Register to reply 
Stressenergy tensor in curved spacetime 
Share this thread: 
#1
Oct2006, 05:48 AM

P: 185

In any textbook on relativity, one finds the classical expression for the stressenergy tensor of a perfect fluid. In generalizing this tensor to curved spacetime, one just replaced the flatspacetime metric tensor by the metric tensor of curved spacetime. It seems logical to do so, but in my view there is no certainty that this is indeed the correct generalisation. Maybe the generalisation could be defined differently. So, how can we know for sure that we use the correct form of the stressenergy tensor in Einsteins Field equations ? Does there exist experimental evidence that we have indeed the correct tensor ? Also for the stressenergy tensor of the electromagnetic field, one could pose the same questions.



#2
Oct2006, 07:41 AM

P: 2,954

Pete 


#3
Oct2006, 10:41 AM

P: 185

In the book Gravity by J. Hartle, one can read on page 480 : "There is some ambiguity but little difficulty in generalizing stressenergy tensors to curved spacetime". The way to generalize seems to be described by "the principal of minimal coupling" which is briefly described in wikipedia : "Also, under the principle of minimal coupling, the physical equations of special relativity can be turned into their general relativity counterparts by replacing the Minkowski metric (ηab) with the relevant metric of spacetime (gab) and by replacing any partial derivatives with covariant derivatives. " Now, I'm not sure whether this "principal of minimal coupling" is to be understood as a recepy (which works most of the time), or as a strict mathematical principle. I found some other internet references where one mentions that there exist other approaches too, such as "conformal coupling", though I must admit that I don't understand this. In any case, there seems to be an ambiguity in the exact form of a stressenergy tensor in GR. If this is the case, why are certain stressenergy tensors (like the one of a perfect fluid) given as facts in most textbooks ? Could it just be wrong ? 


#4
Oct2006, 01:42 PM

Emeritus
Sci Advisor
P: 7,660

Stressenergy tensor in curved spacetime
I'm not sure what Hartle is getting at either.
I've seen the "minimal substitution rule" described as follows (Wald:70) [tex]\eta_{ab} \rightarrow g_{ab} \hspace{.5 in} \partial_a \rightarrow \nabla_a [/tex] I.e. one replaces the Minkowski metric with the spacetime metric, and ordinary partial derivatives with the covariant derviatve. This process usually works, but not always, because covariant derivatives don't in general commute. The above process doesn't guarantee the correct ordering. I could dig up an example of where it fails if there is further interest, but I'm not sure I want to go that deep yet. Unfortunately, if covariant differentiation has anything to do with the stressenergy tensor, I'm not aware of the connection. It would be worthwhile looking at Hartle to see if his notion of "minimal substitution" is the same as Wald's and to see what he has to say about it to gain some illumination on what he's trying to say. Onto the stressenergy tensor. I don't know what Hartle has in mind, but I'll talk about some generalities anyway, in the next post. 


#5
Oct2006, 02:12 PM

Emeritus
Sci Advisor
P: 7,660

The stressenergy tensor
Any curved manifold can be seen as "locally flat". In GR, in a local region, we can always replace the metric g_{ab} with a Minkowskian metric [tex]\eta_{ab}[/tex]. This ties in with the principle of "minimal substitution" as mentioned earlier, so hopefully it's the sense that Hartle hand in mind. We can use the local Minkowskian metric to define the amount of energy and momentum (the energymomentum 4vector) contained in a small volume element at any given time by the rules of special relativity in this locally flat section of the manifold. I'm not aware of any ambiguities here. The volume elements will be very small, so little "gotchas" regarding the relativity of simultaneity in a volume element shouldn't be an issue. The metric will naturally define a local coordinate system which can be used to give the specific components of the energy and momentum. This will in general not be an orthonormal coordinate system of course. The remaining issues boil down to  "how do we naturally specify a volume element on a 4d manifold"? The standard way of doing this is to represent a volume in a 4d manifold as a vector  a vector orthogonal to the volume element. Perhaps this is ambiguous, but I'm not aware of how. It can be justified a bit more formally in terms of geometric algebra. Note: the following is optional and a bit advanaced, but I think it's interesting. If you are already familiar with vectors and their duals, oneforms, it can be quite illuminating. But the whole purpose of this section of the post is only to discuss in more detail how we can represent a volume element by the vector orthogonal to it. In geometric algebra, one can represent a volume element as the interior region of a parallelpiped formed by three vectors. One can also think of it as the interior region of three oneforms. If you have MTW, you should have run across the visual image of oneforms as "stacks of plates'. The idea of a volume element can be expressed as the geometric product, also known as the wedge product http://www.mrao.cam.ac.uk/~clifford/...tro/intro.html of three oneforms. It's actually an oriented volume. This means that volumes can be represented as threeforms, because the wedge product of three oneforms is a threeform. Rather than work with the threeforms, we work with their duals (this is the Hodges dual). There is a long complicated discussion in MTW that's not very intuitive, the geometric picture of a dual as described in http://www.mrao.cam.ac.uk/~clifford/...00000000000000 is a lot easier. Every threeform has a unique "hodges dual', which is an ordinary one form. Thus the "natural" way to represent volumes is with a one form (or equivalently, a vector). This is the Hodges dual of the three form. Note that there are two types of duality here  there is a duality between one forms and vectors, this is a different sort of duality than the Hodges duality, which is the duality between oneforms and threeforms. So, in conclusion  the stressenergy tensor simply takes a vectorvalued volume element and converets it into the local energymomentum 4vector. It can therefore be regarded as the "density" of energy and momentum. 


#6
Oct2006, 02:58 PM

P: 2,954

Best wishes Pete ps  I just emailed the author for clarification. Most authors are often willing to respond and clarify. Is this a text which is in depth (i.e. tensors, differential geometry, etc.)? I think I may pick up a copy. 


