# Relativistic Lorentz force

### Requirements for invariant force laws

To correct the Newtonian equations of motion - or force laws - so that they are the same in all reference frames, we need to write them in terms of four-vectors - meaning we must write them in the form $$F^\mu = \frac{\mathrm{d} P^\mu}{\mathrm{d} \tau}$$ where the components of a certain four-vector, which we could call the proper force, are equal to components of the derivative of the proper momentium with respect to the proper time.

All invariant force laws must satisfy certain conditions:

#### The proper force must be normal to the proper velocity.

That is, $$F^\mu U_\mu = 0$$ where $U_\mu = \eta_{\mu \nu}U^\nu$ are the covariant components of the four-velocity. This stems from the fact that proper velocity has unit length, $U^\mu U_\mu = 1$, from which it follows that $$F^\mu U_\mu = \frac{\mathrm{d} P^\mu}{\mathrm{d} t} U_\mu = m\frac{\mathrm{d} U^\mu}{\mathrm{d} t} U_\mu = m\frac{\mathrm{d} }{\mathrm{d} t} \left(U^\mu U_\mu\right) = 0$$

#### The law must tend to the Newtonian limit in non-relativistic situations.

Given a particular force, when velocities are small compared to the speed of light, the proper force must approximate the Newtonian version of the law, $$F^m \approx \frac{\mathrm{d} P^m}{\mathrm{d} t}$$ where $m$ ranges over the spatial coordinates and the derivative is now with respect to the coordinate time, rather than the proper time.

#### The law must be the same in all reference frames.

Finally, the law must be invariant under Lorentz transformations. Given any other (primed) reference frame, we must have $$F^{\mu'} = \frac{\mathrm{d} P^{\mu'}}{\mathrm{d} \tau}$$
This is the basic premise of the exercise - to convert non-relativistic equations of motion into relativistic ones.

### The relativistic law of motion of a charge particle in an electromagnetic field

We shall show that the invariant version of Lorentz's law of motion of a charge particle in an electromagnetic field is $$\frac{\mathrm{d} P^\mu}{\mathrm{d} \tau} = q F^{\mu \nu} U_\nu$$ where $F^{\mu \nu}$ is the anti-symmetric, rank-two tensor called the Faraday tensor which has six independent components in the coordinate frame (where the three-velocity of the particle is $\vec{v}$) - the three electric field vector components, $$E_m \text{ where } m=1,2,3$$ and the three independent components of the anti-symmetric magnetic field tensor given by $$B_{n k} = -B_m \text{ where } (m,n,k) \text{ is a cycle of } (1,2,3)$$

Thus, in a coordinate frame, $$F^{\mu \nu} = \begin{bmatrix}0 & -E_1 & -E_2 & -E_3\\E_1 & 0 & -B_3 & B_2\\E_2 & B_3 & 0 & -B_1\\E_3 & -B_2 & B_1 & 0\end{bmatrix}$$

Nb. There is (at least, I think) one other tensor that satisfies the conditions, and we shall need to define it when reformulating Maxwell's equations in the next lecture.

#### Force normal to velocity

The first condition is easily satisfied, because $F^{\mu \nu}$ being anti-symmetric means $$F^{\mu \nu} U_\mu U_\nu = 0$$

#### Newtonian limit

Now, the left-hand side of the force equation - in the coordinate frame - is $$\frac{\mathrm{d} P^\mu}{\mathrm{d} \tau} = \gamma \frac{\mathrm{d} P^\mu}{\mathrm{d} t}$$ and we can calculate each of the spatial components of the right-hand side, remembering that the covariant four-velocity looks like $$U_\mu = \left(\gamma, -\gamma v_1, -\gamma v_2, -\gamma v_3\right)$$

Thus, \begin{align*}\gamma \frac{\mathrm{d} P^1}{\mathrm{d} t} &= q F^{1 \nu} U_\nu\\&= q\left( (E_1)(\gamma) + (0)( -\gamma v_1) + (-B_3)( -\gamma v_2) + (B_2)( -\gamma v_3)\right)\\&= \gamma q \left(E_1 + v_2 B_3 - v_3 B_2 \right)\\&= \gamma q \left[\vec{E} + \vec{v} \times \vec{B}\right]_1\end{align*}

