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I am a beginner in QFT, and I have been reading about the energy-momentum tensor from making "infinitesimal transformations" in the Poincare group. (not variation wrt metric tensor). But I think I got caught in a few technical details that I cannot solve on my own. I would appreciate it if you could use some simple Lagrangians as an example.

  1. What is invariant under infinitesimal transformations, especially translation? Fradkin's QFT an integrated approach requires that the action remains unchanged, so are a bunch of other books on gauge theory, but Greiner's relativistic quantum mechanics as well as Peskin all requires invariance of the Lagrangian itself. The problem is that I can show that the energy-momentum tensor is locally conserved if I require the invariance of Lagrangian, but I cannot show that if I require the invariance of action. I think the invariance of action is more general but I failed to see how I can get a locally conserved energy-momentum tensor from that. Here is my derivation: For Lagrangian $\mathcal{L}=\left(D_{\mu} \phi(x)\right)^{*}\left(D^{\mu} \phi(x)\right)-m_{0}^{2}\phi(x)\phi^*(x)-\frac{\lambda}{2}\left(\phi(x)\phi^*(x)\right)^2-\frac{1}{4} F^{\mu \nu} F_{\mu \nu}$, Requiring the system being transnational invariant, $\delta S =0$, we have for arbitrary uniform displacement \begin{align*} \int d^4 x \partial_\mu \left[ \left( g ^\mu_\nu \mathcal L - \frac{\delta{\mathcal L }}{\delta \partial_\mu \phi} \partial_\nu \phi - \frac{\delta{\mathcal L }}{\delta \partial_\mu \phi^*}\partial_\nu \phi^* + \frac{\delta{\mathcal L }}{\delta \partial_\nu A_\sigma} \partial_\nu A_\sigma \right) \delta x^\nu\right] = 0 \end{align*} But that does not guarantee the conservation of the Neother current, because the $ \delta x^\nu$ is a constant (uniform diaplacement), and at best we can say that the integral is zero, but not the integrand. I can still arrive at a conserved charge by doing Stokes theorem on that integral, but I lost the locally conserved tensor. This problem is not present when we consider rotations, since in that case, $\delta x^\nu$ would be truly arbitrary.

  2. How do we do the variation exactly? In Fradkin's book as well as some gauge theory books, they all talk about the variation of integration measure $d^4 x$ under a coordinate transformation. Since the derivative of a determinant is gven by \begin{align*} \newcommand{\D}[2]{\frac{d #1}{d #2}} \newcommand{\tr}{\mathrm tr} \D{\det J(x)}{x} = \det J \tr \left(J^{-1} \D{J(x)}{x} \right) \end{align*} And if I do the variation of this measure, seeing the integral measure as a functional of coordinate unction $x_\mu$, which is a function of some fixed coordinate $\chi_\lambda$, is given by \begin{align*} d^4 x [x_\mu] = \det J[x_\mu] d^4 x[\chi] = \det J[x_\mu (\chi)] d^4 \chi \end{align*}

Then the functional derivative of the measure with respect to the coordinate function is

\begin{align*} \newcommand{\P}[2]{\frac{\partial #1}{\partial #2}} \frac{\delta d^4 x [x_\mu]}{\delta x_\sigma} =& \frac{\delta \det J[x_\mu (\chi)] d^4 \chi}{\delta x_\sigma (\chi)} \\ =& \det J[x_\mu] \tr\left(J^{-1}[x_\mu] \frac{\delta J[x_\mu]}{\delta x_\sigma (\chi)} \right) d^4 \chi\\ =& \det J[x_\mu] \delta_{\nu \lambda}\P{\chi^\nu}{x_\mu} \left( \frac{J^{\mu\lambda}[x_\mu]}{\delta x_\sigma (\chi)} \right) d^4 \chi\\ =&\delta_{\nu \lambda}\P{\chi^\nu}{x_\mu} \left( \frac{\delta}{\delta x_\sigma (\chi)} \P{x^\mu}{\chi_\lambda}\right) \det J[x_\mu] d^4 \chi\\ =&\P{\chi^\nu}{x_\mu} \left( \frac{\delta}{\delta x_\sigma (\chi)} \P{x^\mu}{\chi_\nu}\right) \det J[x_\mu] d^4 \chi\\ =&\P{\chi^\nu}{x_\mu} \P{ g^{\mu \sigma}}{\chi_\nu} \det J[x_\mu] d^4 \chi\\ =&\P{g^{\mu \sigma}}{x_\mu} \det J[x_\mu] d^4 \chi\\ =& \left(\P{}{x_\sigma} \right)\det J[x_\mu] d^4 \chi \end{align*}

But I am having a hard time understanding this result. If I write \begin{align*} \delta d^4 x [x_\mu] = \left(\P{}{x_\sigma} \right) \delta x_\sigma \det J[x_\mu] d^4 \chi = (\partial^\sigma \delta x_\sigma) d^4 x_\mu \end{align*} I got confused. Normally we do $\delta \partial = \partial \delta$, and that would give me nonsense. I am not sure how to properly interpret this result. I think that $\delta x$ here might not be a functional change so I am not allowed to do that? But I am not sure.

