Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘energy density’

A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

Advertisements

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | 1 Comment »

PHY452H1S Basic Statistical Mechanics. Lecture 20: Bosons. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 2, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Bosons

In order to maintain a conservation of particles in a Bose condensate as we decrease temperature, we are forced to change the chemical potential to compensate. This is illustrated in fig. 1.1.

Fig 1.1: Chemical potential in Bose condensation region

 

Bose condensatation occurs for T < T_{\mathrm{BEC}}. At this point our number density becomes (except at \mathbf{k} = 0)

\begin{aligned}n(\mathbf{k}) = \frac{1}{{e^{\beta \epsilon_\mathbf{k}} - 1}}.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Except for \mathbf{k} = 0, n(\mathbf{k}) is well defined, and not described by this distribution. We are forced to say that

\begin{aligned}N = N_0 + \sum_{\mathbf{k} \ne 0} n(\mathbf{k}) = N_0 + V\int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{ e^{\beta \epsilon_\mathbf{k}} - 1 }}.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Introducing the density of states, our density is

\begin{aligned}\rho = \rho_0 + \int_0^\infty d\epsilon \frac{N(\epsilon)}{e^{\beta \epsilon} - 1 },\end{aligned} \hspace{\stretch{1}}(1.2.3)

where

\begin{aligned}N(\epsilon) = \frac{1}{{4 \pi^2}} \left( \frac{2m}{\hbar} \right)^{3/2} \epsilon^{1/2}.\end{aligned} \hspace{\stretch{1}}(1.2.4)

We worked out last time that

\begin{aligned}\rho = \rho_0 + \rho \left( \frac{T}{T_{\mathrm{BEC}}} \right)^{3/2},\end{aligned} \hspace{\stretch{1}}(1.2.4)

or

\begin{aligned}\rho_0 = \rho \left( 1 - \left( \frac{T}{T_{\mathrm{BEC}}} \right) ^{3/2} \right).\end{aligned} \hspace{\stretch{1}}(1.2.6)

This is plotted in fig. 1.2.

Fig 1.2: Density variation with temperature for Bosons

 

\begin{aligned}\rho_0 = \frac{N_{\mathbf{k} = 0}}{V}.\end{aligned} \hspace{\stretch{1}}(1.7)

For T \ge T_{\mathrm{BEC}}, we have \rho_0 = 0. This condensation temperature is

\begin{aligned}T_{\mathrm{BEC}} \propto \rho^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.8)

This is plotted in fig. 1.3.

Fig 1.3: Temperature vs pressure demarkation by T_BEC curve

 

There is a line for each density that marks the boundary temperature for which we have or do not have this condensation phenomina where \mathbf{k} = 0 states start filling up.

Specific heat: T < T_{\mathrm{BEC}}

\begin{aligned}\frac{E}{V} &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{ e^{\beta \hbar^2 k^2/2m} - 1}}\frac{\hbar^2 k^2}{2m} \\ &= \int_0^\infty d\epsilon N(\epsilon) \frac{1}{{ e^{\beta \epsilon} - 1 }} \epsilon \\ &\propto \int_0^\infty d\epsilon \frac{\epsilon^{3/2}}{ e^{\beta \epsilon} - 1 } \\ &\propto \left( k_{\mathrm{B}} T \right)^{5/2},\end{aligned} \hspace{\stretch{1}}(1.9)

so that

\begin{aligned}\frac{C}{V} \propto \left( k_{\mathrm{B}} T \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.10)

Compare this to the classical and Fermionic specific heat as plotted in fig. 1.4.

Fig 1.4: Specific heat for Bosons, Fermions, and classical ideal gases

 

One can measure the specific heat in this Bose condensation phenomina for materials such as Helium-4 (spin 0). However, it turns out that Helium-4 is actually quite far from an ideal Bose gas.

Photon gas

A system that is much closer to an ideal Bose gas is that of a gas of photons. To a large extent, photons do not interact with each other. This allows us to calculate black body phenomina and the low temperature (cosmic) background radiation in the universe.

An important distinction between a photon sea and some of these other systems is that the photon number is actually not fixed.

Photon numbers are not “conserved”.

If a photon interacts with an atom, it can impart energy and disappear. An excited atom can emit a photon and change its energy level. In a thermodynamic system we can generally expect that introducing heat will generate more photons, whereas a cold sink will tend to generate fewer photons.

We have a few special details that distinguish photons that we’ll have to consider.

  1. spin 1.
  2. massless, moving at the speed of light.
  3. have two polarization states.

Because we do not have a constraint on the number of particles, we essentially have no chemical potential, even in the grand canonical scheme.

