What is a Covariant Relation? GR Problem Solution Explained

[Haorong Wu]

TL;DR Summary

what relation is to be defined as covariant and why it would be generalized to any coordinate system?

I am solving some GR problems. In one problem, some relation between a second covariant derivative and the Riemann tensor is to be proven.

In the solution, the relation is first proven in a local flat coordinate system, followed by a statement that, since this relation is covariant it is true in any coordinate system.

I am not sure what relation is to be defined as covariant and why it would be generalized to any coordinate system.

For now I just guess relations with basic algebras and covatiant derivatives are covariant. Maybe I am wrong?

(sorry for some typos. I will correct them when I get access to my PC.)

 

[Dale]

Summary:: what relation is to be defined as covariant and why it would be generalized to any coordinate system?

I am not sure what relation is to be defined as covariant and why it would be generalized to any coordinate system.

A relation is covariant if it is true in any coordinate system, and it would be generalized to any coordinate system by definition.

In addition, if an expression is a proper tensor equation then it is what is called "manifestly covariant". In other words you can tell just by looking at the expression that it is covariant.

 

[romsofia]

I like how William L Burke thought about covariance, so I'll post an excerpt from him

 

[Haorong Wu]

Thanks, @Dale , and @romsofia. So, anytime I have a equation of tensors, with indices matching both side, I could treat them as covariant.

But I am still worried about the proof. Could it be rigorously proven in mathematics?

@romsofia, are there any proof about this statement in the book?

 

[Dale]

Could it be rigorously proven in mathematics?

Sure. Such proofs are common in introductory classes on Riemannian geometry.

Basically they start out by defining tensors in such a way that their components transform covariantly, then you can prove that sums, products, and contractions of tensors are also tensors (meaning that they transform correctly). And then you define a covariant differentiation operator and prove that it is indeed covariant.

With those in place then you write the physical laws in covariant form, often starting with Maxwell’s equations and then going to the Einstein field equations.

 

[pervect]

Which definition of a tensor does the OP, Haorong, use?

Is it the definition being used that of a linear map, which is simper to manipulate (but more abstract), i.e. a rank r,s tensor is a map from r dual vectors and s vectors to a scalar? Or is the definition in terms of the transformation properties used?

One fine point - one will typically find proofs that tensors are independent of the choice of basis in mathematics textbooks. The proofs are a bit tedious, but mostly straightforwards.

To go from the idea that tensors are independent of basis to the idea that tensors are independent of coordinate choices, one needs to associate coordinate choices with basis choices. There's a natural way to do this, called the coordinate basis. Sometimes this fundamental idea gets lost in the shuffle, though.

Probably the most interesting relation to convince oneself of is that the contraction of a tensor is another tensor.

 

[Haorong Wu]

Sure. Such proofs are common in introductory classes on Riemannian geometry.

Basically they start out by defining tensors in such a way that their components transform covariantly, then you can prove that sums, products, and contractions of tensors are also tensors (meaning that they transform correctly). And then you define a covariant differentiation operator and prove that it is indeed covariant.

With those in place then you write the physical laws in covariant form, often starting with Maxwell’s equations and then going to the Einstein field equations.

Thanks, Dale. From you explanation, could I say that tensors in different coordinate systems are different representations for them, so they can "covariantly" transform between representations without changing tensors themselves?

 

[Haorong Wu]

Thanks, @pervect . When I learned tensor the first time, I learned it with the latter definition. Recently, when learning GR, I learned that the map definition of tensor. I trust that the two definition are equivalent. Now I know that the covariant thing is related to transformation between coordinate, so I can convince myself that a equation with tensors does transform covariantly!

:P

 

[martinbn]

What is useful is that if tensor equation written in some coordinates holds, then it holds in any coordinate system.

 

[Dale]

they can "covariantly" transform between representations without changing tensors themselves?

Yes. The idea is that the transformations give you the same underlying geometric object written in the new components.

 

[vanhees71]

Thanks, Dale. From you explanation, could I say that tensors in different coordinate systems are different representations for them, so they can "covariantly" transform between representations without changing tensors themselves?

