# Lagrangian for inhomogeneous Maxwell’s equations

References: Amitabha Lahiri & P. B. Pal, A First Book of Quantum Field Theory, Second Edition (Alpha Science International, 2004) – Chapter 2, Problem 2.1.

The Euler-Lagrange equations for a classical field are

$\displaystyle \frac{\partial\mathcal{L}}{\partial\phi^{r}}-\frac{\partial}{\partial q^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\mu}^{r}}\right)=0 \ \ \ \ \ (1)$

where ${\mathcal{L}}$ is the Lagrangian density, ${\phi^{r}}$ is the ${r}$th field, ${q^{\mu}}$ are the generalized coordinates and the notation ${\phi_{,\mu}\equiv\frac{\partial\phi}{\partial q^{\mu}}}$. As an example of how these equations can give rise to physical equations that are more familiar, we’ll look at the Lagrangian (I’ll leave off the ‘density’ to save space, but when talking about fields instead of particles, we’ll always mean Lagrangian density) for the electromagnetic field. At this point, we won’t worry about where this Lagrangian comes from; we’ll just state it as

$\displaystyle \mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-j_{\mu}A^{\mu} \ \ \ \ \ (2)$

where the electromagnetic field tensor is (This definition is taken from Moore’s book on general relativity and is actually the negative of Lahiri & Pal’s definition in their equation 8.6. We don’t actually use this form of the tensor in what follows so it doesn’t affect the result, but it’s important to realize that different authors use different definitions.)

$\displaystyle F_{\mu\nu}=\left[\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & B_{z} & -B_{y}\\ E_{y} & -B_{z} & 0 & B_{x}\\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (3)$

the four-current is

$\displaystyle j^{\mu}=\left[\rho,\mathbf{J}\right] \ \ \ \ \ (4)$

and ${A^{\mu}}$ is the four-potential

$\displaystyle A^{\mu}\equiv\left[V,\mathbf{A}\right] \ \ \ \ \ (5)$

In applying 1, we take the fields ${\phi^{r}}$ to be the potentials ${A^{\mu}}$. What do the Euler-Lagrange equations give us for these fields?

The first term in 1 is just

$\displaystyle \frac{\partial\mathcal{L}}{\partial A^{\mu}}=-j_{\mu} \ \ \ \ \ (6)$

The second term is a bit more involved. First, we need to write ${F_{\mu\nu}}$ in terms of the four-potential, which is

$\displaystyle F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} \ \ \ \ \ (7)$

In calculating the derivatives, we need to be careful to keep track of the positions (up or down) of the various indices. The ${\phi^{r}}$ in 1 are represented by ${A^{\mu}}$ with raised indices. The derivative becomes

$\displaystyle \phi_{,\alpha}^{r}=A_{,\alpha}^{\mu}=\frac{\partial A^{\mu}}{\partial q^{\alpha}}=\partial_{\alpha}A^{\mu} \ \ \ \ \ (8)$

That is, the index on the derivative is lowered and the index on the field is raised. To take the derivatives of ${\mathcal{L}}$ we need to express ${F^{\mu\nu}F_{\mu\nu}}$ in this form, so we get

 $\displaystyle F^{\mu\nu}F_{\mu\nu}$ $\displaystyle =$ $\displaystyle \left(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right)\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(g^{\mu\pi}\partial_{\pi}A^{\nu}-g^{\nu\pi}\partial_{\pi}A^{\mu}\right)\left(g_{\nu\rho}\partial_{\mu}A^{\rho}-g_{\mu\rho}\partial_{\nu}A^{\rho}\right) \ \ \ \ \ (10)$

where ${g^{\mu\nu}=g_{\mu\nu}=\mbox{diag}\left(+1,-1,-1,-1\right)}$ using Lahiri and Pal’s convention. Now we can calculate the inner derivative in the second term of 1. First, we’ll consider the field ${\phi^{r}=A^{0}}$. Using the product rule we get (I’ll use ${\alpha}$ for the coordinate index to avoid confusion with ${\mu}$ which is used to denote the component of ${A^{\mu}}$; remember also that ${g^{\mu\nu}}$ is diagonal, and there is no implied sum on ${\alpha}$):

 $\displaystyle -4\frac{\partial\mathcal{L}}{\partial A_{,\alpha}^{0}}$ $\displaystyle =$ $\displaystyle g^{\alpha\alpha}\left[g_{00}\partial_{\alpha}A^{0}-g_{\alpha\alpha}\partial_{0}A^{\alpha}\right]-g^{\alpha\alpha}\left[g_{\alpha\alpha}\partial_{0}A^{\alpha}-g_{00}\partial_{\alpha}A^{0}\right]+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle g_{00}\left[g^{\alpha\alpha}\partial_{\alpha}A^{0}-g^{00}\partial_{0}A^{\alpha}\right]-g_{00}\left[g^{00}\partial_{0}A^{\alpha}-g^{\alpha\alpha}\partial_{\alpha}A^{0}\right]\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\left[g^{\alpha\alpha}\partial_{\alpha}A^{0}-\partial_{0}A^{\alpha}\right] \ \ \ \ \ (12)$

When ${\alpha=0}$, the RHS is zero, and for ${\alpha=1,2,3}$, ${g^{\alpha\alpha}=-1}$ so

$\displaystyle \frac{\partial\mathcal{L}}{\partial A_{,\alpha}^{0}}=\begin{cases} 0 & \alpha=0\\ \partial_{i}A^{0}+\partial_{0}A^{i} & i=1,2,3 \end{cases} \ \ \ \ \ (13)$

Plugging this into 1 we get, together with 6

 $\displaystyle -j^{0}$ $\displaystyle =$ $\displaystyle \sum_{i=1}^{3}\partial_{i}\left(\partial_{i}A^{0}+\partial_{0}A^{i}\right)\ \ \ \ \ (14)$ $\displaystyle -\rho$ $\displaystyle =$ $\displaystyle \nabla^{2}V+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right) \ \ \ \ \ (15)$

This is the potential equivalent of the Maxwell equation ${\nabla\cdot\mathbf{E}=\rho/\epsilon_{0}}$ (the system of units used in Lahiri & Pal takes ${\epsilon_{0}=1}$).

