Geodesic in hyperbolic plane
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I'm trying to show that $gamma(t) = (0,t)$ is a geodesic in the hyperbolic plane, that is for $mathbb{R}^2$ equipped with the metric $g_{11}=g_{22} = frac{1}{y^2}$, $g_{12}=0$.
The way I was trying to do this was by computing the associated Christoffel symbols and then show that $gamma$ satisfies the geodesic equation. The Christoffel symbols I computed are:
$Gamma_{11}^1 = Gamma_{12}^2 = Gamma_{22}^1 = 0$, $Gamma_{11}^2 = frac{1}{y}$, $Gamma_{12}^1 = Gamma_{22}^2 = frac{-1}{y}$.
Clearly $gamma'(t) = (0,1), gamma''(t) = (0,0)$, so for the second geodesic equation I believe reduces to:
$Gamma_{22}^2gamma_2'(t)gamma_2'(t) neq 0$
I'm sure there's just a sign error or something in there but I can't spot it at all.
differential-geometry
$endgroup$
add a comment |
$begingroup$
I'm trying to show that $gamma(t) = (0,t)$ is a geodesic in the hyperbolic plane, that is for $mathbb{R}^2$ equipped with the metric $g_{11}=g_{22} = frac{1}{y^2}$, $g_{12}=0$.
The way I was trying to do this was by computing the associated Christoffel symbols and then show that $gamma$ satisfies the geodesic equation. The Christoffel symbols I computed are:
$Gamma_{11}^1 = Gamma_{12}^2 = Gamma_{22}^1 = 0$, $Gamma_{11}^2 = frac{1}{y}$, $Gamma_{12}^1 = Gamma_{22}^2 = frac{-1}{y}$.
Clearly $gamma'(t) = (0,1), gamma''(t) = (0,0)$, so for the second geodesic equation I believe reduces to:
$Gamma_{22}^2gamma_2'(t)gamma_2'(t) neq 0$
I'm sure there's just a sign error or something in there but I can't spot it at all.
differential-geometry
$endgroup$
1
$begingroup$
Perhaps the calculations on p. 92 of my differential geometry text will help.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 16:58
add a comment |
$begingroup$
I'm trying to show that $gamma(t) = (0,t)$ is a geodesic in the hyperbolic plane, that is for $mathbb{R}^2$ equipped with the metric $g_{11}=g_{22} = frac{1}{y^2}$, $g_{12}=0$.
The way I was trying to do this was by computing the associated Christoffel symbols and then show that $gamma$ satisfies the geodesic equation. The Christoffel symbols I computed are:
$Gamma_{11}^1 = Gamma_{12}^2 = Gamma_{22}^1 = 0$, $Gamma_{11}^2 = frac{1}{y}$, $Gamma_{12}^1 = Gamma_{22}^2 = frac{-1}{y}$.
Clearly $gamma'(t) = (0,1), gamma''(t) = (0,0)$, so for the second geodesic equation I believe reduces to:
$Gamma_{22}^2gamma_2'(t)gamma_2'(t) neq 0$
I'm sure there's just a sign error or something in there but I can't spot it at all.
differential-geometry
$endgroup$
I'm trying to show that $gamma(t) = (0,t)$ is a geodesic in the hyperbolic plane, that is for $mathbb{R}^2$ equipped with the metric $g_{11}=g_{22} = frac{1}{y^2}$, $g_{12}=0$.
The way I was trying to do this was by computing the associated Christoffel symbols and then show that $gamma$ satisfies the geodesic equation. The Christoffel symbols I computed are:
$Gamma_{11}^1 = Gamma_{12}^2 = Gamma_{22}^1 = 0$, $Gamma_{11}^2 = frac{1}{y}$, $Gamma_{12}^1 = Gamma_{22}^2 = frac{-1}{y}$.
