In Relativistic Quantum Mechanics, an Angular Momentum Tensor is defined ($J_{\nu \mu}$). This tensor is divided into two terms, one responsible for the spin angular momentum ($S_{\nu \mu}$) and one responsible for the orbital angular momentum ($L_{\nu \mu}$). In essence:

$J_{\nu \mu}$ = $L_{\nu \mu}$ + $S_{\nu \mu}$

I read that the zeroth-components (the time components) of $S_{\nu \mu}$ is zero. In other words, I read that:

$S_{00}$ = $S_{01}$ = $S_{02}$ = $S_{03}$ = 0.

Why is this the case?


  • This question arose when I was concerned about the equality between equation (9.36) and equation (9.37) on page 81 of the following set of notes (http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf).
  • A possible explanation I have: the zeroth-components of the Angular Momentum Tensor corresponds to boosts and there's no boosts when dealing with spin, so there's no zeroth-components. Am I correct?

1 Answer 1


If you want $S^{i0}$ to be zero, then the spin $S^{\mu\nu}$ is the angular momentum of the body about its energy centroid in the lab frame.

Let's see how this comes about.

Suppose that the we have a conserved and symmetric energy-momentum tensor: $$ \partial_\mu T^{\mu\nu}=0, \quad T^{\mu\nu}=T^{\nu\mu} $$ which is non-zero only within the body of interest. Let $x_{\rm A}^\mu$ be a space-time event, $\Sigma$ a spacelike surface, and define the angular momentum about $x_{\rm A}$ by $$ M^{\mu\nu}_{\rm A} = \int_\Sigma\left\{(x^\mu-x_{\rm A}^\mu)T^{\gamma\nu}- (x^\nu-x_{\rm A}^\nu)T^{\gamma\mu}\right\}d\Sigma_\gamma $$ Then $M^{\mu\nu}_{\rm A} $ is a tensor and independent of the choice of the choice of $\Sigma$.

We now choose a lab frame and define the mass-centroid $X^i_{\rm L}$ in that frame by
$$ \left\{\int_{t=\rm const.} T^{00}d^3x \right\}\,X^i_{\rm L}= \int_{t=\rm const.} x^iT^{00}d^3x. $$ Note that $$ \partial_t \int T^{00}d^3x = \int \partial_0 T^{00}d^3x = - \int \partial_j T^{j0}d^3x=0, $$ and $$ \partial_t \int x^i T^{00}d^3x = \int x^i\,\partial_0 T^{00}d^3x = -\int x^i\,\partial_j T^{j0}d^3x= \int \delta^i_j \,T^{j0}d^3x= p^i. $$ So, differentiating its definition with respect to $t$, we read off that the ordinary three-velocity of the centroid is $$ \dot {\bf X}_{\rm L}= {\bf p}/E. $$ Here $$ E= \int T^{00}d^3x, \quad p^i = \int T^{0i}d^3x. $$

Now take $\Sigma$ to be the lab-frame surface $t=$const with $x^\mu_{\rm A}$ a point in that surface. Then $$ M^{i0}_{\rm A} = \int_\Sigma\left\{(x^i-x_{\rm A}^i)T^{00}- (x^0-x_{\rm A}^0)T^{0i}\right\}d^3x\\ = (X^i_{\rm L}-x^i_{\rm A})E. $$ (The second term is zero because $x^0- x^0_{\rm A}$ is zero everywhere in the integral.) Thus $M^{i0}_A$ is zero when ${\rm A}$ is the centroid in the lab frame. If we replace the lab frame with a frame having four-velocity $v^\mu$ we have that $M^{\mu\nu}_{\rm A} v_\nu=0$ if and only if ${\rm A}$ is the mass centroid in that frame.

Define the centre of mass $X^i_{\rm CM}$ to be the mass-centroid in the frame where the three momentum $p^i=0$, then the spin or intrinsic angular momentum $S^{\mu\nu}$ is usually defined to be the angular momentum about the centre of mass. Thus a more natural condition is $S^{\mu\nu}p_\nu=0$.

The total angular momentum about an arbitrary point $x_A$ is then $$ M^{\mu\nu}= (x_{\rm CM}-x_A)^\mu p^\nu- (x_{\rm CM}-x_a)^\nu p^\mu + S^{\mu\nu}\\ = L^{\mu\nu}_A+S^{\mu\nu}. $$

A rather useful combination is the totally antisymmetric Pauli-Lubanski tensor (invented by Myron Matthisson) $$ S^{\lambda\mu\nu}= p^\lambda S^{\mu\nu}+ p^\mu S^{\nu\lambda}+p^\nu S^{\lambda\mu}. $$ This object turns out to be independendent of the choice of $x_A$ and is also sometimes called the intrinsic spin.

  • $\begingroup$ I'm asking this question in terms of relativistic quantum mechanics. I thought the tensors is in terms of operators, right? $\endgroup$
    – user242977
    Commented Dec 7, 2019 at 4:36
  • 2
    $\begingroup$ I have down-voted this answer because it is very clear that the asker is seeking an answer from the point of view of relativistic quantum mechanics (i.e. the tensor being in terms of operators). $\endgroup$ Commented Dec 7, 2019 at 5:11
  • $\begingroup$ Same... I believe he should answer the question with regards to operators and relativistic quantum mechanics. $\endgroup$ Commented Dec 7, 2019 at 6:23
  • $\begingroup$ @mike_stone from a Group Theory or quantum mechanical point of view, am I right that the zeroth-components of the Angular Momentum Tensor corresponds to boosts and there's no boosts when dealing with spin, so there's no zeroth-components? $\endgroup$
    – user242977
    Commented Dec 7, 2019 at 6:28
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    $\begingroup$ The operator picture requires an understanding of the separation between $L$ and $S$, whic, for massless particle in particular,is not a Lorentz invariant decomposition. I know no way to understand this that does not involve the classical definition in terms of energy and momentum. It is not true that $S^{i0}=0$ in general frames and it it is not true that the there are no boosts for spin. $\endgroup$
    – mike stone
    Commented Dec 7, 2019 at 12:50

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