Diagonal Matrix of Stress-Energy Tensor: Why?

In summary, the conversation discusses the diagonal matrix in all frames and its properties. It is mentioned that the diagonal elements of a diagonal matrix are the eigenvalues and that in the MCRF frame, the stress energy tensor of a continuum of massive particles with no pressure is diagonal. However, in any other frame, the stress energy tensor is non-diagonal. The conversation also discusses the case of a perfect fluid and its isotropic pressure. It is finally stated that the stress energy tensor is not always diagonal, as in the case of a dust model where it depends on the medium's rest frame.
  • #1
GR191511
76
6
I came across a statement in《A First Course in General Relativity》:“The only matrix diagonal in all frames is a multiple of the identity:all its diagonal terms are equal.”Why?I don’t remember this conclusion in linear algebra.The preceding part of this sentence is:Viscosity is a force parallel to the interface between particles.Its absence means that the forces should always be peependicular to the interface,i.e.that ##T^i{^j}####\;## should be zero unless i=j.This means that ##T^i{^j}####\;## should be a diagonal matrix.Moreover,it must be diagonal in all MCRF frame,since“no viscosity”is a statement independent of the spatial axes.(T is stress-energy tensor)
 
Physics news on Phys.org
  • #2
Have you tried to verify this, by say, doing a Lorentz boost in the x direction?
 
  • #3
I guess, "MCRF" is an acronym synonymous with "LRF" (local rest frame of the fluid cell), because only than it's a correct statement about a perfect fluid energy-momentum-stress tensor. In this frame it must necessarily be of the form ##(T^{\mu \nu})=\mathrm{diag}(\epsilon,P,P,P)##. Here ##\epsilon## is the energy density and ##P## the pressure as measured in this local restframe of the fluid cell. This can be written in the form (using the west-coast signature (+---))
$$T{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu} - P \eta^{\mu \nu}$$
in a local inertial frame, where the four-velocity of the fluid element is ##u^{\mu}##. So finally in an arbitrary frame you have
$$T^{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu}-P g^{\mu \nu}.$$
Note that ##\epsilon## and ##P## are scalars since they are defined in a specific "preferred" frame of reference, i.e., the local rest frame of the fluid cell.
 
  • Like
Likes GR191511
  • #4
Momentarily comoving rest frame
 
  • Like
  • Informative
Likes GR191511, dextercioby and vanhees71
  • #5
GR191511 said:
The only matrix diagonal in all frames is a multiple of the identity: all its diagonal terms are equal.
The diagonal elements of a diagonal matrix are the eigenvalues. If it is diagonal in all frames it means that any basis of the vector spaces will consist of eigen vectors. If you have two different eigenvalues, say ##\alpha\not=\beta## and ##v## and ##w## are corresponding eingenvectors, then choose a basis with first basis vector ##v+w##. When you apply the matrix to it you get ##\alpha v+\beta w##, which is not a multiple of ##v+w##.
 
  • Like
Likes GR191511 and vanhees71
  • #6
I'm not sure what the author is trying to say. Consider a stress energy tensor of a continuum of massive particles with some density ##\rho## and no pressure. Then in the MCRF frame, we'll have.

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

This is diagonal, but it'll be non-diagonal if you boost it - i.e. in any other frame than the MCRF frame.

Now let's consider a case with pressure.

If the whole stress energy tensor is T[0:3, 0:3], the submatrix representing pressure, also called the stress tensor, will be T[1:3, 1:3]. If the pressure is isotropic, the pressure will be diagonal under rotational transformations. But if the pressure is not isotropic, I.e. if we have

$$\begin{bmatrix} p1 & 0 & 0 \\ 0 & p2 & 0 \\ 0 & 0 & p3 \end{bmatrix}$$

where p1, p2, and p3 are not all equal, a rotational transformation will make the stress tensor non-diagonal.

A perfect fluid, which was discussed in a previous post, will always have an isotropic pressure by definition.

