Recent content by DuckAmuck

If you have a magnetic monopole with non-zero spin, would this result in an electric dipole? Just like an electric charge with spin results in a magnetic dipole?

Let's say Bob falls into a blackhole, and Alice is sufficiently far away that she is not falling in. She sees Bob's clock stop and his image fade away and all that. However, from Alice's perspective Bob never actually crosses the horizon. From her perspective, Bob is sitting frozen and invisible...

From the Born Approximation, you can relate the potential to the scattering amplitude. So it follows that a potential can be derived from the scattering amplitude from Delbruck scattering. I tried to solve this myself, and get a scattering amplitude with only angular dependence, no momentum...

Chapter 7 of Griffith's particle book

The sigma tensor composed of the commutator of gamma matrices is said to be able to represent any anti-symmetric tensor. \sigma_{\mu\nu} = i/2 [\gamma_\mu,\gamma_\nu] However, it is not clear how one can arrive at something like the electromagnetic tensor. F_{\mu\nu} = a \bar{\psi}...

Looking to calculate the amplitude and cross section of the process: electron + positron to photon + Z boson. Basically the annihilation resulting in Z + gamma rather than gamma +gamma. My question is mainly about how to deal with the polarization states with the Z boson, since there are 3 and...

Yes, of course they are acting on different spaces in most cases. Was trying to keep things very generalized in an attempt to "rescue" invariance, but I think that may be overkill. And, Psi is indeed a dirac spinor. My question still remains on what to do about T being non-unitary. As ##A_\mu...

You see in the literature that the vector potentials in a gauge covariant derivative transform like: A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1} Where T is not necessarily unitary. (In the case that it is ##T^{-1} = T^\dagger##) My question is then if T is not unitary, how is...

You’re right. I am just trying to figure out *how* this could be zero at this point, as in what conditions. Otherwise I’m stumped.

ok i think i have solid reasoning here: Suppose ##C^{ij} = M^{ij} + N^{ij}## From symmetry and antisymmetry we have: ##\epsilon_{ijkl} C^{ij}C^{kl} = 0## Also if you foil the CC product in terms of M and N you get ##C^{ij}C^{kl} = M^{ij}M^{kl} + N^{ij}N^{kl} + M^{ij}N^{kl} + N^{ij}M^{kl}##...

ep_{ijkl} M^{ij} N^{kl} + ep_{ijkl}N^{ij} M^{kl} The second term can be rewritten with indices swapped ep_{klij} N^{kl}M^{ij} Shuffle indices around in epsilon ep{klij} = ep{ijkl} Therefore the expression becomes 2ep_{ijkl}M^{ij}N^{kl} Not zero. What is wrong here?

Okay so the Lagrangian behavior is straightforward then. What about the Lagrangian density? Where rho is the mass density of a particle cloud. $$ \mathcal{L} = -\rho(y) \sqrt{\dot{y}_\mu \dot{y}^\mu} - A_\mu J^\mu -\frac{1}{4} F_{\rho\sigma} F^{\rho\sigma}$$ $$ \frac{\partial...

The last paragraph is basically asking, how do I write the full Lagrangian of a massive charged particle in an electromagnetic field? From what you've said, I gathered that it would be written like: $$ L = -m\sqrt{\dot{y}_\mu \dot{y}^\mu} - q A_\mu (y) \dot{y}^\mu - \frac{1}{4} \int d^3 x...

How would you unify the two Lagrangians you see in electrodynamics? Namely the field Lagrangian: Lem = -1/4 Fμν Fμν - Aμ Jμ and the particle Lagrangian: Lp = -m/γ - q Aμ vμ The latter here gives you the Lorentz force equation. fμ = q Fμν vν It seems the terms - q Aμ vμ and - Aμ Jμ account for...

Amazing. Thank you.