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?
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?
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' ??
(try using the X^{2} button just above the Reply box ) because (from the ampere-maxwell law) ∂E/∂t = J … 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 3-vectors [itex]\mathbf{E}[/itex] and [itex]\mathbf{B}[/itex] into the 6-component Faraday 2-form [itex](\mathbf{E};\mathbf{B})[/itex], or its dual, the Maxwell 2-form [itex](\mathbf{E};\mathbf{B})^*[/itex]. And we may define the current 4-vector J as [itex](Q_f,\mathbf{j}_f)[/itex]. Then the differential versions of Gauss' Law and the Ampère-Maxwell Law can be combined as: [tex]\nabla \times (\mathbf{E};\mathbf{B})^*\,=\,(\nabla \cdot \mathbf{E}\ ,\ \frac{\partial\mathbf{E}}{\partial t}\,+\,\nabla\times\mathbf{B})^*\,=\,J^*[/tex] and those of Gauss' Law for Magnetism and Faraday's Law can be combined as: [tex]\nabla \times (\mathbf{E};\mathbf{B}) = (\nabla \cdot \mathbf{B}\ ,\ \frac{\partial\mathbf{B}}{\partial t}\,+\,\nabla\times\mathbf{E})^*\,=\,0[/tex]
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'?
sorry, not following you 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) (ρ,J_{x},J_{y},J_{z}) becomes ρt + J_{x}x + J_{y}y + J_{z}z (and similarly for div) and you use aΛb = -bΛa, aΛa = 0, xΛyΛz = t*, yΛzΛt = x* etc ρ is part of the 4-vector (ρ,J_{x},J_{y},J_{z})
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 tx-plane [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.)