- #1

notknowing

- 185

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter notknowing
- Start date

In summary, the stress-energy tensor is a geometrical object that can be found in any textbook on relativity. It is used to describe the amount of energy and momentum in a perfect fluid. In generalizing this tensor to curved spacetime, one replaces the flat-spacetime metric tensor with the metric tensor of curved spacetime, using the principle of minimal coupling. However, there is some ambiguity in this process and other approaches, such as conformal coupling, may also exist. This can lead to different forms of the stress-energy tensor in general relativity. The stress-energy tensor is usually given as a fact in textbooks, but it is important to understand the process of generalization and the potential for ambiguity.

- #1

notknowing

- 185

- 0

Physics news on Phys.org

- #2

pmb_phy

- 2,952

- 1

Why would you need to 'generalize' it? A tensor is a geometrical object which exists independantly of the presence of spacetime curvature. The definition of the tensor doesn't change when going from flat spacetime to curved spacetime. Can you please clarify this "generalizing" that you're referring to? Thanks.notknowing said:

Pete

- #3

notknowing

- 185

- 0

I'm afraid it is not so simple as you indicated.pmb_phy said:Why would you need to 'generalize' it? A tensor is a geometrical object which exists independantly of the presence of spacetime curvature. The definition of the tensor doesn't change when going from flat spacetime to curved spacetime. Can you please clarify this "generalizing" that you're referring to? Thanks.

Pete

In the book Gravity by J. Hartle, one can read on page 480 : "There is some ambiguity but little difficulty in generalizing stress-energy 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 stress-energy tensor in GR. If this is the case, why are certain stress-energy tensors (like the one of a perfect fluid) given as facts in most textbooks ? Could it just be wrong ?

- #4

- 10,330

- 1,508

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 space-time 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 stress-energy 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 stress-energy tensor. I don't know what Hartle has in mind, but I'll talk about some generalities anyway, in the next post.

- #5

- 10,330

- 1,508

Any curved manifold can be seen as "locally flat". In GR, in a local region, we can always replace the metric g

We can use the local Minkowskian metric to define the amount of energy and momentum (the energy-momentum 4-vector) 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 4-d manifold"?

The standard way of doing this is to represent a volume in a 4-d 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, one-forms, 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 one-forms. If you have MTW, you should have run across the visual image of one-forms 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/introduction/intro/intro.html

of three one-forms. It's actually an oriented volume. This means that volumes can be represented as three-forms, because the wedge product of three one-forms is a three-form.

Rather than work with the three-forms, 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/introduction/intro/node8.html#SECTION00026000000000000000

is a lot easier. Every three-form 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 one-forms and three-forms.

So, in conclusion - the stress-energy tensor simply takes a vector-valued volume element and converets it into the local energy-momentum 4-vector. It can therefore be regarded as the "density" of energy and momentum.

- #6

pmb_phy

- 2,952

- 1

Do you have a scanner? If so then please scan in that page and any other page which requires to be included for context. I find it hard to believe that Hartle is saying what you think he is. Thanks. You can either post the pages here or e-mail them to me.notknowing said:I'm afraid it is not so simple as you indicated.

In the book Gravity by J. Hartle, one can read on page 480 : "There is some ambiguity but little difficulty in generalizing stress-energy tensors to curved spacetime".

That applies to tensorThe 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. "

No ambiguity exists. A tensor can be defined as a multi-linear function of vectors and 1-forms to real numbers. This function does not depend on the components of the vectors/1-forms nor does it depend on the components of the tensor itself. You can evaluate the tensor in any system of coordinates to your liking. The locally Minkowski coordinates are usually the easiest since it has the same form as in flat time in globally Minkowski coordinates. As such the components of the tensor are well-defined making the tensor well-defined.In any case, there seems to be an ambiguity in the exact form of a stress-energy tensor in GR. If this is the case, why are certain stress-energy tensors (like the one of a perfect fluid) given as facts in most textbooks ? Could it just be wrong ?

Best wishes

Pete

ps - I just e-mailed 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.

