View Full Version : A Question about GR revealing my ignorance
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nHi,\n\nI posted this on .relativity, but I think it was ignored. Probably\nbecause I had the wrong title. I have a question about GR that goes\nlike this:\n\nI have a question about general relativity. It is based off of the\nFeynman Lectures. Suppose you have a tower on a planet. And at the\ntop of the planet you drop a photon to the Earth. As it leaves, the\nphoton kicks the tower in one direction and then speeds to the\nsurface. At the surface, the photon strikes with an increse in\nenergy, a higher frequency, and thus a higher momentum. It kicks the\ntower/Earth in the opposite direction than before. But this kick is\nstronger than the one at the tower when it left. So there is a net\nmomentum transfer in the opposite direction. But this doesn\'t make\nsense. you could make the Earth more compact, so that the difference\nin kick gets more and more. Until finally you could build a space\nship where all you do is shine a bit of light in one direction, and go\ncareening off into space somewhere. But this net "motion exnihilo" is\nweird. Clearly I\'m missing something. You have the same thing in\nNewtonian mechanics. Where on the tower you drop a ball and it falls\nto the earth, striking with greater momentum than when it left. But\nwhen you look at this from a non-inertial reference frame, you see\nwhat\'s happening. The momentum picked up by the ball is exactly\ncounterballanced by the momentum picked up by the Earth/tower. This\nis because the force acting on the ball is exactly equal in magnitude\nto the force of the ball on the Earth, but opposite in direction, and\nthe impulse of force over time is the same, and they both cancel out.\nBut with a photon, you don\'t have that luxury. For a photon to\n"accelerate" the Earth upwards, there would have to be some "advanced\nwarning" that the photon was coming, travelling faster than light.\nAnd this apparently, is excluded. How do you resolve this [apparent]\nparadox? One thing is by the equivalence principle, this is the same\nas a rocket accelerating upwards in outerspace, and firing a photon\nfrom the top to the bottom. I haven\'t thought enough about this case.\nAm I looking at the Earth/tower system from the\nwrong frame? Am I misunderstanding the conservation of three momentum\nwith four momentum? What\'s going on here? Thanks...\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi,
I posted this on .relativity, but I think it was ignored. Probably
because I had the wrong title. I have a question about GR that goes
like this:
I have a question about general relativity. It is based off of the
Feynman Lectures. Suppose you have a tower on a planet. And at the
top of the planet you drop a photon to the Earth. As it leaves, the
photon kicks the tower in one direction and then speeds to the
surface. At the surface, the photon strikes with an increse in
energy, a higher frequency, and thus a higher momentum. It kicks the
tower/Earth in the opposite direction than before. But this kick is
stronger than the one at the tower when it left. So there is a net
momentum transfer in the opposite direction. But this doesn't make
sense. you could make the Earth more compact, so that the difference
in kick gets more and more. Until finally you could build a space
ship where all you do is shine a bit of light in one direction, and go
careening off into space somewhere. But this net "motion exnihilo" is
weird. Clearly I'm missing something. You have the same thing in
Newtonian mechanics. Where on the tower you drop a ball and it falls
to the earth, striking with greater momentum than when it left. But
when you look at this from a non-inertial reference frame, you see
what's happening. The momentum picked up by the ball is exactly
counterballanced by the momentum picked up by the Earth/tower. This
is because the force acting on the ball is exactly equal in magnitude
to the force of the ball on the Earth, but opposite in direction, and
the impulse of force over time is the same, and they both cancel out.
But with a photon, you don't have that luxury. For a photon to
"accelerate" the Earth upwards, there would have to be some "advanced
warning" that the photon was coming, travelling faster than light.
And this apparently, is excluded. How do you resolve this [apparent]
paradox? One thing is by the equivalence principle, this is the same
as a rocket accelerating upwards in outerspace, and firing a photon
from the top to the bottom. I haven't thought enough about this case.
Am I looking at the Earth/tower system from the
wrong frame? Am I misunderstanding the conservation of three momentum
with four momentum? What's going on here? Thanks...
Uncle Al
Oct8-04, 06:20 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Joe wrote:\n>\n> Hi,\n>\n> I posted this on .relativity, but I think it was ignored. Probably\n> because I had the wrong title. I have a question about GR that goes\n> like this:\n>\n> I have a question about general relativity. It is based off of the\n> Feynman Lectures. Suppose you have a tower on a planet. And at the\n> top of the planet you drop a photon to the Earth. As it leaves, the\n> photon kicks the tower in one direction and then speeds to the\n> surface. At the surface, the photon strikes with an increse in\n> energy, a higher frequency, and thus a higher momentum. It kicks the\n> tower/Earth in the opposite direction than before. But this kick is\n> stronger than the one at the tower when it left. So there is a net\n> momentum transfer in the opposite direction. But this doesn\'t make\n> sense.\n[snip]\n\nMomentum and energy are locally conserved. The closed system gains\nnothing. The attraction of the photon to the center of mass is\nexactly balanced by the attraction of the center of mass to the\nphoton. The center of mass is more massive.\n\n--\nUncle Al\nhttp://www.mazepath.com/uncleal/\n(Toxic URL! Unsafe for children and most mammals)\nhttp://www.mazepath.com/uncleal/qz.pdf\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Joe wrote:
>
> Hi,
>
> I posted this on .relativity, but I think it was ignored. Probably
> because I had the wrong title. I have a question about GR that goes
> like this:
>
> I have a question about general relativity. It is based off of the
> Feynman Lectures. Suppose you have a tower on a planet. And at the
> top of the planet you drop a photon to the Earth. As it leaves, the
> photon kicks the tower in one direction and then speeds to the
> surface. At the surface, the photon strikes with an increse in
> energy, a higher frequency, and thus a higher momentum. It kicks the
> tower/Earth in the opposite direction than before. But this kick is
> stronger than the one at the tower when it left. So there is a net
> momentum transfer in the opposite direction. But this doesn't make
> sense.
