# Parallelism between a vector and a covector

1. Dec 16, 2013

### bcrowell

Staff Emeritus
The notion of parallelism between a vector and a covector comes up naturally in the following context. Say you have a scalar field T that measures the temperature of the CMB. Then $\nabla_a T$ is a cosmologically preferred covector field. I would think of it as being "parallel" to the velocity vector field $v^a$ of the Hubble flow, but since vectors and covectors live in different vector space, there is no natural notion of parallelism between them.

One thing I could do would be to raise or lower an index, so I could say that $v^a$ was parallel to $\nabla^a T$. Or I could say that $v^a$ maximizes $v^a \nabla_a T$ subject to the constraint $v^av_a=1$. (I'm using a +--- metric.)

Are these equivalent?

2. Dec 16, 2013

### ChrisVer

I think they are equivalent, since you are working in a metric space, and so you can always do the map from the vector space to the dual space, and inversely...

3. Dec 16, 2013

### Mentz114

I'm not sure if $t_a=\partial_a T$ is a 4-vector. It has norm $g_{ab}t^a t^b$ and I don't see why this should be any particular value. It could still be 'parallel' to the cosmic flow but I'm not seeing how to express this right now.

The Einstein tensor in the FLRW spacetimes is that of a perfect fluid or dust. These do not permit any heat conduction or convection.

Sorry this hasn't been much help, but your question is thought provoking.

4. Dec 16, 2013

### ChrisVer

what kind of d-vector could it be?
Nevertheless, is it 4,5 or N as long as you define a metric (as the isomorphism map from the one vector space to the other) it's all fine

5. Dec 16, 2013

### Mentz114

Yes, the test is that it has the right transformation properties.

6. Dec 16, 2013

### WannabeNewton

Assume $\nabla^a T = \alpha v^a$ then such an equivalence would imply that at any given point in space-time, $v^a \nabla_a T = \alpha$ is maximized amongst the set of all vectors $w^a$ at that point such that $w^a w_a = 1$; clearly this implies $\alpha > 0$ because if $\alpha < 0$ then the vector field $-v^a$, which is also of unit norm, would satisfy $-v^a\nabla_a T = -\alpha > \alpha$ which trivially contradicts the equivalence of the statements.

Let $v^a$ be part of an orthonormal basis $\{v^a,x^a,y^a,z^a\}$ at said point then $w^a = \beta v^a + \gamma x^a + \mu y^a + \nu z^a$ hence $w^a \nabla_a T = \alpha \beta$. The unit norm condition implies (using your sign convention) $\beta^2 - \gamma^2 - \mu^2 - \nu^2 = 1$ i.e. either $\beta \geq 1$ or $\beta \leq -1$. If $\beta > 1$ then $w^a\nabla_a T > \alpha$ at this point which would contradict the claim that $v^a \nabla_a T = \alpha$ is maximized at any point amongst all unit norm vectors at that point. But we can easily construct such a $w^a$: take $\beta = 2$ and $\gamma = \mu = \nu = 1$.

7. Dec 16, 2013

### bcrowell

Staff Emeritus
The covariant gradient of a scalar field is always a valid covector. (It has to be a covariant derivative, not just an ordinary partial derivative.)

8. Dec 16, 2013

### Staff: Mentor

Is there any difference for a scalar field? Since the field has no indexes, there are no terms in the connection coefficients.

9. Dec 17, 2013

### stevendaryl

Staff Emeritus
That's not the way I think of it. To me, a 4-vector is by definition the tangent to a parametrized path passing through a point. To me, a 4-covector is by definition the gradient to a scalar field. The stuff about components transforming in the right way is saying that if you have 4 numbers, then those 4 numbers can be thought of as components of a 4-vector (or covector) if they transform the right way.

If temperature is a scalar field, then $\partial_i T$ are necessarily components of a covector. I think that temperature is a scalar.

10. Dec 17, 2013

### stevendaryl

Staff Emeritus
You're right---for a scalar, there is no difference between covariant derivative and partial derivative.