Noether states that any continuous symmetry corresponds to a conserved quantity (Noether’s current).
Noether’s argument is very easily confused with those leading up to the Classical equation of motion (EOM) (least action/variation principle). First let’s see a simple example where EOM looks like Noether’s current.
Example Theory
Consider the action $$ S[a]=\int_M d^4x \frac{1}{2}\partial_\mu a(x) \eta^{\mu\nu} \partial_\nu a(x),$$ which is clearly invariant to the transformation \(a(x)\rightarrow a(x)+\epsilon\) where \(\epsilon\) is a scalar.
Equation of Motion
The EOM for the \(a(x)\) field can be derived by making an arbitrary variation
$$ \delta S=\int_M d^4x \delta \partial_\mu a(x) \eta^{\mu\nu} \partial_\nu a(x).$$
Assuming \(a(x)\) is continuous on \(M\), we can exchange \(\delta\) and \(\partial_\mu\),
$$ \delta S=\int_M d^4x \partial_\mu \delta a(x) \eta^{\mu\nu} \partial_\nu a(x).$$
Now use product rule in reverse (integration by parts), $$ \delta S=\int_M d^4x \partial_\mu (\delta a(x) \eta^{\mu\nu} \partial_\nu a(x))-\delta a(x) \eta^{\mu\nu} \partial_\mu\partial_\nu a(x), $$ and use Stoke’s theorem to bring the total derivative to the boundary
$$ \delta S=\int_{\partial M} d^3x n_\mu\delta a(x) \eta^{\mu\nu} \partial_\nu a(x)-\int_M d^4x\delta a(x) \eta^{\mu\nu} \partial_\mu\partial_\nu a(x). $$
Assuming the field \(a(x)\) vanishes on the boundary of space-time \(\partial M\), we may drop the boundary term.
$$ \delta S=-\int_M d^4x\delta a(x) \eta^{\mu\nu} \partial_\mu\partial_\nu a(x). $$
If we believe in the least action principle, then \(a(x)\) better follow the classical path on which the variation of the action vanishs. Since \(\delta a(x)\) is arbitrary, we require
$$-\partial_\mu\partial^\mu a(x) = 0, $$
at all points \(x\) on space-time \(M\).
Noether’s Current
Suppose we are concerned with a local contribution to the action, which is integrated over a section of space-time \(R\) instead of the entire manifold \(M\). I will call it Noether’s S or
$$ nS[a]= \int_R d^4x \frac{1}{2}\partial_\mu a(x) \eta^{\mu\nu} \partial_\nu a(x). $$
This time the boundary term in an arbitrary variation will not vanish
$$ \delta nS=\int_{\partial R} d^3x n_\mu\delta a(x) \eta^{\mu\nu} \partial_\nu a(x)-\int_R d^4x\delta a(x) \partial_\mu\partial^\mu a(x). $$
Noether’s genius is in recognizing that by replacing the arbitrary variation \(\delta a(x)\) with a specific one where the change on the action is eye-ballable, we may learn something about our system. In this example, the variation \(\delta a(x)=\epsilon\) leaves the action invariant. Therefore
$$ 0=\int_{\partial R} d^3x n_\mu\epsilon \eta^{\mu\nu} \partial_\nu a(x)-\int_R d^4x\epsilon \partial_\mu\partial^\mu a(x). $$
Since we already agreed that the field \(a(x)\) should follow the classical EOM, the bulk term vanishes and
$$\epsilon \int_{\partial R} d^3x n_\mu \eta^{\mu\nu} \partial_\nu a(x)=0, $$
which tells us that for any region \(R\) that we pick, the total “flow” of the “current” \(\eta^{\mu\nu} \partial_\nu a(x)\) on the boundary \(\partial R\) is zero. The above boundary form of the conservation law can be written in bulk form by running Stoke’s theorem in reverse
$$\epsilon \int_{R} d^4x \partial_\mu \eta^{\mu\nu} \partial_\nu a(x)=0. $$
In addition, since the region \(R\) is arbitrary, it must be that
$$\partial_\mu \partial^\mu a(x)=0, $$
at all points in space-time.
Subtleties
Notice the bulk form of Noether’s current conservation law derived above looks exactly like the EOM. Also note how I did not drop the minus sign on the EOM, it is important to trace back the sign different. Hint: Neother’s current came from the boundary term whereas EOM came from the bulk. To further illuminate the subtleties in deriving the EOM vs. Noether’s current, I will enumerate the key differences in the two arguments.
- Variation:
For EOM, an arbitrary variation of the field is made.
For Noether’s argument, a specific variation is defined by the symmetry. - Integration region:
For EOM, integration is performed over the entire space-time manifold.
For Neother, integration is performed over an arbitrary section in space-time.
In deriving the EOM, we use integration by parts to turn any derivative of the arbitrary variation \(\partial_\mu \delta a(x)\) to expose the arbitrary variation \(\delta a(x)\) such that any term multiplying it has to vanish. In Noether’s argument, we make terms vanish by restricting the variation and using the EOM.