Question about Algebraic classification and Compatibility of Energy Momentum Tensors

Jay R. Yablon

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I am studying Stephani et al.'s "Exact Solutions of Einstein's Field
Equations" and had a few question about the Algebraic classifications of
the Energy tensors discussed in Chapter 5, which are still new to me.

Because of the Einstein equations, the Ricci R_uv tensor takes on the
same algebraic type as the energy tensor T_uv to which it is linked.
This, by implication, seems to have consequences regarding the
"compatibility" of various energy tensors with one another.
Specifically:

The non-null Maxwell tensor T^uv (Maxwell) is of the Segre type
[(11),(1,1)] / Plebanski type [2S-2T]. It is also a function of the
metric tensor g_uv and the field strength tensor F^uv, that is,
T^uv(Maxwell) = T^uv(g_uv, F_uv)

The Perfect fluid tensor T^uv(Euler) is of the Segre Type [(111),1] /
Plebanski type [3S-T].

My question is about compatibility between energy tensors. Suppose one
wanted to somehow understand whether there exists an electrodynamic
basis underlying a perfect fluid. That is, suppose one wanted to
explore the possibility that T^uv(Euler) = T^uv'(g_uv, F_uv), similarly
to the Maxwell tensor (I have used a prime ' to denote that the
combination of fields T^uv'(g_uv, F_uv) may be different from that in
T^uv(g_uv, F_uv) = T^uv(Maxwell)).

At some juncture, one would need to set T^uv(Euler) = T^uv'(g_uv, F_uv).
In this situation, would it be necessary for T^uv(Euler) to be of the
same algebraic type as T^uv'(g_uv, F_uv)? That is, would one need to
find a T^uv'(g_uv, F_uv) in the Segre Type [(111),1] / Plebanski type
[3S-T] class rather than in the [(11),(1,1)] / Plebanski type [2S-2T]
class for this to be valid? If one tried to set a Segre Type [(111),1]
/ Plebanski type [3S-T] tensor equal to a [(11),(1,1)] / Plebanski type
[2S-2T] class tensor, would this be an apples to oranges equation that
is effectively trying to set the Ricci tensor to two different algebraic
type simultaneously? (It is to be understood that in all cases these
are energy tensors which would also be set to the R^uv-(1/2)g^uvR and so
would have vanishing divergence and establish a second order non-linear
differential equation for the metric tensor).

More generally, how precisely must the algebraic classes of two tensors
match if one is to equate them in a valid manner to one another as well
as to R^uv-(1/2)g^uvR?

Finally, if one were to take T^uv(Maxwell) and augment it with a further
term "S g^uv" where S is a scalar *field* (not a cosmological constant),
that is, if one were to migrate T^uv(Maxwell) --> T^uv(Maxwell) + S
g^uv, which would add a non-zero trace, what would that do to the class
of T^uv(Maxwell)? What class would T^uv(Maxwell) + S g^uv fall into?
Would it by chance move from the Segre type [(11),(1,1)] / Plebanski
type [2S-2T] into the [(111),1] / Plebanski type [3S-T] class of a
perfect fluid tensor? If not, what class would T^uv(Maxwell) + S g^uv
fall into?

Thanks.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com

tessel@um.bot

On Sat, 25 Mar 2006, Jay R. Yablon wrote:

> I am studying Stephani et al.'s "Exact Solutions of Einstein's Field
> Equations" and had a few question about the Algebraic classifications of
> the Energy tensors discussed in Chapter 5, which are still new to me.

Good. Although an approach using a coframe Ansatz is usually a much
better way to solve the EFE "to order", if that is your wish.

BTW, it will probably be helpful to first study a more elementary
eigenthing approach to studying the invariants of tensors. For example,
what is the characteristic polynomial of the contributions from perfect
fluid, dust, EM field, minimally coupled massless scalar field, and
combinations? The ingredients you need are Newton's rules for passing
between powersums of the roots of a polynomial (think eigenvalues) and
coefficients of the polynomial, and also an easy observation about
computing the traces of powers of a linear operator (think second rank
tensor written with "mixed indices") in index gymnastic formalism.