Now look at the field ${A^{1}}$. We have

 $\displaystyle -4\frac{\partial\mathcal{L}}{\partial A_{,\alpha}^{1}}$ $\displaystyle =$ $\displaystyle g^{\alpha\alpha}\left[g_{11}\partial_{\alpha}A^{1}-g_{\alpha\alpha}\partial_{1}A^{\alpha}\right]-g^{\alpha\alpha}\left[g_{\alpha\alpha}\partial_{1}A^{\alpha}-g_{11}\partial_{\alpha}A^{1}\right]+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle g_{11}\left[g^{\alpha\alpha}\partial_{\alpha}A^{1}-g^{11}\partial_{1}A^{\alpha}\right]-g_{11}\left[g^{11}\partial_{1}A^{\alpha}-g^{\alpha\alpha}\partial_{\alpha}A^{1}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -4\left[g^{\alpha\alpha}\partial_{\alpha}A^{1}+\partial_{1}A^{\alpha}\right] \ \ \ \ \ (17)$

We get

$\displaystyle \frac{\partial\mathcal{L}}{\partial A_{,\alpha}^{1}}=\begin{cases} \partial_{0}A^{1}+\partial_{1}A^{0} & \alpha=0\\ -\partial_{i}A^{1}+\partial_{1}A^{i} & i=1,2,3 \end{cases} \ \ \ \ \ (18)$

Plugging this into 1 we have

 $\displaystyle \partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial A_{,\alpha}^{1}}\right)$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}A^{1}}{\partial t^{2}}+\partial_{x}\left(\partial_{t}A^{0}\right)-\nabla^{2}A^{1}+\partial_{x}\left(\nabla\cdot\mathbf{A}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}A_{x}}{\partial t^{2}}+\left[\nabla\left(\partial_{t}V\right)\right]_{x}-\nabla^{2}A_{x}+\left[\nabla\left(\nabla\cdot\mathbf{A}\right)\right]_{x} \ \ \ \ \ (20)$

Combining this with 6 gives

$\displaystyle \frac{\partial^{2}A_{x}}{\partial t^{2}}+\left[\nabla\left(\partial_{t}V\right)\right]_{x}-\nabla^{2}A_{x}+\left[\nabla\left(\nabla\cdot\mathbf{A}\right)\right]_{x}=-j_{1}=+j^{1}=j_{x} \ \ \ \ \ (21)$

Doing the same calculation for ${A^{2}}$ and ${A^{3}}$ yields the ${y}$ and ${z}$ components, so we get

$\displaystyle \frac{\partial^{2}\mathbf{A}}{\partial t^{2}}+\nabla\left(\partial_{t}V\right)-\nabla^{2}\mathbf{A}+\nabla\left(\nabla\cdot\mathbf{A}\right)=\mathbf{J} \ \ \ \ \ (22)$

which is the other inhomogeneous Maxwell equation in potential form.

# Maxwell’s equations using the electromagnetic field tensor

References: Amitabha Lahiri & P. B. Pal, A First Book of Quantum Field Theory, Second Edition (Alpha Science International, 2004) – Chapter 1, Problem 1.7.

We can summarize the electromagnetic field in tensor form by means of the field tensor

$\displaystyle F_{\mu\nu}=\left[\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & B_{z} & -B_{y}\\ E_{y} & -B_{z} & 0 & B_{x}\\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

[This tensor is written using relativistic units with ${c=1}$ so that ${\mathbf{E}}$ and ${\mathbf{B}}$ have the same dimensions.]

We’ve already seen that the pair of homogeneous Maxwell’s equations can be written in terms of this tensor as follows:

$\displaystyle \partial_{\mu}F_{\nu\sigma}+\partial_{\sigma}F_{\mu\nu}+\partial_{\nu}F_{\sigma\mu}=0 \ \ \ \ \ (2)$

With the usual ordering of coordinates ${\left(x^{0},x^{1},x^{2},x^{3}\right)=\left(t,x,y,z\right)}$, if we set ${\mu=2}$, ${\nu=1}$ and ${\sigma=3}$ we get

 $\displaystyle -\partial_{y}B_{y}-\partial_{z}B_{z}-\partial_{x}B_{x}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

Selecting ${\mu=2}$, ${\nu=1}$ and ${\sigma=0}$ gives

$\displaystyle \partial_{y}E_{x}-\partial_{t}B_{z}-\partial_{x}E_{y}=0 \ \ \ \ \ (5)$

This is the ${z}$ component of

$\displaystyle \nabla\times\mathbf{E}+\frac{\partial\mathbf{B}}{\partial t}=0 \ \ \ \ \ (6)$

We can get the ${x}$ component by choosing ${\mu=0}$, ${\nu=2}$ and ${\sigma=3}$:

$\displaystyle \partial_{t}B_{x}-\partial_{z}E_{y}+\partial_{y}E_{z}=0 \ \ \ \ \ (7)$

The ${y}$ component comes from ${\mu=0}$, ${\nu=1}$ and ${\sigma=3}$:

$\displaystyle -\partial_{t}B_{y}-\partial_{z}E_{x}+\partial_{x}E_{z}=0 \ \ \ \ \ (8)$

The two inhomogenous Maxwell’s equations are

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\rho}{\epsilon_{0}}\ \ \ \ \ (9)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}\left(\frac{\mathbf{J}}{\epsilon_{0}}+\frac{\partial\mathbf{E}}{\partial t}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mathbf{J}}{\epsilon_{0}}+\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (12)$

where we used ${\mu_{0}\epsilon_{0}=1/c^{2}}$ and the last line uses relativistic units with ${c=1}$.

