18
$\begingroup$

When books or various references interpret the meaning of Maxwell equations, they typically state that the source (origin of the phenomena) is the right part of the formula, and the resulting effect is on the left part of the formula.

For example, for Maxwell-Faraday law, $\vec{\nabla} \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$ one states

"a time varying magnetic field creates ("induces") an electric field." (see for example : https://en.wikipedia.org/wiki/Maxwell%27s_equations#Faraday's_law )

It seems to me that this is not true. One could interpret in both direction.

For the example above, we could also state that a change of direction of the electric field will create a temporal change of the magnetic field.

Is it true that Maxwell equations should be interpreted by taking right side of formula as the "origin" and the left part as "consequence"? Or could we take also the left side as the origin?

$\endgroup$
26
$\begingroup$

You are basically correct but I think I can elucidate existing answers by pointing out that there are two issues here: a physical one and a mathematical one.

Maxwell's equations are making both mathematical and physical statements. The relationship between the left hand side and the right hand side is not a cause-effect relationship. But when we use the equations to find out how the field at any given place comes about, then we do find a cause-effect relationship: the field at any given place can be expressed in terms of the charge density and current on the surface of the past light cone of that event.

Mathematically the Maxwell equations have the form of differential equations. Looking at the first one, we would normally regard it as telling us something about the electric field if the charge density is given, but you can equally well regard it as telling us the charge density if the electric field is given. The difference between these two perspectives is that the second is mathematically not challenging and does not require any great analysis: if $\bf E$ is known then to find $\rho$ you just do some differentiation and a multiplication by a constant: all fairly simple. But if you have a known charge density and want to find the electric field, you have a lot more work to do, and indeed the problem cannot be solved at all unless you know quite a lot: to get the electric field at one point you need to know the charge density and current on the entire past light cone. Since this calculation is harder it earns some mathematical respect and there is terminology associated with it. We say (mathematically speaking) that $\rho$ is a 'source term' in a differential equation for $\bf E$. This is somewhat reminiscent of cause and effect but strictly speaking it is only indirectly related to cause and effect as I already said.

$\endgroup$
2
  • 1
    $\begingroup$ "the field at any given place can be expressed in terms of the charge density and current on the surface of the past light cone of that event." Maybe I am missing something, but what about free electromagnetic field? Or do you assume something like Sommerfeld radiation condition? $\endgroup$
    – akhmeteli
    Aug 20 '21 at 8:15
  • $\begingroup$ @akhmeteli thanks: a helpful comment. I don't think one can tell what boundary condition at past infinity ought to be taken, but you are correct to point out that if there are already propagating fields in the distant past then one might have simply to take that as part of a boundary condition. $\endgroup$ Aug 27 '21 at 16:34
13
$\begingroup$

Maxwell's equations, in some sense, are better understood by looking at them as two sets of equations. The dynamical variables in classical electrodynamics are the fields. We can thus group the equations such that there are two homogenous PDEs for the dynamical variables - the fields. These are:

$$\nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 ​$$

$$\nabla \cdot \vec{B} = 0$$

The above equations are generally valid.

Additionally we have the following two equations:

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ $$\nabla \times \vec{B} = \mu_0\left(J + \epsilon_0\frac{\partial \vec{E}}{\partial t}\right)$$

In these equations, the RHS are normally said to contain the 'source' terms. Why? The presence of a (static) charge configuration generates an electric field. The presence of a current density or a time varying electric field, can generate a magnetic field. In that sense, the terms on the right generate the fields and their dynamics.

Of course, in the context of boundary value problems and so on, one can have electric fields in a charge free region. But the idea is that there are fundamental physical objects such as charges and currents which generate the dynamical variables of electrodynamics.

Interestingly, in a relativistically covariant formalism of electrodynamics, the four Maxwell equations reduce to two tensor equations, one which contains the first two homogenous equations, and the other which contains the second two equations containing the sources.

$\endgroup$
2
8
$\begingroup$

You are correct. Maxwell-Faraday states that $\vec{\nabla} \times \vec{E}$ is the same thing as $-\frac{\partial \vec{B}}{\partial t}$. Both quantities express the same phenomenon. There is no cause and effect in either direction.

Once you express the fields in terms of derivatives of the vector potential the two expressions become identical. As the Coulomb potential is rotation free we can set it to zero for this excercise. Then $\vec{E} = -\partial_t \vec{A}$ and since $\vec{B} = \vec{\nabla} \times \vec{A}$ both sides of the Maxwell-Faraday are seen to be identical. In this sense Maxwell-Faraday expresses the fact that the fundamental quantity of electromagnetism is the vector potential.

$\endgroup$
8
$\begingroup$

Is it true that Maxwell equations should be interpreted by taking right side of formula as the "origin" and the left part as "consequence" ?

No. If you want to describe something as a consequence then it must happen later than the thing that it is a consequence of. This relationship is usually called cause and effect or causality. Maxwell’s equations do not express a cause and effect relationship.

In electromagnetism the equation that describes the cause of electromagnetic fields is called Jefimenko’s equations: https://en.m.wikipedia.org/wiki/Jefimenko's_equations

Note that these equations do describe a true cause and effect relationship since the right hand side happens at the so-called retarded time, which is earlier than the left hand side. The causes of the fields are the charges and currents, not the other fields.

$\endgroup$
5
  • $\begingroup$ I believe the reason that we think of cause-action relationships as that the action must happen after than its cause is that we perceive time as running forward. I don’t see why relativistic cause-action relationships of our physical universe, such as a magnet field corresponding to an electric field, should also obey this rule as time plays a more fundamental role here, possibly being orthogonal to our perception of „before“ and „after“. Therefore I wouldn’t take „No“ as a general answer to the question. $\endgroup$ Aug 19 '21 at 23:02
  • $\begingroup$ @returntrue Maxwell’s equations are fully relativistic, so my answer applies perfectly fine considering all relativistic effects $\endgroup$
    – Dale
    Aug 20 '21 at 0:00
  • $\begingroup$ I tried to say that a cause-action effect does not necessarily need the action to occur after the cause, especially in a relativistic setting. So my point stands. I only object to your answer's first two sentences. $\endgroup$ Aug 20 '21 at 8:46
  • 1
    $\begingroup$ @returntrue it definitely does. In fact, in relativity the requirement is stronger, not weaker. In relativity the effect must occur in the future light cone of the cause, which is not only after the cause, but also close enough to the cause that light or something slower than light could travel from the cause to the effect. The first two sentences hold in relativity. $\endgroup$
    – Dale
    Aug 20 '21 at 11:14
  • 1
    $\begingroup$ That makes sense! $\endgroup$ Aug 20 '21 at 15:49
3
$\begingroup$

Although the answer to your main question is no, there is a causal relationship contained within Maxwell's equations through the partial derivatives wrt time terms. For example, you can find the 'effect' $\vec{B} (t+dt)$ in the future from the 'causes' in the present $\vec{B}(t)$ and ${\nabla} \times \vec{E}(t)$ via:

$\vec{B}(t+dt)= \vec{B(t)}-dt{\nabla} \times \vec{E(t)}$

There are other ways of modelling CED causality such as via the Lienard-Wiechert fields of a moving charge, or through Jefimenko's equations; but nevertheless Maxwell's equations are causal if you look in the right place.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.