Exploring the Curvature of Space-Time in Relation to Black Holes

[JesseM]

Sure it would curve it around the cylinder, but parallel to the cylinder it would be horizontally flat. What if we replaced the cylinder with flat body with "almost" infinite horizontal dimensions?

I don't quite understand what you mean by "parallel to the cylinder". Do you mean looking at a 2D section of 3D space in which all the points in this section are at the same radius from the cylinder? And for a very large flat body, the curvature of spacetime may be negligible if you choose a region of space where the distance between the bottom of the region and the top is very small compared to the size of the body, so that in Newtonian terms the gravitational force can be treated as pretty much constant in the region, but you'd still see curvature if you picked a much larger region. That's my point, the equivalence principle is all about picking a region of curved spacetime that's small enough that the curvature can be treated as negligible in that region.

So why does GR predict that a particle moving horizontally falls faster than a particle without horizontal motion?

We've discussed this before, but I'm not convinced that it does fall faster. And even if it does, this wouldn't be incompatible with the Equivalence principle if you could show that if you have two inertial bodies above a platform that's accelerating upwards in flat SR spacetime, with one body moving horizontally relative to the platform while the other is not, then in the frame of the body moving horizontally relative to the platform, the surface of the platform would accelerate up to meet it more quickly than for the other body (this would only be true if in the frame of the body moving horizontally, different parts of the platform are accelerating at different rates, which might or might not be true, as I've said before we'd really need to do the math to check).

 

[yuiop]

GR predicts that a horizontally moving particle falls faster than a purely vertically falling particle.

That implies that a falling spinning gyrosope (nearly all particles moving horizontally when the spin axis is vertical) will fall faster than a none spinning gyrosope.

An actual experiment to test this showed this did not happen. Does that invalidate GR?

Or does it validate the EP and invalidate the incorrect assumption that GR predicts a horizontally moving particle will fall faster in a locally flat space?

 

[JesseM]

GR predicts that a horizontally moving particle falls faster than a purely vertically falling particle.

Are you just relying on the arguments you made on the other thread, or have you found a reference for this? Like I said, I wasn't convinced by those arguments, so if you have the reference please provide it, I'd like to see what it says exactly (in particular, whether it's talking about a small local region of spacetime or a larger one that can't be treated as equivalent to SR).

That implies that a falling spinning gyrosope (nearly all particles moving horizontally when the spin axis is vertical) will fall faster than a none spinning gyrosope.

An actual experiment to test this showed this did not happen. Does that invalidate GR?

Pretty sure physicists would have noticed if it went against GR, and we'd have heard more about it. And is the implication here your own conclusion, or do you have a reference for that too? I imagine most experiments with gyroscopes would only involve comparing them in the same local region, so if your statement about a horizontally moving particle falling faster was based on looking at motion in a large region of spacetime where curvature was significant, then that would explain why the statement about horizontal motion doesn't lead to the implication about spinning gyroscopes.

 

[yuiop]

We've discussed this before, but I'm not convinced that it does fall faster. And even if it does, this wouldn't be incompatible with the Equivalence principle if you could show that if you have two inertial bodies above a platform that's accelerating upwards in flat SR spacetime, with one body moving horizontally relative to the platform while the other is not, then in the frame of the body moving horizontally relative to the platform, the surface of the platform would accelerate up to meet it more quickly than for the other body (this would only be true if in the frame of the body moving horizontally, different parts of the platform are accelerating at different rates, which might or might not be true, as I've said before we'd really need to do the math to check).

"this would only be true if in the frame of the body moving horizontally" ... exactly! In the frame of the platform the left and right sides of the platform are moving upwards at the same rate and collide with both particles simultaneously as far as the observer on the platform is concerned. To him, the particles fall at exactly the same rate irrespetive of horizontal motion.

Two clocks spatially separated at the top of the platorm could be synchronised and will remain in sync as the platform accelerates upward. The same is true for two clocks horizontally separated at the bottom of the platform. The fact that the top clocks are not exactly in sync with the lower clocks is irrelevant. All that matters is that to the platform based observer all the particles left simultaneously, and landed simultaneously. To him, the particles fall at at exactly the same rate. The EP requires the same is true for a gravitational body within a region where the spacetime is negligably curved within the locality of the experiment.

