Given the Lagrangian density for a real scalar field $\mathcal{L}(\phi, \partial_\mu \phi)$, we obtain from Noether's theorem the canonical stress-energy tensor $$ T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\nu \phi - g^{\mu \nu} \mathcal{L} $$ Can anyone tell me, or provide a (hopefully detailed) reference, why there is always (I think?) a rank $3$ tensor $X^{\lambda \mu \nu}$ such that $$\widetilde{T}^{\mu \nu} = T^{\mu \nu} + \partial_{\lambda} X^{\lambda \mu \nu}$$ is symmetric? I'm fine with assuming $g$ is the Minkowski metric, but I can only find this done in particular cases of $\mathcal{L}$ and no general argument.
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?
35312
I have have set my self a challenge to learn all the maths behind the Einstein field equation (EFE), and from reading it seems that the Metric tensor is the thing we are trying to find (from the 10 equations that forms the EFE). but since nearly all the terms in it are functions of this metric tensor, this seems very hard. What type of maths do we use to do this? (I have copied and pasted the EFE just for reference):  $$R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R+g_{\mu\nu}\Lambda=\frac{8 \pi G}{c^4}T_{\mu \nu}$$  Edit: Thanks for your comments. Just as I have no one else to ask, is this the equation that we try to solve:  $$\left(\frac{\partial}{\partial x^{\lambda}}\Gamma^{\lambda}_{\mu \nu}- \frac{\partial}{\partial x^{\nu}}\Gamma^{\lambda}_{\lambda \mu}+\Gamma^{\lambda}_{\lambda \rho}\Gamma^{\rho}_{\nu \mu}-\Gamma^{\lambda}_{\nu \rho}\Gamma^{\rho}_{\lambda \mu}\right)-\frac{1}{2}g_{\mu \nu}g^{ab}\left(\Gamma^c_{\enspace ab,c}-\Gamma^c_{\enspace ac,b}+\Gamma^d_{\enspace ab}\Gamma^c_{\enspace cd}-\Gamma^d_{\enspace ac}\Gamma^c_{\enspace bd} \right)+g_{\mu \nu}\Lambda=\frac{8 \pi G}{c^4}T_{\mu \nu}$$  with the $R_{μν}$ and $R$ terms expanded and where $g_{μν}$ and $T_{μν}$ are the μν th components from the related tensors? (a yes or no answer will be fine, thanks).
122938
For Minkowski space with a weak gravitational field the metric takes the form $$ ds^2 = (1+2\phi/c^2)c^2dt^2 -(1-2\phi/c^2)(dx^2+dy^2+dz^2), $$ where $\phi$ is the Newtonian gravitational potential. You can get the $(1+2\phi/c^2)$ factor in front of the $dt^2$ by starting with the geodesic equation and going to the &quot;Newtonian limit&quot; of slow speeds and a slowly changing field. But is there a way to get the $(1-2\phi/c^2)$ factor for the spatial part of the metric by a similar procedure? What about some clever thought experiment?
65962
I'm trying to grasp Levi-civita and Kronecker del notation to use when evaluating geophysical tensors, but I came across a few problems in the book I'm reading that have me stumped.   1) $\delta_{i\,j}\delta_{i\,j}$  2) $\delta_{i\,j}  \epsilon_{i\,j\,k}$  I have no idea how to approach evaluating these properties. Without knowing i, j, or k, how would I approach? I don't feel confident using the notation further until I can understand these properties.
40385
Maybe my understanding is just off, but the  is just a scalar, right? What are it's units? Why a scalar? - was a tensor 'cosmological constant' ever considered or is it just not conformable to the math or physics of the field equations?