But make sure you understand how the spectral theorem changes in case of
indefinite signature; e.g. "diagonalization" of symmetric matrices will
typically fail. This is precisely why a more sophisticated approach is
neccessary here.

> Because of the Einstein equations, the Ricci R_uv tensor takes on the
> same algebraic type as the energy tensor T_uv to which it is linked.

Don't forget that in gtr, T_uv is proportional to G_uv, not to R_uv.
G_uv is the "tracereverse" of R_uv.

Trace reversal is an algebraic "duality" operation, popularized by
Einstein, in which given a tensor P_uv one produces a tensor Q_uv such
that the traces obey P=-Q (with generalizations to higher rank tensors).

> The non-null Maxwell tensor T^uv (Maxwell) is of the Segre type
> [(11),(1,1)] / Plebanski type [2S-2T]. It is also a function of the
> metric tensor g_uv and the field strength tensor F^uv, that is,
> T^uv(Maxwell) = T^uv(g_uv, F_uv)
>
> The Perfect fluid tensor T^uv(Euler) is of the Segre Type [(111),1] /
> Plebanski type [3S-T].

Which just means that possible contributions arising from the matter in
a perfect fluid and from the field energy-momentum of an EM field will
have different algebraic properties. Such contributions can be added to
yield legal stress-energy tensors in appropriate exact solutions of the
EFE (e.g. for charged dusts), although it may be difficult to find
examples of such models which "solve the EFE" simply by guessing.
Fortunately, systematic and elementary approaches are available, if
sometimes delicate or challenging to realize.

> My question is about compatibility between energy tensors. Suppose one
> wanted to somehow understand whether there exists an electrodynamic
> basis underlying a perfect fluid. That is, suppose one wanted to
> explore the possibility that T^uv(Euler) = T^uv'(g_uv, F_uv), similarly
> to the Maxwell tensor (I have used a prime ' to denote that the
> combination of fields T^uv'(g_uv, F_uv) may be different from that in
> T^uv(g_uv, F_uv) = T^uv(Maxwell)).

Not sure I understand what you mean by "electrodynamic basis underlying
a perfect fluid", but the contribution to the stress energy tensor of a
spacetime model which comes from the field/energy an EM field is for
example traceless, while the contribution from the mass-energy density,
momentum, and pressure of a fluid (perfect or otherwise) is not, so
certainly one cannot hope that an EM field -is- a perfect fluid or vice
versa.

> More generally, how precisely must the algebraic classes of two tensors
> match if one is to equate them in a valid manner to one another as well
> as to R^uv-(1/2)g^uvR?
>
> Finally, if one were to take T^uv(Maxwell) and augment it with a further
> term "S g^uv" where S is a scalar *field* (not a cosmological constant),

Contributions to the Lagrangian governing your spacetime model should be
associated with corresponding contributions to the net energy-momentum
tensor, and this is usually the best way to proceed.

For example, if you add a massless scalar field to your model, that adds
terms to the Lagrangian (describing the field itself and its interaction
with matter if any), and this gives rises in a "canonical" way to a new
contribution to the stress-energy tensor in your model. Similarly for a
Lagrangian describing a hypothetical field which has the nature of a
"variable Lambda", which seems to be what you have in mind.

In general, you should expect that even if various contributions from
various hypothetical "physical fields" in your model happen to have
particular Ricci types, the stress-energy tensor obtained by adding up
all these contributions will not. Consider adding nonzero symmetric and
nonzero antisymmetric matrices; the result is neither symmetric nor
antisymmetric.

Does this help?

"T. Essel"

Jay R. Yablon

>
>> I am studying Stephani et al.'s "Exact Solutions of Einstein's Field
>> Equations" and had a few question about the Algebraic classifications of
>> the Energy tensors discussed in Chapter 5, which are still new to me.
>
> Good. Although an approach using a coframe Ansatz is usually a much
> better way to solve the EFE "to order", if that is your wish.