 

[JesseM]

"this would only be true if in the frame of the body moving horizontally" ... exactly! In the frame of the platform the left and right sides of the platform are moving upwards at the same rate and collide with both particles simultaneously as far as the observer on the platform is concerned. To him, the particles fall at exactly the same rate irrespetive of horizontal motion.

Two clocks spatially separated at the top of the platorm could be synchronised and will remain in sync as the platform accelerates upward. The same is true for two clocks horizontally separated at the bottom of the platform. The fact that the top clocks are not exactly in sync with the lower clocks is irrelevant. All that matters is that to the platform based observer all the particles left simultaneously, and landed simultaneously. To him, the particles fall at at exactly the same rate. The EP requires the same is true for a gravitational body within a region where the spacetime is negligably curved within the locality of the experiment.

Well, it requires it to be true for an observer on a gravitational body who is not moving horizontally relative to the body at least. So, do you think there is any reason to believe that the EP doesn't work here? Isn't it in fact true that if you drop a bunch of objects from the same height and simultaneously in this guy's frame, they'll hit the ground simultaneously in his frame too, regardless of their horizontal velocity?

 

[yuiop]

Are you just relying on the arguments you made on the other thread, or have you found a reference for this? Like I said, I wasn't convinced by those arguments, so if you have the reference please provide it, I'd like to see what it says exactly (in particular, whether it's talking about a small local region of spacetime or a larger one that can't be treated as equivalent to SR).

I can not find a reference that specifically addresses the problem. That is why I asked the memebrs of this forum what would happen, and several members have asserted that GR predicts a horizontally moving particle will fall faster in locally flat spacetime. I disagree with that view, mainly because it contradicts the EP.

The gyroscope experiment I referred to is here http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111069v1.pdf

 

[yuiop]

Well, it requires it to be true for an observer on a gravitational body who is not moving horizontally relative to the body at least. So, do you think there is any reason to believe that the EP doesn't work here? Isn't it in fact true that if you drop a bunch of objects from the same height and simultaneously in this guy's frame, they'll hit the ground simultaneously in his frame too, regardless of their horizontal velocity?

That is exactly my argument, and I am glad we have come to some agreement here. :)

The problem is that in other thread an equation of GR (by Pervect) that is relevant to gravitational acceleration $d^2 r /dt ^2$ in Schwarzschild geometry, when converted to flat spacetime predicts the horizontally moving object falls faster. It seems some members are inclined to believe that conclusion, even though it appears to contradict the EP.

Maybe the conversion to flat spacetime was not done correctly?

 

[JesseM]

That is exactly my argument, and I am glad we have come to some agreement here. :)

The problem is that in other thread an equation of GR (by Pervect) that is relevant to gravitational acceleration $d^2 r /dt ^2$ in Schwarzschild geometry, when converted to flat spacetime predicts the horizontally moving object falls faster. It seems some members are inclined to believe that conclusion, even though it appears to contradict the EP.

Maybe the conversion to flat spacetime was not done correctly?

I thought that was just Jorrie's interpretation of what pervect was saying, rather than something pervect had clearly said himself, although I didn't follow that part of the thread very closely. I suspect that Jorrie was either misunderstanding, or just talking about the behavior of particles in a large region of curved spacetime rather than in a very small local region.

 

[tiny-tim]

"space directions"

I can't think of any way to interpret this statement {"the time-like geodesics (inside an event horizon) now do not go in every space direction, but are confined within a cone"} in a way that doesn't make it nonsense.

The time-like geodesics quite clearly are confined within a cone … in fact, the cone in the very diagrams you keep referring to!

Outside an event horizon, the projection of a light-cone is an expanding sphere … which makes "light-cone" a really stupid name!.

But inside an event horizon, the name is more sensible, because it is a surface which actually does expand outwards inside a cone!

When I say "projected onto three-dimensional space", I mean any projection in which one can draw those cones!

If you insist on my specifying a coordinate system, I choose the following: a series of spheres of test particles fall together through the event horizon. Each zeroes its clock as it passes, say, 1000GM. They fall radially inwards, so each can be given a latitude and longitude. The radius coordinate of any event inside the event horizon is defined as the (proper) time (on its own clock) of the test particle going through that event.

(We could use photons instead of test particles, and count their wavelengths in lieu of a clock.)

(I have a vague recollection that the tortoise coordinate system would do as well.)

ok … in that system, if we choose a fixed time, then we have a fairly ordinary three-dimensional sphere, inside which we can draw these cones of yours.