#7
Oct2006, 05:15 PM

P: 2,954

This is the response I got from the author
Do you like this text? I'm thinking about picking up a new text on GR. Best wishes Pete 


#8
Oct2006, 11:03 PM

Sci Advisor
P: 674

It is useful to think of the decomposition of a stressenergy tensor into density and pressure terms as an eigenvalue problem. In particular, look at the solutions of
[tex] T^{a}{}_{b} v^{b} = \lambda_{v} v^{a} [/tex] There will be 4 mutually orthogonal eigenvectors. For reasonable matter, one will be timelike, and the others spacelike. Now the eigenvalue associated with the timelike eigenvector is just the (negative of) the density [itex]\rho[/itex]. The eigenvectors associated with the three spacelike eigenvectors are called the principle pressures (or stresses). A perfect fluid is defined to be an object where all principle pressures are identical. Calling this [itex]p[/itex], it's easy to show that the only way of writing a stressenergy tensor with these properties is [tex] T^{a}{}_{b} = \rho u^{a} u_{b} + p (\delta^{a}_{b} + u^{a} u_{b} ) [/tex] Incidentally, the fourvelocity [itex]u^{a}[/itex] is the timelike eigenvector. 


#9
Oct2106, 02:41 AM

P: 185




#10
Oct2106, 04:34 AM

Emeritus
Sci Advisor
P: 7,660

I'm wondering about Hartle's approach. Is he perhaps deriving the stressenergy tensor from a Lagrangian density formalism? If he is, this might give some insight into his comments. There's nothing in particular in his resonse to indicate if he is doing this  but that would be one logical way to define energy and momentum.
The stressenergy tensor one gets via this procedure isn't unique, one can add a divergence term to it. I believe that this divergence term is usually used to make the tensor symmetric and gaugeinvariant, which doesn't always happen automatically. http://bolvan.ph.utexas.edu/~vadim/C...works/hw01.pdf talks about this some. 


#11
Oct2106, 07:51 AM

P: 2,954

Pete 


#12
Oct2106, 08:34 AM

Emeritus
PF Gold
P: 8,147

But look, the tensor multiplication with the metric tensor is what goes through, just replacing Minkowski's diagonal metric with the fiull tensor of pseudoRiemannian geometry. That will automatically fix the energymomentum tensor. 


#13
Oct2106, 08:51 AM

P: 2,954

Best wishes Pete 


#14
Oct2106, 09:11 AM

P: 185




#15
Oct2106, 11:06 AM

Sci Advisor
P: 674

But that may not completely specify the stressenergy tensor. For example, you might want to find the equations of motion of the system from [itex]\nabla_{a} T^{ab}=0[/itex]. The perfectfluid expression isn't quite sufficient for that. There are more variables that you need to solve for than there are equations. So you need to bring in an "equation of state" relating the pressure to the density. More complicated materials naturally need much more involved constitutive relations, and it may be natural to write these in terms of (say) the Riemann tensor, some potentials, etc. You also might not want to decompose the stressenergy tensor into a density and pressures. For example, it would look a little odd to describe longrange fields in that way (despite that it's possible). The "natural variables" which characterize the system may be related to the stressenergy tensor in a complicated way. In general, most people prefer to define the behavior of an object through a Lagrangian formalism. You get a different material for each action you write down, but there isn't necessarily a unique way of going from a flatspacetime Lagrangian to a curvedspacetime one. 


#16
Oct2106, 06:38 PM

Sci Advisor
P: 924

[QUOTE=notknowing]
I just can not understand the reason for inventing such confusing name for "the equivalence principle" Every noninertial frame is locally equivalent to some gravitational field. Or, At every point in an arbitrary gravitational field one can choose a locally inertial frame in which the laws of physics take the same form as in SR. So, the "commagoestosemicolon" rule is the mathematical form of the equivalence principle. Note that, [tex]\eta_{ab} \rightarrow g_{ab}[/tex] is nothing but a switching between referece frames. These frames are connected by smooth transformations. So, as long as the two metrics have the same signature, the "generalization" [tex]T^{ab..}(\eta) \rightarrow T^{ab..}(g)[/tex] is unique for any tensor T. Gravity enters through, and only through, the covariant derivative of curved spacetime (i.e. spacetime with nonvanishing Riemann tensor). Here, in applying the [tex]\partial_{a} \rightarrow \nabla_{a}[/tex] rule of equivalence principle, one, sometimes, encounters factorordering problems. I believe, the MTW explains the "Equivalence principle as a tool to mesh nongravitational laws with gravity". It also gives "rulesofthumb" which resolve most factorordering problems. Anyway, one only needs to remember that interchanging covariant derivatives produces terms that couple to curvature. regards sam 


#17
Oct2106, 07:59 PM

Sci Advisor
P: 674

The equivalence principle is also a ruleofthumb (there is no precise nontrivial definition which is known to be correct). But it has so much historical baggage associated with it that it means different things to almost everyone. Some of those are similar to minimal coupling, but many are not. It's better to use a term which everyone agrees on. "Every noninertial frame is locally equivalent to some gravitational field" is only true if you can ignore all derivatives of the metric of order 2 or higher. That's often possible, but it certainly isn't universal. 


#18
Oct2106, 08:39 PM

Emeritus
Sci Advisor
P: 7,660




Register to reply 
Related Discussions  
The stress energy tensor again  Special & General Relativity  9  
Stressenergy tensor of a wire under stress  Special & General Relativity  178  
StressEnergy Tensor  Special & General Relativity  17  
Energy Tensor Divergences and Gauss / Stokes in Curved Spacetime  General Physics  0  
I can't see how stressenergy tensor meets the minumum tensor requirement  Special & General Relativity  4 