# Are 4-vectors vectors?

1. Jul 6, 2011

### shrubber

I am trying to figure out the exact meaning of the concepts of 4-vector and relativistic tensor in the Minkowski spacetime. In my understanding, a tensor is a map that assigns an array of numbers to each basis in such a way that certain transformation rules apply. A vector can be viewed as a special case. Relativists define a 4-vector or a relativistic tensor as an object that transforms correctly under the Lorentz transformations.

So far so good. I pick a basis and assign it an array of numbers. I then pick another basis that can be obtained by a Lorentz transform and compute a new array of numbers.

However, what happens if I pick a basis that cannot be obtained through a Lorentz transform? Is the array undefined? Or is it arbitrary? One way or other, can the relativistic tensor (or 4-vector) be called a mathematical tensor (or vector), considering it does not transform correctly under GL(n,R), but only under a subgroup?

2. Jul 6, 2011

### Fredrik

Staff Emeritus
Yes, the "something that transforms as" definition of a 4-vector only requires that a 4-tuple of numbers is associated with each inertial coordinate system, and that when you switch from one inertial coordinate system to another, the 4-tuple gets multiplied by the matrix that defines the coordinate change.

I consider this way of thinking of 4-vectors stupid and obsolete. Just define a 4-vector as a tangent vector of Minkowski spacetime (defined as a manifold) and be done with it. That stuff about how the components transform under Lorentz transformations is included automatically as a special case of the general rule about how components of tangent vectors change under a change of coordinates.