Writing

\begin{aligned}\lambda = \left\{\begin{array}{l l}+1 & \quad \mbox{Right circular polarization} \\ -1 & \quad \mbox{Left circular polarization}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.11)

Our number density, since we have no chemical potential, is of the form

\begin{aligned}n_{\mathbf{k}, \lambda}= \frac{1}{{e^{\beta \epsilon_{\mathbf{k}, \lambda}} - 1 }},\end{aligned} \hspace{\stretch{1}}(1.12)

Observe that the average number of photons in this system is temperature dependent. Because this chemical potential is not there, it can be quite easy to work out a number of the thermodynamic results.

Photon average energy density

We’ll now calculate the average energy density of the photons. The energy of a single photon is

\begin{aligned}\epsilon_{\mathbf{k}, \lambda} = \hbar c k = \hbar \omega,\end{aligned} \hspace{\stretch{1}}(1.2.13)

so that the average energy density is

\begin{aligned}\frac{E}{V} &= \sum_{\mathbf{k}, \lambda} \frac{1}{{ e^{ \beta \epsilon_\mathbf{k}} - 1}} \epsilon_\mathbf{k}\rightarrow\underbrace{2}_{\text{number of polarizations}}\int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\frac{ \hbar c k}{ e^{ \beta \epsilon_\mathbf{k}} - 1} \\ &= 2 \int_0^\infty d\epsilon \underbrace{\frac{1}{{(2 \pi)^3}} 4 \pi \frac{\epsilon^2}{(\hbar c)^3} }_{\text{Photon density of states}}\frac{\epsilon}{e^{\beta \epsilon} - 1} \\ &= \frac{1}{{\pi^2}} \frac{1}{{ (\hbar c)^3 }} \int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1}\end{aligned} \hspace{\stretch{1}}(1.2.13)

Mathematica tells us that this integral is

\begin{aligned}\int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1} =\frac{\pi ^4}{15 \beta ^4},\end{aligned} \hspace{\stretch{1}}(1.2.13)

for an end result of

\begin{aligned}\frac{E}{V} =\frac{\pi^2}{15} \frac{1}{{(\hbar c)^3}} \left( k_{\mathrm{B}} T \right)^4.\end{aligned} \hspace{\stretch{1}}(1.2.13)

Phonons and other systems

There is a very similar phenomina in matter. We can discuss lattice vibrations in a solid. These are called phonon modes, and will have the same distribution function where the only difference is that the speed of light is replaced by the speed of the sound wave in the solid. Once we understand the photon system, we are able to look at other Bose distributions such as these phonon systems. We’ll touch on this very briefly next time.

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , | Leave a Comment »

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?

For $latex r b$ we have \mathbf{E} = 0 since there is no charge enclosed by any Gaussian surface that we choose.

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{For latex 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

[1] D. Fleisch. A Student’s Guide to Maxwell’s Equations. Cambridge University Press, 2007. “http://www4.wittenberg.edu/maxwell/index.html“.

[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.

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , | Leave a Comment »

Energy and momentum for Complex electric and magnetic field phasors.

Posted by peeterjoot on December 15, 2009

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

Motivation.

In [1] a complex phasor representations of the electric and magnetic fields is used

\begin{aligned}\mathbf{E} &= \boldsymbol{\mathcal{E}} e^{-i\omega t} \\ \mathbf{B} &= \mathbf{B} e^{-i\omega t}.\end{aligned} \quad\quad\quad(1)

Here the vectors \boldsymbol{\mathcal{E}} and \mathbf{B} are allowed to take on complex values. Jackson uses the real part of these complex vectors as the true fields, so one is really interested in just these quantities

\begin{aligned}\text{Real} \mathbf{E} &= \boldsymbol{\mathcal{E}}_r \cos(\omega t) + \boldsymbol{\mathcal{E}}_i \sin(\omega t) \\ \text{Real} \mathbf{B} &= \mathbf{B}_r \cos(\omega t) + \mathbf{B}_i \sin(\omega t),\end{aligned} \quad\quad\quad(3)

but carry the whole thing in manipulations to make things simpler. It is stated that the energy for such complex vector fields takes the form (ignoring constant scaling factors and units)

\begin{aligned}\text{Energy} \propto \mathbf{E} \cdot {\mathbf{E}}^{*} + \mathbf{B} \cdot {\mathbf{B}}^{*}.\end{aligned} \quad\quad\quad(5)

In some ways this is an obvious generalization. Less obvious is how this and the Poynting vector are related in their corresponding conservation relationships.

