Query about nonsymmetric energy tensorsby Jay R. Yablon Tags: energy, nonsymmetric, query, tensors 
#1
Mar2308, 05:41 AM

P: n/a

Can someone please explain or provide some links which explain how the
physics of a nonsymmetric energy tensor would be different from that of a symmetric energy tensor. Especially, with the Poynting components T^0k <> T^k0, k=1,2,3, how would one interpret T^0k versus T^k0 and the physics of the energy flux associated with each? Thanks, Jay, ____________________________ Jay R. Yablon Email: jyablon@nycap.rr.com comoderator: sci.physics.foundations Weblog: http://jayryablon.wordpress.com/ Web Site: http://home.nycap.rr.com/jry/FermionMass.htm 


#2
Mar2608, 05:00 AM

P: n/a

On Mar 23, 6:36 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> Can someone please explain or provide some links which explain how the > physics of a nonsymmetric energy tensor would be different from that of > a symmetric energy tensor. > > Especially, with the Poynting components > > T^0k <> T^k0, k=1,2,3, > > how would one interpret T^0k versus T^k0 and the physics of the energy > flux associated with each? > > Thanks, > > Jay, > ____________________________ > Jay R. Yablon > Email: jyab...@nycap.rr.com > comoderator: sci.physics.foundations > Weblog:http://jayryablon.wordpress.com/ > Web Site:http://home.nycap.rr.com/jry/FermionMass.htm Nonsymmetric energy tensors lead to 1) reducible theories and 2) non conservation of angular momentum. Any asymmetry is physically irrelevant in any case, because by definition the energymomentum of a system is obtained by differentiating the action integral with respect to g_mn (Hamiltonian derivative, dL/dg_mn  d/dx_k dL/dg_mn,k) which is manifestly symmetric. The construction of the symmetric energy tensor for any system was given definitively by Belinfante and Rosenfeld. drl 



#3
Mar2608, 12:12 PM

Sci Advisor
P: 5,468

'When there are no assigned couples and no couplestresses, a necessary and sufficient condition for the balance of moment of momentum in a body where linear momentum is balanced is T^[mk] = 0, i.e. the stress tensor is symmetric.' We had a thread about this a while ago on sci.physics IIRC, if a body can internally produce angular momentum, the stress tensor is nonsymmetric. I don't think we discussed situations where there is a material boundary surface, which could potentially effect the results. 


