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Poincare conserved currents : Energy-momentum and Angular-momentum tensors

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crackjack
#1
Sep12-09, 11:57 PM
P: 117
Not sure if this is the right place to ask, but this doubt originated when reading on string theory and so here it goes...

The general canonical energy-momentum tensor (as derived from translation invariance), [tex]T^{\mu\nu}_{C}[/tex] is not symmetric. Also, the general angular momentum conserved current (as derived from lorentz invariance) consists of two parts to it - the orbital angular momentum component and the spin angular momentum component...
[tex]j^{\mu\nu\rho} = T^{\mu\nu}_{C} x^\rho - T^{\mu\rho}_{C} x^\nu + S^{\mu\nu\rho}[/tex]

But, by taking clues from the above angular momentum expression, we can append a suitable term to the generally non-symmetric canonical energy-momentum tensor and modify it (without breaking its conservation) into a symmetric Belinfante tensor...
[tex]T^{\mu\nu}_{C} \to T^{\mu\nu}_{B} [/tex]


Further, the angular momentum tensor can also be modified to absorb the spin term in its orbital momentum term by rewriting it as...
[tex]j^{\mu\nu\rho} = T^{\mu\nu}_{B} x^\rho - T^{\mu\rho}_{B} x^\nu[/tex]

Now, there is no spin operator at all now - at least not explicitly. And we haven't really modified the physics at all. What is then the spin of the system now?

In light cone gauge of bosonic strings, the above (original) spin operator is used to calculate the spins of the massless fields (photon, graviton)...
So, what would the above disappearance of the spin operator mean then? If you say that its not done away with but is just hidden in the last expression above, what rule do we follow in separating the orbital and spin components of angular momentum?
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samalkhaiat
#2
Sep25-09, 04:30 PM
Sci Advisor
P: 884
[QUOTE]
Quote Quote by crackjack View Post
Not sure if this is the right place to ask,
No, this is not the right place! Don't mention string theory in this part of the PF
People in here are too busy, with the LOOPY "quantum gravity" stuff of Rovelli and Smolin, to reply to your question! Any way let us do it.

Let us write

[tex]J^{\rho \mu \nu} = L^{\rho \mu \nu} + S^{\rho \mu \nu}[/tex]

where

[tex]
L^{\rho\mu\nu} = x^{\mu}T_{C}^{\rho\nu} - x^{\nu}T_{C}^{\rho\mu} \ \ \ (1)
[/tex]

and

[tex]
S^{\rho\mu\nu} = \frac{\partial \mathcal{L}}{\partial \partial_{\rho}\phi^{r}} (\Sigma^{\mu\nu})^{r}{}_{s}\phi^{s}
[/tex]

The object L is called by Belinfate the orbital angular momentum density "tensor" because its space components have the form of non-relativistic orbital momentum density. Because the object S clearly expresses the transformation of the internal degree of freedom of the field, it is called (by Belinfate) the "tensor" of internal angular momentum density or spin density.
Notice that [itex]S^{\rho\mu\nu}[/itex] does not have the typical form of a classical angular momentum like [itex]L^{\rho\mu\nu}[/itex]. In order to show that it can be written in such a form, Belifate (and independently Rosenfeld) introduced the so-called symmetric energy-momentum tensor;

[tex]T_{B}^{\mu\nu} = T_{C}^{\mu\nu} + t^{\mu\nu}[/tex]

with

[tex]
t^{\mu\nu} = (1/2) \partial_{\rho}( S^{\mu\nu\rho} + S^{\rho\mu\nu} + S^{\nu\mu\rho})
[/tex]

Because of the following properties

[tex]T_{B}^{\mu\nu} = T_{B}^{\nu\mu}[/tex]

[tex]\partial_{\mu}T_{B}^{\mu\nu} = \partial_{\mu}T_{C}^{\mu\nu} = 0[/tex]

[tex]P^{\mu} = \int T_{B}^{0\mu} \ d^{3}x = \int T_{C}^{0\mu} \ d^{3}x ,[/tex]

