# A series of questions regarding Gravity:

1. Apr 23, 2013

### VeryConfusedP

#1 Is gravity a force? If it is not, can it be described as the geometrical structure of spacetime?

#2 Do gravitational fields "compete" with each other? Okay, here's what I mean by that. The sun has its gravity and the earth has its gravity. From my understanding, I'm standing upright on earth (attached to the earth's surface) because I'm in a gravitation field (earth's) and space is pushing me down. If I hop on a broomstick and accelerate into outer-space, towards the sun, is there a point at which I will no longer be drawn in by the earth's gravity and will be gravitating towards the sun? Perhaps a better way of asking this is: Is there a middle point, perhaps calculated by Einstein's Field Equations, that says I will be tugged in both directions (towards the sun and towards the earth--stuck between gravitational fields) if I'm in outer-space?

#3 Are Einstein's theories of relativity showing us that what is of greater significance is not what is relative but what isn't relative? Examples: constants that define the universe, laws of physics, speed of light, my birth, scalar fields, etc.?

#4 True or false: Gravity and acceleration are locally indistinguishable. If that is the case, under what conditions are gravity and acceleration not equivalent?

#5 If an object is in freefall, are all other objects around it in freefall?

#6 Are Einstein's relativity theories essentially saying that the only things that are real are things that don't depend on point of view? Or, in other words, if any phenomenoa is made up of both magnitude and direction and is dependent upon a frame of reference, does that suggest, in some sense, that it's not real?

#7 Is tidal force (gravity) the "force" people think of when they think of gravitational "force"?

#8 Is this essentially the "take-home" point of general relativity: The presence of mass and energy changes the geometry of spacetime to be a curved spacetime, and objects not subjected to forces move in the straighest possible path in curved spacetime.

If any of my questions are inappropriate, I want to apologize in advance. I've done some studying on General Relativity and I'm trying my hardest to understand it, without having any kind of scientific background. I've been watching some tapes made by Richard Wolfson, and that's where some of my questions stem from.

2. Apr 23, 2013

### Staff: Mentor

It depends on which theory you are using. In Newtonian gravity, yes, it is. In GR, no, it's not.

In GR, yes.

It's good to get used to thinking of spacetime, not "space". Also, it's not really that spacetime is pushing you down; the Earth is pushing you *up*, away from the path in spacetime that you would naturally follow if there were no force (remember that in GR, gravity is not a force!) acting on you (i.e., the freely falling path).

In GR, gravity isn't a force, so thinking of the Earth and the Sun as both "tugging" on you isn't really correct. However, in Newtonian gravity, where gravity is a force, you can indeed calculate such a "middle point"; and the motion of objects at or near such a point will be almost the same when calculated using GR as they are in Newtonian gravity (there are small corrections with GR). Another way of saying this is that, although gravity in GR is not a force, in situations like the solar system you can adopt coordinates that let you describe most of its effects as a force similar to the Newtonian force (but again with small corrections).

I would say yes, and Einstein said similar things in many of his writings. Some physicists say that the theory of relativity is misnamed; it should really be called "the theory of invariants".

This is a somewhat sloppy way of stating it. A better way is: being at rest in an accelerating frame (such as a rocket whose engine is on) is locally indistinguishable from being at rest in a static gravitational field. Stated this way, yes, it's true.

With the restatement I gave above, they are not equivalent under any conditions where tidal effects, i.e., changes in the gravitational field from place to place, are detectable. (Or changes from time to time--"local" means local in spacetime, not just in space. This is an important point that many people forget.)

Not necessarily. Why would they be? If you drop a rock in free fall next to you, why would you have to be in free fall just because the rock is?

Scientific theories in general don't use the word "real", because it's very difficult to define precisely what "real" means, and whatever definition you pick, people will argue that you picked the wrong one. So scientific theories focus on predicting observable quantities, because those can be objectively measured and the numbers that are measured can be objectively compared with the predicted numbers from the theory. Relativity has an excellent track record of doing that. What, if anything, that says about what is or is not "real" is left to your interpretation.

Probably not; when most people think of gravitational "force" they think of a "force" that pulls things towards a large mass like the Earth, and that isn't tidal gravity. In GR, tidal gravity isn't a force any more than any other aspect of gravity is a force.

Also, tidal "force" itself is not quite the same as tidal gravity. If I take two rocks and drop them from slightly different altitudes high above the Earth, their spatial separation will gradually increase due to tidal gravity; but neither rock feels a force, they are both in free fall. On the other hand, if I take a single rod oriented radially and drop it, the rod will develop a tensile stress due to "tidal force", because it is resisting being stretched by tidal gravity, even if its length doesn't change at all.

