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PHY450H1S. Relativistic Electrodynamics Tutorial 5 (TA: Simon Freedman). Angular momentum of EM fields

Posted by peeterjoot on March 10, 2011

[Click here for a PDF of this post with nicer formatting]

Motivation.

Long solenoid of radius $R$, n turns per unit length, current $I$. Coaxial with with solenoid are two long cylindrical shells of length $l$ and $(\text{radius},\text{charge})$ of $(a, Q)$, and $(b, -Q)$ respectively, where $a < b$.

When current is gradually reduced what happens?

The initial fields.

Initial Magnetic field.

For the initial static conditions where we have only a (constant) magnetic field, the Maxwell-Ampere equation takes the form

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} = \frac{4 \pi}{c} \mathbf{j}\end{aligned} \hspace{\stretch{1}}(1.1)

\paragraph{On the name of this equation}. In notes from one of the lectures I had this called Maxwell-Faraday equation, despite the fact that this isn’t the one that Maxwell made his displacement current addition. Did the Professor call it that, or was this my addition? In [2] Faraday’s law is also called the Maxwell-Faraday equation. [1] calls this the Ampere-Maxwell equation, which makes more sense.

Put into integral form by integrating over an open surface we have

\begin{aligned}\int_A (\boldsymbol{\nabla} \times \mathbf{B}) \cdot d\mathbf{a} = \frac{4 \pi}{c} \int_A \mathbf{j} \cdot d\mathbf{a}\end{aligned} \hspace{\stretch{1}}(1.2)

The current density passing through the surface is defined as the enclosed current, circulating around the bounding loop

\begin{aligned}I_{\text{enc}} = \int_A \mathbf{j} \cdot d\mathbf{a},\end{aligned} \hspace{\stretch{1}}(1.3)

so by Stokes Theorem we write

\begin{aligned}\int_{\partial A} \mathbf{B} \cdot d\mathbf{l} = \frac{4 \pi}{c} I_{\text{enc}}\end{aligned} \hspace{\stretch{1}}(1.4)

Now consider separately the regions inside and outside the cylinder. Inside we have

\begin{aligned}\int_{\partial A} B \cdot d \mathbf{l} = \frac{4 \pi I }{c} = 0,\end{aligned} \hspace{\stretch{1}}(1.5)

Outside of the cylinder we have the equivalent of $n$ loops, each with current $I$, so we have

\begin{aligned}\int \mathbf{B} \cdot d\mathbf{l} = \frac{4 \pi n I L}{c} = B L.\end{aligned} \hspace{\stretch{1}}(1.6)

Our magnetic field is constant while $I$ is constant, and in vector form this is

\begin{aligned}\mathbf{B} = \frac{4 \pi n I}{c} \hat{\mathbf{z}}\end{aligned} \hspace{\stretch{1}}(1.7)

Initial Electric field.

How about the electric fields?

Between $a$ and $b$ we have, for a Gaussian surface of height $l$ (assuming that $l \gg a$)

\begin{aligned}E (2 \pi r) l = 4 \pi (+Q),\end{aligned} \hspace{\stretch{1}}(1.8)

so we have

\begin{aligned}\mathbf{E} = \frac{2 Q }{r l} \hat{\mathbf{r}}.\end{aligned} \hspace{\stretch{1}}(1.9)

Poynting vector before the current changes.

Our Poynting vector, the energy flux per unit time, is

\begin{aligned}\mathbf{S} = \frac{c}{4 \pi} (\mathbf{E} \times \mathbf{B})\end{aligned} \hspace{\stretch{1}}(1.10)

This is non-zero only in the region both between the solenoid and the enclosing cylinder (radius $b$) since that’s the only place where both $\mathbf{E}$ and $\mathbf{B}$ are non-zero. That is

\begin{aligned}\mathbf{S} &= \frac{c}{4 \pi} (\mathbf{E} \times \mathbf{B}) \\ &=\frac{c}{4 \pi} \frac{2 Q }{r l} \frac{4 \pi n I}{c} \hat{\mathbf{r}} \times \hat{\mathbf{z}} \\ &= -\frac{2 Q n I}{r l} \hat{\boldsymbol{\phi}}\end{aligned}

(since $\hat{\mathbf{r}} \times \hat{\boldsymbol{\phi}} = \hat{\mathbf{z}}$, so $\hat{\mathbf{z}} \times \hat{\mathbf{r}} = \hat{\boldsymbol{\phi}}$ after cyclic permutation)

A motivational aside: Momentum density.

