Confusion in Proof of Noether's theorem

Question

This question is related to this Noether's theorem under arbitrary coordinate transformation and this Transformation of $d^4x$ under translation disregarded?
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $x\mapsto x^\prime$is given by
$$\delta S=\int \mathscr{L}'(\phi^\prime,\partial_\mu\phi^\prime)d^4x^\prime-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$$

For me this variation should always be zero since this is just a change of variables so there is nothing new. Suppose that in one dimension flat space we have $\phi(x)=\sin(x)$ and our action is given by $$S=\int{\sin(x)dx}$$ under a trasnlation $x\mapsto x+a$ we should have $\phi(x) \mapsto \phi'(x)=\sin(x-a)$.

Under a change of coordinates $x\mapsto y=x+a$, $\phi(x) \mapsto \phi'(y)=\sin(y-a)$ , since $x$ and $y$ are dummy index the action is the same in both case.

Now assuming we have $\sqrt g= x^2$. Translating will give us an action
$$S=\int{ x^2\sin(x-a)dx}$$

But change of variables will give us an action $$S=\int{ (y-a)^2\sin(y-a)dx}$$ witch is different.

Shouldn't the variation of the action be written like this $$\delta S=\int \mathscr{L}'(\phi^\prime,\partial_\mu\phi^\prime)d^4x^-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$$

What about the Jacobian arising in front of the measure when changing variables? — Valter Moretti, May 02 '19 at 07:14
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help. — Manvendra Somvanshi, May 02 '19 at 07:16
Don't we make a transformation of the fields themselves, i.e., $\phi(x)\to\phi(x)+\delta\phi(x)$ where the variation $\delta\phi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $\delta\phi(x)\neq\frac{\partial\phi(x)}{\partial x^\mu}dx^\mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken. — , May 02 '19 at 07:17
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action — amilton moreira, May 02 '19 at 07:21
@amiltonmoreira See, Eq. $2.42$ of Greiner. There are two terms. Only the second term $\frac{\partial \phi(x)}{\partial x^\mu}\delta x^\mu$ arises out of the change of variables. The first term $\delta\phi(x)$ is an actual variation in the field in addition to the variation induced in it due to the change of variables. If this first term were absent, only then the variation in the action could have been expected to trivially vanish--as the only transformation would have been a change of variables in the variable of integration--again, that too, only if the measure were also invariant. — , May 02 '19 at 07:36
@ Dvij Mankad my problem is equation 2.45. Transformation 2.38 induces transformation 2.39, wich give us eq 2.45..... — amilton moreira, May 02 '19 at 07:43
For me eq 2.45 should be written as $\delta S=\int \mathscr{L}'(\phi^\prime\partial_\mu\phi^\prime)d^4x-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$ — amilton moreira, May 02 '19 at 07:54
The physical volume is $\sqrt{|\det g|} dx^1 \cdots dx^4$ where $g$ is the matrix of the metric. If you are using Minkowskian coordinates $\sqrt{|\det g|}=1$, but when you transform the coordinates in general $\sqrt{|\det g'|}\neq 1$. The true invariance condition (under change of coordinates) is $\int L dx^1dx^2dx^3dx^4 = \int L' \sqrt{|\det g'|} dx'^1dx'^2dx'^3dx'^4$. where the former coordinates are Minkowskian and the latter generic. For Jacobian factor I meant $\sqrt{|\det g|}$. — Valter Moretti, May 02 '19 at 08:09
So the condition you wrote is automatically true if $|det g'|=1$ and this is in fact true if the transformation $x\to x'$ stays in Poincaré group. — Valter Moretti, May 02 '19 at 08:14
@Valter Moretti you right but even in arbitrary coordinates we should always have $\delta S=\int \mathscr{L}'(\phi^\prime,\partial_\mu\phi^\prime)d^4x^\prime-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$ — amilton moreira, May 02 '19 at 08:21
No, it is wrong, a jacobian factor would appear. Noether condition just concerns that factor. — Valter Moretti, May 02 '19 at 08:30

Qmechanic · Answer 1 · 2019-05-02T09:07:50.690

In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$ $$ \delta S~:=~S_{V^{\prime}}[\phi^{\prime}]- S_{V}[\phi]\tag{A}$$ describes an active (as opposed to passive) infinitesimal transformation $$\phi(x)\to\phi^{\prime}(x^{\prime}), \tag{B}$$ $$ x\to x^{\prime}, \tag{C}$$ of the action functional $$ S_{V}[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}(\phi(x),\partial\phi(x),x).\tag{D}$$ The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $\phi$ variation does vanish cf. my Phys.SE answer here.

--

$^1$ The integration region $V\to V^{\prime}$ is conventionally moved according to the horizontal variation (C).

knzhou · Accepted Answer · 2019-05-02T15:05:30.717

2

This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.

For example, consider translational symmetry. Sometimes this is written as $$x \to x' = x + a, \quad \phi \to \phi'(x') = \phi(x).$$ However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $\phi$ with a $\phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have $$\int_{-\infty}^\infty x^2 f(x) \, dx = \int_{-\infty}^\infty (x+a)^2 f(x+a) \, dx$$ even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant.

What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $\phi$, and end up with the field $\phi'$. Indeed in general, we have $$\int_{-\infty}^\infty x^2 f(x) \, dx \neq \int_{-\infty}^\infty x^2 f(x+a) \, dx.$$ You can derive the expression for $\phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.

edited May 02 '19 at 15:05

answered May 02 '19 at 08:40

knzhou

101,976

you are right. But my question is why the text books insist in this presentation? – amilton moreira May 02 '19 at 08:53
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves. – knzhou May 02 '19 at 08:55
could you give me a reference where this proof is made right? – amilton moreira May 02 '19 at 08:56
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good. – knzhou May 02 '19 at 08:59
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $\phi(x)\to\phi(x)+\delta\phi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing. – May 02 '19 at 09:07
@DvijMankad I wasn't aware! I do know that Srednicki is less sloppy than a lot of other QFT texts but I haven't gone through it in detail. – knzhou May 02 '19 at 10:18

score 0 · Answer 3 · answered May 02 '19 at 08:26

Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $\delta \cal L=0$ for a coordinate transformation is a choice.

score 0 · Answer 4 · answered May 02 '19 at 14:05

To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $x\mapsto x²$ is given by ...

I have here only one book where to search for Noether's theorem. Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94)) $$\mathscr L(x+a) \equiv \mathscr L(\phi_i(x+a), \partial_\mu\phi_i(x+a))$$ where $a$ may also depend on $x$. Variation of fields is written $$\delta\phi_i = \delta a^\mu(x)\,\partial_\mu\phi_i(x)$$ etc. No variation of coordinates in the action integral.

A final note may help you to clarify the matter. Variational formulation in relativistic field theory is but an extension of the one in ordinary mechanics. The analogy runs as follows:

$$\matrix{ \hfill \rm time & \rightarrow & \rm spacetime\ coordinates \hfill\cr \hfill \rm lagrangian\ coordinates & \rightarrow & \rm fields. \hfill \cr}$$ So the relevant variation is the one of fields, not of coordinates.

Confusion in Proof of Noether's theorem

4 Answers4