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@@ -6,70 +6,70 @@ The laws of physics exhibit temporal, spatial, and angular symmetry. That is, th
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The conservation of chargee is another symmetry-based conservation law, derived from "gauge invariance". In this section, this is conservation law is assumed to be valid.
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Recall the *continuity equation*, that is, $\frac{\partial \rho(\vb{r})}{\partial t} = - \div \vb{J}(\vb{r})$. That is, the charge density at any point in space is equal to to the divergence of the current. From this, we can integrate to find $Q(t) = \int_V \rho(\vb{r}, t) dV$, and $\frac{dQ}{dt} = -\int_{\partial V} \vb{J}(\vb{r}, t) \vdot \vu{n} dS$. This is a *local conservation law*, because it does not address situations in which charge decreases in one region and increases in another without the flow of current.
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Recall the *continuity equation*, that is, $\frac{\partial \rho(\mathbf{r})}{\partial t} = - \nabla \cdot \mathbf{J}(\mathbf{r})$. That is, the charge density at any point in space is equal to to the divergence of the current. From this, we can integrate to find $Q(t) = \int_V \rho(\mathbf{r}, t) dV$, and $\frac{dQ}{dt} = -\int_{\partial V} \mathbf{J}(\mathbf{r}, t) \cdot \hat{\mathbf{n}} dS$. This is a *local conservation law*, because it does not address situations in which charge decreases in one region and increases in another without the flow of current.
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## Section 9.2 - Conservation of Electomagnetic Energy
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Consider a volume $V$ with surface $\partial V$, that encloses some magnetic and electric point charges. Then, for any electric charge $q_{ei}$ or magnetic charge $q_{mj}$, the force on each charge is
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$$\begin{align}
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\vb{F}_i &= q_{ei} (\vb{E}_i + \vb{v}_i \times \vb{B}_i) \\
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\vb{F}_j &= q_{mi} (\vb{H}_j - \vb{v}_j \times \vb{D}_j)
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\mathbf{F}_i &= q_{ei} (\mathbf{E}_i + \mathbf{v}_i \times \mathbf{B}_i) \\
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\mathbf{F}_j &= q_{mi} (\mathbf{H}_j - \mathbf{v}_j \times \mathbf{D}_j)
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\end{align}$$
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We also know that the rate at which energy changes due to changing fields is $\frac{dw_i}{dt} = \vb{F}_i \vdot \vb{v}_i$ and $\frac{dw_m}{dt} = \vb{F}_j \vdot \vb{v}_j$. This allows us to conclude that at any point $\vb{r}$ inside the volume, the rate at which the mechanical energy density changes is
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We also know that the rate at which energy changes due to changing fields is $\frac{dw_i}{dt} = \mathbf{F}_i \cdot \mathbf{v}_i$ and $\frac{dw_m}{dt} = \mathbf{F}_j \cdot \mathbf{v}_j$. This allows us to conclude that at any point $\mathbf{r}$ inside the volume, the rate at which the mechanical energy density changes is
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$$\frac{du_{mech}}{dt} = \sum_i \delta(\vb{r} - \vb{r}_i) \vb{F}_i \vdot \vb{v}_i + \delta(\vb{r} - \vb{r}_j) \vb{F}_j \vdot \vb{v}_j$$
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$$\frac{du_{mech}}{dt} = \sum_i \delta(\mathbf{r} - \mathbf{r}_i) \mathbf{F}_i \cdot \mathbf{v}_i + \delta(\mathbf{r} - \mathbf{r}_j) \mathbf{F}_j \cdot \mathbf{v}_j$$
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Since $(\vb{v}_i \times \vb{B}_i) \vdot \vb{v}_i = 0$ and $(\vb{v}_j \times \vb{D}_j) \vdot \vb{v}_j = 0$, and the current densities are given as $\vb{J}_e(\vb{r}) = \sum_i \vb{v}_i q_{ei}\delta(\vb{r} - \vb{r}_i)$ and $\vb{J}_m(\vb{r}) = \sum_j \vb{v}_j q_{mj}\delta(\vb{r} - \vb{r}_j)$, we can rewrite this as
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Since $(\mathbf{v}_i \times \mathbf{B}_i) \cdot \mathbf{v}_i = 0$ and $(\mathbf{v}_j \times \mathbf{D}_j) \cdot \mathbf{v}_j = 0$, and the current densities are given as $\mathbf{J}_e(\mathbf{r}) = \sum_i \mathbf{v}_i q_{ei}\delta(\mathbf{r} - \mathbf{r}_i)$ and $\mathbf{J}_m(\mathbf{r}) = \sum_j \mathbf{v}_j q_{mj}\delta(\mathbf{r} - \mathbf{r}_j)$, we can rewrite this as
