# How to visualize the physical examples of contravariant and covariant vectors?

I am a beginner to tensors. For many days,I have been stuck with the contravariant and covariant tensors. I searched everywhere in Internet for its meanings and examples but everywhere there are mathematics for transformation of one form to other. It has become very difficult for me to understand and visualize the contravariant and covariant vectors. Without understanding,I am unable to move forward to other topics. In wikipedia, i found an example of it which says

For example, if v consists of the x-, y-, and z-components of velocity, then v is a contravariant vector: if the coordinates of space are stretched, rotated, or twisted, then the components of the velocity transform in the same way. Examples of contravariant vectors include displacement, velocity and acceleration. On the other hand, for instance, a triple consisting of the length, width, and height of a rectangular box could make up the three components of an abstract vector, but this vector would not be contravariant, since a change in coordinates on the space does not change the box's length, width, and height: instead these are scalars.

I have trouble understanding for the covariant part. What does the change of basis mean in tensor? I have a very poor background in linear algebra. May be the problem is trying to find a physical meaning for mathematical definition. But still maybe anyone who can come up with an example to distinguish between contra and covariant vectors and explain why it works that way and help me.

• have you checked the wikipedia site on covariance and contravariance of vectors? – Crimson Dec 7 '17 at 10:00
• yes the above example is from that site. But i have trouble understanding it. en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors – Rima Dec 7 '17 at 10:04
• Contravariance & covariance is confusing, so its not surprised you're confused; the traditional explanation is touched on in Diracs small book General Relativity; but unfortunately he's not thorough about it; its further expanded by Weinberg in his text on Gravitation & Cosmology. – Mozibur Ullah Dec 7 '17 at 10:09
• The traditional explanation is based around a 'change of basis' in a vector space, so if you're confused about this you might need to focus on why this is important and what it means. – Mozibur Ullah Dec 7 '17 at 10:12
• Covariant vectors change in the same way as a change of basis, and contravariant change in the opposite way; thats why they're called contravariant. – Mozibur Ullah Dec 7 '17 at 10:24

Preamble

Covariant and contravariant vectors are so tied up with the formalism of tensors as used in general relativity that its quite hard to disentangle the notion and look at in a striaght-forward manner; and it is straight-forward despite the way that it is described in many places.

Dirac writes in his book, The Theory of General Relativity:

before passing to the formalism it is convenient to consider an intermediate formalism – special relativity referred to oblique axes.

'Referred to oblique axes' is his way of saying its useful to refer to linear algebra - there was no such subject around when he was writing so he could not refer to it by name; unfortunately his exposition only just touches upon this; also special relativity is not important – at least in order to understand the concept of covariance & contravariance.

The reason why linear algebra is important is simply because when we look at a small part of a curved surface, it looks like its Euclidean; and in fact the definition of a manifold relies upon this in a very direct manner – a manifold is topological space that is locally Euclidean. To see this physically, look at a small part of a curve – it looks like a tiny, straight rod. In terms of vector spaces this will be the 1d vector space; when we look at a map between one curve and another curve, then again looking locally (this is the usual phrase to say we are looking at a small part of the curve), we see we are mapping between one straight rod and another; in terms of vector spaces we are mapping between a 1d vector space and another – and this is a linear transformation. We generalise and say that a local transformation between any manifold and another is a linear transformation.

Intuitive & physical description of contravariance

The simplest way of describing it physically and intuitively is to first take a sheet of paper and draw an arrow on it. This is our vector; now also draw a coordinate axes, these need not be an orthogonal basis, the axes can be 'oblique'; however its simplest to stick to an orthogonal axes - so let us do that. We can do two things: either we can rotate the vector or we can rotate the axes; if we rotate the vector, obviously the vector itself is changing and also its coordinates change; if we rotate the coordinate axes then the vector itself is not changing, but its coordinates is changing; the first is called an active change and the second is called a passive change and we can see that there is a real distinction between a active and passive change despite the fact the end result looks the same.