The sphere is entirely spatial, and in particular the tangential directions, and generally all out-of-cone directions, are spatial.

It's a three-dimensional space … what else can they be?

You are refusing to accept that the lines (to choose a neutral word) which miss the cone are "in space" … a refusal based on your conviction that a local observer would not be aware of them.

I say (a) he knows perfectly well that they're there, even though he has no means of detecting them, and (b) sod him … we know that they're there!

I don't understand your distinction between the "laws of physics" and their "application". Do you agree that any experiment done in a small windowless room over a small period of time will have the same result regardless of whether the room is inside our outside the horizon, provided the region of spacetime is small enough that there are no significant tidal forces?

No.

In particular, inside an event horizon, a photon will always be overtaken by an electron falling next to it.

… in what sense do you think objects can "go in every space direction" outside the horizon but not inside the horizon, and in what sense are they "confined to a cone" inside but not outside?

You are using "space direction" in the sense of those directions detectable by a local inertial observer. Which prevents you from drawing those cones!

I am using it in the sense (and in the coordinate system) described above … it enables me to make the common-sense observation that:

Geodesics are four-dimensional curves (which involve no movement).
They can be projected onto three-dimensional space.
Every free-fall object has a time-like geodesic.
It moves along the three-dimensional projection of that geodesic, but inside an event horizon not all directions are projections of time-like geodesics. ​

 

[Antenna Guy]

Geodesics are four-dimensional curves (which involve no movement).
They can be projected onto three-dimensional space.
​

If I understand waves correctly, a geodesic associated with one would project onto all of space unless a definite time (relative to emission) is specified.

Regards,

Bill

 

[JesseM]

The time-like geodesics quite clearly are confined within a cone … in fact, the cone in the very diagrams you keep referring to!

What diagrams are you talking about? The Eddington-Finkelstein diagrams I posted? Of course in these one of the axes is the time coordinate axis, and timelike geodesics are confined to cones both inside and outside the horizon. You had been talking before about projecting the directions of geodesics onto a purely spatial diagram, and I say again, as long as you foliate the spacetime into a stack of spacelike surfaces (the set of all events at constant t in either Schwarzschild coordinates or Eddington-Finkelstein coordinates wouldn't qualify, because the t-coordinate is spacelike inside the horizon for both systems), then when you project the direction of geodesics onto anyone of these surfaces, you find that the geodesics do go in all directions from every point on the surface.

Outside an event horizon, the projection of a light-cone is an expanding sphere … which makes "light-cone" a really stupid name!.

Obviously you understand that the "cone" refers to diagrams where we only draw 2 (or 1) space dimension and 1 time dimension, like the Eddington-Finkelstein diagrams I kept linking to (where the worldlines of light emanating from any given point formed a visual cone both inside and outside the horizon). And again, if you're talking about projecting light paths onto a purely spacelike surface, then light will go in all directions from every point, and look like an expanding sphere everywhere.

But inside an event horizon, the name is more sensible, because it is a surface which actually does expand outwards inside a cone!

Only if you use a coordinate system where the t-coordinate becomes spacelike inside the horizon, and take a surface of constant t in that system. If you use a coordinate system where the t-coordinate is timelike everywhere, so that a surface of constant t is purely spacelike, then the projection of light geodesics onto this surface will show that light can travel in all spatial directions on the surface from any point, inside or outside the horizon.

When I say "projected onto three-dimensional space", I mean any projection in which one can draw those cones!

So you don't care if the surface you're projecting onto is actually spacelike everywhere? If not, then the phrase "projected onto three-dimensional space" is pretty misleading.

If you insist on my specifying a coordinate system, I choose the following: a series of spheres of test particles fall together through the event horizon. Each zeroes its clock as it passes, say, 1000GM. They fall radially inwards, so each can be given a latitude and longitude. The radius coordinate of any event inside the event horizon is defined as the (proper) time (on its own clock) of the test particle going through that event.

In this coordinate system, a surface of constant t won't be spacelike outside the horizon! After all, if two successive spheres pass the fixed 1000GM sphere and both set their clocks to zero when they do, then a guy on the first sphere can send a message as he passes the 1000GM sphere which will reach a guy on the second sphere before he passes the 1000GM sphere. So, if you take the hypersurface composed of all events that are assigned a time of 0 in this coordinate system, there will be a timelike separation between some of these events.

