How Does Space-Time Curvature Affect Light Near Black Holes?

[JesseM]

erm … no … geodesics can be time-like or light-like or space-like.

I was talking about the geodesics of physical objects--there are no objects which follow spacelike geodesics. And I would interpret the phrase "follow a future time direction" in a way that would cover light-like geodesics (since for any two events which lie on a light-like geodesic, all observers will agree on which came earlier and which came later), though perhaps you would define it differently.

An object moves only through space (otherwise, "movement" is meaningless); and its free-fall path in space-time (in which there is no movement) is a time-like geodesic.

I didn't say anything about an object "moving" through time, did I? And yes, of course I agree that a non-massless particle in free-fall will follow a timelike geodesic. Why are you telling me this? Do you think I was saying something different in what I wrote?

I used the phrase "enforced falling" only to describe the subject-matter: my explanation of that subject is:

the time-like geodesics now do not go in every space direction, but are confined within a cone.

I can't think of any way to interpret this statement in a way that doesn't make it nonsense. Do you agree that different objects passing through a particular point in spacetime can end up at any point in the future light cone of that point in spacetime depending on their velocity, regardless of whether the point is inside or outside the event horizon? Do you also agree that objects going through a given point can never end up at a point outside the light cone of that point, outside the horizon as well as inside? If so, 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 refer to "the three spacelike dimensions" … this is counter-intuitive, since it is not obvious which space directions they correspond to.

An observer is obviously free to orient his three spatial axes in any spacelike direction he wants, the point is that he can come up with some locally inertial coordinate system in his local region that has three spatial coordinates and one time dimension, such that the usual laws of SR apply in this region.

I prefer to refer to "every space direction", with its usual meaning.

What is the "usual meaning"? And do you agree that an observer inside the horizon is indeed free to move in any spatial direction, but there is no spatial direction that takes him further from the singularity in Schwarzschild coordinates?

I was referring to your:

Once an object is inside the event horizon of the black hole, the radial axis becomes the time axis for them …

in which you seemed to be saying that the radially in-falling object is not moving along a space direction.

As I mentioned, I meant "radial axis" to refer to the usual Schwarzschild coordinates.

What if I amended it to say "the radial axis of Schwarzschild coordinates becomes the time axis for them"?

It's still true, but I still don't like it … it's now even further away from reality. I want explanations which use concrete concepts such as directions, not abstract ones like coordinate axes.

It's difficult to make meaningful statements about space and time that don't refer to coordinate systems. And at least when talking about locally inertial coordinate systems, the coordinates do have a very simple physical meaning--they represent measurements on a grid of rulers and clocks moving inertially.

But there are! I entirely accept that there are, and I also understand why GR requires it … but I don't accept that GR denies the existence of tangential (or, more generally, out-of-cone) space directions!

Can you define "tangential" without referring to a coordinate system like Schwarzschild coordinates? I suppose "out-of-cone" is a start, but I haven't claimed that there are no events on the event horizon which lie out of the light cones of an event inside the horizon--of course there are! But that doesn't mean the horizon lies in any particular spatial direction for an observer inside the horizon--for this observer I think it would be a spacelike surface that lies in their past (as defined in whatever coordinate system they're using inside the horizon, not all parts of the surface would lie in their past light cone), much like the spacelike surface consisting of the set of all events that happened precisely 10 billion years after the Big Bang in comoving cosmological coordinates. Do you agree this surface lies in our past, not in any particular spatial direction for us? Do you also agree that there are plenty of events on this surface which don't like in our past light cone?

I don't think the equivalence principle does require space inside an event horizon to be locally indistinguishable from space outside. The laws of physics must be indistinguishable, but their application need not be.

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?

For example, do you accept that material objects inside an event horizon must travel faster than light, and that that alone distinguishes inside from outside, even for an inertial observer?

No. They may have a coordinate speed greater than c in Schwarzschild coordinates (which is different from 'faster than light', since a light beam in the same region will have a greater coordinate speed), and even in SR if you use non-inertial coordinate systems objects can move faster than c, but in any local region it's possible to use freefalling rulers and clocks in that region to create a locally inertial coordinate system in that region, and nothing will move faster than c in this coordinate system.

