Query about non-symmetric energy tensors

Jay R. Yablon

Can someone please explain or provide some links which explain how the
physics of a non-symmetric 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
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

DRLunsford

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 non-symmetric 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
> co-moderator: sci.physics.foundations
> Weblog:http://jayryablon.wordpress.com/
> Web Site:http://home.nycap.rr.com/jry/FermionMass.htm

Non-symmetric 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 energy-momentum 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

Andy Resnick

Truesdell and Toupin, in "The Classical Field Theories" (Encyclopeida of Physics) derive a general expression for the stress tensor in Section 205. Cauchy's second law is a statement of the conditions under which the stress tensor is symmetric:

'When there are no assigned couples and no couple-stresses, 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.

Rock Brentwood

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 non-symmetric 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 3-form. The 3-form 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 3-form 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 3-form 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 space-time, 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 anti-symmetric 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 2-form, 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 1-form dx_b = h_{b mu} dx^{mu} is the above-mentioned 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 flat-space 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 above-mentioned 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 3-potential 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).

Daryl McCullough

DRLunsford says...

>Non-symmetric 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

Daryl McCullough

In article <b9660087-b437-4f2b-a63a-6a770f77229c@p73g2000hsd.googlegroups.com>,
DRLunsford says...

>Non-symmetric 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 energy-momentum 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/gr-qc/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 anti-symmetric
part is an anti-symmetric stress-energy tensor (due to
spin density). So non-symmetric energy tensors are not
completely irrelevant, physically.

--
Daryl McCullough
Ithaca, NY