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$$\frac{du_{mech}}{dt} = \vb{J}_e(\vb{r}) \vdot \vb{E}(\vb{r}) + \vb{J}_m(\vb{r}) \vdot \vb{H}(\vb{r})$$
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$$\frac{du_{mech}}{dt} = \mathbf{J}_e(\mathbf{r}) \cdot \mathbf{E}(\mathbf{r}) + \mathbf{J}_m(\mathbf{r}) \cdot \mathbf{H}(\mathbf{r})$$
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Combine this with the Maxwell equations to remove current densities, we see that
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$$\frac{du_{mech}}{dt} = (\curl \vb{H} - \frac{\partial \vb{H}}{\partial t}) \vdot \vb{E} + (-\curl \vb{E} - \frac{\partial \vb{B}}{\partial t}) \vdot \vb{H}$$
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$$\frac{du_{mech}}{dt} = (\nabla \times \mathbf{H} - \frac{\partial \mathbf{H}}{\partial t}) \cdot \mathbf{E} + (-\nabla \times \mathbf{E} - \frac{\partial \mathbf{B}}{\partial t}) \cdot \mathbf{H}$$
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With a vector identity, this simplifies to
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$$\frac{du_{mech}}{dt} = -\curl (\vb{E} \times \vb{H})$$
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$$\frac{du_{mech}}{dt} = -\nabla \times (\mathbf{E} \times \mathbf{H})$$
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**Definition**. We call $\vb{S} = \vb{E} \times \vb{H}$ the *Poynting vector*.
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**Definition**. We call $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ the *Poynting vector*.
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With this vector, we can define
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$$\frac{du_{mech}}{dt} = -\curl \vb{S}$$
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$$\frac{du_{mech}}{dt} = -\nabla \times \mathbf{S}$$
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### Section 9.2.2 - Energy Density for Linear Materials
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For a simple material, that is, one in which $\vb{D} = \epsilon \vb{E}$ and $\vb{B} = \mu \vb{H}$, we can express the rate of change of electomagnetic energy $\frac{\partial u_{em}}{\partial t}$ as $\frac{\partial u_{em}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} (\vb{E} \vdot \vb{D} + \vb{H} \vdot \vb{B})$. After integration, we can see that $u_{em} = \frac{1}{2}(\epsilon E^2 + \mu H^2)$.
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For a simple material, that is, one in which $\mathbf{D} = \varepsilon \mathbf{E}$ and $\mathbf{B} = \mu \mathbf{H}$, we can express the rate of change of electomagnetic energy $\frac{\partial u_{em}}{\partial t}$ as $\frac{\partial u_{em}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} (\mathbf{E} \cdot \mathbf{D} + \mathbf{H} \cdot \mathbf{B})$. After integration, we can see that $u_{em} = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$.
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**Definition**. A material is said to be an *anisotropic linear material* if $\vb{D} = \overleftrightarrow{\vb{\epsilon}} \vdot \vb{E}$ and $\vb{B} = \overleftrightarrow{\vb{\mu}} \vdot \vb{H}$. Then, the internal electromagnetic energy becones
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**Definition**. A material is said to be an *anisotropic linear material* if $\mathbf{D} = \overleftrightarrow{\mathbf{\varepsilon}} \cdot \mathbf{E}$ and $\mathbf{B} = \overleftrightarrow{\mathbf{\mu}} \cdot \mathbf{H}$. Then, the internal electromagnetic energy becomes
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$$u_{em} = \frac{1}{2}(\vb{E} \vdot \overleftrightarrow{\vb{\epsilon}} \vdot \vb{E}$ and $\vb{B} + \vb{H} + \overleftrightarrow{\vb{\mu}} \vdot \vb{H}) = \frac{1}{2}\sum_{i,j}(\epsilon_{ij}E_iE_j + \mu_{ij}H_iH_j)$$
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$$u_{em} = \frac{1}{2}(\mathbf{E} \cdot \overleftrightarrow{\mathbf{\varepsilon}} \cdot \mathbf{E}+ \mathbf{H} + \overleftrightarrow{\mathbf{\mu}} \cdot \mathbf{H}) = \frac{1}{2}\sum_{i,j}(\varepsilon_{ij}E_iE_j + \mu_{ij}H_iH_j)$$
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If the dyadics are symmetric, the energy functions uniquely specify the energy in terms of fields. In general, the polarization or magnetization of a material may depend on its past history and on time, and as such, the energy density for such materials cannot be expressed entirely in terms of the fields.
