# Covariant and contravariant 4-vector in special relativity

I've just learned about contra- and covariant vector in the context of special relativity (in electrodynamic) and I'm struggling with some concept. From what I found, an intuitive definition of contravariant vector (like position and velocity vector)

"transform as the coordinates do" under changes of coordinates (and so inversely to the transformation of the reference axes). Wikipedia

For example a change in scale from meter to millimeter will change a position for 1 to 1000

For covariant vector it is the opposit:

covariant vector has components that change oppositely to the coordinates or, equivalently, transform like the reference axes. Wikipedia

with the classical example being the gradient.

Now what bothers me is this "lowering and rising index" stuff where one can transform a contravariant vector to a covariant one (and vice-versa) by multiplying by the Minkowski metric tensor in the special relativity case. If one does this operation on a 4-position (contravariant) it will just change some sign of the 4-position but not the dimension (e.g. meter) of the 4-position.

How comes then that it is a covariant vector since I would guess (but here I must apparently be wrong) it will still transform as the contravariant vector (i.e. "transform as the coordinates do") because it is still "meters" and not "1/meters" as the gradient. I would have guessed that it should invert the dimension (meter->1/meter) to be consistent with the intuitive definition (but I don't know it could make any sense at all...).

You can see I'm confused here. In my course the proofs of the above properties doesn't give me any insight of what is really happening.

• Coordinates are not a vector!!! They just transform like $x^\mu\mapsto \Lambda^\mu{}_\nu x^\nu$ (by definition) and contravariant vectors transform like $v^\mu\mapsto \Lambda^\mu{}_\nu v^\nu$. Coordinates are not elements of the tangent space. – Ryan Unger Dec 31 '15 at 11:24

• Thanks for your answer (You should be able to leave comments now I think). Do you have the name of such a book on differential geometry ? – EigenDavid Jan 1 '16 at 18:37
• Yes I am now! And yes, I do. I studied elementary topology and differential geometry from John Lee's books, and I found them simply fantastic. "Introduction to Smooth Manifolds" should contain everything you need to fully understand the mathematics of special relativity. You should really be looking for an introduction to Pseudo-Riemmanian manifolds, but "Smooth Manifolds" has a chapter on Riemannian manifolds which should do as well. Just remember that in special relativity we use a non-Riemannian signature for the metric (i.e. (+---) insted of (++++)). – Giorgio Comitini Jan 1 '16 at 18:49
• This requires a special treatment, but as long as you don't want to go deep in the subject, it shouldn't encounter difficulties. One important subject that Lee doesn't treat in "Smooth Manifolds" is geodesics on Pseudo-Riemannian Manifolds, i.e. for example how we formalize the motion of a point not subject to forces in spacetime. Lee's "Riemannian Manifolds" has some chapters on geodesics on Riemannian Manifolds, which should give you an idea of how we treat them. The pseudo-Riemannian case is not very different from the Riemannian one. – Giorgio Comitini Jan 1 '16 at 18:57
• Full books on pseudo-Riemannian manifolds are usually quite complex, and I would not suggest them unless you want to extend your study to general relativity as well. – Giorgio Comitini Jan 1 '16 at 19:00

The metric is not always in the form

$$g^{\mu\nu}=\begin{pmatrix} -1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$

If you change your coordinate systems by rescaling the $x^0$ axis by $1000$, the metric will be

$$g^{\mu\nu}=\begin{pmatrix} -1000^2&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$

and the inverse metric will be

$$g_{\mu\nu}=\begin{pmatrix} \frac{-1}{1000^2}&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$

And raising and lowering must be done with these objects.

In general if you have an arbitrary coordinate change $q^\mu=F^\mu(x)$, then the components of the vectors will change from $X^\mu$ to $\frac{\partial F^\mu(x)}{\partial x^\nu}X^\nu$, where a sum over indices appearing twice is implied.

The metric and inverse metric also transform to:

$$g^{\alpha\beta}\frac{\partial F^\mu(x)}{\partial x^\alpha}\frac{\partial F^\nu(x)}{\partial x^\beta} \qquad g_{\alpha\beta}\frac{\partial F^\alpha(x)}{\partial x^\mu}\frac{\partial F^\beta(x)}{\partial x^\nu}$$

And you can find, for example if you switch to spherical coordinates, that the metric no longer is constant on your space.

• You correctly show the transformation of the metric for an arbitrary F. Following this, your initial example should read, $g^{00}=-1000^2$ and $g_{00}=\frac{-1}{1000^2}$. You just forgot the squares. – Gary Godfrey Dec 31 '15 at 17:50
• Yes sorry about that – s.harp Dec 31 '15 at 18:42
• Thanks for your answer. So the metric changes, ok. But for the dimension thing to work, shouldn't the "1000^2" be "(1000 sec)^2 so that when transforming the covariant gradient (1/dimension) it leads to the contravariant gradient dimension^2/dimension=dimension which is now in agreement with the intuitive definition? – EigenDavid Jan 1 '16 at 18:45
• Usually one does not write explicitly any units. If you want multiply $g^{\mu\nu}$ by $s^2$ or $m^2$ and $g_{\mu\nu}$ by $\frac{1}{s^2}$ or $\frac{1}{m^2}$ (remember that here length and time are the same unit). I don't know if this addresses your problem. – s.harp Jan 3 '16 at 20:50

Even though we use the terms "contravariant tensor" or "covariant tensor," what we are really referring to are the components of a tensor, and not to the tensor itself. The tensor itself is independent of the coordinate system we are using. The contravariant components of a tensor, for example, are obtained by resolving the tensor into components in terms of the so-called coordinate basis vectors.

