Curvature of Spacetime on Earth

In summary: It's not really clear to me how exactly this would cause anything to go slower, but it does..In summary, spacetime between the Earth and the sun is warped, mainly due to the sun's mass. This explanation replaces the idea of a gravitational pulling force. Because this is all happening in outer space, I can easily accept it. However, I would also like to understand how it works here on earth. The key is that it's spacetime that is curved, not just space.
  • #1
Johninch
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I am trying to improve my understand of the basic elements of GR.

I have read that the Earth orbits the sun because spacetime between the Earth and the sun is warped, mainly due to the sun’s mass.

The Earth follows a geodesic, which is the equivalent of a straight line in curved space. This explanation replaces the idea of a gravitational pulling force.

Because this is all happening in outer space, I can easily accept it. But I would also like to understand how it works here on earth.

I suppose that all dimensions are warped, including an increasing compression of space in the direction of the center of the earth.

First question: If I let go of a rock at the North Pole, why does it fall when there is no rotation causing movement of the rock? Is the distortion of spacetime a kind of slippery slope which the rock has to fall down without being pushed? When we talk GR, do we continue to observe potential energy and its conversion to kinetic energy?

Second question: If I drop a rock 5 meters it accelerates and takes about 1 second to reach the ground. Does the final meter contain less space than the first meter, or is there less time in the final meter, or what?

Third question: Considering that we see the effect of gravity on the falling rock so clearly, why can’t we see any distortion or compression of the space through which the rock is falling? I don’t mean the curved flight path due to the rotation of the earth, I mean the distortion of space through which the rock is travelling.

Fourth question: If the final meter contains less space than the first meter, is this the reason why atmospheric density is greater nearer to the earth’s surface? In other words, does the distortion of space explain the increase in atmospheric density? Likewise, does the warping of space due to the earth’s mass continue below the surface and cause the increasing density of the Earth towards its center?

Fifth question: The other day I saw a TV program where it was stated that the Apollo program used Newtonian physics alone. Is this true? Or did they mean that for journeys to the moon, the errors in Newtonian physics are immaterial?
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  • #2
Johninch said:
First question: If I let go of a rock at the North Pole, why does it fall when there is no rotation causing movement of the rock?

Check out the frist picture here:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

Johninch said:
Second question: If I drop a rock 5 meters it accelerates and takes about 1 second to reach the ground. Does the final meter contain less space than the first meter, or is there less time in the final meter, or what?
No, a meter is a meter. But if you use circumferences divided by 2*pi to make the "meter" marks on your radial ruler, then the lowest "meter" will contain more space. Comapre ds to dr in the frist picture here:
http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Black_Holes.htm
Johninch said:
Third question: Considering that we see the effect of gravity on the falling rock so clearly, why can’t we see any distortion or compression of the space through which the rock is falling? I don’t mean the curved flight path due to the rotation of the earth, I mean the distortion of space through which the rock is travelling.
Distortion of space has little effect on a rock. It is mostly distrotion of time. See the first link above.

Johninch said:
Fourth question: If the final meter contains less space than the first meter, is this the reason why atmospheric density is greater nearer to the earth’s surface?
No

Johninch said:
Fifth question: The other day I saw a TV program where it was stated that the Apollo program used Newtonian physics alone. Is this true? Or did they mean that for journeys to the moon, the errors in Newtonian physics are immaterial?
Yes
 
  • #3
Johninch said:
I have read that the Earth orbits the sun because spacetime between the Earth and the sun is warped, mainly due to the sun’s mass.

The key is that it's spacetime that is curved, not just space. It's important to keep this in mind; I'll refer to it below.

Johninch said:
The Earth follows a geodesic, which is the equivalent of a straight line in curved space. This explanation replaces the idea of a gravitational pulling force.

Yes, but again, the geodesic is a straight line in curved spacetime. In all the cases you talk about, "time curvature" accounts for basically all of the observed effect. The reason for this is that the speed of light is so large in ordinary units: 1 second of time is equivalent to 300 million meters of space. So for ordinary problems where all the objects are moving much slower than light, space curvature is negligible compared to "time curvature".

