# Energy momentum tensor from Noether's theorem

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?

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Related: physics.stackexchange.com/q/99853/2451 and links therein. – Qmechanic Jan 3 '15 at 15:51

This is a clever method used to derive Noether's current for any global symmetry; for the translational symmetry, it produces the stress-energy tensor.

We have to consider a local transformation because the variation of the action, $\delta S$, vanishes for the global transformation because the global transformation is by definition a symmetry: $$\delta S = 0$$ This value of $\delta S$ would follow "tautologically" and we couldn't deduce anything new out of it.

It follows that if we "generalize" the symmetry transformation rules and make them $x$-dependent, i.e. the transformation will be specified by $\delta a(x)$ (and $a$ is a vector for translations, but may be scalar for other symmetries), then $\delta S$ will be nonzero but it will inevitably depend on derivatives of $a$ only; for constant choices of $a$, we must get zero (because it's the global symmetry). For actions that only depend on first derivatives of the fields, the variation of the action will inevitably have the form $$S = \int (\partial_\mu a) j^\mu d^d x$$ where $j^\mu$ is some particular function of the fields or other degrees of freedom (and their derivatives). Note that this form is inevitable: $\delta S$ has to be linear in $a$ and/or its derivatives, but it must vanish for $a={\rm const}$ so there can't be any terms of the form $\int ab\,\,d^dx$, i.e. terms proportional to undifferentiated $a$. There are also no higher-derivative terms if the action didn't have higher derivatives of fields to start with.

Now, the argument that $j^\mu$ is a current is simple. When equations of motion are satisfied, $\delta S =0$ for any variation of fields, whether it's a symmetry or not. In particular, $\delta S = 0$ holds for the "generalized" or "localized" global symmetry given by $a(x)$ which is no longer an exact symmetry, so $\delta S$ is the nonzero expression above. But by integration by parts, $\delta S = 0$ means $$0 = \int a(x)\cdot \partial_\mu j^\mu(x)\,\, d^d x$$ which vanishes iff $\partial_\mu j^\mu$ vanishes at each point. This proves that $j^\mu$ obtained in this way is a conserved current; its integral has to be a conserved quantity. You could ask why someone invented this method. He or she invented it because he or she was creative and smart. What's important for everyone else is to check the arguments above and see that one may derive a conserved current in this way. The original author of the method was able to "see" the whole argument in his or her head.

(I say "her" as well to pay some tribute to Noether who didn't quite invent this elegant method – her papers were messy – but she invented the whole relationship between symmetries and conservation laws.)

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This is a trick of variations, it is explained best by Feynman in the Character of Physical law. The point is that the conservation law comes from a symmetry plus a minimum principle. You take the path from A to B, and you translate the path (you translate the endpoints too) to get a path from A' to B'. The path which moves quickly from A to A', then from A' to B', then from B' to B is a variation of the original path, and so has the same action by the principle of stationary action. But the path from A' to B' has the same action as the path from A to B by symmetry. This means that the little bit of action from A to A' must equal the little bit of action from B to B'.

This is Feynman's version of Noether's theorem. Feynman's argument requires evaluating the action of a little jump-kick at the start and the end of a path with jumps. This is mathematically annoying, so you can reformulate the argument in a more mathematically convenient way by using a continuous version.

So instead, consider doing Feynman's argument with many slicings of time, and doing a little independent translation at each slicing. The same argument tells you that the little bit of action you get one each slice from the little translation there will be equal at opposite ends. So the variation of the action with a time-dependent translation will be equal to the conserved quantity times the derivative of the translation parameter with time.

This is the version of Noether's theorem that the books do, and it is the mathematically easiest. The conceptual reasoning is the same as in Feynman's version; the two arguments are interchangable.

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Why do we consider an $x$-dependent local transformation instead of a global transformation?

There is a good reason (see below) why we like to start out by a general $x$-dependent local infinitesimal transformation,

$$x^{\prime \mu}- x^{\mu} ~=~ \delta x^{\mu} ~=~ - \varepsilon^{\mu},$$ $$\phi^{\prime}(x)-\phi(x) ~=~\varepsilon^{\mu} \phi_{,\mu},$$

where $\varepsilon^{\mu}\equiv\delta a^{\mu}$ is a local $x$-dependent infinitesimal parameter, and only later specialize to a global $x$-independent (=rigid) transformation. Going a bit over the top in local transformations (at least from the perspective of Noether's first Theorem), Itzykson & Zuber write on p.23 in the book QFT:

From the vanishing of $\delta I$ for arbitrary $\delta a^{\nu}(x)$, we deduce that the energy momentum flow described by the canonical tensor [...] satisfies the conservation law [...].

