results in
$$ \mathbf{F} =\oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} -\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau $$
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David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p388-390 ↩︎