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# Posts Tagged ‘three vector’

## Energy term of the Lorentz force equation.

Posted by peeterjoot on February 8, 2011

# Motivation.

In class this week, the Lorentz force was derived from an action (the simplest Lorentz invariant, gauge invariant, action that could be constructed)

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds A^i u_i.\end{aligned} \hspace{\stretch{1}}(1.1)

We end up with the familiar equation, with the exception that the momentum includes the relativistically required gamma factor

\begin{aligned}\frac{d (\gamma m \mathbf{v})}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(1.2)

I asked what the energy term of this equation would be and was answered that we would get to it, and it could be obtained by a four vector minimization of the action which produces the Lorentz force equation of the following form

\begin{aligned}\frac{du^i}{d\tau} \propto e F^{ij} u_j.\end{aligned} \hspace{\stretch{1}}(1.3)

Let’s see if we can work this out without the four-vector approach, using the action expressed with an explicit space time split, then also work it out in the four vector form and compare as a consistency check.

# Three vector approach.

## The Lorentz force derivation.

For completeness, let’s work out the Lorentz force equation from the action 1.1. Parameterizing by time we have

\begin{aligned}S &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \gamma \left( 1, \frac{1}{{c}} \mathbf{v}\right) \cdot (\phi, \mathbf{A}) \\ &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \int dt \left( \phi - \frac{1}{{c}} \mathbf{A} \cdot \mathbf{v} \right)\end{aligned}

Our Lagrangian is therefore

\begin{aligned}\mathcal{L}(\mathbf{x}, \mathbf{v}, t) = -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi(\mathbf{x}, t) + \frac{e}{c} \mathbf{A}(\mathbf{x}, t) \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.4)

We can calculate our conjugate momentum easily enough

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.5)

and for the gradient portion of the Euler-Lagrange equations we have

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} = -e \boldsymbol{\nabla} \phi + e \boldsymbol{\nabla} \left( \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right).\end{aligned} \hspace{\stretch{1}}(2.6)

Utilizing the convective derivative (i.e. chain rule in fancy clothes)

\begin{aligned}\frac{d}{dt} = \mathbf{v} \cdot \boldsymbol{\nabla} + \frac{\partial {}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(2.7)

This gives us

\begin{aligned}-e \boldsymbol{\nabla} \phi + e \boldsymbol{\nabla} \left( \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right) = \frac{d(\gamma m \mathbf{v})}{dt} + \frac{e}{c} (\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}+ \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(2.8)

and a final bit of rearranging gives us

\begin{aligned}\frac{d(\gamma m \mathbf{v})}{dt} =e \left( -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}}\right)+ \frac{e}{c} \left( \boldsymbol{\nabla} \left( \mathbf{v} \cdot \mathbf{A} \right) - (\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}\right).\end{aligned} \hspace{\stretch{1}}(2.9)

The first set of derivatives we identify with the electric field $\mathbf{E}$. For the second, utilizing the vector triple product identity [1]

\begin{aligned}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c},\end{aligned} \hspace{\stretch{1}}(2.10)

we recognize as related to the magnetic field $\mathbf{v} \times \mathbf{B} = \mathbf{v} \times (\boldsymbol{\nabla} \times \mathbf{A})$.

## The power (energy) term.

When we start with an action explicitly constructed with Lorentz invariance as a requirement, it is somewhat odd to end up with a result that has only the spatial vector portion of what should logically be a four vector result. We have an equation for the particle momentum, but not one for the energy. In tutorial Simon provided the hint of how to approach this, and asked if we had calculated the Hamiltonian for the Lorentz force. We had only calculated the Hamiltonian for the free particle.

Considering this, we can only actually calculate a Hamiltonian for the case where $\phi(\mathbf{x}, t) = \phi(\mathbf{x})$ and $\mathbf{A}(\mathbf{x}, t) = \mathbf{A}(\mathbf{x})$, because when the potentials have any sort of time dependence we do not have a Lagrangian that is invariant under time translation. Returning to the derivation of the Hamiltonian conservation equation, we see that we must modify the argument slightly when there is a time dependence and get instead

\begin{aligned}\frac{d}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathcal{L} - \mathcal{L} \right) + \frac{\partial {\mathcal{L}}}{\partial {t}} = 0.\end{aligned} \hspace{\stretch{1}}(2.11)

Only when there is no time dependence in the Lagrangian, do we have our conserved quantity, what we label as energy, or Hamiltonian.