Last edited:

- #7

pmb_phy

- 2,952

- 1

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

Now if someone can explain to me what he means when he said that ...Dear Pete: This is an oblique reference to the following issue:

Stress-energy tensors in curved spacetime can have terms in them

proportional to the spacetime curvature (the Einstein tensor for

instance). The spacetime curvature vanishes in flat spacetime. Thus

knowing the SE tensor in flat spacetime doesn't determine how much of

such curvature terms appear in generalizing to curved spacetime (the

ambiguity). On the other hand there is always the straightforward

generalization leaving out such terms (the little difficulty). I

agree it could have been said better.

Jim Hartle

Do you like this text? I'm thinking about picking up a new text on GR.

Best wishes

Pete

Last edited:

- #8

Stingray

Science Advisor

- 678

- 2

It is useful to think of the decomposition of a stress-energy 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 stress-energy 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 four-velocity [itex]u^{a}[/itex] is the timelike eigenvector.

[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

[tex]

T^{a}{}_{b} = \rho u^{a} u_{b} + p (\delta^{a}_{b} + u^{a} u_{b} )

[/tex]

Incidentally, the four-velocity [itex]u^{a}[/itex] is the timelike eigenvector.

Last edited:

- #9

notknowing

- 185

- 0

Thanks for contacting the author. I think it is very clear what he means (how could he explain it better ?). I'm curious about the opinion of the other experts (Pervect, ..)pmb_phy said:This is the response I got from the author

Now if someone can explain to me what he means when he said that ...

Do you like this text? I'm thinking about picking up a new text on GR.

Best wishes

Pete

- #10

- 10,330

- 1,508

The stress-energy 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 gauge-invariant, which doesn't always happen automatically.

http://bolvan.ph.utexas.edu/~vadim/Classes/2004f.homeworks/hw01.pdf

talks about this some.

- #11

pmb_phy

- 2,952

- 1

Yeah. I see what he means now.notknowing said:Thanks for contacting the author. I think it is very clear what he means (how could he explain it better ?). I'm curious about the opinion of the other experts (Pervect, ..)

Pete

- #12

selfAdjoint

Staff Emeritus

Gold Member

Dearly Missed

- 6,894

- 11

pmb_phy said:

Now if someone can explain to me what he means when he said that ...Dear Pete: This is an oblique reference to the following issue:

Stress-energy tensors in curved spacetime can have terms in them

proportional to the spacetime curvature (the Einstein tensor for

instance). The spacetime curvature vanishes in flat spacetime. Thus

knowing the SE tensor in flat spacetime doesn't determine how much of

such curvature terms appear in generalizing to curved spacetime (the

ambiguity). On the other hand there is always the straightforward

generalization leaving out such terms (the little difficulty). I

agree it could have been said better.

Jim Hartle

Do you like this text? I'm thinking about picking up a new text on GR.

Best wishes

Pete

This is clear enough. In flat spacetime there are obviously no terms proportional to the local curvature. In any given coordinate system you will have a bunch of zeros there. But in curved spacetime - curved spacetime whise curvature is a function of this very tensor! - those zeros will become non-zero amounts that depend on the curvature. And you can't tell from the flat version of the tensor which zeros are going to do what.

But look, the tensor multiplication with the metric tensor is what goes through, just replacing Minkowski's diagonal metric with the fiull tensor of pseudo-Riemannian geometry. That will automatically fix the energy-momentum tensor.

- #13

pmb_phy

- 2,952

- 1

We must have posted at the same time because I had already posted that I understood what the author meant. Thanks anyway selfadjoint.selfAdjoint said:This is clear enough. In flat spacetime there are obviously no terms proportional to the local curvature. In any given coordinate system you will have a bunch of zeros there. But in curved spacetime - curved spacetime whise curvature is a function of this very tensor! - those zeros will become non-zero amounts that depend on the curvature. And you can't tell from the flat version of the tensor which zeros are going to do what.

Aplication of the metric tensor to the stress-energy-momentum tensor will not change the tensor itself. Only the components in a given coordinate system. It can change the type of tensor though and perhaps this was what you were thinking. But normally we assign the same name to the tensor A^ub as we do to A^u_v.But look, the tensor multiplication with the metric tensor is what goes through, just replacing Minkowski's diagonal metric with the fiull tensor of pseudo-Riemannian geometry. That will automatically fix the energy-momentum tensor.