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### Section 9.2.3 - Poynting Vector Examples
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Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\vb{H} = \frac{R}{2\pi s R}\vb{\varphi}$. Additionally, we know that
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Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\mathbf{H} = \frac{R}{2\pi s R}\mathbf{\varphi}$. Additionally, we know that
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$$V=\int_a^b \vb{E} \vdot d\vb{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$
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$$V=\int_a^b \mathbf{E} \cdot d\mathbf{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$
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This then implies that $\vb{E} = \frac{V}{\ln(\frac{b}{a})s} \vu(s)$. We can then solve for both the energy density and Poynting vector, as well as $\vb{v} = \vb{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that
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This then implies that $\mathbf{E} = \frac{V}{\ln(\frac{b}{a})s} \hat{\mathbf{s}}$. We can then solve for both the energy density and Poynting vector, as well as $\mathbf{v} = \mathbf{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that
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$$\vb{v} = \frac{2c \vu{z}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$
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$$\mathbf{v} = \frac{2c \hat{\mathbf{z}}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$
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---
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Consider a long cylindrical ohmic wire of radius $a$, length $L$, and resistivity $\rho$ along the $z$-axis. If this wire is carrying a constant current $I$, we know that inside the wire, $E_z = \rho J = \rho \frac{I}{\pi a^2}$. From Ampere's Law, $H_\phi = \frac{s}{2\pi a^2} I$. Then, inside the wire,
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$$\vb{S} = -s \frac{\rho I^2}{2 \pi^2 a^4} \vu{s}$$
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$$\mathbf{S} = -s \frac{\rho I^2}{2 \pi^2 a^4} \hat{\mathbf{s}}$$
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Outside the wire, we know that $\vb{E} = \frac{\rho I}{\pi a^2} \vu{z}$ and $\vb{H} = \frac{I}{2\pi s}\vu{\varphi}$, so
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Outside the wire, we know that $\mathbf{E} = \frac{\rho I}{\pi a^2} \hat{\mathbf{z}}$ and $\mathbf{H} = \frac{I}{2\pi s}\hat{\mathbf{\varphi}}$, so
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$$\vb{S} = -\frac{\rho I}{\pi a^2} \frac{I}{2\pi s} \vu{s}$$
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$$\mathbf{S} = -\frac{\rho I}{\pi a^2} \frac{I}{2\pi s} \hat{\mathbf{s}}$$
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Notably, inside the wire, $\div \vb{S} = \frac{\rho I^2}{\pi^2 a^4}$, but is equal to $0$ outside of the wire.
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Notably, inside the wire, $\nabla \cdot \mathbf{S} = \frac{\rho I^2}{\pi^2 a^4}$, but is equal to $0$ outside of the wire.
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---
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@@ -79,50 +79,50 @@ Further examples are present but omitted.
