# Lorentz force law

### Newton's law

For non-relativistic speeds, Newton's law, $$\vec{F} = m \vec{a}$$ is a very good law.

Forces and accelerations are expressed in terms of three-vectors, which, along with the mass, $m$ are invariants, hence the equation itself is invariant.

We usually write the force in terms of momentum, $\vec{p} = m \vec{v}$, $$\vec{F} = \frac{\mathrm{d} \vec{p}}{\mathrm{d} t}$$ where, since mass is invariant, $$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = \frac{\mathrm{d}}{\mathrm{d} t}\left(m \vec{v}\right) = m \frac{\mathrm{d}\vec{v}}{\mathrm{d} t} = m \vec{a}$$

This equation does transcend Newtonian mechanics but with a significant change - the derivative should be with respect to proper time, rather than coordinate time.

This comes about about when the kinematics involve relativistic speeds, meaning that the relationship between force and acceleration has to be modified.

The four-vector equivalent of this equation should have the form, $$F^\mu = \frac{\mathrm{d} P^\mu}{\mathrm{d} \tau}$$

### Lorentz force

Lorentz would have been thinking in terms of Newtonian physics when he wrote down the laws of motion of an electrically charged particle in an electromagnetic field, $$\vec{F} = q\left(\vec{E} + \vec{v} \times \vec{B}\right)$$ where

• $q$ is the electric charge of the particle
• $\vec{v}$ is the velocity of the charged particle
• $\vec{E}$ is the electric part of the field
• $\vec{B}$ is the magnetic part of the field
• $\vec{v} = \vec{v}\left(t,x,y,z\right), \vec{E} = \vec{E}\left(t,x,y,z\right), \vec{B} = \vec{B}\left(t,x,y,z\right)$ are all functions of time and space

We would like to convert this equation into its correct Lorentz-invariant form so that it remains the same in all reference frames.

If we ignore the magnetic part of the Lorentz force for a moment, then the electric part, $q\vec{E}$, is very similar to the Newtonian gravitational force, $m\vec{G}$, in that they are both versions of inverse-square laws, where the charge, $q$, plays the same role in the Lorentz force as mass, $m$, does in gravity.

So you might think that these two forces would need to be modified in the same way when we re-formulate the laws in terms of four-vectors. But, even ignoring magnetism, the Lorentz force and gravity are very different in relativistic terms. In fact, special relativity does not describe gravity very well at all, and Einstein needed to extend the special theory to the general theory to emcompass it.

Incidentally, the magnetic portion of the force is relativistic itself, since it comes into the law compounded with the velocity - or, more precisely, the velocity divided by the speed of light.

### Mixing of electric and magnetic fields under transformations

It is easy to show that, if we assume that the laws of physics are the same in all reference frames, then the electric and magnetic fields will get mixed up with each other under transformations from frame to frame.

In Newtonian physics, where we consider speeds much, much less than the speed of light, then acceleration is an invariant. It is the same in all reference frames.

Figure 7.1 - Magnetic-only field in coordinate frame

Suppose, at some instant, we have a purely magnetic field, \begin{align*}\vec{E} &= 0\\\vec{B} &= B\hat{y}\end{align*} acting on a charged particle, which - at that instant - is moving with velocity $$\vec{v} = v\hat{x}$$

Then, the force acting on the particle is $$\vec{F} = q\vec{v} \times \vec{B} = qvB\hat{z}$$ and so the acceleration - at that instant and in the coordinate frame - is $$\vec{a} = \frac{qvB}{m}\hat{z}$$

Figure 7.2 - Zero magnetic field in proper frame

Now consider a frame moving in the same speed and direction as the particle.

In this frame, the particle is at rest, $\vec{v}' = 0$ hence there cannot be any contribution to the force on the particle from the magnetic field $$\vec{v}' \times \vec{B}' = 0$$

But, acceleration, being invariant, must be the same in both reference frames, $\vec{a}' = \vec{a}$ and so it must be due to some component of the electric field, $$q\vec{E}' = m\vec{a}' = m\vec{a} = qvB\hat{z}$$

So, even at non-relativistic speeds, mixing of the electric and magnetic fields occurs under transformations.

Nb. At relativistic speeds, acceleration ceases to be an invariant, but remains an absolute quantity. That is, if acceleration is non-zero in one frame, then it will be non-zero in all frames. If it is zero in one frame, then it will be zero in all frames.