Note that we have not considered all possible casses of the stress energy tensor, such as those containing radiation and not matter.
 
  • Like
Likes vanhees71 and GR191511
  • #7
The above is the "dust model", i.e., a collection of non-interacting very dilute "dust particles", where you can also neglect their mutual gravitational interaction. Of course, in an arbitrary frame
$$T^{\mu \nu}=\rho u^{\mu} u^{\nu},$$
where ##u^{\mu}## is the four-velocity field of the dust distribution, which is of course not diagonal. Again the quantity ##\rho## is a scalar field, i.e., the energy density as measured in the rest frame of the medium.
 
  • #8
pervect said:
I'm not sure what the author is trying to say. Consider a stress energy tensor of a continuum of massive particles with some density ##\rho## and no pressure. Then in the MCRF frame, we'll have.

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

This is diagonal, but it'll be non-diagonal if you boost it - i.e. in any other frame than the MCRF frame.

Now let's consider a case with pressure.

If the whole stress energy tensor is T[0:3, 0:3], the submatrix representing pressure, also called the stress tensor, will be T[1:3, 1:3]. If the pressure is isotropic, the pressure will be diagonal under rotational transformations. But if the pressure is not isotropic, I.e. if we have

$$\begin{bmatrix} p1 & 0 & 0 \\ 0 & p2 & 0 \\ 0 & 0 & p3 \end{bmatrix}$$

where p1, p2, and p3 are not all equal, a rotational transformation will make the stress tensor non-diagonal.

A perfect fluid, which was discussed in a previous post, will always have an isotropic pressure by definition.

Note that we have not considered all possible casses of the stress energy tensor, such as those containing radiation and not matter.
I am not sure why you are trying to consider examples! The whole point is that if it is diagonal in all frames it has to be a scalar times the identity.
 
  • Like
Likes GR191511 and vanhees71
  • #9
martinbn said:
I am not sure why you are trying to consider examples!
Beucase it is illustrative to see how examples play out?
 
  • #10
drmalawi said:
Beucase it is illustrative to see how examples play out?
In general yes, but why here? His reply started with "I'm not sure what the author is trying to say." The point is that the matter content is irrelevant.
 
  • #11
So - is the point here that there's a step in the reasoning expressed in the OP that is only implicit? Namely, that all MCRF (i.e. tetrads at some event that share a time axis) are related by purely spatial rotations. Hence we are only interested in finding stress tensors that remain diagonal under arbitrary rotation. In that case, that they are a constant times the identity is fairly trivial to show with a bit of matrix algebra.
 
  • Like
Likes GR191511
  • #12
The reference is section 4.6 Perfect Fluids, pages 100-101 in the second edition.
 
  • Like
Likes GR191511 and malawi_glenn
  • #13
Like
I don't know the book, but for perfect fluids, it's simply Pascal's Law: In a static fluid, on which no external forces act the pressure is homogeneous and isotropic. In GR this applies to each fluid cell in the local restframe of the fluid. Thus in this LRF the components of the energy-momentum-stress tensor in a local inertial frame must be of the form
$$T^{\mu \nu}=\mathrm{diag}(\epsilon,P,P,P).$$
In an arbitrary frame, all you have to express this in a covariant way is the metric tensor and the fluid four-velocity field, i.e.,
$$T^{\mu \nu} = (\epsilon+P) u^{\mu} u^{\nu} - P g^{\mu \nu},$$
using the mainly-minus convention of the metric, i.e., ##(\eta_{\mu \nu})=(1,-1,-1,-1)## in the local inertial frame.
 
  • Like
Likes GR191511
  • #14
martinbn said:
The reference is section 4.6 Perfect Fluids, pages 100-101 in the second edition.
Author and publisher could be helpful
 
  • #16
  • Like
Likes GR191511
  • #17
GR191511 said:
“The only matrix diagonal in all frames is a multiple of the identity:all its diagonal terms are equal.”
One can visualize a symmetric matrix using a quadric (for example, an ellipsoid or hyperboloid).

