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:

$$S=\frac{m}{2}\int\eta_{\mu\nu}u^\mu u^\nu ds-q\int A_\mu u^\mu ds$$

Landau and Lifgarbagez has this action

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

 

[paranormal]

Your derivation is correct, but there is a sign convention difference between the definition of the field strength tensor and the Lorentz force law. In the definition of the field strength tensor, the electric field is defined as the negative of the gradient of the scalar potential, while in the Lorentz force law, the electric field is defined as the negative of the gradient of the electric potential. This leads to a difference in the sign of the electric field component in the Lorentz force law. To reconcile this difference, you can either change the sign of the electric field component in the Lorentz force law, or change the definition of the field strength tensor to match the sign convention used in the Lorentz force law.