Clearly $gamma'(t) = (0,1), gamma''(t) = (0,0)$, so for the second geodesic equation I believe reduces to:
$Gamma_{22}^2gamma_2'(t)gamma_2'(t) neq 0$
I'm sure there's just a sign error or something in there but I can't spot it at all.
differential-geometry
differential-geometry
asked Mar 31 '17 at 13:46
user291678user291678
13510
13510
1
$begingroup$
Perhaps the calculations on p. 92 of my differential geometry text will help.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 16:58
add a comment |
1
$begingroup$
Perhaps the calculations on p. 92 of my differential geometry text will help.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 16:58
1
1
$begingroup$
Perhaps the calculations on p. 92 of my differential geometry text will help.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 16:58
$begingroup$
Perhaps the calculations on p. 92 of my differential geometry text will help.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 16:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can prove this without complicated calculation :
${ dx^2+dy^2over y^2 }geq {dy^2 over y^2}$ implies that given two point on the vertical line $x=0$ and a path $c(t)=(x(t),y(t))$ between these two points, the length of this path is greater that the length of the path $d(t)=(0,y(t))$, which contains the vertical segment between these points. This segment is therefore the unique shortest path between these points, and is a geodesic.
$endgroup$
add a comment |
$begingroup$
I don't understand how, from the generic second geodesic equation
$$ddot{y}+Gamma_{11}^2(dot{x})^2+Gamma_{12}^2(dot{x})^2(dot{y})^2+Gamma_{21}^2(dot{x})^2(dot{y})^2+Gamma_{22}^2(dot{y})^2=0$$
you obtain this single term ; in this equation, the central terms vanish and it remains:
$$ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$$
(Take also a look at (https://physics.stackexchange.com/q/91113))
I take this opportunity to explain a simple physical model that I have never seen explained very clearly.
This model provides the geodesics of $mathbb{H}$ , i.e., the half circles orthogonal to the real axis as optical shortest paths in a medium with a variable optical index: $n=tfrac{1}{y}$ at point $(x,y)$ (this index is thus constant along horizontal lines).
Question: What is the trajectory followed by a light ray starting in $(x_0,y_0)$ with an incidence angle $i_0$ with respect to a vertical reference (see figure below) ?
Solution : Let us consider $mathbb{H}$ as a ''stratified medium'' with an infinity of infinitesimal diopters separating medias with respective indices $dfrac{1}{y+dy}$ and $dfrac{1}{y}$. For such a diopter, the Snell's refraction law (https://en.wikipedia.org/wiki/Snell%27s_law) gives :
$$tag{1}dfrac{1}{y+dy}sin{(i+di)}=dfrac{1}{y}sin{(i)} iff sin{(i+di)}=left(1+dfrac{dy}{y}right)sin{(i)}.$$
Let us expand the LHS of (1) up to the first order:
$$sin(i)+cos{(i)} di=sin{(i)}+dfrac{dy}{y}sin(i).$$
$$dfrac{cos{(i)}}{sin{(i)}} di=dfrac{dy}{y}.$$
This differential equation can be integrated as follows:
$$ln(sin{(i)})=ln(y)+K.$$
Let $K=-ln(R)$. The previous relationship is equivalent to:
$sin{(i)}=dfrac{y}{R} $ with initial conditions $sin{(i_0)}=dfrac{y_0}{R}$ giving
$$R=dfrac{y_0}{sin(i_0)}.$$
Out of which, finally, we get $y=R sin{i}$ : it's, as awaited, a circular arc with radius $R$ centered on the $x$ axis.
Remark 1: It is interesting to see that this law $n=tfrac{1}{y}$ is a kind of "potential" with respect to the law $d=tfrac{1}{y^2}$ expressing the hyperbolic distance to the $x$ axis.
Remark 2: We have not considered specifically here the particular case of the vertical lines.
$endgroup$
$begingroup$
the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
$endgroup$
– user291678
Mar 31 '17 at 16:57
$begingroup$
@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
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2 Answers
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2 Answers
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$begingroup$
You can prove this without complicated calculation :
${ dx^2+dy^2over y^2 }geq {dy^2 over y^2}$ implies that given two point on the vertical line $x=0$ and a path $c(t)=(x(t),y(t))$ between these two points, the length of this path is greater that the length of the path $d(t)=(0,y(t))$, which contains the vertical segment between these points. This segment is therefore the unique shortest path between these points, and is a geodesic.
$endgroup$
add a comment |
$begingroup$
You can prove this without complicated calculation :
${ dx^2+dy^2over y^2 }geq {dy^2 over y^2}$ implies that given two point on the vertical line $x=0$ and a path $c(t)=(x(t),y(t))$ between these two points, the length of this path is greater that the length of the path $d(t)=(0,y(t))$, which contains the vertical segment between these points. This segment is therefore the unique shortest path between these points, and is a geodesic.
