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Let's say we have a field $\phi(x)$ that gets transformed to $\phi(x, \epsilon)$ under some finite transformation. We also define $\phi(x,0)=\phi(x)$. If we Taylor expand our transformation we get: $$\phi(x,\epsilon) = \phi(x) + \frac{\partial \phi(x,\epsilon)}{\partial \epsilon}\Bigr|_{\epsilon = 0}\epsilon+\frac{1}{2}\frac{\partial^2 \phi(x,\epsilon)}{\partial \epsilon^2}\Bigr|_{\epsilon = 0}\epsilon^2+\cdots$$ As this is a symmetry that conserves the action we want $S(\epsilon)=S$, so we can also expand this out using a Taylor series $$S(\epsilon)=S(0)+\frac{d S(\epsilon)}{d \epsilon}\Bigr|_{\epsilon = 0}\epsilon+\frac{1}{2}\frac{d^2 S(\epsilon)}{d\epsilon^2}\Bigr|_{\epsilon = 0}\epsilon^2+\cdots.$$ The definition of the coefficients is the different orders of variation of the function so $$S(\epsilon)=S+\delta S\epsilon+\frac{1}{2}\delta^2 S \epsilon^2+\cdots$$ So for this to be true we can see that each coefficient of $\mathcal{O}(\epsilon)$ must be $0$. So the first-order expression gives the 'normal' Noether's theorem expression. Will the rest give new currents?

Qmechanic
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Joshua Pasa
  • 735

3 Answers3

5

  1. Assuming that we are talking about a single finite 1-parameter global quasisymmetry, it is a flow, which is in 1-to-1 correspondence with a vector field, or equivalently, an infinitesimal 1-parameter global quasisymmetry.

  2. In other words, it counts as 1 and the same quasisymmetry, so Noether's first theorem only yields 1 continuity equation, which in turn leads to just 1 conserved quantity.

Related:

  1. Noether's Theorem: Lie groups vs. Lie algebras; finite vs. infinitesimal symmetries

  2. Why are infinitesimal shifts sufficient to prove that a symmetry holds?

Qmechanic
  • 201,751
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Technically "symmetries" are in general not so much about $S$ not changing at all as $\delta S$ being $0$, or in the language of functional derivatives $\frac{\delta S}{\delta\phi}=0$.

Consider, for example, $\delta\phi=i\epsilon\phi$ of $\mathcal{L}=\frac12\partial^\mu\phi^\ast\partial_\mu\phi-\frac12m^2\phi^\ast\phi$ for a complex scalar $\phi$ and infinitesimal real $\epsilon$. Replacing $\phi$ with $\phi+\delta\phi$, $\mathcal{L}$ is muliplied by $1+\epsilon^2$, as is $S=\int\mathcal{L}d^4x$. But $S$ is still an action with the above symmetry, because we don't care about $o(\epsilon)$ terms. If we used your logic, we'd conclude a spurious "symmetry" $S=0$.

(The example above is only a global symmetry; for a local one, feel free to replace $\partial$ with a gauge covariant derivative as an exercise.)

J.G.
  • 24,837
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The conserved current (under a transformation) is defined to be the variation in the Lagrangian that keeps the equations of motion invariant, under the latter transformation. It's infinitesimal form is derived by following the procedure you describe (roughly at least), to first order. Since you are still studying how the Lagrangian (or action if you prefer) varies with respect to the same transformation, there is no reason to think about other currents. You can, if you like think the whole "going up to second (or third and so on) order in the Taylor expansion" as a procedure that helps you derive the infinitesimal form of the conserved current more accurately (I think)...

So, having said that, there exists only one (maximum one) conserved current for every transformation. However, there might exist infinite dimensional symmetries if the transformation depends on a multivalued function (for instance see Large Gauge Symmetries in QED in which there exists a conserved current for every value of an angle dependent function!)

Hopefully I am not missinforming you in any way. If there are any questions, please comment...

schris38
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