# Curvature of Spacetime on Earth

1. Jan 21, 2013

### Johninch

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. Jan 21, 2013

### A.T.

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

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

Distortion of space has little effect on a rock. It is mostly distrotion of time. See the first link above.

No

Yes

3. Jan 21, 2013

### Staff: Mentor

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.

"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".

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.

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.

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.

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.

"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.

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

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

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.

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

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.

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.

4. Jan 21, 2013

### Naty1

5. Jan 22, 2013

### Johninch

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.
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?

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. Jan 22, 2013

### kweba

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?

So in GR, there are no such things as Potential and Kinetic energies?

7. Jan 22, 2013

### A.T.

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. Jan 22, 2013

### A.T.

That's a disscussion about gravity inside Earth in terms of Newton. I think OP here wants to know about gravity on Earth in terms of GR. But here is a GR visualization for both: inside and outside of a uniform sphere:

Note, that the Earth is not actually uniform, so the geometry would be a bit different.

9. Jan 22, 2013

### A.T.

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. Jan 22, 2013

### Johninch

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.

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. Jan 22, 2013

### Staff: Mentor

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.

Last edited: Jan 22, 2013
12. Jan 22, 2013

### A.T.

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. Jan 22, 2013

### Staff: Mentor

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. Jan 22, 2013

### A.T.

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.
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. Jan 22, 2013

### Johninch

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. Jan 22, 2013

### A.T.

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.
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.

Last edited: Jan 22, 2013
17. Jan 22, 2013

### Staff: Mentor

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. Jan 22, 2013

### Staff: Mentor

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.

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

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).

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. Jan 24, 2013

### Staff: Mentor

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. Jan 24, 2013

### Johninch

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?

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