- #1

haushofer

Science Advisor

- 2,319

- 696

## Main Question or Discussion Point

Hi, I have a computational question which concerns forms. I want to compute the variation of the electrodynamic Lagrangian, seen here as an n-form:

[tex]L = -\frac{1}{2}F \wedge *F[/tex]

with F=dA. I want to derive the Noether-current from this Lagrangian. The symmetrytransformation we are concerned with are coordinatetransformations induced by Lie-derivatives acting on A. A general variation of L can be composed as

[tex]\delta L = E \cdot\delta A+ d\Theta[/tex]

where [tex]\Theta[/tex] are the boundary terms and E are the equations of motion for the vector potential A. If we now have a vector field [tex]\xi[/tex] we can construct the Noether current

[tex]

\mathcal{J} \equiv \Theta -\xi\cdot L

[/tex]

(where the dot indicates contraction with the first index of L) such that

[tex]d\mathcal{J} = - E\delta A[/tex]

If the equations of motion hold, then there can be a Noether charge Q such that

[tex] \mathcal{J} = dQ [/tex]

I want to verify this for the electrodynamic Lagrangian given above, and I have the suspicion that for this particular Lagrangian we can't construct this Q ( so that the current [tex]\mathcal{J}[/tex] isn't exact, but it should be closed). But I'm a little stuck with the calculation. A variation of L gives me

[tex] \delta L = -\frac{1}{2} (\delta F \wedge *F + F \wedge \delta *F)[/tex]

which can be worked out, with F=dA, as

[tex]

\delta L = -\frac{1}{2}[d(\delta A \wedge *F) + \delta A \wedge d*F + F \wedge \delta *F ][/tex]

I'm interested in the A-field. I thought that

[tex]

\delta * F = * \delta F + \frac{1}{2}(g^{\alpha\beta}\delta g_{\alpha\beta}) * F

[/tex]

and the metric-part is going to give me the energy-momentum tensor of the electromagnetic field, which we can disregard. I recognize in this variation

[tex]

\Theta = -\frac{1}{2}\delta A \wedge *F

[/tex]

So I would say that my Noether current is given by

[tex]

\mathcal{J} = -\frac{1}{2}\Bigr(\delta A - \xi\cdot F \Bigr)\wedge * F

[/tex]

but if I take the exterior derivative of this, it doesn't give me the form I want; It's not exact if the equations of motion for A hold.

So my questions are :

1)what is the corresponding Noether current for the electrodynamic Lagrangian associated with diffeomorphism-invariance of the action?

2) Is this current exact?

[tex]L = -\frac{1}{2}F \wedge *F[/tex]

with F=dA. I want to derive the Noether-current from this Lagrangian. The symmetrytransformation we are concerned with are coordinatetransformations induced by Lie-derivatives acting on A. A general variation of L can be composed as

[tex]\delta L = E \cdot\delta A+ d\Theta[/tex]

where [tex]\Theta[/tex] are the boundary terms and E are the equations of motion for the vector potential A. If we now have a vector field [tex]\xi[/tex] we can construct the Noether current

[tex]

\mathcal{J} \equiv \Theta -\xi\cdot L

[/tex]

(where the dot indicates contraction with the first index of L) such that

[tex]d\mathcal{J} = - E\delta A[/tex]

If the equations of motion hold, then there can be a Noether charge Q such that

[tex] \mathcal{J} = dQ [/tex]

I want to verify this for the electrodynamic Lagrangian given above, and I have the suspicion that for this particular Lagrangian we can't construct this Q ( so that the current [tex]\mathcal{J}[/tex] isn't exact, but it should be closed). But I'm a little stuck with the calculation. A variation of L gives me

[tex] \delta L = -\frac{1}{2} (\delta F \wedge *F + F \wedge \delta *F)[/tex]

which can be worked out, with F=dA, as

[tex]

\delta L = -\frac{1}{2}[d(\delta A \wedge *F) + \delta A \wedge d*F + F \wedge \delta *F ][/tex]

I'm interested in the A-field. I thought that

[tex]

\delta * F = * \delta F + \frac{1}{2}(g^{\alpha\beta}\delta g_{\alpha\beta}) * F

[/tex]

and the metric-part is going to give me the energy-momentum tensor of the electromagnetic field, which we can disregard. I recognize in this variation

[tex]

\Theta = -\frac{1}{2}\delta A \wedge *F

[/tex]

So I would say that my Noether current is given by

[tex]

\mathcal{J} = -\frac{1}{2}\Bigr(\delta A - \xi\cdot F \Bigr)\wedge * F

[/tex]

but if I take the exterior derivative of this, it doesn't give me the form I want; It's not exact if the equations of motion for A hold.

So my questions are :

1)what is the corresponding Noether current for the electrodynamic Lagrangian associated with diffeomorphism-invariance of the action?

2) Is this current exact?