We need to introduce the four-current to put these in four-vector form. This is

$\displaystyle J^{\mu}=\left[\rho,\mathbf{J}\right] \ \ \ \ \ (13)$

where ${\rho}$ is the charge density and ${\mathbf{J}}$ is the three-current. Then if we look at Gauss’s law 9 we see that this can be written as

$\displaystyle \partial_{\nu}F^{0\nu}=\frac{J^{0}}{\epsilon_{0}} \ \ \ \ \ (14)$

where ${F^{\mu\nu}}$ is the raised version of the tensor

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (15)$

If we generalize this formula we get

$\displaystyle \partial_{\nu}F^{\mu\nu}=\frac{J^{\mu}}{\epsilon_{0}} \ \ \ \ \ (16)$

For ${\mu=1}$ we get

$\displaystyle -\partial_{t}E_{x}+\partial_{y}B_{z}-\partial_{z}B_{y}=\frac{J^{x}}{\epsilon_{0}} \ \ \ \ \ (17)$

This is the ${x}$ component of 12. Choosing ${\mu=2}$ and ${\mu=3}$ give the ${y}$ and ${z}$ components respectively.

From our examination of the electromagnetic tensor, we saw the four-vector form of the Lorentz force law for a charge ${q}$:

$\displaystyle \frac{dp^{\mu}}{d\tau}=qF^{\mu\nu}u_{\nu} \ \ \ \ \ (18)$

where ${\tau}$ is the proper time, ${p^{\mu}}$ is the four-momentum and ${u_{\nu}}$ is the four-velocity.

To summarize, Maxwell’s equations can be written as

 $\displaystyle \partial_{\mu}F_{\nu\sigma}+\partial_{\sigma}F_{\mu\nu}+\partial_{\nu}F_{\sigma\mu}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \partial_{\nu}F^{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{J^{\mu}}{\epsilon_{0}} \ \ \ \ \ (20)$

The Lorentz force law can be written as

$\displaystyle \frac{dp^{\mu}}{d\tau}=qF^{\mu\nu}u_{\nu} \ \ \ \ \ (21)$

# Maxwell’s equations in cylindrical coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Problem P17.3.

As an example of using the geodesic equation to calculate Christoffel symbols, we’ll consider Maxwell’s equations in cylindrical coordinates. We compare the geodesic equation:

$\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (1)$

with the expression for the Christoffel symbols:

$\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (2)$

Using cylindrical coordinates to describe flat spacetime, we have

$\displaystyle ds^{2}=-dt^{2}+dr^{2}+r^{2}d\theta^{2}+dz^{2} \ \ \ \ \ (3)$

The geodesic equation is, for each of the four coordinates

 $\displaystyle -\ddot{t}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle \ddot{r}-\frac{1}{2}\left(2r\right)\dot{\theta}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle r^{2}\ddot{\theta}+2r\dot{r}\dot{\theta}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \ddot{z}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

From this we get the Christoffel symbols:

 $\displaystyle \Gamma_{\; ij}^{t}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (8)$ $\displaystyle \Gamma_{\; ij}^{r}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & -r & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\; ij}^{\theta}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\ 0 & \frac{1}{r} & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle \Gamma_{\; ij}^{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (11)$

The electromagnetic field tensor is, in rectangular coordinates:

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (12)$

We can write this tensor in cylindrical coordinates as follows:

$\displaystyle \left(F'\right)^{ij}=\left[\begin{array}{cccc} 0 & E_{r} & E_{\theta} & E_{z}\\ -E_{r} & 0 & B_{z} & -B_{\theta}\\ -E_{\theta} & -B_{z} & 0 & B_{r}\\ -E_{z} & B_{\theta} & -B_{r} & 0 \end{array}\right] \ \ \ \ \ (13)$

Each entry in this tensor is found by the usual transformation rule. For example, since ${t^{\prime}=t}$, ${z^{\prime}=z}$, ${r=\sqrt{x^{2}+y^{2}}}$ and ${\theta=\tan^{-1}\frac{y}{x}}$

 $\displaystyle E_{r}$ $\displaystyle =$ $\displaystyle \left(F'\right)^{tr}=F^{ij}\frac{\partial t}{\partial x{}^{\prime i}}\frac{\partial r}{\partial x^{\prime j}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F^{tx}\frac{x}{r}+F^{ty}\frac{y}{r}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{x}\frac{x}{r}+E_{y}\frac{y}{r} \ \ \ \ \ (16)$

Using the same transformation rule, we get for the other 3 components:

 $\displaystyle E_{z^{\prime}}$ $\displaystyle =$ $\displaystyle E_{z}\ \ \ \ \ (17)$ $\displaystyle E_{\theta}$ $\displaystyle =$ $\displaystyle -E_{x}\frac{y}{r^{2}}+E_{y}\frac{x}{r^{2}} \ \ \ \ \ (18)$

Two of Maxwell’s equationscan be written in terms of the electromagnetic tensor in rectangular coordinates as

$\displaystyle \boxed{\partial_{a}F^{ba}=\frac{1}{\epsilon_{0}}J^{b}} \ \ \ \ \ (19)$

This corresponds to the two Maxwell equations (in units where ${c=1/\sqrt{\mu_{0}\epsilon_{0}}=1}$):

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\rho}{\epsilon_{0}}\ \ \ \ \ (20)$ $\displaystyle \nabla\times\mathbf{B}-\frac{\partial\mathbf{E}}{\partial t}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon_{0}}\mathbf{J} \ \ \ \ \ (21)$

In other coordinate systems, we can write 19 by replacing the ordinary derivative by the covariant derivative:

$\displaystyle \boxed{\nabla_{a}F^{ba}=\frac{1}{\epsilon_{0}}J^{b}} \ \ \ \ \ (22)$

To evaluate the covariant derivative, we write it in terms of the Christoffel symbols. Since there is a sum over the index ${a}$, we first write out the derivative without the sum:

$\displaystyle \nabla_{a}F^{bc}=\partial_{a}F^{bc}+\Gamma_{\; aj}^{b}F^{jc}+\Gamma_{\; aj}^{c}F^{bj} \ \ \ \ \ (23)$