The other spatial components follow the same pattern, so we can write $$\gamma \frac{\mathrm{d} P^m}{\mathrm{d} t} = q F^{m \nu} U_\nu = \gamma q \left[\vec{E} + \vec{v} \times \vec{B}\right]_m$$

Dividing everything by $\gamma$, we seem to recover the Newtonian Lorentz force. However, there is a hidden factor of $\gamma$ in the momentum (using $n$ for the index to save confusion with the mass $m$), $$P^n = m U^n = \gamma m v_n$$ and so, $$\frac{\mathrm{d}}{\mathrm{d} t} \left(\gamma m v_n\right) = q \left[\vec{E} + \vec{v} \times \vec{B}\right]_n$$

Now, if $v = \sqrt{v_n v_n}$ is small compared to the speed of light, then $\gamma \approx 1$, and $$\frac{\mathrm{d}}{\mathrm{d} t} \left(m v_n\right) = q \left[\vec{E} + \vec{v} \times \vec{B}\right]_n$$

Since $$\frac{\mathrm{d}}{\mathrm{d} t} \left(m v_n\right) = \left[\frac{\mathrm{d}}{\mathrm{d} t} \left(m \vec{v}\right)\right]_n = \left[\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}\right]_n$$ we can drop the indices and completely recover the Newtonian form $$\frac{\mathrm{d}\vec{p}}{\mathrm{d} t} = q \left(\vec{E} + \vec{v} \times \vec{B}\right)$$

#### Invariance

The last, and most important requirement for this formulation is that is the same in all reference frames. To show that this is indeed the case, we need a result from the next section, which tells us how covariant four-vectors transform.

In short, if $L_\nu^\mu$ are the components of a given Lorentz transformation matrix and $X_\mu$ the covariant components of some four-vector, then $$X_{\mu'} = \tilde{L}_{\mu'}^\nu X_\nu$$ where $\tilde{L}_\mu^\nu$ are components of the inverse transformation matrix, so that $$\tilde{L}_\rho^\mu L_\nu^\rho = \delta_\nu^\mu = \left\{\begin{matrix}1 & \text{ if } \mu = \nu\\0 & \text{ otherwise }\end{matrix}\right.$$ where $\delta_\nu^\mu$ is a four-dimensional version of the Kronecker delta and has the same components as the identity matrix.

We want to show that the equation, $$\frac{\mathrm{d} P^\mu}{\mathrm{d} \tau} = q F^{\mu \nu} U_\nu$$ is the same in all reference frames. So, consider \begin{align*}q F^{\mu' \nu'} U_{\nu'} &= q \left(L_\rho^{\mu'} L_\sigma^{\nu'} F^{\rho \sigma} \right)\left(\tilde{L}_{\nu'}^{\omega} X_\omega\right)\\&= q L_\rho^{\mu'} \left(L_\sigma^{\nu'} \tilde{L}_{\nu'}^{\omega}\right) F^{\rho \sigma} X_\omega\\&= q L_\rho^{\mu'} \delta_\sigma^\omega F^{\rho \sigma} X_\omega\\&= q L_\rho^{\mu'} F^{\rho \sigma} X_\sigma\\&= L_\rho^{\mu'} \left(q F^{\rho\sigma} X_\sigma\right)\\&= L_\rho^{\mu'} \frac{\mathrm{d} P^\rho}{\mathrm{d} \tau}\\&= \frac{\mathrm{d} P^{\mu'}}{\mathrm{d} \tau}\end{align*}

Thus, we have shown that the four-vector force law for charged particles in electromagnetic fiields satisfies the three requirements of an invariant force law.