From 2.5, we have

\begin{aligned}0 &= \frac{d}{dt} \left( \left( \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A} \right) \cdot \mathbf{v} +m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + e \phi - \frac{e}{c} \mathbf{A} \cdot \mathbf{v}\right) - e \frac{\partial {\phi}}{\partial {t}} + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ \end{aligned}

Our $\mathbf{A} \cdot \mathbf{v}$ terms cancel, and we can combine the $\gamma$ and $\gamma^{-1}$ terms, then apply the convective derivative again

\begin{aligned}\frac{d}{dt} \left( \gamma m c^2 \right) &= - e \left( \mathbf{v} \cdot \boldsymbol{\nabla} + \frac{\partial {}}{\partial {t}} \right) \phi + e \frac{\partial {\phi}}{\partial {t}} - \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ &= - e \mathbf{v} \cdot \boldsymbol{\nabla} \phi - \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ &= + e \mathbf{v} \cdot \left( - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right).\end{aligned}

This is just

\begin{aligned}\frac{d}{dt} \left( \gamma m c^2 \right) = e \mathbf{v} \cdot \mathbf{E},\end{aligned} \hspace{\stretch{1}}(2.12)

and we find the rate of change of energy term of our four momentum equation

\begin{aligned}\frac{d}{dt}\left( \frac{E}{c}, \mathbf{p}\right) = e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.13)

Specified explicilty, this is

\begin{aligned}\frac{d}{dt}\left( \gamma m \left( c, \mathbf{v} \right) \right)= e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.14)

While this was the result I was looking for, once written it now stands out as incomplete relativistically. We have an equation that specifies the time derivative of a four vector. What about the spatial derivatives? We really ought to have a rank two tensor result, and not a four vector result relating the fields and the energy and momentum of the particle. The Lorentz force equation, even when expanded to four vector form, does not seem complete relativistically.

With $u^i = dx^i/ds$, we can rewrite 2.14 as

\begin{aligned}\partial_0 (\gamma m u^i) = e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.15)

If we were to vary the action with respect to a spatial coordinate instead of time, we should end up with a similar equation of the form $\partial_i (\gamma m u^i) = ?$. Having been pointed at the explicitly invariant result, I wonder if those equations are independent. Let’s defer exploring this, until at least after calculating the result using a four vector form of the action.

# Four vector approach.

## The Lorentz force derivation from invariant action.

We can rewrite our action, parameterizing with proper time. This is

\begin{aligned}S = -m c^2 \int d\tau \sqrt{ \frac{dx^i}{d\tau} \frac{dx_i}{d\tau} }- \frac{e}{c} \int d\tau A_i \frac{dx^i}{d\tau}\end{aligned} \hspace{\stretch{1}}(3.16)

Writing $\dot{x}^i = dx^i/d\tau$, our Lagrangian is then

\begin{aligned}\mathcal{L}(x^i, \dot{x^i}, \tau)= -m c^2 \sqrt{ \dot{x}^i \dot{x}_i }- \frac{e}{c} A_i \dot{x}^i\end{aligned} \hspace{\stretch{1}}(3.17)

The Euler-Lagrange equations take the form

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^i}} = \frac{d}{d\tau} \frac{\partial {\mathcal{L}}}{\partial {\dot{x}^i}} .\end{aligned} \hspace{\stretch{1}}(3.18)

Our gradient and conjugate momentum are

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^i}} &= - \frac{e}{c} \frac{\partial {A_j}}{\partial {x^i}} \dot{x}^j \\ \frac{\partial {\mathcal{L}}}{\partial {\dot{x}^i}} &= -m \dot{x}_i - \frac{e}{c} A_i.\end{aligned} \hspace{\stretch{1}}(3.19)

