The  states that  A 3 + 1D QFT quantum field theory with a conserved $4$-vector current $J^\mu$ which is Poincaré covariant does not admit massless particles with helicity  $|h| &gt; 1/2$. A 3 + 1D QFT with a conserved stress–energy tensor $T^{\mu \nu}$ which is Poincaré covariant does not admit massless particles with helicity $|h| &gt; 1$.  This theorem is interpreted to mean that the graviton cannot be a composite particle. However, I do not quite follow this interpretation. My understanding is that the massless graviton of spin $2$ ought to have a Poincare-covariant conserved stress-tensor (and Poincare-covariant conserved $4$-vector currents associated to other symmetries). So the Weinberg-Witten theorem means the graviton is not allowed to exist at all. That the graviton is not allowed to be a composite particle appears to be a weaker conclusion.
121309
The  (disclaimer: I don't know this wikipedia entry) is usually mentioned as the reason why gravitons may not be composite particles. I do understand the proof of the theorem, but not the previous conclusion.   The theorem states that in an interacting and Poincare invariant quantum field theory, there are not massless, spin-2 particles unless there exists a gauge symmetry — which makes the energy-momentum tensor non-covariant (actually covariant up to a gauge transformation) under Lorentz transformations in the Fock space. So the immediate conclusion of the theorem is that the existence of a massless, spin-2 particle (like a graviton) requires linearized diffeomorphisms.  My question is: why do linearized diffeomorphisms imply that gravitons are elementary particles? Or, more in general, why  the particle corresponding to a gauge field must be elementary (I know that a gauge symmetry must be exact, but why this implies that the corresponding particle must be elementary?).
72049
in the book Quantum Field Theory by Itzykson and Zuber the following derivation for the stress-energy tensor is proposed (p.22):  Assume a Lagrangian density depending on the spacetime coordinates $x$ only through fields and their gradients. Under a translation we have $$\mathcal L (x+a)=\mathcal L[\phi_i(x+a),\partial_\mu\phi_i(x+a)].$$  Consider an infinitesimal $x$-dependent transformation $$\delta\phi_i=\delta a^\mu(x)\partial_\mu\phi_i(x),$$ $$\delta\partial_u\phi_i(x)=\delta a^\nu \partial_\nu\partial_\mu\phi_i(x)+\partial_\mu[\delta a^\nu (x)]\partial_\nu \phi_i(x).$$  The proof then proceeds with variation of the action and integration by parts. But why do we consider an $x$-dependent local transformation instead of a global transformation?
75152
I'm trouble with exercise 1.8 of Carroll's Space-Time and Geometry:     If $\partial_\nu T^{\mu \nu} = Q^\mu$, what physically does the spatial vector $Q^i$ represent? Use the dust energy momentum tensor to make your case.    First of all, since the momentum energy tensor of dust $T^{\mu \nu}=\rho U^\mu U^\nu$ is symmetric, one has the energy-momentum conservation law $$ \partial_\nu T^{\mu \nu} = \partial_\nu T^{\nu \mu}  =0.$$ A vector $Q$ is constant in any frame related by a Lorentz transformation, while this isn't the case for the components $Q^\mu$. But here $Q=0$ and hence $Q^\mu=0$ also for every component $\mu =0,\cdots,4$?   Even if a straightforwardly apply the equation: $$ Q^i = \partial_\nu T^{i \nu} = (\partial_\nu \rho) U^i U^\nu + \rho(\partial_\nu U^i)  U^\nu + \rho U^i(\partial_\nu  U^\nu ),$$ doesn't seems to yield something...   Is my first reasoning right ? Otherwise: can someone give me a hint ?  Thanks
6496
I am stuck with an exercise in Sean Carroll's Spacetime and Geometry (Chapter 4, Exercise 3). The goal is to show that the continuity of the energy-momentum tensor, i.e. \begin{equation} \nabla_\mu T^{\mu\nu}=0\tag{1} \end{equation} is equivalent to the geodesic equation in the case of a free particle. The energy-momentum tensor of a free particle with mass $m$ moving along its worldline $x^\mu (\tau )$ is \begin{equation} T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2} \end{equation} Taking the covariant derivative of this tensor gives $$\begin{align} \nabla_\mu T^{\mu\nu}=&amp;m\int d \tau \nabla_\mu\left[ \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right]\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\cr &amp;+m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\nabla_\mu\left[\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\right].\tag{3} \end{align}$$ The first covariant derivative of the right-hand side of the above equation reduces to an ordinary partial derivative, as the argument is a scalar. This allows us to apply partial integration to this term. The second covariant derivative has an argument that is not explicitly dependent on $y^\sigma$, so the covariant derivative can be written as a multiplication of this tensor with the appropriate Christoffel symbols. This finally leads us to $$\begin{align} &amp;-m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{d^2x^\nu}{d\tau^2} \cr &amp;+  m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\left[ \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau} \right].\tag{4} \end{align}$$ The continuity equation requires \begin{equation} -\frac{d^2x^\nu}{d\tau^2} + \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau}=0.\tag{5} \end{equation} This is the geodesic equation with an extra term, i.e. the term in the middle and with an incorrect sign for the first term. Can I get rid of this term in the middle by changing the parameter $\tau$ of the worldline? What about the incorrect sign? What did I do wrong?