Best wishes

Pete

Last edited:

- #14

notknowing

- 185

- 0

pervect said:

The stress-energy 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 gauge-invariant, which doesn't always happen automatically.

http://bolvan.ph.utexas.edu/~vadim/Classes/2004f.homeworks/hw01.pdf

talks about this some.

I also guess that Hartle is working with the Lagrangian density formalism. After reading the various comments, I'm not sure what to think now. Is it the Lagrangian formalism only which leads to the ambiguity, while other approaches lead to a unique stress-energy tensor ?

- #15

Stingray

Science Advisor

- 678

- 2

notknowing said:I also guess that Hartle is working with the Lagrangian density formalism. After reading the various comments, I'm not sure what to think now. Is it the Lagrangian formalism only which leads to the ambiguity, while other approaches lead to a unique stress-energy tensor ?

Even if you just take the point of view I outlined (with eigenvalues/vectors), there can still be some ambiguity. I showed how to define the 4-velocity, density, principle pressures, etc. That's pretty much fixed.

But that may not completely specify the stress-energy tensor. For example, you might want to find the equations of motion of the system from [itex]\nabla_{a} T^{ab}=0[/itex]. The perfect-fluid 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 stress-energy tensor into a density and pressures. For example, it would look a little odd to describe long-range fields in that way (despite that it's possible). The "natural variables" which characterize the system may be related to the stress-energy 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 flat-spacetime Lagrangian to a curved-spacetime one.

- #16

samalkhaiat

Science Advisor

- 1,802

- 1,200

notknowing said:"the principle of minimal coupling"? why not "I'm afraid it is not so simple as you indicated.

In the book Gravity by J. Hartle, one can read on page 480 : "There is some ambiguity but little difficulty in generalizing stress-energy 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,maximal coupling" (is there such a thing? )

I just can not understand the reason for inventing such confusing name for "the equivalence principle"

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. "

This statement is a rephrased version of the equivalence principle which states:

Every non-inertial 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 "comma-goes-to-semicolon" 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 non-vanishing Riemann tensor). Here, in applying the

[tex]\partial_{a} \rightarrow \nabla_{a}[/tex]

rule of equivalence principle, one, sometimes, encounters factor-ordering problems.

I believe, the MTW explains the "Equivalence principle as a tool to mesh nongravitational laws with gravity". It also gives "rules-of-thumb" which resolve most factor-ordering problems. Anyway, one only needs to remember that interchanging covariant derivatives produces terms that couple to curvature.

The equivalence principle has stood the test of time both theoretically and experimentally.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.

regards

sam

Last edited:

- #17

Stingray

Science Advisor

- 678

- 2

samalkhaiat said:"the principle of minimal coupling"? why not "maximal coupling" (is there such a thing? )

I just can not understand the reason for inventing such confusing name for "the equivalence principle"

This statement is a rephrased version of the equivalence principle which states:

Every non-inertial 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 "comma-goes-to-semicolon" rule is the mathematical form of the equivalence principle.

Minimal coupling is not the same as the equivalence principle. It says that the form of an equation that people like to write down in SR is essentially the same as the form of that equation in GR. In other words, the equations involve the curvature as little as possible. This is just a rule-of-thumb.

The equivalence principle is also a rule-of-thumb (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 non-inertial 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

- 10,330

- 1,508

Stingray said:It is useful to think of the decomposition of a stress-energy 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]

What motivates this decomposition? To me this equation is requiring that the energy-momentum 4-vector is a multiple of the 4-velocity? I suspect I'm missing the point here :-(.

- #19

Stingray

Science Advisor

- 678

- 2

pervect said:What motivates this decomposition? To me this equation is requiring that the energy-momentum 4-vector is a multiple of the 4-velocity? I suspect I'm missing the point here :-(.

[itex]v^{a}[/itex] is any eigenvector. So there are actually four of them in any decomposition. If you normalize their magnitudes, you might get a unique set. But for perfect fluids, there will always be a 2-parameter ambiguity representing spatial rotations.

