Grassmann Algebra: Derivative of $\theta_j \theta_k \theta_l$

[fzero]

Got it. Thanks.

Although, one thing is bothering me: when we do the integration by parts -\frac{1}{2} \int d^dx \partial_\mu \lambda^a \Delta^{\mu \nu} A_\nu^a = - \frac{1}{2} \lambda^a \Delta^{\mu \nu} A_\nu^a + \frac{1}{2} \int d^dx \lambda^a \partial_\mu \Delta^{\mu \nu} A_\nu^a
haven't we lost an index on the first term there? So the indices aren't balanced in this equation?

As I mentioned back in post #45, that term should be a surface integral. Look up Stokes' theorem.

 

[latentcorpse]

As I mentioned back in post #45, that term should be a surface integral. Look up Stokes' theorem.

\int_\Omega ( \vec{\nabla} \times \vec{F} ) \cdot \vec{da} = \int_{\partial \Omega} \vec{F} \cdot d \vec{s}

But we don't have a curl in our expression, do we?

 

[fzero]

\int_\Omega ( \vec{\nabla} \times \vec{F} ) \cdot \vec{da} = \int_{\partial \Omega} \vec{F} \cdot d \vec{s}

But we don't have a curl in our expression, do we?

A special case of Stokes' theorem is the divergence theorem, so check that one out.

 

[latentcorpse]

A special case of Stokes' theorem is the divergence theorem, so check that one out.

<br /> -\frac{1}{2} \int d^dx \partial_\mu \lambda^a \Delta^{\mu \nu} A_\nu^a = - \frac{1}{2} \int_{S^{d-1}} n_\mu \lambda^a \Delta^{\mu \nu} A_\nu^a + \frac{1}{2} \int d^dx \lambda^a \partial_\mu \Delta^{\mu \nu} A_\nu^a<br />

where n^\mu are the components of the vector normal to the surface of the S^{d-1}.
Is that correct?

Then do we just say that this term vanishes because either A or \lambda vanishes at infinity. Which one is it though? Am I right in saying it would need to be \lambda since we don't have an A term only derivatives of A?
Even if it is the \lambda term that vanishes - I don't understand physically why it should?

Thanks.

 

[fzero]

<br /> -\frac{1}{2} \int d^dx \partial_\mu \lambda^a \Delta^{\mu \nu} A_\nu^a = - \frac{1}{2} \int_{S^{d-1}} n_\mu \lambda^a \Delta^{\mu \nu} A_\nu^a + \frac{1}{2} \int d^dx \lambda^a \partial_\mu \Delta^{\mu \nu} A_\nu^a<br />

where n^\mu are the components of the vector normal to the surface of the S^{d-1}.
Is that correct?

Then do we just say that this term vanishes because either A or \lambda vanishes at infinity. Which one is it though? Am I right in saying it would need to be \lambda since we don't have an A term only derivatives of A?
Even if it is the \lambda term that vanishes - I don't understand physically why it should?

Thanks.

We just require that the variation of the fields vanishes at the boundary. We're only considering infinitesimal variations, so we can do this.