Recent content by loops496

Hello everyone, So my university opened some Summer Research Opportunities for Undergraduates and part of the application requires a motivation letter. Here is what I wrote, I'm open to any suggestions or improvements, Thanks in advance!Respected Sir/Madam, I was very excited when I learned...

Hello everyone, I'm a senior undergraduate and I'm planing to do a Monograph related to Quantum effects in Gravity (Hawking radiation or something similar(?)) or even some QFT in curved space-time (maybe ambitious since I know this is VERY HARD and I don't have much time), the thing is I can't...

Yes indeed I can Avodyne, actually after some work using both commutators with the Hamiltonian I managed to prove what the original question asked. Thanks for the hints guys!

Yes you're totally right I missed an equals sign! It should be ##[H,F]=-i \partial_o F##

Since both substitutions are suitable for dealing with square roots inside integrals if you can do a problem with normal trig you can do it with hyperbolic trig. In the end it boils down to how well you can manipulate trigonometric identities and integrals vs hyperbolic ones. Even if you are...

Homework Statement Consider the quantum mechanical Hamiltonian ##H##. Using the commutation relations of the fields and conjugate momenta , show that if ##F## is a polynomial of the fields##\Phi## and ##\Pi## then ##[H,F]-i \partial_0 F## Homework Equations For KG we have: ##H=\frac{1}{2} \int...

As Geofleur said start with the simplest but non trivial case, then step it up a little bit, you'll see the pattern right away and convince yourself of the general formula.

I think metric compatibility is a weaker condition, i.e. you can have various Riemannian connections without any being the LC. But that still does not guarantee that the derivative along the Killing of the curvature scalar is 0, or does it?

Hey Mentz114, thank for replying! Since you don't need a connection for the Lie derivative, and Killing Vector Fields depend upon the Lie derivative I suspect you don't need the Levi-Civita Connection, However for the derivation of such identity I used the Bianchi identities which rely on a...

As TSny said \mathbf{E}=-\nabla \phi - \partial_t \mathbf{A} nowusing the triple product identity: \mathbf{v}\times(\nabla \times \mathbf{A}) = \nabla(\mathbf{A} \cdot \mathbf{v}) - \mathbf{A}(\mathbf{v} \cdot \nabla) Which in the Lorentz equation: \mathbf{F} = q \left[-\nabla \phi - \partial_t...

I have a question about the directional derivative of the Ricci scalar along a Killing Vector Field. What conditions are necessary on the connection such that K^\alpha \nabla_\alpha R=0. Is the Levi-Civita connection necessary? I'm not sure about it but I believe since the Lie derivative is...

You're totally rigth fzero, it is v^\mu \nabla_\mu R=0 and I mistyped the sign killing equation (ooops sorry) shame on me :/.

Homework Statement Suppose v^\mu is a Killing Vector field, the prove that: v^\mu \nabla_\alpha R=0 Homework Equations 1) \nabla_\mu \nabla_\nu v^\beta = R{^\beta_{\mu \nu \alpha}} v^\alpha 2) The second Bianchi Identity. 3) If v^\mu is Killing the it satisfies then Killing equation, viz...

Oh my algebra was off on that one! thanks.

Hey! I need to calculate the electric field on the axis of a circular plate of radius a with the following charge distribution: \sigma_0 \frac{r^2}{a^2} \delta (z), \; r\leq a0, \; r>a where \sigma_0 is a constant. I've already calculated the potential and taken its gradient to get the...