[snip]
Momentum and energy are locally conserved. The closed system gains
nothing. The attraction of the photon to the center of mass is
exactly balanced by the attraction of the center of mass to the
photon. The center of mass is more massive.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\njhelfand@umd.edu writes:\n\n>Hi,\n\nHi.\n\n[interesting thought experiment snipped - summary below]\n\n>And this apparently, is excluded. How do you resolve this [apparent]\n>paradox? One thing is by the equivalence principle, this is the same\n>as a rocket accelerating upwards in outerspace, and firing a photon\n>from the top to the bottom. I haven\'t thought enough about this case.\n> Am I looking at the Earth/tower system from the\n>wrong frame? Am I misunderstanding the conservation of three momentum\n>with four momentum? What\'s going on here? Thanks...\n\nOkay, here\'s my condensation of this thought experiment: We start with two\nmirrors at different heights, and we bounce a photon between them. Due to\nblue/redshifting, the impulse imparted to the mirror will be greater at the\nlower one than at the higher one, leading to apparent nonconservation of\nmomentum. Is that about right?\n\nAs you rightly note, in the Newtonian case with a ball, equal action/reaction\nmakes conservation work out since the force of gravity on the Earth is the same\nas on the ball. But with a photon, we can\'t necessarily say that the photon\nexerts the same "force" on the Earth as the Earth exerts on it (re: the "what\nis the Schwarzschild radius of a photon" discussion that comes up frequently).\n\nI really like this thought experiment for its illusive simplicity. We can be\nsure, of course, that there is no nonconservation of momentum happening here -\nhowever, this is not as simple as noting that the covariant divergence of the\nstress energy tensor is always zero (sorry, I don\'t know how much GR you\nknow...I\'ll at least use no equations, even if I slip into jargon). For\nasymptotically flat spacetimes, there are useful generalizations of the law of\nenergy/momentum conservation which include the "energy" stored in the\ngravitational field, and these will give us confidence that (in some weird\nsense, perhaps) action equals reaction in the long run.\n\nBut this doesn\'t shed much light on the question you asked. Unfortunately I\nwould highly doubt that an explicit solution to the Einstein field equations\nexists for the situation you\'re talking about, as it would necessarily include\nthe gravitational radiation (however negligible) emitted by the system. It\'s\npossible that we could handle the problem through perturbation of a\nSchwarzschild metric, but even this seems like a lot to chew for such a simple\nproblem. Let me start by commenting on something you said:\n\n>For a photon to\n>"accelerate" the Earth upwards, there would have to be some "advanced\n>warning" that the photon was coming, travelling faster than light.\n\nLet me point out that this argument is insufficient to prove that the photon\ncannot affect the Earth while in flight. If that were the case, then we could\nextend that to say that ANY massless particle cannot accelerate ANYTHING,\nbecause there would need to be some "advanced warning" that it was coming,\ntravelling faster than light. In particular, this would imply that a gluon\ncannot possess color charge, which we certainly know is false.\n\nI\'m sort of torn two ways on this problem, though:\n\n1) one part of my brain says that the action/reaction balancing has to occur in\nflight\n2) the other part says that it has to occur in the mirror "bounce"\n\nSomeone else might have to clear this up, but let me explain my dilemma. My\nmotivation for 1) is that we could replace this thought experiment by the\nfollowing one: instead of bouncing the photon off of a mirror, we could simply\nset it into some highly elliptical orbit around the Earth, but keeping the\nmirror at the top. If we idealize the Earth as a point mass, then (keeping\naway from the horizon) we could make the orbit just about as eccentric as we\nwanted, and we\'d have very close to the same problem, the only difference being\nthat instead of bouncing off the lower mirror, the photon whips around the\n"Earth." Hence just about the same thing happens, but there is no lower bounce\n- we\'d expect the physical result to be almost identical, however, thus the\nphoton must have the same net effect during flight around the planet as it does\nduring collision with a mirror on the surface.\n\nBut my motivation for 2) is the following: exchanging a bit of momentum at one\nlocation is not necessarily physically equivalent to exchanging an identical\n(as locally measured) bit of momentum at another location. Ideally we\'d like\nto compare the exchanged bits of momentum as measured at infinity, where space\nis flat. Think of the example of tugging on a string at infinity - depending\nwhere the object you\'re tugging on is in the gravitational field, it will take\nvarying amounts of "pull" to change its momentum locally by some given amount -\nthis is almost identical to the idea of redshift. I haven\'t worked it out (it\nshould be pretty simple if you try, though), but I suspect that you\'ll find\nthat the scaling of momentum by redshift may make this thought experiment\nresolve itself.\n\nThat still doesn\'t satisfy the 1) side of my brain, though (this is starting to\nremind me of Feynman\'s bent-hose spewing/sucking water story), which says that\nthe photon damn well better have some effect on the Earth even in flight...I\nthink the resolution of that may be that yes, it does have some effect, but\nsince it has (neglecting gravitational radiation) identical time reversed\nmotions going both ways, the effect of the in-flight photon on the Earth will\ncancel itself out in the long run.\n\nOr perhaps I\'m just babbling on...maybe someone can clarify if I\'m wrong.\nActually, come to think of it, as long as we neglect gravitational radiation,\nthe time reversal invariance of Einstein\'s equations tells us immediately that\nthere is no net acceleration of the system - the system is time symmetric about\neither of the reflections off the mirrors, and I\'ll leave it to you to convince\nyourself that this implies no net acceleration (suppose it did, then flip the\nsituation in time...).\n\n-Eric\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>jhelfand@umd.edu writes:
>Hi,
Hi.
[interesting thought experiment snipped - summary below]
>And this apparently, is excluded. How do you resolve this [apparent]
>paradox? One thing is by the equivalence principle, this is the same
>as a rocket accelerating upwards in outerspace, and firing a photon
>from the top to the bottom. I haven't thought enough about this case.
> Am I looking at the Earth/tower system from the
>wrong frame? Am I misunderstanding the conservation of three momentum
>with four momentum? What's going on here? Thanks...
Okay, here's my condensation of this thought experiment: We start with two
mirrors at different heights, and we bounce a photon between them. Due to
blue/redshifting, the impulse imparted to the mirror will be greater at the
lower one than at the higher one, leading to apparent nonconservation of
momentum. Is that about right?
As you rightly note, in the Newtonian case with a ball, equal action/reaction
makes conservation work out since the force of gravity on the Earth is the same
as on the ball. But with a photon, we can't necessarily say that the photon
exerts the same "force" on the Earth as the Earth exerts on it (re: the "what
is the Schwarzschild radius of a photon" discussion that comes up frequently).
I really like this thought experiment for its illusive simplicity. We can be
sure, of course, that there is no nonconservation of momentum happening here -
however, this is not as simple as noting that the covariant divergence of the
stress energy tensor is always zero (sorry, I don't know how much GR you
know...I'll at least use no equations, even if I slip into jargon). For
asymptotically flat spacetimes, there are useful generalizations of the law of
energy/momentum conservation which include the "energy" stored in the
gravitational field, and these will give us confidence that (in some weird
sense, perhaps) action equals reaction in the long run.