Well, as you can see from my new draft at
http://home.nycap.rr.com/jry/Papers/...ce%20Paper.pdf, I
needed a good off-the shelf stationary, axially symmetric perfect fluid
solution which I found in equation (21.61). Chapter 5 was also very
helpful. I actually like this book better than others, because it is very
mathematical but not abstractly so, and my own personal approach is to try
to understand nature by understanding how the various mathematical
constructs we use to describe nature all fit together in the most
mutually-compatible manner. I also like that Stephani et al. did not think
it beneath them to use tensors which in the end is still where the rubber
meets the road.

>
> BTW, it will probably be helpful to first study a more elementary
> eigenthing approach to studying the invariants of tensors. For example,
> what is the characteristic polynomial of the contributions from perfect
> fluid, dust, EM field, minimally coupled massless scalar field, and
> combinations? The ingredients you need are Newton's rules for passing
> between powersums of the roots of a polynomial (think eigenvalues) and
> coefficients of the polynomial, and also an easy observation about
> computing the traces of powers of a linear operator (think second rank
> tensor written with "mixed indices") in index gymnastic formalism.
>
> But make sure you understand how the spectral theorem changes in case of
> indefinite signature; e.g. "diagonalization" of symmetric matrices will
> typically fail. This is precisely why a more sophisticated approach is
> neccessary here.
>
>> Because of the Einstein equations, the Ricci R_uv tensor takes on the
>> same algebraic type as the energy tensor T_uv to which it is linked.
>
> Don't forget that in gtr, T_uv is proportional to G_uv, not to R_uv.
> G_uv is the "tracereverse" of R_uv.

Yes, however, Stephani et al. do point out that a term proportional to g_uv
merely shifts all eigenvalues by the same amount.

>
> Trace reversal is an algebraic "duality" operation, popularized by
> Einstein, in which given a tensor P_uv one produces a tensor Q_uv such
> that the traces obey P=-Q (with generalizations to higher rank tensors).
>
>> The non-null Maxwell tensor T^uv (Maxwell) is of the Segre type
>> [(11),(1,1)] / Plebanski type [2S-2T]. It is also a function of the
>> metric tensor g_uv and the field strength tensor F^uv, that is,
>> T^uv(Maxwell) = T^uv(g_uv, F_uv)
>>
>> The Perfect fluid tensor T^uv(Euler) is of the Segre Type [(111),1] /
>> Plebanski type [3S-T].
>
> Which just means that possible contributions arising from the matter in
> a perfect fluid and from the field energy-momentum of an EM field will
> have different algebraic properties. Such contributions can be added to
> yield legal stress-energy tensors

Yes, Stephani et al. note this and point out however that it is not a
generally-used practice. But my question at the time was not in *adding*
two tensors in different algebraic classes but *setting them equal to one
another* and thereby constraining the properties of one or both, and whether
that was "legal." I have not seen this done before (though that does not
means it has not been done), but I have done just that in my paper and
carried out a full solution of the resulting sixteen simultaneous "mixed
index" equations.

in appropriate exact solutions of the
> EFE (e.g. for charged dusts), although it may be difficult to find
> examples of such models which "solve the EFE" simply by guessing.
> Fortunately, systematic and elementary approaches are available, if
> sometimes delicate or challenging to realize.
>
>> My question is about compatibility between energy tensors. Suppose one
>> wanted to somehow understand whether there exists an electrodynamic
>> basis underlying a perfect fluid. That is, suppose one wanted to
>> explore the possibility that T^uv(Euler) = T^uv'(g_uv, F_uv), similarly
>> to the Maxwell tensor (I have used a prime ' to denote that the
>> combination of fields T^uv'(g_uv, F_uv) may be different from that in
>> T^uv(g_uv, F_uv) = T^uv(Maxwell)).
>
> Not sure I understand what you mean by "electrodynamic basis underlying
> a perfect fluid", but the contribution to the stress energy tensor of a
> spacetime model which comes from the field/energy an EM field is for
> example traceless, while the contribution from the mass-energy density,
> momentum, and pressure of a fluid (perfect or otherwise) is not, so
> certainly one cannot hope that an EM field -is- a perfect fluid or vice
> versa.

