- #1
LAHLH
- 409
- 1
Hi,
I've been trying to work out exactly why the t=const and r=const lines look like they do in the Minkowski conformal diagram.
I started with the usual Minkowski metric in polar coords (t,r) then go into null coords, then pull in the infinities by using arctan transformations, finally I then go back to non-null coords (T,R).
I find that the coords must be related as: [tex] t=\frac{1}{2}\left(\tan{\left(\frac{T-R}{2}\right)}+\tan{\left(\frac{T+R}{2}\right)}\right)=\frac{1}{2}\frac{\sin{(T)}}{\cos{(T)}+\cos{(R)}}[/tex]
The last equality following from a few sum product trig idents etc. Thus it would seem to me that t=constant lines satisfy:
[tex] \cos{(T)}+\cos{(R)}=c\sin{(T)} [/tex] and there is also the condition [tex] 0\leq R < \pi [/tex] and [tex] |T|+R < \pi [/tex]
Having Maple solve these, they kind of look correct if I choose to piece together the correct +/- solutions it spits out, but I'm not sure, and I thought finding these t=const lines would be simpler some how. Is there another way?
I've been trying to work out exactly why the t=const and r=const lines look like they do in the Minkowski conformal diagram.
I started with the usual Minkowski metric in polar coords (t,r) then go into null coords, then pull in the infinities by using arctan transformations, finally I then go back to non-null coords (T,R).
I find that the coords must be related as: [tex] t=\frac{1}{2}\left(\tan{\left(\frac{T-R}{2}\right)}+\tan{\left(\frac{T+R}{2}\right)}\right)=\frac{1}{2}\frac{\sin{(T)}}{\cos{(T)}+\cos{(R)}}[/tex]
The last equality following from a few sum product trig idents etc. Thus it would seem to me that t=constant lines satisfy:
[tex] \cos{(T)}+\cos{(R)}=c\sin{(T)} [/tex] and there is also the condition [tex] 0\leq R < \pi [/tex] and [tex] |T|+R < \pi [/tex]
Having Maple solve these, they kind of look correct if I choose to piece together the correct +/- solutions it spits out, but I'm not sure, and I thought finding these t=const lines would be simpler some how. Is there another way?