But this doesn't shed much light on the question you asked. Unfortunately I
would highly doubt that an explicit solution to the Einstein field equations
exists for the situation you're talking about, as it would necessarily include
the gravitational radiation (however negligible) emitted by the system. It's
possible that we could handle the problem through perturbation of a
Schwarzschild metric, but even this seems like a lot to chew for such a simple
problem. Let me start by commenting on something you said:
>For a photon to
>"accelerate" the Earth upwards, there would have to be some "advanced
>warning" that the photon was coming, travelling faster than light.
Let me point out that this argument is insufficient to prove that the photon
cannot affect the Earth while in flight. If that were the case, then we could
extend that to say that ANY massless particle cannot accelerate ANYTHING,
because there would need to be some "advanced warning" that it was coming,
travelling faster than light. In particular, this would imply that a gluon
cannot possess color charge, which we certainly know is false.
I'm sort of torn two ways on this problem, though:
1) one part of my brain says that the action/reaction balancing has to occur in
flight
2) the other part says that it has to occur in the mirror "bounce"
Someone else might have to clear this up, but let me explain my dilemma. My
motivation for 1) is that we could replace this thought experiment by the
following one: instead of bouncing the photon off of a mirror, we could simply
set it into some highly elliptical orbit around the Earth, but keeping the
mirror at the top. If we idealize the Earth as a point mass, then (keeping
away from the horizon) we could make the orbit just about as eccentric as we
wanted, and we'd have very close to the same problem, the only difference being
that instead of bouncing off the lower mirror, the photon whips around the
"Earth." Hence just about the same thing happens, but there is no lower bounce
- we'd expect the physical result to be almost identical, however, thus the
photon must have the same net effect during flight around the planet as it does
during collision with a mirror on the surface.
But my motivation for 2) is the following: exchanging a bit of momentum at one
location is not necessarily physically equivalent to exchanging an identical
(as locally measured) bit of momentum at another location. Ideally we'd like
to compare the exchanged bits of momentum as measured at infinity, where space
is flat. Think of the example of tugging on a string at infinity - depending
where the object you're tugging on is in the gravitational field, it will take
varying amounts of "pull" to change its momentum locally by some given amount -
this is almost identical to the idea of redshift. I haven't worked it out (it
should be pretty simple if you try, though), but I suspect that you'll find
that the scaling of momentum by redshift may make this thought experiment
resolve itself.
That still doesn't satisfy the 1) side of my brain, though (this is starting to
remind me of Feynman's bent-hose spewing/sucking water story), which says that
the photon damn well better have some effect on the Earth even in flight...I
think the resolution of that may be that yes, it does have some effect, but
since it has (neglecting gravitational radiation) identical time reversed
motions going both ways, the effect of the in-flight photon on the Earth will
cancel itself out in the long run.
Or perhaps I'm just babbling on...maybe someone can clarify if I'm wrong.
Actually, come to think of it, as long as we neglect gravitational radiation,
the time reversal invariance of Einstein's equations tells us immediately that
there is no net acceleration of the system - the system is time symmetric about
either of the reflections off the mirrors, and I'll leave it to you to convince
yourself that this implies no net acceleration (suppose it did, then flip the
situation in time...).
-Eric
Blake Winter
Oct8-04, 06:20 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nThe photon itself will attract the earth gravitationally, and so\naccelerate the earth towards itself slightly to begin with before it\nstrikes it and pushes it the other way. You can see this from\nNewtonian gravity as well using a very small object. I think this\nprovides a basic answer to your question...\nBut beware of naively trying conserve quantities in GR! Conservation\nof global momentum and energy aren\'t really well defined in GR,\nbecause there\'s no global time relative to which you can conserve\nthem. In fact, since GR can be done using an arbitrary foliation (a\ntechincal term for slicing) of spacetime, conservation of quantities\nis far more complicated than you might even think, since time can in\nfact run at arbitrary rates along the hypersurface slicing that we\nhave chosen.\n\nTo give an example of this let\'s say that I\'m working in a flat\n(Minkowski) spacetime, but I choose to let time run much more quickly\non one part of my coordinate system than the other - let\'s say that I\ngive dt/dtau=r^2, where tau is my parameter marking which hypersurface\nI\'m at. In other words I\'m going to let my coordinate system in\nspacetime have time run much faster the further it is away from me. I\nnow have to be extraordinarily careful about conservation, because\ntime is running differently out there. As I recall in GR one cannot\nactually define a global momentum and energy for the whole universe as\na result, and hence one can\'t conserve what\'s not defined. What is\nmeaningful still is the local conservation of energy and momentum,\nwhich still holds.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>The photon itself will attract the earth gravitationally, and so
accelerate the earth towards itself slightly to begin with before it
strikes it and pushes it the other way. You can see this from
Newtonian gravity as well using a very small object. I think this
provides a basic answer to your question...
But beware of naively trying conserve quantities in GR! Conservation
of global momentum and energy aren't really well defined in GR,
because there's no global time relative to which you can conserve
them. In fact, since GR can be done using an arbitrary foliation (a
techincal term for slicing) of spacetime, conservation of quantities
is far more complicated than you might even think, since time can in
fact run at arbitrary rates along the hypersurface slicing that we
have chosen.
To give an example of this let's say that I'm working in a flat
(Minkowski) spacetime, but I choose to let time run much more quickly
on one part of my coordinate system than the other - let's say that I
give dt/dtau=r^2, where \tau is my parameter marking which hypersurface
I'm at. In other words I'm going to let my coordinate system in
spacetime have time run much faster the further it is away from me. I
now have to be extraordinarily careful about conservation, because
time is running differently out there. As I recall in GR one cannot
actually define a global momentum and energy for the whole universe as
a result, and hence one can't conserve what's not defined. What is
meaningful still is the local conservation of energy and momentum,
which still holds.
