Finding the function from ds=[2/(1-r^2)]√ (dr^2+(rd∅)^2)

[nafizamin]

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 !

 

[robphy]

rapidity space?
http://abacus.bates.edu/~msemon/RhodesSemonFinal.pdf

 

[Mentz114]

I think all you need is to show that $\ddot{\phi}=0$. With $L=(1/2)(g_{rr}\dot{r}^2+g_{\phi\phi}\dot{\phi}^2)$. $m$ is set to 1 since it cancels out of the EOM.

From $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)=\frac{\partial L}{\partial \phi}$ is evident that $\ddot{\phi}=0$.

 

[nafizamin]

I think all you need is to show that $\ddot{\phi}=0$. With $L=(1/2)(g_{rr}\dot{r}^2+g_{\phi\phi}\dot{\phi}^2)$. $m$ is set to 1 since it cancels out of the EOM.

From $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)=\frac{\partial L}{\partial \phi}$ is evident that $\ddot{\phi}=0$.

i didn't understand. could you elaborate, please ?

 

[Mentz114]

i didn't understand. could you elaborate, please ?

Which piece did you not understand ?

I have to admit that the extremization here is to find equations of motion and I don't know if that corresponds to what you are trying to do. Also, my assertion about the sufficiency of $\ddot{\phi}=0$ is probably not right.

I need help too. Maybe someone else can put us right.

Did you read the reference given by robphy ? The section on geodesics may be relevant.

 

[PAllen]

See page 949 in Robphy's link. The derivation given is clear and simple.

 

[nafizamin]

yes found it on rolphy. thank you everyone.