When rotating the axes to make the matrix diagonal, these axes are along the axes of symmetry of the quadric.
(Think about the moment of inertia tensor.)

If the matrix is diagonal in all frames, then the quadric is a sphere.
 
  • Like
Likes GR191511, vanhees71 and martinbn
  • #18
robphy said:
One can visualize a symmetric matrix using a quadric (for example, an ellipsoid or hyperboloid).

When rotating the axes to make the matrix diagonal, these axes are along the axes of symmetry of the quadric.
(Think about the moment of inertia tensor.)

If the matrix is diagonal in all frames, then the quadric is a sphere.

Yes, the stress-energy tensor is a symmetric bilinear form, which means that it can be represented by a quadratic form. And the principal axis theorem applies to quadratic forms, thus one can always diagonalize them. (This is from memory, I might have missed a few subtle points).

In the diagonal form, the diagonal elemetns are the eigenvalues, and if all three eignevalues are the same, one has the spherical symmetry you note.

Since the digaonal form of the stress-energy tensor always exists, it's simplest to imagine choosing a preferred coordinate system where the stress-energy tensor is diagonal. This makes it a lot easier to grasp, at least for me. Various techniques exist to visualize diagonal matrices, such as representing them via ellispoids.

One then can eventually realize that if the three eigenvalues are identical, the form has spherical symmetry, and rotation won't change the diagonalization property. If the three eigenvalues are not identical, spherical symmetry is lacking, and rotational transformations will not preserve the diagonalization.

Because the context of the original text is apparently restricted to perfect fluids, spherical symmetry exists. But this won't necessarily be true for stress-energy tensors that are not perfect fluids.

As for examples, I always find it useful to consider specific examples of abstract theorems. Your mileage may vary, I suppose.
 
  • Like
Likes GR191511, martinbn, vanhees71 and 1 other person

Related to Diagonal Matrix of Stress-Energy Tensor: Why?

1. What is a diagonal matrix of stress-energy tensor?

A diagonal matrix of stress-energy tensor is a mathematical representation of the stress-energy tensor, which describes the distribution of energy and momentum in a given space. It is called diagonal because its elements are arranged in a diagonal pattern, with zeros in all off-diagonal positions.

2. Why is a diagonal matrix of stress-energy tensor important?

A diagonal matrix of stress-energy tensor is important because it provides a concise and efficient way to represent the complex distribution of energy and momentum in a space. It also allows for easier mathematical calculations and analysis of physical systems.

3. How is a diagonal matrix of stress-energy tensor calculated?

A diagonal matrix of stress-energy tensor is calculated by taking the stress-energy tensor, which is a 4x4 matrix, and transforming it into a diagonal form using mathematical operations such as eigenvalue decomposition or diagonalization.

4. What are the applications of a diagonal matrix of stress-energy tensor?

A diagonal matrix of stress-energy tensor has various applications in physics, particularly in the study of general relativity and the behavior of matter and energy in curved spacetime. It is also used in the analysis of fluid dynamics and electromagnetism.

5. Are there any limitations to using a diagonal matrix of stress-energy tensor?

One limitation of using a diagonal matrix of stress-energy tensor is that it can only be applied to systems that are in a state of equilibrium. It also assumes that the distribution of energy and momentum is isotropic, meaning it is the same in all directions. Additionally, it is not suitable for describing systems with strong gravitational fields or high energy densities.

Similar threads

  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
10
Views
1K
  • Special and General Relativity
Replies
21
Views
2K
  • Special and General Relativity
Replies
33
Views
3K
  • Special and General Relativity
Replies
4
Views
359
Replies
40
Views
2K
Replies
15
Views
1K
  • Special and General Relativity
Replies
4
Views
811
  • Special and General Relativity
Replies
13
Views
1K
  • Special and General Relativity
Replies
9
Views
2K
Back
Top