92820
I have heard that in QFT, the vacuum has is postulated to have zero four-momentum, $$P^\mu |\Omega \rangle = 0.$$ However, I also know that it's supposed to have vacuum energy, $$ \langle \Omega | H | \Omega \rangle = E_0 \neq 0.$$ In textbooks, we patch up the latter equation by subtracting an infinite constant, and are told not to worry about it. In any case, the vacuum has to have zero four-momentum, as a fundamental axiom.   I am confused since  on this site talk about vacuum energy as a real thing that has a definite value. If it's real, then why can we subtract it off to get $P^\mu |\Omega \rangle = 0$? If it's not real, why do we talk about it as if it is?
101271
I was going through my notes on the unitary irreducible representations of the Poincare group and the subsequent construction of one particle states and I stumbled across the following steps in the method of induced representations: ($\mathscr{H}$ is the Hilbert space and $\mathscr{H}_p$ is the subspace of one-particle states with momenta p. $\mathcal{S}_k$ is just the little group of the standard momentum $k$. $L(p)$ is the standard Lorentz boost; $L(p)k=p$)   (1) Let $|k,\sigma\rangle$ be a basis for $\mathscr{H}_k$ and let $\mathcal{D}$ be the (finite) unitary irrep of $\mathcal{S}_k$ induced by the unitary irrep $\mathcal{U}$. Then $\mathcal{D}$ acts on $\mathscr{H}_k$ by mixing the spin polarization states,  \begin{align} \mathcal{U}(h)|k,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}(h)|k,\sigma'\rangle,~~~ \forall h\in\mathcal{S}_k,~~~\mathcal{U}(h)\equiv\mathcal{U}(h,0).  \end{align} (2) Define the basis for $\mathscr{H}_p$ by \begin{align} |p,\sigma\rangle=\mathcal{U}(L(p))|k,\sigma\rangle. \end{align} (3) The action of Lorentz transformations on these states is then \begin{align} \mathcal{U}(\Lambda)|p,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}\big(h(\Lambda,p)\big)|\Lambda p,\sigma'\rangle  \end{align} where $h(\Lambda,p)=L^{-1}(\Lambda p)\Lambda L(p)\in \mathcal{S}_k$ is known as the Wigner rotation.  (4) Finally, we extend this action to an arbitrary state in the Hilbert space $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$ by, \begin{align} \mathcal{U}(\Lambda)|\Psi\rangle&amp;=\sum_{p,\sigma}\big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big( p\big)| p,\sigma\rangle,\notag\\ \text{ where } \big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big(p\big)&amp;=\sum_{\sigma'}\mathcal{D}_{\sigma\sigma'}\big(h(\Lambda,\Lambda^{-1}p)\big)\Psi_{\sigma'}(\Lambda^{-1}p). \end{align}  These are my own notes which are largely influenced by Weinbergs construction, so they may be incorrect. I understand all of the steps except for one part of step 4; The part where I say $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$. What are these coefficients? Are they the wavefunctions? Do they depend continuously on $p$, if so should the sum over $p$ be replaced by an integral? Should these coefficients be square-integrable; $\Psi_{\sigma}(p)\in L^2(\Sigma_m,\mu,\mathbb{C})$ where $\Sigma_m$ is the mass shell and $\mu$ a Lorentz invariant measure? Is this expansion of an arbitrary state even correct?
68412
I am trying to prove the Lorentz invariance of the (left-handed) Weyl Lagrangian: $$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi$$  A Lorentz transformation is realized as $\psi\to M\psi$, where $M=\exp[-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}]$, and $S$ are the generators of the representation, i.e., $S^{0i}=-\frac{i}{2}\sigma^i$ and $S^{ij}=\frac{1}{2}\epsilon^{ijk}\sigma^k$. Therefore, the Lagrangian transforms as $$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi\to i\psi^\dagger M^\dagger\bar\sigma^\mu M\partial_\mu \psi$$ so that if it actually is invariant, we must have $M^\dagger\bar\sigma^\mu M=\bar\sigma^\mu$ (and, in particular, $M^\dagger M=\mathbb I$). I think this is a nice relation, which expresses the invariance of the symbol $\bar\sigma^\mu$.  Now the thing is, I think this relation is not true. Take, for example, a boost in the $z$-direction with rapidity $\omega$. This means $$M=\exp[-i\omega S^{03}]=\exp\left[\frac{\omega}{2}\begin{pmatrix}-1&amp; 0\\0&amp;1\end{pmatrix}\right]=\begin{pmatrix}e^{-\omega/2}&amp;0\\0&amp;e^{\omega/2}\end{pmatrix}$$  In this case, it is easly seen that $M^\dagger M\neq\mathbb I$. My question is: what went wrong in my argument?
