# Tensor formalism in GR

## Main Question or Discussion Point

In explanations of the importance the tensors I often see people refer to transformation properties, general covariance and the like. Now, I have also often read that in principle any physical theory, e.g. classical mechanics and special relativity, can be written in a generally covariant form, but this would just make the formulation unnecessarily complicated. I'm now looking for a statement along the lines

"CM has property X, which allows us to use vectors instead of the full tensor calculus."
or
"GR does not have property X, which necessitates the use of tensors."

I assume "X" will have something to do with the no prior geometry feature of GR, but I'd like to make this more precise.

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phyzguy
I'm not sure I understand your question. Vectors are simply tensors of rank 1, so a vector formulation is still a tensor formulation. Also, your statement that expressing non-GR physics in tensor form is more complicated is also questionable. Maxwell's equations are actually simpler when expressed in tensor form, not more complicated. The single rank two E-M field tensor F replaces the two vectors E and B and makes the relation between E and B much clearer.

• guitarphysics and vanhees71
Ah yes, I was afraid using "vector" might lead to confusion (which is why I wrote "full tensor calculus"). My question would then be why we need only this special fragment of tensor calculus for CM, whereas we need the general tensor concept in GR.

As for your other comment, isn't what you're describing the Lorentz covariant formulation of EM as opposed to a generally covariant formulation?

For it being more complicated, the first paragraph of the last answer here

http://physics.stackexchange.com/questions/105223/what-transformation-is-the-metric-of-general-relativity-invariant-under

basically says what I want to convey. Rereading it, perhaps I should have not used the term "tensor formalism", but more generally "differential geometry". Does this make it clearer?

Ah yes, I was afraid using "vector" might lead to confusion (which is why I wrote "full tensor calculus"). My question would then be why we need only this special fragment of tensor calculus for CM, whereas we need the general tensor concept in GR.
Tensors are needed in classical mechanics. Rigid body dynamics requires a rank 2 tensor to quantify the rotational inertia of a general solid object, for example. Tensors are also required for both linear and nonlinear models of elasticity.

• vanhees71
jambaugh
Gold Member
There are some ambiguities in your question. There's a broader context of the phrase "Generally Covariant form" when it comes to formulation of physical theories. Typically one may deal with a simple non-relativistic one dimensional theory say a mass-spring in a generally covariant form which is to use an arbitrary time parameterization. This does connect back to the meaning in GR where one begins formulation under arbitrary space-time coordinates. It ends up being a matter of asking how general the group of relativity transformations is that your application necessitates.

Let me add also that there are vectors and tensors, and then there are vector fields and tensor fields. If you want to discuss rotation of rigid bodies about arbitrary axes you will need to work with the moment of inertia tensor. Also note that vectors are a form of tensor so you almost always are "working with tensors" so the question becomes how deep into that pond you must swim. Here are a few points to consider in that context:

We think of rotations in 3-dimensions as being "about an axis" and thus we can express the rotation parameter as an angle vector. However we see that this ability is concidental to our use of 3 dimensions and rotations are properly expressed using rank 2 antisymmetric tensors which are d(d-1)/2 dimensional objects when rank 1 tensors = vectors are d-dimensional. For d=3 d(1-d)/2 = 3. Hodge duality lets us map between these two 3-dimensinoal spaces. But when we raise the spatial dimension the coincidence goes away and we realize that we must work with the proper tensor format. (you rotate one direction toward another direction so you must express the plane of rotation with a "bivector" = "rank 2 antisymmetric tensor".

So in short: Increasing or generalizing the dimension may require implicit tensor objects to be treated explicitly as tensors.