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We know that the rate of change of momentum for any given particle is simply the force acting on it. To calculate this, recall the force density:
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$$\vb{f} = \sum_i \vb{F}_i\delta(\vb{r}-\vb{r}_i) + \sum_j \vb{F}_j\delta(\vb{r}-\vb{r}_j) = \sum_i q_{ei}\delta(\vb{r}-\vb{r}_i)(\vb{E} + \vb{v}_i \cross \vb{B}) + \sum_j q_{ej} \delta(\vb{r}-\vb{r}_i) (\vb{H} - \vb{v}_j \times \vb{D})$$
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$$\mathbf{f} = \sum_i \mathbf{F}_i\delta(\mathbf{r}-\mathbf{r}_i) + \sum_j \mathbf{F}_j\delta(\mathbf{r}-\mathbf{r}_j) = \sum_i q_{ei}\delta(\mathbf{r}-\mathbf{r}_i)(\mathbf{E} + \mathbf{v}_i \times \mathbf{B}) + \sum_j q_{ej} \delta(\mathbf{r}-\mathbf{r}_i) (\mathbf{H} - \mathbf{v}_j \times \mathbf{D})$$
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Converting to currents, we see that
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$$\vb{f}(\vb{r}) = \rho_e(\vb{r})\vb{E}(\vb{r}) + \vb{J}_e(\vb{r}) \times \vb{B}(\vb{r}) + \rho_m(\vb{r}) + \vb{H}(\vb{r}) - \vb{J}_m(\vb{r}) \times \vb{D}(\vb{r})$$
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$$\mathbf{f}(\mathbf{r}) = \rho_e(\mathbf{r})\mathbf{E}(\mathbf{r}) + \mathbf{J}_e(\mathbf{r}) \times \mathbf{B}(\mathbf{r}) + \rho_m(\mathbf{r}) + \mathbf{H}(\mathbf{r}) - \mathbf{J}_m(\mathbf{r}) \times \mathbf{D}(\mathbf{r})$$
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Substituting in Maxwell's Equations, we see that
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$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \varepsilon_0 (\div \vb{E})\vb{E} + (\curl \vb{E})\times\vb{D} + \mu_0(\div \vb{H})\vb{H} + (\curl \vb{H})\times\vb{B}$$
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$$\mathbf{f}(\mathbf{r}) + \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = \varepsilon_0 (\nabla \cdot \mathbf{E})\mathbf{E} + (\nabla \times \mathbf{E})\times\mathbf{D} + \mu_0(\nabla \cdot \mathbf{H})\mathbf{H} + (\nabla \times \mathbf{H})\times\mathbf{B}$$
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Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\vb{T}}$, so that
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Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\mathbf{T}}$, so that
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$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \div \overleftrightarrow{\vb{T}}$$
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$$\mathbf{f}(\mathbf{r}) + \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = \nabla \cdot \overleftrightarrow{\mathbf{T}}$$
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This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\div \overleftrightarrow{\vb{T}} = \div \overleftrightarrow{\vb{T}}_e + \div \overleftrightarrow{\vb{T}}_m$. Then, we can state
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This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\nabla \cdot \overleftrightarrow{\mathbf{T}} = \nabla \cdot \overleftrightarrow{\mathbf{T}}_e + \nabla \cdot \overleftrightarrow{\mathbf{T}}_m$. Then, we can state
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$$\begin{align}
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\div \overleftrightarrow{\vb{T}}_e &= \varepsilon_0 [(\div \vb{E})\vb{E} + (\curl \vb{E})\times \vb{E}] \\
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\div \overleftrightarrow{\vb{T}}_m &= \varepsilon_0 [(\div \vb{H})\vb{H} + (\curl \vb{H})\times \vb{H}]
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\nabla \cdot \overleftrightarrow{\mathbf{T}}_e &= \varepsilon_0 [(\nabla \cdot \mathbf{E})\mathbf{E} + (\nabla \times \mathbf{E})\times \mathbf{E}] \\
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\nabla \cdot \overleftrightarrow{\mathbf{T}}_m &= \varepsilon_0 [(\nabla \cdot \mathbf{H})\mathbf{H} + (\nabla \times \mathbf{H})\times \mathbf{H}]
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\end{align}$$
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We know that $\div(\vb{EE}) = (\div \vb{E})\vb{E} + (\vb{E} \vdot \nabla)\vb{E}$ and $\div(\overleftrightarrow{\vb{I}}f) = \grad f$. If we let $f = \frac{1}{2}\vb{E} \vdot \vb{E}$, we see that $\grad(\frac{1}{2}\vb{E} \vdot \vb{E}) = (\vb{E} \vdot \nabla)\vb{E} + (\curl \vb{E})\vb{E}$. Then, we see that
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We know that $\nabla \cdot(\mathbf{EE}) = (\nabla \cdot \mathbf{E})\mathbf{E} + (\mathbf{E} \cdot \nabla)\mathbf{E}$ and $\nabla \cdot(\overleftrightarrow{\mathbf{I}}f) = \nabla f$. If we let $f = \frac{1}{2}\mathbf{E} \cdot \mathbf{E}$, we see that $\nabla(\frac{1}{2}\mathbf{E} \cdot \mathbf{E}) = (\mathbf{E} \cdot \nabla)\mathbf{E} + (\nabla \times \mathbf{E})\mathbf{E}$. Then, we see that