You seemed to be suggesting … while I'm saying …{you wrote "spacelike" - I assume you meant "timelike"?} … Is there any of this you disagree with?

We're both correct! I'm using three-dimensional space directions to explain why geodesics end in the singularity, and you're using four-dimensional time-like directions for the same purpose.

When I wrote "while I'm saying that he can move in any spacelike direction, but he can't avoid the singularity because it lies in a timelike direction", I did mean "any spacelike direction"; in other words, if he constructs a locally inertial coordinate system, he can move along any of the three orthogonal rulers, his movements are not restricted to a cone in space as I was thinking your quote was suggesting.

My only issue is with "the singularity … lies in a timelike direction" … that makes it look as if the singularity is a point in space-time … but it's a line, isn't it, with different bits of it in different timelike directions?

Yes, but that's why I emphasized the part in Egan's quote about the "approach to the singularity" looking like a collapsing hypercylinder from the perspective of an observer inside. If you picture a 2D universe on the surface of a regular cylinder, and the radius of the cylinder is shrinking until it hits zero at some moment, then this is a line singularity rather than a point, but it still lies in a timelike direction for a flatlander living on the cylinder...before the cylinder has collapsed, there's no spatial direction the flatlander on the surface can point to and say "singularity that-a-way".

To summarise my approach:

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. ​

"not all directions are projections of time-like geodesics" is wrong if "directions" is meant to refer to spatial directions--if you foliate a black hole spacetime into a stack of spacelike hypersurfaces, then all directions in a given hypersurface will be a projection of a time-like geodesic. I think the issue with Schwarzschild coordinates is that the set of all events at a particular coordinate time t does not represent a spacelike hypersurface, only the portion outside the event horizon would be spacelike.

When you say "projected onto three-dimensional space" this is just too vague without a particular coordinate system and a particular definition of simultaneity (since you are obviously talking about position in space changing over time, which requires us to have a meaningful notion of what space looks like at a particular time). But if you do pick a coordinate system which assigns every event in the spacetime a time-coordinate, then there are two possibilities:

1. the set of all events at a single time-coordinate is always "spacelike" in the physical sense (no event in the set lies within the light cone of any other in the set), in which case every event will have valid timelike geodesics going in every direction in space.

2. The set of all events at a single time-coordinate is not a spacelike surface, so it doesn't make sense to say that projections of geodesics onto this surface qualifies as projecting the geodesics "onto three-dimensional space".

So, either way, I think I disagree with your summary above.

 

[yuiop]

A cilinder is not a curved but a flat space.

Hi again Jennifer,

Does that mean an almost infinitely long body with most of its mass in a central cylindrical core parallel to the main axis (similar cross section to that of the Earth) can be handled by Special relativity (eg using Minkowski spacetime) ?

I do not want to hijack this thread so could you reply to thread I started here https://www.physicsforums.com/showthread.php?t=225573&page=3 which is where my question relates to? (basically it asks if a horizontally moving object will fall at the same rate as a purely vertically falling object as measured by an observer at rest with such a gravitational body)

Thanks :)

 

[MeJennifer]

Does that mean an almost infinitely long body with most of its mass in a central cylindrical core parallel to the main axis (similar cross section to that of the Earth) can be handled by Special relativity (eg using Minkowski spacetime) ?

I would say no, it seems to me that such a configuration would still give a curved spacetime.

 

[yuiop]

I would say no, it seems to me that such a configuration would still give a curved spacetime.

Is that because we are talking about concentric cylinders and moving from one to the other rather than staying on the surface of one cylinder?

I am, by the way only talking about motion parallel to the main axis of the cylinder and not motion around it. I am also talking about a fall distance dr that is infinitessimal compared to radius (R) of the massive body, so we can consider R to be aproximately constant. For example on Earth, considering the acceleration to be a constant 1g is a reasonable aproximation when we are talking about a fall of a few meters.

 

[JesseM]

Hi again Jennifer,

Does that mean an almost infinitely long body with most of its mass in a central cylindrical core parallel to the main axis (similar cross section to that of the Earth) can be handled by Special relativity (eg using Minkowski spacetime) ?

