# The energy-momentum tensor and the equivalence principle

1. Apr 26, 2005

### hellfire

Is it correct that the only way to have a theory of gravitation that fulfills the equivalence principle is to make use of a tensor as the source of gravity (and not a scalar or a vector, for example)? How can this be proven?

2. Apr 26, 2005

### Garth

If gravity is to be described by the Einstein tensor Gab then its source must be a tensor as well.

Most tests of GR to date are in vacuo, that is all they are testing is
Gab = 0 ; it seems to work pretty well!

Garth

3. Apr 26, 2005

### dextercioby

Dirac's short book gives a nice insight to this subject:couplings of gravity and matter.No wonder,Dirac was a field theorist,like Pauli,Feynman,Weinberg...

Daniel.

4. Apr 26, 2005

### pmb_phy

A scalar and a vector are both tensors. From what I recall there is a theory by Dicke (Brans too???) which is a relativistic theory of gravity which is consistent with the equivalence principle.

Pete

5. Apr 27, 2005

### Garth

Yes, the Brans Dicke theory takes the Einstein field equation and adds a scalar field coupled to the (rest) mass density of the universe that endows particles with inertial mass. It retains the equivalence principle and so the effect of this scalar field is to vary the 'Gravitational 'constant''. Although many attempts to integrate QT with GR would like such a scalar field the BD theory is not verified observationally in solar system experiments; the scalar field perturbs space-time.

For information SCC modifies this theory by allowing particle masses to vary (with gravitational potential energy) and G then becomes observationally constant. It breaks the equivalence principle but nevertheless is consistent with solar system tests.

Garth

Last edited: Apr 27, 2005
6. Apr 27, 2005

### hellfire

As far as I know, scalar (rank 0 tensor) theories of gravity in which the source of gravity is the trace of the energy-momentum tensor do not fulfill the equivalence principle, since light does not couple to gravity (the trace of the electromagnetic energy-momentum tensor is zero). The question is whether a vector (rank 1 tensor) theory of gravity may fulfill the equivalence principle or whether only theories in which the source of gravity is at least a rank 2 tensor do.

7. Apr 27, 2005

### Chronos

There is no way to avoid a tensor description of gravity geometrically.

8. Apr 27, 2005

### hellfire

That's the kind of answer I am looking for… but with a proof.

9. Apr 27, 2005

### pmb_phy

You're incorrectly expecting proof where there is only postulate. The principle of general covariance requires the laws of nature to be the same in all coordinate systems. We give the name "tensor" to those objects which satisfy this property of covariance.

Pete

10. Apr 27, 2005

### hellfire

But a theory in which gravity couples to the trace of the energy-momentum tensor would be also covariant, as the trace is scalar (a rank 0 tensor). But it does not fulfill the equivalence principle.

11. Apr 27, 2005

### pmb_phy

Since when????????????????????????????

Pete

12. Apr 27, 2005

### hellfire

I am not aware of any error in what I wrote, but If I wrote something wrong, please correct me. That's the best way for me to learn.

13. Apr 27, 2005

### Chronos

What Pete said - GR treats spacetime as a four dimensional manifold. To describe the geometry of such a manifold without introducing frame dependence, you must use a rank 2 tensor.

14. Apr 27, 2005

### pmb_phy

That is not true. The Branse Dicke theory treats spacetime as a 4-d manifold and the description of the geometry is not frame dependant. However the field equations need not be a second rank tensor (e.g. Brans Dicke).

Pete

15. Apr 27, 2005

### Garth

Hmmm... The Brans Dicke theory has two separate field equations and an equation of state, The gravitational field equation certainly is an equation of second rank tensors, the scalar field equation is an equation in which each term is a scalar, as is the equation of state, however, that is no different to GR, which also requires a 'scalar' equation of state. [Though no scalar field equation]

Garth

16. Apr 27, 2005

### pmb_phy

What are these two seperate field equations you speak of??

Note: I made an error above. The trace can't be the source of gravity because for a beam of directed light the trace is zero and since mass is equivalent to energy the trace can't be a source.

I believe that in the Brans Dicke theory there is a scalar field but the field equation is second rank. The d'Lanbertian of the scalar field is proportional to the energy momentum tensor in that theory.

Is that correct pervect? You seem to know more about it than I do.

