Measuring angle on positively curved space
$begingroup$
Suppose you are a 2D being, living on the surface of a sphere with radius R. An object of width $ds << R$ is at a distance $r$ from you. What angular width
$dtheta$ will you measure for the object? Explain behavior of $dtheta$ as r goes to $pi$R
I write the metric first
$$ds^2=dr^2+R^2sin^2(r/R)dtheta^2$$
Then I thought I can write
$$dtheta^2=frac {ds^2-dr^2} {R^2sin^2(r/R)}$$
or
$$dtheta=frac {1} {sin(r/R)} sqrt {ds^2R^2-dr^2R^2}$$
After this point I thought that we can say
$$ds^2R^2=0$$ since $ds<<R$ but in this case the $dtheta$ becomes
$$dtheta=frac {idr} {Rsin(r/R)}$$
I couldnt be sure that this makes sense since theres "i" in the equation?
If we cannot say $ds^2R^2=0$ why ?
spherical-geometry
asked Dec 30 '18 at 7:59
ReignReign
14918
$endgroup$
add a comment |
$begingroup$
Suppose you are a 2D being, living on the surface of a sphere with radius R. An object of width $ds << R$ is at a distance $r$ from you. What angular width
$dtheta$ will you measure for the object? Explain behavior of $dtheta$ as r goes to $pi$R
I write the metric first
$$ds^2=dr^2+R^2sin^2(r/R)dtheta^2$$
Then I thought I can write
$$dtheta^2=frac {ds^2-dr^2} {R^2sin^2(r/R)}$$
or
$$dtheta=frac {1} {sin(r/R)} sqrt {ds^2R^2-dr^2R^2}$$
After this point I thought that we can say
$$ds^2R^2=0$$ since $ds<<R$ but in this case the $dtheta$ becomes
$$dtheta=frac {idr} {Rsin(r/R)}$$
I couldnt be sure that this makes sense since theres "i" in the equation?
If we cannot say $ds^2R^2=0$ why ?
spherical-geometry
asked Dec 30 '18 at 7:59
ReignReign
14918
$endgroup$
add a comment |
$begingroup$
Suppose you are a 2D being, living on the surface of a sphere with radius R. An object of width $ds << R$ is at a distance $r$ from you. What angular width
$dtheta$ will you measure for the object? Explain behavior of $dtheta$ as r goes to $pi$R
I write the metric first
$$ds^2=dr^2+R^2sin^2(r/R)dtheta^2$$
Then I thought I can write
$$dtheta^2=frac {ds^2-dr^2} {R^2sin^2(r/R)}$$
or
$$dtheta=frac {1} {sin(r/R)} sqrt {ds^2R^2-dr^2R^2}$$
After this point I thought that we can say
$$ds^2R^2=0$$ since $ds<<R$ but in this case the $dtheta$ becomes
$$dtheta=frac {idr} {Rsin(r/R)}$$
I couldnt be sure that this makes sense since theres "i" in the equation?
If we cannot say $ds^2R^2=0$ why ?
spherical-geometry
asked Dec 30 '18 at 7:59
ReignReign
14918
$endgroup$
Suppose you are a 2D being, living on the surface of a sphere with radius R. An object of width $ds << R$ is at a distance $r$ from you. What angular width
$dtheta$ will you measure for the object? Explain behavior of $dtheta$ as r goes to $pi$R
I write the metric first
$$ds^2=dr^2+R^2sin^2(r/R)dtheta^2$$
Then I thought I can write
$$dtheta^2=frac {ds^2-dr^2} {R^2sin^2(r/R)}$$
or
$$dtheta=frac {1} {sin(r/R)} sqrt {ds^2R^2-dr^2R^2}$$
After this point I thought that we can say
$$ds^2R^2=0$$ since $ds<<R$ but in this case the $dtheta$ becomes
$$dtheta=frac {idr} {Rsin(r/R)}$$
I couldnt be sure that this makes sense since theres "i" in the equation?
If we cannot say $ds^2R^2=0$ why ?
spherical-geometry
spherical-geometry
asked Dec 30 '18 at 7:59
ReignReign
14918
asked Dec 30 '18 at 7:59
ReignReign
14918
edited Dec 30 '18 at 8:15
Reign
asked Dec 30 '18 at 7:59
ReignReign
14918
asked Dec 30 '18 at 7:59
ReignReign
14918
asked Dec 30 '18 at 7:59
ReignReign
14918
14918
add a comment |
add a comment |
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