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I'm fine with $U(1)$ symmetry and Noether's Theorem, but struggling with the translations of the field; namely

$$\phi'(x^{\mu})=\phi(x^{\mu}-a^{\mu}),$$ where $a^{\mu}$ constant four-vector

$$x^{\mu}=x^{\mu}+a^{\mu},$$

and the Lagrangian density

$${\cal L}=\frac{1}{2}\partial_{\mu}\phi^*\partial^{\mu}\phi-V(\phi^*\phi).$$

So a few questions:

  1. I can't show the Lagrangian is invariant under this transformation. Is it just a case that as $a^{\mu}$ is constant then the first term in the Lagrangian will obviously stay the same? But what about $V$? How I can show that's invariant?

  2. Infinitesimally, is the transformation $\phi'(x^{\mu})=\phi(x^{\mu})-a^{\mu}\partial_{\mu}\phi(x^{\mu})?$

  3. If I'm right in point 2., how can I apply Noether's Theorem to this?

JamalS
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Phibert
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  • For the corresponding question in 0+1 dimensions (aka. point mechanics), see e.g. http://physics.stackexchange.com/q/94381/2451 – Qmechanic May 17 '14 at 15:51

1 Answers1

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A translation by $x^\nu \to x^\nu - \epsilon^\nu$ corresponds to an infinitesimal transformation of the fields, by

$$\phi \to \phi + \epsilon^\nu \partial_\nu \phi$$

as we are performing an active rather than passive transformation. The Lagrangian transforms as,

$$\mathcal{L}\to \mathcal{L}+\epsilon^\nu \partial_\nu \mathcal{L}$$

by substituting $\phi$ into the Lagrangian. Notice the change is up to a total derivative, and hence Noether's theorem is applicable to the symmetry. The conserved current density is given by,

$$j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}X(\phi)-F^\mu(\phi)$$

where $X=\delta\phi$ and $F^\mu$ is such that $\partial_\mu F^\mu=\delta \mathcal{L}$ infinitesimally. For our case, we obtain the symmetric stress-energy tensor (analogous to that of general relativity),

$$T^\mu_\nu=\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_\nu \mathcal{L}$$

where the Kronecker delta is raised with the Minkowski metric. The current satisfies, $\partial_\mu T^{\mu}_\nu = 0$, and the corresponding Noether charge,

$$E=\int \mathrm{d}^3 x \, T^{00}$$

is the total energy of the system, whereas,

$$P^i = \int \mathrm{d}^3 x \, T^{0i}$$

is the $i$th component of the total momentum of the field, where $i=(x,y,z)$ only. A caveat: the stress-energy tensor derived by Noether's theorem is not always symmetric, and may require the addition of a term which satisfies the continuity equation, and ensures symmetry in the indices.

Alternate Method

Recall to obtain the Einstein field equations in general relativity, we may vary the Einstein-Hilbert action,

$$S\sim \int \mathrm{d}^4 x \, \sqrt{-g} \, \left( R + \mathcal{L}\right)$$

Similarly, in quantum field theory, we may promote our Minkowski metric to a generic metric tensor, thereby replacing the kinetic term of the Lagrangian with covariant derivatives. Up to some constants, the stress-energy tensor is given by

$$T^{\mu\nu} \sim \frac{1}{\sqrt{-g}} \frac{\partial (\sqrt{-g}\mathcal{L})}{\partial g^{\mu\nu}}$$

evaluated at $g_{\mu\nu}=\eta_{\mu\nu}$, which is precisely the definition we implement when obtaining the Einstein field equations for general relativity.

JamalS
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  • I'm not sure why this gap in my physics education exists, but I don't really understand what you mean by passive and active transformation. As in why have you changed signs in the transformation of $x$ and the transformation of $\phi$? So the Lagrangian $\textit{isn't}$ invariant under this translation? And again, when trying to show the transformation of the Lagrangian which you've stated, what I am doing about the $V$ function which we're not given explicitly, just told it's a function of $\phi^*\phi$? – Phibert May 17 '14 at 16:03
  • @user13223423: No, the action is definitely invariant under translations. To illustrate the difference between passive and active, consider rotation transformations. If we take our field $\phi$, and literally rotate it, then that is an active transformation. However, if instead we relabel our coordinates, e.g. rotate our coordinate axis, then that's a passive transformation. Back to our case: the field in terms of $x'^\mu=x^\mu-\epsilon^\mu$ means that for it to treat it as the same, untranslated point, we need to add $\epsilon^\mu$, so $\phi \to \phi + \epsilon^\mu \partial_\mu \phi$. – JamalS May 17 '14 at 16:43
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    A passive transformation is where you left the field alone and just move a foot to the left. An active transformation is where you stay fixed and shift the field a foot to the right. – mmesser314 May 17 '14 at 16:45