I am confused, in the proof of Noether's theorem, by the change of boundary in the action integral during the transformation of coordinates. I have seen on Wikipedia that along with the change of Field, they also change $\Omega$ to $\Omega'$, where $\Omega$ is the space-time boundary of the action integral.

If we change the fields and boundaries both due to coordinate transformations then wouldn't that constitute a zero change? (I am keeping the intrinsic changes in the field apart)

Don't we consider a fixed region (arbitrary but unchanging during the flow) of space-time and then see the changes on Lagrangian due to only the flow of fields and some intrinsic change of fields, before and after the flow? (as shown below) The coordinates should be treated as dummy variables. $$\int_\Omega \delta L\ d^4x$$ I don't think we should move our boundary with the flow, am I right? Moreover, in the proof shown by joshphysics he didn't consider action at all. He worked only with the variation of Lagrangian and so there was no integral and hence, no boundary.

So, why do some proofs change the boundary and some do not? I mean how are these equivalent?

Another question: If we prove Noether's Theorem as joshphysics did, using only Lagrangian and not action, do we miss some conservations compared to the proof done in Wikipedia using the action integral?

  • $\begingroup$ When you changed coordinate system, your boundary as geometric object are the same, but its description is different. Think of ordinary integral and change of the variables. Does integration boundaries change in this case? Of course it changes ( for example from X,You cartesian to r, phi - cylindrical, so you have change boundary from x(1,10), y(1,10) to something different for r,phi - because phi for example can be only in (0,2pi). ) $\endgroup$
    – kakaz
    Aug 25, 2018 at 10:37
  • $\begingroup$ @kakaz Oh I understood your point, but then why are we changing coordinates instead of changing the field only? Because if we change the coordinates, action will not change as we are changing only the dummy variables. I think that we should keep the boundary the same and only L should be varied using Lie derivatives, instead of varying the coordinates. Please correct me if I am wrong somewhere. $\endgroup$ Aug 25, 2018 at 11:17

2 Answers 2

  1. Note that the transformations$^1$ in Noether's (first) theorem are generally a combination of (vertical) transformations of target space fields $\phi^{\alpha}$ and (horizontal) transformations of spacetime coordinates $x^{\mu}$.

  2. Recall that the action $S_{\Omega}[\phi] =\int_{\Omega} \! d^nx~{\cal L}$ is the Lagrangian density ${\cal L}$ integrated over a spacetime integration region $\Omega$. The more general formulation of Noether's theorem is in term of a (quasi)symmetry of the action rather than the Lagrangian density.

  3. The transformation of the spacetime integration region $\Omega$ is induced by the horizontal transformation $\delta x^{\mu}$.

  4. The Phys.SE post that joshphysics answered does not consider horizontal transformations, and hence has no transformation of the integration region $\Omega$.


$^1$ Noether's theorem can be formulated for finite transformations, but let us only consider infinitesimal transformations in this answer for simplicity.

  • $\begingroup$ In point no. 3 you say that integration region changes because of horizontal transformation. So, we let the boundary flow with the coordinate transformations right? Now, if we also let the fields flow with the coordinate transformation, wouldn't that constitute a zero change. Because then we are changing both the boundary and the fields because of the coordinate transformations. Shouldn't we move only one of them? $\endgroup$ Feb 26, 2019 at 16:46
  • 1
    $\begingroup$ A situation with zero change is a special case, which is not always fulfilled. $\endgroup$
    – Qmechanic
    Feb 27, 2019 at 8:36

So, why do some proofs change the boundary and some do not? I mean how are these equivalent?

They aren't: Common proofs of 'Noether's theorem' often only consider certain limits of it. There's also a tendency to hide intricacies behind notation.

In the following, I'll present an elementary proof of a simple version of Noether's first theorem in 1 dimension in a way that should generalize to what's called the the field-theoretic version on Wikipedia by going from $t$ to $x^\mu$ and $q$ to $\varphi^A$.

The Lagrangian will be a function $$ L = L(x,v,t) $$ and the action a functional $$ S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt $$

Proposition. If the transformation $$ t\to t'(t) = t + \epsilon T(t) $$ $$ x\to x'(x,t) = x + \epsilon X(t)$$ $$ q'(t') = q(t(t')) + \epsilon X(t(t')) $$ is a quasi-symmetry of the action $$ \delta S \approx \Delta K $$ on-shell (ie assuming the equations of motion), then there is a conserved quantity $$ \frac{d}{dt} \left( \frac{\partial L}{\partial v} (X - \dot q T) + LT - K \right) \approx 0 $$ Here, $$ \delta S = \frac{d}{d\epsilon}\Big|_{\epsilon=0} S[q'] $$ $$ \Delta K = K(t_2)-K(t_1) = \int_{t_1}^{t_2} \frac{dK}{dt} dt $$

Proof. \begin{align} \delta S &= \frac{d}{d\epsilon}\Big|_{\epsilon=0} \int_{t'(t_1)}^{t'(t_2)} L(q'(t'),\frac{d}{dt'}q'(t'),t')\,dt' \\&= \frac{d}{d\epsilon}\Big|_{\epsilon=0} \int_{t_1}^{t_2} L(q(t) + \epsilon X(t),\left( \frac{dt'}{dt} \right)^{-1}\frac{d}{dt}(q(t) + \epsilon X(t)),t + \epsilon T(t)) \,dt'(t) \\&= \int_{t_1}^{t_2}\left[ \left( \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v}\frac{d}{d\epsilon}\Big|_{\epsilon=0}\frac{\dot q + \epsilon \dot X}{1 + \epsilon \dot T} + \frac{\partial L}{\partial t} T \right)\,dt + L \frac{d}{d\epsilon}\Big|_{\epsilon=0}d(t + \epsilon T) \right] \\&= \int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v} (\dot X - \dot q \dot T) + \frac{\partial L}{\partial t} T + L \dot T\right]\,dt \end{align}

Using $$ \frac{\partial L}{\partial v} \dot X = \frac{d}{dt}\left( \frac{\partial L}{\partial v} X \right) - \left( \frac{d}{dt} \frac{\partial L}{\partial v} \right) X $$ $$ \frac{\partial L}{\partial t} T + L \dot T = \frac{d}{dt}\left( LT \right) - \frac{\partial L}{\partial x} \dot q T - \frac{d}{dt}\left( \frac{\partial L}{\partial v} \dot q T \right) + \left( \frac{d}{dt}\frac{\partial L}{\partial v} \right) \dot q T + \frac{\partial L}{\partial v}\dot q \dot T $$ we arrive at $$ \delta S = \int_{t_1}^{t_2}\left[ \left( \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial v} \right)(X - \dot q T) + \frac{d}{dt} \left( \frac{\partial L}{\partial v}(X - \dot q T) + LT \right) \right]\,dt $$

The first term vanishes if we asume the Euler-Lagrange equations, the second term yields our conservation law once we move $K$ to this side of the equation. This concludes the proof. $\square$

Note the change of region of intergration in the first step. The new time coordinate was by no means a dummy variable - the transformation is 'active', a one-parameter group of diffeomorphisms.


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