# The main obstacle in learning relativity related theories

1. Sep 3, 2010

### wdlang

i really cannot understand what contravariant and covariant mean

why do we need the machinery of raising and lowering the indices?

i think at least in pure mathematics, we do not need this type of trashes

i cannot see how this notation will not cause any inconsistency

it is really uneasy to see that L_r^s does not mean the same thing as L^r_s

why not just use the notation in pure maths, putting all the indices as subscripts?

2. Sep 4, 2010

### Fredrik

Staff Emeritus
I'm with you as far as elementary SR is concerned. I prefer to state results in special relativity as matrix equations and write all the indices downstairs. No need to even mention tangent spaces and cotangent spaces (if we define spacetime as a vector space instead of a manifold). Even the electromagnetic "field tensor" can be defined as a matrix-valued field.

However, in GR, and other areas of physics which require differential geometry, such as Yang-Mills theory and the most general version of Hamiltonian mechanics, there's no way to avoid talking about tangent spaces, cotangent spaces, and so on.

3. Sep 4, 2010

### dx

Because of the presence of a metric, people who learn differential geometry through relativity sometimes do not understand the very important conceptual differences between vectors and covectors, and cannot see the usefulness of the distinction.

This difference is not only very important in relativity, but in pretty much all areas of physics. For example, generalized force is a covector and not a vector, and this is required for a proper understanding of the principle of virtual work. In classical mechanics, to see the geometric meaning of 'canonical momentum' also requires one to understand this difference.

Last edited: Sep 4, 2010
4. Sep 4, 2010

### matheinste

I think perhaps one reason that the difference between the two is not appreciated is that in the usual rectangular Cartesian coordinate system that we are brought up with they have the same numerical components.

Matheinste

5. Sep 4, 2010

### HallsofIvy

Staff Emeritus
Exactly right. But, of course, in most general relativity problems, the curvature is crucial so we cannot use a rectangular Cartesian coordinate system. That's why, to answer the original question,we must distinguish between "covariant" and "contravariant" components of vectors and tensors.

6. Sep 4, 2010

### stevenb

Yes, and even in SR, where there is no curvature, the notation is needed because raising and lowering the time components generates a negative sign.

Perhaps a very simple way to appreciate the notation is to work first in Euclidean space (even two dimensions is enough with no time involved). Use skewed axes rather than orthogonal ones. Generate a metric and reciprocal base vectors. Then work out dot products and cross products.

7. Sep 4, 2010

### jcsd

Yep it just comes down to the metric. When the metric in some basis has the form of the kronecker delta (i.e. an identity matrix when written out in matrix form) then raising or lowering an index will not change the numerical value of the components making the distinction between covariant and contravariant indices non-obvious.

Even in Minkowski coordinates the distinction isn't so obvious as the only distinction between a vector in a Minkowski basis and it's covector in the associated Minkowski cobasis is that the time components have opposite signs.

8. Sep 4, 2010

### Mentz114

( I think someone's a bit tired )

wdlang, in physics we only observe scalars, and to get scalars, there must be an invariant inner product operation for vectors, and this requires a dual vector space. Can't be avoided.

Last edited: Sep 4, 2010
9. Sep 4, 2010

### wdlang

yes, i now realize that there are two different vectors under coordinate transform

yes, the gradient of the temperature field transforms in a different way than the flow velocity field

that is the point!

but in quantum field theory and relativity, why g_{\mu \nu} F^{\nu} transforms differently than F^{\nu} ? what is the effect of the metric g?

Last edited by a moderator: May 4, 2017
10. Sep 4, 2010

### Fredrik

Staff Emeritus
If V is a vector space, you can define the dual space V* without defining an inner product, and you can define an inner product without defining a dual space.