With our convective derivative taking the form

\begin{aligned}\frac{d}{d\tau} = \dot{x}^i \frac{\partial {}}{\partial {x^i}},\end{aligned} \hspace{\stretch{1}}(3.21)

we have

\begin{aligned}m \frac{d^2 x_i}{d\tau^2} &= \frac{e}{c} \frac{\partial {A_j}}{\partial {x^i}} \dot{x}^j- \frac{e}{c} \dot{x}^j \frac{\partial {A_i}}{\partial {x^j}} \\ &=\frac{e}{c} \dot{x}^j \left( \frac{\partial {A_j}}{\partial {x^i}} -\frac{\partial {A_i}}{\partial {x^j}} \right) \\ &=\frac{e}{c} \dot{x}^j \left( \partial_i A_j - \partial_j A_i\right) \\ &=\frac{e}{c} \dot{x}^j F_{ij}\end{aligned}

Our Prof wrote this with indexes raised and lowered respectively

\begin{aligned}m \frac{d^2 x^i}{d\tau^2} = \frac{e}{c} F^{ij} \dot{x}_j .\end{aligned} \hspace{\stretch{1}}(3.22)

Following the text [2] he also writes $u^i = dx^i/ds = (1/c) dx^i/d\tau$, and in that form we have

\begin{aligned}\frac{d (m c u^i)}{ds} = \frac{e}{c} F^{ij} u_j.\end{aligned} \hspace{\stretch{1}}(3.23)

## Expressed explicitly in terms of the three vector fields.

### The power term.

From 3.23, lets extract the $i=0$ term, relating the rate of change of energy to the field and particle velocity. With

\begin{aligned}\frac{d{{}}}{d\tau} = \frac{dt}{d\tau} \frac{d}{dt} = \gamma \frac{d{{}}}{dt},\end{aligned} \hspace{\stretch{1}}(3.24)

we have

\begin{aligned}\frac{d{{(m \gamma \frac{dx^i}{dt})}}}{dt} = \frac{e}{c} F^{ij} \frac{d{{x_j}}}{dt}.\end{aligned} \hspace{\stretch{1}}(3.25)

For $i=0$ we have

\begin{aligned}F^{0j} \frac{d{{x_j}}}{dt} = -F^{0\alpha} \frac{d{{x^\alpha}}}{dt} \end{aligned} \hspace{\stretch{1}}(3.26)

That component of the field is

\begin{aligned}F^{\alpha 0} &=\partial^\alpha A^0 - \partial^0 A^\alpha \\ &=-\frac{\partial {\phi}}{\partial {x^\alpha}} - \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} \\ &= \left( -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right)^\alpha.\end{aligned}

This verifies the result obtained with considerably more difficulty, using the Hamiltonian like conservation relation obtained for a time translation of a time dependent Lagrangian

\begin{aligned}\frac{d{{(m \gamma c^2 )}}}{dt} = e \mathbf{E} \cdot \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(3.27)

### The Lorentz force terms.

Let’s also verify the signs for the $i > 0$ terms. For those we have

\begin{aligned}\frac{d{{(m \gamma \frac{dx^\alpha}{dt})}}}{dt} &= \frac{e}{c} F^{\alpha j} \frac{d{{x_j}}}{dt} \\ &= \frac{e}{c} F^{\alpha 0} \frac{d{{x_0}}}{dt}+\frac{e}{c} F^{\alpha \beta} \frac{d{{x_\beta}}}{dt} \\ &= e E^\alpha- \sum_{\alpha \beta} \frac{e}{c} \left( \partial^\alpha A^\beta - \partial^\beta A^\alpha\right)v^\beta \\ \end{aligned}

Since we have only spatial indexes left, lets be sloppy and imply summation over all repeated indexes, even if unmatched upper and lower. This leaves us with

\begin{aligned}-\left( \partial^\alpha A^\beta - \partial^\beta A^\alpha \right) v^\beta &=\left( \partial_\alpha A^\beta - \partial_\beta A^\alpha \right) v^\beta \\ &=\epsilon_{\alpha \beta \gamma} B^\gamma\end{aligned}

With the $v^\beta$ contraction we have

\begin{aligned}\epsilon_{\alpha \beta \gamma} B^\gamma v^\beta = (\mathbf{v} \times \mathbf{B})^\alpha,\end{aligned} \hspace{\stretch{1}}(3.28)

leaving our first result obtained by the time parameterization of the Lagrangian

\begin{aligned}\frac{d{{(m \gamma \mathbf{v})}}}{dt} = e \left(\mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(3.29)

This now has a nice symmetrical form. It’s slightly disappointing not to have a rank two tensor on the LHS like we have with the symmetric stress tensor with Poynting Vector and energy and other similar terms that relates field energy and momentum with $\mathbf{E} \cdot \mathbf{J}$ and the charge density equivalents of the Lorentz force equation. Is there such a symmetric relationship for particles too?