[itex]T_{B}^{\mu\nu}[/itex] appears as an equivalent rnergy-momentum tensor. The corresponding new total angular momentum tensor is defined by;

[tex]J_{B}^{\rho\mu\nu} = x^{\mu}T_{B}^{\rho\nu} - x^{\nu}T_{B}^{\rho\mu}[/tex]

This can also be written as

[tex]J_{B}^{\rho\mu\nu} = L^{\rho\mu\nu} + s^{\rho\mu\nu}[/tex]

with

[tex]s^{\rho\mu\nu} = x^{\mu}t^{\rho\nu} - x^{\nu}t^{\rho\mu}[/tex]

having the required form of an angular momentum.

Again

[tex]\partial_{\rho}J_{B}^{\rho\mu\nu} = 0[/tex]

[tex]M^{\mu\nu} = \int J_{B}^{0\mu\nu}\ d^{3}x = \int J^{0\mu\nu} \ d^{3}x[/tex]

and

[tex]S^{\mu\nu} = \int d^{3}x s^{0\mu\nu} = \int d^{3}x S^{0\mu\nu}[/tex]

consequently, the tensor [itex]s^{\rho\mu\nu}[/itex] could equivalently be considered as the spin density instead of [itex]S^{\rho\mu\nu}[/itex]; it is the angular momentum of a "spin" energy-momentum density [itex]t^{\mu\nu}[/itex], which does not contribute to the total energy-momentum vector:

[tex]\int d^{3}x \ t^{0\mu} = 0.[/tex]

In this way, the canonical energy-momentum tensor would then represent an "orbital" energy-momentum tensor.
Although it does not change the physics ( [itex]M^{\mu\nu}[/itex] and [itex]P^{\mu}[/itex] remain the same and generate the same Poincare algebra) the above Blinfate-Rosenfeld method has at least two weaknesses. The 1st one lies in the fact that the total spin and the total orbital angular momentum are, in general, not covariant quantities. Indeed, the quantities

[tex]S^{\mu\nu} = \int d^{3}x \ s^{0\mu\nu}[/tex]

and

[tex]L^{\mu\nu} = \int d^{3}x \ L^{0\mu\nu}[/tex]

are tensorial only if the corrsponding densities [itex]s^{\rho\mu\nu}[/itex] and [itex]L^{\rho\mu\nu}[/itex] are conserved. This is however the case, if and only if [itex]T_{C}^{\mu\nu}[/itex] is symmetric. That is, the Belinfate spin and orbital angular momentum are sensible quantities only in those cases where the entire symmetrization procedure seems to be superfluous!
The 2nd weakness of the B-R method is that it leads some people (I believe you are included) to believe (incorrectly) that a field can only posses non-zero spin if [itex]T_{C}^{\mu\nu}[/itex] is non-symmetric.

regards

sam
crackjack
#3
Sep26-09, 08:18 AM
P: 117
Thanks Sam, for the detailed explanation!

Quote Quote by samalkhaiat View Post
The 2nd weakness of the B-R method is that it leads some people (I believe you are included) to believe (incorrectly) that a field can only posses non-zero spin if [itex]T_{C}^{\mu\nu}[/itex] is non-symmetric.
Ya, my confusion is kind of related to this...

Why should the orbital and spin angular momentum components be separately covariant?

