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PHY450H1S. Relativistic Electrodynamics Lecture 23 (Taught by Prof. Erich Poppitz). Energy Momentum Tensor.

Posted by peeterjoot on April 10, 2011

FIXME: Covering chapter X material from the text [1].

Covering lecture notes pp. 173-178: the force on a surface element of a body (177-178); a plane wave example (179-180).

pp. 181-195: [read on your own: on diagonalizability of energy-momentum tensor (181)]; the Lagrangian for a system of nonrelativistic charged particles to zeroth order in $(v/c)$: electrostatic energy of a system of charges and .mass renormalization. (182-189) [Tuesday, Mar. 29]; the EM potentials to order $(v/c)^2$ (190-193); the “Darwin Lagrangian. and Hamiltonian for a system of nonrelativistic charged particles to order $(v/c)^2$ and its many uses in physics (194-195) [Wednesday, Mar. 30]

Next week (last topic): attempt to go to the next order $(v/c)^3$ – radiation damping, the limitations of classical electrodynamics, and the relevant time/length/energy scales.

Recap.

Last time we found that spacetime translation invariance led to the four conservation relations

\begin{aligned}\partial_k T^{k m} = 0\end{aligned} \hspace{\stretch{1}}(2.1)

where

\begin{aligned}T^{k m} = \frac{1}{{4 \pi}} \left( F^{k j} F^{m l} g_{j l} + \frac{1}{{4}} g^{k m} F_{i j} F^{i j} \right)\end{aligned} \hspace{\stretch{1}}(2.2)

last time we found for $m = 0$

\begin{aligned}\frac{1}{{c}} \frac{\partial {}}{\partial {t}} T^{0 0} + \frac{\partial {}}{\partial {x^\alpha}} T^{\alpha 0} = 0\end{aligned} \hspace{\stretch{1}}(2.3)

Here

\begin{aligned}\begin{array}{l l l}T^{0 0} &= \frac{1}{{8 \pi}} (\mathbf{E}^2 + \mathbf{B}^2) &= \mbox{energy density} \\ c T^{\alpha 0} &= \mathbf{S}^\alpha &= \mbox{energy flux}\end{array}\end{aligned} \hspace{\stretch{1}}(2.4)

Spatial components of $T^{k m}$

Now for $m = 1,2,3$ we write

\begin{aligned}\partial_k T^{k \alpha} = 0\end{aligned} \hspace{\stretch{1}}(3.5)

so we write

\begin{aligned}\frac{\partial {}}{\partial {t}} \frac{\mathbf{S}^\alpha}{c^2} + \partial_\beta T^{\beta \alpha} = 0\end{aligned} \hspace{\stretch{1}}(3.6)

Recall that we argued that

\begin{aligned}\frac{\mathbf{S}}{c^2} = \text{momentum density}\end{aligned} \hspace{\stretch{1}}(3.7)

(it also comes from Noether’s theorem).

\begin{aligned}\frac{\partial {}}{\partial {t}} \left( \frac{T^{0\alpha}}{c} \right) + \frac{\partial {}}{\partial {x^\beta}} T^{\beta \alpha} = 0\end{aligned} \hspace{\stretch{1}}(3.8)

or

\begin{aligned}\frac{\partial {}}{\partial {t}} \left( \frac{\mathbf{S}^{\alpha}}{c^2} \right) + \frac{\partial {}}{\partial {x^\beta}} T^{\beta \alpha} = 0.\end{aligned} \hspace{\stretch{1}}(3.9)

Integrating over $V$ we have

\begin{aligned}\frac{\partial {}}{\partial {t}} \int_V d^3\mathbf{x} \left( \frac{\mathbf{S}^{\alpha}}{c^2} \right) &= -\int_V d^3\mathbf{x} \frac{\partial {}}{\partial {x^\beta}} T^{\beta \alpha} \\ &= -\int_V d^3\mathbf{x} \boldsymbol{\nabla} \cdot (\mathbf{e}_\beta T^{\beta \alpha}) \\ &= -\int_{\partial V} d^2 \sigma \mathbf{n} \cdot \mathbf{e}_\beta) T^{\beta \alpha} \\ &\equiv -\int_{\partial V} d^2 \sigma^\beta T^{\beta \alpha} \\ \end{aligned}

