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 →A=Axˆx+Ayˆy+Azˆz 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 Aμ are the components of a vector in the coordinate system xμ, then the components of the vector in the coordinate system xμ′ are Aμ′=∂xμ′∂xμAμ (this all comes from the fact that the operators ∂∂xμ=∂μ 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 xμ as an example of a vector, when, on an arbitrary transformation, xμ′≠∂xμ′∂xμxμ 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?
Monday, July 3, 2017
general relativity - Does spacetime position not form a four-vector?
Subscribe to:
Post Comments (Atom)
classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?
I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...
-
Are C1, C2 and C3 connected in parallel, or C2, C3 in parallel and C1 in series with C23? Btw it appeared as a question in the basic physics...
-
I have read the radiation chapter, where I have been introduced with the terms emissivity and absorptivity. emissivity tells about the abili...
-
A charged particle undergoing an acceleration radiates photons. Let's consider a charge in a freely falling frame of reference. In such ...
No comments:
Post a Comment