I think it is easier to think of tensors as independent of any choice of a basis (and co-basis) or coordinates (in general relativity).

It's the tensor components wrt. a basis and co-basis which change when changing the basis and co-basis, not the tensors themselves. So if you deal with tensor components and want to be sure to write down relations between tensors you must make sure that you can express these components in a mainfestly covariant way. E.g., when you calculate some derivative of tensor components you must make sure that it leads to a new tensor component, and that's why you introduce the covariant derivatives of tensor components (or more precisely tensor-field components), which needs the introduction of an affine connection on the manifold under consideration; in GR, which is a pseudo-Riemannian space, the connection by assumption is defined as the unique pseudometric-comopatible torsion free affine connection.

 

[Haorong Wu]

@ martinbn , lol. It is a weapon too powerful to hold.

@Dale . OK. I think I understand tensor better than this morning. Also good for my group theory class.

@vanhees71. Thanks for clear that up. I have a little question. In the problem book I read. It seems never talking about torsion-free connections, or metric-compatible connections. The book is published in 1979. Is this assumed to be true back then?

 

[vanhees71]

Yes, it is. In a coordinate basis the affine connection is represented by the Christoffel symbols $\Gamma_{\mu \nu}^{\rho}$ and the covariant derivative of vector-field components are defined as
$$\nabla_{\mu} A^{\rho}=\partial_{\mu} A^{\rho} + \Gamma_{\mu \nu}^{\rho} A^{\nu}.$$
One should note that from this definition and the transformation properties of vector components and $\partial_{\mu}$ it follows that the $\Gamma_{\mu \nu}^{\rho}$ are not tensor components. One can however show that the antisymmetric part
$$\tau_{\mu \nu}^{\rho} = \Gamma_{\mu \nu}^{\rho} - \Gamma_{\nu \mu}^{\rho}$$
are tensor components. This is called the torsion tensor. A differentiable manifold where the affine connection is defined such that the torsion tensor vanishes, i.e., if the Christoffel symbols are symmetric in the lower indices,
$$\Gamma_{\mu \nu}^{\rho}=G_{\nu \mu}^{\rho}.$$

An affine connection is called metric compatible, if
$$\nabla_{\mu} g_{\rho \sigma}=0.$$
If the connection is torsion free, i.e., the Christoffel symbols are symmetric in the lower indices, this then uniquely defines the Christoffel symbols in terms of derivatives of $g_{\mu \nu}$, as given in any GR textbook.

 

[Haorong Wu]

@vanhees71 , so in GR, we only consider metric-compatible connections? Therefore, without explicit statement, we do not need to consider other situatiob, because it is not physically possible?

 

[pervect]

Thanks, @pervect . When I learned tensor the first time, I learned it with the latter definition. Recently, when learning GR, I learned that the map definition of tensor. I trust that the two definition are equivalent. Now I know that the covariant thing is related to transformation between coordinate, so I can convince myself that a equation with tensors does transform covariantly!

:P

With the map definition, things are pretty straightforwards. Scalars need to be defined as objects that must be coordinate and basis independent, as distinguished from objects that might be repreetned by a real number field, but are dependent on the choice of coordinates or basis.

Dual vectors also need to be defined as maps from vectors to scalars, with scalars being coordinate independent as above.

Once you have met those requirements, basically that scalars must be coordinate and basis independent, you have by definition that the scalar result of any tensor manipulation is coordinate independent.

In General relativity, you can basically identify this coordinate and basis independent quantity as Lorentz interval. More physically, timelike Lorentz interval represents proper time intervals, as measured by physical clocks, realized by the SI definition. Similarly, proper distances can be realized by the SI meter, which is based on the constancy of the speed of light, or by the spacelike Lorentz interval.

In the alternate defintion of tensors, things are not nearly so straightforwards.

But in the map view, you are guaranteed by construction that any physical quantity, i.e. Lorentz interval, is independent of the basis or coordinates chosen.

Misner's "Precis of General Relativity", https://arxiv.org/abs/gr-qc/9508043, might be helpful for some background about the idea of what I call a physical quantity, and the authors call "measuring instruments", but it may be a digression.