$endgroup$
add a comment |
$begingroup$
You can prove this without complicated calculation :
${ dx^2+dy^2over y^2 }geq {dy^2 over y^2}$ implies that given two point on the vertical line $x=0$ and a path $c(t)=(x(t),y(t))$ between these two points, the length of this path is greater that the length of the path $d(t)=(0,y(t))$, which contains the vertical segment between these points. This segment is therefore the unique shortest path between these points, and is a geodesic.
$endgroup$
You can prove this without complicated calculation :
${ dx^2+dy^2over y^2 }geq {dy^2 over y^2}$ implies that given two point on the vertical line $x=0$ and a path $c(t)=(x(t),y(t))$ between these two points, the length of this path is greater that the length of the path $d(t)=(0,y(t))$, which contains the vertical segment between these points. This segment is therefore the unique shortest path between these points, and is a geodesic.
answered Apr 2 '17 at 5:42
ThomasThomas
4,082510
4,082510
add a comment |
add a comment |
$begingroup$
I don't understand how, from the generic second geodesic equation
$$ddot{y}+Gamma_{11}^2(dot{x})^2+Gamma_{12}^2(dot{x})^2(dot{y})^2+Gamma_{21}^2(dot{x})^2(dot{y})^2+Gamma_{22}^2(dot{y})^2=0$$
you obtain this single term ; in this equation, the central terms vanish and it remains:
$$ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$$
(Take also a look at (https://physics.stackexchange.com/q/91113))
I take this opportunity to explain a simple physical model that I have never seen explained very clearly.
This model provides the geodesics of $mathbb{H}$ , i.e., the half circles orthogonal to the real axis as optical shortest paths in a medium with a variable optical index: $n=tfrac{1}{y}$ at point $(x,y)$ (this index is thus constant along horizontal lines).
Question: What is the trajectory followed by a light ray starting in $(x_0,y_0)$ with an incidence angle $i_0$ with respect to a vertical reference (see figure below) ?
Solution : Let us consider $mathbb{H}$ as a ''stratified medium'' with an infinity of infinitesimal diopters separating medias with respective indices $dfrac{1}{y+dy}$ and $dfrac{1}{y}$. For such a diopter, the Snell's refraction law (https://en.wikipedia.org/wiki/Snell%27s_law) gives :
$$tag{1}dfrac{1}{y+dy}sin{(i+di)}=dfrac{1}{y}sin{(i)} iff sin{(i+di)}=left(1+dfrac{dy}{y}right)sin{(i)}.$$
Let us expand the LHS of (1) up to the first order:
$$sin(i)+cos{(i)} di=sin{(i)}+dfrac{dy}{y}sin(i).$$
$$dfrac{cos{(i)}}{sin{(i)}} di=dfrac{dy}{y}.$$
This differential equation can be integrated as follows:
$$ln(sin{(i)})=ln(y)+K.$$
Let $K=-ln(R)$. The previous relationship is equivalent to:
$sin{(i)}=dfrac{y}{R} $ with initial conditions $sin{(i_0)}=dfrac{y_0}{R}$ giving
$$R=dfrac{y_0}{sin(i_0)}.$$
Out of which, finally, we get $y=R sin{i}$ : it's, as awaited, a circular arc with radius $R$ centered on the $x$ axis.
Remark 1: It is interesting to see that this law $n=tfrac{1}{y}$ is a kind of "potential" with respect to the law $d=tfrac{1}{y^2}$ expressing the hyperbolic distance to the $x$ axis.
Remark 2: We have not considered specifically here the particular case of the vertical lines.
$endgroup$
$begingroup$
the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
$endgroup$
– user291678
Mar 31 '17 at 16:57
$begingroup$
@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
|
show 3 more comments
$begingroup$
I don't understand how, from the generic second geodesic equation
$$ddot{y}+Gamma_{11}^2(dot{x})^2+Gamma_{12}^2(dot{x})^2(dot{y})^2+Gamma_{21}^2(dot{x})^2(dot{y})^2+Gamma_{22}^2(dot{y})^2=0$$
you obtain this single term ; in this equation, the central terms vanish and it remains:
$$ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$$
(Take also a look at (https://physics.stackexchange.com/q/91113))
I take this opportunity to explain a simple physical model that I have never seen explained very clearly.
