# Tag Info

1

Once you fix a coordinate system $X$, then for any two points A and B the distance $d(A,B)$ between the two can be defined and calculated from the metric in that particular coordinate system, allowing you to define limits of this type: For a sequence of points $P_n$ and a point $Q$, $P_n\rightarrow Q$ if and only if $d(P_n,Q)\rightarrow0$. It's true that ...

1

I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the $p_3$ direction so that $p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$ $$\delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3}$$ $$= ... 0 Any Lorentz transformation will leave p^2 unchanged, hence the mass of the Dirac \delta will remain at m^2. Observe that, for physical reasons, one usually only considers the component of the (full) Lorentz group that is connected to the identity. Any transformation in this proper subgroup leaves the sign of p_0 (which by the spectral condition is ... 0 I think I have came up with the answer, which I hope is correct. Although, neither internal nor the gauge symmetry operations affect the space-time coordinates, there is a big difference: Internal symmetry is an actual symmetry of the system (field): two physically distinct field configurations (or in QM, two physically distinct states in Hilbert space) ... 0 This is indeed the same problem, from different viewpoints. The deep source of the problem is that the number of A_\mu fields is higher than the actual physical degrees of freedom. Here are a few remarks: The operator you are trying to invert does not have an inverse because it is just a projection operator(you can check this by direct computation, the ... 0 The answer to this question is no assuming dimensions higher than 1+1. This can be seen observing that the equation of motion for the fermionic field is just the limit of mass going to infinity of a scalar field coupled to a fermionic field. This can be seen in the following way. Consider the Lagrangian$$ L = ...

0

As I expected, this simple question calls for no more than a little sleight of hand, as pointed out by the sole comment above. It looks as if @Meng Cheng won't make it an answer. Thanks to him. And here I confirm that it works well.

1

The functional derivative $\frac{\delta}{\delta \phi}$ acts on functionals, things that map functions to real numbers. That is, they act on actions $S$, not lagrangians $L$. I don't know where you got your original question, but there indeed should be a minus sign! Altogether, I think what you're asking is why: $$\frac{\delta}{\delta \phi} \int d^4x ... 1 In this answer we will just make a general conceptional remark about variational/functional derivative (FD), which hopefully implicitly answers OP's specific questions. OP is apparently considering the 'same-spacetime' FD,$$\tag{A}\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} ...

3

The lagrangian you're dealing with is $\mathcal{L}= \frac12 (\partial_{\mu} \phi)^2 - \frac12 m^2 \phi^2$. When you take the partial with respect to $\partial_{\mu}\phi$, you should be getting $2 * (\frac12 \partial^{\mu}\phi)$. This would make the first term in your expression $\dot{\phi}^2$ instead of $\frac12 \dot{\phi}^2$ and things would work out. If ...

0

Just add a source term of the form $\int d^4 x J(x) \phi(x)$, or $\int d^4 x J^{\mu}(x) A_{\mu}(x)$ for a scalar field or $U(1)$ Maxwell field. $J$ is a function that you are free to specify.

4

Because that way the Euler-Lagrange equations turn out to be linear and thus the superposition principle holds. Superposition principle means "free theory".

2

Comments to the question (v5): If an action functional $S$ is invariant under a Lie algebra $L$ of symmetries, the corresponding Noether currents & charges do not always form a representation of the Lie algebra $L$. There could be (classical) anomalies. In some cases such (classical) anomalies appear as central extensions, cf. e.g. Ref. 1-3 and this ...

2

I) Let $\overline{\mathbb{R}^{p,q}}$ denote the conformal compactification of $\mathbb{R}^{p,q}$. Let $n:=p+q$ denote the dimension. (The pseudo-Riemannian generalization of) Liouville's theorem states that if $n\geq 3$, then all local conformal transformations of $\overline{\mathbb{R}^{p,q}}$ can be extended to global conformal transformations. II) ...

0

This comes from the idea that the action $S=\int d^4x\,\mathcal{L}$ is a scalar. From planck's law it is easy to see that the energy $E=\hbar \omega$ has the dimension of inverse time. From de-broglie relation you can see that momentum has the inverse dimension of space. Hence $\mathcal{L}$ should have a dimension of energy ${[\mathcal{L}]=E^{4}}$. This ...

Top 50 recent answers are included