Here I explore this, employing a Geometric Algebra representation of the energy momentum tensor based on the real field representation found in [2]. Given the complex valued fields and a requirement that both the real and imaginary parts of the field satisfy Maxwell’s equation, it should be possible to derive the conservation relationship between the energy density and Poynting vector from first principles.

Review of GA formalism for real fields.

In SI units the Geometric algebra form of Maxwell’s equation is

\begin{aligned}\nabla F &= J/\epsilon_0 c,\end{aligned} \quad\quad\quad(6)

where one has for the symbols

\begin{aligned}F &= \mathbf{E} + c I \mathbf{B} \\ I &= \gamma_0 \gamma_1 \gamma_2 \gamma_3 \\ \mathbf{E} &= E^k \gamma_k \gamma_0  \\ \mathbf{B} &= B^k \gamma_k \gamma_0  \\ (\gamma^0)^2 &= -(\gamma^k)^2 = 1 \\ \gamma^\mu \cdot \gamma_\nu &= {\delta^\mu}_\nu \\ J &= c \rho \gamma_0 + J^k \gamma_k \\ \nabla &= \gamma^\mu \partial_\mu = \gamma^\mu {\partial {}}/{\partial {x^\mu}}.\end{aligned} \quad\quad\quad(7)

The symmetric electrodynamic energy momentum tensor for real fields \mathbf{E} and \mathbf{B} is

\begin{aligned}T(a) &= \frac{-\epsilon_0}{2} F a F = \frac{\epsilon_0}{2} F a \tilde{F}.\end{aligned} \quad\quad\quad(15)

It may not be obvious that this is in fact a four vector, but this can be seen since it can only have grade one and three components, and also equals its reverse implying that the grade three terms are all zero. To illustrate this explicitly consider the components of T^{\mu 0}

\begin{aligned}\frac{2}{\epsilon_0} T(\gamma^0) &= -(\mathbf{E} + c I \mathbf{B}) \gamma^0 (\mathbf{E} + c I \mathbf{B}) \\ &= (\mathbf{E} + c I \mathbf{B}) (\mathbf{E} - c I \mathbf{B}) \gamma^0 \\ &= (\mathbf{E}^2 + c^2 \mathbf{B}^2 + c I (\mathbf{B} \mathbf{E} - \mathbf{E} \mathbf{B})) \gamma^0 \\ &= (\mathbf{E}^2 + c^2 \mathbf{B}^2) \gamma^0 + 2 c I ( \mathbf{B} \wedge \mathbf{E} ) \gamma^0 \\ &= (\mathbf{E}^2 + c^2 \mathbf{B}^2) \gamma^0 + 2 c ( \mathbf{E} \times \mathbf{B} ) \gamma^0 \\ \end{aligned}

Our result is a four vector in the Dirac basis as expected

\begin{aligned}T(\gamma^0) &= T^{\mu 0} \gamma_\mu \\ T^{0 0} &= \frac{\epsilon_0}{2} (\mathbf{E}^2 + c^2 \mathbf{B}^2) \\ T^{k 0} &= c \epsilon_0 (\mathbf{E} \times \mathbf{B})_k \end{aligned} \quad\quad\quad(16)

Similar expansions are possible for the general tensor components T^{\mu\nu} but lets defer this more general expansion until considering complex valued fields. The main point here is to remind oneself how to express the energy momentum tensor in a fashion that is natural in a GA context. We also know that one has a conservation relationship associated with the divergence of this tensor \nabla \cdot T(a) (ie. \partial_\mu T^{\mu\nu}), and want to rederive this relationship after guessing what form the GA expression for the energy momentum tensor takes when one allows the field vectors to take complex values.

Computing the conservation relationship for complex field vectors.

As in 5, if one wants

\begin{aligned}T^{0 0} \propto \mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*},\end{aligned} \quad\quad\quad(19)

it is reasonable to assume that our energy momentum tensor will take the form

\begin{aligned}T(a) &= \frac{\epsilon_0}{4} \left( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \right)= \frac{\epsilon_0}{2} \text{Real} \left( {{F}}^{*} a \tilde{F} \right)\end{aligned} \quad\quad\quad(20)

For real vector fields this reduces to the previous results and should produce the desired mix of real and imaginary dot products for the energy density term of the tensor. This is also a real four vector even when the field is complex, so the energy density and power density terms will all be real valued, which seems desirable.

Expanding the tensor. Easy parts.

