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Conserved Noether charges for Lorentz symmetry of the action

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider the infinitesimal form of the Lorentz tranformation: ##x^{\mu} \rightarrow x^{\mu}+{\omega^{\mu}}_{\nu}x^{\nu}##.

    Show that a scalar field transforms as ##\phi(x) \rightarrow \phi'(x) = \phi(x)-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\phi(x)## and hence show that the variation of the Lagrangian density is a total derivative ##\delta \mathcal{L}=-\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L})##.

    Using Noether's theorem deduce the existence of the conserved current ##j^{\mu}=-{\omega^{\rho}}_{\nu}[{T^{\mu}}_{\rho}x^{\nu}]##.

    The three conserved charges arising from spatial rotational invariance define the total angular momentum of the field. Show that these charges are given by ##Q_{i}=\epsilon_{ijk}\int d^{3}x (x^{j}T^{0k}-x^{k}T^{0j})##.

    Derive the conserved charges arising from invariance under Lorentz boosts. Show that they imply ##\frac{d}{dt}\int d^{3}x (x^{i}T^{00})= \text{constant}## and interpret this equation.

    2. Relevant equations

    3. The attempt at a solution

    Under the infinitesimal form of the Lorentz tranformation given by

    ##x^{\mu} \rightarrow x^{\mu}+{\omega^{\mu}}_{\nu}x^{\nu}={\delta^{\mu}}_{\nu}x^{\nu}+{\omega^{\mu}}_{\nu}x^{\nu}=({\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu})x^{\nu}={(\Lambda)^{\mu}}_{\nu}x^{\nu}##, so that ##{(\Lambda^{-1})^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}-{\omega^{\mu}}_{\nu},##

    a scalar field transforms as

    ##\phi(x) \rightarrow \phi'(x) = \phi(\Lambda^{-1}x) = \phi({(\Lambda^{-1})^{\mu}}_{\nu}x^{\nu}) = \phi(({\delta^{\mu}}_{\nu}-{\omega^{\mu}}_{\nu})x^{\nu}) = \phi(x^{\mu}-{\omega^{\mu}}_{\nu}x^{\nu}) = \phi(x)-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\phi(x),##

    where in the last step, I used the following Taylor expansion of ##f(x+\delta x)## (where ##f## is a scalar and ##x## is a ##4##-vector):

    ##f(x+\delta x) = f(x) + x^{\nu}\frac{\partial f}{\partial x^{\nu}}+\dots##

    and we ignore all the higher order terms since the transformation ##{\omega^{\mu}}_{\nu}## is infinitesimal.

    Similarly,
    ##\mathcal{L}(x) \rightarrow \mathcal{L}'(x) = \mathcal{L}(\Lambda^{-1}x) = \mathcal{L}({(\Lambda^{-1})^{\mu}}_{\nu}x^{\nu}) = \mathcal{L}(({\delta^{\mu}}_{\nu}-{\omega^{\mu}}_{\nu})x^{\nu}) = \mathcal{L}(x^{\mu}-{\omega^{\mu}}_{\nu}x^{\nu}) = \mathcal{L}(x)-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\mathcal{L}(x)##

    so that

    ##\delta \mathcal{L} = -{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\mathcal{L}(x)##
    ##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))+{\omega^{\mu}}_{\nu}\mathcal{L}(x)\partial_{\mu}x^{\nu}##
    ##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))+{\omega^{\mu}}_{\nu}\mathcal{L}(x)\delta^{\nu}_{\mu}##
    ##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))+{\omega^{\mu}}_{\mu}\mathcal{L}(x)##
    ##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))##,

    where the last line follows from the antisymmetry ##{\omega^{\mu}}_{\nu}=-{\omega^{\nu}}_{\mu}## of ##{\omega^{\mu}}_{\nu}## so that the diagonal elements, and by extension the sum of the diagonal elements ##{\omega^{\mu}}_{\mu}##, is ##0##.

    Am I correct so far?
     
  2. jcsd
  3. Apr 24, 2016 #2

    strangerep

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    Looks ok to me. (A mathematican would barf on seeing ##\phi(x) = \phi(x^\mu)##, but physicists will know what you mean.)

    (BTW, you should probably include the general formula for the energy-momentum tensor ##T## in the "relevant equations" section, since you'll need it for the next part.)
     
