# Contra and covariance question

1. Apr 29, 2015

### Coffee_

From what I understood, a vector can be either covariant or contravariant. Which one this is will depend on how the coordinates of this vector transform under a coordinate transformation. Let's take a look at the electric field then:

$\vec{E}=-\nabla{V}$, so here it looks as if the electric field is covariant.

However if we have a particle in space, it will accelerate due to this electric field and now the electric field can be expressed as :

$\vec{E}=c\vec{a}$ where $\vec{a}$ is contravariant , and so here it looks like the electric field is contravariant.

So is my understanding about vectors being either covariant or contravariant incorrect?

2. Apr 29, 2015

### bcrowell

Staff Emeritus
A relativist's answer would be that the electric field is not a vector at all, because it doesn't transform as any kind of vector. The electric and magnetic fields together make up a tensor.

Anyway, as long as you have a metric, any vector can have its indices raised or lowered, so it can exist in either covariant or contravariant form. One can worry about whether it's more natural to consider a certain vector as covariant or contravariant, but such a worry is pointless in the context of relativity, where we always have a metric.

3. Apr 29, 2015

### Mentz114

It is the same vector expressed in a different way. We can write a vector V either as $V^\mu$ or $V_\mu$.

4. Apr 29, 2015

### Coffee_

Is this an alright way to think about it?

''Vectors themselves are either solely contra or covariant. However with physical quantities like the electric field, come two different vectors, one covariant and one contravariant which can be transformed into eachother with the metric?''

5. Apr 29, 2015

### Coffee_

This is bad new for my few hours of studying this subject, my understanding of the concepts is all wrong then. So we cannot call a vector ''covariant'' or ''contravariant'' any vector however has contravariant and covariant components?

6. Apr 29, 2015

### Mentz114

Last edited: Apr 29, 2015
7. Apr 29, 2015

### stevendaryl

Staff Emeritus
I would not put it that way. Any vector of the one type can be converted into a vector of the other type using the metric tensor: $V_\mu = \sum_\nu g_{\mu \nu} V^\nu$. But for 3-vectors in Cartesian coordinates, the metric tensor is trivial: $g_{ij} =0$ (if $i \neq j$) and $g_{ii} = 1$. So for 3-vectors in Cartesian coordinates, $V^j = V_j$

8. Apr 29, 2015

### Mentz114

Why not ? When an index is raised or lowered the components change.

9. Apr 29, 2015

### bcrowell

Staff Emeritus
Not true. I can for example take Cartesian coordinates and rescale them uniformly. The result is then a different set of Cartesian coordinates with a different metric tensor.

In any case, the reason all of these concepts were invented was precisely so that we could be permissive about coordinate systems rather than prescribing a preferred set of coordinates.

10. Apr 29, 2015

### stevendaryl

Staff Emeritus
Why not what? Why not say that a vector with components $V^\mu$ is the same as a covector with components $V_\mu$?

Because they are fundamentally different geometric objects. If you have a metric tensor (which you usually do for vectors in space), then you can convert between them, but they are very different types of objects.

The example I like to give is atmospheric pressure, which is a function of temperature and altitude on the Earth. So you can make an abstract space with one variable, $T$ being temperature and another variable, $H$ being altitude. Pressure would be a scalar field $P(T,H)$ on this abstract space.

Now, in this abstract space, there are two very different types of vector-like objects. Given a scalar field such as $P$, we can construct a vector describing how $P$ varies with "location" within the 2-D space:

$\nabla P$ has components $P_T = \frac{\partial P}{\partial T}$ and $P_H = \frac{\partial P}{\partial H}$.

On the other hand, if both $H$ and $T$ are varying with time (because you're in a car traveling into the mountains), then you can track your position through the abstract space using an abstract "velocity" $V$ with components $V^T = \frac{dT}{dt}$ and $V^H = \frac{dH}{dt}$

$V$ can ONLY be a tangent vector, which is something with contravariant components. $\nabla P$ can ONLY be a covector, which is something with covariant components. There is no "metric" that would allow you to convert one type of vector to another.

11. Apr 29, 2015

### stevendaryl

Staff Emeritus
Well, it is true that in 3D you can often get away with not distinguishing between vectors and covectors. They don't typically teach the distinction until Special Relativity and sometimes not until General Relativity.

Anyway, I think we're on the same side here. I think that in general tangent vectors and covectors (or one-forms) are very different objects, and they only seem interconvertible because we have a metric.

12. Apr 30, 2015

### Coffee_

You don't seem to have such similar views to me. So a vector is either covariant or contravariant and can be converted to a different vector that is the opposite when you have a metric?

13. Apr 30, 2015

### Mentz114

stevendaryl correctly points out that when we come to give numbers to the vectors, tangent space vectors are contravariant and their derivatives are contravariant. If I measure a velocity in some x-direction, I'm measuring the x component of a contravariant vector.

But in any expressions we are free to write a covariant vector with its contravariant components if we want to.

So $V^\mu V_\mu$ still makes sense whether $V$ is a tangent space vector or not. So, in these terms. $V^\mu$ and $V_\mu$ represent the same vector.

I don't think we disagree at all.

Last edited: Apr 30, 2015
14. Apr 30, 2015

### robphy

"Getting away with not distinguishing vectors and covectors" is not really about the dimensionality of space, but more about the Euclidean metric (that is, the Pythagorean theorem) in arbitrary dimensions. If you have a Minkowski-signature metric in 3D, the rectangular components of a vector differ from that of its metric-dual covector.

Dimensionality comes when one tries to think of, e.g., the "cross-product of two vectors" as another vector [of the same type]. In that example, there will be issues of polar vs axial vectors.... which we never add together.

While there may be a metric $g_{ab}$ lying around that takes a vector $V^a$
to form the metric-dual covector $g_{ab}V^b$ (which is written in shorthand as $V_a$),
one still has to distinguish them because they cannot be meaningfully added together... it's like adding a column vector to a row vector (which MATLAB might not let you do).
While we might get away with that in Euclidean space, we'll run into trouble in other spaces.

Last edited: Apr 30, 2015
15. Apr 30, 2015

### stevendaryl

Staff Emeritus
Right. I just brought up 3D because the 3D part of the metric for flat spacetime is Euclidean. There's nothing special about the number 3, except that that's the number that's relevant in most physics classes, which is where vectors are typically introduced.

Right, the fact that in 3D, antisymmetric bivectors can be thought of as vectors is another source of confusion about the nature of different mathematical objects.

Well, the original poster was asking about this very issue: If $\frac{d^2 x^j}{dt^2}$ is naturally considered a vector (with contravariant components), and $\frac{\partial}{\partial x^j} V$ is naturally considered a covector (with covariant components), then how does it make sense to equate them, which seems to be the case when we write:

$m \frac{d^2 x^j}{dt^2} = - q \frac{\partial}{\partial x^j} V$