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Energy-momentum tensor for the Dirac spinor

  1. Jan 2, 2012 #1
    Hi there, i'm having a problem calculating the energy momentum tensor for the dirac spinor [tex]\psi (x) =\left(\begin{align}\psi_{L1}\\ \psi_{L2}\\\psi_{R1}\\ \psi_{R2}\end{align}\right)[/tex](free theory).

    So, with the dirac lagrangian [tex]\mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi-m\bar{\psi}\psi[/tex]in hand i should be able to figure out the energy momentum tensor by using the formula[tex]T^\mu{}_\nu=\frac{\delta\mathcal{L}}{\delta \partial u_A}\partial_\nu u_A-\delta^\mu_\nu\mathcal{L}[/tex]
    and since we're assuming the equations of motion to be true we can forget of the latter term in the above equation, and focus on the former.

    Here come the probelms.
    First of all, i'm not sure on which values does the [itex]A[/itex] index in the [itex]T^\mu{}_\nu[/itex] forumula run: are they 1L, 2L, 1R, 2R?

    If so, how can you tell me explicitly how to do the functional derivation [itex]\frac{\delta\mathcal{L}}{\delta\partial u_A}[/itex]?

    By writing explicitly the lagrangian (forgetting about the mass term) i get to
    [tex]\mathcal{L}=i(\psi^*_{L1},\psi^*_{L2},\psi^*_{1R}, \psi^*_{2R})\gamma^0\partial_0\left(\begin{align} \psi_{L1}\\ \psi_{L2}\\ \psi_{R1}\\ \psi_{R2}\end{align}\right) +i(\psi^*_{L1},\psi^*_{L2},\psi^*_{1R},\psi^*_{2R})\gamma^1\partial_1\left(\begin{align}\psi_{L1}\\ \psi_{L2}\\\psi_{R1}\\ \psi_{R2}\end{align}\right)+i(\psi^*_{L1},\psi^*_{L2}, \psi^*_{1R},\psi^*_{2R})\gamma^2\partial_2\left (\begin{align}\psi_{L1}\\\psi_{L2}\\\psi_{R1}\\ \psi_{R2}\end{align}\right)+ i(\psi^*_{L1},\psi^*_{L2},\psi^*_{1R},\psi^*_{2R})\gamma^3\partial_3\left( \begin{align}\psi_{L1}\\\psi_{L2}\\\psi_{R1}\\ \psi_{R2}\end{align}\right)[/tex]and I stop here becaouse those gamma matrices makes the calculation ridiculously complicated, which bring me nowhere.

    Can you folks help me??
    Last edited: Jan 2, 2012
  2. jcsd
  3. Jan 2, 2012 #2


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    Science Advisor

    You don't need to vary each component of the 4-spinor separately, but you do need to independently vary both ψ and ψ, because these are the two values of uA. So you'd better first integrate by parts and write the Lagrangian in its symmetrized form: ½ψγμ(∂μψ) - ½(∂μψμψ + mψψ.

    Taking the functional derivative δL/δ∂μψ is easy, because L is linear in ∂μψ.
  4. Jan 2, 2012 #3
    That was my first guess.

    But can you explain me the following issue then?

    The spin angular momentum [itex]S^{\mu\lambda\kappa}[/itex] is defined to be [tex]S^{\mu\lambda\kappa}=-i\frac{\delta\mathcal{L}}{\delta\partial_\mu u_A}(S^{\lambda\kappa})_{AB}u_B[/tex]where [itex]S^{\lambda\kappa}=\frac{i}{4}[\gamma^\lambda ,\gamma^\kappa][/itex].

    Now here [itex]S^{\lambda\kappa}=\frac{i}{4}[\gamma^\lambda ,\gamma^\kappa][/itex] is a 4x4 matrix that acts on the 4 components of [itex]\psi (x) =\left(\begin{align}\psi_{L1}\\ \psi_{L2}\\\psi_{R1}\\ \psi_{R2}\end{align}\right)[/itex], so i'm pushed to say that those A,B indicies on the [itex](S^{\lambda\kappa})_{AB}[/itex] actually run over the 4 values (1L 2L 1R 2R), and so do the ones of the [itex]u_A[/itex]

    Where am i mistaking?
    Thanks for your time!
  5. Jan 3, 2012 #4
    By the way, since the question has actually changed should I open a new post??
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