Regarding the shortest path connecting two points on a sphere of radius $R$
$begingroup$
Consider a path on the surface of the sphere given by a function $phi(theta)$. The line element along the path $phi$ is given by
$$
mathrm{d}s=Rmathrm{d}thetasqrt{1+left (frac{ mathrm{d} phi}{mathrm{d} theta}right )^2sin^2theta}.
$$
Deriving the Euler-Lagrange equation for $phi(theta)$, they lead to the differential equation
$$
frac{mathrm{d} phi}{mathrm{d} theta}=frac{c}{sinthetasqrt{sin^2theta-c^2}}tag{1}
$$
with $c$ some constant. Now, there are two things I'd like to figure out. How do I show that Euler-Lagrange equation is solved by
$$
phi=arcsin(C_1cottheta),
$$
where $C_1$ is constant? Determining such an integral is difficult for me. I think I should apply the identites identites $(arcsin x)'=frac{1}{sqrt{1-x^2}}$ and $(cot x)'=-frac{1}{sin^2(x)}$. Secondly, I want to transform $(1)$ into cartesian coordinates, but I am not sure how to do it. Should I use the trigonometric addition formula here?
integration mathematical-physics euler-lagrange-equation
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
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migrated from physics.stackexchange.com Dec 14 '18 at 21:05
This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
Consider a path on the surface of the sphere given by a function $phi(theta)$. The line element along the path $phi$ is given by
$$
mathrm{d}s=Rmathrm{d}thetasqrt{1+left (frac{ mathrm{d} phi}{mathrm{d} theta}right )^2sin^2theta}.
$$
Deriving the Euler-Lagrange equation for $phi(theta)$, they lead to the differential equation
$$
frac{mathrm{d} phi}{mathrm{d} theta}=frac{c}{sinthetasqrt{sin^2theta-c^2}}tag{1}
$$
with $c$ some constant. Now, there are two things I'd like to figure out. How do I show that Euler-Lagrange equation is solved by
$$
phi=arcsin(C_1cottheta),
$$
where $C_1$ is constant? Determining such an integral is difficult for me. I think I should apply the identites identites $(arcsin x)'=frac{1}{sqrt{1-x^2}}$ and $(cot x)'=-frac{1}{sin^2(x)}$. Secondly, I want to transform $(1)$ into cartesian coordinates, but I am not sure how to do it. Should I use the trigonometric addition formula here?
integration mathematical-physics euler-lagrange-equation
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
$endgroup$
migrated from physics.stackexchange.com Dec 14 '18 at 21:05
This question came from our site for active researchers, academics and students of physics.
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic
Dec 14 '18 at 20:41
$begingroup$
Yes, this is a pure math question. If it were in the context of a physics problem, like geodesics in GR, or the path of a moving particle I'd say otherwise.
$endgroup$
– ggcg
Dec 14 '18 at 20:45
add a comment |
$begingroup$
Consider a path on the surface of the sphere given by a function $phi(theta)$. The line element along the path $phi$ is given by
$$
mathrm{d}s=Rmathrm{d}thetasqrt{1+left (frac{ mathrm{d} phi}{mathrm{d} theta}right )^2sin^2theta}.
$$
Deriving the Euler-Lagrange equation for $phi(theta)$, they lead to the differential equation
$$
frac{mathrm{d} phi}{mathrm{d} theta}=frac{c}{sinthetasqrt{sin^2theta-c^2}}tag{1}
$$
with $c$ some constant. Now, there are two things I'd like to figure out. How do I show that Euler-Lagrange equation is solved by
$$
phi=arcsin(C_1cottheta),
$$
where $C_1$ is constant? Determining such an integral is difficult for me. I think I should apply the identites identites $(arcsin x)'=frac{1}{sqrt{1-x^2}}$ and $(cot x)'=-frac{1}{sin^2(x)}$. Secondly, I want to transform $(1)$ into cartesian coordinates, but I am not sure how to do it. Should I use the trigonometric addition formula here?
integration mathematical-physics euler-lagrange-equation
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
$endgroup$
Consider a path on the surface of the sphere given by a function $phi(theta)$. The line element along the path $phi$ is given by
$$
mathrm{d}s=Rmathrm{d}thetasqrt{1+left (frac{ mathrm{d} phi}{mathrm{d} theta}right )^2sin^2theta}.
$$
Deriving the Euler-Lagrange equation for $phi(theta)$, they lead to the differential equation
$$
frac{mathrm{d} phi}{mathrm{d} theta}=frac{c}{sinthetasqrt{sin^2theta-c^2}}tag{1}
$$
with $c$ some constant. Now, there are two things I'd like to figure out. How do I show that Euler-Lagrange equation is solved by
$$
phi=arcsin(C_1cottheta),
$$
where $C_1$ is constant? Determining such an integral is difficult for me. I think I should apply the identites identites $(arcsin x)'=frac{1}{sqrt{1-x^2}}$ and $(cot x)'=-frac{1}{sin^2(x)}$. Secondly, I want to transform $(1)$ into cartesian coordinates, but I am not sure how to do it. Should I use the trigonometric addition formula here?
integration mathematical-physics euler-lagrange-equation
integration mathematical-physics euler-lagrange-equation
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
edited Dec 14 '18 at 21:07
UnknownW
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
asked Dec 14 '18 at 20:34
UnknownWUnknownW
1,025922
1,025922
migrated from physics.stackexchange.com Dec 14 '18 at 21:05
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Dec 14 '18 at 21:05
This question came from our site for active researchers, academics and students of physics.