Suppose we have a vector $$\vec{v}$$ and suppose we have two sets of basis vectors:

$$\vec{v}=v^1\hat{e_1}+v^2\hat{e_2}+v^3\hat{e_3}$$ where the superscripts are not exponents but indicate the nth component of the vector and the subscripts are the corresponding unit vectors.

In a different basis, we might have $$\vec{v}=u^1\hat{\epsilon_1}+u^2\hat{\epsilon_2}+u^3\hat{\epsilon_3}$$

$$\vec{v}$$ is the same object no matter what coordinate system its in. It is the same geometric object.

There is a process by which to take the components in the $$v^\alpha$$ coordinate system and derive the components in the $$u^\nu$$ coordinate system. This operation is linear in the $$v$$ coordinates and has a corresponding inverse transformation. The operation can be represented as a Matrix product with a vector made up of the coordinates.

There is a separate, but related, transformation rule for the basis vectors. It is essentially a matrix by a matrix multiplication as opposed to the previous matrix by vector multiplication. Further whereas the previous transformation mapped coordinates to coordinates, that is real numbers to real numbers, this transformation maps vectors to vectors. The objects to be transformed are different. Any vector, including the unit vector in one coordinate system, can be represented as a linear combination of the unit basis vectors of the other coordinate system.

Coordinates are not the vectors they represent. They have to multiply the associated unit vector to get the full vector.

Let paired up and down indices represent repeated multiplication:

$$\sum_{i=1}^3v^i\hat{e_i}=v^i\hat{e_i}$$

An lets let latin scripts represent coordinates in one system and greek indices coordiantes in another, then the vector v has the representations:

$$v^i\vec{e_i}=v^\mu\vec{e_\mu}=\vec{v}$$

To go from one coordinate system to another we can have $$v^a=\Lambda_\mu^av^\mu$$, where again repeated raised and lowered indices imply repeated summation and

$$\Lambda^a_\mu=\frac{\partial x^a}{\partial x^\mu}$$

Where $$x^a$$ is the $$a_{th}$$ coordinate of one coordinate system and the $$x^\mu$$ is the $$\mu_{th}$$ coordinate of the other.

Keep track of the latin vs. greek indices.

The transformation for basis vectors is $$\Lambda^\mu_a$$. The indices are flipped, but this isn't always the inverse of the transformation. It is the "opposite direction" from the coordinate transform procedure.

If an object transforms in the same way as basis vectors, its a Covariant vector also called a 1-form. If it transforms as coordinates, and therefore in the opposite direction as basis vectors, it's a Contravariant vector, or just a vector.

A 1-form can also be thought of geometrically as a series of infinitestimal planes placed parallel to each other at some interval. A vector of one length will in general pierce more of the planes if its parallel to the planes' normal than if it strikes the planes obliquely. We can associated the usual notion of an inner product with how many planes are pierced by a given vector.

In this formalism we don't say that you take two vectors to get an inner product. You take a vector and a one form to get an inner product.

If you have $$a^i$$ and $$b^j$$, $$a^ib^j$$ is not their inner product even if $$i=j$$ in a general coordinate system.Remember we need a raised and a lowered matching index to perform a sum. We only have an inner product with a specified Metric, $$g_{ij}$$ (though in Minkowski space the one in which all coordinates are 1 along the diagona, zero elsewhere accept the time/time cordiante which is -1. ).

$$\vec{a} \cdot \vec{b}=g_{ij}a^ib^j$$

This time a double sum is implied by the repeated indices. We have matching raised and lowered indices, so we can perform the summation get the inner product.

Now $$g_{ij}v^i$$ has its owns pecial meaning. You might recognize this as the procedure for lowering an index. It becomes $$v_i$$, a coordinate to be associated with a basis 1-form. A single upper index is the coordinate of a vector, a single lower index represents the coordinate of a 1-form.

See he for more on one forms

The dimension is not changed in a covariant/contravariant transformation because, if we think of covectors as the dual space on contravectors (and vice versa), we would use the multiplication $$X^\mu X_\mu$$ to calculate the norm of the two vectors - $$||X_\mu||=\sqrt{X_\mu X^\mu}$$, thus $$[||X_\mu||]=\left[\sqrt{x_\mu X^\mu}\right]=\sqrt{\left[X_\mu\right]}\sqrt{\left[X^\mu\right]}$$ and, because we require $$[||X_\mu||]=[X_\mu]$$ this gives $$[X_\mu]=[X^\mu]$$, hence the covariant distance 4-vector still has units of length.