Here's what "time curvature" means: the "rate of time flow" gets slower as you get closer to a gravitating body like the Sun or the Earth. This is called gravitational time dilation, and it has been experimentally confirmed. What we normally think of as the "gravitational force" is due to the *gradient* of the rate of time flow; the force "points" in the direction of slower time flow, and its magnitude is proportional to how fast the rate changes. (Technical point: the "curvature" is actually the *second* derivative of the rate of time flow, not the first; the first derivative is called the "connection coefficient". But that doesn't change the key point, which is that all of this has to do with *time*; there's no space curvature involved at all.)

Why does the rate of time flow matter? Because the "straight line" path in spacetime is the path of maximal *proper time*--i.e., maximum time experienced by an observer traveling along the path, compared to other nearby paths that have the same endpoints. I won't go into the details of this right now (mainly since it would take quite a bit more exposition), but this is the basic thing to keep in mind.

Johninch said:
I suppose that all dimensions are warped, including an increasing compression of space in the direction of the center of the earth.

"Compression of space" is not really a good description; see further comments below. But bear in mind, as above, that this effect is negligible compared to "time curvature".

Johninch said:
If I let go of a rock at the North Pole, why does it fall when there is no rotation causing movement of the rock?

Rotation has nothing to do with it anywhere on the surface of the Earth; rotation would cause a *sideways* motion, if it caused any relative motion at all. It has no effect on vertical motion.

To answer why the rock falls when it's released, first consider: why does the rock *not* fall *before* it's released? The answer is that there is a force pushing upwards on the rock: to be concrete, let's suppose that the force is exerted by a platform that is slid out from under the rock. Before the platform is slid out, it pushes up on the rock; i.e,. the rock has a force on it and is not in free fall. Another way of saying this is that the rock has weight.

After the platform is slid out, there is no force on the rock and it is in free fall--weightless. And the fact that it took an *upward* force to keep the rock at the same height shows that the free-fall paths in the rock's vicinity--the weightless paths--are downwards.

Johninch said:
Is the distortion of spacetime a kind of slippery slope which the rock has to fall down without being pushed?

Some people think of it this way, but the "slope" analogy is limited because it leads to the question of why the rock goes *down* the "slope". Basically you're appealing to an intuitive notion of "gravity" to explain "gravity", which isn't very satisfying logically. The key is that the "straight line" paths in spacetime correspond to *free fall*: inertial motion; weightless motion.

Johninch said:
When we talk GR, do we continue to observe potential energy and its conversion to kinetic energy?

Only in certain situations. Single gravitating bodies like the Earth or the Sun are such situations: for many purposes, like analyzing the motion of falling rocks or orbiting planets, we can idealize them as static, non-rotating, spherically symmetric masses, and since the system is static (unchanging in time), we can define a useful notion of "potential energy" (it basically corresponds to the "rate of time flow" I talked about above, though there are some further technicalities), and we can show that for freely falling objects, potential and kinetic energy get inter-converted to keep total energy constant, just as in Newtonian mechanics.

But as soon as you try to deal with non-static systems, all that breaks down; and there are plenty of GR problems that deal with non-static systems.

Johninch said:
If I drop a rock 5 meters it accelerates and takes about 1 second to reach the ground. Does the final meter contain less space than the first meter

A "meter" is a *definition* of a certain spatial length; it's always the same no matter where you are. As I said above, for this problem "space curvature" is negligible, but I'll digress briefly to explain how it works.

For the idealized case of a static, spherically symmetric massive body, the only "space curvature" is in the radial direction; so we can view "space" as a family of nested 2-spheres, each with a slightly larger area (the degenerate "2-sphere" at the center has zero area, and the area increases monotonically from there). "Space curvature" then manifests itself this way: suppose I take two adjacent 2-spheres, with areas A and A + dA, and I measure the radial distance between them using a meter stick. Euclidean geometry would lead me to expect that the distance between the two 2-spheres, as a function of A and dA, will be given by the standard Euclidean formulas; and in flat spacetime that is what we would find. But in the spacetime around a gravitating body, we find that the distance between the two 2-spheres is *larger* than the Euclidean formula would predict. That doesn't mean the length of the meter sticks is changed: it means that it takes *more meter sticks* to cover the distance between the 2-spheres than the Euclidean formula would predict.

Johninch said:
or is there less time in the final meter, or what?

"Less time" isn't a meaningful statement as it stands; less time relative to which observer? The falling rock doesn't see "less time"; a given object always experiences its own time to "flow" at the same rate. It's only when looking at *other* objects that differences in "rate of time flow" can be observed.