As with real fields expansion of T(a) in terms of \mathbf{E} and \mathbf{B} is simplest for a = \gamma^0. Let’s start with that.

\begin{aligned}\frac{4}{\epsilon_0} T(\gamma^0) \gamma_0&=-({\mathbf{E}}^{*} + c I {\mathbf{B}}^{*} )\gamma^0 (\mathbf{E} + c I \mathbf{B}) \gamma_0-(\mathbf{E} + c I \mathbf{B} )\gamma^0 ({\mathbf{E}}^{*} + c I {\mathbf{B}}^{*} ) \gamma_0 \\ &=({\mathbf{E}}^{*} + c I {\mathbf{B}}^{*} ) (\mathbf{E} - c I \mathbf{B}) +(\mathbf{E} + c I \mathbf{B} ) ({\mathbf{E}}^{*} - c I {\mathbf{B}}^{*} ) \\ &={\mathbf{E}}^{*} \mathbf{E} + \mathbf{E} {\mathbf{E}}^{*} + c^2 ({\mathbf{B}}^{*} \mathbf{B} + \mathbf{B} {\mathbf{B}}^{*} ) + c I ( {\mathbf{B}}^{*} \mathbf{E} - {\mathbf{E}}^{*} \mathbf{B} + \mathbf{B} {\mathbf{E}}^{*} - \mathbf{E} {\mathbf{B}}^{*} ) \\ &=2 \mathbf{E} \cdot {\mathbf{E}}^{*} + 2 c^2 \mathbf{B} \cdot {\mathbf{B}}^{*}+ 2 c ( \mathbf{E} \times {\mathbf{B}}^{*} + {\mathbf{E}}^{*} \times \mathbf{B} ).\end{aligned}

This gives

\begin{aligned}T(\gamma^0) &=\frac{\epsilon_0}{2} \left( \mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*} \right) \gamma^0+ \frac{\epsilon_0 c}{2} ( \mathbf{E} \times {\mathbf{B}}^{*} + {\mathbf{E}}^{*} \times \mathbf{B} ) \gamma^0\end{aligned} \quad\quad\quad(21)

The sum of {{F}}^{*} a F and its conjugate has produced the desired energy density expression. An implication of this is that one can form and take real parts of a complex Poynting vector \mathbf{S} \propto \mathbf{E} \times {\mathbf{B}}^{*} to calculate the momentum density. This is stated but not demonstrated in Jackson, perhaps considered too obvious or messy to derive.

Observe that the a choice to work with complex valued vector fields gives a nice consistency, and one has the same factor of 1/2 in both the energy and momentum terms. While the energy term is obviously real, the momentum terms can be written in an explicitly real notation as well since one has a quantity plus its conjugate. Using a more conventional four vector notation (omitting the explicit Dirac basis vectors), one can write this out as a strictly real quantity.

\begin{aligned}T(\gamma^0) &=\epsilon_0 \Bigl( \frac{1}{{2}}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*}),c \text{Real}( \mathbf{E} \times {\mathbf{B}}^{*} ) \Bigr)\end{aligned} \quad\quad\quad(22)

Observe that when the vector fields are restricted to real quantities, the conjugate and real part operators can be dropped and the real vector field result 16 is recovered.

Expanding the tensor. Messier parts.

I intended here to compute T(\gamma^k), and my starting point was a decomposition of the field vectors into components that anticommute or commute with \gamma^k

\begin{aligned}\mathbf{E} &= \mathbf{E}_\parallel + \mathbf{E}_\perp \\ \mathbf{B} &= \mathbf{B}_\parallel + \mathbf{B}_\perp.\end{aligned} \quad\quad\quad(23)

The components parallel to the spatial vector \sigma_k = \gamma_k \gamma_0 are anticommuting \gamma^k \mathbf{E}_\parallel = -\mathbf{E}_\parallel \gamma^k, whereas the perpendicular components commute \gamma^k \mathbf{E}_\perp = \mathbf{E}_\perp \gamma^k. The expansion of the tensor products is then

\begin{aligned}({{F}}^{*} \gamma^k \tilde{F} + \tilde{F} \gamma^k {{F}}^{*}) \gamma_k&= - ({\mathbf{E}}^{*} + I c {\mathbf{B}}^{*}) \gamma^k ( \mathbf{E}_\parallel + \mathbf{E}_\perp + c I ( \mathbf{B}_\parallel + \mathbf{B}_\perp ) ) \gamma_k \\ &- (\mathbf{E} + I c \mathbf{B}) \gamma^k ( {\mathbf{E}_\parallel}^{*} + {\mathbf{E}_\perp}^{*} + c I ( {\mathbf{B}_\parallel}^{*} + {\mathbf{B}_\perp}^{*} ) ) \gamma_k \\ &=  ({\mathbf{E}}^{*} + I c {\mathbf{B}}^{*}) ( \mathbf{E}_\parallel - \mathbf{E}_\perp + c I ( -\mathbf{B}_\parallel + \mathbf{B}_\perp ) ) \\ &+ (\mathbf{E} + I c \mathbf{B}) ( {\mathbf{E}_\parallel}^{*} - {\mathbf{E}_\perp}^{*} + c I ( -{\mathbf{B}_\parallel}^{*} + {\mathbf{B}_\perp}^{*} ) ) \\ \end{aligned}

This isn’t particularly pretty to expand out. I did attempt it, but my result looked wrong. For the application I have in mind I do not actually need anything more than T^{\mu 0}, so rather than show something wrong, I’ll just omit it (at least for now).