It is an important point to stress (as OP seems aware of) that only global symmetry is necessary in Noether's first Theorem.

So let us demonstrate this in the case at hand. If one starts with a global transformation, one derives

$$0~=~ \delta S ~=~ S[\phi^{\prime}]- S[\phi] ~=~ \varepsilon^{\mu} \int_{V} {\rm d}^dx \left(\frac{\partial \cal L}{\partial \phi}\phi_{,\mu}+\frac{\partial \cal L}{\partial \phi_{,\nu}}\phi_{,\mu\nu}-d_{\mu}{\cal L}\right), \qquad\qquad (A)$$

where $V$ is some integration region, and $\varepsilon^{\mu}$ is a global $x$-independent infinitesimal parameter. Let us take $V$ to be $\subseteq\mathbb{R}^d$ for simplicity. One can proceed in three cases:

1. If the integration region $V$ is fixed, and since eq. $(A)$ by assumption holds for all off-shell configurations of the $\phi$ field, then it is possible to deduce that the integrand $(A)$ is a total divergence, $$\frac{\partial \cal L}{\partial \phi}\phi_{,\mu}+\frac{\partial \cal L}{\partial \phi_{,\nu}}\phi_{,\mu\nu}-d_{\mu}{\cal L}~=~d_{\nu} f^{\nu}_{\mu}. \qquad\qquad (B)$$ [The words on-shell and off-shell refer to whether the equations of motion are satisfied or not. We use the symbol $d_{\mu}$ (rather than $\partial_{\mu}$) to stress the fact that the derivative $d_{\mu}$ is a total derivative, which involves both implicit differentiation through the field variable $\phi(x)$, and explicit differentiation wrt. $x^{\mu}$.]

2. If one assumes (as Noether did in 1918) that the symmetry $(A)$ holds for arbitrary integration regions $V$, then one deduces that the integrand $(A)$ vanishes identically $$\frac{\partial \cal L}{\partial \phi}\phi_{,\mu}+\frac{\partial \cal L}{\partial \phi_{,\nu}}\phi_{,\mu\nu}-d_{\mu}{\cal L}~=~0.$$ This corresponds to eq. $(B)$ with $f^{\nu}_{\mu}=0$.

3. If one assumes (as Itzykson & Zuber) that the Lagrangian density ${\cal L}$ has no explicit $x^{\mu}$ dependence, then the symmetry $(A)$ holds for arbitrary integration regions $V$, and one is back in case 2.

Next define the full Noether current as

$$T^{\nu}_{\mu}~:=~\frac{\partial \cal L}{\partial \phi_{,\nu}}\phi_{,\mu}-\delta^{\nu}_{\mu}{\cal L} -f^{\nu}_{\mu}.\qquad\qquad (C)$$

It is not hard to deduce the conservation law

$$d_{\nu}T^{\nu}_{\mu}~=~\left(d_{\nu}\frac{\partial \cal L}{\partial \phi_{,\nu}}-\frac{\partial \cal L}{\partial \phi}\right)\phi_{,\mu}~\approx~0,$$

with the help of eqs. $(B)$, $(C)$, and Euler-Lagrange equation. [We use the $\approx$ sign to stress that an equation is an on-shell equation.]

Now let us return to the original question. The standard reason to starts with a local variation is, that one does not have to guess/remember/pull-out-of-the-hat the bare Noether current

$$t^{\nu}_{\mu}~:=~ \frac{\partial \cal L}{\partial \phi_{,\nu}}\phi_{,\mu}-\delta^{\nu}_{\mu}{\cal L}.$$

It simply comes out as the term that multiplies $d_{\nu}\varepsilon^{\mu}$ in the local variation, as Lubos Motl also explains in his answer.

Finally, note that the full Noether current $T^{\nu}_{\mu}$ may still contain a $f^{\nu}_{\mu}$ piece. This final piece may be determined from the total divergence term $d_{\nu}f^{\nu}_{\mu}$ that is multiplied by the undifferentiated $\varepsilon^{\mu}$ in the local variation.

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