### Putting the speed of light back in

We can abuse matrix notation somewhat and write the equations more succinctly in terms of the three-velocity $\vec{v}$, the electric field $\vec{E}$ and the magnetic tensor $\mathbf{B}$, $$\gamma \frac{\mathrm{d}}{\mathrm{d} t}\begin{bmatrix} \gamma m \\ \gamma m \vec{v}\end{bmatrix} = q \begin{bmatrix}0 & -\vec{E} \\ \vec{E} & \mathbf{B}\end{bmatrix} \begin{bmatrix} \gamma \\ -\gamma \vec{v}\end{bmatrix}$$ where here, \begin{align*}U_\mu &\rightarrow \begin{bmatrix} \gamma \\ -\gamma \vec{v}\end{bmatrix}\\P^\mu &\rightarrow \begin{bmatrix} \gamma m \\ \gamma m \vec{v}\end{bmatrix}\\F^{\mu \nu} &\rightarrow \begin{bmatrix}0 & -\vec{E} \\ \vec{E} & \mathbf{B}\end{bmatrix}\end{align*}

Nb. Covariant four-vectors form column-vectors (as do contravariant four-vectors).

Setting $c \neq 1$ means we multiply the time components of the two four-vectors by $c$, \begin{align*}U_\mu &\rightarrow \begin{bmatrix} \gamma c \\ -\gamma \vec{v}\end{bmatrix}\\P^\mu &\rightarrow \begin{bmatrix} \gamma m c \\ \gamma m \vec{v}\end{bmatrix}\end{align*} where, as usual, \begin{align*}\gamma &\rightarrow \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\\v^2 &= \vec{v} \cdot \vec{v}\end{align*}

But what about the electromagnetic tensor (Faraday tensor)? We need to ensure that the equations still tend to the Newtonian limit as $v$ gets small compared to $c$. We can, naively, just use the current version and then see how we need to correct it, $$\gamma \frac{\mathrm{d}}{\mathrm{d} t}\begin{bmatrix} \gamma m c \\ \gamma m \vec{v}\end{bmatrix} = q \begin{bmatrix}0 & -\vec{E} \\ \vec{E} & \mathbf{B}\end{bmatrix} \begin{bmatrix} \gamma c\\ -\gamma \vec{v}\end{bmatrix}= \gamma \begin{bmatrix}q \vec{E} \cdot \vec{v}\\q \left(c \vec{E} + \vec{v} \times \vec{B}\right)\end{bmatrix} \text{ wrong !!}$$

In order to get the correct answer for the Newtonian momentum, we need to divide the electric field vector by the speed of light in the Faraday tensor. That is, $$F^{\mu \nu} \rightarrow \begin{bmatrix}0 & -\vec{E}/c \\ \vec{E}/c & \mathbf{B}\end{bmatrix}$$ in which case, the correct law when $c \neq 1$ is $$\gamma \frac{\mathrm{d}}{\mathrm{d} t}\begin{bmatrix} \gamma m c \\ \gamma m \vec{v}\end{bmatrix} = q \begin{bmatrix}0 & -\vec{E}/c \\ \vec{E}/c & \mathbf{B}\end{bmatrix} \begin{bmatrix} \gamma c\\ -\gamma \vec{v}\end{bmatrix} = \gamma \begin{bmatrix}q \frac{\vec{E}}{c} \cdot \vec{v}\\q \left(\vec{E} + \vec{v} \times \vec{B}\right)\end{bmatrix}$$

The spatial components haven't changed (except for $\gamma$ of course), but the time component on the right-hand side has been divided by $c$. We can show that this works out correctly by multiplying both sides of the first equation by $c$ to get $$\frac{\mathrm{d}}{\mathrm{d} t}\left(\gamma m c^2\right) = q \vec{E} \cdot \vec{v} = q \left(\vec{E} + \vec{v} \times \vec{B}\right) \cdot \vec{v} = \vec{F} \cdot \vec{v} = \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{1}{2}mv^2\right)$$ which shows that both sides of the equation have the same dimensions.

