- #1
nafizamin
- 5
- 0
in relativity theory, in a certain "rapid space" the distance between two neighboring points is given by-
ds=[2/(1-r^2)]√ (dr^2+(rd∅)^2)
(considering a 2D space )
use the euler lagrange equation to show that the shortest distance from origin to any point is straight line.
attempt-
i don't know what's a rapid space, but i took the function in the euler lagrange equation as
f= [2/(1-r^2)]√ (+(r∅')^2) [∅'=d∅/dr]
then since ∂f/∂∅=0, so ∂f/∂∅'= const= r^2/ [(1-r^2)]√ (dr^2+(rd∅)^2)]
but how does the integration lead to a straight line ? could anybody do that for me ?
p.s: this isn't a homework !
ds=[2/(1-r^2)]√ (dr^2+(rd∅)^2)
(considering a 2D space )
use the euler lagrange equation to show that the shortest distance from origin to any point is straight line.
attempt-
i don't know what's a rapid space, but i took the function in the euler lagrange equation as
f= [2/(1-r^2)]√ (+(r∅')^2) [∅'=d∅/dr]
then since ∂f/∂∅=0, so ∂f/∂∅'= const= r^2/ [(1-r^2)]√ (dr^2+(rd∅)^2)]
but how does the integration lead to a straight line ? could anybody do that for me ?
p.s: this isn't a homework !