Geodesic on a sphere and on a plane in 2D

  • #1
MatinSAR
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Homework Statement
Find the geodesic on a sphere and on a plane in 2D.
Relevant Equations
Calculus of variation.
I start with the 2D plane. Suppose y(x) is the curve that connects these two points. Its length is given by:
$$S=\int_1^2 \, ds=\int_1^2 (1+y'^2)^{\frac {1}{2}} \, dx$$ Applying Euler's equation we get:$$\frac {\partial f} {\partial y'}=A$$$$\dfrac {y'}{(1+y'^2)^{\frac {1}{2}}}=A$$ $$y'^{2}=\dfrac {A^2}{1-A^2}$$$$\dfrac {dy}{dx}=\pm (\dfrac {A^2}{1-A^2})^{\frac 1 2}=B$$$$dy=Bdx$$$$y=Bx+C$$
So the shortest distance is a line that connects the points. I know it is one of the easiest questions in calculus of variation, I just wanted to know If I learnt it correctly. Any suggustion would be appreciated.

For the sphere, My answer is completely similar to this:
1711129926847.png

My problem is that I don't know how to obtain 5.10.5 using 5.10.4 ...
Many thanks.
 

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  • #2
The image sends me to the PF black hole …
 
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  • #4
Solve algebraically for ##1/\theta’##?
 
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  • #5
Orodruin said:
Solve algebraically for ##1/\theta’##?
1711130679017.png

I've done that before. I forget to post it.

Don't you have any idea for the 1st question @Orodruin?
 
  • #6
docnet said:
$$\begin{align*}
\sin^2\theta &= a\sqrt{\theta'^2+\sin^2\theta}\\
\sin^4\theta &= a^2(\theta'^2+\sin^2\theta)
\end{align*}$$
Thanks, But I don't see any problem ...
 
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  • #7
MatinSAR said:
Thanks, But I don't see any problem ...
That's because there isn't a problem.. I'm sorry I hit post before I was finished typing.
 
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  • #8
MatinSAR said:
I've done that before. I forget to post it.
So where is the problem?

MatinSAR said:
Don't you have any idea for the 1st question @Orodruin?
What is the first question? The straight line equation? That looks fine to me so I don't see a problem there either.
 
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  • #9
Orodruin said:
So where is the problem?


What is the first question? The straight line equation? That looks fine to me so I don't see a problem there either.
According to me:
$$\dfrac {1}{\theta'}=\dfrac {a\csc \theta}{(1-a^2 \csc^2 \theta)^{\frac 1 2}}$$
Which is not true.


I've just found my mistake in above equation.

Orodruin said:
What is the first question? The straight line equation?
Yes.
Orodruin said:
That looks fine to me so I don't see a problem there either.
Thanks.


Edit:
Many thanks @Orodruin and @docnet ...
 
  • #10
If you still wanted to know how to get (5.10.5) from (5.10.4):

\begin{align*}
\sin^2\theta&= a\sqrt{\theta'^2+\sin^2\theta} \\
\sin^4\theta &= a^2 (\theta'^2+\sin^2\theta) &\text{squared both sides} \\
\sin^4\theta &= a^2\theta'^2+a^2\sin^2\theta\\
\theta'^2 &= \frac{\sin^4\theta - a^2\sin^2\theta}{a^2} &\text{solved for } \theta'^2\\
\theta'^2 &= \frac{1-\frac{a^2}{\sin^2\theta}}{\frac{a^2}{\sin^4\theta}} &\text{divided numerator and denominator by } \sin^4\theta\\
\theta' &= \frac{\sqrt{1-\frac{a^2}{\sin^2\theta}}}{\frac{a}{\sin^2\theta}}& \text{square rooted both sides}\\
\theta' &= \frac{\sqrt{1-a^2\csc^2\theta}}{ a\csc^2\theta}\\
\frac{1}{\theta'}&= \frac{ a\csc^2\theta}{\sqrt{1-a^2\csc^2\theta}}.
\end{align*}
 
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  • #11
The easier way of doing the sphere:

Assume a curve parameter ##t##. It is then clear that ##\dot\theta## constant with ##\dot\varphi=0## solves the geodesic equations. By rotational symmetry, all geodesics are great circles.
 
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  • #12
docnet said:
If you still wanted to know how to get (5.10.5) from (5.10.4):
Thanks for your time.
Orodruin said:
The easier way of doing the sphere:

Assume a curve parameter ##t##. It is then clear that ##\dot\theta## constant with ##\dot\varphi=0## solves the geodesic equations. By rotational symmetry, all geodesics are great circles.
I'll try it. Many thanks.
 
  • #13
For the plane, I would start from [tex]
L = (\dot x^2 + \dot y^2)^{1/2}[/tex] which also gets the [itex]x = \mbox{constant}[/itex] solution.
 
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  • #14
pasmith said:
For the plane, I would start from [tex]
L = (\dot x^2 + \dot y^2)^{1/2}[/tex] which also gets the [itex]x = \mbox{constant}[/itex] solution.
Thank you @pasmith for your help. Is that ##L## lagrangian? And why ##x## and ##y## are functions of time? I haven't completed lagrangian and hamiltonian mechanics yet ...

I am currently reading calculus of variation and I wanted to solve using euler equation.

Edit:
I think you are dealing with the problem as motion of a particle in a plane. Then we should show that it is moving on a straight line with constant speed, right?
 
Last edited:

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