Question about noether's theorem argument

[painfive]

Given a lagrangian $L[\phi]$, where $\phi$ is a generic label for all the fields of the system, a transformation $\phi(x) \rightarrow \phi(x) + \epsilon \delta \phi(x)$ that leaves the lagrangian invariant corresponds to a conserved current by the following argument.

If we were to send $\phi(x) \rightarrow \phi(x) + \epsilon(x) \delta \phi(x)$, this would not in general be a symmetry, but would be in the special case that $\epsilon$ is constant. Therefore*, the change in the lagrangian must be proportional to the derivative of $\epsilon(x)$, that is:

$$\delta L = j^\mu (x) \partial_\mu \epsilon(x)$$

Now when the equations of motion are satisfied, all infinitessimal variations, symmetries or not, leave the action unchanges, so in this case we must have:

$0 = \delta S = \int d^4x j^\mu (x) \partial_\mu \epsilon(x)$

or, integrating by parts:


$0 = \int d^4x \epsilon(x) \partial_\mu j^\mu (x)$

Since $\epsilon(x)$ is arbitrary, this implies $\partial_\mu j^\mu(x) = 0$.

My problem is with the part marked by a *. Just because something vanishes when $\epsilon$ is constant, why should we expect the thing to be proportional to the dderivative of $\epsilon(x)$? I could imagine other dependences. For example, the following things all vanish when $\epsilon(x)$ is constant:

$$(j^\mu \partial_\mu \epsilon(x) )^2$$

$$\partial^2 \epsilon(x)$$

$$\epsilon(x+1) - \epsilon(x)$$

I could go on. Granted, you could eliminate these examples individually, eg, the variation should be linear and local in $\epsilon(x)$, and by integrating by parts we can turn the middle one into the desired form. But there are other examples, and I'm wondering how we can argue for the form $\partial_\mu \epsilon$ directly rather than eliminating these other possibilities one by one.

 

[Ben Niehoff]

Try expanding L as a Taylor series in $(\epsilon \delta \phi)$. Then because $\delta \phi$ is by definition infinitesimal, the higher-order terms vanish.

Now, because you are looking at the change in L under infinitesimal variations, the zeroth-order term also vanishes. What you should be left with is the first-order term.

 

[painfive]

So the taylor expansion would have to be a sum over all terms of the form:

$$\Pi_{i, m_i, n_i} ({\partial_i}^{m_i} \epsilon(x) )^{n_i}$$

where the product is over some subset of the values i, m_i, and n_i could take, and the sum would be over all such subsets. This is a pretty extreme generalization of taylors theorem. If it is true, then I guess we could argue we can ignore terms except those of the form:

$${\partial_i}^{m_i} \epsilon (x)$$

and then integrate by parts. This might be what they mean, but if so, it isn't very satisfying or intuitive, and they really skipped over a lot of details.

 

[StatusX]

I'm actually the OP, I illegally had two user names, painfive is now gone. But I'd still like someone to tell me if this taylor's theorem argument is the way to go.

 

[samalkhaiat]

My problem is with the part marked by a *. Just because something vanishes when $\epsilon$ is constant, why should we expect the thing to be proportional to the dderivative of $\epsilon(x)$? I could imagine other dependences. For example, the following things all vanish when $\epsilon(x)$ is constant:

$$(j^\mu \partial_\mu \epsilon(x) )^2$$

$$\partial^2 \epsilon(x)$$

$$\epsilon(x+1) - \epsilon(x)$$

I could go on. Granted, you could eliminate these examples individually, eg, the variation should be linear and local in $\epsilon(x)$, and by integrating by parts we can turn the middle one into the desired form. But there are other examples, and I'm wondering how we can argue for the form $\partial_\mu \epsilon$ directly rather than eliminating these other possibilities one by one.

I hate to see the beautiful theory made ugly by some textbooks! And I also hate the method that I will describe below!
Let us consider the infinitesimal transformation

$$\phi \rightarrow \phi + \delta \phi ,$$

$$\delta \phi = F( \phi ) \epsilon (x) , \ \ | \epsilon | \ll 1$$

Infinitesimal means that

$$\epsilon^{n} \approx 0, \ \forall n > 1$$

This transformation induces an infinitesimal change in the Lagrangian according to

$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi} F( \phi ) \epsilon + \frac{\partial \mathcal{L}}{\partial \partial_{a} \phi} \epsilon \partial_{a} F + \frac{\partial \mathcal{L}}{\partial \partial_{a} \phi} F \partial_{a} \epsilon$$

Now, we define the objects

$$J^{a} \equiv \frac{\partial \mathcal{L}}{\partial \partial_{a} \phi} \ F( \phi )$$

and

$$E( \phi ) = \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{a} \left( \frac{\partial \mathcal{L}}{\partial \partial_{a} \phi } \right)$$

and rewrite the change in the Lagrangian as

$$\delta \mathcal{L} = \left( E( \phi ) F + \partial_{a} J^{a} \right) \epsilon + J^{a} \partial_{a} \epsilon$$

Notice that the first term represents the infinitesimal change in $\mathcal{L}$ for constant $\epsilon$. So, we can write

$$\delta \mathcal{L} = \delta \mathcal{L}|_{\epsilon = \mbox{const.}} + J^{a} \partial_{a}\epsilon$$

Now, if for CONSTANT $\epsilon$, the transformation above is a symmetry, i.e., if the (Noether) identity

$$\delta \mathcal{L}|_{\epsilon = \mbox{conct.}} = E( \phi ) F( \phi ) + \partial_{a}J^{a} = 0$$

is satisfied, then the change in the lagrangian will be given by

$$\delta \mathcal{L} = J^{a}\partial_{a}\epsilon$$

where $J^{a}$ now is the conserved current of the corresponding GLOBAL symmetry; notice that the Noether identity above implies $\partial_{a}J^{a} = 0$ ON-SHELL, i.e., when the field satisfies the equation of motion $E( \phi ) = 0$.

Regards

sam