Pete

17. Apr 27, 2005

### dextercioby

I definitely know it's Jean le Rond d'Alembert and,consequently,the operator's name is "d'Alembertian".

Daniel.

18. Apr 27, 2005

Staff Emeritus
Is that scalar, in the equation or the Lagrangian, multiplied by something (metric tensor?) to bring it up to second rank and make the whole thing homogeneous? As you know that is how the cosmological constant is brought into the field equation in GR.

19. Apr 28, 2005

### Garth

Pete - As I explained - (although obviously rather obscurely!) The two BD field equations are:
1. The gravitational field equation; Einstein's with G replaced by Phi-1 and the stress-energy-momentum tensor of the scalar field TPhi ab added to that of normal matter-energy. The result is a homogeneous second rank tensor equation.

2. The scalar field equation; The d'Alembertian of Phi coupled to the trace of the matter stress-energy-momentum tensor. In this equation each term is a scalar and therefore selfAdjoint homogeneous.

Garth

20. Apr 28, 2005

### hellfire

I am sorry but I still don’t get it, so please be patient with me.

A theory in which gravity would couple only to the trace of the energy-momentum tensor (such a theory was considered by Einstein before general relativity):

$$\square \phi = - 4 \pi G T_{\mu}^{\mu}$$

With $$\inline g_{\mu \nu}$$ being diagonal with $$\inline \phi$$ or $$\inline - \phi$$ as diagonal elements...

a)...is a covariant theory, isn’t it?

b) but, however, it does not satisfy the equivalence principle, since light would not couple to a gravitational field. A photon would not be redshifted or blueshifted in a gravitational field, but it would be redshifted or blueshifted when emitted from an accelerated frame.

If all above is correct, what are the reasons, if any, for the need of having a rank 2 tensor in order to satisfy the equivalence principle?

Last edited: Apr 28, 2005
21. Apr 28, 2005

### Stingray

Yes. As a side note, almost any theory can be put into covariant form. Whether it looks nice when written that way is another question.

Before saying that, you have to prescribe how matter couples to the field. In GR, this is automatic (due to the Bianchi identities). Here, you have to introduce additional postulates.

I'll assume for simplicity that you want light to move along null geodesics. If so, then the coupling of $$\phi$$ to $$g_{\mu \nu}$$ automatically implies that light is affected by gravity. The only difference is that electromagnetic field is not a (direct) source for the gravitational one. I'm sure there would be different predictions for light bending etc., but I don't want to work them out.

Also, to nitpick a little, the metric should have components like $$1+ \phi$$, not $$\phi$$ all by itself. Otherwise, spacetime wouldn't be well-defined

There aren't any. The statements that scalar and vector gravity don't work are really saying that specific theories with simple Lagrangians have been found to disagree with experiment. That doesn't mean that you couldn't make up another Lagrangian that would pass all the tests.

Also, any theory can be written in an infinite number of different ways. You shouldn't put too much emphasis on the most popular one. For example, Ashtekar's formulation of GR looks completely different from Einstein's/Hilbert's, but it's the same thing.

Even if you wanted to stay with something of the form $$G_{\mu \nu} = (\ldots)$$, you can still come up with plenty of source terms other than the (standard) stress-energy tensor.

22. Apr 28, 2005

### hellfire

Thank you Stingray. I have the impression I was mixing two things. The field equation which relates the gravitational field to the energy-momentum tensor makes only a statement about how the gravitational field is produced by matter, but makes no statement about the way how matter or other fields behave inside a gravitational field. For that, additional conditions are necessary (i.e. the Lagrangian for the matter fields has to be defined). This second aspect is relevant for the equivalence principle, but not the first.

23. Apr 28, 2005

### Garth

Also Black Holes would not be coupled to a gravitational field. As the matter of a collapsed star became degenerate and relativistic its gravitational field would disappear!
Garth

Last edited: Apr 28, 2005
24. Apr 28, 2005

### pmb_phy

Thanks for the converasation folks. For the most part I think I'll have to nod off on the Brans-Dicke stuff at this point. At least for now. I've never rigorously learned it up to this point and as I've read it in Wienberg the last few days I've made mistakes in the reading. That's not all that uncommon for the average gent but it appears that the old peepers are now unable to read a book without glasses.

Please note that there are two different symbols used to express the D'Alambertion operator. One is a little box and the other is the square of a little box.

Pete