I think you could solve this problem by having a set of ordinary clocks fixed at the 1000GM sphere, and then each successive falling sphere sets its own clocks to match the current readings on these fixed-radius clocks at the moment it passes the fixed 1000GM sphere, instead of zeroing its clocks as it passes the fixed sphere like you suggested. In this case I would think a surface of constant t would be spacelike both inside and outside the horizon, though I'm not sure. If it is, though, then I'm sure that if you project the direction of geodesics emanating from an event inside the horizon onto the surface corresponding to the t-coordinate of that event, then the projected geodesics would go in all directions on the surface from that event!

ok … in that system, if we choose a fixed time, then we have a fairly ordinary three-dimensional sphere, inside which we can draw these cones of yours.

A sphere? No, even in the system you described, if you pick a set of points at constant t it wouldn't just be a sphere--say we pick t=0, then since you said every sphere zeroes its clocks as it passes 1000GM, then there will be events on every sphere assigned a time of t=0. But like I said, your coordinate system is problematic because a surface of constant t won't be spacelike, not even outside the horizon; it will contain events which lie in one another's light cones.

The sphere is entirely spatial, and in particular the tangential directions, and generally all out-of-cone directions, are spatial.

Like I said, you seem to be pretty confused, a surface of constant t would not be a sphere in your coordinate system, nor would it be spacelike (which is what I guess you mean by 'entirely spatial').

I don't understand your distinction between the "laws of physics" and their "application". Do you agree that any experiment done in a small windowless room over a small period of time will have the same result regardless of whether the room is inside our outside the horizon, provided the region of spacetime is small enough that there are no significant tidal forces?

No.

In particular, inside an event horizon, a photon will always be overtaken by an electron falling next to it.

Just to be clear, do you mean an electron will overtake a photon even in a locally inertial coordinate system constructed out of freefalling rulers and clocks using the same procedure as inertial coordinate systems constructed out of inertial rulers and clocks in SR? If you are claiming that, you're badly mistaken--this would amount to a denial of the equivalence principle.

You are using "space direction" in the sense of those directions detectable by a local inertial observer. Which prevents you from drawing those cones!

I am using it in the sense (and in the coordinate system) described above … it enables me to make the common-sense observation that:

Geodesics are four-dimensional curves (which involve no movement).
They can be projected onto three-dimensional space.
Every free-fall object has a time-like geodesic.
It moves along the three-dimensional projection of that geodesic, but inside an event horizon not all directions are projections of time-like geodesics. ​

Your "common-sense observation" is highly misleading if you are using "projected onto three-dimensional space" to mean anything other than "projected onto a purely spacelike hypersurface". And if you do mean to project onto a spacelike hypersurface, then you're wrong that not all directions on this surface would correspond to projections of time-like geodesics, they certainly would, inside the horizon as well as outside.

 

[tiny-tim]

… a short post at last … !

If I understand waves correctly, a geodesic associated with one would project onto all of space unless a definite time (relative to emission) is specified.

Hi Bill!

I don't understand how you'd associate a geodesic to a wave.

So you don't care if the surface you're projecting onto is actually spacelike everywhere? If not, then the phrase "projected onto three-dimensional space" is pretty misleading.

You've got it!

And it's only misleading if you confuse "space" with "space-like".

In this coordinate system, a surface of constant t …

erm … I didn't define a t coordinate … I left it to the reader to choose one … yours looks ok to me …

Just to be clear, do you mean an electron will overtake a photon even in a locally inertial coordinate system constructed out of freefalling rulers and clocks using the same procedure as inertial coordinate systems constructed out of inertial rulers and clocks in SR?

Bingo!

It will hit the singularity first.

 

[Antenna Guy]

I don't understand how you'd associate a geodesic to a wave.

Consider that (EM) waves propogate at c, but don't go anywhere. n.b. that's not to say that a wave is not changing "position" with respect to time - just that the change in position with respect to any given spatial direction is 0 ($\delta r(\hat{v})=\pm c\delta t$).

Regards,

Bill

 

[tiny-tim]

Consider that (EM) waves propogate at c, but don't go anywhere. n.b. that's not to say that a wave is not changing "position" with respect to time - just that the change in position with respect to any given spatial direction is 0 (δr(v^) = ±cδt).

Hi Bill!

