1
$\begingroup$

In most presentations of general-relativity I see the following statement,

We can change from a covariant vector to a contravariant vector by using the metric as follows, ${ A }^{ \mu }={ g }^{ \mu \nu }{ A }_{ \nu }$

My questions are,

  1. What is the need to do this particular change in relativity?
  2. The covariant components represent the components of a vector the contravariant components represent the components of a dual-vector, for finite dimensional vector spaces the two spaces are isomorphic. What is the significance of representing a quantity in contravariant or convariant forms? Is the need purely mathematical?
$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/105347/2451 and links therein. $\endgroup$ – Qmechanic Jan 13 '18 at 6:22
  • $\begingroup$ The first answer in the link there says the samething the I mean to ask, the author says that they are isomorphic which does not mean they are the same. $\endgroup$ – Abhikumbale Jan 13 '18 at 6:40
  • 2
    $\begingroup$ When the metric is fixed and constant, we get the illusion that there is an automatic, obvious identification of vectors and covectors, and also that things like dot products are just obviously right when expressed in a certain way. That isn't so, and nothing works right in relativity until you recognize that those manipulations depend on the metric and don't work the same way when the metric varies. When you take a dot product in freshman physics, you are actually converting a vector into a covector. You just don't realize you're doing it, because the metric happens to be diag(1,1,1). $\endgroup$ – Ben Crowell Jan 13 '18 at 19:37
0
$\begingroup$

It matters of two different concepts. Given a manifold, a vector is a geometric object associated to each point in the manifold. It can be decomposed into components with respect to a set of basis vectors.
$A = A^\mu \hat e_{(\mu)}$
where:
$\mu = 0, 1, 2, 3$
$A$ vector
$A^\mu$ contravariant components
$\hat e_{(\mu)}$ basis vectors
The geometric object is a reality independently of the coordinate system. A characterization is given by its square.
$A^2 = A \cdot A = A^\mu \hat e_{(\mu)} \cdot A^\nu \hat e_{(\nu)} = \hat e_{(\mu)} \cdot \hat e_{(\nu)} A^\mu A^\nu = g_{\mu\nu} A^\mu A^\nu$
where:
$\cdot$ scalar (dot) product
$g_{\mu\nu} = \hat e_{(\mu)} \cdot \hat e_{(\nu)}$ metric tensor
The square can also be written as
$A^2 = A_\mu A^\mu$
where:
$A_\mu = g_{\mu\nu} A^\nu$
As per above, we can define the dual vector.
$\tilde A = A_\mu \hat \theta^{(\mu)}$
where:
$\tilde A$ dual vector
$A_\mu$ covariant components
$\hat \theta^{(\mu)}$ basis dual vectors
By demanding
$\hat \theta^{(\mu)} (\hat e_{(\nu)}) = \delta^\mu_\nu$
where:
$\delta^\mu_\nu$ Kronecker delta
we can write the action of the dual vector on the vector as
$\tilde A (A) = A_\mu \hat \theta^{(\mu)} (A^\nu \hat e_{(\nu)}) = A_\mu A^\nu \hat \theta^{(\mu)} (\hat e_{(\nu)}) = A_\mu A^\nu \delta^\mu_\nu = A_\mu A^\mu$
Hence, a dual vector is a linear map from the vector space to the real numbers.
By defining the inverse metric tensor as
$g^{\mu\lambda} g_{\lambda\nu} = \delta^\mu_\nu$
where:
$g^{\mu\nu}$ inverse metric tensor
we have also
$A^\mu = g^{\mu\nu} A_\nu$

$\endgroup$
  • $\begingroup$ This is just definitions $\endgroup$ – Bellem Jan 14 '18 at 12:56
  • $\begingroup$ By contracting a vector with its dual you get a scalar, the norm of the vector (its square in reality), which is an invariant. If the vector is given by the coordinates you get the squared distance in spacetime, which is fundamental in both special and general relativity. That is why you need both contravariant and covariant components, however the latter have a different meaning as they define a linear application. $\endgroup$ – Michele Grosso Jan 14 '18 at 17:08
  • $\begingroup$ Square distance? In gr you don't get square dinstance by squaring a vector, you can do that obly in sr. Then you can obtain the same thing through the definition of the pairing product, which gives you exactly the same results. You simply want to stress the importamce of having a metric manifold, which I agree with you, but the question was different. Of course if you want to have a metric manifold raising and lowering indices are a consequence... $\endgroup$ – Bellem Jan 14 '18 at 18:22
0
$\begingroup$
  1. Because you need to identify vectors with covectors and that is possible only through the metric. If you think about components: you have upper and lower indices which reflect the way in which a tensor transforms under a change of coordinates. The position of indices reflects only that, thus I want something which allows me to indentify tensor with different disposition of indices.

  2. The difference is that if you lower an upper index you change the quantities it contains by the metric. If you have the momentum four-vector $P^\mu=(E, \textbf{p})$ and you lower the index, in particular, $P_o$ is not the energy anymore, since you get $P_o=g_{o\mu}P^\mu$.

This is just a rough treating though, cheers!

$\endgroup$
  • $\begingroup$ Vectors and co-vectors can be identified without the metric. $\endgroup$ – Abhikumbale Jan 13 '18 at 17:52
  • $\begingroup$ I mean identify vectors with covectors of course... Now I edit $\endgroup$ – Bellem Jan 13 '18 at 18:07
  • $\begingroup$ The point is that the position of indices gives you only information about the way under a tensor transforms and no more. So there should be something which allows me to move them... $\endgroup$ – Bellem Jan 13 '18 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.