# Contravariant and Covariant Vectors

1. Jan 28, 2014

### kderakhshani

I remember I have read somewhere that contravariant/covariant vectors correspond to polar/axial vectors in physics, respectively. Examples for polar/axial vectors are position, velocity,... and angular momentum, torque,..., respectively.

Is this right?

Can I prove that, say, any axial vector in physics transforms like a covariant vector under a coordinate transformation?

2. Jan 28, 2014

### Staff: Mentor

3. Jan 28, 2014

### kderakhshani

OK! Now remember that the angular momentum is an axial vector (it gains no minus sign under inversions).
On the other hand, it is the cross product of x and p, both of which are contravariant vectors.
In the component notations, it is the product of the Levi-Civita symbol and two contravariant components. The result is the covariant component of the angular momentum. Thus, if x and p are contravariant vectors, L is necessarily a covariant one.
This line of reasoning is also true for every other cross product, say, the curl.

Anything wrong here? What?

Thanks

4. Jan 28, 2014

### Staff: Mentor

Yes.

First, an important note: if you have a metric, you can raise or lower indexes on any vector (or more generally on any tensor), so there is a 1-to-1 correspondence between contravariant and covariant vectors. So whether a vector is treated as contravariant or covariant is a matter of choice.

Physically, I've always seen position, $\vec{x}$, treated as a contravariant vector, yes. But momentum, $\vec{p}$, is not so simple. If you look at momentum as the time derivative of position, $\vec{p} = m d \vec{x} / dt$, then yes, momentum is naturally a contravariant vector. However, if you look at force as the time rate of change of momentum, $\vec{F} = d \vec{p} / dt$, then momentum is naturally a covariant vector, because force is naturally a covariant vector: work done is the dot product of force with displacement, $W = \vec{F} \cdot \vec{x}$, and a dot product naturally multiplies one contravariant and one covariant vector (in component notation this is obvious: we would have $W = F_a x^a$). Since position is contravariant, as above, force must be covariant, so on this view, momentum is naturally covariant as well.

Again, since you can always use the metric to raise or lower indexes, both of these views of momentum are consistent with each other; but if you're talking about physical interpretation, it's worth keeping in mind the different ways in which various vectors enter into the equations that describe the physics.

First, another important note: the cross product of two vectors is not actually a vector, strictly speaking; it's an antisymmetric 2nd-rank tensor. It just so happens that in 3-dimensional space, there is a 1-to-1 correspondence between antisymmetric 2nd-rank tensors and pseudovectors. That's where pseudovectors actually come from, strictly speaking; they're a way of simplifying things by trading 2 indexes for one. The component notation you are talking about is the realization of this correspondence between antisymmetric 2nd-rank tensors and pseudovectors.

But, as above, vectors are not "necessarily" contravariant or covariant, since you can use the metric to raise or lower indexes. Also, as I noted above, momentum can be naturally viewed as a covariant vector. The cross product of a contravariant vector and a covariant vector can be either covariant or contravariant, depending on whether you choose to raise or lower an index.

Yes, including everything I said above. Also, cross products are not the only way that pseudovectors can arise (although they're the most common way in 3 dimensions). The bottom line is, as I said before, that the distinction between vectors and pseudovectors is not the same as the distinction between contravariant and covariant vectors; you can have contravariant vectors or pseudovectors, and you can have covariant vectors or pseudovectors. There's no necessary connection between the two.

5. Jan 29, 2014

### kderakhshani

PeterDonis,

Right. A metric can do that, but it is not just a trivial task of index gymnastics! A metric does not convert the vector to itself, but to its dual which is a one-form. There IS a 1-1 correspondence here, but vectors and their corresponding 1-forms are NOT the same. In a little more mathematical words, the metric is a linear mapping between a vector space and its dual vector space.
The inner product is also a multiplication of a vector and a 1-form, a mapping from a vector space and its dual space onto the set of real/complex numbers. Remember the Dirac's <bra|ket> notation in QM.

No, it is not a matter of choice, but a matter of transformation.
Contravariance or covariance of vectors are defined according to their behaviors under coordinate transformations.

Just to remind:
If under the coordinate transformation: $x^{'i}=x^{′i}(x^1,x^2,x^3,⋯,x^n)$ a vector transforms as $V^{'j}=V^i∂x^{′j}/∂x^i$ then it is a $\textbf{contravariant vector}$, and if it transforms as $V'_j=V_i∂x^{′j}/∂x^i$ then it is a $\textbf{covariant vector}$ or a $\textbf{covector}$. This definitions are easily generalized to higher rank tensors.(Any reference needed?)