Now we can take the sum by setting ${a=c}$:

$\displaystyle \nabla_{a}F^{ba}=\partial_{a}F^{ba}+\Gamma_{\; aj}^{b}F^{ja}+\Gamma_{\; aj}^{a}F^{bj} \ \ \ \ \ (24)$

Since ${\Gamma_{\; aj}^{b}=\Gamma_{\; ja}^{b}}$ and ${F^{ja}=-F^{aj}}$, the double sum ${\Gamma_{\; aj}^{b}F^{ja}}$ is always zero, so the second term on the RHS vanishes. We are left with

$\displaystyle \nabla_{a}F^{ba}=\partial_{a}F^{ba}+\Gamma_{\; aj}^{a}F^{bj} \ \ \ \ \ (25)$

From 8 to 11, there are only 3 non-zero ${\Gamma_{\; aj}^{b}}$, so the expression on the RHS isn’t terribly complicated when the sums are expanded. Considering the ${t}$, ${r}$, ${\theta}$ and ${z}$ components in that order, we get

 $\displaystyle \partial_{a}F^{ta}+\Gamma_{\; aj}^{a}F^{tj}$ $\displaystyle =$ $\displaystyle \partial_{r}E_{r}+\partial_{\theta}E_{\theta}+\partial_{z}E_{z}+\frac{1}{r}E_{r}\ \ \ \ \ (26)$ $\displaystyle \partial_{a}F^{ra}+\Gamma_{\; aj}^{a}F^{rj}$ $\displaystyle =$ $\displaystyle -\partial_{t}E_{r}+\partial_{\theta}B_{z}-\partial_{z}B_{\theta}\ \ \ \ \ (27)$ $\displaystyle \partial_{a}F^{\theta a}+\Gamma_{\; aj}^{a}F^{\theta j}$ $\displaystyle =$ $\displaystyle -\partial_{t}E_{\theta}-\partial_{r}B_{z}+\partial_{z}B_{r}-\frac{1}{r}B_{z}\ \ \ \ \ (28)$ $\displaystyle \partial_{a}F^{za}+\Gamma_{\; aj}^{a}F^{zj}$ $\displaystyle =$ $\displaystyle -\partial_{t}E_{z}+\partial_{r}B_{\theta}-\partial_{\theta}B_{r}+\frac{1}{r}B_{\theta} \ \ \ \ \ (29)$

As Moore points out, these are not the same components that we’d get if we converted Maxwell’s equations to the usual cylindrical coordinate system in which the basis vectors are unit vectors, since in the cylindrical coordinate basis, the ${\theta}$ basis vector is ${\mathbf{e}_{\theta}=r\hat{\boldsymbol{\theta}}}$. However, I did check that if you substitute 16 and 18 into 26 and work out all the derivatives, it is indeed true that

$\displaystyle \partial_{r}E_{r}+\partial_{\theta}E_{\theta}+\partial_{z}E_{z}+\frac{1}{r}E_{r}=\partial_{x}E_{x}+\partial_{y}E_{y}+\partial_{z}E_{z} \ \ \ \ \ (30)$

Presumably the other 3 equations also work out to the 3 components of 21 if we work out the components of ${B_{i}}$ in the cylindrical basis and then substitute them in and do the derivatives, though I’m too lazy to check this.

# Electromagnetic stress-energy tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problem 7.8; Chapter 20, Problem 20.5.

Since energy is the time component of the four-momentum and the density of a scalar quantity such as mass or charge is the time component of a four-vector (four-current in the case of charge), the energy density of a collection of particles would appear to transform as the ${tt}$ component of a second-rank tensor (since we have one Lorentz transformation for the energy and another for the density). If we then take a slight leap in logic and assume that the energy density of electromagnetic fields transforms in the same way, then we can assume that this energy density is the ${tt}$ component of a tensor known as the stress-energy tensor.

To deduce the form of this tensor, we use the usual technique of trying to convert an expression from classical physics into tensor form. In this case the classical expression for energy density is

$\displaystyle \rho_{E}=\frac{1}{8\pi k}\left(E^{2}+B^{2}\right) \ \ \ \ \ (1)$

The idea, then, is to find a second-rank tensor from the electromagnetic field tensor that has this as its ${tt}$ component. We can get a clue as to how to do this by looking at the quantity ${F_{ij}F^{ij}=2\left(B^{2}-E^{2}\right)}$. During the derivation of this, we found that the ${tt}$ (that is, the 00 component) component of the tensor product ${F^{0j}F_{j0}=E^{2}}$. Combining this with ${B^{2}=\frac{1}{2}F_{ij}F^{ij}+E^{2}=\frac{1}{2}F_{ij}F^{ij}+F^{0j}F_{j0}}$ we get

$\displaystyle E^{2}+B^{2}=2F^{0j}F_{j0}+\frac{1}{2}F_{ij}F^{ij} \ \ \ \ \ (2)$

If we raise the 0 index in the first term on the RHS, we get

$\displaystyle E^{2}+B^{2}=-2F^{0j}F_{jk}\eta^{k0}+\frac{1}{2}F_{ij}F^{ij} \ \ \ \ \ (3)$

where we’ve introduced the minus sign in the first term since raising the index changes the sign in the 0 column of ${F_{jk}}$.