We write this as

\begin{aligned}\frac{\partial {}}{\partial {t}} \left( \text{momentum of EM fields in V} \right)^\alpha = - \int_{\partial V} d^2 \sigma^\beta T^{\beta \alpha}\end{aligned} \hspace{\stretch{1}}(3.10)

and describe our spatial tensor components as

\begin{aligned}T^{\beta\alpha} = \text{flux oflatex \alpha-th momentum through a unit area $\perp \beta$},\end{aligned} \hspace{\stretch{1}}(3.11)

where

\begin{aligned}T^{\alpha\beta} &= \frac{1}{{4\pi}} \left( -F^{\alpha j} F^{\beta m} g_{m j} + \frac{1}{{4}} g^{\alpha\beta} F^{ij}F_{ij} \right) \\ &= \frac{1}{{4\pi}} \left( -F^{\alpha 0} F^{\beta 0} +F^{\alpha \sigma} F^{\beta \sigma} - \frac{1}{{4}} \delta^{\alpha\beta} 2 (\mathbf{B}^2 - \mathbf{E}^2) \right) \\ &= \frac{1}{{4\pi}} \left( -E^\alpha E^\beta+\sum_\sigma (\epsilon^{\mu\alpha\sigma} B^\mu)(\epsilon^{\nu\beta\sigma} B^\nu)- \frac{1}{{2}} \delta^{\alpha\beta} (\mathbf{B}^2 - \mathbf{E}^2) \right) \\ &= \frac{1}{{4\pi}} \left( -E^\alpha E^\beta+\sum_{\mu,\nu} \underbrace{\delta^{\mu\alpha}_{[\nu\beta]}B^\mu B^\nu}_{(\delta^{\alpha \beta} \delta^{\mu \nu} - \delta^{\alpha \nu} \delta^{\beta \mu} ) B^\mu B^\nu=\delta^{\alpha\beta}\mathbf{B}^2 - B^\alpha B^\beta} + \frac{1}{{2}} \delta^{\alpha\beta} (\mathbf{E}^2 - \mathbf{B}^2) \right) \\ &=- \frac{1}{{4\pi}} \left(E^\alpha E^\beta+B^\alpha B^\beta+ \delta^{\alpha\beta} \left(-\frac{\mathbf{E}^2}{2} + \frac{\mathbf{B}^2}{2} - \mathbf{B}^2 \right)\right) \\ &=- \frac{1}{{4\pi}} \left(E^\alpha E^\beta+B^\alpha B^\beta- \frac{\delta^{\alpha\beta} }{2}\left(\mathbf{E}^2 + \mathbf{B}^2 \right)\right) \\ \end{aligned}

We define

\begin{aligned}\sigma^{\alpha\beta}=-T^{\alpha\beta}= \frac{1}{{4\pi}} \left(E^\alpha E^\beta+B^\alpha B^\beta- \frac{\delta^{\alpha\beta} }{2}\left(\mathbf{E}^2 + \mathbf{B}^2 \right)\right) \end{aligned} \hspace{\stretch{1}}(3.12)

This is the Maxwell stress tensor. Maxwell apparently derived this without any use of four vectors or symmetry arguments. I’d be curious what his arguments were and how he related this to the Lorentz force?

In gigantic matrix form, where symmetric opposites are omitted, we have for ${\left\lVert{T^{ij}}\right\rVert}$

