# Galilean transform and the maxwell equations

1. Jun 23, 2012

### GarageDweller

So I keep hearing that the maxwell equations are variant under Galilean transform. Tired of simply accepting it without seeing the maths, I decided to do the transformation on my own.

To make things easy, I only tried Gauss' law, furthermore I constricted the field to the x axis only. So I have E(x,t).

∇°E(x,t)=ρ(x)/ε

So now I will transform to another inertial frame x' that is moving with speed u with respect to the original frame x.
x'=x-ut
t'=t

What originally was ∂E/∂x=ρ(x)/ε became ∂E/∂x'-(1/u)∂E/∂t=ρ'(x')/ε.
Is this basically what they mean when they say it isn't invariant?

I looked at this again, and noticed that if the electric field is independent of time, then the Galilean transform of this turns out to be invariant, coincidence?

Last edited: Jun 23, 2012
2. Jun 23, 2012

### tiny-tim

Welcome to PF!

Hi GarageDweller! Welcome to PF!
Yup!
Not really … if there's no time, then there's no difference between galilean and relativistic, is there?

3. Jun 23, 2012

### GarageDweller

oops, missed that

4. Jun 23, 2012

### GarageDweller

However, I tried transforming the equation by the Lorentz transform, and yet I'm still getting something different, I think I may have a conceptual error here, my transformation process was:

∂E/∂x=∂E/∂x' * ∂x'/∂x + ∂E/∂t' * ∂t'/∂x

x'=γ(x-ut)
t'=γ(1-ux/c^2)

Which gives me..

∂E/∂x=∂E/∂x' * γ + ∂E/∂t' * (-γu/c^2)

Exactly how does this reduce to ∂E/∂x' ??

5. Jun 23, 2012

### tiny-tim

(try using the X2 button just above the Reply box )
because (from the ampere-maxwell law) ∂E/∂t = J …

so the RHS ρ * γ + J * (-γu/c2) = ρ'

from the pf library on Maxwell's equations

(on the RHS, * denotes a pseudovector: a "curl" must be a pseudovector, the dual of a vector)

Changing to units in which $\varepsilon_0$ $\mu_0$ and $c$ are 1, we may combine the two 3-vectors $\mathbf{E}$ and $\mathbf{B}$ into the 6-component Faraday 2-form $(\mathbf{E};\mathbf{B})$, or its dual, the Maxwell 2-form $(\mathbf{E};\mathbf{B})^*$.

And we may define the current 4-vector J as $(Q_f,\mathbf{j}_f)$.

Then the differential versions of Gauss' Law and the Ampère-Maxwell Law can be combined as:

$$\nabla \times (\mathbf{E};\mathbf{B})^*\,=\,(\nabla \cdot \mathbf{E}\ ,\ \frac{\partial\mathbf{E}}{\partial t}\,+\,\nabla\times\mathbf{B})^*\,=\,J^*$$

and those of Gauss' Law for Magnetism and Faraday's Law can be combined as:

$$\nabla \times (\mathbf{E};\mathbf{B}) = (\nabla \cdot \mathbf{B}\ ,\ \frac{\partial\mathbf{B}}{\partial t}\,+\,\nabla\times\mathbf{E})^*\,=\,0$$​

6. Jun 23, 2012

### GarageDweller

Ooh right forgot bout the charge density term, thx lol
So basically the lorentz factors all get canceled and the J terms go away on either side?

Oh and one more thing, what's the exact process of changing p into p'?

Last edited: Jun 23, 2012
7. Jun 23, 2012

### tiny-tim

sorry, not following you

i always prefer to translate everything into wedge (Λ) products when dealing with maxwell's laws

(Ex,Ey,Ez;Bx,By,Bz) becomes Ex(tΛx) + Ey(tΛy) + Ez(tΛz) + Bx(yΛz) + By(zΛx) + Bz(xΛy)

(ρ,Jx,Jy,Jz) becomes ρt + Jxx + Jyy + Jzz (and similarly for div)

and you use aΛb = -bΛa, aΛa = 0, xΛyΛz = t*, yΛzΛt = x* etc​
ρ is part of the 4-vector (ρ,Jx,Jy,Jz)

8. Jun 23, 2012

### Muphrid

To elaborate on what Tim is saying: once you get into relativity and EM, it's helpful to get used to the idea that the EM field isn't a simple vector field. Rather, just as a vector field is a combination of directions, there are bivector fields which are combinations of planes. That's what the EM field is. You can write its six components as

$$F = E_x e_t \wedge e_x + E_y e_t \wedge e_y + E_z e_t \wedge e_z + B_x e_y \wedge e_z + B_y e_z \wedge e_x + B_z e_x \wedge e_y$$

Each of the $e_\mu \wedge e_\nu$ represents a plane spanning the $e_\mu, e_\nu$ directions.

Because the EM field is a set of planes, the Lorentz transformation works a little differently. It acts on each basis vector in a wedge, so for example, under a Lorentz transformation $\underline L$, we have the tx-plane $e_t \wedge e_x \mapsto \underline L(e_t) \wedge \underline L(e_x)$. If the boost itself is in the tx plane, however, it must leave that plane invariant, even though both vectors get rotated. Similarly, since both $e_y, e_z$ are out of the plane, neither get transformed, and they're left invariant.

What's nice about using the Faraday bivector $F$ is that, without matter, Maxwell's equations boil down to a single expression:

$$\nabla F = - \mu_0 j$$

(The sign depends on your metric convention and how you assemble $F$, but this is the convention I prefer.)