
#1
Jun2312, 12:51 AM

P: 104

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'=xut 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? 



#2
Jun2312, 03:18 AM

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Hi GarageDweller! Welcome to PF!




#3
Jun2312, 03:21 AM

P: 104

oops, missed that




#4
Jun2312, 03:41 AM

P: 104

Galilean transform and the maxwell equations
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(xut) t'=γ(1ux/c^2) Which gives me.. ∂E/∂x=∂E/∂x' * γ + ∂E/∂t' * (γu/c^2) Exactly how does this reduce to ∂E/∂x' ?? 



#5
Jun2312, 04:32 AM

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(try using the X^{2} button just above the Reply box )
so the RHS ρ * γ + J * (γu/c^{2}) = 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 [itex]\varepsilon_0[/itex] [itex]\mu_0[/itex] and [itex]c[/itex] are 1, we may combine the two 3vectors [itex]\mathbf{E}[/itex] and [itex]\mathbf{B}[/itex] into the 6component Faraday 2form [itex](\mathbf{E};\mathbf{B})[/itex], or its dual, the Maxwell 2form [itex](\mathbf{E};\mathbf{B})^*[/itex]. 



#6
Jun2312, 04:38 AM

P: 104

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'? 



#7
Jun2312, 05:00 AM

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i always prefer to translate everything into wedge (Λ) products when dealing with maxwell's laws (E_{x},E_{y},E_{z};B_{x},B_{y},B_{z}) becomes E_{x}(tΛx) + E_{y}(tΛy) + E_{z}(tΛz) + B_{x}(yΛz) + B_{y}(zΛx) + B_{z}(xΛy) 



#8
Jun2312, 11:09 AM

P: 834

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
[tex]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[/tex] Each of the [itex]e_\mu \wedge e_\nu[/itex] represents a plane spanning the [itex]e_\mu, e_\nu[/itex] 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 [itex]\underline L[/itex], we have the txplane [itex]e_t \wedge e_x \mapsto \underline L(e_t) \wedge \underline L(e_x)[/itex]. 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 [itex]e_y, e_z[/itex] are out of the plane, neither get transformed, and they're left invariant. What's nice about using the Faraday bivector [itex]F[/itex] is that, without matter, Maxwell's equations boil down to a single expression: [tex]\nabla F =  \mu_0 j[/tex] (The sign depends on your metric convention and how you assemble [itex]F[/itex], but this is the convention I prefer.) 


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