Anyway, the point of the decomposition is to extract intuitive information from the stress-energy tensor. As you mentioned, in the case where [itex]v^{a}=u^{a}[/itex], the eigenvalue equation is basically stating that the momentum is a multiple of the velocity. The proportionality "constant" is usually called [itex]\rho[/itex]. But if you don't know [itex]u^{a}[/itex] or [itex]\rho[/itex] beforehand, you can extract both of them uniquely from [itex]T^{ab}[/itex].

The spacelike eigenvectors represent the directions of principal stress. These are a carryover from the Newtonian treatment of (3x3) stress tensors. There, one often discusses the frame which diagonalizes the stress tensor. In other words, what frame removes all shear stresses (at a point)? The associated eigenvalues are the magnitudes of the stress in each of these directions.

In the end you get 4 unique and easily interpretable scalars. You also get 4 vectors, at least one of which is unique. Also, people sometimes state energy conditions in terms of the eigenvalues (e.g. Wald p. 219).

Last edited:

- #20

armandowww

- 78

- 0

In flat spacetime the energy tensor T_(ab) is defined as the conserved charges of the translation invariance, that's not valid in curved spacetime.

The general definiton of T_(ab) is simply the derivative of the matter part of the action S with respect to the metric g_(ab).

These words describe only a part of the question so...to be continued.

- #21

samalkhaiat

Science Advisor

- 1,802

- 1,200

Stingray said:The post I was replying to, defind "minimal couplind" by the substitutionMinimal coupling is not the same as the equivalence principle.

[tex]\partial \rightarrow \nabla[/tex]

I have no problem with this, because it is consistent with THE minimal coupling that we use in gauge field theories.

Now, CLASSIC textbooks on GR tell you that this substitution is nothing but the mathematical representation of the equivalence principle (EP).(see MTW P.386). So, Yes, defind this way, minimal coupling does have the same meaning as the EP.

In general, minimal coupling is an expression forformal simplicity. According to Einstein, formal simplicity is achievable bygeneral covariance(tensorial equations). But general covariance in GR is a mathematical statement about the EP.Therefore, formal simplicity (minimal coupling) is achievable by the above substitution(i.e. replacing the partial derivative by genuine tensor).

The word "coupling" comes from the fact that the covariant derivative contains an object (connection, gauge field) whichcouplesto matter fields. The word "minimal" is used to mean that such object (connection, gauge field) is thesimplistmathematical object that turns the mathematical formalisim into a physical theory.

In gauge field theories (GR is one of them), the minimal coupling is understood in terms of a local gauge principle (in GR this is the EP):

"a local frame can be found in which the gauge (gravitational) field vanishes"

So, whenever you use the term "minimal coupling" in GR, you are actually using different "name" for the equivalence principle and this is what I don't like.

Stated as bad as this the EP means the same thingIt says that the form of an equation that people like to write down in SR is essentially the same as the form of that equation in GR.

In other words, the equations involve the curvature as little as possible.

Curvature? A "curved" spacetime is conceivable only because the equivalence principle leads to it.

This is an insult to Einstein who described the EP as the happiest thought of his life. It is also an insult to Eotvos(and others) who tested the principle to a great accuracy.The equivalence principle is also a rule-of-thumb

Statement as simple as "(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.Local physics is Lorentz invariant" should not mean different thing to different people. People should not disagree on the meaning of "Inertial and gravitational masses are identical", or "It is impossible to speak of an absolute acceleration"

These three (and many more) statement are just different wording for the same principle, the equivalence principle: the following, frequently used, formulation of the EP, is due to Pauli;

For every sufficiently small spacetime region (i.e. so small that the space- and time-variation of gravity can be neglected) there always exists a coordinate system in which gravitation has no influence on physical processes[Lorentz frame]. In short, in a very small spacetime region every gravitational field can be transformed away [EP as local gauge principle].

Indeed, this "transforming away" is only possible because gravity imparts the same acceleration to all bodies; or, stated differently, because the gravitational mass is always equal to the inerial mass.