  4. Apr 25, 2016 #3
    Thanks!

    The general formula for the energy-momentum tensor ##{T^{\mu}}_{\nu}## is

    ##{T^{\mu}}_{\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial_{\nu}\phi(x)-\delta^{\mu}_{\nu}\mathcal{L}##

    (I am not able to include the general formula for the energy-momentum tensor ##{T^{\mu}}_{\nu}## in the "relevant equations" section, since I don't have the editing privilege for my first post anymore.)


    Next, I need to use Noether's theorem to deduce the existence of the conserved current ##j^{\mu}=-{\omega^{\rho}}_{\nu}[{T^{\mu}}_{\rho}x^{\nu}]##:

    ##\delta\phi=-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\phi(x)## and ##\delta\mathcal{L}=-\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L})##

    so that

    ##\delta\mathcal{L}=\partial_{\mu}\Big(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi\Big)##
    ##\implies -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L})=-\partial_{\mu}\Big(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}{\omega^{\rho}}_{\nu}x^{\nu}\partial_{\rho}\phi(x)\Big)##
    ##\implies \partial_{\mu}\Big(-\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}{\omega^{\rho}}_{\nu}x^{\nu}\partial_{\rho}\phi(x)+{\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}\Big)=0##
    ##\implies \partial_{\mu}j^{\mu}=0##

    where

    ##j^{\mu}=-\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}{\omega^{\rho}}_{\nu}x^{\nu}\partial_{\rho}\phi(x)+{\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}##
    ##\implies j^{\mu}=-\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}{\omega^{\rho}}_{\nu}x^{\nu}\partial_{\rho}\phi(x)+\delta^{\mu}_{\rho}{\omega^{\rho}}_{\nu}x^{\nu}\mathcal{L}##
    ##\implies j^{\mu}=-{\omega^{\rho}}_{\nu}x^{\nu}\Big(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial_{\rho}\phi(x)-\delta^{\mu}_{\rho}\mathcal{L}\Big)##
    ##\implies j^{\mu}=-{\omega^{\rho}}_{\nu}x^{\nu}{T^{\mu}}_{\rho}##
    ##\implies j^{\mu}=-{\omega^{\rho}}_{\nu}[{T^{\mu}}_{\rho}x^{\nu}]##

    What do you think?
     
  5. Apr 26, 2016 #4

    strangerep

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    It's not clear how you got the last line. Which textbook (or other source) are you using?
     
  6. Apr 26, 2016 #5
    I am using David Tong's lecture notes on QFT. In particular, I'm referring to equation (1.37) on page 14.

    I missed the following couple of lines:

    ##\delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial_{\mu}(\delta\phi)##

    ##\implies \delta\mathcal{L}=\Big[\frac{\delta\mathcal{L}}{\delta\phi}-\partial_{\mu}\Big(\frac{\delta\mathcal{L}}{\partial (\partial_{\mu}\phi)}\Big)\Big]\delta\phi+\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi)}\delta\phi\Big)##

    ##\implies \delta\mathcal{L}=\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi)}\delta\phi\Big)##, using the Euler-Lagrange field equations.
     
  7. Apr 26, 2016 #6

    strangerep

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    Oh, OK. It's a long time since I looked at Tong's notes. I should probably refresh my memory.

    I learned this stuff from Greiner & Reinhardt "Field Quantization". But their formulas seem consistent with what you've done, since you're only working with a scalar field.

    Edit: Dammit. Tong's proof of Noether's Thm takes less than a page, whereas G&R take more than 3 pages since they do it far more carefully. Tong just considers symmetries of the Lagrangian but G&R consider invariance of the action integral, which is more involved (and also leads to a slightly larger class of symmetries). Now I'm gonna have to totally refresh my memories of both. :oops:

    But not today.
     
    Last edited: Apr 26, 2016
  8. Apr 28, 2016 #7
    Thanks again, strangerep, for your insightful comments!

    Let me try and solve the remaining parts of the problem.

    The three conserved charges arising from spatial rotational invariance define the total angular momentum of the field. Show that these charges are given by ##Q_{i}=\epsilon_{ijk}\int d^{3}x (x^{j}T^{0k}-x^{k}T^{0j})##.

