# Tensors in GR and in mechanics

1. Jan 28, 2014

### TrickyDicky

Besides the dimensionality (4 vs. 3), how would you go about explaining the difference between tensors in GR and in continuum mechanics?
I was asked by an engineer friend that finds GR too "esoteric" and complex to get into.

2. Jan 28, 2014

### Staff: Mentor

In my judgement, there is no difference. Of course, when you start working with the components of tensors in GR using coordinate systems, you are dealing with a non-Euclidean geometry rather than 3D Euclidean space. But the basic principles of manipulating tensors is exactly the same.

3. Jan 28, 2014

### pervect

Staff Emeritus
Classical tensors tranform between frames using the Galilean transform, t' = t, x' = t - vx. When you replace the above Galilean tensor transformation rules with the Lorentz transform, you get relativistic tensors.

There may be slight differences in notation as well, for instance according to wikki the engineering stress-energy tensor differs from the physicists stress-energy tensor by convective terms.

I'm biased, but I'd say the physicists version is actually simpler and easier to work with.

4. Jan 29, 2014

### WannabeNewton

I agree with Chet in that there is none. Furthermore continuum mechanics is in and of itself a powerful and ubiquitously used calculational and conceptual tool in GR.

Conceptually GR is definitely on the hard side but the mathematics of GR is easy.

5. Jan 29, 2014

### Staff: Mentor

To elaborate on what I said earlier, as an engineer who first used tensors in solid and fluid mechanics and then studied SR and GR, there were some features of tensors in relativity that took some getting used to, mainly related to the fact that, because of the peculiar nature of the time dimension, one or more of the covariant components of the metric tensor is negative. However, aside from that, one is still dealing with tensors using curvilinear coordinates (at least in GR), and one is working with a curved manifold using the same mathematical approaches that Gauss developed for 2D curved surfaces.

Chet

6. Jan 29, 2014

### bapowell

Yes, you are hitting on the fact that there is a different metric signature in Euclidean space (positive diagonals, or (+1,+1,+1,+1)), where ordinary continuum dynamics lives, and Minkowski space (with signature (-1,+1,+1,+1) or (+1,-1,-1,-1) depending on convention), which is the appropriate arena for special relativity. The Lorentzian signature of the Minkowski metric (namely the different sign in front the time-time and space-space components) translates over to general relativity since any more general metric that describes a real gravitational field must reduce, as per the equivalence principle, to the Minkowski metric locally. It is therefore sometimes said that general relativity includes only pseudo-Riemannian manifolds with Lorentzian signature.

Last edited: Jan 29, 2014
7. Jan 29, 2014

### HallsofIvy

Staff Emeritus
Typically, tensors in "continuum mechanics" are "Euclidean tensors". That is, they are constructed on geometries where the metric tensor is diagonal with constants along the diagonal (and so can, by a change of scale, be transformed to the identity matrix). A result of that is that the "covariant" and "contra-variant" representations of a matrix are the same.

8. Jan 29, 2014

### Staff: Mentor

I have lots of experience with tensors in continuum mechanics that are not, as you refer to them, Euclidean tensors. Any time you use a non-orthogonal coordinate system, the off diagonal components of the metric tensor are non-zero. Also, when you are describing the kinematics of the large deformations of materials, you use tensors like the Cauchy Green tensor and the finite strain tensor, both of that have non-diagonal components, and for which the covariant and contravariant components are not the same. When you use an embedded material coordinate system, even if the coordinate lines are initially orthogonal, after the deformation has occurred, the coordinate lines are usually non-orthogonal and curved. Also, in continuum mechanics, even if the coordinate axes are cartesian, the stress tensor generally features non-zero off-diagonal components. Only if the coordinate axes are aligned with the principal directions of stress is the stress tensor diagonal. Even then, the covariant and contravariant components are usually not the same.

9. Jan 29, 2014

### Staff: Mentor

Engineers and physicists use the term stress tensor to describe two different entities. First of all, in engineering we don't use the term "stress-energy tensor." Secondly, what physicists call the stress-energy tensor, we call the momentum flux tensor (plus the isotropic pressure tensor). Third, what engineers call the stress tensor describes the internal state of stresses within a solid or fluid material that is deforming. There is no corresponding entity in relativity (to my knowledge) because, in practice, there are not many practical applications for quantifying the mechanics of deforming materials on the relativistic scale. In 3D fluid mechanics, we describe the internal stresses within a fluid as twice the Newtonian viscosity times the rate of deformation tensor. The latter is equal to the sum of the velocity gradient tensor and the transpose of the velocity gradient tensor, divided by 2. If this representation were to be expanded to relativistic fluid mechanics, I suppose we would replace the velocity gradient tensor by the gradient of the 4 velocity.

Chet

Last edited: Jan 29, 2014
10. Jan 29, 2014

### bapowell

That just makes them non-Cartesian. The underlying geometry is still Euclidean, no?

11. Jan 29, 2014

### TrickyDicky

Thanks for the interesting replies.
Of course a tensor is always obviously a tensor, but it is always more didactic to concentrate on the differences as it is done in the last answers.

Chestermiller makes some very good points, in line with what I had in mind:

I think you and HallsofIvy are referring to Cartesian tensors in general (with the caveat you highlight below). I mean AFAIK(but you surely know more tan I do so correct me if I'm wrong) all engineering applications of tensors not related to relativity, use Euclidean tensors, defined as those set in Euclidean space (vanishing Riemannian tensor).
As you explain, obviously not all of them use cartesian coordinates, some use curvilinear coordinates that basically demand most of the machinery used in non-euclidean tensors(in terms of Christoffel symbols and position-dep. basis, covariance and contravariance...), and some use non-orthogonal coordinate system, etc.