So having drawn the vector and the axes, suppose now we do a passive change of axes where we rotate the axes clockwise by a certain angle \theta; this is equivalent to an active change of the vector anti-clockwise by an angle of also \theta. This is what is meant by saying that the vector is contravariant, it rotates or changes or varies in the opposite way to which the axes rotates or changes or varies; it is contra-varying.

These are changes that vary in the same manner in which the axes changes - the simplest example of this are vectors in the dual space. The dual space is defined algebraically, and so its best tackled algebraically.

Contravariance, algebraically

Weinberg expands on Diracs 'oblique' reference (pun intended) in his book Gravitation and Cosmology; he writes:

$v^{i}\rightarrow v^{i'}=A_{i}^{i'}.v^{i}$ when $e_{i}\rightarrow e_{i'}=A_{i'}^{i}.e_{i}$ as the definition of a contravariant vector.

Here, $e_i$ is the ith vector in a basis $E$. Lets see if we can derive this.

First, we can write the second equation as a matrix equation:

$E'=A.E$

its worth noting we've generalised the usual algebra for matrices since $E$ is actually a column vector of vectors

$(e_1,..., e_n)$

and which we think of as a matrix and so we can invert $E$ and we get that

$A=E'.E^{-1}$.

Now,

$E:\mathbb{R}\rightarrow V$

This is the map which takes a vector and gives its coordinate vector in the basis $E$; and so we have that:

$v_E = E^{-1}.v$ and so $v=E.v_{E}$; and likewise $v=E'.v_{E'}$; so we have $v=E.v_{E}=E'.v_{E'}$; and this means $v_{E'}=E.E'^{-1}v_{E}$; now recalling that $A=E'.E^{-1}$ and so $v_{E'}=A.v_{E}$

This is exactly what Weinberg writes when we identify $v_{E}^{i}$ with $v^{i}$ and $v_{E'}^{i}$ with $v^{i'}$.

Covariance, algebraically

Let $\alpha$ be a covector - I'm assuming you know what a covector is; its just a linear functional on the vector space; and we write its components as $\alpha_{i}^{E}:=\alpha(e_{i})$; and so we get the row vector $\alpha^{E}:=(\alpha_{1}^{E},...,\alpha_{n}^{E})$; and then $\alpha_{i}^{E'}:=\alpha(e_{i'})=\alpha(A_{i'}^{i}e_{i})=A_{i'}^{i}\alpha(e_{i})=A_{i'}^{i}\alpha_{i}^{E}$ and writing this as a matrix equation means $\alpha^{E'}=A\alpha^{E}$; so it is co-varying with the change of basis matrix $A$, and so co-vectors change covariantly under a change of the basis.

• @rima: I thought you might, what I wrote there wasn't so clear; I skipped over a few things. Consider a basis $E$ of basis vectors $e_1,..., e_n$ then we can construct a linear operator $E: R \rightarrow V$ which given a vector $v$ gives us the coordinates of $v$ in the basis $E$ and we write this coordinate vector as $v_E$, ie $v_E:=E.v$. – Mozibur Ullah Dec 9 '17 at 17:27
• @rima: we can see directly that $E$ must be a linear operator since $(u+v)_E=u_E+v_E$ for any two vecters $u$ and $v$; and that $(cu)_E=c.u_E$ for any constant c. I hope this helps rather than confuse the issue. – Mozibur Ullah Dec 9 '17 at 17:28
The visual picture of a "contravariant" vector is the usual arrow picture. The visual picture of a "covariant" vector (dual vector) is a series of surfaces. See Schutz (2009, $\S3.3$, "Picture of a one-form") or Misner, Thorne & Wheeler ($\S2.5$). A [contravariant] vector acts on a dual vector to produce a scalar, and the visual picture of this is the number of surfaces the arrow crosses. Some classic textbook on electromagnetism gave a different picture, with dual vectors represented by arrows also, however as I recall it was a bit misleading. I recommend Schutz for a superb modern introduction to tensors.