As for why the rock accelerates, acceleration is what is caused by the gradient of the "rate of time flow", as I discussed above. If an object already has some downward speed, acceleration adds to it; that's why the rock covers progressively more vertical distance in equal increments of time.

Johninch said:
Considering that we see the effect of gravity on the falling rock so clearly, why can’t we see any distortion or compression of the space through which the rock is falling?

Because space curvature is negligible for this case; the observed effect is all due to time curvature. See above.

Johninch said:
If the final meter contains less space than the first meter

It doesn't, but the rest of your question is still worth a response; see below.

Johninch said:
does the distortion of space explain the increase in atmospheric density?

No. The increase in density is explained the same way you learned in school: the weight of air above a given altitude compresses the air at that altitude. The only difference is the underlying explanation of where the weight comes from: in GR, it comes from the upward force exerted by the air underneath (and ultimately by the surface of the Earth), and that's all; there is no "force of gravity" involved.

Johninch said:
does the warping of space due to the earth’s mass continue below the surface

Yes, but it's too small to have any effect on the density.

Johninch said:
and cause the increasing density of the Earth towards its center?

No, that also works the same as you learned in school: the weight of the Earth above a given depth compresses the material at that depth.

Johninch said:
did they mean that for journeys to the moon, the errors in Newtonian physics are immaterial?

Yes, that's what they meant. The Apollo computers couldn't do GR calculations anyway, so it was a good thing they didn't need to.
 
  • #5
Thanks Peter and A.T. for your exellent answers. I now see that I have to recognize the much higher importance of the time dimension than what I had assumed.
PeterDonis said:
the weight of air above a given altitude compresses the air at that altitude. The only difference is the underlying explanation of where the weight comes from: in GR, it comes from the upward force exerted by the air underneath (and ultimately by the surface of the Earth).

You refer to an "upward force exerted by the air underneath" and you referred previously to the push of the platform supporting the rock. What upward force and push are you talking about?

Would I be correct to interpret it like this: the air molecules and rock are in free fall and want to reach the center of the earth, but other air molecules and rocks are in the way, so they get closer together, which gives the higher density.

But I don't like this interpretation, because I don't see why the molecules should get closer together. What is forcing them to close the gap in spite of their mutual electromagnetic repulsion?

Yes, that's what they meant. The Apollo computers couldn't do GR calculations anyway, so it was a good thing they didn't need to.

What length of spaceflight now uses GR calculations? For example to get to Titan?

Is it easier to use Newtonian calculations and correct the path en route? This would need more fuel, but perhaps you have to correct the path anyway for other reasons?

If you want to intercept and land on a comet, is this a critical case where you have to use GR because otherwise big corrections are more likely?

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  • #6
PeterDonis said:
The key is that it's spacetime that is curved, not just space. It's important to keep this in mind; I'll refer to it below.

Yes, but again, the geodesic is a straight line in curved spacetime. In all the cases you talk about, "time curvature" accounts for basically all of the observed effect. The reason for this is that the speed of light is so large in ordinary units: 1 second of time is equivalent to 300 million meters of space. So for ordinary problems where all the objects are moving much slower than light, space curvature is negligible compared to "time curvature".

Here's what "time curvature" means: the "rate of time flow" gets slower as you get closer to a gravitating body like the Sun or the Earth. This is called gravitational time dilation, and it has been experimentally confirmed. What we normally think of as the "gravitational force" is due to the *gradient* of the rate of time flow; the force "points" in the direction of slower time flow, and its magnitude is proportional to how fast the rate changes. (Technical point: the "curvature" is actually the *second* derivative of the rate of time flow, not the first; the first derivative is called the "connection coefficient". But that doesn't change the key point, which is that all of this has to do with *time*; there's no space curvature involved at all.)

Why does the rate of time flow matter? Because the "straight line" path in spacetime is the path of maximal *proper time*--i.e., maximum time experienced by an observer traveling along the path, compared to other nearby paths that have the same endpoints. I won't go into the details of this right now (mainly since it would take quite a bit more exposition), but this is the basic thing to keep in mind.

PeterDonis said:
As for why the rock accelerates, acceleration is what is caused by the gradient of the "rate of time flow", as I discussed above. If an object already has some downward speed, acceleration adds to it; that's why the rock covers progressively more vertical distance in equal increments of time.

PeterDonis said:
Because space curvature is negligible for this case; the observed effect is all due to time curvature. See above.

I had the same question posted on the SR/GR forum before, "Why we fall, according to GR?", and a lot of people had explained, just like you said above, it's because of time and its curvature. And I still don't fully understand it. It helps for me when I can picture things, but with this, it's hard to visualize and imagine.

They said that in Relativity, an object is never really at rest and there is this thing called "Four Vector", that besides spatial motion, we also always move forward through the time dimension at speed C, at the rate of 1s/s, just like you stated above. It was also explained that when we move through space, we take away some of the speed/velocity from our movement through the time dimension/direction, and that is why time dilation occurs, as according to SR, especially when the velocity gets near or at the speed of light. I would like to confirm, is this correct? Because it still has not sinked in my head and I want fully understand it and clarify it.

But same question I want to ask again and clarify: objects fall, just like the rock in question, all because of the time curvature? So basically the rock may not be in "spatial" motion, but since it still always moves through time, it follows that geodesic in that spacetime curvature, and therefore end up falling on the ground? Is this correct?

PeterDonis said:
Only in certain situations. Single gravitating bodies like the Earth or the Sun are such situations: for many purposes, like analyzing the motion of falling rocks or orbiting planets, we can idealize them as static, non-rotating, spherically symmetric masses, and since the system is static (unchanging in time), we can define a useful notion of "potential energy" (it basically corresponds to the "rate of time flow" I talked about above, though there are some further technicalities), and we can show that for freely falling objects, potential and kinetic energy get inter-converted to keep total energy constant, just as in Newtonian mechanics.

So in GR, there are no such things as Potential and Kinetic energies?
 
  • #7
kweba said:
So basically the rock may not be in "spatial" motion, but since it still always moves through time, it follows that geodesic in that spacetime curvature, and therefore end up falling on the ground? Is this correct?
Yes. Initially it advances only along the time dimension. It advances on a locally straight path (geodesic), but since space time is distorted it ends up deviating from the purely temporal path.

Did you check the links I gave you in the previous thread? They show it quite nicely:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html
http://www.relativitet.se/spacetime1.html
 
  • #8
  • #9
Johninch said:
But I don't like this interpretation, because I don't see why the molecules should get closer together. What is forcing them to close the gap in spite of their mutual electromagnetic repulsion?
It is the electromagnetic repulsion of the matter below them that forces them closer together.

Gravity alone would actually make them move apart (see tidal forces). At least if we ignore their own mass, and consider only Earth's gravity. Otherwise a gas cloud can also create gravity, that makes the contents move together. But that is not the main effect in the Earth's atmosphere.
 
  • #10
A.T. said:
It is the electromagnetic repulsion of the matter below them that forces them closer together.

Sorry, I don't understand your use of language. How can repulsion bring things together?

You appear to be saying the same thing as PeterDonis without explaining it. He wrote that the weight of the air comes from the "upward force exerted by the air underneath". I want to know what this force is. Electromagnetic repulsion will cause the atoms and molecules to stay apart. So if density is increasing with depth, I want to know what is overcoming this repulsion.

Gravity alone would actually make them move apart (see tidal forces). At least if we ignore their own mass, and consider only Earth's gravity. Otherwise a gas cloud can also create gravity, that makes the contents move together. But that is not the main effect in the Earth's atmosphere.

Lost me again. Since when did gravity make things move apart? You then write "a gas cloud can also create gravity, that makes the contents move together." Well, I can agree with that.

Can you explain these points without the apparent contradictions?

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  • #11
Johninch said:
Lost me again. Since when did gravity make things move apart? You then write "a gas cloud can also create gravity, that makes the contents move together." Well, I can agree with that.

Consider tidal forces. If I have two objects near each other, but not at exactly the same point, the gravitational force they feel from a large third object won't be exactly the same - the distance and the direction from each object to the large third object will be slightly different, so the strength and direction of the forces on the two objects will be slightly different. Depending on how I arrange things, the effect can be to cause the two objects to separate from one another or to draw nearer to each other.
 
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  • #12
Johninch said:
How can repulsion bring things together? Since when did gravity make things move apart?

You stand on the ground. An apple is released 10m above. What brings your head and the apple together? It can't be gravity, because it pulls you slightly more than the apple, so gravity actually tries to pull you and the apple apart. It is the repulsion between ground and your feet that forces you and the apple together.
 
  • #13
Johninch said:
You appear to be saying the same thing as PeterDonis without explaining it. He wrote that the weight of the air comes from the "upward force exerted by the air underneath". I want to know what this force is. Electromagnetic repulsion will cause the atoms and molecules to stay apart. So if density is increasing with depth, I want to know what is overcoming this repulsion.

Consider two air molecules, one at the surface of the Earth and another high up. Both are in free fall (at least until they collide with another molecule) so they are both moving on geodesics, straight paths through space-time. But, because space-time is curved, these geodesics are drawing closer to one another (think about how lines of latitude on the Earth's surface are geodesics, but two people traveling along two adjacent lines of latitude will find themselves moving closer to one another and colliding when they meet at the North pole).

So the two molecules are being pushed towards one another just from following their natural path through space-time. The electromagnetic repulsion between molecules wants to keep them apart; it's the interplay of these forces that gives us an atmosphere with higher density and pressure near the surface of the earth, and lower further away where the curvature is forcing molecules together less strongly.
 
  • #14
Nugatory said:
Consider two air molecules, one at the surface of the Earth and another high up. Both are in free fall (at least until they collide with another molecule) so they are both moving on geodesics, straight paths through space-time. But, because space-time is curved, these geodesics are drawing closer to one another
No, if both are free falling outside the earth, one above each-other, the distance between them will increase. The geodesics move apart, not closer.
Nugatory said:
(think about how lines of latitude on the Earth's surface are geodesics, but two people traveling along two adjacent lines of latitude will find themselves moving closer to one another and colliding when they meet at the North pole).
That is not a good analogy for two object stacked vertically outside of the earth. A sphere has positive curvature, so geodesics converge. This is what happens inside uniform mass spehere (at some depth for the non-uniform Earth). But outside the of the mass geodesics can converge or diverge, depending on the arrangement. Vertically arranged particles diverge.
 
  • #15
A.T. said:
You stand on the ground. An apple is released 10m above. What brings your head and the apple together? It can't be gravity, because it pulls you slightly more than the apple, so gravity actually tries to pull you and the apple apart. It is the repulsion between ground and your feet that forces you and the apple together.

Got it! thanks. Could you just continue a bit and describe how it works when gravity is decreasing, such as in the Earth's core. Gravity will try to pull the apple towards me, but electromagnetic repulsion will resist the apple. So does this mean that the Earth's density progressively reduces in the core as you get nearer to the Earth's center? Could there be a hole in the middle?

By the way, I am still talking GR, not Newton.
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  • #16
Johninch said:
So does this mean that the Earth's density progressively reduces in the core as you get nearer to the Earth's center?
No, because of the weight of the mass column above. The electromagnetic repulsion is transmitted through the column and the pressure is maximal in the center.
Johninch said:
By the way, I am still talking GR, not Newton.
In GR the upwards electromagnetic repulsion is not working against weight (force of gravity), but against the inertia (mass) that needs to be overcome, in order to accelerate the mass away from the free falling geodesic path.
 
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  • #17
A.T. said:
No, if both are free falling outside the earth, one above each-other, the distance between them will increase. The geodesics move apart, not closer.

Ah - right - good point. I should have started with the non-geodesic non-freefall worldline of a point on the surface of the Earth - that's what converges on and intersects with the worldline of a free-falling particle.
 
  • #18
Johninch said:
You refer to an "upward force exerted by the air underneath" and you referred previously to the push of the platform supporting the rock. What upward force and push are you talking about?

The rock feels a force pushing on it from the platform; that's what keeps it from freely falling. This force is equal to the rock's weight, and it pushes upward on the rock.

Similarly, if you consider a small volume of air in the middle of the atmosphere, it experiences a net force in the upward direction: the air above it pushes downward, and the air below it pushes upward. Since there is a pressure gradient, the force from below is greater than the force from above, so the net force is upward.

Johninch said:
Would I be correct to interpret it like this: the air molecules and rock are in free fall and want to reach the center of the earth, but other air molecules and rocks are in the way, so they get closer together, which gives the higher density.

Kinda sorta, but this isn't all there is to the picture. See below.

Johninch said:
But I don't like this interpretation, because I don't see why the molecules should get closer together. What is forcing them to close the gap in spite of their mutual electromagnetic repulsion?

If you want to take your modeling to this level of detail, you're no longer looking at macroscopic concepts like pressure or density. You're looking at individual molecules of air and how they move and collide with other molecules. This makes the analysis a lot more complicated, and I don't have time right now to go to that level of detail. However, I can give at least a quick pointer at the answer: air molecules are essentially in free fall between collisions, so if there is a gradient in gravitational potential, there will similarly be a gradient in the average kinetic energy of the molecules (because they gain kinetic energy as they fall and lose it as they rise).

Johninch said:
What length of spaceflight now uses GR calculations? For example to get to Titan?

I don't think GR is routinely used in any spaceflight calculations. The only kind of flight for which I would expect relativistic effects to be significant would be a close approach to the Sun, but as you note, you could also just rely on course corrections. Normally a certain amount of delta v (i.e., rocket power) is budgeted for course corrections for a mission, so there is a margin for error in the calculations. But it probably isn't very large, so it's possible that GR calculations would be needed for a close solar approach. For any other mission (such as missions to Venus, Mars, and out to the outer planets and beyond), I would expect GR corrections to be negligible.
 
  • #19
Johninch said:
Got it! thanks. Could you just continue a bit and describe how it works when gravity is decreasing, such as in the Earth's core. Gravity will try to pull the apple towards me, but electromagnetic repulsion will resist the apple. So does this mean that the Earth's density progressively reduces in the core as you get nearer to the Earth's center?
The same thing happens when gravity is decreasing. The difference is only that the upward acceleration is not as strong.

However, this does not mean that density or pressure is decreasing. Consider a 1 m³ cube of rock at the Earth's surface. It has a mass of 2500 kg and it is accelerating upwards at 9.8 m/s². The force pushing down on the top is 101 kN, so to get the required upwards acceleration needs a force of 125.5 kN which is a pressure of 125.5 kPa.

Now, consider the 1 m³ cube of rock directly below that. Further, let's consider not just the Earth's actual distribution of density, but a distribution so extreme that the acceleration of the next chunk of rock is only 9.7 m/s². This chunk of rock has the same 2500 kg mass, but the force pushing down is 125.5 kN, so to get the required upwards acceleration needs a force of 149.75 kN which is a pressure of 149.75 kPa.

So, even though the gravitational acceleration is decreasing sharply, the pressure remains increasing.
 
  • #20
DaleSpam said:
The same thing happens when gravity is decreasing. The difference is only that the upward acceleration is not as strong ... So, even though the gravitational acceleration is decreasing sharply, the pressure remains increasing.

Thanks, I’m trying to get my mind round it. I haven’t come across upward acceleration in this context before. I always considered that the force of gravity or warping of space only causes downward acceleration. Of course, if we are talking density, then there must be some upward pressure too, I agree.

Can you explain one other point: when I look at those diagrams which show a massive body like the Earth creating a sink in spacetime, it looks like the distortion is increasing, the nearer you get to the massive body. I assumed that this gradient continues to get steeper inside the body. This must be wrong, because if gravity declines to zero at the center of the earth, the distortion of spacetime can't be increasing, it must be going back to normal.

Is this correct?

Is this also what happens in a black hole, i.e. gravity and the distortion of spacetime both decrease once you go below the event horizon, although pressure increases?

Could you give me a link to a diagram which shows what spacetime does below the surface of a massive body?

.
 
  • #21
Johninch said:
I haven’t come across upward acceleration in this context before. I always considered that the force of gravity or warping of space only causes downward acceleration.

That's because you're using a different definition of "acceleration" than the one DaleSpam and I were using. The downward "acceleration due to gravity" (for example, the acceleration of a falling rock as seen by a person standing on the ground) is called "coordinate acceleration", because it depends on the coordinates you adopt; if you use coordinates in which the rock is at rest, it's you who are accelerating, not the rock.

However, the upward acceleration that I referred to, which you feel as weight, is called "proper acceleration", and is independent of coordinates; it's directly observable as weight (or lack thereof). That's why using "acceleration" to mean "proper acceleration" is preferred in GR: because SR and GR have taught us that it's better to focus on things that are independent of coordinates. The weight felt by you standing on the surface of the Earth is the same regardless of which coordinate chart you are using. And the falling rock (in the idealized case where we can neglect air resistance) feels *no* weight, so it has zero proper acceleration; it's in free fall, and again that is true regardless of which coordinate chart you are using.

Johninch said:
I assumed that this gradient continues to get steeper inside the body. This must be wrong, because if gravity declines to zero at the center of the earth, the distortion of spacetime can't be increasing, it must be going back to normal.

Is this correct?

Yes; the interior of the massive body, in the kind of diagram you refer to, looks like a bowl that smoothly connects to the exterior shape. I know there are pictures on the web showing this, but unfortunately Google is not being kind and I can't find a link to one right now.

Johninch said:
Is this also what happens in a black hole, i.e. gravity and the distortion of spacetime both decrease once you go below the event horizon, although pressure increases?

First of all, there is no pressure inside a black hole; a black hole is a vacuum solution of the Einstein Field Equation, so there is no matter present anywhere (except in the past when there was an object collapsing to form the black hole).

Second, the curvature of spacetime continues to increase inside a black hole; at the singularity in the center, the curvature of spacetime goes to infinity. A black hole is not the same as an ordinary gravitating body like the Earth or the Sun.
 
  • #22
Johninch said:
Could you give me a link to a diagram which shows what spacetime does below the surface of a massive body?
Here is an interactive diagram, for outside and inside of a uniform spherical mass, along a line through the center:
http://www.adamtoons.de/physics/gravitation.swf
The green lines mark the surface.
 
  • #23
PeterDonis said:
First of all, there is no pressure inside a black hole; a black hole is a vacuum solution of the Einstein Field Equation, so there is no matter present anywhere (except in the past when there was an object collapsing to form the black hole).

Second, the curvature of spacetime continues to increase inside a black hole; at the singularity in the center, the curvature of spacetime goes to infinity. A black hole is not the same as an ordinary gravitating body like the Earth or the Sun.

The Wikipedia entry for Black Hole says that it has mass. So how can it be a vacuum and how can it have no pressure? When you say that there is no matter in a BH, do you mean that it contains mass only as energy? The Wikipedia entry explains that a BH attracts and absorbs matter, but it does not explain that the matter is then converted to energy. Can you give me a link to a better BH explanation which covers the points you are making?

What do you mean by a BH is not a gravitating body? It exerts gravitational effects on its surroundings. I suppose you are saying that beyond the event horizon the equivalent of gravity (spacetime curvature) increases, whereas it decreases in other massive bodies. Can we say that gravity goes to infinity at the center of a BH or are you saying that it is incorrect to talk about gravity inside a BH?

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  • #24
Johninch said:
Thanks, I’m trying to get my mind round it. I haven’t come across upward acceleration in this context before. I always considered that the force of gravity or warping of space only causes downward acceleration. Of course, if we are talking density, then there must be some upward pressure too, I agree.
As PeterDonis mentioned, the acceleration we talk about in GR is proper acceleration. In addition to his comments above, proper acceleration is the acceleration measured by an accelerometer. When it is in free fall an accelerometer reads 0 acceleration. When it is resting on the surface of the Earth it reads 9.8 m/s² upwards, not downwards.

Johninch said:
Can you explain one other point: when I look at those diagrams which show a massive body like the Earth creating a sink in spacetime, it looks like the distortion is increasing, the nearer you get to the massive body. I assumed that this gradient continues to get steeper inside the body. This must be wrong, because if gravity declines to zero at the center of the earth, the distortion of spacetime can't be increasing, it must be going back to normal.

Is this correct?
I don't like those diagrams in general so I won't link to one. However, you can consider a diagram of the gravitational potential, which is essentially the same as a Newtonian potential for the earth. The first picture here is a good example: http://en.wikipedia.org/wiki/Gravitational_potential

As the caption says, the inflection point is the surface of the body.
 
  • #25
Johninch said:
The Wikipedia entry for Black Hole says that it has mass. So how can it be a vacuum and how can it have no pressure?

Because "mass" as the term is being used here is not the same thing as "matter". "Mass" is an externally measured quantity; the "mass" of the Earth is what you measure if you put an object into orbit about the Earth, measure its orbital parameters, and apply Kepler's Third Law. You can do that with a black hole just as you can with a planet or star, so a black hole has mass in this sense.

How mass in this sense correlates to "matter" being present is a different question. Ultimately, if mass is present in this sense, some form of "matter" must be present somewhere in the spacetime. "Matter" here includes what we would normally call "radiation": it includes anything with a nonzero stress-energy tensor:

http://en.wikipedia.org/wiki/Stress–energy_tensor

The key question is, *where* in the spacetime does matter have to be present for mass in the above sense to be measured? The most general answer is: somewhere in the past light cone of the event at which you are measuring the mass. So if I put an object in orbit about the Earth and measure its orbital parameters, what I measure is determined by the matter inside the Earth that is present in the past light cone of the object's orbit.

Similarly, if I put an object in orbit about a black hole, the mass I measure is determined by matter that is present in the past light cone of the object's orbit; but in the case of a black hole, unlike the case of the Earth, that matter may be very, very far in the past, millions or even billions of years, when some massive object originally collapsed to form the hole. So if we look at the hole "now", we don't see any of the matter that is the ultimate source of the mass we measure; whereas in the case of the Earth we see that matter easily. But matter is the ultimate source of the mass in both cases.

Johninch said:
When you say that there is no matter in a BH, do you mean that it contains mass only as energy?

No. See above.

Johninch said:
Can you give me a link to a better BH explanation which covers the points you are making?

You might try these:

http://en.wikipedia.org/wiki/Mass_in_general_relativity

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_holes.html

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

Johninch said:
What do you mean by a BH is not a gravitating body?

I didn't say a BH wasn't a gravitating body; I said it wasn't the same as an ordinary gravitating body like the Sun or the Earth, because it has no "interior" where ordinary matter is present.

Johninch said:
I suppose you are saying that beyond the event horizon the equivalent of gravity (spacetime curvature) increases, whereas it decreases in other massive bodies.

First, other massive bodies like the Sun or the Earth don't have event horizons. I think you meant to say inside the *surface* of other massive bodies; but the event horizon of a black hole is not really the "surface" of the hole.

Also, what decreases inside an ordinary massive body like the Earth is not spacetime curvature, but the "acceleration due to gravity"--the gradient of the "gravitational potential" or "rate of time flow". The "rate of time flow" itself continues to increase inside the massive body; it just increases more slowly as you go further in. The "rate of time flow" at the center of the Earth is slower than at its surface.

And neither of these things--"rate of time flow" or its gradient--are the same as spacetime curvature; that is actually a tensor, not a single number, so it's more complicated than just "increasing" or "decreasing". I should have made that clear before.

Johninch said:
Can we say that gravity goes to infinity at the center of a BH

If we equate "gravity" to "spacetime curvature", then yes. But the other concepts, "rate of time flow" and its gradient, are meaningless inside the event horizon, since they can only be meaningfully defined in static regions of spacetime, and the region inside the event horizon is not static.

Johninch said:
or are you saying that it is incorrect to talk about gravity inside a BH?

It depends on what you mean by "gravity"; the term has more than one meaning. See above.
 
  • #26
Thanks to all. I feel that my questions have been well answered and that I now have a better understanding of GR and gravity.

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What is the curvature of spacetime on Earth?

The curvature of spacetime on Earth refers to the bending of the fabric of space and time caused by the mass of the Earth. This curvature is what keeps the Earth in orbit around the Sun and affects the movement of objects on its surface.

How is the curvature of spacetime measured on Earth?

The curvature of spacetime on Earth is measured using the theory of general relativity, which was developed by Albert Einstein. This theory explains how mass and energy affect the curvature of spacetime and can be used to make precise calculations of the curvature on Earth.

Does the curvature of spacetime on Earth affect time?

Yes, the curvature of spacetime on Earth affects time. According to the theory of relativity, time moves slower in areas with stronger gravitational fields, such as near the surface of the Earth. This is known as time dilation and has been confirmed through experiments and observations.

Is the curvature of spacetime on Earth constant?

No, the curvature of spacetime on Earth is not constant. It varies depending on the distribution of mass and energy on the Earth's surface, as well as its rotation and orbit around the Sun. The curvature also changes over time as the Earth's mass and energy distribution changes.

How does the curvature of spacetime on Earth affect space travel?

The curvature of spacetime on Earth has a significant impact on space travel. It affects the trajectory and speed of spacecraft, and the amount of fuel needed to reach a certain destination. Scientists must take into account the curvature of spacetime when planning and executing space missions.

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