Why is there an emphasis on tensor equations in GR?

Question

In my understanding the purpose of using tensor equations in GR is to ensure that they are true in all coordinate systems. I understand that writing equations tensorially ensures this will be the case; however, are there not non-tensor equations that would also be true in all coordinate systems?

For example, one can define a tensor by its components and how they transform from one coordinate system to another (the tensor transformation law). It seems to me that you could define some other quantity that transforms according to another transformation law, and that equations written in this quantity would also be valid in all coordinate systems.

I've also seen tensors defined as geometric objects on the manifold that act as linear forms on the tangent and cotangent spaces on the manifold. This geometric definition immediately guarantees coordinate independence. Again, I don't see why we can't define a more general geometric object (i.e., not a tensor) and make that the basis of our coordinate independent equations.

To summarise, why is there an emphasis on tensor equations in GR when it seems to me that there should be plenty of non-tensor equations that are valid in all coordinate systems as well?

EDIT: As an example, consider some arbitrary mapping from the tangent space to the reals that is not linear in the tangent vectors. This is a coordinate independent definition. The only difference between these objects and tensors is that for tensors, the mapping is linear. I suppose non-linearity means that these objects won't have straightforward, easily interpretable 'components' in each coordinate system, but I don't see why we still couldn't make important statements about the geometry of spacetime using them.

Answer 1

There are coordinate-independent objects that aren't tensors.

Connections, densities, spinors, sections of fiber bundles in general etc.

Tensors however are

• related to manifold geometry (contrast this with sections of an arbitrary vector bundle)

• have linear dependence on directions.

I'm going to illustrate this with the same example Wald uses in his GR book. Imagine a magnetic field $B$ permeating space. You have a detector that measures the magnetic field in the direction the detector probe is pointing. How do you measure the magnetic field at point $x$?

You choose and record three linearly independent probe orientations. Since the probe probably uses the same units in all directions, and has the same sensitivity, all three orientations can be taken to be unit vectors.

Let the three directions be $e_1,e_2$ and $e_3$. You take the three measurements, these return the values $$ B_1=B(e_1),\ B_2=B(e_2),\ B_3=B(e_3). $$

As you can see, the magnetic field plays the role of a covector here, instead of a vector. From this, you can assemble the magnetic field as $$ B=B_1 e^1+B_2 e^2+B_3e^3, $$ where $e^i$ is the dual basis element to $e_i$.

The metric tensor is $$ g_{ij}=\left(\begin{matrix} 1 && \cos\alpha_{12} && \cos\alpha_{13} \\ \cos{\alpha_{12}} && 1 && \cos{\alpha_{23}} \\ \cos{\alpha_{13}} && \cos{\alpha_{23}} && 1\end{matrix}\right) $$ where $\alpha_{ij}$ is the angle between $e_i$ and $e_j$.

The magnetic field vector is given by $\sharp B=g^{1i}B_ie_1+g^{2i}B_ie_2+g^{3i}B_ie_3$, the magnitude is given by $||B||=g^{ij}B_iB_j$ etc.

Now, if instead of having a linear dependence on directions, $B$ was some arbitrary smooth function $B:T_xM\rightarrow\mathbb{R}$, then you'd need an infinite amount of measurements (in an infinite amount of directions) to reconstruct it at a point.

Clearly, these "direction-dependent" quantities in physics behave in such a way that you don't need an infinite amount of measurements to measure them at a point. If they did, physics as we know it would not exist! So the reason we use tensors is that physics is measurable.

Answer 2

Good question, and one that isn't commonly gone into in the physics literature when they introduce the tensor transformation law.

Try visualising it geometrically: take a simple example, deforming the surface of a sphere to an ellipsoid; if we take a small (ie infinitesimal) patch of the ball and see how it transforms we see that there is a multilinear dependence; this example can be generalised to arbitrary manifolds in any cartesian space, and we can also, with some thought, drop the embedding.

A multilinear transformation is characterised universally by tensors, and then by taking bases we get the usual coordinate transformation property that characterises tensors common in the physics literature.

The best reference I've seen for this is Lees book on Differential Geometry, and Dodsons Tensor Geometry, though he tends to use some idiosyncratic terminology.

Answer 3

Tensors by themselves don't ensure that a formula is correct in all coordinate systems. Navier-Stokes equations for example can be written in tensor form bu are not coordinate independent. What you need in fact is the covariant derivative property. The tensor form on the other hand is needed to describe stresses on a surface. To describe the shear stress on a surface for example, you need one vector lying in the surface for describing the force on that surface with magnitude and direction. But then you need another vector to describe the position and orientation of the surface itself- hence the two indices of the second rank tensor.