Suppose ${\left\lvert{\mathbf{E}}\right\rvert} = {\left\lvert{\mathbf{B}}\right\rvert}$, then our Poynting vector is

\begin{aligned}\mathbf{S} = \frac{c}{4 \pi} \mathbf{E} \times \mathbf{B} = \frac{ c \hat{\mathbf{k}}}{4 \pi} \mathbf{E}^2,\end{aligned} \hspace{\stretch{1}}(1.11)

but

\begin{aligned}\mathcal{E} = \text{energy density} = \frac{\mathbf{E}^2 + \mathbf{B}^2}{8 \pi} = \frac{\mathbf{E}^2}{4 \pi},\end{aligned} \hspace{\stretch{1}}(1.12)

so

\begin{aligned}\mathbf{S} = c \hat{\mathbf{k}} \mathcal{E} = \mathbf{v} \mathcal{E}.\end{aligned} \hspace{\stretch{1}}(1.13)

Now recall the between (relativistic) mechanical momentum $\mathbf{p} = \gamma m \mathbf{v}$ and energy $\mathcal{E} = \gamma m c^2$

\begin{aligned}\mathbf{p} = \frac{\mathbf{v}}{c^2} \mathcal{E}.\end{aligned} \hspace{\stretch{1}}(1.14)

This justifies calling the quantity

\begin{aligned}\mathbf{P}_{\text{EM}} = \frac{\mathbf{S}}{c^2},\end{aligned} \hspace{\stretch{1}}(1.15)

the momentum density.

Momentum density of the EM fields.

So we label our scaled Poynting vector the momentum density for the field

\begin{aligned}\mathbf{P}_{\text{EM}} = -\frac{2 Q n I}{c^2 r l} \hat{\boldsymbol{\phi}},\end{aligned} \hspace{\stretch{1}}(1.16)

and can now compute an angular momentum density in the field between the solenoid and the outer cylinder prior to changing the currents

\begin{aligned}\mathbf{L}_{\text{EM}}&= \mathbf{r} \times \mathbf{P}_{\text{EM}} \\ &= r \hat{\mathbf{r}} \times \mathbf{P}_{\text{EM}} \\ \end{aligned}

This gives us

\begin{aligned}\mathbf{L}_{\text{EM}} = -\frac{2 Q n I}{c^2 l} \hat{\mathbf{z}} = \text{constant}.\end{aligned} \hspace{\stretch{1}}(1.17)

Note that this is the angular momentum density in the region between the solenoid and the inner cylinder, between $z = 0$ and $z = l$. Outside of this region, the angular momentum density is zero.

After the current is changed

Induced electric field

When we turn off (or change) $I$, some of the magnetic field $\mathbf{B}$ will be converted into electric field $\mathbf{E}$ according to Faraday’s law

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = - \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.18)

In integral form, utilizing an open surface, this is

\begin{aligned}\int_A (\boldsymbol{\nabla} \times \mathbf{l}) \cdot \hat{\mathbf{n}} dA&=\int_{\partial A} \mathbf{E} \cdot d\mathbf{l} \\ &= - \frac{1}{{c}} \int_A \frac{\partial {\mathbf{B}}}{\partial {t}} \cdot d\mathbf{A} \\ &= - \frac{1}{{c}} \frac{\partial {\Phi_B(t)}}{\partial {t}},\end{aligned}

where we introduce the magnetic flux

\begin{aligned}\Phi_B(t) = \int_A \mathbf{B} \cdot d\mathbf{A}.\end{aligned} \hspace{\stretch{1}}(1.19)

We can utilizing a circular surface cutting directly across the cylinder perpendicular to $\hat{\mathbf{z}}$ of radius $r$. Recall that we have the magnetic field 1.7 only inside the solenoid. So for $r < R$ this flux is

\begin{aligned}\Phi_B(t)&= \int_A \mathbf{B} \cdot d\mathbf{A} \\ &= (\pi r^2) \frac{4 \pi n I(t)}{c}.\end{aligned}

For $r > R$ only the portion of the surface with radius $r \le R$ contributes to the flux

\begin{aligned}\Phi_B(t)&= \int_A \mathbf{B} \cdot d\mathbf{A} \\ &= (\pi R^2) \frac{4 \pi n I(t)}{c}.\end{aligned}