    Let me start with the conserved Noether current

    ##j^{\mu}=-{\omega^{\rho}}_{\nu}[{T^{\mu}}_{\rho}x^{\nu}]##

    ##=-\omega_{\rho\nu}[T^{\mu\rho}x^{\nu}]## [I've raised the ##\rho## index on ##T## and lowered the ##\rho## index on ##\omega## since ##T## in the final expression for ##Q_i## has both upper indices.]

    ##=(-\omega_{0i}[T^{\mu 0}x^i]-\omega_{i0}[T^{\mu i}x^0])+(-\omega_{ij}[T^{\mu i}x^j]-\omega_{ji}[T^{\mu j}x^i])##, for ##i<j## in the third and fourth terms

    ##=(-\omega_{0i}[T^{\mu 0}x^i]+\omega_{0i}[T^{\mu i}x^0])+(-\omega_{ij}[T^{\mu i}x^j]+\omega_{ij}[T^{\mu j}x^i])##, for ##i<j## in the third and fourth terms

    ##=(\omega_{0i}[-T^{\mu 0}x^i+T^{\mu i}x^0])+(\omega_{ij}[-T^{\mu i}x^j+T^{\mu j}x^i])##, for ##i<j## in the third and fourth terms

    ##=(\omega_{\rho\sigma}[-T^{\mu\rho}x^{\sigma}+T^{\mu\sigma}x^{\rho}]),## for ##\rho < \sigma##

    Defining ##(\mathcal{J}^{\mu})^{\rho\sigma}=-T^{\mu\rho}x^{\sigma}+T^{\mu\sigma}x^{\rho}## for ##\rho < \sigma##,

    we see that ##\partial_{\mu}(\mathcal{J}^{\mu})^{\rho\sigma}=-(\partial_{\mu}T^{\mu\rho})x^{\sigma}+(\partial_{\mu}T^{\mu\sigma})x^{\rho}=0,## because ##\partial_{\mu}T^{\mu\nu}=0##.

    Therefore, for ##\rho\sigma=\{01,02,03,12,13,23\}##, ##(\mathcal{J}^{\mu})^{\rho\sigma}## corresponds to six conserved Noether currents which sum to give the Noether current ##j^{\mu}##.

    The Noether currents ##(\mathcal{J}^{\mu})^{12},(\mathcal{J}^{\mu})^{13}## and ##(\mathcal{J}^{\mu})^{23}## correspond to the invariance of the action under the spatial rotation of the spacetime coordinates.

    Therefore, the conserved Noether charges ##(\mathcal{J}^{0})^{12},(\mathcal{J}^{0})^{13}## and ##(\mathcal{J}^{0})^{23}## arise as a consequence of the invariance of the action under the spatial rotation of the spacetime coordinates.

    Therefore, the conserved Noether charges are ##Q^{jk}=-T^{0j}x^{k}+T^{0k}x^{j}=x^{j}T^{0k} - x^{k}T^{0j}## for ##j < k##

    so that ##Q_{i}=\epsilon_{ijk}x^{j}T^{0k}## for ##j < k##.

    Am I correct? My answer looks different from what they require and I'm also having difficulty understanding the concept of the total angular momentum.
     
  9. Apr 28, 2016 #8

    strangerep

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    I suspect you're missing something here. I get: $$\partial_{\mu}(\mathcal{J}^{\mu})^{\rho\sigma}~=~ -(\partial_{\mu}T^{\mu\rho})x^{\sigma} - T^{\mu\rho} \delta^\sigma_\mu + (\partial_{\mu}T^{\mu\sigma})x^{\rho} + T^{\mu\sigma} \delta^\sigma_\mu .$$ The derivatives vanish as you said, but that still leaves ##-T^{\sigma\rho} + T^{\rho\sigma},## which only vanishes if ##T## is symmetric. Usually, that requires an additional argument, wherein one adds a particular total derivative to ##T## which preserves the conservation rule, but also guarantees that its integrals over all space are unchanged. (This might also be the source of your uncertainty about "total angular momentum", which involves such an integral.)

    Can you access a copy of Greiner & Reinhardt? They explain quite a bit more detail than what's in Tong's notes -- far more than I can reasonably reproduce in a PF post.
     
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