If $F^\nu$ are the components of a tensor field in a coordinate system, then that tensor field is

$$F^\nu\frac{\partial}{\partial x^\nu}$$

If $g_{\mu \nu} F^{\nu}$ are the components of a tensor field in a coordinate system, then that tensor field is

$$g_{\mu \nu} F^{\nu}dx^{\mu}$$

I would write the values of those two tensor fields at a point p in the manifold as

$$F^\nu(p)\frac{\partial}{\partial x^\nu}\bigg|_p$$

and

$$g_{\mu \nu}(p) F^{\nu}(p)dx^{\mu}|_p$$

The $$\frac{\partial}{\partial x^\nu}\bigg|_p$$ are basis vectors for the tangent space at p, $T_pM$. The $dx^{\mu}|_p$ are basis vectors for its dual space, $T_pM^*$ or $T_p^*M$. Specifically, they are the members of the dual basis of that particular basis of the tangent space.

See this post for the definition of "dual space", and click the link in it for the definition of "tangent space at p".

If $\{e_i\}$ is a basis of a vector space V, then its dual basis, which I will write as $\{e^i\}$, is defined by $e^i(e_j)=\delta^i_j$. Suppose that $\{f_i\}$ is another basis of V, related to the first by $f_i=A_i^j e_j$. (By definition of "basis", the $f_i$ have to be linear combinations of the $e_i$). The $A_i^j$ can be thought of as the components of a matrix A (row j, column i).

$$\delta^i_j=f^i(f_j)=f^i(A_j^k e_k)=A_j^k f^i(e_k)$$

$$(A^{-1})^j_l\delta^i_j=(A^{-1})^j_l A_j^k f^i(e_k)$$

$$\delta^i_l=(A^{-1}A)^k_l f^i(e_k)=\delta^k_l f^i(e_k)=f^i(e_l)$$

$$A^k_i\delta^i_l=A^k_if^i(e_l)$$

$$\delta^k_l=A^k_i f^i(e_l)$$

$$\Rightarrow e^k=A^k_i f^i$$

$$(A^{-1})^j_k e^k=(A^{-1})^j_k A^k_i f^i=(A^{-1}A)^j_i f^i=\delta^j_i f^i=f^j$$

So when the basis "transform" by a matrix A, the dual basis "transforms" by A-1.

Last edited: Sep 4, 2010
11. Sep 4, 2010

### jcsd

Yep, V* is just the set of all the linear functionals on V. Nowhere in the definition do you need a scalar product you just need V.

Though of course on the other hand defining a scalar product on V, in the finite-dimensional case, you also define a 'natural bijection between V and V*. The scalar products for a vector v in V with all the other vectors in V defines a linear functional on V. All linear functionals on V can be defined in this way, so the scalar product defines a 'natural' bjiection between V and V*. From a multilinear algebra point of view this natural bijection would be what is called the metric tensor, though I'll re-emphasize what is written in this paragraph is only true when V is of finite dimension.

Of course, reading your posts which are well-informed on the subject, I'm 100% certain you knew this already and perhaps my post is really semantic quibbling over the word 'define'

12. Sep 4, 2010

### Fredrik

Staff Emeritus
I do, as you can see here, and I could do some semantic quibbling about how you use the word "natural".
(I would say that the isomorphism between V and V** is natural because it can be defined without the metric. Apparently there's also a fancy definition of "natural" that I don't know myself. Link. Hm...I also found this post when I did the search. Looks like I will have to read it.)

Last edited: Sep 4, 2010
13. Sep 4, 2010

### Hurkyl

Staff Emeritus
An interesting oddity about my education -- when I was first introduced to multi-variable calculus, I found it easier to use after I worked out the notion of "covector" on my own, in the form of making a careful distinction between row vectors and column vectors -- e.g. always writing the gradient of a scalar function as a row vector, and always writing vectors representing position or direction as column vectors.

I admit it has left me with the problem of being mildly irritable every time I'm faced with a context that converts everything into vectors or otherwise removes the distinction between the two.

Last edited: Sep 4, 2010
14. Sep 4, 2010

### jcsd

Yep, I was certain you knew this as my observation was hardly Earth-shattering!

I have heard 'natural' used in the context I have used it, I didn't use particularly confidently hencde the use of scare quotes. Thinking about it further, though I don't know much about catergory theory, I'm guessing when I use the term 'natural' to describe the bijection between V and V* given by the scalar product, I'm talking about a natural isomorphism in the catergory of finite-dimensional inner product spaces.