9 added 436 characters in body edited Mar 25 at 1:28 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. People usually think of the Doppler effect in a context like this as changing the color of light, but it is also the factor by which the 10 second period of the flashes will vary. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. As a numerical example, suppose that the observer is in the innermost stable circular orbit (ISCO), with $$r=3$$. Then the Doppler shift factor varies between 0.71 and 2.12, which means that the 10 second period of the signals is perceived as varying from 4.7 s to 14.1 s. Unstable circular orbits are possible for any $$r>1.5$$. As an extreme example, $$r=1.5001$$ gives a period varying from 0.041 s to 820 s. This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. People usually think of the Doppler effect in a context like this as changing the color of light, but it is also the factor by which the 10 second period of the flashes will vary. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. As a numerical example, suppose that the observer is in the innermost stable circular orbit (ISCO), with $$r=3$$. Then the Doppler shift factor varies between 0.71 and 2.12, which means that the 10 second period of the signals is perceived as varying from 4.7 s to 14.1 s. This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. People usually think of the Doppler effect in a context like this as changing the color of light, but it is also the factor by which the 10 second period of the flashes will vary. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. As a numerical example, suppose that the observer is in the innermost stable circular orbit (ISCO), with $$r=3$$. Then the Doppler shift factor varies between 0.71 and 2.12, which means that the 10 second period of the signals is perceived as varying from 4.7 s to 14.1 s. Unstable circular orbits are possible for any $$r>1.5$$. As an extreme example, $$r=1.5001$$ gives a period varying from 0.041 s to 820 s. 8 added 436 characters in body edited Mar 25 at 1:22 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. People usually think of the Doppler effect in a context like this as changing the color of light, but it is also the factor by which the 10 second period of the flashes will vary. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The answer is that the Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. As a numerical example, suppose that the observer is in the innermost stable circular orbit (ISCO), with $$r=3$$. Then the Doppler shift factor varies between 0.71 and 2.12, which means that the 10 second period of the signals is perceived as varying from 4.7 s to 14.1 s. This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The answer is that the Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. People usually think of the Doppler effect in a context like this as changing the color of light, but it is also the factor by which the 10 second period of the flashes will vary. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. As a numerical example, suppose that the observer is in the innermost stable circular orbit (ISCO), with $$r=3$$. Then the Doppler shift factor varies between 0.71 and 2.12, which means that the 10 second period of the signals is perceived as varying from 4.7 s to 14.1 s. 7 added 237 characters in body edited Mar 24 at 21:16 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The answer is that the Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. Here is the graphThe graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The answer is that the Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. Here is the graph of this result as a function of $$r$$: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that. This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The answer is that the Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit. In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift. For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is $$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$. Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be $$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$. The velocity vector of the distant observer emitting the ray is $$u=(1,0,0,0)$$. We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as $$v=(1-1/r,\ldots)$$. The Doppler shift is $$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$. The graph of this result as a function of $$r$$ looks like this: So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit. For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book. 6 deleted 26 characters in body edited Mar 24 at 21:07 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges 5 added 1291 characters in body edited Mar 24 at 20:27 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges 4 added 1291 characters in body edited Mar 24 at 20:22 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges 3 added 33 characters in body edited Mar 23 at 19:15 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges 2 added 33 characters in body edited Mar 23 at 19:05 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges 1 answered Mar 23 at 18:59 Ben Crowell 57.4k66 gold badges172172 silver badges332332 bronze badges