Circular Orbits Around a Reissner-Nordstrom Black Hole Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) 2019 Moderator Election Q&A - Question CollectionSurface gravity of Kerr black holeInfalling light signals seen by a free falling observerDensity of states from the Retarded Green's function for a rotating black holeView of the sky from inside a black holeOn black hole ground statesFinding the minimum angular velocity for the stable orbit to exist (inflection point can be stable orbit?)Covariant Derivative acting on different coordinate systemDoes time speed up or slow down near a black hole?Time dilation, and curvature of space caused by two black holes of unequal massesRelativistic doppler effect vs Gravitational Redshift
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Circular Orbits Around a Reissner-Nordstrom Black Hole
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
2019 Moderator Election Q&A - Question CollectionSurface gravity of Kerr black holeInfalling light signals seen by a free falling observerDensity of states from the Retarded Green's function for a rotating black holeView of the sky from inside a black holeOn black hole ground statesFinding the minimum angular velocity for the stable orbit to exist (inflection point can be stable orbit?)Covariant Derivative acting on different coordinate systemDoes time speed up or slow down near a black hole?Time dilation, and curvature of space caused by two black holes of unequal massesRelativistic doppler effect vs Gravitational Redshift
$begingroup$
I'm doing some practice problems on GR and came across this which I'm totally stuck on:
The worldline of a photon orbiting a RN black hole is desribed by:
beginequation
left(fracdrdsigmaright)^2 = k^2 - l^2 fracalphar^2 \
alpha = 1 - frac2GMr + fracQ^2r^2
endequation
Where $k$ and $l$ are integration constants, $Q$ is the charge radius of the RN black hole $sigma$ is an affine parameter. $r$ has its usual definition in spherical coordinates.
The question asks to find the value(s) of $r$ corresponding to circular orbits by differentiating equation 1 with respect to $sigma$.
My knee jerk reaction to trying to find circular orbits would be to set the derivatives of $r$ to zero and solve for $r$ so I'm a bit stumped by the suggestion that I differentiate. I did try just to see where it went but I end up with a bunch of stuff that doesn't simplify to anything nice/anything I can interpret easily.
My questions are why does $fracdrdsigma=0$ not give you the solutions (or if it does, are there others?) and how to physically interpret $fracddsigmaleft( fracdrdsigmaright)^2$ so I might understand why I'm being asked to do that.
homework-and-exercises general-relativity black-holes
asked 3 hours ago
CT1234CT1234
235
$endgroup$
add a comment |
$begingroup$
I'm doing some practice problems on GR and came across this which I'm totally stuck on:
The worldline of a photon orbiting a RN black hole is desribed by:
beginequation
left(fracdrdsigmaright)^2 = k^2 - l^2 fracalphar^2 \
alpha = 1 - frac2GMr + fracQ^2r^2
endequation
Where $k$ and $l$ are integration constants, $Q$ is the charge radius of the RN black hole $sigma$ is an affine parameter. $r$ has its usual definition in spherical coordinates.
The question asks to find the value(s) of $r$ corresponding to circular orbits by differentiating equation 1 with respect to $sigma$.
My knee jerk reaction to trying to find circular orbits would be to set the derivatives of $r$ to zero and solve for $r$ so I'm a bit stumped by the suggestion that I differentiate. I did try just to see where it went but I end up with a bunch of stuff that doesn't simplify to anything nice/anything I can interpret easily.
My questions are why does $fracdrdsigma=0$ not give you the solutions (or if it does, are there others?) and how to physically interpret $fracddsigmaleft( fracdrdsigmaright)^2$ so I might understand why I'm being asked to do that.
homework-and-exercises general-relativity black-holes
asked 3 hours ago
CT1234CT1234
235
$endgroup$
$begingroup$
Can you give the text book reference?
$endgroup$
– M.N.Raia
2 hours ago
add a comment |
$begingroup$
I'm doing some practice problems on GR and came across this which I'm totally stuck on:
The worldline of a photon orbiting a RN black hole is desribed by:
beginequation
left(fracdrdsigmaright)^2 = k^2 - l^2 fracalphar^2 \
alpha = 1 - frac2GMr + fracQ^2r^2
endequation
Where $k$ and $l$ are integration constants, $Q$ is the charge radius of the RN black hole $sigma$ is an affine parameter. $r$ has its usual definition in spherical coordinates.
The question asks to find the value(s) of $r$ corresponding to circular orbits by differentiating equation 1 with respect to $sigma$.
My knee jerk reaction to trying to find circular orbits would be to set the derivatives of $r$ to zero and solve for $r$ so I'm a bit stumped by the suggestion that I differentiate. I did try just to see where it went but I end up with a bunch of stuff that doesn't simplify to anything nice/anything I can interpret easily.
My questions are why does $fracdrdsigma=0$ not give you the solutions (or if it does, are there others?) and how to physically interpret $fracddsigmaleft( fracdrdsigmaright)^2$ so I might understand why I'm being asked to do that.
homework-and-exercises general-relativity black-holes
asked 3 hours ago
CT1234CT1234
235
$endgroup$
I'm doing some practice problems on GR and came across this which I'm totally stuck on:
The worldline of a photon orbiting a RN black hole is desribed by:
beginequation
left(fracdrdsigmaright)^2 = k^2 - l^2 fracalphar^2 \
alpha = 1 - frac2GMr + fracQ^2r^2
endequation
Where $k$ and $l$ are integration constants, $Q$ is the charge radius of the RN black hole $sigma$ is an affine parameter. $r$ has its usual definition in spherical coordinates.
The question asks to find the value(s) of $r$ corresponding to circular orbits by differentiating equation 1 with respect to $sigma$.
My knee jerk reaction to trying to find circular orbits would be to set the derivatives of $r$ to zero and solve for $r$ so I'm a bit stumped by the suggestion that I differentiate. I did try just to see where it went but I end up with a bunch of stuff that doesn't simplify to anything nice/anything I can interpret easily.
My questions are why does $fracdrdsigma=0$ not give you the solutions (or if it does, are there others?) and how to physically interpret $fracddsigmaleft( fracdrdsigmaright)^2$ so I might understand why I'm being asked to do that.
homework-and-exercises general-relativity black-holes
homework-and-exercises general-relativity black-holes
asked 3 hours ago
CT1234CT1234
235
asked 3 hours ago
CT1234CT1234
235
asked 3 hours ago
CT1234CT1234
235
asked 3 hours ago
CT1234CT1234
235
asked 3 hours ago
CT1234CT1234
235
235
$begingroup$
Can you give the text book reference?
$endgroup$
– M.N.Raia
2 hours ago
add a comment |
$begingroup$
Can you give the text book reference?
$endgroup$
– M.N.Raia
2 hours ago
$begingroup$
Can you give the text book reference?
$endgroup$
– M.N.Raia
2 hours ago
$begingroup$
Can you give the text book reference?
$endgroup$
– M.N.Raia
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Generally speaking, you have to solve both $dr/dsigma =0$ and $d^2r/dsigma^2=0$ at once. The reason for that is that if a ray reaches only $dr/dsigma=0$ momentarily, this can be a radial turning point from which it still goes on to escape to infinity. Consider the following image (source):
Here you see the third ray from top approaching close to the black hole and then flying off. At the point it is closest to the black hole, it will obviously have $dr/dsigma =0$ without being a circular orbit! So to make sure it is a circular orbit, you also have to make sure that the conditions are such that $d^2r/dsigma^2=0$. Since the equations of motion of the light-ray are second order, this is really enough to make sure $r$ stays constant eternally.
Good luck!
answered 1 hour ago
VoidVoid
11.2k11958
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Generally speaking, you have to solve both $dr/dsigma =0$ and $d^2r/dsigma^2=0$ at once. The reason for that is that if a ray reaches only $dr/dsigma=0$ momentarily, this can be a radial turning point from which it still goes on to escape to infinity. Consider the following image (source):
Here you see the third ray from top approaching close to the black hole and then flying off. At the point it is closest to the black hole, it will obviously have $dr/dsigma =0$ without being a circular orbit! So to make sure it is a circular orbit, you also have to make sure that the conditions are such that $d^2r/dsigma^2=0$. Since the equations of motion of the light-ray are second order, this is really enough to make sure $r$ stays constant eternally.
Good luck!
answered 1 hour ago
VoidVoid
11.2k11958
$endgroup$
add a comment |
$begingroup$
Generally speaking, you have to solve both $dr/dsigma =0$ and $d^2r/dsigma^2=0$ at once. The reason for that is that if a ray reaches only $dr/dsigma=0$ momentarily, this can be a radial turning point from which it still goes on to escape to infinity. Consider the following image (source):
Here you see the third ray from top approaching close to the black hole and then flying off. At the point it is closest to the black hole, it will obviously have $dr/dsigma =0$ without being a circular orbit! So to make sure it is a circular orbit, you also have to make sure that the conditions are such that $d^2r/dsigma^2=0$. Since the equations of motion of the light-ray are second order, this is really enough to make sure $r$ stays constant eternally.
Good luck!
answered 1 hour ago
VoidVoid
11.2k11958
$endgroup$
add a comment |
$begingroup$
Generally speaking, you have to solve both $dr/dsigma =0$ and $d^2r/dsigma^2=0$ at once. The reason for that is that if a ray reaches only $dr/dsigma=0$ momentarily, this can be a radial turning point from which it still goes on to escape to infinity. Consider the following image (source):
Here you see the third ray from top approaching close to the black hole and then flying off. At the point it is closest to the black hole, it will obviously have $dr/dsigma =0$ without being a circular orbit! So to make sure it is a circular orbit, you also have to make sure that the conditions are such that $d^2r/dsigma^2=0$. Since the equations of motion of the light-ray are second order, this is really enough to make sure $r$ stays constant eternally.
Good luck!
answered 1 hour ago
VoidVoid
11.2k11958
$endgroup$
Generally speaking, you have to solve both $dr/dsigma =0$ and $d^2r/dsigma^2=0$ at once. The reason for that is that if a ray reaches only $dr/dsigma=0$ momentarily, this can be a radial turning point from which it still goes on to escape to infinity. Consider the following image (source):
Here you see the third ray from top approaching close to the black hole and then flying off. At the point it is closest to the black hole, it will obviously have $dr/dsigma =0$ without being a circular orbit! So to make sure it is a circular orbit, you also have to make sure that the conditions are such that $d^2r/dsigma^2=0$. Since the equations of motion of the light-ray are second order, this is really enough to make sure $r$ stays constant eternally.
Good luck!
answered 1 hour ago
VoidVoid
11.2k11958
answered 1 hour ago
VoidVoid
11.2k11958
answered 1 hour ago
VoidVoid
11.2k11958
answered 1 hour ago
VoidVoid
11.2k11958
11.2k11958
add a comment |
add a comment |
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$begingroup$
Can you give the text book reference?
$endgroup$
– M.N.Raia
2 hours ago