Greg Egan
Oct8-04, 06:20 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <e62610ea.0410060821.162be0ca@posting.google.com>, \njhelfand@umd.edu (Joe) wrote:\n\n[snip]\n>Suppose you have a tower on a planet. And at the\n> top of the planet you drop a photon to the Earth. As it leaves, the\n> photon kicks the tower in one direction and then speeds to the\n> surface. At the surface, the photon strikes with an increse in\n> energy, a higher frequency, and thus a higher momentum. It kicks the\n> tower/Earth in the opposite direction than before. But this kick is\n> stronger than the one at the tower when it left. So there is a net\n> momentum transfer in the opposite direction.\n[snip]\n\nHere\'s an answer in the spirit of "full-strength GR", which might or\nmight not satisfy you.\n\nGenerally, you can\'t expect to be able to compare, or add up, any kind of\nvectors at different points in curved spacetime. In a general curved\nspacetime, there is no conservation law which involves "adding up"\n4-momentum vectors over, say, each spacelike hypersurface in some\nfoliation of the spacetime, and expecting to get "the same" total over\ntime. The reason is that the vectors at each point lie in different\nspaces -- the tangent space at that point -- and to add them up you\'d\nneed to move them all into a single tangent space. But there\'s no\nnatural way to do that. There\'s a notion in curved spacetime called\n"parallel transport" which lets you move a vector at event A to event B,\nbut in general the result of parallel transport from A to B will depend\non the path you follow.\n\nAll is not lost, though. If the region you\'re interested in is embedded\nin an asymptotically flat spacetime, you can limit yourself to\nobservations in that flat spacetime, and then you do find that 4-momentum\nis conserved.\n\nIn your example, imagine that the photon emitted downwards from the\ntower, instead of hitting the surface below, travelled through a narrow\nchannel all the way through the planet, came out the other side, and\nescaped to infinity. At infinity, let its energy be E and its momentum\nbe -E in the z direction (where we can put Cartesian coordinates on our\nasymptotic flat spacetime). I\'m using units where c=1.\n\nNow, imagine that the photon that hits the surface, instead of being\nemitted from the tower, actually came in from infinity. To match the\nenergy and momentum, blue-shifted down to the surface, such a photon\nwould have to have had, at infinity, an energy of E, and a momentum of -E\nin the z-direction.\n\nSo by imagining a closely related situation where we can make all\nmeasurements at infinity, we can see that the emission and absorption of\nthe photon will give the planet equal and opposite kicks in momentum, at\nleast when that momentum is defined as a vector in the asymptotically\nflat spacetime far from the planet.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <e62610ea.0410060821.162be0ca@posting.google.com>,
jhelfand@umd.edu (Joe) wrote:
[snip]
>Suppose you have a tower on a planet. And at the
> top of the planet you drop a photon to the Earth. As it leaves, the
> photon kicks the tower in one direction and then speeds to the
> surface. At the surface, the photon strikes with an increse in
> energy, a higher frequency, and thus a higher momentum. It kicks the
> tower/Earth in the opposite direction than before. But this kick is
> stronger than the one at the tower when it left. So there is a net
> momentum transfer in the opposite direction.
[snip]
Here's an answer in the spirit of "full-strength GR", which might or
might not satisfy you.
Generally, you can't expect to be able to compare, or add up, any kind of
vectors at different points in curved spacetime. In a general curved
spacetime, there is no conservation law which involves "adding up"
4-momentum vectors over, say, each spacelike hypersurface in some
foliation of the spacetime, and expecting to get "the same" total over
time. The reason is that the vectors at each point lie in different
spaces -- the tangent space at that point -- and to add them up you'd
need to move them all into a single tangent space. But there's no
natural way to do that. There's a notion in curved spacetime called
"parallel transport" which lets you move a vector at event A to event B,
but in general the result of parallel transport from A to B will depend
on the path you follow.
All is not lost, though. If the region you're interested in is embedded
in an asymptotically flat spacetime, you can limit yourself to
observations in that flat spacetime, and then you do find that 4-momentum
is conserved.
In your example, imagine that the photon emitted downwards from the
tower, instead of hitting the surface below, travelled through a narrow
channel all the way through the planet, came out the other side, and
escaped to infinity. At infinity, let its energy be E and its momentum
be -E in the z direction (where we can put Cartesian coordinates on our
asymptotic flat spacetime). I'm using units where c=1.
Now, imagine that the photon that hits the surface, instead of being
emitted from the tower, actually came in from infinity. To match the
energy and momentum, blue-shifted down to the surface, such a photon
would have to have had, at infinity, an energy of E, and a momentum of -E
in the z-direction.
So by imagining a closely related situation where we can make all
measurements at infinity, we can see that the emission and absorption of
the photon will give the planet equal and opposite kicks in momentum, at
least when that momentum is defined as a vector in the asymptotically
flat spacetime far from the planet.
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nSo Uncle Al, what you\'re saying is just like Blake? That the photon\n"accelerates" the Earth upwards, to keep the same CM, before it gets\nthere?\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>So Uncle Al, what you're saying is just like Blake? That the photon
"accelerates" the Earth upwards, to keep the same CM, before it gets
there?
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> In your example, imagine that the photon emitted downwards from the\n> tower, instead of hitting the surface below, travelled through a narrow\n> channel all the way through the planet, came out the other side, and\n> escaped to infinity. At infinity, let its energy be E and its momentum\n> be -E in the z direction (where we can put Cartesian coordinates on our\n> asymptotic flat spacetime). I\'m using units where c=1.\n>\n> Now, imagine that the photon that hits the surface, instead of being\n> emitted from the tower, actually came in from infinity. To match the\n> energy and momentum, blue-shifted down to the surface, such a photon\n> would have to have had, at infinity, an energy of E, and a momentum of -E\n> in the z-direction.\n>\n> So by imagining a closely related situation where we can make all\n> measurements at infinity, we can see that the emission and absorption of\n> the photon will give the planet equal and opposite kicks in momentum, at\n> least when that momentum is defined as a vector in the asymptotically\n> flat spacetime far from the planet.\n\nOkay, if I follow this correctly, here is what I get. The 4-vector of\nthe planet and tower is (M,0,0,0) Now when the photon is emmitted\nfrom the tower, the system is (M-E_at_tower,E_at_tower,0,0) +\n(E_at_tower,-E_at_tower,0,0) where E_at_tower is the energy of the\nphoton at the tower, and the first vector is the 4-vector of the\ntower/planet, and the second is the photon. (Note I\'m talking out of\nmy *** here, so I don\'t really know what I\'m saying!) At infinity,\nthe system is:\n\n(M-E_at_infinity,E_at_infinity,0,0) +\n(E_at_infinity,-E_at_infinity,0,0)\n\nwhere E_at_infinity is the energy of the photon at infinity. Now\nsomehow, to maintain conservation of momentum, as the momenum of the\nparticle drops at it leasves the planet, the momentum of the\nplanet/tower also drops from the photon. It is this which I don\'t\nget. Similarly, using the particle now coming in from infinity, in\norder for the system to be symmetric, doesn\'t there have to be a\ncorresponding initial momentum of the planet? In otherwords, wouldn\'t\nthe system be:\n\n(M-E_at_infinity,E_at_infinity,0,0) +\n(E_at_infinity,-E_at_infinity,0,0)\n\nNow, when it reaches the earth you have:\n\n(M-E_at_surface,E_at_surface,0,0) + (E_at_surface,-E_at_surface,0,0)\n\nbut E_at_surface > E_at_tower, so there is still, speaking for the\nphoton a net difference in kick between the photon at the tower and\nthe photon at the surface. But that is precisely what I don\'t get,\nhow does this extra kick get counter-balanced? I think you have the\nsame question as the photon comes in from infinity:\n\n(M-E_at_infinity,E_at_infinity,0,0) +\n(E_at_infinity,-E_at_infinity,0,0)\n\ndoesn\'t this stay maintained throughout the trajectory of the photon?\nIn otherwords, at all points the sum corresponds to (M,0,0,0). But\nthis means the photon is "influencing" the Earth at a spooky distance.\nThere must be some mistake in my picture of GR!\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In your example, imagine that the photon emitted downwards from the
> tower, instead of hitting the surface below, travelled through a narrow
> channel all the way through the planet, came out the other side, and
> escaped to infinity. At infinity, let its energy be E and its momentum
> be -E in the z direction (where we can put Cartesian coordinates on our
> asymptotic flat spacetime). I'm using units where c=1.
>
> Now, imagine that the photon that hits the surface, instead of being
> emitted from the tower, actually came in from infinity. To match the
> energy and momentum, blue-shifted down to the surface, such a photon
> would have to have had, at infinity, an energy of E, and a momentum of -E
> in the z-direction.
>
> So by imagining a closely related situation where we can make all
> measurements at infinity, we can see that the emission and absorption of
> the photon will give the planet equal and opposite kicks in momentum, at
> least when that momentum is defined as a vector in the asymptotically
> flat spacetime far from the planet.
Okay, if I follow this correctly, here is what I get. The 4-vector of
the planet and tower is (M,0,0,0) Now when the photon is emmitted
from the tower, the system is (M-E_{at_tower},E_{at_tower},0,0) +(E_{at_tower},-E_{at_tower},0,0) where E_{at_tower} is the energy of the
photon at the tower, and the first vector is the 4-vector of the
tower/planet, and the second is the photon. (Note I'm talking out of
my *** here, so I don't really know what I'm saying!) At infinity,
the system is:
(M-E_{at_infinity},E_{at_infinity},0,0) +(E_{at_infinity},-E_{at_infinity},0,0)
where E_{at_infinity} is the energy of the photon at infinity. Now
somehow, to maintain conservation of momentum, as the momenum of the
particle drops at it leasves the planet, the momentum of the
planet/tower also drops from the photon. It is this which I don't
get. Similarly, using the particle now coming in from infinity, in
order for the system to be symmetric, doesn't there have to be a
corresponding initial momentum of the planet? In otherwords, wouldn't
the system be:
(M-E_{at_infinity},E_{at_infinity},0,0) +(E_{at_infinity},-E_{at_infinity},0,0)
Now, when it reaches the earth you have:
(M-E_{at_surface},E_{at_surface},0,0) + (E_{at_surface},-E_{at_surface},0,0)
but E_{at_surface} > E_{at_tower}, so there is still, speaking for the
photon a net difference in kick between the photon at the tower and
the photon at the surface. But that is precisely what I don't get,
how does this extra kick get counter-balanced? I think you have the
same question as the photon comes in from infinity:
(M-E_{at_infinity},E_{at_infinity},0,0) +(E_{at_infinity},-E_{at_infinity},0,0)
doesn't this stay maintained throughout the trajectory of the photon?
In otherwords, at all points the sum corresponds to (M,0,0,0). But
this means the photon is "influencing" the Earth at a spooky distance.
There must be some mistake in my picture of GR!
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nblake.winter@houghton.edu (Blake Winter) wrote in message news:<87423d2a.0410071607.3b3ea771@posting.google. com>...\n> The photon itself will attract the earth gravitationally, and so\n> accelerate the earth towards itself slightly to begin with before it\n> strikes it and pushes it the other way. You can see this from\n> Newtonian gravity as well using a very small object. I think this\n> provides a basic answer to your question...\n\nIt does! But what I don\'t get, is doesn\'t this mean that the effect\nof the falling photon on the Earth, is "faster" than the photon\nitself? For example, a sensitive accelerometer on the Earth could\npick up it was "accelerating" before the photon ever got there. This\nwould mean that the "speed of gravity" is faster than light. The\neffect that happens at one point, like an emitted photon, effects\nanother point, which is "outside the light cone" of the emitted\nphoton. This is fine by me, but people seemed to argue that this\nwould violate a precious law of relativity.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>blake.winter@houghton.edu (Blake Winter) wrote in message news:<87423d2a.0410071607.3b3ea771@posting.google.com>...
> The photon itself will attract the earth gravitationally, and so
> accelerate the earth towards itself slightly to begin with before it
> strikes it and pushes it the other way. You can see this from
> Newtonian gravity as well using a very small object. I think this
> provides a basic answer to your question...
It does! But what I don't get, is doesn't this mean that the effect
of the falling photon on the Earth, is "faster" than the photon
itself? For example, a sensitive accelerometer on the Earth could
pick up it was "accelerating" before the photon ever got there. This
would mean that the "speed of gravity" is faster than light. The
effect that happens at one point, like an emitted photon, effects
another point, which is "outside the light cone" of the emitted
photon. This is fine by me, but people seemed to argue that this
would violate a precious law of relativity.
tipertin
Oct11-04, 03:52 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\njhelfand@umd.edu (Joe) wrote in message news:<e62610ea.0410060821.162be0ca@posting.google. com>...\n\n>\n[an interesting puzzle]\n>\n\nHi Joe, I enjoyed thinking about this question. Let me know if what I\nthought makes sense:\n\nLet\'s say we have a spherical box (mirror) full of photons bouncing\naround (assume the box itself has negligable mass). Say there are so\nmany photons that the situation is almost spherically symmetric. These\ncrazy photons will have a gravitational field associated with them,\nwhich, outside the box, will be like that of any other spherically\nsymmetric thing (Schwarzchild/Birkhoff), at least approximately.\n\nNow hang this box at the top of a tower. The box will pull the Earth,\nthe Earth will pull the box. Equal and opposite. With a little bit of\nthought one can see how the gravitational effect of the Earth on the\nphotons (red/blue shift as they move) gets transferred through the\nbounces on the box to the tower, so that there is an equilibrium.\n\nNow quickly open and close a small door at the bottom of the box, and\nassume a photon moving directly downwards escapes, and shoots towards\nthe Earth. Now that it can\'t transfer the momentum it had through the\nwall, the equilibrium is broken.\n\nThings in this scenario are very similar to case you described: a\nphoton was shot downwards, and there is a recoil at the top of the\ntower, transferred to the Earth-tower system, which in turn starts\nmoving up.\n\nHowever, in this case we can see something perhaps a bit more clearly:\nthat there was a gravitational effect of the aforementioned photon all\nalong, pulling the Earth upwards; it doesn\'t just spring into\nexistence the moment the photon is "created". The Earth will not\n"know" that something has been changed, and will continue being pulled\nup by the "old" gravitational field. Without the balancing effect of\nour photon bouncing in the box, the Earth-tower system will\naccelerate, rather than moving up with constant speed. That is, until\nthe change in the gravitational field propagates all the way down to\nthe Earth, which happens at the speed of light.\n\nIn the perhaps more "conventional" case of a photon being radiated by\nan excited atom, one should think that the gravitational pull on the\nEarth due to the energy of the excited state is there, before the\nphoton is sent out. However, up until it *is* sent out, it is being\nbalanced by the pull of the Earth on the excited atom. Once the photon\nstarts moving down, the old gravitational field near the Earth is\nstill there, accelerating the latter upwards, until the change in the\ngravitational field propagates to the Earth.\n\nIt might be neat to calculate and check whether the acceleration I\nmentioned above actually exactly takes care of effect of the blueshift\nyou mentioned. Let me know if you think it doesn\'t.\n\narkadas ozakin\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>jhelfand@umd.edu (Joe) wrote in message news:<e62610ea.0410060821.162be0ca@posting.google.com>...