So, what are the "trivial" or/and inacurate aspects of Pauli's formulation of the EP?

If you do not understand the EP, then you have NO chance of understanding GR.

The fact that a gravitational field can be considered locally equivalent to an accelerated frame [the EP], implies that SR cann't be valid in an extended region where gravitational fields are present. A wider than Lorentz group is needed and all laws should be covariant under the most general coordinate transformation[general covariance].

"The principle of equivalence (which necessarily leads to the introduction of a curved spacetime), plus the assumption of general covariance, is"mostof what is needed to generate Einstein's general relativity

M. Carmeli

"Theory of Spinors"

What is it that gives us the right to

1) describe gravity in terms of the metric tensor of some "curved" spacetime?

2) couple matter to geometry? and,

3) give the metric tensor any physical meaning?

Well the answer always is "because of the EP".

Let me see what Pauli has to say about this in his book "Theory of relativity", OK, on Page150, I read:

"The generally covariant formlulation of the physical lawsacquires a physical contentonly through the principle of equivalence,in consequence of whichgravitation is describedsoleyby theg...only for this reasoncan thegbe described as physical quantities"

So you really need to understand the EP before you start to play with the action integral of Hilbert & Einstein.

The meaning of the EP becomes involved only when inertia is analysed in terms of Mach's principle. But who cares about Mach's, GR can live happily without it.

NON of the classic textbooks and papers on GR use the term "minimal coupling". So Who is "everyone"? and When did "everyone" agree on using this "minimal coupling" thing?It's better to use a term which everyone agrees on.

"You need to understand the meaning of "locally" in this statement. The only difference, between "actual" gravitational fields and the fields to which noninertial reference frames are equivalent, is the behavior at infinity. Actual gravitational field vanishes at infinity while noninertial fields(accelerations) don't. This is why the phrase "locally equivalent" appears in many different (but equivalent) formulation of the EP.Every non-inertial frame is locally equivalent to some gravitational field" is only true if you can ignore all derivatives of the metric of order 2

That's often possible, but it certainly isn't universal

The statement, that the word line is a geodesic, is invariant and therefore holds generally.

If that is not universally true, then astronauts would feel their weights and you do not need to fasten your seat-belt every time you step into a car.

regards

sam

Last edited:

- #22

Stingray

Science Advisor

- 678

- 2

samalkhaiat said:Now, CLASSIC textbooks on GR tell you that this substitution is nothing but the mathematical representation of the equivalence principle (EP).(see MTW P.386).

Synge said:Perhaps they speak of the Principle of Equivalence. If so, it is my turn to have a blank mind, for I have never been able to understand this Principle. Does it mean that the signature of the metric is +2 (or -2 if you prefer that convention)? If so, it is important, but hardly a Principle. Does it mean that the effects of a gravitational field are indistinguishable from the effects of an observer's acceleration? If so, it is false. In Einstein's theory, either there is a gravitational field or there is none, according as the Riemann tensor does not or does vanish. This is an absolute property; it has nothing to do with any observer's world-line. Space-time is either flat or curved, and in several places in this book I have been at considerable pains to separate truly gravitational effects due to the curvature of space-time from those due to curvature of the observer's world-line (in most ordinary cases the latter predominate). The Principle of Equivalence performed the essential office of midwife at the birth of general relativity, but, as Einstein remarked, the infant would never have got beyond its long-clothes had it not been for Minkowski's concept. I suggest that the midwife be now buried with appropriate honours and the facts of absolute space-time faced.

The EP was of fundamental importance to Einstein, as it guided his thoughts to the right direction. But we have much better ways of thinking about things nowadays. It is mainly of historical interest at this point, although it can still be helpful as a rule-of-thumb. I don't know of any field of science besides relativity where people still try to cling to the methods of its founders as gospel. Einstein was certainly brilliant, but progress has been made.

Anyway, I'll agree that MTW's description of the EP is identical to minimal coupling. But it is not the only one floating around.

Curvature? A "curved" spacetime is conceivable only because the equivalence principle leads to it.

Hardly. You can write down theories of gravity on curved spacetimes which violate the EP in (say) the Newtonian limit. Just because the EP was the historical motivation for curved spacetime doesn't mean that it can't be useful if it were false.

These three (and many more) statement are just different wording for the same principle, the equivalence principle: the following, frequently used, formulation of the EP, is due to Pauli;

For every sufficiently small spacetime region (i.e. so small that the space- and time-variation of gravity can be neglected) there always exists a coordinate system in which gravitation has no influence on physical processes[Lorentz frame]. In short, in a very small spacetime region every gravitational field can be transformed away [EP as local gauge principle].

Indeed, this "transforming away" is only possible because gravity imparts the same acceleration to all bodies; or, stated differently, because the gravitational mass is always equal to the inerial mass.

Pauli's version of the principle is wrong, or at best a tautology. Consider a rotating particle. Even in the test-mass limit, such an object falling freely will have a worldline with a nonzero 4-acceleration. That doesn't happen in flat spacetime, so there is already a difference. Measure the stresses in a small solid material (via changes in proper volume, optical properties, etc). Put some weights and viscous fluid in a jar, and have a gravitational wave pass through. Measure the temperature change in the fluid due to those weights moving around. Let these experiments fall freely from different places and see if identical results are obtained. According to GR, that won't happen. But GR does satisfy the EP, doesn't it?

Huh? The metric can be made Minkowski at a point, and its first derivatives can be made to vanish at the same time. But that's it. You still have second and higher derivatives of the metric no matter what coordinates you choose. Those do directly couple to some physical processes. They also comprise what most relativists call the "gravitational field" (the Riemann tensor or occasionally the Weyl tensor). If the ability to transform the metric into a nice form is the "EP," that's really just a simple mathematical result on pseudo-Riemannian manifolds. Why not state it as such.You need to understand the meaning of "locally" in this statement. The only difference, between "actual" gravitational fields and the fields to which noninertial reference frames are equivalent, is the behavior at infinity. Actual gravitational field vanishes at infinity while noninertial fields(accelerations) don't. This is why the phrase "locally equivalent" appears in many different (but equivalent) formulation of the EP.

The statement, that the word line is a geodesic, is invariant and therefore holds generally.

If that is not universally true, then astronauts would feel their weights and you do not need to fasten your seat-belt every time you step into a car.

I never mentioned geodesics... But geodesic motion isn't universal in GR anyway. Of course it's an excellent approximation in the cases you mentioned (and many others).

Last edited:

- #23

Chris Hillman

Science Advisor

- 2,355

- 10

notknowing said:In the book Gravity by J. Hartle, one can read on page 480 : "There is some ambiguity but little difficulty in generalizing stress-energy tensors to curved spacetime". The way to generalize seems to be described by "the principal of minimal coupling" ... Now, I'm not sure whether this "principal of minimal coupling" is to be understood as a recipe (which works most of the time), or as a strict mathematical principle.

Sounds like you already indirectly received a response from Hartle, so that you know that the short answer is: "the former".

notknowing said:I found some other internet references where one mentions that there exist other approaches too, such as "conformal coupling"

Well, be careful of InterNet sources, including this one. Textbooks are more reliable. This might be the exception which proves the rule, since you obtained an answer to your question from Hartle! Anyway, I just wanted to point out that the principle of minimal coupling is discussed in section 16.3 of MTW. In particular, the connection with "factor ordering" arises because interchaning covariant derivatives introduces a term involving the Riemann tensor (in fact, this is taken as the definining property of the Riemann tensor in many textbooks). In Box 16.1, example1 should answer your original question, and example2 discusses the problem of determining the correct contribution from the EM field to the stress-energy tensor.

One place where curvature coupling can arise is in theories in which one postulates a field equation for a scalar field. In Hawking and Ellis you can find a discussion of the Lagrangian, the field equation, and the stress-energy contribution for a massless or massive scalar field. Although they don't bother to mention this, their example is minimally coupled. You can modify the Lagrangian for the massless scalar field to obtain a theory which is conformally invariant (like Maxwell's theory of EM) and this is called "conformal coupling" of the scalar field to the gravitational field. For example, this changes the field equation from the standard (curved spacetime) wave equation [tex]\Box h = 0[/tex] to [tex]\Box h = \frac{h}{6} \, R[/tex]

samalkhaiat said:The post I was replying to, defind "minimal couplind" by the substitution

LaTeX graphic is being generated. Reload this page in a moment.

I have no problem with this, because it is consistent with THE minimal coupling that we use in gauge field theories.

Now, CLASSIC textbooks on GR tell you that this substitution is nothing but the mathematical representation of the equivalence principle (EP).(see MTW P.386). S

Oh dear, I think you might have misunderstood MTW. In their customary helical approach, they say something naive in section 16.2 and immediately provide an important caveat in section 16.3. Their point, I think, was actually that in some situations, the EP should be supplemented by NON-minimal coupling--- but that this possibility is unlikely to arise except in Track 3! :-/

Chris Hillman

- #24

notknowing

- 185

- 0

Chris Hillman said:Sounds like you already indirectly received a response from Hartle, so that you know that the short answer is: "the former".

Well, be careful of InterNet sources, including this one. Textbooks are more reliable. This might be the exception which proves the rule, since you obtained an answer to your question from Hartle! Anyway, I just wanted to point out that the principle of minimal coupling is discussed in section 16.3 of MTW. In particular, the connection with "factor ordering" arises because interchaning covariant derivatives introduces a term involving the Riemann tensor (in fact, this is taken as the definining property of the Riemann tensor in many textbooks). In Box 16.1, example1 should answer your original question, and example2 discusses the problem of determining the correct contribution from the EM field to the stress-energy tensor.

One place where curvature coupling can arise is in theories in which one postulates a field equation for a scalar field. In Hawking and Ellis you can find a discussion of the Lagrangian, the field equation, and the stress-energy contribution for a massless or massive scalar field. Although they don't bother to mention this, their example is minimally coupled. You can modify the Lagrangian for the massless scalar field to obtain a theory which is conformally invariant (like Maxwell's theory of EM) and this is called "conformal coupling" of the scalar field to the gravitational field. For example, this changes the field equation from the standard (curved spacetime) wave equation [tex]\Box h = 0[/tex] to [tex]\Box h = \frac{h}{6} \, R[/tex]

Oh dear, I think you might have misunderstood MTW. In their customary helical approach, they say something naive in section 16.2 and immediately provide an important caveat in section 16.3. Their point, I think, was actually that in some situations, the EP should be supplemented by NON-minimal coupling--- but that this possibility is unlikely to arise except in Track 3! :-/

Chris Hillman

Hi Chris, thanks for all your comments. I noticed now that you commented on some previous topics as well (I still have to check out some ..)

The stress-energy tensor is a mathematical object that describes the distribution of energy and momentum in a particular region of curved spacetime. It accounts for the effects of both matter and non-matter sources, such as gravitational fields, on the curvature of spacetime.

The stress-energy tensor can be calculated using Einstein's field equations of general relativity, which relate the curvature of spacetime to the distribution of matter and energy. It involves complicated mathematical equations and is often done using computer simulations.

The stress-energy tensor is a crucial component of general relativity, as it allows us to understand how matter and energy interact with the fabric of spacetime. It is essential for understanding the effects of gravity on the motion of objects and the structure of the universe.

The stress-energy tensor is a more general concept than the energy-momentum tensor. While the energy-momentum tensor only accounts for the energy and momentum of matter sources, the stress-energy tensor accounts for the effects of all sources, including non-matter sources like gravitational fields.

Yes, the stress-energy tensor is a fundamental tool for predicting the behavior of matter and energy in curved spacetime. By using Einstein's field equations to calculate the stress-energy tensor, we can make accurate predictions about the motion of objects, the bending of light, and the overall structure of the universe.

- Replies
- 11

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 13

- Views
- 1K

- Replies
- 33

- Views
- 3K

- Replies
- 57

- Views
- 2K

- Replies
- 10

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 186

- Views
- 8K

- Replies
- 6

- Views
- 3K

- Replies
- 11

- Views
- 1K

Share: