Spacetime curvature

at the center of a planet/star/etc

Study matter and its motion through spacetime...

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Spacetime curvature

#1  Postby Xaihe » Feb 27, 2011 1:01 am

Since gravity is stronger at the surface of a planet than at the center of a planet, does that also mean spacetime is curved less at the center of the planet, or just that the gravitational effect is canceled? In the same effect, does time go slower or faster at the center of the earth than at the earth's surface? If spacetime is curved less due to this cancellation, what does this tell us about the goings on inside black holes?

In the same spirit, what happens to the shape and position of the event horizons of two black holes as they approach each other? I'd imagine an extending of the event horizons outward and a decrease of the event horizon between the two masses.

My own understanding of GR is too limited to figure this out, nor could I find the answer elsewhere. I suspect I'm employing some kind of naive view of spacetime curvature here.
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Re: Spacetime curvature

#2  Postby CdesignProponentsist » Feb 27, 2011 1:55 am

Gravity can cancel out curvature of spacetime. So I would imagine that time would run at the same pace as in zero G. Also I believe you are correct about a bridging effect with colliding black holes where the two event horizons merge before the Schwarzschild radius' meet.

Edit: Just notice what you said there about merging black holes.

There would be an extension of the event horizon, not a decrease. Like a saddle, photons between the two black holes would be in a gravitational equilibrium between the two black holes but not with any other direction. So it would be in the event horizon of both, like a sitting in a saddle. Flat space between the holes but highly curved in any other direction.
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Re: Spacetime curvature

#3  Postby james1v » Feb 27, 2011 2:31 am

The question requires a book length reply, but the Samurai did well! :)
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Re: Spacetime curvature

#4  Postby Xaihe » Feb 27, 2011 3:36 am

Thanks, I appreciate it. :)
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Re: Spacetime curvature

#5  Postby twistor59 » Feb 27, 2011 12:49 pm

Xaihe wrote:Since gravity is stronger at the surface of a planet than at the center of a planet, does that also mean spacetime is curved less at the center of the planet, or just that the gravitational effect is canceled? In the same effect, does time go slower or faster at the center of the earth than at the earth's surface? If spacetime is curved less due to this cancellation, what does this tell us about the goings on inside black holes?


Imagine you're in space and you watch a clock drifting towards the earth. You would see it ticking slower and slower as it approached the earth. If it carried onwards, sinking into the earth, then it would continue to tick slower and slower until it reached the centre. One way to think of it is that effectively it's sinking deeper and deeper into the potential well, and photons would have to work harder and harder to climb out of it, hence get more redshifted. The ticks of the clock are equivalent to the cycles of the photons.

The strength of the gravity is more related to the gradient of this potential, whereas for comparing clocks we want to compare the values of the potential at two separate points. So the local strength of the gravitational force is not relevant.

To do this quantitatively, you'd have to look at the Schwarzschild Interior solution and look at the behaviour of g00


Xaihe wrote:
In the same spirit, what happens to the shape and position of the event horizons of two black holes as they approach each other? I'd imagine an extending of the event horizons outward and a decrease of the event horizon between the two masses.

My own understanding of GR is too limited to figure this out, nor could I find the answer elsewhere. I suspect I'm employing some kind of naive view of spacetime curvature here.


If you google something like "merging black holes" you should find loads of computer simulations and animations of this.
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Re: Spacetime curvature

#6  Postby Xaihe » Feb 27, 2011 3:20 pm

CdesignProponentsist's answer seems to be in conflict with your answer. Is it too forward to ask you two to fight it out? :grin:

twistor59 wrote:
Xaihe wrote:Since gravity is stronger at the surface of a planet than at the center of a planet, does that also mean spacetime is curved less at the center of the planet, or just that the gravitational effect is canceled? In the same effect, does time go slower or faster at the center of the earth than at the earth's surface? If spacetime is curved less due to this cancellation, what does this tell us about the goings on inside black holes?


Imagine you're in space and you watch a clock drifting towards the earth. You would see it ticking slower and slower as it approached the earth. If it carried onwards, sinking into the earth, then it would continue to tick slower and slower until it reached the centre. One way to think of it is that effectively it's sinking deeper and deeper into the potential well, and photons would have to work harder and harder to climb out of it, hence get more redshifted. The ticks of the clock are equivalent to the cycles of the photons.

The strength of the gravity is more related to the gradient of this potential, whereas for comparing clocks we want to compare the values of the potential at two separate points. So the local strength of the gravitational force is not relevant.


It makes sense, but isn't the redshift only dependent on the gradient of the potential? For instance, in the case of 2 equal mass bodies, one smaller than its Schwarzschild radius, the other larger. Would then the redshift be different for both objects, while they have the same total potential?

To do this quantitatively, you'd have to look at the Schwarzschild Interior solution and look at the behaviour of g00


:lol: I may want to get into the math's of this stuff...At some point.
Xaihe wrote:
In the same spirit, what happens to the shape and position of the event horizons of two black holes as they approach each other? I'd imagine an extending of the event horizons outward and a decrease of the event horizon between the two masses.

My own understanding of GR is too limited to figure this out, nor could I find the answer elsewhere. I suspect I'm employing some kind of naive view of spacetime curvature here.


If you google something like "merging black holes" you should find loads of computer simulations and animations of this.


I watched one simulation a while ago, but I see there's a lot more available nowadays. :cheers:

p.s. I always thought of gravity as being nothing other than the curvature of spacetime (due to the presence of mass). Gravity being just our interpretation of the curvature, not an effect of it.
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Re: Spacetime curvature

#7  Postby twistor59 » Feb 27, 2011 4:32 pm

Xaihe wrote:CdesignProponentsist's answer seems to be in conflict with your answer. Is it too forward to ask you two to fight it out? :grin:


Well, it's 25 years since I did any relativity, but I'm pretty certain that at the centre of a massive body, clocks would be running slower compared to ones at a distance from the body.

Xaihe wrote:
twistor59 wrote:
Xaihe wrote:Since gravity is stronger at the surface of a planet than at the center of a planet, does that also mean spacetime is curved less at the center of the planet, or just that the gravitational effect is canceled? In the same effect, does time go slower or faster at the center of the earth than at the earth's surface? If spacetime is curved less due to this cancellation, what does this tell us about the goings on inside black holes?


Imagine you're in space and you watch a clock drifting towards the earth. You would see it ticking slower and slower as it approached the earth. If it carried onwards, sinking into the earth, then it would continue to tick slower and slower until it reached the centre. One way to think of it is that effectively it's sinking deeper and deeper into the potential well, and photons would have to work harder and harder to climb out of it, hence get more redshifted. The ticks of the clock are equivalent to the cycles of the photons.

The strength of the gravity is more related to the gradient of this potential, whereas for comparing clocks we want to compare the values of the potential at two separate points. So the local strength of the gravitational force is not relevant.



Xaihe wrote:
It makes sense, but isn't the redshift only dependent on the gradient of the potential? For instance, in the case of 2 equal mass bodies, one smaller than its Schwarzschild radius, the other larger. Would then the redshift be different for both objects, while they have the same total potential?


They might have the same potential in a Newtonian sense, but the potential we're interested in here is g00, the time component of the metric. The redshift, or the slowness of the ticking relative to clocks in the asymptotic region, becomes infinite at the horizon. That sure ain't the same in the 2 cases !
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Re: Spacetime curvature

#8  Postby hackenslash » Feb 27, 2011 4:46 pm

twistor59 wrote:Well, it's 25 years since I did any relativity, but I'm pretty certain that at the centre of a massive body, clocks would be running slower compared to ones at a distance from the body.


Indeed, that would be my understanding as well. The effect that a body would feel would be equivalent to having been cancelled out, because the gravitational attraction is coming from all directions rather than being unidirectional, but the immersion in a gravitational field would be precisely the same as it was at the surface. All that has changed is the direction of mass from the perspective of the observer, because now the observer is surrounded by it as opposed to being outside it. The relativistic time dilation would be the same though, I think.
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Re: Spacetime curvature

#9  Postby CdesignProponentsist » Feb 28, 2011 4:58 am

hackenslash wrote:
twistor59 wrote:Well, it's 25 years since I did any relativity, but I'm pretty certain that at the centre of a massive body, clocks would be running slower compared to ones at a distance from the body.


Indeed, that would be my understanding as well. The effect that a body would feel would be equivalent to having been cancelled out, because the gravitational attraction is coming from all directions rather than being unidirectional, but the immersion in a gravitational field would be precisely the same as it was at the surface. All that has changed is the direction of mass from the perspective of the observer, because now the observer is surrounded by it as opposed to being outside it. The relativistic time dilation would be the same though, I think.


I disagree. I am pretty sure it is all dependent on what frame you are in. Let me explain.

Just as in the twin paradox, it is always the one who matches the acceleration of the other that travels a shorter path through time. Imagine a planet that moves close to the speed of light passes the earth. Both planets have sets of twins, all four are the same age at the time of passing. Let's call them twins 1a, 1b and 2a and 2 b. One twin-a from each planet decides to visit other twin-b on the opposite planet. Both will get there and find the opposite twin-b older than him and will return home to find his own twin-b even more older than him.

Why is this? You would think since from the perspective of twin 1a, twin 2b is traveling close to the speed of light and see his clock running slow. But it isn't. He is only seeing his time skewed from his frame of reference, just as twin 2a sees twin 1b's time skewed from his perspective.

Only when one accelerate to the other's velocity does the time difference resolve, and it can be resolve either way depending on who does the accelerating.

I am almost certain that this is the very same with gravitational fields. In order to reach the center of the earth you MUST experience an deceleration in order to reach a stable position in the center of a gravitational field, otherwise you will just yo-yo back and forth past the center of gravity.

The opposite is also true. You must accelerate in order to escape the gravity well.

Now take Rockman and Starman. Rockman is sitting in the center of a gravity well, he is in an inertial frame of reference because he is not experiencing any acceleration (gravity). Starman is in orbit and can say the same thing. Both are in inertial frames, neither is accelerating.

This next part I am having difficulty with, which is what does each see the other's clock doing. Do they both see the other's clock moving slowly? I'm guessing yes, but I would have to do some mental gymnastics to figure out why, but the fact that Rockman isn't in an accelerated state has got to do something with the way perceives Starman's clock as apposed to someone standing on the surface feeling the full force of gravity.

I believe if Starman does the acceleration necessary to visit Rockman, Rockman will be older than expected. And if Rockman does the acceleration nessisary to visit Starman, Starman will be older than expected.

I'll bet my lunch money on this.
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Re: Spacetime curvature

#10  Postby CdesignProponentsist » Feb 28, 2011 8:05 am

Now, after thinking this over, I'm pretty sure I'm wrong on this (rescinds bet). I think I wrongly concluded that a falling or orbiting frame is in an inertial frame. This must be wrong as you can determine that you are falling from tidal forces or the acceleration between bodies dependent on their location within your space. objects at the center of the gravity well would also experience a tidal force in all directions.

I am right about the twins paradox though.
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Re: Spacetime curvature

#11  Postby hackenslash » Feb 28, 2011 8:15 am

:thumbup:
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Re: Spacetime curvature

#12  Postby twistor59 » Feb 28, 2011 9:20 am

You can apply the twin paradox type stuff when there are accelerations present, no problem, but you can't apply it when spacetime is curved as it is in this example. When this happens, you can't escape using GR. Fortunately for this type of question, we just need the metric, and don't need to compute the connection or curvature.


OK, let's have a crack at it:


I don't bring my old relativity books to work, so I had to resort to Google: Post number 5 in this thread gives the Schwarzschild interior metric (in units where c = G = 1):

ds2 = (3/2sqrt(1-2M/R)-1/2sqrt(1-2Mr2/R3)2)dt2 - (1/(1-2Mr2/R3)2))dr2 - r22 - r2sin2θdφ2

Consider two points - one at the centre of the earth (r=0), and one at the surface θ = π/2 (to make the maths easy). dr = 0, dθ=0, dφ = 0.

So ds12 = (3/2sqrt(1-2M/R)-1/2)2dt2
and ds22 = (sqrt(1-2M/R)-1/2)2dt2

where ds1 is the interval at a point at the earths centre, and ds2 is the interval at a point at the earths surface.

Putting the c's and G's back I get, where 𝜏1 and 𝜏2 are the proper times at the centre and surface respectively

d𝜏1 / d𝜏2 = (3/2sqrt(1-2GM/c2R)-1/2) / (sqrt(1-2GM/c2R)-1/2)

Shoving in the G, M, R numbers from my Kepler thread I get

d𝜏1 / d𝜏2 = 0.999999999653

So the centre clock is ticking slower (but not by much !!!)
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Re: Spacetime curvature

#13  Postby CdesignProponentsist » Feb 28, 2011 11:04 pm

That has as much effect on me as voodoo, but I'll take your word for it, unless you can put that in mathematically retarded terms for me.
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Re: Spacetime curvature

#14  Postby twistor59 » Mar 01, 2011 8:03 am

OK I'll have a go at translating the maths to English:

You probably know a lot of this already, but I need to mention it for completeness.

Much of special relativity can be encapsulated in the statement that the "Minkowski metric"

ds2 = c2dt2 - dx2- dy2- dz2 is "invariant under Lorentz transformations".

ds2 is the spacetime distance between two points (t, x, y, z) and (t+dt, x+dx, y+dy, z+dz).

"Invariant under Lorentz transformations" means that if I transform to a new frame with coordinates t', x', y', z' using a Lorentz transformation, I get the same number for ds2

In general relativity, you have a similar concept except this time, the metric components making up ds2 are functions of the coordinates, and are not just +1, -1, -1, -1 like in the Minkowski case. Also, you have to consider general coordinate transformations, not just Lorentz transformations.

The metric appropriate for describing the gravitational field from a spherically symmetric non rotating mass is called the Schwarzschild solution. This describes the metric outside the object. The field inside the object is given by the Schwarzschild Interior Solution which is that complicated expression for ds2 I gave in the above post. That expression applies when the density of the object is constant. The metric in that post was given in spherical polar form for the spatial coordinates.

Now if you choose points where the two objects whose times we want to compare are at rest, you can set dr=0, dθ=0, dφ=0, so you're left with

ds2 = f(r)c2dt2

Now we can also write ds2 = c2d𝜏2 where 𝜏 is the proper time. Note t is the "coordinate time" - t=const is a spacelike slice of all of spacetime, i.e. it's a picture of space at a given time. So I can write

c2d𝜏12 = f(r1)c2dt2 and

c2d𝜏22 = f(r2)c2dt2

where 𝜏1 is the proper time measured by a clock at r1, and 𝜏2 is the proper time measured by a clock at r2. By dividing these equations the result follows.
Last edited by twistor59 on Mar 01, 2011 2:59 pm, edited 1 time in total.
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Re: Spacetime curvature

#15  Postby hackenslash » Mar 01, 2011 1:45 pm

It might be worth explaining why '-'dx2, etc, in other words explaining why the minus solution is employed rather than the plus. That makes the math more understandable, I reckon. It's how I got how Minkowski spacetime works.
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Re: Spacetime curvature

#16  Postby twistor59 » Mar 01, 2011 2:04 pm

hackenslash wrote:It might be worth explaining why '-'dx2, etc, in other words explaining why the minus solution is employed rather than the plus. That makes the math more understandable, I reckon. It's how I got how Minkowski spacetime works.


Since all my d(spatial coordinate) 's vanished, I didn't bother !
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Re: Spacetime curvature

#17  Postby hackenslash » Mar 01, 2011 2:22 pm

Actually, thinking about it, it isn't really relevant to the question at hand, although it could possibly firm up how it works. I'd be happy to explain it if any thinks it worth while. It does have very profound implications for causality, though.
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Re: Spacetime curvature

#18  Postby twistor59 » Mar 01, 2011 2:58 pm

hackenslash wrote:Actually, thinking about it, it isn't really relevant to the question at hand, although it could possibly firm up how it works. I'd be happy to explain it if any thinks it worth while. It does have very profound implications for causality, though.


yeah go ahead, it's probably worthwhile. Maybe this will turn into a relativity expository thread....
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Re: Spacetime curvature

#19  Postby hackenslash » Mar 01, 2011 3:34 pm

OK, I just found this video, which doesn't really go into explaining the minus sign, but it does say what it implies:

[youtube]http://www.youtube.com/watch?v=LfCv9GLwvYE[/youtube]

I'll borrow directly from Cox on this, because the example he gave is quite elegant. Here's how it works:

If we take two events in spacetime, say, getting out of bed and eating breakfast, we can plot them on a spacetime graph. For simplicity, we'll say that the kitchen is 10 metres from your bed and you ate breakfast 1 hour after getting out of bed.

Using the simpler Pythagorean formulation of Twistor's above calculation, we can give the equation for the distance in spacetime as s2=(ct)2+x2, where c is our exchange rate, t is the distance through time and x is the distance through space.

We'll call 'getting out of bed' 'O' and 'eating breakfast' 'A', and we plot them on our spacetime graph thus:

Image

Here. you'll see that s is the distance in spacetime between the two events. On the graph, we have plotted a circle, the radius of which represents all the points in spacetime that are the same distance s from event O. There are also plotted two events, A' and A", which are also the same distance in spacetime from event O. For A', nothing very interesting has happened, but if you look at event A", you can see that something very strange has happened, namely that it occurs in the past of O, which means that yoou ate your breakfast before you got out of bed! Causality is violated.

However, if we use the minus sign version s2=(ct)2-x2, as stated in the video by Cox above, we get a slightly different picture of spacetime, namely a non-Euclidean, hyperbolic spacetime.

Image

The hyperbola still represents all the points that lie the same distance s in spacetime from event O, but you can now see that they all lie within the future light cone of O, thus protecting causality.

So, that's how the minus sign protects the past from the future.

If anybody wants to see this explained in greater detail, I recommend Why Doe E=mc2, by Brian Cox and Jeff Forshaw, from which this example is borrowed, as are the diagrams.

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Re: Spacetime curvature

#20  Postby twistor59 » Mar 01, 2011 4:25 pm

hackenslash wrote:
If anybody wants to see this explained in greater detail, I recommend Why Doe E=mc2, by Brian Cox and Jeff Forshaw, from which this example is borrowed, as are the diagrams.

Edit: Tags


Nice explanation :thumbup:
I also (after your suggestion) read Cox & Forshaw's book, and can heartily recommend it to anybody who's interested in relativity.
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