Ostensible contradiction between the continuity equation and Cartan's magic formulaby mma Tags: cartan's magic, continuity equation 

#1
Mar112, 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 timedependent 2form and a timedependent 3form on the 3dimensional 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 4dimensional spacetime [itex]\mathbb R\times M[/itex] instead of timedependent forms on [itex]M[/itex], than the continuity equation tells that the integral of the [itex]J:=\rho+dt\wedge j[/itex] 3form on the boundary of any 4dimensional 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 timedependent 3form [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? 



#2
Mar112, 11:22 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

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 3form on a 3manifold, 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. 



#3
Mar212, 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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