This looks promising, but the expression on the RHS is the sum of the ${tt}$ component of a tensor (the first term) and a scalar. We can convert this to be the ${tt}$ component of a tensor by multiplying the last term by ${-\eta^{00}=+1}$ (the Kronecker delta):

$\displaystyle E^{2}+B^{2}=-2F^{0j}F_{jk}\eta^{k0}-\frac{1}{2}F_{ij}F^{ij}\eta^{00} \ \ \ \ \ (4)$

If we now propose this is the ${tt}$ term of the stress-energy tensor ${8\pi kT^{ml}}$ we can postulate

 $\displaystyle T^{ml}$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{mj}F_{jk}\eta^{kl}+\frac{1}{2}F_{ij}F^{ij}\eta^{ml}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon_{0}F^{mj}F_{jk}\eta^{kl}-\frac{\epsilon_{0}}{4}F_{ij}F^{ij}\eta^{ml} \ \ \ \ \ (6)$

The other elements of ${T^{ml}}$ have meanings that we’ll get to eventually, but we can note that for ${i\ne0}$

$\displaystyle T^{0i}=-\epsilon_{0}F^{0j}F_{jk}\eta^{ki} \ \ \ \ \ (7)$

For ${i=1}$ for example, we get

 $\displaystyle T^{01}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}F^{0j}F_{j1}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(E_{y}B_{z}-E_{z}B_{y}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\mathbf{E}\times\mathbf{B}\right)_{x} \ \ \ \ \ (10)$

This is the ${x}$ component of the Poynting vector, which describes the rate of energy transfer per unit area of an electromagnetic field. Again, something we’ll get to in EM theory eventually.

# Electromagnetic field tensor: invariance of inner product

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problem 7.7.

We’ve worked out earlier the scalar quantity ${F_{ij}F^{ij}=2\left(B^{2}-E^{2}\right)}$. We can check this for the case of a Lorentz transformation in flat space-time. We found there that

 $\displaystyle E_{x}^{\prime}$ $\displaystyle =$ $\displaystyle E_{x}\ \ \ \ \ (1)$ $\displaystyle E_{y}^{\prime}$ $\displaystyle =$ $\displaystyle \gamma E_{y}-\gamma\beta B_{z}\ \ \ \ \ (2)$ $\displaystyle E_{z}^{\prime}$ $\displaystyle =$ $\displaystyle \gamma E_{z}+\gamma\beta B_{y}\ \ \ \ \ (3)$ $\displaystyle B_{x}^{\prime}$ $\displaystyle =$ $\displaystyle B_{x}\ \ \ \ \ (4)$ $\displaystyle B_{y}^{\prime}$ $\displaystyle =$ $\displaystyle -\gamma\beta E_{y}+\gamma B_{z}\ \ \ \ \ (5)$ $\displaystyle B_{z}^{\prime}$ $\displaystyle =$ $\displaystyle -\gamma\beta E_{z}-\gamma B_{y} \ \ \ \ \ (6)$

Calculating the invariant in the new system we get

 $\displaystyle B^{\prime2}-E^{\prime2}$ $\displaystyle =$ $\displaystyle B_{x}^{\prime2}+B_{y}^{\prime2}+B_{z}^{\prime2}+E_{x}^{\prime2}+E_{y}^{\prime2}+E_{z}^{\prime2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B_{x}^{2}+\left(-\gamma\beta E_{y}+\gamma B_{z}\right)^{2}+\left(-\gamma\beta E_{z}-\gamma B_{y}\right)^{2}-\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle$ $\displaystyle E_{x}^{2}-\left(\gamma E_{y}-\gamma\beta B_{z}\right)^{2}-\left(\gamma E_{z}+\gamma\beta B_{y}\right)^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B_{x}^{2}+\left(B_{y}^{2}+B_{z}^{2}\right)\gamma^{2}\left(1-\beta^{2}\right)-E_{x}^{2}-\left(E_{y}^{2}+E_{z}^{2}\right)\gamma^{2}\left(1-\beta^{2}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B^{2}-E^{2} \ \ \ \ \ (11)$

since ${\gamma=1/\sqrt{1-\beta^{2}}}$. All the cross terms involving the product of a component of ${E}$ and one of ${B}$ cancel out between lines 2/3 and 4.

# Electromagnetic field tensor: cyclic derivative relation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problem 7.6.

We’ve used the following relation between the derivatives of the electromagnetic field tensor ${F^{ij}}$ to get several of Maxwell’s equations.

$\displaystyle \partial_{i}F_{jk}+\partial_{k}F_{ij}+\partial_{j}F_{ki}=0 \ \ \ \ \ (1)$

Here we verify that this relation is true when ${F^{ij}}$ is written in terms of the four-potential, that is

$\displaystyle F^{ij}=\partial^{i}A^{j}-\partial^{j}A^{i} \ \ \ \ \ (2)$

We can lower both indices in this equation and the plug it into the first equation:

 $\displaystyle \partial_{i}F_{jk}+\partial_{k}F_{ij}+\partial_{j}F_{ki}$ $\displaystyle =$ $\displaystyle \partial_{i}\left(\partial_{j}A_{k}-\partial_{k}A_{j}\right)+\partial_{k}\left(\partial_{i}A_{j}-\partial_{j}A_{i}\right)+\partial_{j}\left(\partial_{k}A_{i}-\partial_{i}A_{k}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{i}\partial_{j}A_{k}-\partial_{j}\partial_{i}A_{k}+\partial_{k}\partial_{i}A_{j}-\partial_{i}\partial_{k}A_{j}+\partial_{j}\partial_{k}A_{i}-\partial_{k}\partial_{j}A_{i}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

The terms in the second line cancel in pairs since the order of the partials doesn’t matter.

# Electromagnetic field tensor: Lorentz transformations

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problems 7.1.

Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.48.

To apply this to Griffiths problem 12.48, use

$\displaystyle t^{ij}=\left[\begin{array}{cccc} 0 & t^{01} & t^{02} & t^{03}\\ -t^{01} & 0 & t^{12} & t^{13}\\ -t^{02} & -t^{12} & 0 & t^{23}\\ -t^{03} & -t^{13} & -t^{23} & 0 \end{array}\right] \ \ \ \ \ (1)$

in place of ${F^{ij}}$ in what follows. The results are the same.