This model provides the geodesics of $mathbb{H}$ , i.e., the half circles orthogonal to the real axis as optical shortest paths in a medium with a variable optical index: $n=tfrac{1}{y}$ at point $(x,y)$ (this index is thus constant along horizontal lines).
Question: What is the trajectory followed by a light ray starting in $(x_0,y_0)$ with an incidence angle $i_0$ with respect to a vertical reference (see figure below) ?
Solution : Let us consider $mathbb{H}$ as a ''stratified medium'' with an infinity of infinitesimal diopters separating medias with respective indices $dfrac{1}{y+dy}$ and $dfrac{1}{y}$. For such a diopter, the Snell's refraction law (https://en.wikipedia.org/wiki/Snell%27s_law) gives :
$$tag{1}dfrac{1}{y+dy}sin{(i+di)}=dfrac{1}{y}sin{(i)} iff sin{(i+di)}=left(1+dfrac{dy}{y}right)sin{(i)}.$$
Let us expand the LHS of (1) up to the first order:
$$sin(i)+cos{(i)} di=sin{(i)}+dfrac{dy}{y}sin(i).$$
$$dfrac{cos{(i)}}{sin{(i)}} di=dfrac{dy}{y}.$$
This differential equation can be integrated as follows:
$$ln(sin{(i)})=ln(y)+K.$$
Let $K=-ln(R)$. The previous relationship is equivalent to:
$sin{(i)}=dfrac{y}{R} $ with initial conditions $sin{(i_0)}=dfrac{y_0}{R}$ giving
$$R=dfrac{y_0}{sin(i_0)}.$$
Out of which, finally, we get $y=R sin{i}$ : it's, as awaited, a circular arc with radius $R$ centered on the $x$ axis.
Remark 1: It is interesting to see that this law $n=tfrac{1}{y}$ is a kind of "potential" with respect to the law $d=tfrac{1}{y^2}$ expressing the hyperbolic distance to the $x$ axis.
Remark 2: We have not considered specifically here the particular case of the vertical lines.
$endgroup$
$begingroup$
the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
$endgroup$
– user291678
Mar 31 '17 at 16:57
$begingroup$
@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
|
show 3 more comments
$begingroup$
I don't understand how, from the generic second geodesic equation
$$ddot{y}+Gamma_{11}^2(dot{x})^2+Gamma_{12}^2(dot{x})^2(dot{y})^2+Gamma_{21}^2(dot{x})^2(dot{y})^2+Gamma_{22}^2(dot{y})^2=0$$
you obtain this single term ; in this equation, the central terms vanish and it remains:
$$ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$$
(Take also a look at (https://physics.stackexchange.com/q/91113))
I take this opportunity to explain a simple physical model that I have never seen explained very clearly.
This model provides the geodesics of $mathbb{H}$ , i.e., the half circles orthogonal to the real axis as optical shortest paths in a medium with a variable optical index: $n=tfrac{1}{y}$ at point $(x,y)$ (this index is thus constant along horizontal lines).
Question: What is the trajectory followed by a light ray starting in $(x_0,y_0)$ with an incidence angle $i_0$ with respect to a vertical reference (see figure below) ?
Solution : Let us consider $mathbb{H}$ as a ''stratified medium'' with an infinity of infinitesimal diopters separating medias with respective indices $dfrac{1}{y+dy}$ and $dfrac{1}{y}$. For such a diopter, the Snell's refraction law (https://en.wikipedia.org/wiki/Snell%27s_law) gives :
$$tag{1}dfrac{1}{y+dy}sin{(i+di)}=dfrac{1}{y}sin{(i)} iff sin{(i+di)}=left(1+dfrac{dy}{y}right)sin{(i)}.$$
Let us expand the LHS of (1) up to the first order:
$$sin(i)+cos{(i)} di=sin{(i)}+dfrac{dy}{y}sin(i).$$
$$dfrac{cos{(i)}}{sin{(i)}} di=dfrac{dy}{y}.$$
This differential equation can be integrated as follows:
$$ln(sin{(i)})=ln(y)+K.$$
Let $K=-ln(R)$. The previous relationship is equivalent to:
$sin{(i)}=dfrac{y}{R} $ with initial conditions $sin{(i_0)}=dfrac{y_0}{R}$ giving
$$R=dfrac{y_0}{sin(i_0)}.$$
Out of which, finally, we get $y=R sin{i}$ : it's, as awaited, a circular arc with radius $R$ centered on the $x$ axis.