# References

[1] Wikipedia. Triple product — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 7-February-2011]. http://en.wikipedia.org/w/index.php?title=Triple_product&oldid=407455209.

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

## PHY450H1S. Relativistic Electrodynamics Tutorial 1 (TA: Simon Freedman).

Posted by peeterjoot on January 21, 2011

# Worked question.

The TA blasted through a problem from Hartle [1], section 5.17 (all the while apologizing for going so slow). I’m going to have to look these notes over carefully to figure out what on Earth he was doing.

At one point he asked if anybody was completely lost. Nobody said yes, but given the class title, I had the urge to say “No, just relatively lost”.

\paragraph{Q:}
In a source’s rest frame $S$ emits radiation isotropically with a frequency $\omega$ with number flux $f(\text{photons}/\text{cm}^2 s)$. Moves along x’-axis with speed $V$ in an observer frame ($O$). What does the energy flux in $O$ look like?

## A brief intro with four vectors

A 3-vector:

\begin{aligned}\mathbf{a} &= (a_x, a_y, a_z) = (a^1, a^2, a^3) \\ \mathbf{b} &= (b_x, b_y, b_z) = (b^1, b^2, b^3)\end{aligned} \hspace{\stretch{1}}(1.1)

For this we have the dot product

\begin{aligned}\mathbf{a} \cdot \mathbf{b} = \sum_{\alpha=1}^3 a^\alpha b^\alpha\end{aligned} \hspace{\stretch{1}}(1.3)

Greek letters in this course (opposite to everybody else in the world, because of Landau and Lifshitz) run from 1 to 3, whereas roman letters run through the set $\{0,1,2,3\}$.

We want to put space and time on an equal footing and form the composite quantity (four vector)

\begin{aligned}x^i = (ct, \mathbf{r}) = (x^0, x^1, x^2, x^3),\end{aligned} \hspace{\stretch{1}}(1.4)

where

\begin{aligned}x^0 &= ct \\ x^1 &= x \\ x^2 &= y \\ x^3 &= z.\end{aligned} \hspace{\stretch{1}}(1.5)

It will also be convenient to drop indexes when referring to all the components of a four vector and we will use lower or upper case non-bold letters to represent such four vectors. For example

\begin{aligned}X = (ct, \mathbf{r}),\end{aligned} \hspace{\stretch{1}}(1.9)

or

\begin{aligned}v = \gamma \left(c, \mathbf{v} \right).\end{aligned} \hspace{\stretch{1}}(1.10)

Three vectors will be represented as letters with over arrows $\vec{a}$ or (in text) bold face $\mathbf{a}$.

Recall that the squared spacetime interval between two events $X_1$ and $X_2$ is defined as

\begin{aligned}{S_{X_1, X_2}}^2 = (ct_1 - c t_2)^2 - (\mathbf{x}_1 - \mathbf{x}_2)^2.\end{aligned} \hspace{\stretch{1}}(1.11)

In particular, with one of these zero, we have an operator which takes a single four vector and spits out a scalar, measuring a “distance” from the origin

\begin{aligned}s^2 = (ct)^2 - \mathbf{r}^2.\end{aligned} \hspace{\stretch{1}}(1.12)

This motivates the introduction of a dot product for our four vector space.

\begin{aligned}X \cdot X = (ct)^2 - \mathbf{r}^2 = (x^0)^2 - \sum_{\alpha=1}^3 (x^\alpha)^2\end{aligned} \hspace{\stretch{1}}(1.13)

