Curvature of Spacetime on Earth

PeterDonis

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.

DaleSpam

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.

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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PeterDonis

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.

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.

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.

A.T.

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.

Johninch

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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DaleSpam

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.

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.

PeterDonis

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%...3energy_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.

No. See above.

You might try these:

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

http://math.ucr.edu/home/baez/physic...ack_holes.html

http://math.ucr.edu/home/baez/physic...k_gravity.html

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.

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.

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.

It depends on what you mean by "gravity"; the term has more than one meaning. See above.

Johninch

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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