Recent content by Woolyabyss

Thanks for the reply, I managed to work out the answer my issue turned out to be I wasn't taking into account that covariant derivatives don't commute.

I'm confused by this question, from minimal coupling shouldn't the answer simply be ## \nabla^a \nabla_a F_{bc} = 0 ##? Any help would be appreciated. EDIT: I should also point out ##F_{ab}## is the EM tensor.

##U_1 \otimes U_2 = (1- i H_1 \ dt) \otimes (1- i H_2 \ dt)## We can write ## | \phi_i(t) > \ = U_i(t) | \phi_i(0)>## where i can be 1 or 2 depending on the subsystem. The ## U ##'s are unitary time evolution operators. Writing as tensor product we get ## |\phi_1 \phi_2> = (1- i H_1 \ dt) |...

Much appreciated guys. I understand now thanks.

I was studying linearized GR where we make the following coordinate transformation ## \tilde{x}^{a} = x^{a} + \epsilon y^{a}(x) ## This coordinate transformation is then meant to imply ## g_{ab}(x) = \tilde{g}_{ab}(x) + \epsilon \mathcal{L}_{Y} g_{ab} ## Would anyone be kind enough to explain...

Since in 2D the riemman curvature tensor has only one independent component, ## R = R_{ab} g^{ab} ## can be reversed to get the riemmann curvature tensor. Write ## R_{ab} = R g_{ab} ## Now ## R g_{ab} = R_{acbd} g^{cd}## Rewrite this as ## R_{acbd} = Rg_{ab} g_{cd} ## My issue is I'm not...

I've just realized my mistake. the partial of q A_b with respect to lambda isn't zero, you get an extra term from the chain rule and it works out nicely.

I've also attached my attempt as a pdf file. My main issue seems to be I only get one A partial term. Any help would be appreciated.

Thanks for the replies, I've managed to get out (p^2 -m ) term of (6.9) but am still unsure for the second term. It appears as though its being split in two and they are flipping the momentum variables, But why?

I'm having trouble understanding a specific line in my lecturers notes about the path integral approach to deriving the Klein Gordon propagator. I've attached the notes as an image to this post. In particular my main issue comes with (6.9). I can see that at some point he integrates over x to...

Thanks a lot to the both of you. I made an error when I wrote down the Lagrangian, the term on the right was actually $$ \frac{1}{2} ( \partial_{\mu} A^{\mu} )^2 $$ Regardless, I managed to get the right answer in the end. It never occurred to me that if my derivative had indices matching...

Homework Statement $$ L = -\frac{1}{2} (\partial_{\mu} A_v) (\partial^{\mu} A^v) + \frac{1}{2} (\partial_{\mu} A^v)^2$$ calculate $$\frac{\partial L}{\partial(\partial_{\mu} A_v)}$$ Homework Equations $$ A^{\mu} = \eta^{\mu v} A_v, \ and \ \partial^{\mu} = \eta^{\mu v} \partial_{v}$$ The...

Yes. Also I know that the events X = x of a Poisson distribution are independent of one another but surely P(X >= 70) and P(X >= 80) for example can't be, because given at least 70 events happen, the probability that at least 80 events happen would be 10 no?

Hi guys, I have a question about computing conditional probabilities of a Poisson distribution. Say we have a Poisson distribution P(X = x) = e^(−λ)(λx)/(x!) where X is some event. My question is how would we compute P(X > x1 | X > x2), or more specifically P(X> x1 ∩ X > x2) with x1 > x2? I...

Thanks again, using Gaussian elimination as you suggested I found X1,X2 and X3 all to be equal to d(1-αt)/(1-(α^3)(t^3)). I suspect this to be the correct answer as its easy to see why when t approaches α^-1 this motion is unrealistic.