Another factor is that one may be working with components of a tensor and not realizing it until one considers a more unified setting. Unification often goes hand in hand with Relativization both of which affect your transformation groups. Let's take a simple example. When considering ocean sailing you would treat vertical distanced different from horizontal. Note nautical depths are measured in fathoms while travel distances are measured in leagues. Note also that you have separate terms for rotations in the various directions, pitch, yaw, and roll of the ship. Now move to a space-ship and without the ocean surface to break the symmetry we see a more generalized description of motion. We unify the three directions and express rotation using a three dimensional bi-vector (rank 2 tensor) or cheat by mapping it to its Hodge dualized vectors. We take the principle axis components of the ocean ships three moments of inertia for the three cardinal rotations and combine them into the moment of inertia tensor. We may even recognize time as a 4-th dimension and unify to the Gallilean relativity group but since it is a graded group (with non-trivial normal subgroups of velocity boosts and spatial translations) we still break certain unified tensors up into components. We have a scalar electric potential and 3-vector magnetic potential. We have 3-vector electric fields (which boost charges' velocities) and 3-bivector magnetic tensor fields (which rotate charges' velocities).
With Einstein's SR the relativity group simplifies. Component object mix when transformed and we are almost forced to work with unified object which are typically higher rank tensors. Scalar and 3-vector potentials unify to a 4-vector, 3-vector Electric and 3-bivector Magnetic forces become the six dimensional (4(4-1)/2 = 6) Electro-Magnetic field tensor.

So in short: Unification and Relativization will tend to combine components we used to treat as vectors or scalars forming higher rank tensor objects.

Finally returning to the general covariance issue, one may describe global relativity transformations but to be generally covariant we would want to consider relativity transformations applied in a way depending on location. We then have fields of local relativity transformations and we must consider their interconnection. How local relativity at one point relates to such at another. We are then forced to work with the tensor nature of the frame connections. You have objects which contain a component of translation and a component of other group transformation properties (boosts or rotations or internal gauge transformations) combined together telling us how transforms there map to transforms here.

Considering the generally covariant case of local vs global transformations requires us to account for more structure, typically expressed with vector and tensor fields.

Ultimately by "tensor" we mean a mathematical object that does not transform trivially under general linear transformations. In a physical application we specifically mean the general linear group of vector isomorphisms of the local tangent and cotangent spaces to our physical manifold. (GL(4) for space-time GL(3) for just space.) So a tensor is an element of in irrep of the general linear group.

You'll need to use tensors when ever you need to consider arbitrary ways things change under general linear transformations of space or space-time components.

• Logic Cloud
stevendaryl
Staff Emeritus
Any laws of physics that are expressible using smooth functions of the spatial and time coordinates can be written in a covariant way, using tensors. However, if you try to make Newtonian physics into a covariant theory, you find that there are two additional bits of structure that don't exist in General Relativity:
1. A universal time coordinate, $t$.
2. A purely spatial metric, to be used for determining distances between events that are simultaneous, according to the universal time coordinate.
The first bit of structure, universal time, can be easily written in a covariant form: You just let universal time be a scalar field. The second bit of structure, the spatial metric, is a little bit complicated to express in a coordinate-free manner, in terms of tensors.

• Logic Cloud
Nugatory
Mentor
My question would then be why we need only this special fragment of tensor calculus for CM, whereas we need the general tensor concept in GR.
The general tensor concept is not needed because we can work with Cartesian (or Minkowski, in SR) coordinates. In these friendly coordinates in the metric tensor takes such a simple form that you don't see any of the fun stuff: $A^i=A_i$ so we ignore the distinction between contravariant and covariant, the covariant derivative is just the ordinary derivative, the scalar product is just the dot product $\vec{A}\cdot\vec{B}=g_{uv}A^uB^v=A^uB_u$ so you don't see the metric there, basis vectors are always orthonormal, coordinate velocities can be taken at face value....

In the curved spacetime of GR we can't use such nice coordinates. No matter what reasonable coordinates we choose, we'll end up with something with a non-trivial metric tensor that exposes us to the full tensor machinery.