61942
I recently came across a video by prof Fredrick Schuller on general relativity where he defines the leibniz rule to be,  $\nabla_X (T(\omega,Y))=\nabla_XT(\omega,Y)+T(\nabla_{X} \omega,Y)+T(\omega,\nabla_{X} Y)$  Where $X$ and $Y$ are vector fields, $\omega$ is a covector field and $T$ is a $(1,1)$ tensor. The rule can be generalised for $(p,q)$ tensors similarly.  I cannot find a way to show that it is equivalent to the leibniz rule expressed as $\nabla_X$ acting on tensor product of two tensor fields.  i.e. $\nabla_{X}(T \otimes S) = \nabla_XT\otimes S + T\otimes \nabla_XS$  How do I proceed to show the equivalence between the two?
40940
When I learned about the Minkowski Space and it's coordinates, it was explained such that the metric turns out to be $$ ds^{2} = -(c^{2}dx^{0})^{2} +(dx^{1})^{2} + (dx^{2})^{2} + (dx^{3})^{2}   $$ where $ x^0,x^1,x^2,x^3 $ come from $ x^{\mu} : \mu = 0,1,2,3 $, and $ c $ is the speed of light. The first resource I had access to--I have to do a bit of digging for the exact paper--of course addressed that this invariant takes the tensor form: $$ ds^2 = g_{\mu \nu}dx^{\mu}dx^{\nu} $$ as well. This two things I've seen in all of my texts and online resources regarding resources. The element in question that varies between authors, is the time coordinate, $ x^0 $. When it was first explained to me, it was using the standard Cartesian representation for the spatial portion of the coordinates, and the time coordinate was labeled as $ x^0 = ict $. Squaring gives $ (x^0)^2 = -c^2t^2 $, and applying differential calculus, we get $ (dx^0)^2 = -c^2dt^2 $. Sensible and expected to have the tensor formula spit out the Minkowski Metric. The author then later explicitly states the coordinates are $ x^0 = ict, x^1=x,x^2=y,$ and $x^3 = z$.  My questions is then, why is it most authors on the subject omit the imaginary unit on the time coordinate? For example, .   The only reason I can fathom omission is if the author is using metric signature $ [+,-,-,-] $, where I started off learning the theory with signature $ [-,+,+,+] $ which may be the reason seeing the time coord with no imaginary unit seems dissonant to me. All help appreciated!  Edit: After reading the other answers, my questions is now one of why and how (mathematically) do we obtain the Minkowski Metric Signature. More specifically the one element with a different sign.
77202
Einstein assumed in the derivation of his field equations that the covariant derivative (defined through the metric) of the energy momentum tensor of matter is zero, i.e. $$\nabla_{\mu}T^{\mu\nu}=0,$$  no matter how distorted the spacetime is.  It seems such a conclusion cannot be reached at via Nöther's theorem (since the metric is itself changing over spacetime). So my question is how did he have this insight then, and how would one justify the correctness of this assumption in retrospect from a Lagrangian point of view now?   If it's based on the existence of infinitesimal coordinate transformations that fix the boundary of some volume in spacetime, then would this also imply that the energy momentum tensor of any field will always have zero covariant derivative with or without gravity?    Edit: Previous form of the question had an abuse of language, viz., that I referred to the "vanishing of the covariant derivative of the energy-momentum tensor" as "energy momentum tensor is covariantly conserved". This was rightly pointed out by Ben Crowell in his answer and, I've edited the question and made it mathematically precise for the sake of clarity.
6911
I am reading through this article   about the special case of the positive mass theorem in general relativity. I do not understand the section below:  In particular what does it mean to isometrically embed an $n$-dimensional Riemannian manifold $(M,g)$ in a higher dimensional space, and where does the embedding given for the  $n=3$ Schwarzschild metric come from? I assume we are referring to the standard definition of isometric embeddings using pullbacks of covariant tensor fields. I assume the author refers to the map between Riemannian manifolds $f: \mathbb{R}^3-B_{2m}\to\mathbb{R}^4$ given by $f(x,y,z)=(x,y,z,\sqrt{8m(r-2m)})$. I have tried to check if the Schwarzchild metric on $\mathbb{R}^3$ is the pullback of the flat metric on $\mathbb{R}^{4}$, however this doesn't appear to be true...