I don't think that's possible … surely even a plane EM wave, at fixed time, will "go up and down" along the spatial direction of its velocity?

 

[Antenna Guy]

I don't think that's possible … surely even a plane EM wave, at fixed time, will "go up and down" along the spatial direction of its velocity?

There's no such thing as a "plane wave". You're referring to a local approximation of a spherical wave at very large radius.

Regards,

Bill

 

[harryjoon]

There are number of points which I believe suggest that it is not that simple;
1)-Worldline of objects may or may not intersect. If it does it is given that they will meet.
2)-Worldline of an object may or may not be along the geodesic line of the curved space-time field produced by Earth's mass.
3)-A free-falling object travels along a geodesic of the curved space-time field produced by Earth's mass, which is also its worldline.
4)-The world line of the Earth is NOT along a geodesic of its curved space-time field (produced by Earth's mass).
5)- Earth must carry its curved space-time field produced by its mass, along its worldline, which means that the world line of an object orbitting say one meter above Earth surface, i.e following a geodesic of the field which is a meter above the surface of the earth, will always be one meter above the Earth surface, independent of the position of Earth along its worldline. A contrarry sugestion would mean objects will be left behind as Earth travels along its worldline, which is contrary to our observation.

the point I was trying to make was this; A condition of geodesic line is that; A timelike geodesic is a worldline which parallel transports its own tangent vector and maintains the magnitude of its tangent as a constant.
An object such as Earth produces acurved space time in which each geodesic line is a great circle of the sphereical surfaces of constant curvature along which the above condition is satisfied and a free falling object in orbit travells. An object falling towards the center of the Earth passes from one surface to another thus moves from one curvature to another in the increasing direction of curvature. How can this object be regarded as following a geodesic of the curved space-time of Earth.

 

[Dale]

An object such as Earth produces acurved space time in which each geodesic line is a great circle of the sphereical surfaces of constant curvature along which the above condition is satisfied and a free falling object in orbit travells.

This is incorrect. The geodesics are more like distorted helixes (except for the ones that follow escape trajectories).

 

[harryjoon]

This is incorrect. The geodesics are more like distorted helixes (except for the ones that follow escape trajectories).

Can you give me reference to that definition?

 

[DrGreg]

This is incorrect. The geodesics are more like distorted helixes (except for the ones that follow escape trajectories).

I agree.

harryjoon, the geodesics are lines in spacetime, not in space. Therefore as you move along a geodesic, you move forward in time, and so a geodesic can't be a circle. Geodesics are the wordlines of small test particles that are falling freely without any other forces acting on them. So, as it is possible for a particle to have a perfectly circular orbit around a planet, and the worldline of that orbit is a spacetime helix, then helixes are geodesics.

There are lots of other geodesics, too, including almost-elliptical helixes, and almost-hyperbolic helixes. (I say "almost" because GR differs from Newtonian gravitiation.) Through any event in spacetime there are an infinite number of geodesics, each one the worldline of a particle with a different velocity.

Note that, whenever there is zero spacetime curvature (gravity is negligible), every straight worldline through any event is a geodesic.

 

[harryjoon]

I agree.

DaleSpam, the geodesics are lines in spacetime, not in space. Therefore as you move along a geodesic, you move forward in time, and so a geodesic can't be a circle. .

you are correct in the first part, but not on the second part. You are picturing the world line of the particle on a (time-space) sketch in two or three dimention, as viewed by someone like yourself. However, the wordline path of a particle ( test particle or otherwise) follows a three dimensional spatial path in time, such as a great circle of a spherical surface in hyperspace of the curved space-time, where the time dimension is "observale" only by the use of appropriate measuring instrument called "clocks". The helix which you are taking about can only be observed as a great circle in 3-D space. However, what I am asking is why the path of "some" objects in 3-D space is helical while others are on a closed cirle (I agree with you that they are all helical paths in 4-D spacetime). Furthermore, what makes those objects in 3-D helical path to deviate from their circular path along a great circle, and thus necessarily, move from one great circle to another on a neighbouring spherical surface of the hypersurface.

 

[Antenna Guy]

Can you give me reference to that definition?

Let a circular orbit lie in the z=0 plane.

Replace the z-axis with a time axis, and the orbit follows a helix.

Regards,

Bill

 

[Dale]

Can you give me reference to that definition?