• Logic Cloud
Dale
Mentor
"CM has property X, which allows us to use vectors instead of the full tensor calculus."
or
"GR does not have property X, which necessitates the use of tensors."
I don't think that X exists.

pervect
Staff Emeritus
In explanations of the importance the tensors I often see people refer to transformation properties, general covariance and the like. Now, I have also often read that in principle any physical theory, e.g. classical mechanics and special relativity, can be written in a generally covariant form, but this would just make the formulation unnecessarily complicated. I'm now looking for a statement along the lines

"CM has property X, which allows us to use vectors instead of the full tensor calculus."
or
"GR does not have property X, which necessitates the use of tensors."

I assume "X" will have something to do with the no prior geometry feature of GR, but I'd like to make this more precise.
I'm not sure if I understand what you're trying to do. So I'm going to ramble a bit and hope that it's useful.

Let's talk about some cases where you don't need tensors first. If you have a flat plane, you can assign coordinates to it in such a manner that the distance between points (x1,y1) and (x2,y2) is $\sqrt{(x2-x1)^2 + (y2-y1)^2}$. Then you can use non-tensor methods to do geometry on the plane that I presume you're familiar with.

Now, sometimes coordinates like (x1,y1) are not convenient. You might want to use polar coordinates, for instance, $r, \theta$. You can still do geometry, but the rules are a bit different. Typically you initially learn how to do this all as special cases for simple special coordinates, like cartesian and polar coordinates.

But there's lots of coordinates you could use on a plane. You might have occasion (rarely) to use oddball coordinte systems like elliptical coordinates, hyperbolic coordinates, or others.

The rule of covariance, says, basically, that we must get the same answers to any question we ask regardless of how we assign our coordinates. We can use cartesian coordinates, polar coordinates, whatever we like - but we can convert our answers from one coordinate system to another coordinate system, and they have to agree.

The general formula for the distance between two arbitrary points p1 and p2 with general coordinates will be given by a more general formula that is difficult to write down. If the two points are "close together", then we can use methods from differential geometry, where we write distances in terms of differentials like dx and dy - though instead of using x and y, we'll often introduce subscripts such as $dx_1, dx_2$ , which is just a matter of notation. I was going to get into that more, but this post is long enough and it's not an important issue.

When we use differentials, we come up with different formulas for distance depending on our coordinate choices, such as $d^2 = dx^2 + dy^2$ for cartesian coordinates, or $d^2 = dr^2 + r^2 d\theta^2$ for polar coordinates.

Differential geometry is an important tool in general relativity in it's own right, but you were asking about tensors. Tensors arise naturally in differential geometry. If we wanted to write the distance between two nearby points via the diferentials, we might write a general 2d formula as something like: $d^2 = g_{11} dx^2 + 2\,g_{12}dx\,dy + g_{22}dy^2$. Here the various values of g are in general functions of the coordinates x, y. The entity "g" follows certain transformation rules that make it a tensor. So we are strongly motivated to learn about tensors at this point, so we can learn differential geometry, because the metric tensor is so important to this topic. There are some other, more exotic tensors that are important to differential geometry and General realtivity too, but for the purposes of this post, the metric tensor is enough.

So far it's all been a matter of choice - we aren't forced to use anything but cartesian coordinates, though we may find it convenient. Are there any cases where we wouldn't be able to use cartesian coordinates?

The answer to this is yes. Suppose we want to do geometry on the surface of a sphere. Try as we might, we cannot find coordinates that make the distance between two nearby points as simple as $dp^2 + dq^2$. If we use coordinates like lattitude and longitude, though, we can find the metric tensor g (I won't write it down), and use the methods of differential geoemtry via the use of this metric tensor.

What prevents us from using cartesian coordinates when we want to do geometry on the surface of the sphere is the fact that the surface of the sphere is curved, and the cartesian coordinates only work on flat geometries in a plane.

