- #1
binbagsss
- 1,266
- 11
I have a question which asks show that a null geodesic to get to r> R , r some constant, given the space time metric etc, takes infinite coordinate time but finite proper time. ( It may be vice versa ).
I just want to confirm that, ofc there is no affine parameter for a null geodesic and so you could take coordinate time and proper time to be the same, and so that this question is referring to the time as observed by a timelike observer.
So if I denote the Lagrangian by L , you would proceed as follows :
1)set L=1 to find the relationship between proper time and coordinate time for the timelike observer ( only ##\dot{t} \neq 0 ## since at rest to get proper time, all other coordinates 0 )
2) set L=0 solve for the relationship between r(t) and (t) - assume radial for simplicity and then
a) not substituting in the expression from 1) will give you coordinate time
b) substituting the expression in from 1) will give you the proper time.
Can I confirm these ideas are correct - basically the question does not specify, but it should be the proper time of a timelike observer ?
Many thanks
I just want to confirm that, ofc there is no affine parameter for a null geodesic and so you could take coordinate time and proper time to be the same, and so that this question is referring to the time as observed by a timelike observer.
So if I denote the Lagrangian by L , you would proceed as follows :
1)set L=1 to find the relationship between proper time and coordinate time for the timelike observer ( only ##\dot{t} \neq 0 ## since at rest to get proper time, all other coordinates 0 )
2) set L=0 solve for the relationship between r(t) and (t) - assume radial for simplicity and then
a) not substituting in the expression from 1) will give you coordinate time
b) substituting the expression in from 1) will give you the proper time.
Can I confirm these ideas are correct - basically the question does not specify, but it should be the proper time of a timelike observer ?
Many thanks