Thus, physical vectors such as position, velocity, acceleration, etc. are contravariant vectors, and others,especially those which are the gradient or curl of scalars, such as electric, magnetic, gravitational fields, are covariant vectors. This is why

I think this discrepancy is removed in special relativity, where there is no physical distinction between contravariant and covariant quantities.

Anyway, now that physical position $\vec{x}$ and velocity $\vec{v} = d \vec{x} / dt$ are intrinsically contravariant, how about the angular momentum per unit mass: $L_i = ε_{ijk}x^jv^k$ ? Is it intrinsically covariant?
Can I claim in this way that every 3-D vector which is a cross product of two contravariant vectors is a covariant one?

Thanks a lot

6. Jan 29, 2014

### Staff: Mentor

Agreed. (I note, btw, that you have shifted terminology; what you are now calling a "vector" is what you were previously calling a "contravariant vector", and what you are now calling a "1-form"--another term would be "covector"--is what you were previously calling a "covariant vector". I prefer the new terminology.) When I said a physical quantity like position is "naturally" a vector, I meant that physically, its representation as a vector is the one that directly represents the physics; its representation as a covector (the dual of the "natural" vector) brings in the issues you raise.

Yes, but again, since the metric provides a 1-to-1 mapping between vectors and covectors, we can always find a pair of objects, one with each of the 2 types of transformation properties, that represents a physical quantity. The question is which representation is the "natural" one, as above.

I agree that the "natural" representation of the gradient is a covector (MTW goes into some detail about this in an early chapter). But I note that you did not include momentum in your list of "natural" vectors. That's good, because, as I noted before, momentum has both a vector and a covector interpretation. See below.

Huh? There certainly is; it's the same distinction you described, to do with transformation properties. Covectors are not the same as vectors in SR, any more than they are in ordinary vector analysis.

But the momentum $\vec{p}$ is not; more precisely, whether or not it is depends on which physical interpretation of momentum you are using, as I described before.

This would be a covector as you've written it, yes. (Note that I'm glossing over the issue I mentioned before, that actually what we have here is an antisymmetric 2nd-rank tensor, which is mapped to a covector using the Levi-Civita symbol. This mapping only works in 3 dimensions.) But note that it only has a direct physical interpretation for test objects whose rest mass is constant. The more general formula for angular momentum that you gave before (the cross product of position and momentum, rather than position and velocity) works for any object. But, as I noted before, if momentum is "naturally" a covector (which it is if it is defined such that force is the time derivative of momentum), then the cross product of position and momentum is an antisymmetric 2nd-rank tensor with one upper index and one lower index, which can't be mapped "naturally" to either a covector or a vector; you have to either raise or lower an index before contracting with the Levi-Civita symbol.

Phrased this way, yes, the claim is true--though you have used your terminology inconsistently; a better phrasing would be "the cross product of two vectors is a covector" (and an even better phrasing would be "the cross product of two vectors is an antisymmetric 2nd-rank tensor, which can be mapped to a covector using the Levi-Civita symbol"). But it still leaves open the question of which physical quantities are "naturally" vectors vs. covectors.

7. Jan 29, 2014

### WannabeNewton

No. An expression like $\epsilon_{\mu\nu\alpha\beta}u^{\nu}x^{\alpha}u^{\beta}$ is manifestly covariant. If you represent it in the rest frame of the particle in question then it reduces to what you wrote. The same goes for the curl of a vector field: the manifestly covariant form is $\epsilon_{\mu\nu\alpha\beta}u^{\nu}\nabla^{\alpha}u^{\beta}$ which in the momentarily comoving local inertial frame reduces to $\vec{\nabla}\times \vec{v}$.

8. Jan 29, 2014

### WannabeNewton

By the way we can cast the above in a very general context. Recall that antisymmetric 4-tensors are irreducible representations of $SO(3,1)$. However they are reducible under $SO(3)$ into two $j = 1$ angular momentum representations. So given such an $J^{\mu\nu}$ we can say that $J^{\mu\nu}\in 1 \oplus 1$. Given a 4-velocity $u^{\mu}$ we can therefore write $J^{\mu\nu} = 2K^{[\mu}u^{\nu]} + \frac{1}{2}\epsilon^{\mu\nu\alpha\beta}u_{\alpha}J_{\beta}$. If the $J^{\mu\nu}$ correspond to the generators of the Lorentz algebra then $K^{i}$ is just the boost operator and $J^i$ is the angular momentum operator. If we instead consider an electromagnetic field $F_{\mu\nu}$ then we get the electric and magnetic fields. Yet another example would be the vorticity tensor $\omega_{\mu\nu}$ which would give us the vorticity vector $\omega^i$ mentioned above, as well as the acceleration $a^i$. This is what Peter was referring to.