Utilizing the spacetime dot product of 1.13 we have for the dot product of the difference between two events

\begin{aligned}(X - Y) \cdot (X - Y)&=(x^0 - y^0)^2 - \sum_{\alpha =1}^3 (x^\alpha - y^\alpha)^2 \\ &=X \cdot X + Y \cdot Y - 2 x^0 y^0 + 2 \sum_{\alpha =1}^3 x^\alpha y^\alpha.\end{aligned}

From this, assuming our dot product 1.13 is both linear and symmetric, we have for any pair of spacetime events

\begin{aligned}X \cdot Y = x^0 y^0 - \sum_{\alpha =1}^3 x^\alpha y^\alpha.\end{aligned} \hspace{\stretch{1}}(1.14)

How do our four vectors transform? This is really just a notational issue, since this has already been discussed. In this new notation we have

\begin{aligned}{x^0}' &= ct' = \gamma ( ct - \beta x) = \gamma ( x^0 - \beta x^1 ) \\ {x^1}' &= x' = \gamma ( x - \beta ct ) = \gamma ( x^1 - \beta x^0 ) \\ {x^2}' &= x^2 \\ {x^3}' &= x^3\end{aligned} \hspace{\stretch{1}}(1.15)

where $\beta = V/c$, and $\gamma^{-2} = 1 - \beta^2$.

In order to put some structure to this, it can be helpful to express this dot product as a quadratic form. We write

\begin{aligned}A \cdot B = \begin{bmatrix}a^0 & \mathbf{a}^\text{T} \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\begin{bmatrix}b^0 \\ \mathbf{b}\end{bmatrix}= A^\text{T} G B.\end{aligned} \hspace{\stretch{1}}(1.19)

We can write our Lorentz boost as a matrix

\begin{aligned}\begin{bmatrix}\gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.20)

so that the dot product between two transformed four vectors takes the form

\begin{aligned}A' \cdot B' = A^\text{T} O^\text{T} G O B\end{aligned} \hspace{\stretch{1}}(1.21)

## Back to the problem.

We will work in momentum space, where we have

\begin{aligned}p^i &= (p^0, \mathbf{p}) = \left( \frac{E}{c}, \mathbf{p}\right) \\ p^2 &= \frac{E^2}{c^2} -\mathbf{p}^2 \\ \mathbf{p} &= \hbar \mathbf{k} \\ E &= \hbar \omega \\ p^i &= \hbar k^i \\ k^i &= \left(\frac{\omega}{c}, \mathbf{k}\right)\end{aligned} \hspace{\stretch{1}}(1.22)

### Justifying this.

Now, the TA blurted all this out. We know some of it from the QM context, and if we’ve been reading ahead know a bit of this from our text [2] (the energy momentum four vector relationships). Let’s go back to the classical electromagnetism and recall what we know about the relation of frequency and wave numbers for continuous fields. We want solutions to Maxwell’s equation in vacuum and can show that such solution also implies that our fields obey a wave equation

\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \Psi}{\partial t^2} - \boldsymbol{\nabla}^2 \Psi = 0,\end{aligned} \hspace{\stretch{1}}(1.28)

where $\Psi$ is one of $\mathbf{E}$ or $\mathbf{B}$. We have other constraints imposed on the solutions by Maxwell’s equations, but require that they at least obey 1.28 in addition to these constraints.

With application of a spatial Fourier transformation of the wave equation, we find that our solution takes the form

\begin{aligned}\Psi = (2 \pi)^{-3/2} \int \tilde{\Psi}(\mathbf{k}, 0) e^{i (\omega t \pm \mathbf{k} \cdot \mathbf{x}) } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.29)

If one takes this as a given and applies the wave equation operator to this as a test solution, one finds without doing the Fourier transform work that we also have a constraint. That is

\begin{aligned}\frac{1}{{c^2}} (i \omega)^2 \Psi - (\pm i \mathbf{k})^2 \Psi = 0.\end{aligned} \hspace{\stretch{1}}(1.30)

So even in the continuous field domain, we have a relationship between frequency and wave number. We see that this also happens to have the form of a lightlike spacetime interval

\begin{aligned}\frac{\omega^2}{c^2} - \mathbf{k}^2 = 0.\end{aligned} \hspace{\stretch{1}}(1.31)

