How to make sense of this definition of a reference frame?
Answers
Note that a central idea of special relativity is that you can't define a frame of reference with respect to anything but another frame of reference. Just keep in mind that this doesn't make the space any less general. Provided that (vi) is an acceptable basis, this is precisely the correct definition. Lets take P=(0,0,0,0), where the coordinates are like (ct,x,y,z). Then take v1=(γ,βγ,0,0), v2=(βγ,γ,0,0), v3=(0,0,1,0), v4=(0,0,0,1). Then, since P is zero, we can let (vi) be a nonaffine linear transformation. As usual with a change of basis ("the new coordinates for a column vector v are given by the matrix product M−1v"), the matrix that will change our basis is the inverse of M=(v1,v2,v3,v4) where each basis vector is a column vector, or: M−1=⎛⎝⎜⎜⎜γ−βγ00−βγγ0000100001⎞⎠⎟⎟⎟ the usual Lorentz transform. (In general an acceptable basis vi would be the composition of a rotation and a lorentz transformation) So what gives us the right to say this represents a frame moving with respect to another? If a particle in coordinate system (vi) is at position s=(ct′,0,0,0), then its position in the standard basis would be Ms=(ct′γ,ct′βγ,0,0)=(ct,x,y,z). So since the position is linear, to find its velocity we find: xt=ct′βγt′γ=cβ. (I calculated xt because its meaning is physically clearer than xct.) Therefore the basis (vi) given represents a coordinate system moving at velocity βc with respect to the current coordinate system.