Start with the wikipedia article on the http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity" . As it says "The motion of the lighter body (called the "particle" below) is described by the space-time geodesics of the Schwarzschild solution.". It then goes on to derive the geodesic equations representing a precessing elliptical orbit, which is what I was referring to by a "distorted helix" in spacetime.

If the distorted helix (geodesic) worldline intersects the worldline of the ground then your object crashes, if not it orbits. It really is as simple as that.

 

[Antenna Guy]

If the distorted helix (geodesic) worldline intersects the worldline of the ground then your object crashes, if not it orbits. It really is as simple as that.

In my model, the area of "the ground" (at z=0) is assumed constant with respect to time.

I agree with your assessment, but see it in a different way.

Regards,

Bill

 

[Dale]

In my model, the area of "the ground" (at z=0) is assumed constant with respect to time.

Yes, that is correct. The worldline of the ground is not a geodesic. Neglecting the rotation of the Earth the worldline is parallel to the time axis (constant wrt time).

 

[Antenna Guy]

The worldline of the ground is not a geodesic.

If not a geodesic, what is it? A family of geodesics that constitute a world-area? (IMO - yes)

Regards,

Bill

 

[MeJennifer]

A point on the surface of an ideal spherical mass is in a state of constant acceleration. Points that accelerate are obviously not on a geodesic.

 

[Antenna Guy]

A point on the surface of an ideal spherical mass is in a state of constant acceleration. Points that accelerate are obviously not on a geodesic.

If points on a geodesic cannot accelerate, how can a geodesic change direction?

Regards,

Bill

 

[snoopies622]

Maybe MeJennifer meant that a point on a mass-surface is constantly being pushed (away from it) so it isn't in freefall and therefore isn't moving on a geodesic.

 

[Antenna Guy]

Maybe MeJennifer meant that a point on a mass-surface is constantly being pushed (away from it) so it isn't in freefall and therefore isn't moving on a geodesic.

If that was was she meant, would that make it (the point on a mass-surface) stationary with respect to time?

Regards,

Bill

 

[snoopies622]

No. I didn't mean to imply that. (Guess I should read the whole thread before jumping in again...)

 

[Dale]

An accelerometer placed on the ground reads g, not 0. Therefore the ground does not follow a geodesic.

If you have a stationary, non-rotating, spherical mass then the spacetime has the Swartzschild metric. In such a spacetime a worldline which is parallel to the time axis (stationary wrt time) is not a geodesic.

 

[A.T.]

If points on a geodesic cannot accelerate, how can a geodesic change direction?

You could put it that way: The geodesic doesn't change direction. But the direction of the dimensions relative to the geodesic changes, in curved spacetime. If you then assume the curved spacetime to be flat, it apears that the geodesic changes direction.

Analogy: Every great circle on the Earth is a geodesic. And when you put a toy car on a globus, you will see that it follows these geodesics without changing direction (steering). But if draw them on a flat map of the earth, some of them are curved, because the flat map of the world is distorted.

In this applet: http://www.adamtoons.de/physics/gravitation.swf you can see both views:
On the left: The curved spacetime forced to be flat by distorting it, also distorting the geodesic which seems to change direction.
On the right: The curved spacetime undistorted, where the geodesic doesn't change direction.

 

[JesseM]

So you don't care if the surface you're projecting onto is actually spacelike everywhere? If not, then the phrase "projected onto three-dimensional space" is pretty misleading.

You've got it!

And it's only misleading if you confuse "space" with "space-like".

But what you are doing is saying we can call absolutely any slice through spacetime "space", which makes the term fairly meaningless. For example, I could take a standard Minkowski coordinate system, relabel the t coordinate as "x" and relabel the x coordinate as "t", and then take a surface of constant t, and if we call this surface "space" and project geodesics onto it, it will be true here as well that geodesics can only radiate in certain directions from an event, forming a cone in "space". But calling this "space" is obviously pretty misleading, since clearly x is a time dimension!

If you insist on my specifying a coordinate system, I choose the following: a series of spheres of test particles fall together through the event horizon. Each zeroes its clock as it passes, say, 1000GM. They fall radially inwards, so each can be given a latitude and longitude. The radius coordinate of any event inside the event horizon is defined as the (proper) time (on its own clock) of the test particle going through that event.

