# Physical meaning of index gymnastics

1. Dec 21, 2009

### gnieddu

What is the physical meaning of raising/lowering indexes?

From a mathematical standpoint, I clearly understand what an expression like $$v_a = v^bg_{ab}$$ means. But let's assume that $$v^a$$ is, say, a 4-velocity: can I say that $$v_a$$ is a 4-velocity as well? Or is it something different?

Not to speak of things like $$\nabla^av_a$$, which can be obtained with some index maths from $$\nabla_av^a$$. In my mind, $$\nabla_a$$ is associated with the idea of covariant derivative, but what about $$\nabla^a$$?

Thanks to whoever could shed some light on this.

2. Dec 22, 2009

### Mentz114

There is a geometrical meaning that is straightforward. Think of a two orthogonal axes (x, y) in two dimensions. The coordinates of a point are defined unambiguously by drawing a perpendicular to each axis. Now let the axes bend, and perhaps no longer intersect at a right angle.

We can define a set of coordinates by the method above, but they suffer from the failure to make the distance between points invariant under coordinate transformations.

We can define a second set of coordinates for a point by projecting a line through the point parallel to one axis and taking our value where this line cuts the other axis. So every point can be given 2 sets of coords, the perpendicular and the parallel-projection type.

It happens that if we have two points A, B then the quantity

$$(x_{A,para}-x_{B,para})(x_{A,perp}-x_{B,perp})+(y_{A,para}-y_{B,para})(y_{A,perp}-y_{B,perp})$$

is invariant under coordinate transformations. You'll have worked out that the perpendicular and parallel-projection coords are respectively the components of a covariant and contravariant vector. Those terms come from the fact that the inner product defined as above varies with the covariant components, but against the contravariant components. If a coordinate transformation increases the value of the covariant components, then the corresponding contravariant component must decrease, in order that the length stays the same.

The physical usefulness comes from the invariance of scalars formed by contracting tensors.

3. Dec 22, 2009

### gnieddu

Thanks for the clarification. So, if I wanted to rephrase it using my terms $$v^a$$ and $$v_a$$ are just two different ways of expressing the same entity (in my example, a 4-velocity), as viewed under different coordinate types. If the reference frame is cartesian and orthogonal, the two "views" coincide, and $$v^a=v_a$$.

Did I get it right?

4. Dec 22, 2009

### Mentz114

Yes.

Of course, the inner-product of $$v^a$$ and $$v_a$$ is $$c^2$$ the best known invariant.

5. Dec 22, 2009

### haushofer

I never really got into the precise geometrical picture of "contravariant" and "covariant" components, altough I've seen how you could depict them. I like to think of them mathematically; if you use vectors in a certain vector space to describe physical things, then there is also a corresponding dual vector space. If you want to describe scalar things, you need a vector from the vector space, and a covector from the covector space.

Because these two spaces are diffeomorphic, there is a correspondence between the two. That's the metric.

6. Dec 22, 2009

### atyy

Roughly the down and up index vectors are row and column vectors, which are different spaces. They are different spaces. You can multiply row and column vectors to get a number, even without a metric. You cannot multiple two column vectors unless you have a metric. If you have a metric, then you can multiply two column vectors.

I am not sure about this, but I believe that without a metric, there is no correspondence between row and column vectors. However, if you specify a basis for the column space, even without a metric, that specifies a dual basis for the row space, which changes with every different basis for the column space. In the presence of a metric, there is a correpondence between row and vector spaces. Upon Rn, one can put many different metrics.

7. Dec 22, 2009

### bcrowell

Staff Emeritus
But keep in mind that there can be a useful distinction between $v^a$ and $v_a$ even in one dimension, even though there is no useful distinction between row vectors and column vectors in one dimension.

The way I like to think about it is this. Suppose you have a one-dimensional space labeled by a coordinate x, with differences in x not necessarily corresponding to metric distances in any simple or linear way. Suppose you have points P and Q, separated by an infinitesimally small difference in x, dx. Now given dx, you want to find out the squared metric distance $ds^2$ between P and Q. There has to be some conversion factor that fills in the blank in the relationship $ds^2=(\ldots)dx$. Since we're in one dimension, the superscripts and subscripts can only take on a single value, say 1. Then the thing we've been referring to as dx can be notated as $dx^1$, and the mysterious conversion factor can be notated as $dx_1$.

In principle, the distinction between a vector $u_a$ and its dual $u^a$ is completely arbitrary. You could take every single equation in a general relativity book and swap the upper indices to lower and lower indices to upper, and everything would be just as valid as before. The only thing is that there's a convention that if we start talking about coordinates like t, x, etc., these are to be construed as components of the upper-index version of the vector.

8. Dec 23, 2009

### gnieddu

Thanks everybody for your comments. There is still a point, though, on which I'm not so sure, and it was hinted at in the last part of my original post.

Let's start with $$\nabla_av^a$$. The effect of the co-variant derivative $$\nabla_a$$ is to turn the vector $$v^a$$ into a (1,1) tensor. At this point, I can raise and lower indexes, and get another (1,1) tensor, for instance:

$$\nabla_av^a=\nabla_av^ag^{ab}g_{ab}=\nabla^bv_b$$

From a tensor perspective, this looks fine. But is $$\nabla^b$$ still a derivative? Perhaps a contra-variant derivative (not sure if this term has any meaning at all)?. Or has the index gym turned $$\nabla_a$$ into something completely different?

9. Dec 23, 2009

### bcrowell

Staff Emeritus
You may be getting confused by the confusing way the term "covariant" is used. In the term "covariant derivative," what "covariant" means is that it respects the fundamental invariance of GR under arbitrary smooth, one-to-one coordinate transformations; this property of the theory is known as general covariance. The use of the terms "covariant" and "contravariant" for upper and lower indices is only vaguely related to the former usage. Yes, $$\nabla^b$$ is a perfectly reasonable derivative operator.