Ah, now we are converging here. The tracelessness of the usual EM tensor
matching up against the non-zero trace of another tensor is, for example,
precisely the problem that Einstein had when he wrote his 1919 paper "Do the
gravitational fields play an essential role in the structure of the
Elementary particles of matter?" In that case, he was dealing with G_uv and
the Maxwell tensor. That is why we must use a more general tensor than the
EM tensor, and Kerrighan's work on uniqueness in the early 1980s presents
the perfect candidates, because they have a trace, but T^uv is still a
function only of F^uv and g_uv, i.e., T^uv=T^uv (g_uv) . The next step is
to get to g_uv = g_uv(F^uv), so that T^uv = T^uv (F^uv only). By doing this,
we overcome Kerrighan's limitation of the "variable lambda" you mention
below having to be a "constant lambda," because the metric tensor g_uv is
not longer fixed independently of the F^uv.

>
>> More generally, how precisely must the algebraic classes of two tensors
>> match if one is to equate them in a valid manner to one another as well
>> as to R^uv-(1/2)g^uvR?
>>
>> Finally, if one were to take T^uv(Maxwell) and augment it with a further
>> term "S g^uv" where S is a scalar *field* (not a cosmological constant),
>
> Contributions to the Lagrangian governing your spacetime model should be
> associated with corresponding contributions to the net energy-momentum
> tensor, and this is usually the best way to proceed.

Agreed.

>
> For example, if you add a massless scalar field to your model, that adds
> terms to the Lagrangian (describing the field itself and its interaction
> with matter if any), and this gives rises in a "canonical" way to a new
> contribution to the stress-energy tensor in your model. Similarly for a
> Lagrangian describing a hypothetical field which has the nature of a
> "variable Lambda", which seems to be what you have in mind.

Bingo, and very penetrating! That is just what I discuss in section 2 of
the paper linked above, and such a "variable Lambda" (actually the EM term E
dot B) is very central to the model I am working on, and I believe, a
physical field of very fundamental import the real physical universe as
well. It is, in my view -- and the paper show why in a quantitative
manner -- the foundation of gravitating matter.

>
> In general, you should expect that even if various contributions from
> various hypothetical "physical fields" in your model happen to have
> particular Ricci types, the stress-energy tensor obtained by adding up
> all these contributions will not. Consider adding nonzero symmetric and
> nonzero antisymmetric matrices; the result is neither symmetric nor
> antisymmetric.

Agreed. But again, the approach I am taking -- and again, it may take some
getting used to and I certainly want to make sure it is at most unusual but
not "illegal" -- is not to add tensor properties, but to set one tensor
equal to another which thereby constrains each one. In fact, I am setting
three tensors all to one another: 1) The Maxwell tensor with an extra
"variable Lambda" term based on Kerrighan, 2) the perfect fluid tensor with
rho = mu + p (i.e., specific enthalpy = 1), and 3) G^uv. The first
derivative of this equality is the Lorentz force law; the solution to
Einstein's equations for this equality yields the spacetime geometry of "a
perfect fluid which moves according to the Lorentz force law." Perhaps when
fully developed, we will find that such "a perfect fluid which moves
according to the Lorentz force law" actually has the properties of some real
physical entities already known in nature. With any luck, maybe these
properties will even have some aspects which are reminiscent of quantum
mechanics, but this last point is still speculative and so not raised in my
paper. I learned that from Igor last time through :-).

> Does this help?

Very much so, and I appreciate the time you have taken to offer serious and
helpful replies to several of my recent posts.

>
> "T. Essel"
>