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (2)$

We can use the usual tensor transformation rules to see how the electric and magnetic fields transform under a Lorentz transformation. We get

 $\displaystyle F^{\prime ij}$ $\displaystyle =$ $\displaystyle \frac{\partial x'^{i}}{\partial x^{k}}\frac{\partial x'^{j}}{\partial x^{l}}F^{kl}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Lambda_{\;\; k}^{i}\Lambda_{\;\; l}^{j}F^{kl} \ \ \ \ \ (4)$

where the Lorentz transformation matrix is

$\displaystyle \Lambda_{\;\; k}^{i}=\left[\begin{array}{cccc} \gamma & -\gamma\beta & 0 & 0\\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (5)$

As we saw when discussing the inertia tensor, we can write this transformation as a matrix equation

$\displaystyle F^{\prime}=\Lambda F\Lambda^{T} \ \ \ \ \ (6)$

The first product is

$\displaystyle \Lambda F=\left[\begin{array}{cccc} \gamma\beta E_{x} & \gamma E_{x} & \gamma E_{y}-\gamma\beta B_{z} & \gamma E_{z}+\gamma\beta B_{y}\\ -\gamma E_{x} & -\gamma\beta E_{x} & -\gamma\beta E_{y}+\gamma B_{z} & -\gamma\beta E_{z}-\gamma B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (7)$

The final product is

$\displaystyle F^{\prime}=\Lambda F\Lambda^{T}=\left[\begin{array}{cccc} 0 & \gamma^{2}\left(1-\beta^{2}\right)E_{x} & \gamma E_{y}-\gamma\beta B_{z} & \gamma E_{z}+\gamma\beta B_{y}\\ -\gamma^{2}\left(1-\beta^{2}\right)E_{x} & 0 & -\gamma\beta E_{y}+\gamma B_{z} & -\gamma\beta E_{z}-\gamma B_{y}\\ -\gamma E_{y}+\gamma\beta B_{z} & \gamma\beta E_{y}-\gamma B_{z} & 0 & B_{x}\\ -\gamma E_{z}-\gamma\beta B_{y} & \gamma\beta E_{z}+\gamma B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (8)$

Using ${\gamma=1/\sqrt{1-\beta^{2}}}$ we get

$\displaystyle F^{\prime}=\left[\begin{array}{cccc} 0 & E_{x} & \gamma E_{y}-\gamma\beta B_{z} & \gamma E_{z}+\gamma\beta B_{y}\\ -E_{x} & 0 & -\gamma\beta E_{y}+\gamma B_{z} & -\gamma\beta E_{z}-\gamma B_{y}\\ -\gamma E_{y}+\gamma\beta B_{z} & \gamma\beta E_{y}-\gamma B_{z} & 0 & B_{x}\\ -\gamma E_{z}-\gamma\beta B_{y} & \gamma\beta E_{z}+\gamma B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (9)$

From this, we see that

 $\displaystyle E_{x}^{\prime}$ $\displaystyle =$ $\displaystyle E_{x}\ \ \ \ \ (10)$ $\displaystyle E_{y}^{\prime}$ $\displaystyle =$ $\displaystyle \gamma E_{y}-\gamma\beta B_{z}\ \ \ \ \ (11)$ $\displaystyle E_{z}^{\prime}$ $\displaystyle =$ $\displaystyle \gamma E_{z}+\gamma\beta B_{y}\ \ \ \ \ (12)$ $\displaystyle B_{x}^{\prime}$ $\displaystyle =$ $\displaystyle B_{x}\ \ \ \ \ (13)$ $\displaystyle B_{y}^{\prime}$ $\displaystyle =$ $\displaystyle \gamma\beta E_{z}+\gamma B_{y}\ \ \ \ \ (14)$ $\displaystyle B_{z}^{\prime}$ $\displaystyle =$ $\displaystyle -\gamma\beta E_{y}+\gamma B_{z} \ \ \ \ \ (15)$

Unlike lengths, the components of ${E}$ and ${B}$ in the direction of motion are unchanged, while those perpendicular to the motion are altered.

# Electromagnetic field tensor: justification

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problems 7.5.

We’ve been using the electromagnetic (EM) field tensor ${F^{ij}}$ in several problems without saying where it comes from, so it’s time to fill in the gap with a look at how the form of this tensor was deduced.

To say this is a derivation of the EM tensor is probably stretching things a bit; it’s more of a plausibility argument. As always in relativity, the idea is to generalize the equations of classical physics by putting them in tensor form.

We start with the charge density and current. As we’ve already seen, densities (of either mass or charge) are not invariant under a Lorentz transformation because of length contraction. If we start with a charge density ${\rho}$ at rest and transform to an inertial frame moving at velocity ${\boldsymbol{\beta}}$, the density becomes ${\gamma\rho}$. However, we also generate a current due to the motion of the charge, which is ${\rho\boldsymbol{\beta}}$.

The four-current ${\mathbf{J}}$ is defined as a four-vector whose ${t}$ component is the charge density and whose spatial components are the current. That is, in a frame moving with velocity ${\boldsymbol{\beta}}$ relative to the charge, we have

$\displaystyle \mathbf{J}=\left[\gamma\rho,\gamma\beta_{x}\rho,\gamma\beta_{y}\rho,\gamma\beta_{z}\rho\right] \ \ \ \ \ (1)$

In the charge’s rest frame, this is

$\displaystyle \mathbf{J}=\left[\rho,0,0,0\right] \ \ \ \ \ (2)$

so the invariant square of the four-current is

$\displaystyle J^{2}=-\rho^{2} \ \ \ \ \ (3)$

The next step is to look at Gauss’s law in differential form. This has the form

$\displaystyle \nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_{0}} \ \ \ \ \ (4)$

Moore uses the alternative form

$\displaystyle \nabla\cdot\mathbf{E}=4\pi k\rho \ \ \ \ \ (5)$

where ${k=1/4\pi\epsilon_{0}}$. If we generalize the RHS to the four-current (which is a four-vector), then we need to make the LHS a four-vector as well. In its current form, ${\nabla\cdot\mathbf{E}}$ is a scalar, so we need to find some four-vector of which this is the 0 component (since ${\rho}$ is the 0 component of ${\mathbf{J}}$). Since we need to take the derivative on the LHS, we can try the form

$\displaystyle \partial_{j}F^{ij}=4\pi kJ^{i} \ \ \ \ \ (6)$

where ${F^{ij}}$ is a rank-2 tensor to be determined.