Mentz114
#4
Sep27-09, 02:12 AM
PF Gold
P: 4,087
Poincare conserved currents : Energy-momentum and Angular-momentum tensors

You may find this interesting

Symmetric energy-momentum tensor in Maxwell, Yang-Mills, and Proca theories
obtained using only Noether’s theorem

Merced Montesinos and Ernesto Flores

(Dated: February 1, 2008)

Abstract:
The symmetric and gauge-invariant energy-momentum tensors for source-free Maxwell and Yang-
Mills theories are obtained by means of translations in spacetime via a systematic implementation
of Noether’s theorem. For the source-free neutral Proca field, the same procedure yields also the
symmetric energy-momentum tensor. In all cases, the key point to get the right expressions for
the energy-momentum tensors is the appropriate handling of their equations of motion and the
Bianchi identities. It must be stressed that these results are obtained without using Belinfante’s
symmetrization techniques which are usually employed to this end.

arXiv:hep-th/0602190v1 20 Feb 2006
The emphasis is mine.
crackjack
#5
Oct5-09, 10:00 AM
P: 117
Thanks for the link! That is a neat way of deriving the symmetric stress tensor using Bianchi identity.

But I actually don't find Belinfante's procedure to be too hotch potch. In Belinfante's procedure, the spin momentum component can be totally done away with, leaving only the orbital momentum component (in the procedure employed in the above paper, we don't ever arrive at any spin component).
But, from what I have read, it is this spin momentum component that is used to determine the (integer) spin in bosonic string theory. This is what is confusing me.

Additional note...
The above paper also shows that we could arrive at one 'pure orbital' angular momentum tensor without first going through one which has both orbital and spin components (as with Belinfante's procedure). So then, we dont have any rule to split the 'pure orbital' angular momentum tensor to get 'back' the spin component (as samalkhaiat did above).
Ed Rex
#6
Oct5-09, 08:46 PM
P: 1
These papers, arXiv:0905.4529 by Andrew Randono and Dave Sloan and the follow-up paper arXiv:0906.1385 by Andrew Randono, may help if you are interested in the way spin is encoded in the gravitational field. The first emphasizes some algebraic properties of intrinsic spin which are not usually covered, and shows how these algebraic properties are encoded in the gravitational field (tetrad) at asymptotic infinity. The algebraic properties reveal how the two tensors are separately covariant, as you asked. The second paper uses the example of a spinor in linearized gravity and shows explicitly how the intrinsic spin is buried in the symmetric stress energy tensor. To unbury it, the paper uses a "gravitational Gordon decomposition" which is a direct analogue of the procedure to extract the spin in electromagnetism.
crackjack
#7
Oct19-09, 05:24 PM
P: 117
Thanks Ed Rex. I will take a look at them.
They seem to be specialized to the case of gravitational fields, but I will see what all generic information I can decouple from this.
Federation 2005
#8
Oct4-10, 06:18 AM
P: 16
Montesinos and Flores is the standard way of handling the stress tensor for form-valued fields, rather than canonical + Belinfante. A much more comprehensive treatment of Noether symmetry was already, long before then, covered by Hehl in "Two Lectures on Fermions and Gravity", Hehl. et al. in Geometry and Theoretical Physics; (J. Debrus, A. C. Hirshfeld (Eds.)), Springer-Verlag 1991; sections 6-12 cover the general issue.

If a Lagrangian 4-form is a function L(q,dq,x,dx) of form-valued fields q, their exterior differentials dq, the coordinates x and coordinate differentials dx, then you can write the total variational as DL = (Dq)^f + (D(dq))^p + (Dx)^K + (D(dx))^T, which applies generally as D ranges over all natural derivations. The case D = L_m (the Lie derivative of the coordinate vector field d/dx^m) yields an expression for K. the case D = i_m (the contraction of d/dx^m) yields an expression for T. Substituting these in eliminates K and T.

Then, the general case D = L_X + delta, where L_X is the infinitesimal diffeomorphism generated by a vector field X and delta an internal symmetry yields the weak form of Noether's theorem for this combined symmetry. You can read off the stress tensor, almost directly, from the result. And that's the correct way to handle the problem.

Both an on-shell and off-shell Noether theorem can be easily formulated. The general assumption made is delta(L) = d(L_d) for some 3-form L_d. That generalize the Noether theorem, which takes L_d = 0.