In this coordinate system, a surface of constant t won't be spacelike outside the horizon! After all, if two successive spheres pass the fixed 1000GM sphere and both set their clocks to zero when they do, then a guy on the first sphere can send a message as he passes the 1000GM sphere which will reach a guy on the second sphere before he passes the 1000GM sphere. So, if you take the hypersurface composed of all events that are assigned a time of 0 in this coordinate system, there will be a timelike separation between some of these events.

erm … I didn't define a t coordinate … I left it to the reader to choose one … yours looks ok to me …

You did define a t coordinate, see the quote above. Anyway, if the "space" you're projecting geodesics onto doesn't have to be spacelike, just a surface of constant t, this means that your statement that "Outside an event horizon, the projection of a light-cone is an expanding sphere" while "inside an event horizon, the name is more sensible, because it is a surface which actually does expand outwards inside a cone!" is then not physical statement at all, but just a statement about certain coordinate systems which have the property that t is timelike outside the horizon and spacelike inside. Do you agree that, if we pick a coordinate system where t is timelike both inside and outside the horizon, then the projection of a light-cone onto a surface of constant t will look like an expanding sphere both inside and outside the horizon? And likewise, if we picked a weird coordinate system where t is timelike inside the horizon but not outside, then the projection of a light cone would look like a cone outside but it would look like an expanding sphere inside?

Also, what do you mean when you say mine "looks ok" to you? If you're referring to the one I described here:

I think you could solve this problem by having a set of ordinary clocks fixed at the 1000GM sphere, and then each successive falling sphere sets its own clocks to match the current readings on these fixed-radius clocks at the moment it passes the fixed 1000GM sphere, instead of zeroing its clocks as it passes the fixed sphere like you suggested.

...then as I said, I think the t coordinate would be timelike both inside and outside the horizon in this coordinate system, and therefore projecting the geodesics of light rays from an event onto a surface of constant t would yield expanding spheres both inside and outside the horizon.

Just to be clear, do you mean an electron will overtake a photon even in a locally inertial coordinate system constructed out of freefalling rulers and clocks using the same procedure as inertial coordinate systems constructed out of inertial rulers and clocks in SR? If you are claiming that, you're badly mistaken--this would amount to a denial of the equivalence principle.

Bingo!

It will hit the singularity first.

I didn't ask which would hit the singularity first. I asked whether you're saying the electron's velocity at any given time, as measured using this system of freefalling rulers and clocks in a local region, would be lower than the photon's velocity at the same time. If you are saying that, this would be a violation of the equivalence principle, so I'm confident this would be wrong. Of course, it might still be true that the electron's spatial velocity as measured in this freefalling locally inertial system was lower than the photon's, yet it hits the singularity first in Schwarzschild coordinates or some other system.

 

[tiny-tim]

… a fairly short post …

Hi JesseM!

Do you agree that, if we pick a coordinate system where t is timelike both inside and outside the horizon …

But t is spacelike inside the horizon … that's obvious from the sign of the coefficient of dt² in the metric.

The other three dimensions … which are space … happen to be two space-like and one time-like … again from the signs of the coefficients of the metric.

You did define a t coordinate, see the quote {below}.

Goodness … that was 7 days ago …

If you insist on my specifying a coordinate system, I choose the following: a series of spheres of test particles fall together through the event horizon. Each zeroes its clock as it passes, say, 1000GM. They fall radially inwards, so each can be given a latitude and longitude. The radius coordinate of any event inside the event horizon is defined as the (proper) time (on its own clock) of the test particle going through that event.

No … I used those clocks only to define an r coordinate … with that r coordinate, one could choose t or t + r or t - r or various other possibilities as the t coordinate, and I left it unspecified!

My "t coordinate" that seems to have impressed you so much gives a t coordinate of 0 to all events at 1000GM !

I didn't ask which would hit the singularity first …

I maintain that any realistic coordinate system will have faster particles hitting the singularity first.

 

[JesseM]

But t is spacelike inside the horizon … that's obvious from the sign of the coefficient of dt² in the metric.

t of what coordinate system? Schwarzschild coordinates? Of course I agree that t is spacelike inside the horizon in Schwarzschild coordinates, I've said this several times before. The point is that this is not a physical fact about black holes, it is just an artifact of the way Schwarzschild coordinates are constructed, much like the coordinate singularity at the event horizon of Schwarzschild coordinates. Physicists can and do construct other coordinate systems where t is timelike inside the horizon as well as outside. Do you disagree? If not, do you agree with my statement that if we pick a surface of constant t in such a coordinate system, then the projection of light geodesics emanating from an event will go in all directions rather than being confined to a cone, regardless of whether we choose an event inside the horizon or outside?