\begin{aligned}\begin{bmatrix}\frac{1}{{8 \pi}}(\mathbf{E}^2 + \mathbf{B}^2) & \frac{1}{{4\pi}} (\mathbf{E} \times \mathbf{B})^x & \frac{1}{{4\pi}} (\mathbf{E} \times \mathbf{B})^y & \frac{1}{{4\pi}} (\mathbf{E} \times \mathbf{B})^z \\ . & -\frac{1}{{4 \pi}}\left(E_x^2 + B_x^2 - \frac{1}{{2}}(\mathbf{E}^2 + \mathbf{B}^2)\right) & -\frac{1}{{4 \pi}}\left(E_x E_y + B_x B_y \right)& -\frac{1}{{4 \pi}}\left(E_x E_z + B_x B_z \right) \\ . & . & -\frac{1}{{4 \pi}}\left(E_y^2 + B_y^2 - \frac{1}{{2}}(\mathbf{E}^2 + \mathbf{B}^2)\right) & -\frac{1}{{4 \pi}}\left(E_y E_z + B_y B_z \right) \\ . & . & . & -\frac{1}{{4 \pi}}\left(E_z^2 + B_z^2 - \frac{1}{{2}}(\mathbf{E}^2 + \mathbf{B}^2)\right) \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.13)

In words this matrix is

\begin{aligned}\begin{bmatrix}\text{energy density} & \frac{1}{{c}} (\text{energy flux inlatex \hat{\mathbf{x}}}) & \frac{1}{{c}} (\text{energy flux in $\hat{\mathbf{y}}$}) & \frac{1}{{c}} (\text{energy flux in $\hat{\mathbf{z}}$}) \\ c \times (\text{momentum density})^x& (\text{momentum})^x \text{flux in $\hat{\mathbf{x}}$}& (\text{momentum})^x \text{flux in $\hat{\mathbf{y}}$}& (\text{momentum})^x \text{flux in $\hat{\mathbf{z}}$} \\ c \times (\text{momentum density})^y& (\text{momentum})^y \text{flux in $\hat{\mathbf{x}}$}& (\text{momentum})^y \text{flux in $\hat{\mathbf{y}}$}& (\text{momentum})^y \text{flux in $\hat{\mathbf{z}}$} \\ c \times (\text{momentum density})^z& (\text{momentum})^z \text{flux in $\hat{\mathbf{x}}$}& (\text{momentum})^z \text{flux in $\hat{\mathbf{y}}$}& (\text{momentum})^z \text{flux in $\hat{\mathbf{z}}$} \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.14)

On the geometry.

PICTURE: rectangular area with normal $\alphacap$, and area $d^2 \sigma^\alpha \perp \alphacap$.

$T^{\alpha\beta}$ is the amount of $P^\beta$ that goes through unit area $\perp \alphacap$ in unit time. $d^2 \sigma^\alpha T^{\alpha \beta}$ (no sum) is the amount of $P^\beta$ through $d^2 \sigma^\alpha$ in unit time.

For a general surface element

PICTURE: normal $\mathbf{n}$ decomposed into perpendicular components $\hat{\Balpha}$, $\hat{\boldsymbol{\beta}}$, with respective area elements $d^2 \sigma^\alpha$ and $d^2 \sigma^\beta$.

PICTURE: triangulated area element decomposed into three perpendicular areas with their respective normals.

We have

\begin{aligned}\int d^3 \mathbf{x} \frac{\partial {}}{\partial {x^\alpha}} T^{\alpha\beta} = \int d^3 \mathbf{x} \boldsymbol{\nabla} \cdot ( \mathbf{e}_\beta T^{\alpha\beta} ) = \int d^2 \sigma (\mathbf{n} \cdot \mathbf{e}_\beta) T^{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.15)

Write

\begin{aligned}d^2 \boldsymbol{\sigma} = d^2 \sigma \mathbf{n} = \sum_\alpha d^2 \sigma n^\alpha \mathbf{e}_\alpha,\end{aligned} \hspace{\stretch{1}}(4.16)

where $\mathbf{n} = (n^1, n^2, n^3)$. The amount of $\beta$ momentum that goes through $d^2 \sigma$ in unit time is

\begin{aligned}\sum_\alpha d^2\sigma^\alpha T^{\alpha\beta} \end{aligned} \hspace{\stretch{1}}(4.17)

If this is greater than zero, this is a flow in the $\mathbf{n}$ direction, whereas if less than zero the momentum flows in the $-\mathbf{n}$ direction.