Hehl framed "relocalization theorems" in section 10-12. Each one produces a kind of canonical result. One relocalization (mentioned in section 11) is defined as that which zeros out the spin-current. This is one and the same as Belinfante. The second (mentioned in section 12) also zeros out the Noether current associated with scale symmetry. So, you can do a lot more than Belinfante. Belinfante is not the end-all or be-all of relocalization transformations.

The Maxwell field yields a symmetric tensor. The Dirac field (which is also treated in the Hehl article) does not. The electromagnetic field has zero spin tensor. The situation is explained in depth in the Hehl article.

But the best way to understand it, here, is that photon does *not* have spin. It has helicity (and the helicity it is an invariant). The same would be true for Weyl fermions.

The best way to understand this -- and the right way to understand and classify the representations of space-time symmetry groups -- is to work out the coadjoint orbits of the Poincare' group (i.e. write down the symplectic decomposition of the Poisson manifold associated with the Poincare' group).

The symplectic leaves associated with luxons (light-speed representations) fall into 3 families: (1) the spin 0 family, (2) the "helion" and (3) the "continuous spin" family.

The symplectic leaf which photons (and Weyl particles) belong to is what I called the "helion" family. It has a Pauli-Lubanski vector W that is proportional to the momentum vector P. The symplectic leaf is only *6* dimensional (i.e. it has only *3* Heisenberg pairs, not 4).

The helions are spin 0 luxons and spin 0 tardions (both of which have 6-dimensional symplectic leaves) than they are to spin non-zero representations. (Tardion = "slower than light" representations).

In contrast, the subfamily of luxons which have what may more properly be called an extra spin degree of freedom reside on the symplectic leaves that are *8* dimensional are what are also known as the "continuous spin" luxons. There, the spatial part of the Pauli Lubanski vector W resides on a cylinder, the width of the cylinder being an invariant.

They're never considered as models for any of the known fundamental fields.

(For spin non-zero tardions, W resides on a sphere, the radius essentially giving you the spin; the Dirac fermions being the prime example).

The class (3) luxons. too, there is no real spin-orbit decomposition, since W isn't on a sphere. They have no position operator, to begin with (a well-known impossibility theorem). But the class is more closely analogous to the spin non-zero particles.

Class (2) luxons sorta have a position operator, because you can apply the Darboux theorem to write down the symplectic form as omega = dp_x ^ dx + dp_y ^ dy + dp_z ^ dz. But the coordinates are singular in the same way that the (phi, theta) coordinates of a sphere are.

The symplectic representation for the helion class, in fact, is similar to that for the magnetic monopole (something that's developed in LNP 188, "Gauge Symmetries and Fibre Bundles") In fact, the construction used in LNP 188 yields an extra angular momentum term which is analogous to how angular momentum arises for the helion sector out of its helicity.

But the angular momentum does *not* come from spin.
Federation 2005
#9
Oct4-10, 06:23 AM
P: 16
"The helions are spin 0 luxons and spin 0 tardions" should read "The helions are more akin to spin 0 luxons and spin 0 tardions"
tom.stoer
#10
Oct5-10, 01:02 AM
Sci Advisor
P: 5,364
Quote Quote by samalkhaiat View Post
No, this is not the right place! Don't mention string theory in this part of the PF
People in here are too busy, with the LOOPY "quantum gravity" stuff of Rovelli and Smolin, to reply to your question!
I can't see how you came to this conclusion.

There are a couple of interesting discussions regarding string theory - and there are members participating in the "beyond forum" able to respond on experts-level on string theory questions.
crackjack
#11
Oct8-10, 12:27 PM
P: 117
Quote Quote by Federation 2005 View Post
A much more comprehensive treatment of Noether symmetry was already, long before then, covered by Hehl in "Two Lectures on Fermions and Gravity", Hehl. et al. in Geometry and Theoretical Physics; (J. Debrus, A. C. Hirshfeld (Eds.)), Springer-Verlag 1991; sections 6-12 cover the general issue.
...
But the angular momentum does *not* come from spin.
Thanks. I understood your post only in patches - will look at the lectures of Hehl et al to fill in my ignorance.


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