Recent content by countable

thanks for the info Bill:)

I'm doing a course which assumes knowledge of Group Theory - unfortunately I don't have very much. Can someone please explain this statement to me (particularly the bits in bold): "there is only one non-trivial irreducible representation of the Cliford algebra, up to conjugacy" FYI The...

thanks that was a good explanation:)

Sorry if this is a stupid question but I really want to understand this. If we have an operator that's defined in the following way: D_\mu \phi^a = \partial_\mu \phi^a - iA^{r}_{\mu}{(T_r)^a}_b\phi^b How would we go about working out: D_\mu D_\nu \phi^a What's confusing me is that the...

Thanks for the reply. So they are both antisymmetric tensors - by this do we simply mean that F_{\mu\nu}=-F_{\nu\mu}? Also would I be right in saying that the square brackets mean cyclic permutations thus: T_{\mu\nu\rho} = \partial_{[\mu}\partial_{\nu}A_{\rho ]} = \partial_\mu \partial_\nu...

bump.

I see where I've gone wrong. Thanks for your help Dexter.

What do the square brackets represent in the expressions below: \partial_{[\mu}\partial_{\nu}A_{\rho ]} \partial_{[\mu}F_{\nu\rho ]} I'm guessing they aren't commutators? thanks.

I have a very superficial understanding of this subject so apologies in advance for what's probably a stupid question. Can someone please explain to me why if we have a Lie Group, G with elements g, the adjoint representation of something, eg g^{-1} A_\mu g takes values in the Lie Algebra of G...

Well, I understand how to get the first term \frac{\partial}{\partial A_\nu}(D_\mu \psi)^* D^\mu \psi= +iq \delta^{\nu}_{\mu} \Psi^{*} D^\mu \psi +(D^\mu \psi)^* . -iq \eta^{\mu\nu} \psi = iq\psi^* D^\nu \psi - iq\psi \eta^{\mu\nu} (D_\nu \psi)^* The second term in the last equality above is...

thanks dexter - very helpful: So to get the second term in (1) - ie. -iq\psi D^\nu \psi^* I have to work out: \frac{\partial }{\partial A_\nu} (D_\mu \psi)^* (-iqA^\mu \psi) = \frac{\partial }{\partial A_\nu} (D_\mu \psi)^* (-iq \eta^{\mu\nu} A_\nu \psi)= -iq\psi \eta^{\mu\nu} (D_\mu \psi)^*...

Hi, Hope some one can help me with a problem I am working on: It involves working out: \frac{\delta L}{\delta A_\nu} of the following Lagrangian: L=\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2} (D_{\mu} \Psi)^{*} D^{\mu}\Psi The solutions show that this is equal to: \frac{\delta...