You did define a t coordinate, see the quote {below}.

Goodness … that was 7 days ago …

Please don't roll your eyes at me, it's rude. I've been away on vacation since Thursday (got back Monday night). But in any case, when you made your statement "erm … I didn't define a t coordinate … I left it to the reader to choose one", that was in response to a post of mine where I quoted the post of yours where you had defined a t coordinate (a post you had written only a few hours earlier), and discussed your choice of coordinates in detail.

My "t coordinate" that seems to have impressed you so much gives a t coordinate of 0 to all events at 1000GM !

It didn't "impress me", in fact I criticized it for exactly this reason: every successive sphere gets assigned a time-coordinate of t=0 when it crosses 1000GM, therefore there will be a timelike separation between different events on a surface of constant t. Please reread my response, noting the part in bold:

If you insist on my specifying a coordinate system, I choose the following: a series of spheres of test particles fall together through the event horizon. Each zeroes its clock as it passes, say, 1000GM. They fall radially inwards, so each can be given a latitude and longitude. The radius coordinate of any event inside the event horizon is defined as the (proper) time (on its own clock) of the test particle going through that event.

In this coordinate system, a surface of constant t won't be spacelike outside the horizon! After all, if two successive spheres pass the fixed 1000GM sphere and both set their clocks to zero when they do, then a guy on the first sphere can send a message as he passes the 1000GM sphere which will reach a guy on the second sphere before he passes the 1000GM sphere. So, if you take the hypersurface composed of all events that are assigned a time of 0 in this coordinate system, there will be a timelike separation between some of these events.

I think you could solve this problem by having a set of ordinary clocks fixed at the 1000GM sphere, and then each successive falling sphere sets its own clocks to match the current readings on these fixed-radius clocks at the moment it passes the fixed 1000GM sphere, instead of zeroing its clocks as it passes the fixed sphere like you suggested. In this case I would think a surface of constant t would be spacelike both inside and outside the horizon, though I'm not sure. If it is, though, then I'm sure that if you project the direction of geodesics emanating from an event inside the horizon onto the surface corresponding to the t-coordinate of that event, then the projected geodesics would go in all directions on the surface from that event!

And you never answered my question of what coordinate system you were talking about when you said "yours looks ok to me"--were you indeed talking about the coordinate system I suggested in the last paragraph of the above quote? If so, can you tell me if you agree with my statement "I think the t coordinate would be timelike both inside and outside the horizon in this coordinate system, and therefore projecting the geodesics of light rays from an event onto a surface of constant t would yield expanding spheres both inside and outside the horizon"?

I maintain that any realistic coordinate system will have faster particles hitting the singularity first.

You're obscuring the issue once again. Even if it's true in Schwarzschild coordinates that the electron hits the singularity before the photon (and this actually depends on which side of the light cone we're looking at), I wasn't asking you about Schwarzschild coordinates, I was asking very specifically about a coordinate system constructed of freefalling rulers and clocks inside the horizon. Just because the electron hits the singularity "first" in Schwarzschild coordinates doesn't mean it'll hit it first in a freefalling rulers/clocks coordinate system, which should be obvious since the t-axis is actually spacelike inside the horizon while a t-coordinate based on freefalling physical clocks must of course be timelike inside the horizon as well as outside (the statement that 'the electron hits the singularity at an earlier t-coordinate in Schwarzschild coordinates' might turn out to be equivalent to something like 'the electron is crushed into a singularity further in the -x direction on the x-axis of the freefalling rulers/clocks system', which makes sense if we keep in mind Greg Egan's point that observers inside the black hole see the approach to the singularity as resembling the collapse of a universe that has the shape of a hypercylinder, the end product of which is a line singularity rather than a point singularity).

There is nothing sacred or holy about Schwarzschild coordinates, they are just one of many possible coordinate systems you can use for dealing with a black hole. And in a local coordinate system constructed out of freefalling rulers and clocks, the laws of physics must look identical to those of SR, so the photon must have a higher speed than the electron. If you deny this, then you have confused yourself with your implicit reliance on Schwarzschild coordinates into denying the Equivalence principle, a pretty serious mistake.