# Does spacetime position not form a four-vector?

When one starts learning about physics, vectors are presented as mathematical quantities in space which have a direction and a magnitude. This geometric point of view has encoded in it the idea that under a change of basis the components of the vector must change contravariantly such that the magnitude and direction remain constant. This restricts what physical ideas may be the components of a vector (something much better explained in Feynman's Lectures), so that three arbitrary functions de not form an honest vector $\vec{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$ in some basis. So, in relativity a vector is defined "geometrically" as directional derivative operators on functions on the manifold $M$ and this implies, if $A^{\mu}$ are the components of a vector in the coordinate system $x^\mu$, then the components of the vector in the coordinate system $x^{\mu'}$ are $$A^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\mu}A^\mu$$ (this all comes from the fact that the operators $\frac{\partial}{\partial x^\mu}=\partial_\mu$ form a basis for the directional derivative operators, see Sean Carrol's Spacetime and Geometry)
My problem is the fact that too many people use the coordinates $x^\mu$ as an example of a vector, when, on an arbitrary transformation, $$x^{\mu'}\neq\frac{\partial x^{\mu'}}{\partial x^\mu}x^\mu$$ I understand that this equation is true if the transformation beween the two coordinates is linear (as is the case of a lorentz transformation between cartesian coordinate systems) but I think it can´t be true in general. Am I correct in that the position does not form a four-vector? If not, can you tell me why my reasoning is flawed?

• Where have you seen $x^\mu$ being treated as a vector? Jul 7, 2015 at 6:03
• See for example Goldstein's Classical Mechanics Jul 7, 2015 at 9:39
• @PraharMitra but when changing from cartesian to polar coordinates (even in SR), we do have a transformation matrix $\partial x^{\mu' }/\partial x^{\mu}$ thus the coordinates transforming as $x^{\mu'} = \partial x^{\mu'}/\partial x^{\mu }x^{\mu}$. Am I wrong Feb 23, 2021 at 5:51
• @Shashaank you are mistaken. The coordinates do not transform like that. Feb 23, 2021 at 8:34
• @PraharMitra don’t we have a matrix (Jacobian matrix) $\partial x^{\mu’}/ \partial x^{\mu}$ when changing coordinates from Cartesian to polar? Feb 23, 2021 at 8:41

You are correct.

Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to $\mathbb{R}^n$, and nonlinear maps do not preserve the linear structure.

On a manifold, there is no sense in attempting to "vectorize" points. A point is a point, an element of the manifold, a vector is a vector, element of a tangent space at a point. Of course you can map points into $n$-tuples, that is part of the definition of a topological manifold, but there is no reason why the inverse of this map should carry the linear structure over to the manifold.

And now, for a purely personal opinion: While Carroll's book is really good, the physicist's way of attempting to categorize everything by "transformation properties" is extremely counterproductive, and leads to such misunderstandings as you have overcome here. If one learns proper manifold theory, this is clear from the start...

• Concise and precise, +1! :) Jul 7, 2015 at 0:15
• @GennaroTedesco: It's called "succinct" :P Jul 7, 2015 at 3:37
• Upvote for your great answer, but I would remark that physicists in general are not mistaking position for a vector. We have spent a couple centuries arguing amongst ourselves about Galilean relativity and even a capable high school physics teacher can explain the difference between a map of points (labels) and physical vectors that live in tangent space without ever needing to go to the formal definition of manifolds. Unfortunately not every teacher is capable and not every student picks up the subtlety. OTOH, the students that don't won't profit from a lecture on Riemann manifolds, either. Jul 7, 2015 at 6:22
• @CuriousOne Your last sentence is an interesting one. It's certainly true that if a student has trouble grasping the difference, then maybe they wouldn't be ready for Riemannian geometry. On the other hand, a bad teacher is like a bad drawing partner in Pictionary: apt to send even the brightest students down the wrong path. Jul 7, 2015 at 7:02
• @Mehrdad True, but the sound of Gennaro's phrase is pretty cool too! Jul 7, 2015 at 7:03

Great reasoning: as in Uldreth's fantastic answer but I would add one more thing that may help cement your good understanding in place.

Co-ordinates are absolutely not vectors, they are labels on charts and are no more vectors than your street address is a vector. Almost certainly the reason people make the implication that you have correctly identified as wrong is this: in flat space (i.e. Euclidean, Minkowski or generally signatured spaces), affine co-ordinates for positions can have two roles: they are both labels and (once one has chosen an origin) superposition weights that combine linear basis tangents to the Euclidean (Minkowski ...) manifold linearly to yield a general tangent to the manifold. If you think about it, what I have just said is a slightly different take on Uldreth's second paragraph that begins "Position is a vector ...".

It's worth saying that I definitely recall the following learning sequence as a teenager. When beginning high school at about age 11, I was first shown co-ordinates (Cartesian of course) as labels. I suspect that this is how they are introduced to all children. I distinctly recall the idea that only two years later was the notion (that only works for Cartesian and generally affine co-ordinates) of a point's co-ordinates as a position vector introduced. Before that I had a very clear idea of a vector as a displacement or link between two points, an idea that, through the appropriate limit, leads to the tangent idea in a general manifold. On reading your question, I laugh when I recall the teacher's implying that the second role of co-ordinates as position vectors was a "new and advanced" way to look at vectors, whereas on the contrary it is a way of thinking that you correctly understand to be very limited and only workable in the affine case.

Here is a bare bones easy way to see that coordinate tuples are not 4-vectors.

Start in an inertial coordinate system in flat spacetime. Change the coordinate system with a constant translation:
$x' = x + A$
$y' = y$
$z' = z$
$t' = t$

Even in this idealistic case, 4-vectors and coordinate tuples transform differently. The components of the 4-vectors don't change at all in this case, while the coordinate tuples do.

• And this let us see another point: a linear transformation that changes the origin point destroys the "position vector" idea, but one could still save the concept of "displacement vector". If the transformation is non-linear that is also lost. May 26, 2017 at 10:45

Right -- vectors in general relativity live in some tangent space. This is the point of differential geometry, and of calculus in general -- you approximate non-linear things, which are not vector spaces (like curvy manifolds) with linear things (like their tangent spaces), which are vector spaces. This is exactly the motivation for defining the basis vectors as $$\partial_\mu$$, as you describe.