106272
In QM, we are generally looking for projective representations of the underlying symmetry group (because of Wigner's theorem). So generally, we should be looking for representations of central extensions of the universal covering group, according to Bargmann. In the case of $SO(n)$ with $n&gt;2$, the universal covering groups happen to coincide with the double coverings. However, for $SO(2)$, this is not the case. Why are we still interested only in the double cover in the case of $SO(2)$ and not in its universal cover? EDIT: To be more specific, let me restrict the question to the case of the $\mathrm{Spin}(2)$ group: Why is this group defined this way and not as the universal cover of $SO(2)$? Is there any physical reason to consider the double cover instead of the universal cover? It seems like this definition potentially excludes interesting physics from consideration.
27511
What is the correspondence between the conserved canonical energy-momentum tensor, which is $$ T^{\mu\nu}_{can} := \sum_{i=1}^N\frac{\delta\mathcal{L}_{Matter}}{\delta(\partial_\mu f_i)}\partial^\nu f_i - \eta^{\mu\nu}\mathcal{L}$$ (the four conserved Noether currents corresponding to four possible spacetime translations)  where $\{f_i\}_{i=1}^N$ are the $N$ matter fields in the theory, and we assume $f_i\mapsto\alpha^\nu\partial_\nu f_i$ for translations,  and stress-energy tensor from the Einstein-Hilbert action, which is: $$ T_{\mu\nu}\equiv-\frac{2}{\sqrt{-g}}\frac{\delta\mathcal{L}_{Matter}}{\delta g^{\mu\nu}} $$   In particular, how do you get that the two are equal (are they?) for Minkowski space, for which there is no variation in the metric?
94808
If we have a complex scalar $\phi$ we know that the gauge-invariant interaction with $A$ is given by $A^\mu J_\mu$, where $J$ is the Noether current of the $U(1)$ symmetry of the Lagrangian $$ J_\mu\sim \phi^\dagger\partial_\mu\phi-\phi\partial_\mu \phi^\dagger \tag{1} $$  If we have a real scalar instead, $J=0$ and the field doesn't couple to $A$: real particles aren't charged.  My question is: what if we took $$ J_\mu\sim \phi\partial_\mu\phi \tag{2} $$ to couple $\phi$ to $A$? The interaction is Lorentz invariant and renormalisable, but it is not gauge-invariant, which is probably a bad thing. At what point would this theory break down?  This is probably very naïve, but I think that the Feynman rules for this theory are straightforward. So I guess the theory makes sense at least perturbatively. The theory is probably flawed at a more fundamental level, but I cant't seem to find where (some kind of gauge anomaly perhaps?)
57592
In the second chapter of Peskin and Schroeder, An Introduction to Quantum Field Theory, it is said that the action is invariant if the Lagrangian density changes by a four-divergence. But if we calculate any change in Lagrangian density we observe that under the conditions of equation of motion being satisfied, it only changes by a four-divergence term.  If ${\cal L}(x) $ changes to $ {\cal L}(x) + \alpha \partial_\mu J^{\mu} (x) $ then action is invariant. But isn't this only in the case of extremization of action to obtain Euler-Lagrange equations.   Comparing this to $ \delta {\cal L}$  $$ \alpha \delta {\cal L} = \frac{\partial {\cal L}}{\partial \phi} (\alpha \delta \phi) + \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \partial_{\mu}(\alpha \delta \phi) $$  $$= \alpha \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi \right) + \alpha \left[ \frac{\partial {\cal L}}{\partial \phi} - \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \right) \right] \delta \phi.    $$   Getting the second term to zero assuming application of equations of motion. Doesn't this imply that the noether's current itself is zero, rather than its derivative? That is:  $$J^{\mu} (x) =  \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi .$$  I add that my doubt is why changing ${\cal L}$ by a four divergence term lead to invariance of action globally when that idea itself was derived while extremizing the action which I assume is a local extremization and not a global one.
41757
Recently i read about Loop Quantum Gravity in Wikipedia and found this below statement .     Presently, no semiclassical limit recovering general relativity has been shown to exist.    But i also came across a paper in  whose abstract claims that Einstein's equations arise as a result of Semi-Classical Limit from Covariant Loop Quantum Gravity.     In this paper we explain how 4-dimensional general relativity and in particular, the Einstein equation, emerge from the spinfoam amplitude in loop quantum gravity.    So is the Wikipedia article outdated or is the paper's idea falsified?