#4
Apr108, 05:00 AM

P: n/a

Query about nonsymmetric energy tensors
On Mar 23, 5:36 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> Can someone please explain or provide some links which explain how the > physics of a nonsymmetric energy tensor would be different from that of > a symmetric energy tensor. The natural position of the indices for the energy tensor is with one in the upper position, one in the lower. You would then interpret the tensor as components of a 3form. The 3form is the kernel of an integral, the integral giving you the total momentum. Correspondingly, the other index is in the lower position. Thus, denoting the tensor by P^{mu}_a, the total momentum would be p_a = integral (P^{mu}_a (d^3x)_{mu}). and the 3form would be P_a = P^{mu}_a (d^3x)_{mu}. So, P^{mu}_a is actually a tensor density, rather than just a tensor. Consequently, the 3form P_a is interpreted as the continuum representation for momentum. It is the current density for the p_a component of momentum. Each component has its own 3 current, which is how you end up with 4 x 4 indices in all. Written explicitly in Cartesian coordinates with x^0 = t, (x^1, x^2, x^3) = (x,y,z) = r, you have (d^3x)_0 = dxdydz, (d^3x)_1 = dtdydz, (d^3x)_2 = dtdxdz, (d^3x)_3 = dtdydz where I'm using juxtaposition to denote the wedge product. Hence (d^3x)_1 = dtdzdy. For angular momentum, one has a similar tensor and integral and 3 form: s_{ab} = integral (S^{mu}_{ab} (d^3x)_{mu}) S_{ab} = S^{mu}_{ab} (d^3x)_{mu}. This gives you the intrinsic angular momentum. In flat spacetime, the coordinate index mu would no longer be distinguished from the frame indices a, b. This can be exploited to come up with a simple representation for the total angular momentum. It is arrived at by adding in the orbital part: J_{ab} = S_{ab} + x_a P_b  x_b P_a where indices are raised and lowered with the flat space metric eta_{ab}. Each quantity is subject to a conservation law: dP_a = 0; dJ_{ab} = 0. Written explicitly, the second one becomes d/dx^{mu} (S^{mu}_{ab} + x_a P^{mu}_b  x_b P^{mu}_a) = 0 or d/dx^{mu} S^{mu}_{ab} + eta_{a mu} P^{mu}_b  eta_{b mu} P^{mu}_a = 0. Using the convention P_{ab} = eta_{b mu} P^{mu}_a, this becomes P_{ab}  P_{ba} = d/dx^{mu} S^{mu}_{ab}. That's the antisymmetric part of the energy tensor. For curvilinear coordinates or curved spacetimes, one would also have a "frame" tensor h^a_{mu} which converts between the two indices. This is the square root of the metric g_{mn} = eta_{ab} h^a_{mu} h^b_{nu}. You can't write down a simple form for the orbital part of angular momentum, but you can cheat by integrating by parts. This exploits the conservation law dP_a = 0. First, one writes down a "kinetic potential" U_a: dU_a = P_a. This is a 2form, and I've not really seen much mentioned of this anywhere in the literature. Using this, one can write x_a P_b = x_a dU_b = d(x_a U_b)  dx_a U_b. The total differential is not significant in the integral for J_{ab}. Therefore, one has, in effect, J_{ab} = S_{ab} + dx_b U_a  dx_a U_b. The 1form dx_b = h_{b mu} dx^{mu} is the abovementioned frame tensor. So, now you have something that ultimately will work in curvilinear coordinates and curved spacetime. The symmetric part the energy tensor comes straight out the momentum conservation law. One writes T^{mu nu} = P^{mu nu}  1/2 d/dx^{rho} (S^{mu nu rho} + S^{rho mu nu} + S^{rho nu mu}). I'm using the convention here that the lower index for S is placed 3rd on top. This works fine in flatspace form, where the frame indices and coordinate indices can be equated. I'm not sure which indices should be which, when generalizing this to curvilinear form. The symmetry of this tensor is grounded in the abovementioned anti symmetry relation for P. The correction term is the Belinfante correction. It's the standard way to symmetrize the stress tensor for a field. To use an example: For a general electromagnetic field, one could write down a Lagrangian L = L(E, B, A, phi), in terms of the 3potential A = (A_1, A_2, A_3), the electric potential phi, and the fields E = (E_1, E_2, E_3) = grad phi  dA/dt B = (B^1, B^2, B^3) = curl A. Express the variational of the Lagrangian as delta L = (delta E).D  (delta B).H + deltaA.J  (delta phi) rho. This introduces the "conjugate" fields D, H and (external) "sources" J and rho. For an explicitly gauge invariant Lagrangian there is no dependence on A or phi, so that the external sources J and rho are 0. The energy tensor is P^{mu}_{nu} = dL/(dA_{i,mu}) A_{i,nu} + dL/(dphi_{,mu}) phi_{,nu}  delta^{mu}_{nu} L. where I'm using (,mu) denote denote the coordinate derivative operator d/dx^{mu}. The components are P^0_0 = D^i dA_i/dt  L = D.dA/dt  L P^0_j = D^i dA_i/dx^j = D.dA/dx^j P^i_0 = (H x dA/dt + D^i d(phi)/dt)^i P^i_j = (H x dA/dx^j + D^i d(phi)/dx^j)^i  delta^i_j L. From this, with the addition of an appropriate divergence, one may derive the tensor T^0_0 = D.E  L T^0_j = (D x B)_j T^i_0 = (H x E)^i T^i_j = D^i E_j + H_j B^i  delta^i_j (B.H + L). This is only symmetric with respect to the metric eta_{ij} = delta_{ij}; eta_{i0} = 0 = eta_{0j}, eta_{00} = c^2 if and only if a constitutive law of the form holds D = epsilon E + theta B H = epsilon c^2 B  theta E where epsilon and theta may be constant or functions of the fields. In fact, these coefficients are simply the derivatives of the Lagrangian with respect to the Lorentz invariants epsilon = dL/dI_1; theta = dL/dI_2 where I_1 = (E^2  B^2 c^2)/2, I_2 = E.B. So, if you also require the Lagrangian, itself, to be Lorentz invariant this will then imply that epsilon and theta, themselves, can have functional dependence on E and B only through I_1 and I_2: epsilon = epsilon_0 + e1 I_1 + e2 I_2 + 1/2 e3 (I_1)^2 + e4 I_1 I_2 + 1/2 e5 (I_2)^2 + ... theta = theta_0 + t1 I_1 + t2 I_2 + 1/2 t3 (I_1)^2 + t4 I_1 I_2 + 1/2 t5 (I_2)^2 + ... Without loss of generality one can also always take theta_0 = 0, since theta is only determined up to a plus or minus coefficient (i.e. Maxwell's equations are invariant under transformation D > D  k B; H > H + k E ... a vestige of complexion symmetry). Finally, the trace of the stress tensor will be 0 if and only if the Lagrangian is homogeneous to the first order in I_1 and I_2. This would make the coefficients epsilon and theta independent of the fields (though possibly still variable, and subject to their own dynamic laws). I'll leave it you to express the difference T^{mu}_{nu}  P^{mu}_{nu} as a divergence. An even more interesting exercise is to write down P^{mu}_{nu}, itself, as a divergence (which, as implied above, could be done ... here, it's doable when the field is a free field. For a coupled field, one has to write out the entire stress tensor for everybody and to the construction on that, instead of piecemeal on the electromagnetic "part" of the stress tensor). 


#5
Apr108, 05:00 AM

P: n/a

DRLunsford says...
>Nonsymmetric energy tensors lead to 1) reducible theories and 2) non >conservation of angular momentum. I don't think that's necessarily true. According to this article http://en.wikipedia.org/wiki/Spin_tensor the difference between T_ij and T_ji is equal to the rate at which orbital angular momentum is converted to spin angular momentum. *Total* angular momentum can still be conserved even with a nonsymmetric T_ij.  Daryl McCullough Ithaca, NY 


#6
Apr108, 05:00 AM

P: n/a

In article <b9660087b4374f2ba63a6a770f77229c@p73g2000hsd.googlegroups.com>,
DRLunsford says... >Nonsymmetric energy tensors lead to 1) reducible theories and 2) non >conservation of angular momentum. Any asymmetry is physically >irrelevant in any case, because by definition the energymomentum of a >system is obtained by differentiating the action integral with respect >to g_mn (Hamiltonian derivative, dL/dg_mn  d/dx_k dL/dg_mn,k) which >is manifestly symmetric. The construction of the symmetric energy >tensor for any system was given definitively by Belinfante and >Rosenfeld. I've only briefly look at it, but this article http://arxiv.org/PS_cache/grqc/pdf/0409/0409029v1.pdf describes a generalization of General Relativity that allows for torsion (the connection coefficients are not symmetric). With torsion, the Einstein tensor G_uv becomes nonsymmetric and the source of the antisymmetric part is an antisymmetric stressenergy tensor (due to spin density). So nonsymmetric energy tensors are not completely irrelevant, physically.  Daryl McCullough Ithaca, NY 


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