Remark 1: It is interesting to see that this law $n=tfrac{1}{y}$ is a kind of "potential" with respect to the law $d=tfrac{1}{y^2}$ expressing the hyperbolic distance to the $x$ axis.
Remark 2: We have not considered specifically here the particular case of the vertical lines.
$endgroup$
I don't understand how, from the generic second geodesic equation
$$ddot{y}+Gamma_{11}^2(dot{x})^2+Gamma_{12}^2(dot{x})^2(dot{y})^2+Gamma_{21}^2(dot{x})^2(dot{y})^2+Gamma_{22}^2(dot{y})^2=0$$
you obtain this single term ; in this equation, the central terms vanish and it remains:
$$ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$$
(Take also a look at (https://physics.stackexchange.com/q/91113))
I take this opportunity to explain a simple physical model that I have never seen explained very clearly.
This model provides the geodesics of $mathbb{H}$ , i.e., the half circles orthogonal to the real axis as optical shortest paths in a medium with a variable optical index: $n=tfrac{1}{y}$ at point $(x,y)$ (this index is thus constant along horizontal lines).
Question: What is the trajectory followed by a light ray starting in $(x_0,y_0)$ with an incidence angle $i_0$ with respect to a vertical reference (see figure below) ?
Solution : Let us consider $mathbb{H}$ as a ''stratified medium'' with an infinity of infinitesimal diopters separating medias with respective indices $dfrac{1}{y+dy}$ and $dfrac{1}{y}$. For such a diopter, the Snell's refraction law (https://en.wikipedia.org/wiki/Snell%27s_law) gives :
$$tag{1}dfrac{1}{y+dy}sin{(i+di)}=dfrac{1}{y}sin{(i)} iff sin{(i+di)}=left(1+dfrac{dy}{y}right)sin{(i)}.$$
Let us expand the LHS of (1) up to the first order:
$$sin(i)+cos{(i)} di=sin{(i)}+dfrac{dy}{y}sin(i).$$
$$dfrac{cos{(i)}}{sin{(i)}} di=dfrac{dy}{y}.$$
This differential equation can be integrated as follows:
$$ln(sin{(i)})=ln(y)+K.$$
Let $K=-ln(R)$. The previous relationship is equivalent to:
$sin{(i)}=dfrac{y}{R} $ with initial conditions $sin{(i_0)}=dfrac{y_0}{R}$ giving
$$R=dfrac{y_0}{sin(i_0)}.$$
Out of which, finally, we get $y=R sin{i}$ : it's, as awaited, a circular arc with radius $R$ centered on the $x$ axis.
Remark 1: It is interesting to see that this law $n=tfrac{1}{y}$ is a kind of "potential" with respect to the law $d=tfrac{1}{y^2}$ expressing the hyperbolic distance to the $x$ axis.
Remark 2: We have not considered specifically here the particular case of the vertical lines.
edited Dec 27 '18 at 23:33
answered Mar 31 '17 at 14:45
Jean MarieJean Marie
30.3k42051
30.3k42051
$begingroup$
the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
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– user291678
Mar 31 '17 at 16:57
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@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
|
show 3 more comments
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the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
$endgroup$
– user291678
Mar 31 '17 at 16:57
$begingroup$
@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
$begingroup$
the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is!
$endgroup$
– user291678
Mar 31 '17 at 16:50
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
$endgroup$
– user291678
Mar 31 '17 at 16:57
$begingroup$
The equation you give $ddot{y}=tfrac{(dot{x})^2}{y}-tfrac{(dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic?
$endgroup$
– user291678
Mar 31 '17 at 16:57
$begingroup$
@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 17:57
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
@TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish.
$endgroup$
– user291678
Mar 31 '17 at 18:10
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
$begingroup$
Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :)
$endgroup$
– Ted Shifrin
Mar 31 '17 at 18:28
|
show 3 more comments
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$begingroup$
Perhaps the calculations on p. 92 of my differential geometry text will help.
$endgroup$
– Ted Shifrin
Mar 31 '17 at 16:58