# Causality in electromagnetism

1. Jan 3, 2013

### VantagePoint72

We are usually told in an introduction to general relativity that when special relativity was completed, there was a contradiction with Newtonian physics. The Newtonian gravitational force of M on m is $F_g = -G \frac{M m}{r^2} \hat{r}$ where $\vec{r}$ is the spatial vector from M to m. This equation for the gravitational force on a point mass, plus the principle of superposition, are equivalent to Gauss' law for the gravitational field: $\nabla \cdot \vec{g} = -4\pi G \rho_{mass}$.
We conclude from this that the Newtonian gravitational force is instantaneous since any change in $\vec{r}$ immediately translates into a change in the force. Such an instantaneous action-at-a-distance violates special relativity, which tells us that c is the maximum speed of influence. Otherwise, causality is violated.

Now, in the case of electromagnetism, we know that Maxwell's equations are Lorentz invariant. This is true even when they're written in the form usually first encountered in an undergraduate course (i.e. http://www.physics.udel.edu/~watson/phys208/ending2.html), which is to say, not the covariant form in which Lorentz symmetry is obvious. Hence—a point usually emphasized at the end of such a course—classical electromagnetism is fully consistent with special relativity. However, the first of Maxwell's equations is just Gauss' law for the electric field: $\nabla \cdot \vec{E} = \rho_{charge}/\epsilon_0$. This has precisely the same form as Newton's law of gravity in differential form does. So why does this equation imply action-at-a-distance in the case of the Newtonian field but not in the case of the electric field?

2. Jan 3, 2013

### Simon Bridge

So... could it be that undergraduate/introduction courses in physics are incomplete?

It doesn't - you need the other three equations as well where, for Newtonian Gravity, you only have the one. Are Maxwell's equations individually Lorentz invarient?

3. Jan 3, 2013

### VantagePoint72

My point is that even in that form they are Lorentz invariant.

Ah, of course. Thanks.