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic
Dec 14 '18 at 20:41
$begingroup$
Yes, this is a pure math question. If it were in the context of a physics problem, like geodesics in GR, or the path of a moving particle I'd say otherwise.
$endgroup$
– ggcg
Dec 14 '18 at 20:45
add a comment |
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic
Dec 14 '18 at 20:41
$begingroup$
Yes, this is a pure math question. If it were in the context of a physics problem, like geodesics in GR, or the path of a moving particle I'd say otherwise.
$endgroup$
– ggcg
Dec 14 '18 at 20:45
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic
Dec 14 '18 at 20:41
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic
Dec 14 '18 at 20:41
$begingroup$
Yes, this is a pure math question. If it were in the context of a physics problem, like geodesics in GR, or the path of a moving particle I'd say otherwise.
$endgroup$
– ggcg
Dec 14 '18 at 20:45
$begingroup$
Yes, this is a pure math question. If it were in the context of a physics problem, like geodesics in GR, or the path of a moving particle I'd say otherwise.
$endgroup$
– ggcg
Dec 14 '18 at 20:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
for your line element:
$left(frac{ds}{dt}right)^2=R^2,left(dot{vartheta}right)^2+
R^2,sin^2(vartheta)left(dot{varphi}right)^2
$
I get this differential equations :
$
ddot{vartheta}-frac{c^2cos(vartheta)}{R^4,m^2,sin^3(vartheta)}=0
$
and
$dot{varphi}=frac{c}{R^2,m,sin^2(vartheta)}$
There is no analytical solution ?
answered Dec 15 '18 at 16:00
EliEli
1234
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
for your line element:
$left(frac{ds}{dt}right)^2=R^2,left(dot{vartheta}right)^2+
R^2,sin^2(vartheta)left(dot{varphi}right)^2
$
I get this differential equations :
$
ddot{vartheta}-frac{c^2cos(vartheta)}{R^4,m^2,sin^3(vartheta)}=0
$
and
$dot{varphi}=frac{c}{R^2,m,sin^2(vartheta)}$
There is no analytical solution ?
answered Dec 15 '18 at 16:00
EliEli
1234
$endgroup$
add a comment |
$begingroup$
for your line element:
$left(frac{ds}{dt}right)^2=R^2,left(dot{vartheta}right)^2+
R^2,sin^2(vartheta)left(dot{varphi}right)^2
$
I get this differential equations :
$
ddot{vartheta}-frac{c^2cos(vartheta)}{R^4,m^2,sin^3(vartheta)}=0
$
and
$dot{varphi}=frac{c}{R^2,m,sin^2(vartheta)}$
There is no analytical solution ?
answered Dec 15 '18 at 16:00
EliEli
1234
$endgroup$
add a comment |
$begingroup$
for your line element:
$left(frac{ds}{dt}right)^2=R^2,left(dot{vartheta}right)^2+
R^2,sin^2(vartheta)left(dot{varphi}right)^2
$
I get this differential equations :
$
ddot{vartheta}-frac{c^2cos(vartheta)}{R^4,m^2,sin^3(vartheta)}=0
$
and
$dot{varphi}=frac{c}{R^2,m,sin^2(vartheta)}$
There is no analytical solution ?
answered Dec 15 '18 at 16:00
EliEli
1234
$endgroup$
for your line element:
$left(frac{ds}{dt}right)^2=R^2,left(dot{vartheta}right)^2+
R^2,sin^2(vartheta)left(dot{varphi}right)^2
$
I get this differential equations :
$
ddot{vartheta}-frac{c^2cos(vartheta)}{R^4,m^2,sin^3(vartheta)}=0
$
and
$dot{varphi}=frac{c}{R^2,m,sin^2(vartheta)}$
There is no analytical solution ?
answered Dec 15 '18 at 16:00
EliEli
1234
edited Dec 15 '18 at 20:46
answered Dec 15 '18 at 16:00
EliEli
1234
answered Dec 15 '18 at 16:00
EliEli
1234
answered Dec 15 '18 at 16:00
EliEli
1234
1234
add a comment |
add a comment |
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$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic
Dec 14 '18 at 20:41
$begingroup$
Yes, this is a pure math question. If it were in the context of a physics problem, like geodesics in GR, or the path of a moving particle I'd say otherwise.
$endgroup$
– ggcg
Dec 14 '18 at 20:45