Calculating the divergence.

Working with 20, let’s calculate the divergence and see what one finds for the corresponding conservation relationship.

\begin{aligned}\frac{4}{\epsilon_0} \nabla \cdot T(a) &=\left\langle{{ \nabla ( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} )}}\right\rangle \\ &=-\left\langle{{ F \stackrel{ \leftrightarrow }\nabla {{F}}^{*} a + {{F}}^{*} \stackrel{ \leftrightarrow }\nabla F a }}\right\rangle \\ &=-{\left\langle{{ F \stackrel{ \leftrightarrow }\nabla {{F}}^{*} + {{F}}^{*} \stackrel{ \leftrightarrow }\nabla F }}\right\rangle}_{1} \cdot a \\ &=-{\left\langle{{ F \stackrel{ \rightarrow }\nabla {{F}}^{*} +F \stackrel{ \leftarrow }\nabla {{F}}^{*} + {{F}}^{*} \stackrel{ \leftarrow }\nabla F+ {{F}}^{*} \stackrel{ \rightarrow }\nabla F}}\right\rangle}_{1} \cdot a \\ &=-\frac{1}{{\epsilon_0 c}} {\left\langle{{ F {{J}}^{*} - J {{F}}^{*} - {{J}}^{*} F+ {{F}}^{*} J}}\right\rangle}_{1} \cdot a \\ &= \frac{2}{\epsilon_0 c} a \cdot ( J \cdot {{F}}^{*} + {{J}}^{*} \cdot F) \\ &= \frac{4}{\epsilon_0 c} a \cdot \text{Real} ( J \cdot {{F}}^{*} ).\end{aligned}

We have then for the divergence

\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(25)

Lets write out J \cdot {{F}}^{*} in the (stationary) observer frame where J = (c\rho + \mathbf{J}) \gamma_0. This is

\begin{aligned}J \cdot {{F}}^{*} &={\left\langle{{ (c\rho + \mathbf{J}) \gamma_0 ( {\mathbf{E}}^{*} + I c {\mathbf{B}}^{*} ) }}\right\rangle}_{1} \\ &=- (\mathbf{J} \cdot {\mathbf{E}}^{*} ) \gamma_0- c \left( \rho {\mathbf{E}}^{*} + \mathbf{J} \times {\mathbf{B}}^{*}\right) \gamma_0\end{aligned}

Writing out the four divergence relationships in full one has

\begin{aligned}\nabla \cdot T(\gamma^0) &= - \frac{1}{{ c }} \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) \\ \nabla \cdot T(\gamma^k) &= - \text{Real} \left( \rho {{(E^k)}}^{*} + (\mathbf{J} \times {\mathbf{B}}^{*})_k \right)\end{aligned} \quad\quad\quad(26)

Just as in the real field case one has a nice relativistic split into energy density and force (momentum change) components, but one has to take real parts and conjugate half the terms appropriately when one has complex fields.

Combining the divergence relation for T(\gamma^0) with 22 the conservation relation for this subset of the energy momentum tensor becomes

\begin{aligned}\frac{1}{{c}} \frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ c \epsilon_0 \text{Real} \boldsymbol{\nabla} \cdot (\mathbf{E} \times {\mathbf{B}}^{*} )=- \frac{1}{{c}} \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) \end{aligned} \quad\quad\quad(28)

Or

\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \text{Real} \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0\end{aligned} \quad\quad\quad(29)

It is this last term that puts some meaning behind Jackson’s treatment since we now know how the energy and momentum are related as a four vector quantity in this complex formalism.

While I’ve used geometric algebra to get to this final result, I would be interested to compare how the intermediate mess compares with the same complex field vector result obtained via traditional vector techniques. I am sure I could try this myself, but am not interested enough to attempt it.

Instead, now that this result is obtained, proceeding on to application is now possible. My intention is to try the vacuum electromagnetic energy density example from [3] using complex exponential Fourier series instead of the doubled sum of sines and cosines that Bohm used.

References

[1] JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

Posted in Math and Physics Learning. | Tagged: , , , , , | 3 Comments »