We can now compute the circulation of the electric field

\begin{aligned}\int_{\partial A} \mathbf{E} \cdot d\mathbf{l} = - \frac{1}{{c}} \frac{\partial {\Phi_B(t)}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(1.20)

by taking the derivatives of the magnetic flux. For $r > R$ this is

\begin{aligned}\int_{\partial A} \mathbf{E} \cdot d\mathbf{l}&= (2 \pi r) E \\ &=-(\pi R^2) \frac{4 \pi n \dot{I}(t)}{c^2}.\end{aligned}

This gives us the magnitude of the induced electric field

\begin{aligned}E&= -(\pi R^2) \frac{4 \pi n \dot{I}(t)}{2 \pi r c^2} \\ &= -\frac{2 \pi R^2 n \dot{I}(t)}{r c^2}.\end{aligned}

Similarly for $r < R$ we have

\begin{aligned}E = -\frac{2 \pi r n \dot{I}(t)}{c^2}\end{aligned} \hspace{\stretch{1}}(1.21)

Summarizing we have

\begin{aligned}\mathbf{E} =\left\{\begin{array}{l l}-\frac{2 \pi r n \dot{I}(t)}{c^2} \hat{\boldsymbol{\phi}} & \mbox{Forlatex r R}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.22)

Torque and angular momentum induced by the fields.

Our torque $\mathbf{N} = \mathbf{r} \times \mathbf{F} = d\mathbf{L}/dt$ on the outer cylinder (radius $b$) that is induced by changing the current is

\begin{aligned}\mathbf{N}_b&= (b \hat{\mathbf{r}}) \times (-Q \mathbf{E}_{r = b}) \\ &= b Q \frac{2 \pi R^2 n \dot{I}(t)}{b c^2} \hat{\mathbf{r}} \times \hat{\boldsymbol{\phi}} \\ &= \frac{1}{{c^2}} 2 \pi R^2 n Q \dot{I} \hat{\mathbf{z}}.\end{aligned}

This provides the induced angular momentum on the outer cylinder

\begin{aligned}\mathbf{L}_b&= \int dt \mathbf{N}_b = \frac{ 2 \pi n R^2 Q}{c^2} \int_I^0 \frac{dI}{dt} dt \\ &= -\frac{2 \pi n R^2 Q}{c^2} I.\end{aligned}

This is the angular momentum of $b$ induced by changing the current or changing the magnetic field.

On the inner cylinder we have

\begin{aligned}\mathbf{N}_a&= (a \hat{\mathbf{r}} ) \times (Q \mathbf{E}_{r = a}) \\ &= a Q \left(- \frac{2 \pi}{c} n a \dot{I} \right) \hat{\mathbf{r}} \times \hat{\boldsymbol{\phi}} \\ &= -\frac{2 \pi n a^2 Q \dot{I}}{c^2} \hat{\mathbf{z}}.\end{aligned}

So our induced angular momentum on the inner cylinder is

\begin{aligned}\mathbf{L}_a = \frac{2 \pi n a^2 Q I}{c^2} \hat{\mathbf{z}}.\end{aligned} \hspace{\stretch{1}}(1.23)

The total angular momentum in the system has to be conserved, and we must have

\begin{aligned}\mathbf{L}_a + \mathbf{L}_b = -\frac{2 n I Q}{c^2} \pi (R^2 - a^2) \hat{\mathbf{z}}.\end{aligned} \hspace{\stretch{1}}(1.24)

At the end of the tutorial, this sum was equated with the field angular momentum density $\mathbf{L}_{\text{EM}}$, but this has different dimensions. In fact, observe that the volume in which this angular momentum density is non-zero is the difference between the volume of the solenoid and the inner cylinder

\begin{aligned}V = \pi R^2 l - \pi a^2 l,\end{aligned} \hspace{\stretch{1}}(1.25)

so if we are to integrate the angular momentum density 1.17 over this region we have

\begin{aligned}\int \mathbf{L}_{\text{EM}} dV = -\frac{2 Q n I}{c^2} \pi (R^2 - a^2) \hat{\mathbf{z}}\end{aligned} \hspace{\stretch{1}}(1.26)

which does match with the sum of the mechanical angular momentum densities 1.24 as expected.

References

[2] Wikipedia. Faraday’s law of induction — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 10-March-2011]. http://en.wikipedia.org/w/index.php?title=Faraday\%27s_law_of_induction&oldid=416715237.