Onto some remarks about physics. Both Newtonian mechanics and special relativity can be done in Cartesian coordinates, because the geometries are basically flat. The geometry of Newton is a flat spatial geometry, which we can quickly summarize via the Cartesian expression for distance, $dx^2 + dy^2 + dz^2$. The geometry of special relativity is a flat space-time geometry. Because it's flat , we don't need tensors yet, but the 4-d geometry involves a quanitity called the "Lorentz interval" that has the form $dx^2 + dy^2 + dz^2 - c^2 dt^2$. The minus sign is a definite complication - it turns out we don't need tensors to handle it, but our notion of "geometry" has shifted towards the abstract, rather than the geometry of "distances" we used in Eucliedean geometyr, we have a geometry of the "Loretnz Interval" in special relativity.

When we want to move on to the geometry of General Relativity, it's just not flat at all, and we can't find Cartesian coordinates at all because of the fact that it's not flat. So we are more or less forced to use differential geometry with a general metric tensor. And this means we need to understand tensors.

Once we understand tensors, it turns out they are very useful, and at the graduate level of physics one often re-learns how to use them as a powerful tool, that's useful in areas other than differential geometry that was our initial motivation for learning them.

It may be convenient to do things in a different order, though. One might learn electromagnetism without tensors, for instance, then learn about how tensors can be used in E&M in a familiar setting, before setting sail for the seas of differential geometry and General relativity.

Thank you all for your replies. I believe I understand the general idea. What confused me initially was that I wanted to equate my "property X" with the flatness of spacetime in non-GR theories. But this does not seem quite correct, since even in flat spacetime we might still require differential geometry for the formulation of our theories, if we wish to allow the use of exotic coordinate systems (as described in pervect's post).

The crucial feature thus seems to be the fact that flat spacetime allows us to use Cartesian coordiantes, which reduces differential geometry to the simpler, classical formalism. Is this description something you can get behind?

Tensors are needed in classical mechanics. Rigid body dynamics requires a rank 2 tensor to quantify the rotational inertia of a general solid object, for example. Tensors are also required for both linear and nonlinear models of elasticity.
Yes, I have indeed encountered tensors in the context you mention. However, since every introduction to GR I've seen deems it necessary explicitly introduce the notion of tensors, I've always had the idea that the general tensor concept is somehow inherent to GR (due to their transformation properties) whereas in the classical case I've always seen them more as bookkeeping devices.

Any laws of physics that are expressible using smooth functions of the spatial and time coordinates can be written in a covariant way, using tensors. However, if you try to make Newtonian physics into a covariant theory, you find that there are two additional bits of structure that don't exist in General Relativity:
1. A universal time coordinate, $t$.
2. A purely spatial metric, to be used for determining distances between events that are simultaneous, according to the universal time coordinate.
The first bit of structure, universal time, can be easily written in a covariant form: You just let universal time be a scalar field. The second bit of structure, the spatial metric, is a little bit complicated to express in a coordinate-free manner, in terms of tensors.
I see. Could you elaborate on why we let time be a scalar field? Couldn't we, for instance, simply have one spatial metric and one temporal metric on flat, 4-dimensional spacetime?

I don't think that X exists.
Could you elaborate on why you think this? It now seems to me I can simply put X=permits Cartesian coordinates.

stevendaryl
Staff Emeritus
Could you elaborate on why we let time be a scalar field? Couldn't we, for instance, simply have one spatial metric and one temporal metric on flat, 4-dimensional spacetime?
Well, a scalar field is just a way of assigning a real number to each point in spacetime. So if you have a temporal metric, then you can just arbitrarily pick one event, $e_0$ and call it $t=0$, and for any other event $e$, let $t$ be the temporal distance between that point and $e_0$. (For events before $e_0$, use the negative of the temporal distance).

Dale
Mentor
Could you elaborate on why you think this? It now seems to me I can simply put X=permits Cartesian coordinates.
I don't think that tensors are ever necessary. You can always expand tensor equations out and write them explicitly in terms of coordinates. So I don't think that the second statement is true for any X.

haushofer
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