  1. There are other ways to do this variation as well, what is the right one to use? Some book says that \begin{align*} \delta S =& \int d^4 x' \mathcal L'(\phi ' (x'))-\int d^4 x \mathcal L(\phi (x))\\ =&\int d^4 x \mathcal L'(\phi ' (x))-\int d^4 x \mathcal L(\phi (x))\\ =&\int d^4 x (\mathcal L'(\phi ' (x))- \mathcal L(\phi (x)))\\ \end{align*} because the $x'$ in the first integral is a dummy index. Therefore there's no variation on the integration measure at all. I am not sure if that works since the boundary might be shaped differently. But the result is the same if we allow that.

There is also another way of doing it by allowing a $\partial \mathcal L/\partial x$ term. Some book says that this term arises from the fact that the Lagrangian density is a scaler itself and should transform, others avoided it, and some other book says that is comes from the freedom of choice of an total derivative of the Lagrangian (and hints super symmetry). I am not sure if they are all equivalent. But allowing a $\partial \mathcal L/\partial x$ term doesn't make sense to me.

I can provide more details if needed.

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  • $\begingroup$ I recommend first understanding what happens in a simpler theory, say a free, real scalar field, before trying to tackle something more complicated. Also, spacetime symmetries generically transform the Lagrangian by a total derivative. $\endgroup$ Oct 2 at 7:38
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There is a lot of notation above that I'm having trouble parsing but an infinitesimal translation is $x^\prime_\mu = x_\mu + \delta a_\mu$. To compute $\delta S$ under this transformation, you change $x^\mu$ to $x^\mu + \delta a^\mu$ everywhere, expand in $\delta a^\mu$ and keep the linear term. Let's start with the massless free scalar. \begin{align} S &= \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \\ &\to \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x + \delta a) \partial_\mu \phi(x + \delta a) \\ &= \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) + \int d^4 x \, \partial_\nu \partial_\mu \phi(x) \partial_\mu \phi(x) \delta a_\nu + O(\delta a^2) \\ &= \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) + \int d^4 x \, \partial_\nu \left [ \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \right ] \delta a_\nu + O(\delta a^2) \end{align} Therefore $\delta S$ is the integral of a total derivative, in other words zero. This is reassuring (since we expect translation to be a symmetry) but it answers one of your questions. The variation of the Lagrangian isn't zero, only the variation of the action.

Now how do we get the Noether current for this symmetry? There is a helpful trick due to either Glashow or Gel-Mann which involves promoting the constant $\delta a_\mu$ to a local $\delta a_\mu(x)$. This is not a symmetry anymore (unless we are working with a diffeomorphism invariant theory like GR) but it is still helpful for studying the original symmetry. Proceeding from the last line above, we have \begin{align} \delta S &= \int d^4 x \, \partial_\nu \left [ \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \right ] \delta a_\nu(x) \\ &= \int d^4 x \, \partial_\nu \left [ \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \delta a_\nu(x) \right ] - \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \partial_\nu \delta a_\nu(x) \\ &= - \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \partial_\nu \delta a_\nu(x) \\ &= \int d^4 x \, \partial_\mu \partial_\nu \phi(x) \partial_\mu \phi(x) \delta a_\nu(x) \\ &= \int d^4 x \, \partial_\mu \left [ \frac{1}{2} \partial_\nu \phi(x) \partial_\mu \phi(x) \right ] \delta a_\nu(x) \end{align} We have now written this more general $\delta S$ as the integral of $\partial_\mu T_{\mu\nu} \delta a_\nu$. Even though $\delta S$ need not be zero because we have promoted our symmetry to a non-symmetry, $\delta S$ on shell does need to be zero since the equations of motion are defined by leading to stationary actions under any variation. Then, since $\delta a_\nu$ is arbitrary, $\partial_\mu T_{\mu\nu} = 0$ on shell. In other words, the $T_{\mu\nu}$ is indeed the stress tensor and we can read it off above.

I hope this explains your confusion in point number 1. It doesn't have to do with translations vs rotations. Translations are $x^\prime_\mu = x_\mu + \delta a_\mu$ specified by four constant parameters and have Noether current $T_{\mu\nu}$. Rotations are $x^\prime_\mu = x_\mu + \delta \omega_{[\mu\nu]} x_\nu$ specified by six constant parameters and have Noether current $M_{\mu\nu\rho} = x_\mu T_{\nu\rho} - x_\nu T_{\mu\rho}$. It's just that, for either one of these symmetries, the constant parameter becomes arbitrary when we want to use the above trick.

These calculations were done for the massless free scalar but a good exercise is to repeat them for the scalar QED theory you wrote or indeed abstract theories where $\mathcal{L}$ is a polynomial in a bunch of fields and their derivatives. All of these will have conserved stress tensors derivable with the same steps. Conversely, Lagrangians that involve further integrals of the fields, such as the famous \begin{align} \mathcal{L}[\phi(x)] = \int d^4 y\, \frac{\phi(x) \phi(y)}{|x - y|^\gamma} \end{align} will lead to the phenomenon you worried about: global but not local conservation. In other words, translation is still a symmetry but there is no longer an associated conserved current. This is the hallmark of a nonlocal field theory.

Once you are comfortable with this Noether procedure for getting the stress tensor, there are optional steps which involve adding improvement terms which vanish on shell but make the symmetries more manifest. These are usually accounted for automatically by the Hilbert procedure you mention which comes from varying with respect to the metric.

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