Right, the shear stresses. However I would have thought in the last case you mention there wouldn't be covariant and contravariant distinction(unless for instance non-orthogonal basis were used).

12. Jan 29, 2014

### Staff: Mentor

Yes. The underlying geometry is still Euclidean. So, in all these cases, it is possible to find a cartesian coordinate system in which the metric tensor is diagonal, with components +1.

However, if you have a curved (non-Euclidean) 2D surface immersed in flat 3D space (for example, the interior contour of an automobile tire or a saddle), the 2D surface would feature a metric tensor which could not be reduced to the Euclidean representation. It would exhibit mathematical features completely analogous to those of curved space-time, except, of course, for negative components of the metric tensor.

13. Jan 29, 2014

### Staff: Mentor

Not if you are using curvilinear coordinates. In orthogonal spherical and cylindrical coordinates, for example, the contravariant and the covariant components are not all equal.

14. Jan 29, 2014

### WannabeNewton

Actually there is. Given a 4-velocity field $\xi^{\mu}$ describing a fluid flow, we can write $\nabla_{\mu}\xi_{\nu} = \omega_{\mu\nu} + \theta_{\mu\nu} - \xi_{\mu}a_{\nu}$. If we let $h_{\mu\nu} = g_{\mu\nu} + \xi_{\mu}\xi_{\nu}$ denote the orthogonal projector relative to $\xi^{\mu}$ then $\theta_{\mu\nu} = h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{(\alpha}\xi_{\beta)}$ is called the expansion tensor. It's trace $\theta = \nabla_{\mu}\xi^{\mu}$ has the usual physical interpretation in terms of volume transport by fluid elements. It's trace-free part $\sigma_{\mu\nu} = \theta_{\mu\nu} - \frac{1}{3}h_{\mu\nu}\theta$ is the shear stress tensor.

$\omega_{\mu\nu} = h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{[\alpha}\xi_{\beta]}$ is the vorticity tensor from which the vorticity vector $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta}$ can be defined. $\omega^{\mu}$ and $\sigma_{\mu\nu}$ reduce to the usual Newtonian expressions for fluids if we evaluate them in a momentarily comoving local inertial frame of a fluid element. Most standard GR textbooks will develop this hydrodynamical formulation of relativistic kinematics. See, for example, chapter 22 of MTW and chapter 9 of Wald.

Actually on the contrary it's very important when analyzing relativistic problems. Both the Ehrenfest paradox and the Bell paradox can be explained in very simple terms using the hydrodynamical formulation of relativistic kinematics.

Last edited: Jan 29, 2014
15. Jan 29, 2014

### stevendaryl

Staff Emeritus
The distinction between covariant and contra-variant only disappears for Cartesian coordinates. If you are using curvilinear coordinates $r, \theta, \phi$, then there is a distinction even for Euclidean geometry.

16. Jan 29, 2014

### WannabeNewton

The differences are purely conventional. Special and general relativistic fluid dynamics are only different from the non-relativistic counter-part in terms of the physical consequences of SR and GR. For example clocks comoving with elements of a fluid with non-vanishing vorticity cannot be Einstein synchronized globally. However the formalism is no different-we just make physical quantities generally covariant.

17. Jan 29, 2014

### stevendaryl

Staff Emeritus
A difference in what is typically done in General Relativity compared with Euclidean space is the choice of basis vectors.

Suppose you are using the curvilinear coordinates $r, \theta, \phi$. If you make an infinitesimal change in coordinates, then the displacement vector in the GR convention is typically written:

$\delta \vec{r} = \delta r\ \textbf{e}_r + \delta \theta\ \textbf{e}_{\theta} + \delta \phi\ \textbf{e}_{\phi}$

This gives different magnitudes to the different basis vectors:
$\textbf{e}_r \cdot \textbf{e}_r = 1$
$\textbf{e}_\theta \cdot \textbf{e}_\theta = r^2$
$\textbf{e}_\phi\cdot \textbf{e}_\phi= r^2 sin^2(\theta)$

In contrast, people working with Euclidean geometry typically choose "unit vectors". So they would right:

$\delta \vec{r} = \delta r\ \hat{r} + r\ \delta \theta\ \hat{\theta} + r\ sin(\theta)\ \hat{\phi}$

where $\hat{r} \cdot \hat{r} = \hat{\theta} \cdot \hat{\theta} = \hat{\phi} \cdot \hat{\phi} = 1$

In GR, because of the indefinite metric, you can't always get a "unit vector" in an arbitrary direction.

18. Jan 29, 2014

### TrickyDicky

Sure.

But you would still use an Euclidean stress or deformation 3D tensor, have you found any situation in wich you needed to model a curved surface object using a 2D stress tensor(if that can be done at all)?

19. Jan 29, 2014

### WannabeNewton

We can and very frequently do GR problems using frame fields which are sections of the vector bundle that assign an orthonormal frame to each event in space-time. The integrals curves of the time-like basis vector of a frame field represent a family of observers and the spatial basis vectors represent their meter sticks, gyroscopes etc. so that distinction is rather non-existent.

Last edited: Jan 29, 2014
20. Jan 29, 2014

### stevendaryl

Staff Emeritus
But you can't have unit vectors for an arbitrary coordinate system, though.