88173
I am reading Schwartz's book on Quantum field theory. In the chapter anomalous magnetic moment (chapter 17) he mentions that Dirac equation naturally implies that the electron spin $g$-factor is 2. Can anyone please explain why does the Dirac equation implies that $g_s=2$?
82583
Probably a silly question, but something that came to mind yesterday. I couldn't find anything when searching.  Why is there an Energy mass equivalence principle but not an Energy charge equivalence principle?   In other words, why do our field theories have a Gauge invariance which allow for charge conservation, but not mass conservation, or why is it that charge has escaped from being a term in the energy mass equivalence? Is it just because we have observed these to be the case and made our field theories around this, or because that's just how the mathematics works out?
22381
From Landau &amp; Lifshitz The Classical Theory Of Fields it is said:     To determine the action integral for a free material particle (a particle not under the influence of any external force), we note that this integral must not depend on our choice of reference system, that is, it must be invariant under Lorentz transformations.   This seems understandable. But in comments on  Ján Lalinský says that "there is no good physical reason why action should be invariant". Further he suggests another Lagrangian than that given in L&amp;L, namely, if we denote L&amp;L Lagrangian as $L_0$, then the example could be $L_0+Cv_x$, which is clearly anisotropic. Clearly, the equations of motion must not change with this Lagrangian, because $Cv_x$ is a total time derivative (of $Cx$).  On the other hand, in  Luboš Motl says that "the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. [snip] If $S$ depended on the inertial system, so would the terms in the equations $\delta S=0$, and these laws of motion couldn't be Lorentz-covariant".  How could I connect L&amp;L and Luboš's arguments with Ján's example? Can both sides be simultaneously right? They seem to contradict each other.
2433
I've had a brief look through similar threads on this topic to see if my question has already been answered, but I didn't find quite what I was looking for, perhaps it is because I'm finding it hard to put words on my question and I hope that you will be able to help me ask it clearly.  I'm trying to link what I know from mathematics to what we are writing in physics for Noether's theorem. If I understand correctly we are looking at the symmetries of the action i.e. under which symmetry groups is it invariant. Noether's theorem allows us to calculate a conserved current in the case of a continuous symmetry (Lie groups), by means of so-called infinitesimal symmetries which I believe to be elements of the Lie algebra (so the tangent space to the neutral element) of the Lie group.  I'm guessing that if the action is invariant under the symmetry then it's "variation" should be 0 when we vary the system (space-time coordinate, or field) using this symmetry; it is exactly this step that I would like to understand better, how can I formalize this step correctly mathematically? How should I understand this variation, and how does it's calculation give rise to the elements of the Lie algebra?
20218
I've been reading about connections on fibre bundles recently and it's made me think about the exact nature of the Christoffel symbols in GR.  If we have a vector bundle $E$ over $M$ and put a connection $D$ on it, in some local trivialisation we can choose a fixed connection $D^0$ and a vector potential $A\in\Gamma(\textrm{End}(E)\otimes T^*M)$ such that $D=D^0+A$. (Typically $D^0$ is the flat connection where allowed.) Then, if we have a section $s\in\Gamma(E)$ written in our local trivialisation as $s^ie_i$, and similarly $A=A_{\mu j}^ie_i\otimes e^j\otimes dx^\mu$, the covariant derivative of $s$ is given by $$D_\mu s=(D_\mu^0s^i+A_{\mu j}^is^j)e_i$$ Now this will immediately recall the covariant derivative of Riemannian geometry/GR, which, on a vector field $v=v^\mu\partial_\mu$, is given by $$D_\mu v=(\partial_\mu v^\nu+\Gamma^\nu_{\mu\lambda}v^\lambda)\partial_\nu$$ Indeed vector fields are just sections of the tangent bundle so this is to be expected. The Christoffels symbol play the role of the vector potential for general bundles.  Now my question is the following: given $E$, $A$ is a bona fide section of the tensor product bundle $\textrm{End}(E)\otimes T^*M$ - but this seems to suggest that $\Gamma$ should be a bona fide section of the bundle $$\textrm{End}(TM)\otimes T^*M\cong TM\otimes T^*M\otimes T^*M.$$ But then should it not be a bona fide type-$(1,2)$ tensor? However it is well-known that it is not a tensor. The only thing I could think of was that $A$ can in general only be defined locally (i.e. on a local trivialisation); but the familiar calculation in GR that shows the Christoffels are not tensors computes their transformation properties, which should be local, so I don't think that the explicit locality of the vector potential should be a problem. In any case, trivial tangent bundles are possible, on which the vector potential can be defined globally. What am I missing?