9129
In General Relativity Einstein's equation implies that stress-energy tensor on its RHS is conserved (has vanishing divergence), due to the Bianchi identity. Considering variational principles leading to Einstein's equation leads to conclusion that this stress tensor is equal to the variational derivative of full action with respect to the metric tensor. However, on several occasions I heard people stating that quite generally one can define stress tensor for a field theory in this way and it is automatically conserved. In flat spacetime and without any coupling to gravity! I wonder if this is true. I don't see a reason why it should be.
27064
The following Lagrangian density describes a charged scalar field. Determine the global U(1) symmetry. Make this global symmetry local and give the Lagrangian density. The Lagrangian density is given by: $$L = \partial_\mu \psi \partial^\mu \psi^* - \frac{1}{2}m^2|\psi|^2 - \frac{1}{4}\lambda|\psi|^2.$$  My first idea was to perform the following transformation (because it is mentioned we are dealing with U(1) invariance): $$\psi \rightarrow e^{i\theta}\psi$$ $$\psi^* \rightarrow e^{-i\theta}\psi^*$$ $$\partial_\mu \psi \rightarrow D_\mu \psi = (\partial_\mu + iqA_\mu)\psi$$ and $$\partial_\mu \psi^* \rightarrow D_\mu \psi^* = (\partial_\mu - iqA_\mu)\psi^*$$ However I think this is what you have to do to obtain a local symmetry out of the global one. I have no idea how to find the global one. In short my questions are:  How do you find the global U(1) symmetry for a given Lagrangian density? Is the way to determine the local symmetry out of the global U(1) symmetry achieved by transforming in the following way: $\psi \rightarrow e^{i\theta}\psi$, $\psi^* \rightarrow e^{-i\theta}\psi^*$, $\partial_\mu \psi \rightarrow D_\mu \psi = (\partial_\mu + iqA_\mu)\psi$ and $\partial_\mu \psi^* \rightarrow D_\mu \psi^* = (\partial_\mu - iqA_\mu)\psi^*$?  Thanks in advance!
102434
When one consider a lagrangian and construct hamiltonian, we expect to be bounded below. While looking to the Hamiltonian formulation of general relativity, I have difficulties to see how it can be bounded.   How this can be shown?  Is it related to the positive energy theorem?
43500
The Lagrangian density for a single real scalar field theory is \begin{equation}\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)\end{equation} I have often seen this written \begin{equation}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)\end{equation} which appears to differ in the kinetic term.  Specifically \begin{equation}(\partial_{\mu}\phi)^{2}=(\partial_{\mu}\phi)(\partial^{\mu}\phi)\neq\partial_{\mu}\phi\partial^{\mu}\phi=(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\phi\partial_{\mu}\partial^{\mu}\phi\end{equation} where the product rule has been used in the last equality. Is this just sloppy notation (which is acceptable because only first derivatives in the field are considered anyway), or is my maths very bad?
30417
This arises from the question , which I'm afraid I answered just by looking it up in . However Schutz (as he frequently does) glosses over details he thinks are irrelevant or too simple to be worth explaining, and I have realised I don't understand an assumption he makes.  Schutz states without proof that if we have an equatorial orbit in a Schwartzschild metric then:     Independence of the metric of the angle $\phi$ about the axis implies that $p_\phi$ is constant.   In the non-relativistic world I assume this corresponds to angular momentum being constant in a central potential. So far so good. But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? The component $p^\phi$ is presumably not constant since (in this case) $p^\phi = p_\phi/r^2$.  Bonus points for also explaining his similar claim that time independance means that $p_t$ is constant rather than $p^t$.  I fear that Schutz didn't explain because it's an insultingly simple question, but if someone can provide a nice intuitive explanation I would be very pleased to read it.