>
[an interesting puzzle]
>
Hi Joe, I enjoyed thinking about this question. Let me know if what I
thought makes sense:
Let's say we have a spherical box (mirror) full of photons bouncing
around (assume the box itself has negligable mass). Say there are so
many photons that the situation is almost spherically symmetric. These
crazy photons will have a gravitational field associated with them,
which, outside the box, will be like that of any other spherically
symmetric thing (Schwarzchild/Birkhoff), at least approximately.
Now hang this box at the top of a tower. The box will pull the Earth,
the Earth will pull the box. Equal and opposite. With a little bit of
thought one can see how the gravitational effect of the Earth on the
photons (red/blue shift as they move) gets transferred through the
bounces on the box to the tower, so that there is an equilibrium.
Now quickly open and close a small door at the bottom of the box, and
assume a photon moving directly downwards escapes, and shoots towards
the Earth. Now that it can't transfer the momentum it had through the
wall, the equilibrium is broken.
Things in this scenario are very similar to case you described: a
photon was shot downwards, and there is a recoil at the top of the
tower, transferred to the Earth-tower system, which in turn starts
moving up.
However, in this case we can see something perhaps a bit more clearly:
that there was a gravitational effect of the aforementioned photon all
along, pulling the Earth upwards; it doesn't just spring into
existence the moment the photon is "created". The Earth will not
"know" that something has been changed, and will continue being pulled
up by the "old" gravitational field. Without the balancing effect of
our photon bouncing in the box, the Earth-tower system will
accelerate, rather than moving up with constant speed. That is, until
the change in the gravitational field propagates all the way down to
the Earth, which happens at the speed of light.
In the perhaps more "conventional" case of a photon being radiated by
an excited atom, one should think that the gravitational pull on the
Earth due to the energy of the excited state is there, before the
photon is sent out. However, up until it *is* sent out, it is being
balanced by the pull of the Earth on the excited atom. Once the photon
starts moving down, the old gravitational field near the Earth is
still there, accelerating the latter upwards, until the change in the
gravitational field propagates to the Earth.
It might be neat to calculate and check whether the acceleration I
mentioned above actually exactly takes care of effect of the blueshift
you mentioned. Let me know if you think it doesn't.
arkadas ozakin
Blake Winter
Oct11-04, 01:06 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\njhelfand@umd.edu (Joe) wrote in message news:<e62610ea.0410091419.2af4d3a@posting.google.c om>...\n> blake.winter@houghton.edu (Blake Winter) wrote in message news:<87423d2a.0410071607.3b3ea771@posting.google. com>...\n> > The photon itself will attract the earth gravitationally, and so\n> > accelerate the earth towards itself slightly to begin with before it\n> > strikes it and pushes it the other way. You can see this from\n> > Newtonian gravity as well using a very small object. I think this\n> > provides a basic answer to your question...\n>\n> It does! But what I don\'t get, is doesn\'t this mean that the effect\n> of the falling photon on the Earth, is "faster" than the photon\n> itself? For example, a sensitive accelerometer on the Earth could\n> pick up it was "accelerating" before the photon ever got there. This\n> would mean that the "speed of gravity" is faster than light. The\n> effect that happens at one point, like an emitted photon, effects\n> another point, which is "outside the light cone" of the emitted\n> photon. This is fine by me, but people seemed to argue that this\n> would violate a precious law of relativity.\nRemember that the energy which is now a photon falling to earth was\noriginally part of the emitter, and hence has been creating a\ncurvature in spacetime for much longer than just when it is emitted.\nYou are correct in noting, however, that such effects are restricted\nto <=c.\nThis is compensated for because the rate at which the upwards\nacceleration of the emitter is transmitted is also going to be\nrestricted to <=c. So a careful analysis would reveal that relativity\nis fully consistent. But again, you have be careful because you can\'t\njust naively assume things will be conserved in GR: often there\'s no\nmeaningful way to define the quantities which you want to conserve\nwhen spacetime is curved.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>jhelfand@umd.edu (Joe) wrote in message news:<e62610ea.0410091419.2af4d3a@posting.google.com>...
> blake.winter@houghton.edu (Blake Winter) wrote in message news:<87423d2a.0410071607.3b3ea771@posting.google.com>...
> > The photon itself will attract the earth gravitationally, and so
> > accelerate the earth towards itself slightly to begin with before it
> > strikes it and pushes it the other way. You can see this from
> > Newtonian gravity as well using a very small object. I think this
> > provides a basic answer to your question...
>
> It does! But what I don't get, is doesn't this mean that the effect
> of the falling photon on the Earth, is "faster" than the photon
> itself? For example, a sensitive accelerometer on the Earth could
> pick up it was "accelerating" before the photon ever got there. This
> would mean that the "speed of gravity" is faster than light. The
> effect that happens at one point, like an emitted photon, effects
> another point, which is "outside the light cone" of the emitted
> photon. This is fine by me, but people seemed to argue that this
> would violate a precious law of relativity.
Remember that the energy which is now a photon falling to earth was
originally part of the emitter, and hence has been creating a
curvature in spacetime for much longer than just when it is emitted.
You are correct in noting, however, that such effects are restricted
to <=c.
This is compensated for because the rate at which the upwards
acceleration of the emitter is transmitted is also going to be
restricted to <=c. So a careful analysis would reveal that relativity
is fully consistent. But again, you have be careful because you can't
just naively assume things will be conserved in GR: often there's no
meaningful way to define the quantities which you want to conserve
when spacetime is curved.