You have to distinguish between the notion of an ordinary physical cylinder in 3D space, and the notion of 3D space itself being represented as a hypercylinder in a 4D embedding diagram. MeJennifer was talking about 3D space having a hypercylindrical shape (which as she said involves no intrinsic curvature, so it's really just flat space with an unusual topology that makes it finite in one or two directions), I would think a physical cylinder would cause some intrinsic curvature in GR.

 

[MeJennifer]

MeJennifer was talking about 3D space having a hypercylindrical shape (which as she said involves no intrinsic curvature, so it's really just flat space with an unusual topology that makes it finite in one or two directions), I would think a physical cylinder would cause some intrinsic curvature in GR.

Basically correct, but in you explanation you are one dimension short.

GR curves a 4D not a 3D space.

 

[JesseM]

Basically correct, but in you explanation you are one dimension short.

GR curves a 4D not a 3D space.

GR does talk about curvature of 4D spacetime, but the cylinder represents a topology for flat 3D space, similar to the different possible finite topologies for space discussed in this article (though the topologies discussed there are finite in all directions, while the a space with the topology of a cylinder would be finite in some directions and infinite in others).

 

[yuiop]

You have to distinguish between the notion of an ordinary physical cylinder in 3D space, and the notion of 3D space itself being represented as a hypercylinder in a 4D embedding diagram. MeJennifer was talking about 3D space having a hypercylindrical shape (which as she said involves no intrinsic curvature, so it's really just flat space with an unusual topology that makes it finite in one or two directions), I would think a physical cylinder would cause some intrinsic curvature in GR.

Surely, if Minkowski spacetime can handle the case of an accelerating rocket, then it can handle the motion of a falling particle in the idealised cylindrical planet?

Does not the Equivalence principle require that they are the equivalent?

An accelerating rocket can not duplicate the the gravity of an spherical massive body, but surely it can duplicate an idealised gravitational flat gravitational body that has planar symmetry horizontally? Otherwise, what is the point of the EP?

[EDIT] Maybe I should phrase it another way. What hypothetical gravitational body is equivalent to an acccelerating rocket? If the answer is none, it makes the EP invalid.

 

[JesseM]

Surely, if Minkowski spacetime can handle the case of an accelerating rocket, then it can handle the motion of a falling particle in the idealised cylindrical planet?

An accelerating rocket doesn't curve spacetime to any significant degree, a cylindrical planet would.

Does not the Equivalence principle require that they are the equivalent?

Only if you zoom in on a very small region of the curved spacetime where the curvature was negligible.

An accelerating rocket can not duplicate the the gravity of an spherical massive body, but surely it can duplicate an idealised gravitational flat gravitational body that has planar symmetry horizontally? Otherwise, what is the point of the EP?

[EDIT] Maybe I should phrase it another way. What hypothetical gravitational body is equivalent to an acccelerating rocket? If the answer is none, it makes the EP invalid.

The point of the equivalence principle is that the laws of physics in the local spacetime neighborhood of a freefalling observer in curved spacetime must reduce to the laws of physics in inertial frames in flat SR spacetime. There's no way the laws of physics in a large region of curved spacetime where tidal forces are significant can be treated as equivalent to the laws of SR.

 

[yuiop]

An accelerating rocket doesn't curve spacetime to any significant degree, a cylindrical planet would.

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?

Only if you zoom in on a very small region of the curved spacetime where the curvature was negligible.

The point of the equivalence principle is that the laws of physics in the local spacetime neighborhood of a freefalling observer in curved spacetime must reduce to the laws of flat SR spacetime.

A vertically free falling observer in a falling elevator would observe that a horizontal light beam, a horizontally moving particle and a released stationary particle do not fall relative to the elevator. They behave as if the elevator was far away from any gravitational body. An observer that was not free falling, would observe that the particles and the light beam and the elevator all appear to be falling at the same rate. So why does GR predict that a particle moving horizontally falls faster than a particle without horizontal motion, even when we consider a flat gravitational body?


P.S Does that mean Rindler spacetime is only valid for accelerating rockets and cannot be applied to even hypothetical gravitational bodies?

 

[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 propagate 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 propagate 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 dimension, 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.