Ostensible contradiction between the continuity equation and Cartan's magic formula


by mma
Tags: cartan's magic, continuity equation
mma
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#1
Mar1-12, 02:13 AM
P: 230
Continuity equation is

[itex]dj+\partial_t\rho_t=0[/itex]

where [itex]j[/itex] and [itex]\rho[/itex] are a time-dependent 2-form and a time-dependent 3-form on the 3-dimensional space [itex]M[/itex] respectively. (see e.g. A gentle introduction to
the foundations of classical electrodynamics
(2.5))

If we use differential forms on the 4-dimensional space-time [itex]\mathbb R\times M[/itex] instead of time-dependent forms on [itex]M[/itex], than the continuity equation tells that the integral of the [itex]J:=\rho+dt\wedge j[/itex] 3-form on the boundary of any 4-dimensional cube is 0, hence [itex]dJ=0[/itex].

If we apply Cartan's magic formula to [itex]J[/itex] and the vector field [itex]v:=\partial_t[/itex] then we get:

[itex]L_vJ=\iota_vdJ+d(\iota_vJ)=d(\iota_vJ)=dj[/itex]

On the other hand, [itex]L_vJ=\frac{\partial}{\partial t}\tilde\rho_t[/itex]

where [itex]\tilde\rho_t=\varphi_t^*\rho[/itex], where [itex]\varphi[/itex] is the flow of [itex]v(=\partial_t)[/itex], i.e. [itex]\tilde\rho_t[/itex] is the same time-dependent 3-form [itex]\rho_t[/itex] on [itex]\{0\}\times M\simeq M[/itex] as appear in the starting continuity equation.

Consequenty, from Cartan's magic formula we get [itex]\partial_t\rho_t=dj[/itex], i.e.

[itex]dj-\partial_t\rho_t=0[/itex]

So, there is a sign difference between this equation an the continuity equation. Were is the error?
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quasar987
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#2
Mar1-12, 11:22 AM
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I think the problem is the definition of J! If I compute dJ, I get

[tex]dJ=d_{sp}\rho+dt\wedge\partial_t\rho - dt\wedge d_{sp}j[/tex]

where d_{sp} means the exterior differential wrt spatial coordinates only. Now, [itex]d_{sp}\rho=0[/itex] since rho is a 3-form on a 3-manifold, and so we see that dJ=0 iff [itex]\partial_t\rho - d_{sp}j = 0[/itex] which is not the conservation equation. On the other hand, with [itex]J:= -\rho +dt\wedge j[/itex] we do get dJ=0, and your little playing around with Cartan's formula gives [tex]dj+\partial_t\rho=0[/tex] at the end.
mma
mma is offline
#3
Mar2-12, 01:10 AM
P: 230
Oh, yes, this solves the problem. I think that I got lost beause the four current vector in Physics is defined with +rho, but now I recognised that it means a - sign when I turn it to differential form because of the - sign in the Minkowski metric.

Thank you very much, Quasar!


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