Igor Khavkine
Oct11-04, 01:06 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOn Mon, 11 Oct 2004 08:52:00 +0000, Joe wrote:\n\n> blake.winter@houghton.edu (Blake Winter) wrote in message\n> news:<87423d2a.0410071607.3b3ea771@posting.google. com>...\n>> The photon itself will attract the earth gravitationally, and so\n>> accelerate the earth towards itself slightly to begin with before it\n>> strikes it and pushes it the other way. You can see this from Newtonian\n>> gravity as well using a very small object. I think this provides a\n>> basic answer to your question...\n>\n> It does! But what I don\'t get, is doesn\'t this mean that the effect of\n> the falling photon on the Earth, is "faster" than the photon itself? For\n> example, a sensitive accelerometer on the Earth could pick up it was\n> "accelerating" before the photon ever got there. This would mean that the\n> "speed of gravity" is faster than light. The effect that happens at one\n> point, like an emitted photon, effects another point, which is "outside\n> the light cone" of the emitted photon. This is fine by me, but people\n> seemed to argue that this would violate a precious law of relativity.\n\nRelativity is not violated and the emission of the photon is not detected\nbefore it reaches the ground. You seem to be assuming that because the\ntower that emitted the photon is solid, as soon as the top of it receives\na push, the bottom should feel the push instantaneously. This is not the\ncase.\n\nTake a piece of wood, put your ear to one end and ask someone to knock on\nthe other end. How long will it take the sound to reach your ear? To you\nit may seem instantaneous, but in reality it takes an amount of time equal\nto the length of the piece of wood divided by the speed of sound in it.\nThe speed of sound in any material cannot be greater than the speed of\nlight.\n\nThus, when the photon leaves the top of your tower, along with it\nan elastic pulse travels down the height of the tower "telling" the rest\nof it that the top has received a push. However, this elastic pulse will\nnot arrive to the ground faster than the photon. At best they will arrive\nat the same time.\n\nHope this helps.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Mon, 11 Oct 2004 08:52:00 +0000, Joe wrote:
> blake.winter@houghton.edu (Blake Winter) wrote in message
> news:<87423d2a.0410071607.3b3ea771@posting.google.com>...
>> The photon itself will attract the earth gravitationally, and so
>> accelerate the earth towards itself slightly to begin with before it
>> strikes it and pushes it the other way. You can see this from Newtonian
>> gravity as well using a very small object. I think this provides a
>> basic answer to your question...
>
> It does! But what I don't get, is doesn't this mean that the effect of
> the falling photon on the Earth, is "faster" than the photon itself? For
> example, a sensitive accelerometer on the Earth could pick up it was
> "accelerating" before the photon ever got there. This would mean that the
> "speed of gravity" is faster than light. The effect that happens at one
> point, like an emitted photon, effects another point, which is "outside
> the light cone" of the emitted photon. This is fine by me, but people
> seemed to argue that this would violate a precious law of relativity.
Relativity is not violated and the emission of the photon is not detected
before it reaches the ground. You seem to be assuming that because the
tower that emitted the photon is solid, as soon as the top of it receives
a push, the bottom should feel the push instantaneously. This is not the
case.
Take a piece of wood, put your ear to one end and ask someone to knock on
the other end. How long will it take the sound to reach your ear? To you
it may seem instantaneous, but in reality it takes an amount of time equal
to the length of the piece of wood divided by the speed of sound in it.
The speed of sound in any material cannot be greater than the speed of
light.
Thus, when the photon leaves the top of your tower, along with it
an elastic pulse travels down the height of the tower "telling" the rest
of it that the top has received a push. However, this elastic pulse will
not arrive to the ground faster than the photon. At best they will arrive
at the same time.
Hope this helps.
Igor
Greg Egan
Oct12-04, 10:49 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn article <e62610ea.0410101149.2a5ec7c6@posting.google.com>, \njhelfand@umd.edu (Joe) wrote:\n\n> doesn\'t this stay maintained throughout the trajectory of the photon?\n> In otherwords, at all points the sum corresponds to (M,0,0,0). But\n> this means the photon is "influencing" the Earth at a spooky distance.\n\nI promise you, according to GR nothing influences anything with "spooky\naction at a distance".\n\n> There must be some mistake in my picture of GR!\n\nThere is, in as much as you\'re ignoring the first part of my earlier\npost. I\'ll repeat it:\n\n*** Generally, you can\'t expect to be able to compare, or add up, any\nkind of vectors at different points in curved spacetime. In a general\ncurved spacetime, there is no conservation law which involves "adding up"\n4-momentum vectors over, say, each spacelike hypersurface in some\nfoliation of the spacetime, and expecting to get "the same" total over\ntime. ***\n\nWhat GR certainly will (at least in principle) let you do is perform a\ndetailed analysis of your thought experiment, and predict what a variety\nof observers at a variety of places and times would measure. These\npredictions certainly *will* obey causality, and any gravitational\neffects will propagate at the speed of light. But if you think that in\ngeneral you can expect to add up the momentum 4-vectors in different\nplaces and "at different times" in curved spacetime and be guaranteed to\nfind conservation of 4-momentum as you would in SR, you\'re mistaken.\n\nNow, it might turn out that a careful, detailed, but *approximate*\nanalysis of your thought experiment can be found that gives a satisfying\nanswer that (a) obeys casuality (it certainly should!) and (b) gives a\nresult that looks like conservation of 4-momentum throughout the motion\nof the photon ... while using an approximation that glosses over the\ncurvature-related inability to meaningfully talk about such a thing. In\nother words, if you pretend you\'re doing everything in *flat* spacetime,\nand that gravity is just another field like the EM field rather than an\neffect due to the curvature of spacetime, you might find a way to get\neverything to add up.\n\nI\'m certainly not trying to discourage you or anyone else from looking\nfor such an approximate analysis. I just want you to be aware that "full\nstrength GR" says that you shouldn\'t expect, in general, to be able to do\nsuch an analysis, and so that even if you *can* for this particular\nthought experiment, it\'s very important that you understand that there\nwill be other scenarios where it\'s just not possible.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <e62610ea.0410101149.2a5ec7c6@posting.google.com>,
jhelfand@umd.edu (Joe) wrote:
> doesn't this stay maintained throughout the trajectory of the photon?
> In otherwords, at all points the sum corresponds to (M,0,0,0). But
> this means the photon is "influencing" the Earth at a spooky distance.
I promise you, according to GR nothing influences anything with "spooky
action at a distance".
> There must be some mistake in my picture of GR!