Note that there is one slight snag in this argument. In general, the derivative of a tensor is not another tensor, so the quantity on the LHS is not a tensor in a general coordinate system. However, if the transformation between coordinate systems involves partial derivatives that are constants (that is, ${\partial x^{\prime i}/\partial x^{j}=\mbox{constant}}$), then the derivative of a tensor is a tensor. For Lorentz transformations, this condition is true, since the transformation depends only on the (constant) relative velocity of the inertial frames. Thus in special relativity, the above equation is valid.

To make this equation consistent with Gauss’s law, therefore, we need

$\displaystyle F^{ij}=\left[\begin{array}{cccc} - & E_{x} & E_{y} & E_{z}\\ - & - & - & -\\ - & - & - & -\\ - & - & - & - \end{array}\right] \ \ \ \ \ (7)$

where the dashed entries are to be determined.

Next we look at the electrostatic force, which in Newtonian terms is

$\displaystyle \frac{d\mathbf{p}}{dt}=q\mathbf{E} \ \ \ \ \ (8)$

If we generalize the LHS to ${dp^{i}/d\tau}$, we again need the RHS to be a four-vector. Since ${F^{ij}}$ is a rank-2 tensor, we need to contract it with something to give us a four-vector. Taking partial derivatives won’t work here, since the force equation involves ${\mathbf{E}}$ on its own, not its derivatives. At this point, we make a leap of logic and propose that we contract ${F^{ij}}$ with the four-velocity ${u_{j}}$. This certainly isn’t a derivation; its only justification is that it works. So we get the relativistic form of the force law:

$\displaystyle \frac{dp^{i}}{d\tau}=qF^{ij}u_{j} \ \ \ \ \ (9)$

Note that ${u_{j}}$ is the covariant version of the four-velocity, so its time component ${u_{0}=-u^{0}}$, while the other components are the same in both forms.

From here, we can use the following identity:

 $\displaystyle \frac{d\left(\mathbf{p}\cdot\mathbf{p}\right)}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{d\left(p_{i}p^{i}\right)}{d\tau}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{d\left(\eta_{ij}p^{j}p^{i}\right)}{d\tau}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{ij}\left(\frac{dp^{i}}{d\tau}p^{j}+\frac{dp^{j}}{d\tau}p^{i}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\frac{dp^{i}}{d\tau}\eta_{ij}p^{j}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2p_{i}\frac{dp^{i}}{d\tau} \ \ \ \ \ (14)$

In line 3 we used the fact that in flat space ${\eta_{ij}}$ is a constant. In line 4, we used the symmetry of the flat space metric: ${\eta_{ij}=\eta_{ji}}$ to swap the indices on the second term.

Now the invariant ${\mathbf{p}\cdot\mathbf{p}=-m^{2}}$ is a constant, so ${\frac{d\left(\mathbf{p}\cdot\mathbf{p}\right)}{d\tau}=0}$. Therefore

 $\displaystyle 2p_{i}\frac{dp^{i}}{d\tau}$ $\displaystyle =$ $\displaystyle 2\left(mu_{i}\right)\left(qF^{ij}u_{j}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2qmu_{i}u_{j}F^{ij}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (17)$

In order for this to be true in general, that is for all four-velocities, we need to impose a condition on ${F^{ij}}$. We could just require ${F^{ij}=0}$, but that wouldn’t get us anywhere, since that would mean the electric field would have to be zero. The trick lies in the summation; if we require ${F^{ij}=-F^{ji}}$ then

 $\displaystyle u_{i}u_{j}F^{ij}$ $\displaystyle =$ $\displaystyle -u_{i}u_{j}F^{ji}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -u_{i}u_{j}F^{ij} \ \ \ \ \ (19)$

where in the last line we swapped the two dummy indices. Thus if ${F^{ij}}$ is anti-symmetric, the condition is automatically satisfied. This allows us to fill out the tensor a bit more:

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & - & -\\ -E_{y} & - & 0 & -\\ -E_{z} & - & - & 0 \end{array}\right] \ \ \ \ \ (20)$

For a particle at rest, ${u_{j}=0}$ for ${j=1,2,3}$ and ${u_{0}=-1}$, so for ${i=1,2,3}$ from above we have

 $\displaystyle \frac{dp^{i}}{d\tau}$ $\displaystyle =$ $\displaystyle qF^{i0}u_{0}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -qF^{i0} \ \ \ \ \ (22)$

which is the same as 8, since for a particle at rest ${t=\tau}$. So far, so good. However, we can now use 9 to see what happens when the particle is not at rest. We will fill in the tensor with the (admittedly suggestive) symbols shown:

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (23)$

At this point, we don’t know what the ${B}$s are; we’re just using them as placeholders in the tensor. Then from 9 we get for the ${x}$ component

 $\displaystyle \frac{dp^{x}}{d\tau}$ $\displaystyle =$ $\displaystyle qF^{xj}u_{j}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q\left(-E_{x}u_{0}+0+u_{y}B_{z}-u_{z}B_{y}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q\left(-E_{x}u_{0}+\left[\mathbf{u}\times\mathbf{B}\right]_{x}\right) \ \ \ \ \ (26)$

This looks a lot like the Lorentz (this guy gets around) force law. If we use ${p^{i}=mu^{i}}$, ${u^{0}=-u_{0}=\gamma}$ and ${u^{i}=u_{i}=\gamma v^{i}}$ for ${i=1,2,3}$ we get

$\displaystyle m\frac{dv^{x}}{d\tau}=q\gamma\left(E_{x}+\left[\mathbf{v}\times\mathbf{B}\right]_{x}\right) \ \ \ \ \ (27)$

Finally, we note that

 $\displaystyle \frac{dv^{i}}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{dv^{i}}{dt}\frac{dt}{d\tau}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dv^{i}}{dt}\gamma \ \ \ \ \ (29)$

The final form is therefore

$\displaystyle m\frac{dv^{x}}{dt}=q\left(E_{x}+\left[\mathbf{v}\times\mathbf{B}\right]_{x}\right) \ \ \ \ \ (30)$

with similar forms for the ${y}$ and ${z}$ components. This really is the Lorentz force law, and as we’ve seen from the derivation, it is valid in relativity as well as Newtonian physics, since we used relativistic four-vectors throughout, and never made the approximation of small velocities.