If $d^2 \boldsymbol{\sigma}$ is at the surface of the body, the rate of flow of $(\text{momentum})^\beta$ through $d^2 \boldsymbol{\sigma}$ is the $(\text{force})^\beta$ that acts on this element.

PICTURE: arbitrary surface depicted with an inwards normal $\mathbf{n}$.

For this surface with the inwards normal we can write

\begin{aligned}df^\beta = \sum_\alpha d^2\sigma^\alpha T^{\alpha\beta}\end{aligned} \hspace{\stretch{1}}(4.18)

The $(\text{force})^\beta$ acting on the $d^2\boldsymbol{\sigma}$ surface element. With an outwards normal we can write this in terms of the Maxwell stress tensor, which has an inverted sign

\begin{aligned}df^\beta = \sum_\alpha d^2\sigma^\alpha \sigma^{\alpha\beta}\end{aligned} \hspace{\stretch{1}}(4.19)

To find the force on the body we want

\begin{aligned}F^\beta = \oint_{\text{surface of body with inwards normal orientation}} d^2 \sigma^\alpha T^{\alpha\beta}\end{aligned} \hspace{\stretch{1}}(4.20)

We can calculate the EM force on any body. We need to know $T^{\alpha\beta}$ on the surface, so we need the EM field on this boundary.

Example. Wall absorbing all radiation hitting it.

With propagation direction $\mathbf{p}$ along the $\hat{\mathbf{x}}$ direction, and mutually perpendicular $\mathbf{E}$ and $\mathbf{B}$.

\begin{aligned}c T^{xx} = \text{amount oflatex P^xgoing in $\hat{\mathbf{x}}$ unit area $\perp \hat{\mathbf{x}}$ in unit time}\end{aligned} \hspace{\stretch{1}}(4.21)

\begin{aligned}df^x &= T^{x x} d^2 \sigma^x \\ df^y &= T^{y y} d^2 \sigma^y\end{aligned} \hspace{\stretch{1}}(4.22)

with $c p = \omega$, our fields are

\begin{aligned}E_y &= p \beta \sin( c p t - p x ) \\ B_z &= p \beta \sin( c p t - p x ) \end{aligned} \hspace{\stretch{1}}(4.24)

\begin{aligned}T^{x x} &= -\frac{1}{{4 \pi}} \left( {(E^x)^2}+{(B^x)^2}-\frac{1}{{2}} (\mathbf{E}^2 + \mathbf{B}^2) \right) \\ &= \frac{1}{{8 \pi}} \left( (E^y)^2+(B^z)^2 \right) \\ &= \frac{p^2 \beta^2}{8 \pi} \sin^2 ( c p t - p x )\end{aligned}

\begin{aligned}T^{y x} = - \frac{1}{{4 \pi}} \left( { E^x } E^y + { B^x B^y } \right) = 0\end{aligned} \hspace{\stretch{1}}(4.26)

The off diagonal $T^{\alpha \beta}$ components vanish since we have no non-zero pair of $E_\alpha E_\beta$ or $B_\alpha B_\beta$. Our other two diagonal terms are also zero

\begin{aligned}T^{y y}&= -\frac{1}{{4 \pi}} \left( (E^y)^2+{(B^y)^2}-\frac{1}{{2}} (\mathbf{E}^2 + \mathbf{B}^2) \right) \\ &= - \frac{1}{{4 \pi}} p^2 \beta^2 \sin^2 ( c p t - p x ) \left( 1 - \frac{1}{{2}} - \frac{1}{{2}} \right) \\ &= 0\end{aligned}

\begin{aligned}T^{y y}&= -\frac{1}{{4 \pi}} \left( +{(E^z)^2}(B^z)^2-\frac{1}{{2}} (\mathbf{E}^2 + \mathbf{B}^2) \right) \\ &= - \frac{1}{{4 \pi}} p^2 \beta^2 \sin^2 ( c p t - p x ) \left( 1 - \frac{1}{{2}} - \frac{1}{{2}} \right) \\ &= 0\end{aligned}

For non-perpendicular reflection we have the same deal.

PICTURE: reflection off of a wall, with reflection coefficient $R$.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.