- #1
jdstokes
- 523
- 1
Hi all,
I had a rather poor introduction to special relativity and right now I'm refreshing myself in order to study quantum field theory.
In particular, I've always found the concept of four-vectors confusing. The problem is that from the mathematical point of view 4-vectors are nothing other than 4-tuples of real numbers, indeed the tangent space is always going to be isomorphic to [itex]\mathbb{R}^4[/itex]. So it seems like everything is a 4-vector.
On the other hand, physicists define them completely differently as creatures which transform according to the Lorentz transformation. So mathematically we have our smooth manifold (ie spacetime) and to each point we assign a bunch of tangent vectors which make up our tangent space.
I'd like to try to reconcile this discrepancy by explaining it in words. I'd appreciate your comments as to whether or not this is a good way of thinking about.
Fix a point p in spacetime and consider any physical quantity which can be described by a real variable, eg energy, or the x-component of momentum. Now consider the set of coordinate charts [itex]\mathcal{A}_p[/itex] about p. We then get a function [itex]f : \mathbb{R} \times \mathcal{A}_p \times \mathcal{A}_p \to \mathbb{R}[/itex] which describes how the value of the quantity changes when we transform from one coordinate chart to another. So now we have a completely mathematical definition of e.g. scalars as constant functions [itex]id : \mathbb{R} \times \mathcal{A}_p \times \mathcal{A}_p \to \mathbb{R}[/itex].
If we take four of these functions and line them up in a row [itex](f^1,f^2,f^3,f^4)[/itex] then we don't necesarily get a 4-vector. The condition that the result be a 4-vector is that [itex]f^\lambda(q,x^\mu,x^{\nu'}) = \sum_\mu\frac{\partial x^{\lambda'}}{\partial x^\mu}f^{\mu}(q)[/itex] for each [itex]\lambda \in \{1,2,3,4 \}[/itex]. This is basically saying that the physical quantity in each entry of the 4-vector must be related to the corresponding coordinate function, otherwise there is no chance of the thing being a 4-vector.
I had a rather poor introduction to special relativity and right now I'm refreshing myself in order to study quantum field theory.
In particular, I've always found the concept of four-vectors confusing. The problem is that from the mathematical point of view 4-vectors are nothing other than 4-tuples of real numbers, indeed the tangent space is always going to be isomorphic to [itex]\mathbb{R}^4[/itex]. So it seems like everything is a 4-vector.
On the other hand, physicists define them completely differently as creatures which transform according to the Lorentz transformation. So mathematically we have our smooth manifold (ie spacetime) and to each point we assign a bunch of tangent vectors which make up our tangent space.
I'd like to try to reconcile this discrepancy by explaining it in words. I'd appreciate your comments as to whether or not this is a good way of thinking about.
Fix a point p in spacetime and consider any physical quantity which can be described by a real variable, eg energy, or the x-component of momentum. Now consider the set of coordinate charts [itex]\mathcal{A}_p[/itex] about p. We then get a function [itex]f : \mathbb{R} \times \mathcal{A}_p \times \mathcal{A}_p \to \mathbb{R}[/itex] which describes how the value of the quantity changes when we transform from one coordinate chart to another. So now we have a completely mathematical definition of e.g. scalars as constant functions [itex]id : \mathbb{R} \times \mathcal{A}_p \times \mathcal{A}_p \to \mathbb{R}[/itex].
If we take four of these functions and line them up in a row [itex](f^1,f^2,f^3,f^4)[/itex] then we don't necesarily get a 4-vector. The condition that the result be a 4-vector is that [itex]f^\lambda(q,x^\mu,x^{\nu'}) = \sum_\mu\frac{\partial x^{\lambda'}}{\partial x^\mu}f^{\mu}(q)[/itex] for each [itex]\lambda \in \{1,2,3,4 \}[/itex]. This is basically saying that the physical quantity in each entry of the 4-vector must be related to the corresponding coordinate function, otherwise there is no chance of the thing being a 4-vector.
Last edited: