# Max Eq in Cmplex Form

## Main Question or Discussion Point

I recently read that Maxwell's equations can be written in a more concise form as the following:

$$\nabla\times\vec{M} = \frac{-i}{c}\frac{\partial\vec{M}}{\partial t}$$
$$\nabla\cdot\vec{M} = 0$$

where $$\vec{M} = c\vec{B} + i\vec{E}$$.

Why is this form not popular?

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marcusl
Gold Member
What does it allow you to do that you can't do with the regular form?

Its easy to remember for one. And I guess it can simplify certain calculations because now you would be able to compress everything into fewer equations...

marcusl
Gold Member
Ok, that makes sense. BTW, these aren't the full eqns, they're only valid in source-free regions. Maybe that's why they aren't popular.

Hurkyl
Staff Emeritus
Gold Member
That's easy to fix -- just apply the ordinary Maxwell's equations to compute what Div M and Curl M should be. I think it just amounts to adding in (electric) current density to the first and (electric) charge density to the second.

P.S. have you tried computing $M \cdot M^*$ and $M \times M^*$?

P.S. have you tried computing $M \cdot M^*$ and $M \times M^*$?
I tried but I it doesn't make sense. How can you dot two imaginary numbers? Or cross them for that matter? How is such a thing defined?

$$\vec{M}\cdot\vec{M}^* = (c\vec{B} + i\vec{E})\cdot(c\vec{B} - i\vec{E}) = ??$$

Hurkyl
Staff Emeritus
Gold Member
You're not crossing and dotting imaginary numbers -- you're crossing and dotting complex vectors. It's defined in exactly the same way as you would for real vectors. In particular, it satisfies the distributive law, and you can always pull out scalar factors. (Any complex number is a scalar!)

I see. So basically,
$$\vec{M}\cdot\vec{M}^* = (c\vec{B} + i\vec{E})\cdot(c\vec{B} - i\vec{E})$$
$$= c\vec{B}\cdot(c\vec{B} - i\vec{E}) + i\vec{E}\cdot(c\vec{B} - i\vec{E})$$

$$= c^2 (\vec{B}\cdot\vec{B}) - ic(\vec{B}\cdot\vec{E}) + ic(\vec{E}\cdot\vec{B}) - i^2(\vec{E}\cdot\vec{E}) = c^2{||\vec{B}||}^2 + {||\vec{E}||}^2$$

So what's the significance of this number?

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Hurkyl
Staff Emeritus
Gold Member
It's proportional to electromagnetic energy density!

The cross product I mentioned turns out to be proportional to electromagnetic power flux.

I guess the real reason is that, while E and B fields are known to be different manifestations of really the same object (nicely represented by the complex 3-vector M, or alternatively by the electromagnetic field tensor), they are measured in very different ways and have radically different effects (in a particular frame). So it makes sense to separate out the equations for B and E fields as it may sometimes cloud the problem. E.g. in a simple use of Ampere's law/Gauss' law, is there a need for the complex form of the equations?

In any case, I prefer using the field tensor to tidy up and condense equations. This is mainly because the transformation rules of the field tensor are glaringly obvious, whereas they are fairly complicated in the case of the complex 3-vector. Furthermore, useful invariants are also easily obtained. And things like equations of motion of a charged particle in an EM field is easily written with regard to the field tensor as opposed to the complex 3-vector M (as far as I can see).

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Swapn,

Remember that the most concise form is the following:

d F = 0
d*F = *J

where

*F is the dual of F,
d is the exterior derivative

This form is very popular too in the study of relativity.
Moreover, this compact system has a simple physical interpretation (see MTW for full details).

Michel

Remember that the most concise form is the following:

d F = 0
d*F = *J

where

*F is the dual of F,
d is the exterior derivative

This form is very popular too in the study of relativity.
Thanks Michel and masudr. I did not know that before.

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Here's is a little more general form of the Maxwell's equations in complex form:

$$\nabla\cdot\vec{M} = \frac{i\rho}{\epsilon_0}$$
$$\nabla\times\vec{M} = c\mu_0\vec{J} + \frac{-i}{c}\frac{\partial\vec{M}}{\partial t}$$

where $$\vec{M} = c\vec{B} + i\vec{E}$$.

I am pretty sure this is correct.

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For either of the forms that Swapnil gave (i.e. with or without sources), what would the Lorentz force equation look like? I mean the one that normally looks like F = q(E + v X B)? I guess it'd be some sort of F=q*f(M), but ... ?

I don't see any particular advantages using this "complex vectors"... the main thing about using complex number is the Euler's equation (as far as what I've learnt)
$$e^{ix}=\cos x+i\sin x$$

unless you can think of an analogous equation for M... hmmm

samalkhaiat
Remember that the most concise form is the following:

d F = 0
d*F = *J

Nicer and more compact form of Maxwell's equations can be found as follow;
Use

$$M_{0}=0$$

$$M_{i}=H_{i} +i E_{i}$$

to define the 2-component (column) objects;

$$\Psi_{1}= (M_{0}+M_{3} , M_{1}+iM_{2})^{T}$$

$$\Psi_{2}= (M_{1}-iM_{2} , M_{0}-M_{3})^T$$

Also, write

$$\mathcal{J}_{1}= 4\pi (\rho +j_{3} , j_{1}+ij_{2})^T$$

$$\mathcal{J}_{2}= 4\pi (j_{1}-ij_{2} ,\rho - j_{3})^T$$

Then you could write the 8 real Maxwell equations in the spinor form (two 2-componen spinor equation):

$$i \sigma_{\mu} \partial^{\mu} \Psi_{a} = \mathcal{J}_{a}$$

Notice the similarity with the Dirac equation in its 2-component form.

Also, for $$\mathcal{J}_{a}=0$$

the equation reduces to Weyl's equation for massless "fermion" field.

regards

sam

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