Variational principle & lorentz force law

[Brian-san]

Homework Statement

Show that the Lorentz force law follows from the following variational principle:
S=\frac{m}{2}\int\eta_{\mu\nu}u^\mu u^\nu ds-q\int A_\mu u^\mu ds

Homework Equations

Definition of Field Strength Tensor
Integration by Parts
Chain Rule & Product Rule for Derivatives

The Attempt at a Solution

In order to find the force equation, we need to vary the action and find when \delta S=0.

So, we begin by varying the action integral,
\delta S=\frac{m}{2}\int\eta_{\mu\nu}\delta(u^\mu u^\nu)ds-q\int\delta(A_\mu u^\mu)ds

Then apply product rule to expand all the terms,
\delta S=m\int\eta_{\mu\nu}u^\nu\frac{d(\delta x^\mu)}{ds}ds-q\int\left(\partial_\nu A_\mu u^\mu\delta x^\nu+A_\mu\frac{d(\delta x^\mu)}{ds}\right)ds

Simplified and renamed indices in the second term,
\delta S=m\int u_\mu\frac{d(\delta x^\mu)}{ds}ds-q\int\left(\partial_\mu A_\nu u^\nu\delta x^\mu+A_\mu\frac{d(\delta x^\mu)}{ds}\right)ds

Reverse integration by parts to get the variation out of the derivatives,
\delta S=m\int\left(\frac{d}{ds}\left(u_\mu\delta x^\mu\right)-\frac{du_\mu}{ds}\delta x^\mu\right)ds-q\int\left(\partial_\mu A_\nu u^\nu\delta x^\mu+\frac{d}{ds}\left( A_\mu\delta x^\mu\right)-\frac{dA_\mu}{ds}\delta x^\mu\right)ds

Grouping like terms we have,
\delta S=\int\left(\frac{d}{ds}\left(mu_\mu\delta x^\mu-qA_\mu\delta x^\mu\right)\right)ds+\int \left( q\left(\frac{dA_\mu}{ds}\delta x^\mu-\partial_\mu A_\nu u^\nu\delta x^\mu\right)-m\frac{du_\mu}{ds}\delta x^\mu\right)ds

After integration, the first term should be zero since the variations must vanish at the endpoints. Also, using the chain rule on the dA/ds derivative gives me
\delta S=\int \left( q\left(\partial_\nu A_\mu u^\nu-\partial_\mu A_\nu u^\nu\right)-m\frac{du_\mu}{ds}\right)\delta x^\mu ds

Rearranged terms to simplify further,
\delta S=\int \left( -q\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)u^\nu -m\frac{du_\mu}{ds}\right)\delta x^\mu ds

Then I use the definition of the field strength tensor to get,
\delta S=\int \left( -qF_{\mu\nu}u^\nu -m\frac{du_\mu}{ds}\right)\delta x^\mu ds

Then \delta S=0 for any arbitrary variation \delta x^\mu only if the integrand itself is zero, so
qF_{\mu\nu}u^\nu+m\frac{du_\mu}{ds}=0

Rearranging the expression gives
\frac{du_\mu}{ds}=-\frac{q}{m}F_{\mu\nu}u^\nu

This is the correct form for the Lorentz force law, however I have an additional minus sign. It's probably a trivial problem, but I've double checked all the work several times and I can't figure out where it's coming from.

 

[arkajad]

Perhaps the problem is this:

On the right hand side you have u^\nu
On the left hand side u_\mu

The relation between the two is given by \eta_{\mu\nu} and that depends on your signature convention. This signature is expressed in the sign of the first term of your action principle.

 

[Brian-san]

I checked my notes and it's supposed to come out to
\frac{du_\mu}{ds}=\frac{q}{m}F_{\mu\nu}u^\nu

We've been using the convention that the spatial terms have the minus signs in the Minkowski metric (+---). In an earlier step I used u_\mu=\eta_{\mu\nu}u^\nu to get the lower mu index in the first term with the mass. I can't find anything that says how the signature affects the process of raising/lowering indices though.

 

[arkajad]

Question is why do you have minus sign in front of the second (or plus in front of the first) term in:

<br /> S=\frac{m}{2}\int\eta_{\mu\nu}u^\mu u^\nu ds-q\int A_\mu u^\mu ds<br />

Landau and Lifgarbagez has this action

The first term is \sqrt{\eta_{\mu\nu}u^{\mu}u^{\nu}}\,ds but it should not matter.