 

[vanhees71]

@vanhees71 , so in GR, we only consider metric-compatible connections? Therefore, without explicit statement, we do not need to consider other situatiob, because it is not physically possible?

In standard GR yes, though it's likely that in fact we need the extension to an Einstein-Cartan space with torsion, because we have spin-1/2 particles. Having spin 1/2 implies the necessity for torsion.

For a nice review, see

P. Ramond, Field Theory: A Modern Primer, Addison-Wesley, Redwood City, Calif., 2 edn. (1989).

This also explains, why we don't need torsion so far: The observations we make, for which GR is important, are all macroscopic, mostly on astronomical scales and mostly with electromagnetic signals. If you only deal with macroscopic "spin-saturated" matter and electromagnetism (and gravity of course) standard GR without torsion works out well.

 

[haushofer]

Thanks, @Dale , and @romsofia. So, anytime I have a equation of tensors, with indices matching both side, I could treat them as covariant.

But I am still worried about the proof. Could it be rigorously proven in mathematics?

@romsofia, are there any proof about this statement in the book?

Just my 2 cents: how I think about this stuff.

Take a tensor equations A=B, with A and B arbitrary tensors of the same rank. That both A and B are tensors, means that under a coordinate transformation their new components are just linear combinations of their old components. That means, in particular, that if a tensor has only zero components, this will be true in every coordinate system; after all, a linear combination of zeroes gives zero.

So the tensor equation A=B can also be written as (A-B) = 0. I.e., (A-B) is a tensor with only zero components. The nice thing now is that under a coordinate transformation this remains zero. The proof is easy, as described above: a linear combination of zeroes remains zero. This makes tensors very convenient for describing equations of motion.

But strictly speaking, we have to specify under which transformations a tensor transforms as a tensor. Take e.g. Newton's second law for constant mass for a particle traveling along a curve x(t):

$$ma^i = m\frac{d^2x^i}{dt^2} = F^i_{res}$$

This equation is only covariant under a restricted group of coordinate transformations. This group defines the class of inertial observers. If I transform to a rotating frame via a rotation R(t) (an element of SO(3)), i.e.

$$x^{'i}(t) = R^i{}_j(t) x^j(t)$$

then it's a matter of some calculus to show that Newton's second law is not covariant. Instead you get two extra terms (because R(t) depends on t and the 2nd law is a second order differential equation), which are inertial forces: the centrifugal and Coriolis force. To keep the 2nd law invariant, you're only allowed for terms up to linear in the time t in your spatial coordinate transformations, i.e. Galilei boosts. If you want to rotate, you'd better do it once and for all.

GR is kind of special, because it's covariant under general coordinate transformations. I say "kind of", because every theory can be made generally-covariant. GR is special, in that it is a general-covariant theory in which the metric is the only geometric object and there is no "prior geometry". E.g., I can make the Poisson equation for a scalar field phi general covariant by writing

$$\nabla_i \nabla^i \phi = 0$$

and imposing the extra condition for the spatial metric,

$$R^i{}_{jkl} = 0$$

The special thing about GR is that the space(time) geometry is not "trivialized" by such an extra condition. Spacetime has its own dynamics, and one doesn't need extra fields besides the metric. Newtonian gravity can also be made general-covariant a la GR, where Newton's second law is made general-covariant by replacing it by a geodesic equation. But there you need a degenerate metric structure and an extra one-form if you want a theory fully analagous to GR (the connection then is not uniquely defined by metric compatibility). The origin of the one-form is algebraic and can be considered as the gauge field of the central extension of the Galilei algebra, but that's another story.

 

[haushofer]

@vanhees71 , so in GR, we only consider metric-compatible connections? Therefore, without explicit statement, we do not need to consider other situatiob, because it is not physically possible?

In e.g. supergravity the connection is necessarily torsion-full because of the contribution of the gravitino (a spinor field, as Vanhees71 indicates). But until now supergravity is in the realm of "beautiful theories without empirical evidence".