Tensor Questions: Significance & Using Einstein's Equations

[khemist]

I have been watching lecture videos on relativity and I have two questions that have not really been answered yet.

1. What is the physical significance of a contravariant and covariant tensor? I understand the indices are writing either "upstairs" or "downstairs," but in the lecture video, the lecturer seems to claim that it really doesn't matter which is which, except for particular cases.

2. How would one use Einsteins equations to calculate, say, the orbit of mercury? What is the kind of information necessary to find those values? If anyone has any resources about implementing the equations to a problem that would be awesome!

 

[bcrowell]

2. How would one use Einsteins equations to calculate, say, the orbit of mercury? What is the kind of information necessary to find those values? If anyone has any resources about implementing the equations to a problem that would be awesome!

http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2

See subsection 6.2.6.

 

[Mentz114]

1. What is the physical significance of a contravariant and covariant tensor? I understand the indices are writing either "upstairs" or "downstairs," but in the lecture video, the lecturer seems to claim that it really doesn't matter which is which, except for particular cases.

It's very hard to explain this without mathematics.

In flat spacetime we can define a point ( which may be the tip of as vector) using 4 numbers. But in curved spacetime we need two sets of 4 numbers, which are called the contravariant and covariant components. The reason for this is to ensure that the scalars we get by 'contracting' covariant and contravariant tensors are unchanged under coordinate transformations.

So, the physical significance is that observable physical phenomena (which must not depend on a choice of coordinates) must correspond to scalars formed by contracting tensors. Thus the laws of physics must be expressed in tensors or some equivalent formulation.

( see, I said it was hard ...)

 

[khemist]

It's very hard to explain this without mathematics.

In flat spacetime we can define a point ( which may be the tip of as vector) using 4 numbers. But in curved spacetime we need two sets of 4 numbers, which are called the contravariant and covariant components. The reason for this is to ensure that the scalars we get by 'contracting' covariant and contravariant tensors are unchanged under coordinate transformations.

So, the physical significance is that observable physical phenomena (which must not depend on a choice of coordinates) must correspond to scalars formed by contracting tensors. Thus the laws of physics must be expressed in tensors or some equivalent formulation.

( see, I said it was hard ...)

Feel free to throw the math out here.

So are you saying that any tensor has both contravariant and covariant indices, regardless of the fact the index is written up or downstairs? Or any tensor can be converted from contra to co variant?

I guess more specifically, why would one write a tensor with a contravariant index over covariant, or vise versa?

Thanks for the resource, bcrowell.

 

[Bill_K]

The covariant and contravariant components of an object transform the same as long as you restrict yourself to Lorentz transformations, but as soon as you consider more general transformations they transform differently. Contravariant components transform like a position vector while covariant components transform like a gradient. For example if you do a scale transformation x' = α x, contravariant components become larger by a factor α, while covariant components become smaller by a factor 1/α.

 

[khemist]

The covariant and contravariant components of an object transform the same as long as you restrict yourself to Lorentz transformations, but as soon as you consider more general transformations they transform differently. Contravariant components transform like a position vector while covariant components transform like a gradient. For example if you do a scale transformation x' = α x, contravariant components become larger by a factor α, while covariant components become smaller by a factor 1/α.

This is what I was looking for. Thanks very much.

Do all tensors have both types of components? I would assume not.

 

[Mentz114]

Feel free to throw the math out here.

So are you saying that any tensor has both contravariant and covariant indices, regardless of the fact the index is written up or downstairs? Or any tensor can be converted from contra to co variant?
Thanks for the resource, bcrowell.

Yes. The indexes of a tensor are raised or lowered by operating on it with the metric. For a vector

<br /> U^\mu=g^{\mu\alpha}U_\alpha = g^{\mu 0}U_0 + g^{\mu 1}U_1 + g^{\mu 2}U_2 + g^{\mu 3}U_3 <br />

The tensors R^ab, R_ab and R^a_b are the same thing expressed in different components.

I'm not sure what you mean by this

"I guess more specifically, why would one write a tensor with a contravariant index over covariant, or vise versa?"

[Edit : I was misreading that question. ]

You can use whatever is required by the situation. These equations represent the same physics

<br /> f_\mu= A_{\mu\nu}j^\nu,\ \ \ f^\mu= A^{\mu\nu}j_\nu<br />

 

[WannabeNewton]

I think the whole contravariant and covariant concept is outdated nowadays. It is more intuitive to think of tensors as multilinear mappings of vectors and one forms into the reals where, in this case, vectors are what you would recognize as contravariant and one forms as covariant. A vector can be seen, in GR at least, as being part of the set of vectors existing in the tangent space to a 4 - manifold at some one point on the manifold. A one form can then be seen as part of the set of one forms existing in the cotangent space to the 4 - manifold at some point on the manifold. A tensor takes such one forms and vectors and maps them into the reals.

As for the orbit of mercury you could use the schwarzschild metric for the geodesic equation to solve for the geodesic along which the object will move as a function of say proper time.

 

[khemist]

Thanks for the excellent responses.