Also recall that the photoelectric effect imposes an experimental constraint on photon energy, where we have

\begin{aligned}E = h \nu = \frac{h}{2\pi} 2 \pi \nu = \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.32)

Therefore if we impose a mechanics like $P = (E/c, \mathbf{p})$ relativistic energy-momentum relationship on light, it then makes sense to form a nilpotent (lightlike) four vector for our photon energy. This combines our special relativistic expectations, with the constraints on the fields imposed by classical electromagnetism. We can then write for the photon four momentum

\begin{aligned}P = \left( \frac{\hbar \omega}{c}, \hbar k \right)\end{aligned} \hspace{\stretch{1}}(1.33)

### Back to the TA’s formula blitz.

Utilizing spherical polar coordinates in momentum (wave number) space, measuring the polar angle from the $k^1$ (x-like) axis, we can compute this polar angle in both pairs of frames,

\begin{aligned} \cos \alpha &= \frac{k^1}{{\left\lvert{\mathbf{k}}\right\rvert}} = \frac{k^1}{\omega/c} \\ \cos \alpha' &= \frac{{k^1}'}{\omega'/c} = \frac{\gamma (k^1 + \beta \omega/c)}{\gamma(\omega/c + \beta k^1)}\end{aligned} \hspace{\stretch{1}}(1.34)

Note that this requires us to assume that wave number four vectors transform in the same fashion as regular mechanical position and momentum four vectors. Also note that we have the primed frame moving negatively along the x-axis, instead of the usual positive origin shift. The question is vague enough to allow this since it only requires motion.

\paragraph{check 1}

as $\beta \rightarrow 1$ (ie: our primed frame velocity approaches the speed of light relative to the rest frame), $\cos \alpha' \rightarrow 1$, $\alpha' = 0$. The surface gets more and more compressed.

In the original reference frame the radiation was isotropic. In the new frame how does it change with respect to the angle? This is really a question to find this number flux rate

\begin{aligned}f'(\alpha') = ?\end{aligned} \hspace{\stretch{1}}(1.36)

In our rest frame the total number of photons traveling through the surface in a given interval of time is

\begin{aligned}N &= \int d\Omega dt f(\alpha) = \int d \phi \sin \alpha d\alpha = -2 \pi \int d(\cos\alpha) dt f(\alpha) \\ \end{aligned} \hspace{\stretch{1}}(1.37)

Here we utilize the spherical solid angle $d\Omega = \sin \alpha d\alpha d\phi = - d(\cos\alpha) d\phi$, and integrate $\phi$ over the $[0, 2\pi]$ interval. We also have to assume that our number flux density is not a function of horizontal angle $\phi$ in the rest frame.

In the moving frame we similarly have

\begin{aligned}N' &= -2 \pi \int d(\cos\alpha') dt' f'(\alpha'),\end{aligned} \hspace{\stretch{1}}(1.39)

and we again have had to assume that our transformed number flux density is not a function of the horizontal angle $\phi$. This seems like a reasonable move since ${k^2}' = k^2$ and ${k^3}' = k^3$ as they are perpendicular to the boost direction.

\begin{aligned}f'(\alpha') = \frac{d(\cos\alpha)}{d(\cos\alpha')} \left( \frac{dt}{dt'} \right) f(\alpha)\end{aligned} \hspace{\stretch{1}}(1.40)

Now, utilizing a conservation of mass argument, we can argue that $N = N'$. Regardless of the motion of the frame, the same number of particles move through the surface. Taking ratios, and examining an infinitesimal time interval, and the associated flux through a small patch, we have

\begin{aligned}\left( \frac{d(\cos\alpha)}{d(\cos\alpha')} \right) = \left( \frac{d(\cos\alpha')}{d(\cos\alpha)} \right)^{-1} = \gamma^2 ( 1 + \beta \cos\alpha)^2\end{aligned} \hspace{\stretch{1}}(1.41)

Part of the statement above was a do-it-yourself. First recall that $c t' = \gamma ( c t + \beta x )$, so $dt/dt'$ evaluated at $x=0$ is $1/\gamma$.