59245
The electromagnetic field tensor is given by $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$, and it appears in the Lagrangian as  $$L = -\frac{1}{4}F_{\mu\nu}^2 - A_{\mu}J_{\mu}.$$ Schwartz's QFT textbook says that (chapter 8, page 116) $$F_{\mu\nu}^2 = 2(\partial_{\mu}A_{\nu})^2 - 2(\partial_{\mu}A_{\mu})^2,$$ but I do not see how that is correct.   The first step of the expansion gives $F_{\mu\nu}^2 = (\partial_{\mu}A_{\nu})^2 - \partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu} - \partial_{\nu}A_{\mu}\partial_{\mu}A_{\nu} + (\partial_{\nu}A_{\mu})^2$, and I see that the first and last term add up to produce $2(\partial_{\mu}A_{\nu})^2$. However, the second term $\partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu}$ when expanded has a term like $\partial_{0}A_{1}\partial_{1}A_{0}$, and I don't see how this is present in $(\partial_{\mu}A_{\mu})^2 = (\partial_{0}A_{0} + \partial_{1}A_{1} + \partial_{2}A_{2} + \partial_{3}A_{3})^2$.
70187
Noether's theorem usually considers coordinate/field transformations which leave the Lagrangian invariant up to a divergence term, i.e.  $$\mathcal{L} \rightarrow \mathcal{L} + \partial_{\mu}f^{\mu}$$  However there is a more general class of transformations which leave the equations of motion invariant, and that is a divergence term along with an overall scaling:  $$\mathcal{L} \rightarrow \alpha\mathcal{L} + \partial_{\mu}f^{\mu}$$  but Noether's theorem does not seem to deal with these types of symmetries, which are exhibited by, for example, the Klein-Gordon Lagrangian:  $$\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - \frac{1}{2}m^2\phi^2$$  under the transformation $\phi \rightarrow \alpha\phi$ (giving $\mathcal{L} \rightarrow \alpha^2\mathcal{L}$).    The action similarly transforms as $S \rightarrow \alpha^2 S$.  I feel this is important to emphasize as there are cases in which the Lagrangian scales by a factor but the action remains strictly invariant.  This is not the case here, as the action scales by $\alpha^2$ but the equations of motion and extrema of the action are invariant, which is ultimately the only physical thing that matters.    Is there a generalization to Noether's theorem for these "scaling-type" symmetries?  I.e. is there a conserved quantity/current arising from scale invariance?
99563
I'm doing some classical field theory exercises with the Lagrangian $$\mathscr{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. To find the conjugate momenta $\pi^\mu_{\ \ \ \nu} = \partial \mathscr{L} / \partial(\partial_\mu A^\nu)$, I can use two methods.  First method: directly apply this to $\mathscr{L}$. We get a a factor of $2$ since there are two $F$'s, and another factor of $2$ since each $F$ contains two $\partial_\mu A_\nu$ terms, giving $$\pi^\mu_{\ \ \ \nu} = -F^\mu_{\ \ \ \nu}.$$  Second method: get $\mathscr{L}$ in terms of $A$ by expanding and integrating by parts, yielding $$\mathscr{L} = \frac{1}{2}(\partial_\mu A^\mu)^2 - \frac{1}{2}(\partial_\mu A^\nu)^2.$$ Differentiating this gets factors of $2$ and gives $$\pi^\mu_{\ \ \ \nu} = \partial_\rho A^\rho \delta^\mu_\nu - \partial^\mu A_\nu.$$  These two answers are different! (They do give the same equations of motion, at least.) I guess that means doing the integration by parts changed the canonical momenta.  Is this something I should be worried about? In particular, I have another exercise that wants me to show that one of the canonical momenta vanishes -- this isn't true for the ones I get from the second method! Plus, my stress-energy tensor is changed too. When a problem asks for "the" canonical momenta, am I forbidden from integrating by parts?