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$$\begin{align}
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\overleftrightarrow{\vb{T}}_e &= \varepsilon_0 \vb{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\vb{I}}(\vb{E} \vdot \vb{E}) \\
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\overleftrightarrow{\vb{T}}_m &= \mu_0 \vb{HH} - \frac{\mu_0}{2} \overleftrightarrow{\vb{I}}(\vb{H} \vdot \vb{H})
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\overleftrightarrow{\mathbf{T}}_e &= \varepsilon_0 \mathbf{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\mathbf{I}}(\mathbf{E} \cdot \mathbf{E}) \\
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\overleftrightarrow{\mathbf{T}}_m &= \mu_0 \mathbf{HH} - \frac{\mu_0}{2} \overleftrightarrow{\mathbf{I}}(\mathbf{H} \cdot \mathbf{H})
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\end{align}$$
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Knowing that $\overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}}_e + \overleftrightarrow{\vb{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum,
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Knowing that $\overleftrightarrow{\mathbf{T}} = \overleftrightarrow{\mathbf{T}}_e + \overleftrightarrow{\mathbf{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum,
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$$\overleftrightarrow{\vb{T}} = \varepsilon_0 \vb{EE} + \mu_0 \vb{HH} - \overleftrightarrow{\vb{I}}u$$
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$$\overleftrightarrow{\mathbf{T}} = \varepsilon_0 \mathbf{EE} + \mu_0 \mathbf{HH} - \overleftrightarrow{\mathbf{I}}u$$
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Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\vb{g}(\vb{r}) = \vb{D}(\vb{r}) \times \vb{B}(\vb{r})$. Thus,
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Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\mathbf{g}(\mathbf{r}) = \mathbf{D}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})$. Thus,
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$$\vb{f}(\vb{r}) = \frac{\partial}{\partial t}\vb{g} = \div \overleftrightarrow{\vb{T}}$$
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$$\mathbf{f}(\mathbf{r}) = \frac{\partial}{\partial t}\mathbf{g} = \nabla \cdot \overleftrightarrow{\mathbf{T}}$$
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---
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The Divergence Theorem states that $\int_V(\div \overleftrightarrow{\vb{T}}) dV = \int_{SofV} dS \vu{n} \vdot \overleftrightarrow{\vb{T}}$. We can prove this by expanding the left-hand side over a cube. Note that as $\overleftrightarrow{\vb{T}}$ is symmetric, $\vu{n} \vdot \overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}} \vdot \vu{n}$.
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The Divergence Theorem states that $\int_V(\nabla \cdot \overleftrightarrow{\mathbf{T}}) dV = \int_{SofV} dS \hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}}$. We can prove this by expanding the left-hand side over a cube. Note that as $\overleftrightarrow{\mathbf{T}}$ is symmetric, $\hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}} = \overleftrightarrow{\mathbf{T}} \cdot \hat{\mathbf{n}}$.
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Given a static field, the momentum does not change in time. That is,
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$$\frac{\partial}{\partial t} \vb{g} = \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = 0$$
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$$\frac{\partial}{\partial t} \mathbf{g} = \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = 0$$
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Then, we can see that $\vb{f}(\vb{r}) = \div \overleftrightarrow{\vb{T}}$. We can thus integrate over the volume to find force on an object.
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Then, we can see that $\mathbf{f}(\mathbf{r}) = \nabla \cdot \overleftrightarrow{\mathbf{T}}$. We can thus integrate over the volume to find force on an object.
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$$\vb{F} = \int_V \vb{f}(\vb{r}) dV = \int_V \div \overleftrightarrow{\vb{T}} dV = \int_{\partial V} dS \vu{n} \vdot \overleftrightarrow{\vb{T}} = \int_{\partial V} \overleftrightarrow{\vb{T}} \vdot \vu{n} dS$$
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$$\mathbf{F} = \int_V \mathbf{f}(\mathbf{r}) dV = \int_V \nabla \cdot \overleftrightarrow{\mathbf{T}} dV = \int_{\partial V} dS \hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}} = \int_{\partial V} \overleftrightarrow{\mathbf{T}} \cdot \hat{\mathbf{n}} dS$$
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