In one sense, we can take this as a prediction of the magnetic field and of the Lorentz force law, since these came out of our generalization of the equations for electrostatics (without any reference to magnetic fields).

There is a lot of ‘try it and see’ in this derivation, but it seems that’s the way a lot of physics works. It’s not like mathematics where we specify a minimal set of axioms and then rigorously derive every other result from them.

# Electromagnetic field tensor: a couple of Maxwell’s equations

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.10, 7.2.

The electromagnetic field tensor ${F^{ij}}$ is

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

The various electromagnetic laws can be expressed by the equation

$\displaystyle \partial_{i}F_{jk}+\partial_{k}F_{ij}+\partial_{j}F_{ki}=0 \ \ \ \ \ (2)$

The lowered version ${F_{ij}}$ of ${F^{ij}}$ in the flat metric of special relativity is

$\displaystyle F_{ij}=\left[\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & B_{z} & -B_{y}\\ E_{y} & -B_{z} & 0 & B_{x}\\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (3)$

Because ${F_{ij}=-F_{ji}}$, if we set two of the indices equal in this equation, the LHS is identically zero. For example if ${i=j}$:

 $\displaystyle \partial_{i}F_{ik}+\partial_{k}F_{ii}+\partial_{i}F_{ki}$ $\displaystyle =$ $\displaystyle -\partial_{i}F_{ki}+0+\partial_{i}F_{ki}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

Since the three terms are cyclic permutations of each other, setting any other pair of indices equal gives the same result.

If we choose ${i=y}$, ${j=x}$, ${k=z}$, we get

$\displaystyle -\partial_{y}B_{y}-\partial_{z}B_{z}-\partial_{x}B_{x}=0 \ \ \ \ \ (6)$

This is the law ${\nabla\cdot\mathbf{B}=0}$.

If we now choose ${i=y}$, ${j=x}$, ${k=t}$, we get

$\displaystyle -\partial_{y}E_{x}+\partial_{t}B_{z}+\partial_{x}E_{y}=0 \ \ \ \ \ (7)$

This is the ${z}$ component of Faraday’s law ${\nabla\times\mathbf{E}+\frac{\partial\mathbf{B}}{\partial t}=0}$.

Other choices give the remaining components of Faraday’s law.

# Electromagnetic field tensor: invariance under Lorentz transformations

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; Problems 4.10.

The electromagnetic field tensor ${F^{ij}}$ transforms between inertial frames by using 2 Lorentz transformations (we’ll see why in a future post):

$\displaystyle F^{\prime ij}=\Lambda_{\;\; a}^{i}\Lambda_{\;\; b}^{j}F^{ab} \ \ \ \ \ (1)$

Using this, let’s see how the quantity ${\eta_{ia}\eta_{jb}F^{ij}F^{ab}}$ that we considered in the last post transforms. Since the metric tensor is invariant under Lorentz transformations

$\displaystyle \eta_{ia}\eta_{jb}F^{\prime ij}F^{\prime ab}=\eta_{ia}\eta_{jb}\Lambda_{\;\; c}^{i}\Lambda_{\;\; d}^{j}F^{cd}\Lambda_{\;\; e}^{a}\Lambda_{\;\; f}^{b}F^{ef} \ \ \ \ \ (2)$

We can now use

$\displaystyle \left(\Lambda^{-1}\right)_{\;\; k}^{a}\eta_{ab}=\eta_{kj}\Lambda_{\;\; b}^{j} \ \ \ \ \ (3)$

So we substitute in 2

 $\displaystyle \eta_{ia}\Lambda_{\;\; c}^{i}$ $\displaystyle =$ $\displaystyle \left(\Lambda^{-1}\right)_{\;\; a}^{i}\eta_{ic}\ \ \ \ \ (4)$ $\displaystyle \eta_{jb}\Lambda_{\;\; d}^{j}$ $\displaystyle =$ $\displaystyle \left(\Lambda^{-1}\right)_{\;\; b}^{j}\eta_{jd} \ \ \ \ \ (5)$
 $\displaystyle \eta_{ia}\eta_{jb}\Lambda_{\;\; c}^{i}\Lambda_{\;\; d}^{j}F^{cd}\Lambda_{\;\; e}^{a}\Lambda_{\;\; f}^{b}F^{ef}$ $\displaystyle =$ $\displaystyle \left(\Lambda^{-1}\right)_{\;\; a}^{i}\eta_{ic}\left(\Lambda^{-1}\right)_{\;\; b}^{j}\eta_{jd}F^{cd}\Lambda_{\;\; e}^{a}\Lambda_{\;\; f}^{b}F^{ef}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\Lambda^{-1}\right)_{\;\; a}^{i}\Lambda_{\;\; e}^{a}\eta_{ic}\left(\Lambda^{-1}\right)_{\;\; b}^{j}\Lambda_{\;\; f}^{b}\eta_{jd}F^{cd}F^{ef}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{\;\; e}^{i}\eta_{ic}\delta_{\;\; f}^{j}\eta_{jd}F^{cd}F^{ef}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{ec}\eta_{fd}F^{cd}F^{ef} \ \ \ \ \ (9)$

Thus the quantity ${\eta_{ia}\eta_{jb}F^{ij}F^{ab}}$ is invariant under Lorentz transformations. As we saw in the last post, this quantity is

$\displaystyle \eta_{ia}\eta_{jb}F^{ij}F^{ab}=2\left(B^{2}-E^{2}\right) \ \ \ \ \ (10)$