There is, in as much as you're ignoring the first part of my earlier
post. I'll repeat it:
*** Generally, you can't expect to be able to compare, or add up, any
kind of vectors at different points in curved spacetime. In a general
curved spacetime, there is no conservation law which involves "adding up"
4-momentum vectors over, say, each spacelike hypersurface in some
foliation of the spacetime, and expecting to get "the same" total over
time. ***
What GR certainly will (at least in principle) let you do is perform a
detailed analysis of your thought experiment, and predict what a variety
of observers at a variety of places and times would measure. These
predictions certainly *will* obey causality, and any gravitational
effects will propagate at the speed of light. But if you think that in
general you can expect to add up the momentum 4-vectors in different
places and "at different times" in curved spacetime and be guaranteed to
find conservation of 4-momentum as you would in SR, you're mistaken.
Now, it might turn out that a careful, detailed, but *approximate*
analysis of your thought experiment can be found that gives a satisfying
answer that (a) obeys casuality (it certainly should!) and (b) gives a
result that looks like conservation of 4-momentum throughout the motion
of the photon ... while using an approximation that glosses over the
curvature-related inability to meaningfully talk about such a thing. In
other words, if you pretend you're doing everything in *flat* spacetime,
and that gravity is just another field like the EM field rather than an
effect due to the curvature of spacetime, you might find a way to get
everything to add up.
I'm certainly not trying to discourage you or anyone else from looking
for such an approximate analysis. I just want you to be aware that "full
strength GR" says that you shouldn't expect, in general, to be able to do
such an analysis, and so that even if you *can* for this particular
thought experiment, it's very important that you understand that there
will be other scenarios where it's just not possible.
The notion that the surface of the earth, or the tower, do not move is an approximation.
When an onbject falls in gravity, linear momentum is conserved. In everyday practice, on the surface of earth, and with everyday objects, the mass of the earth is so large that the effect on the earth is small enough to be ignored when measuring results. The mass of the earth is about 6\times10^{24} Kg so even dropping something that weighs 10^{10} Kg (10,000,000) tons is only going to accelerate the earth about 10^{-13} m/s^2 which is small enough that it's nearly impossible to measure. For systems where both objects have comparable mass - e.g. earth-moon interaction - it's something that needs to be kept track of.
Now, because if we account for this acceleration, the tower is no longer an inertial reference frame so if you keep that reference frame, you would see the tower accelerating upward while the ball accelerates downward so that the net linear momentum remains constant.
As a throught experiment, you could consider two objects with massess m_1 and m_2 influencing each other in gravity, and seeing what happens as m_2 gets much much larger than m_1 to get an idea of what's going on.
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nJoe <jhelfand@umd.edu> writes\n>I have a question about general relativity. It is based off of the\n>Feynman Lectures. Suppose you have a tower on a planet. And at the\n>top of the planet you drop a photon to the Earth. As it leaves, the\n>photon kicks the tower in one direction and then speeds to the\n>surface. At the surface, the photon strikes with an increse in\n>energy, a higher frequency, and thus a higher momentum. It kicks the\n>tower/Earth in the opposite direction than before. But this kick is\n>stronger than the one at the tower when it left. So there is a net\n>momentum transfer in the opposite direction. But this doesn\'t make\n>sense. you could make the Earth more compact, so that the difference\n>in kick gets more and more.\n\n1) I am NOT an expert.\n\n2) How about the following:\n[Which may be wrong, in which case I will be corrected forcefully.]\n\nStood on the top of the tower you observe that everything below you is\n\'going slower\' due to time \'going slower\' in a higher gravitational\npotential.\n\nSo I think this results in (if you could see it) the photon that is\nfired down does not seem to change momentum as it does so. That is as\nviewed from the bottom it has a faster frequency than at the top, but\nviewed from the top the increase in frequency at the bottom is precisely\nbalanced by the \'slower time\' at the bottom.\n\nThe same applies in reverse for the photon fired up.\n\nViewed from the bottom, of course, you get the same result, but for a\nhigher frequency photon, because the one launched from the top (where\ntime is \'going faster\') starts off with a higher frequency.\n\nSo the exchanged photons do not increase in frequency as they are\nbounced between mirrors. That is, in everyones frame no net momentum\nchange occurs for the system as a whole.\n\nOf course if you just reflected photons back and forth you would get the\nequivalent of the inertial space drive where spaceships bounce balls off\neach other (without changing total momentum). However, unlike the\nnewtonian situation it becomes increasing inefficient as the refelected\nphotons have lower and lower momentum the faster the spaceships travel\naway from each other. There is a limit set by c. There is a limit set by\nc for your example, which is when the bottom is at the event horizon\nwhen seen from the top.\n\nc) If I am wrong, the corrections will be good information.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Joe <jhelfand@umd.edu> writes
>I have a question about general relativity. It is based off of the
>Feynman Lectures. Suppose you have a tower on a planet. And at the
>top of the planet you drop a photon to the Earth. As it leaves, the
>photon kicks the tower in one direction and then speeds to the
>surface. At the surface, the photon strikes with an increse in
>energy, a higher frequency, and thus a higher momentum. It kicks the
>tower/Earth in the opposite direction than before. But this kick is
>stronger than the one at the tower when it left. So there is a net
>momentum transfer in the opposite direction. But this doesn't make
>sense. you could make the Earth more compact, so that the difference
>in kick gets more and more.
1) I am NOT an expert.
2) How about the following:
[Which may be wrong, in which case I will be corrected forcefully.]
Stood on the top of the tower you observe that everything below you is
'going slower' due to time 'going slower' in a higher gravitational
potential.
So I think this results in (if you could see it) the photon that is
fired down does not seem to change momentum as it does so. That is as
viewed from the bottom it has a faster frequency than at the top, but
viewed from the top the increase in frequency at the bottom is precisely
balanced by the 'slower time' at the bottom.
The same applies in reverse for the photon fired up.
Viewed from the bottom, of course, you get the same result, but for a
higher frequency photon, because the one launched from the top (where
time is 'going faster') starts off with a higher frequency.
So the exchanged photons do not increase in frequency as they are
bounced between mirrors. That is, in everyones frame no net momentum
change occurs for the system as a whole.
Of course if you just reflected photons back and forth you would get the
equivalent of the inertial space drive where spaceships bounce balls off
each other (without changing total momentum). However, unlike the
newtonian situation it becomes increasing inefficient as the refelected
photons have lower and lower momentum the faster the spaceships travel
away from each other. There is a limit set by c. There is a limit set by
c for your example, which is when the bottom is at the event horizon
when seen from the top.
c) If I am wrong, the corrections will be good information.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.