The derivation of the Belinfante-Rosenfeld tensor - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-17T19:10:15Z https://physics.stackexchange.com/feeds/question/27048 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/27048 12 The derivation of the Belinfante-Rosenfeld tensor user6818 https://physics.stackexchange.com/users/1637 2011-10-02T01:50:20Z 2013-06-19T14:34:29Z <ul> <li><p>It seems me that there is a "difference" (at least apparently) in how the Belinfante-Rosenfeld tensor is thought of in section 7.4 of Volume 1 of Weinberg's QFT book and in section 2.5.1 of the conformal field theory book by di Francesco et. al. </p> <p>I would be glad if someone can help reconcile the issue. </p></li> </ul> <p>Schematically the main issue as I see is this -</p> <ul> <li><p>If the action and the lagrangian density is writable as $I = \int d^4x L$ and $\omega_{\mu \nu}$ be the small parameter of Lorentz transformation then Weinberg is thinking of $\omega_{\mu \nu}$ to be space-time independent and he is varying the action to write it in the form, $\delta I = \int d^4x (\delta L = (A^{\mu \nu})\omega_{\mu \nu})$ Then some symmetrized form of whatever this $A^{\mu \nu}$ comes out to be is what is being called the Belinfante tensor. ( Its conservation needs the fields to satisfy the equations of motion) </p></li> <li><p>But following Francesco et.al's set-up I am inclined to think of $\omega_{\mu \nu}$ as being space-time dependent and then the variation of the action will also pick up terms from the Jacobian and the calculation roughly goes as saying, $\delta I = \int (\delta(d^4x)) L + \int d^4x (\delta L)$. Since under rigid Lorentz transformations the volume element is invariant the coefficient of $\omega_{\mu \nu}$ in the first variation will vanish but the second variation will produce a coefficient say $B^{\mu \nu}$. But both the variations will produce a coefficient for the derivative of $\omega_{\mu \nu}$ and let them be $C^{\mu \nu \lambda}$ and $D^{\mu \nu \lambda}$ respectively.<br> Now the argument will be that since the original action was to start off invariant under Lorentz transformations, when evaluated about them, the $B^{\mu \nu}$ should be $0$ and on shifting partial derivatives the <em>sum</em> of $C^{\mu \nu \lambda}$ and $D^{\mu \nu \lambda}$ is the conserved current (..and its not clear whether their conservation needs the field to satisfy the equation of motion..) </p></li> </ul> <p>So by the first way $A^{\mu \nu}$ will be the conserved current and in the second the conserved current will be, $C^{\mu \nu \lambda} + D^{\mu \nu \lambda}$ (the C tensor will basically look like $-x^\nu \eta^{\lambda \mu}L$) </p> <p>Is the above argument correct? </p> <p>If yes then are the two arguments equivalent?</p> <p>How or is Weinberg's argument taking into account the contribution to the conserved current from the variation of the Jacobian of the transformation? </p> https://physics.stackexchange.com/questions/27048/-/27049#27049 10 Answer by José Figueroa-O'Farrill for The derivation of the Belinfante-Rosenfeld tensor José Figueroa-O'Farrill https://physics.stackexchange.com/users/2782 2011-10-02T13:37:15Z 2011-10-02T13:37:15Z <p>The two derivations are indeed different, but the resulting object should be the same: it should be symmetric and conserved on-shell.</p> <p>In fact, perhaps the cleanest way to derive it is to couple the theory to gravity and then vary the resulting action with respect to the metric. If we let $$S = \int_M d^4x \sqrt{-g} \mathcal{L}$$ denote the action of the theory coupled to gravity (i.e., put the theory on a lorentzian manifold by covariantising derivatives,...) then the Belinfante stress-energy tensor is defined (up to perhaps a constant) by $$T_{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu}}~.$$</p> <p>This form has the virtue that it is easy to see that if the theory is invariant under homotheties -- $\delta g_{\mu\nu} = \lambda g_{\mu\nu}$ for some constant $\lambda$ -- the stress-energy tensor is traceless.</p> https://physics.stackexchange.com/questions/27048/-/27050#27050 9 Answer by David Bar Moshe for The derivation of the Belinfante-Rosenfeld tensor David Bar Moshe https://physics.stackexchange.com/users/2190 2011-10-05T10:41:34Z 2011-10-05T11:05:10Z <p>The two derivations are actually identical, except for the fact that Weinberg didn't have the general form of the Noether theorem for symmetries acting on the space-time coordinates as well as on the fields (Equation 2.141 in Di Francesco, Mathieu and Sénéchal's book).<br> As a consquence, Weinberg had to compute the variation of the action with respect to the Lorentz generators from scratch (including the substitution of the equations of motion). </p> <p>Furthermor, I wanted to remark that the term depending on the variation of the space time coordinates in the general form of the Noether theorem is not due to the noninvariance of the Minkowski space time measure $d^4x$ as this measure is invariant under both translations and Lorentz transformations. The extra term is due to the dependence of the Lagrangian on the space time coordinates through its dependence on the fields.</p> <p>Now, both authors use the derivation as a means of computation of the Belinfante &amp; Rosenfeld 3-tensor whose divergence is to be added to the canonical energy momentum tensor to obtain the symmetric Belinfante energy momentum tensor. The principle upon which this computation is based is that the orbital part of the canonical conserved current corresponding the Lorentz symmetry must have the form:</p> <p>$M^{\mu\nu\rho} = x^{\nu} T_B^{\mu\rho} - x^{\rho} T_B^{\mu\nu}$</p> <p>with $T_B$ both conserved and symmetric (as can be checked by a direct computation), therefore, they arrange the extra-terms they obtained to bring the Lorentz canonical current to this form and as a consequence they obtain the required tensor to be added.</p> <p>I wanted to add that both authors use the derivatives of the symmetry group parameters in their intermediate computations, but this is not required. The same currents can be obtained for variation with respect to global constant parameters. If the action were locally invariant (with respect to variable parameters), then the currents would have been conserved off-shell. This is the Noether's second theorem. </p> <p>Finally, I want to refer you to the <a href="http://authors.library.caltech.edu/19366/1/GoMa1992.pdf">this article</a> of Gotay and Marsden describing a method of obtaining a symmetric and (gauge invariant) energy-momentum tensor directly based on Noether's theory.</p>