Actually Newton's second law is better stated as $$F=\frac{dp}{dt}$$ and this is even valid in relativity, both SR and GR, expressed in the right way $$f^\mu = \frac{dp^\mu}{d\tau}=m\frac{du^\mu}{d\... View answer 18 votes As previous answers have correctly noted gamma matrices do not forma a basis of M(4,\mathbb{C}). Nevertheless you can construct one from them in the following way 1 the identity matrix \mathbb{1} ... View answer Accepted answer 9 votes The theory of a chiral fermion \psi coupled to a Maxwell field in four dimensions has a famous anomaly in the chiral transformations$$ \psi \rightarrow \psi'=e^{i \gamma_5 \theta} \psi $$which ... View answer Accepted answer 6 votes |\Omega\rangle is the vacuum of the full interacting theory and |0\rangle is the vacuum of the free theory. They are related in the following way$$ |\Omega\rangle = \lim_{T\rightarrow(1-i\epsilon)...

Again assuming it only has a zero $x^i=x_0^i$ what you have is $$\delta(f(x^i)) = \frac{\delta(x^1-x_0^1)}{\left|\frac{\partial f}{\partial x^1}\right|_{x^i=x_0^i}} \frac{\delta(x^2-x_0^2)}{\left|\... View answer Accepted answer 4 votes I believe they are just describing in words the following equation$$\frac{1}{p^2- m^2 \pm i\epsilon} = P\frac{1}{p^2- m^2} \,\mp\,i\pi \delta(p^2- m^2) $$where P denotes the principal part. ... View answer Accepted answer 4 votes Let \lambda be an affine parameter of the integral curves of \xi^{\nu} then you question translates as$$ \frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = \xi^\mu \nabla_\mu(\xi_{\nu}\xi^{\nu}) = (\xi^\mu ...

“Monopole operators” only exist as local operators in three dimensions. For free Maxwell theory in that dimension an easy way to think about them is in terms of the dual compact scalar, as you ...

I don't think such a thing exists. Usually reps only help you classify the kind of objects you have (i.e the quantum numbers that identify them) and how they transform under the corresponding group. ...

That factor is there because they quantize the field in a box and not in the continuum. When you take the continuum limit one gets $$\int d^3k \rightarrow \sum_k \sqrt{\frac{\hbar c^2}{V}}$$ by ...

I don't know is there is a historical reason I just assumed that was just no to mistake it with $\gamma^3$ which would be called $\gamma^4$ if your Lorentz indices run from $\mu=1,2,3,4$ instead of ...

The field is not interpreted as a wave function but as an operator $\hat{\psi}$ which creates/annihilates particles. This quantisation procedure is done by expanding the field in its Fourier ...

What you found is precisely that the amplitude factorizes since the correct statement of factorization is $$\lim_{P^2\rightarrow0} M(1,\cdots,n) = \sum_{ i\in \text{helicities}} M(1,\cdots,k,P^i)\... View answer Accepted answer 2 votes Take a look at the beautiful TASI lectures by Polchinski on Effective Field Theory and the Fermi Surface. In Lecture 2 he explains the EFT point of view on BCS condensation and he derives the one-loop ... View answer 2 votes Weinberg's argument relies on the existence of Lagrangian/weak coupling regime. Genuine \mathcal{N}=3 theories were recently found in this paper. As expected, they are strongly coupled and believed ... View answer 2 votes People sometimes talk about on-shell symmetries: symmetries of the equations of motion or the S-matrix, which do not hold off-shell (i.e. at the level of the action, path integral, correlators, etc). ... View answer 2 votes The question of what is the velocity of a photon relative to another photon does not make sense. Neither it does asking what is the velocity of anything relative to a photon. This is because in ... View answer 2 votes @ChrisWhite is right I believe you don't need to worry about that since using the tensor product notation |x s\rangle = |x \rangle \otimes |s\rangle and the angular momentum and spin operators just ... View answer 2 votes Usually the Newtonian limit is described as taking v << c but a much better way to express it is saying that the kinetic energy is much less than the rest energy$$ \frac{1}{2}m v^2 << m ...

Let me answer the second part of your question: In the Abelian case there is no trace at all. The term $$\frac{1}{8\pi^2} \int F\wedge F$$ is indeed the second Chern number of the $U(1)$ bundle. This ...

Nowadays, less and less people use Feynman diagrams for precision calculations (that is, anything beyond a tree level 4 or 5 point amplitude). There is a whole field dedicated to finding better ...

Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width $\Gamma$ that is related to its lifetime $\tau$ by $$\Gamma=\frac{\hbar}{\tau}$$ when ...

All Lorentz indices are summed over, but it is more subtle than you might think. The amplitude for particles of helicity/spin greater than zero is not a scalar. The reason for this is that the ...