Thus, the invariant version of Lorentz's law of motion of a charge particle in an electromagnetic field, even when $c \neq 1$, is $$\frac{\mathrm{d} P^\mu}{\mathrm{d} \tau} = q F^{\mu \nu} U_\nu$$ (going back to normal index notation) and the speed of light only shows up in time components (in a coordinate frame) of the constituent tensors, \begin{align*}U_\mu &\rightarrow \begin{bmatrix} \gamma c \\ -\gamma v_x \\ -\gamma v_y \\ -\gamma v_z\end{bmatrix}\\P^\mu &\rightarrow \begin{bmatrix} \gamma m c \\ \gamma m v_x \\ \gamma m v_y \\ \gamma m v_z\end{bmatrix}\\F^{\mu \nu} &= \begin{bmatrix}0 & -E_x/c & -E_y/c & -E_z/c\\E_x/c & 0 & -B_z & B_y\\E_y/c & B_z & 0 & -B_x\\E_z/c & -B_y & B_x & 0\end{bmatrix}\end{align*} and in $\gamma$, of course.

### Transformation properties of the Faraday tensor

Given we know the components of the Faraday tensor in one frame, we would like to know its components in another (primed) frame, moving at speed $v$ relative to the first, along their common $x$-axes - the usual setup. Also, let's go back to $c \equiv 1$.

Nb. This is not the speed, $v$ of the particle in the Lorentz force - we are not considering the equation now, just the tensor.

Thus, since it's a rank-two tensor, the Faraday tensor in the same way as a product of two four-vectors, $$F^{\mu' \nu'} = L_\rho^{\mu'} L_\sigma^{\nu'} F^{\rho \sigma}$$ where $L_\nu^\mu$ are the components of the given Lorentz transformation.

Now, explicitly, a single four-vector transforms as \begin{align*}X^{0'} &= \gamma\left(X^0 - v X^1\right)\\X^{1'} &= \gamma\left(-v X^0 + X^1\right)\\X^{2'} &= X^2\\X^{3'} &= X^3\end{align*}

Suppose we wanted to know how the $x$-component of the electric field transforms. Well, $E_{x'} = F^{1' 0'}$ which transforms in the same as the product of two vectors, \begin{align*}X^{1'} Y^{0'} &= \gamma \left(-v X^0 + X^1\right) \gamma\left(Y^0 - v Y^1\right)\\&= \gamma^2 \left(-v X^0 Y^0 + v^2 X^0 Y^1 + X^1 Y^0 - v X^1 Y^1\right)\end{align*}

So, using the same pattern, \begin{align*}E_{x'} &= F^{1' 0'}\\&= \gamma^2 \left(-v F^{0 0} + v^2 F^{0 1} + F^{1 0} - v F^{1 1}\right)\\&= \gamma^2 \left(-v \cdot 0 + v^2 (-E_x) + E_x - v \cdot 0\right)\\&= \gamma^2 \left(1 - v^2\right) E_x\\&= E_x\end{align*}

This means that two observers moving at any relative speed along their common $x$-axes will measure exactly the same value for the $x$-component of the electric field.

How about the $y$-component of the electric field? $E_{y'} = F^{2' 0'}$ transforms in the same way as $$X^{2'} Y^{0'} = \left(X^2\right) \gamma\left(Y^0 - v Y^1\right) = \gamma \left(X^2 Y^0 - v X^2 Y^1\right)$$ and so \begin{align*}E_{y'} &= F^{2' 0'}\\&= \gamma \left(F^{2 0} - v F^{2 1}\right)\\&= \gamma \left(E_y - v B_z\right)\\&= \frac{E_y - v B_z}{\sqrt{1 - v^2}}\end{align*}

This means that, for the same two observers, the relative speed is now important, because the $y$-component of the electric field in one frame gets mixed up with the $z$-component of the magnetic field in the other frame.

The full set of transformations are \begin{align*}E_{x'} &= E_x\\E_{y'} &= \frac{E_y - v B_z}{\sqrt{1 - v^2}}\\E_{z'} &= \frac{E_z + v B_y}{\sqrt{1 - v^2}}\end{align*} and \begin{align*}B_{x'} &= B_x\\B_{y'} &= \frac{B_y + v E_z}{\sqrt{1 - v^2}}\\B_{z'} &= \frac{B_z - v E_y}{\sqrt{1 - v^2}}\end{align*}