The rest is messier. We can calculate the $d(\cos)$ values in the ratio above using 1.34. For example, for $d(\cos(\alpha))$ we have

\begin{aligned}d(\cos\alpha) &= d \left( \frac{k^1}{\omega/c} \right) \\ &= dk^1 \frac{1}{{\omega/c}} - c \frac{1}{{\omega^2}} d\omega.\end{aligned}

If one does the same thing for $d(\cos\alpha')$, after a whole whack of messy algebra one finds that the differential terms and a whole lot more mystically cancels, leaving just

\begin{aligned}\frac{d\cos\alpha'}{d\cos\alpha} = \frac{\omega^2/c^2}{(\omega/c + \beta k^1)^2} (1 - \beta^2)\end{aligned} \hspace{\stretch{1}}(1.42)

A bit more reduction with reference back to 1.34 verifies 1.41.

Also note that again from 1.34 we have

\begin{aligned}\cos\alpha' = \frac{\cos\alpha + \beta}{1 + \beta \cos\alpha}\end{aligned} \hspace{\stretch{1}}(1.43)

and rearranging this for $\cos\alpha'$ gives us

\begin{aligned}\cos\alpha = \frac{\cos\alpha' - \beta}{1 - \beta \cos\alpha'},\end{aligned} \hspace{\stretch{1}}(1.44)

which we can sum to find that

\begin{aligned}1 + \beta \cos\alpha = \frac{1}{{\gamma^2 (1 - \beta \cos \alpha')^2 }},\end{aligned} \hspace{\stretch{1}}(1.45)

so putting all the pieces together we have

\begin{aligned}f'(\alpha') = \frac{1}{{\gamma}} \frac{f(\alpha)}{(\gamma (1-\beta \cos\alpha'))^2}\end{aligned} \hspace{\stretch{1}}(1.46)

The question asks for the energy flux density. We get this by multiplying the number density by the frequency of the light in question. This is, as a function of the polar angle, in each of the frames.

\begin{aligned}L(\alpha) &= \hbar \omega(\alpha) f(\alpha) = \hbar \omega f \\ L'(\alpha') &= \hbar \omega'(\alpha') f'(\alpha') = \hbar \omega' f'\end{aligned} \hspace{\stretch{1}}(1.47)

But we have

\begin{aligned}\omega'(\alpha')/c = \gamma( \omega/c + \beta k^1 ) = \gamma \omega/c ( 1 + \beta \cos\alpha )\end{aligned} \hspace{\stretch{1}}(1.49)

Aside, $\beta << 1$,

\begin{aligned}\omega' = \omega ( 1 + \beta \cos\alpha) + O(\beta^2) = \omega + \delta \omega\end{aligned} \hspace{\stretch{1}}(1.50)

\begin{aligned}\delta \omega &= \beta, \alpha = 0 \qquad \text{blue shift} \\ \delta \omega &= -\beta, \alpha = \pi \qquad \text{red shift}\end{aligned} \hspace{\stretch{1}}(1.51)

The TA then writes

\begin{aligned}L'(\alpha') = \frac{L/\gamma}{(\gamma (1 - \beta \cos\alpha'))^3}\end{aligned} \hspace{\stretch{1}}(1.53)

although, I calculate

\begin{aligned}L'(\alpha') = \frac{L}{\gamma^4 (\gamma (1 - \beta \cos\alpha'))^4}\end{aligned} \hspace{\stretch{1}}(1.54)

He then says, the forward backward ratio is

\begin{aligned}L'(0)/L'(\pi) = {\left( \frac{ 1 + \beta }{1-\beta} \right)}^3\end{aligned} \hspace{\stretch{1}}(1.55)

For this I get:

\begin{aligned}L'(0)/L'(\pi) = {\left( \frac{ 1 + \beta }{1-\beta} \right)}^4\end{aligned} \hspace{\stretch{1}}(1.56)

It is still bigger for $\beta$ positive, which I think is the point.

If I can somehow manage to keep my signs right as I do this course I may survive. Why did he pick a positive sign way back in 1.34?

# References

[1] J.B. Hartle and T. Dray. Gravity: an introduction to Einsteins general relativity, volume 71. 2003.

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