This is a pretty good quick summary of what GR says, yes. The only thing I would take issue with is that the presence of mass and energy doesn't "change" the geometry of spacetime; it *determines* the geometry of spacetime, through the Einstein Field Equation. There is no pre-existing "geometry of spacetime" that the presence of mass and energy changes into something else.

No need to apologize. These are all good questions.

3. Apr 23, 2013

### VeryConfusedP

Peter,

This is the part that keeps throwing me, and I know it's just a matter of perspective. "...the Earth is pushing you *up*, away from the path in spacetime that you would naturally follow if there were no force (remember that in GR, gravity is not a force!) acting on you (i.e., the freely falling path)."

I believe it's the shape of the human body that is throwing me. My--our--standing upright is implying direction, *upwards*, and perhaps that's what's confusing me. Let's use a tennis ball. The tennis ball is lying on the surface of the earth. How is the earth's curvature of spacetime causing this to happen? Are you saying that the earth is "getting in the way" of the tennis ball's free fall through spacetime?

4. Apr 23, 2013

### Staff: Mentor

Welcome to Physics Forums!

Just as general piece of advice, in most forums including PF, it's not a good idea to ask eight questions in one post. Each question potentially can start a discussion of its own, and it would be kind of confusing to have eight discussions going in parallel, with the responses all mixed together in the same thread.

In practice, most of the discussion is probably going to focus on one or two questions, and the rest will be more or less ignored. It's best to use separate threads, spaced judiciously in time, and keep them focused. Answers to earlier questions may cause you to modify or even not bother with some of the later questions.

5. Apr 23, 2013

### Staff: Mentor

No, you've got it backwards. Your body stands upright because that's the direction the Earth is pushing you; the direction "up" is determined by the force of the Earth pushing on you, not the direction your body is aligned. Your body aligns itself with that direction because that's the easiest way for it to remain stable; in other words, your body's direction is a consequence of the direction of the force the Earth exerts.

Yes; the tennis ball can't pass through the Earth, so it can't freely fall due to the curvature of spacetime in the Earth's vicinity.

6. Apr 23, 2013

### Staff: Mentor

One piece of advice (free, and worth not less than you paid for it): Don't take on general relativity until you have classical Newtonian mechanics down cold. Your questions 1, 2, 5, and 7 can be answered using just the classical theory, and it will be very hard to make sense of relativity if you don't have that base to build on.

7. Apr 23, 2013

### VeryConfusedP

Yes; the tennis ball can't pass through the Earth, so it can't freely fall due to the curvature of spacetime in the Earth's vicinity.[/QUOTE] (This quote is from Peter. I don't know how to use the shaded quote copying tool.)

Is the earth the actual "curvature of spacetime" or are you saying the curvature of spacetime, which is causing the tennis ball not to be able to freely fall, is independent from the earth?

I cannot decipher--and trust me, it's my ignorance; it's not you--this quote of yours. Is it earth's matter that is blocking the tennis ball from carrying on with its free fall through spacetime? Or is it the curvature of spacetime caused by earth that is causing the tennis ball not to carry on with its free fall through spacetime? Or is that the same thing?

Edit: Perhaps I'm misunderstanding the idea of free fall. In space, free fall doesn't mean you're falling like you would be when you jump out of a plane in earth's atmosphere, right? All free fall is saying is that you're moving along geodesics in spacetime, but when you get near a large mass like the earth you fall towards it because of the the curves it's creating in spacetime.

Is that correct? Again, in outer-space free fall, you're not really falling downwards or upwards or away from planets. You're just not being affected by the curvatures in spacetime that are being caused by these large bodies of mass (unless you get near them).

Last edited: Apr 23, 2013
8. Apr 23, 2013

### Staff: Mentor

No; the Earth is the source of the spacetime curvature, but the curvature of the Earth itself is not the same as the curvature of spacetime around the Earth. As far as the tennis ball is concerned, the Earth is just an object that is in the way of its free-fall path.

Yes, because solid objects can't pass through other solid objects.

No.

No. See above.

They're very similar, but actually in space you can achieve free fall more easily, because there's no atmosphere. Even inside a spacecraft like the International Space Station (ISS), where there is obviously air present, the astronauts are in free fall without having to move relative to the air (unlike when jumping out of a plane in an atmosphere--see below), so they are weightless.

When you jump out of a plane in Earth's atmosphere, you start out in free fall, i.e., weightless, but air resistance creates drag as you go faster because you're moving through the air, so you aren't in free fall any more; you feel weight.

If nothing obstructs your path, yes. But bear in mind that "falling towards it" is not the only way to free-fall. Consider the ISS again: it's in orbit about the Earth, but it's in free fall; there is nothing obstructing its motion, and no force is acting on it; it's moving on a geodesic through spacetime.

If you're far enough away from all of them, yes.

9. Apr 23, 2013

### VeryConfusedP

Okay, my next question should be a pretty obvious one. If you are far, far away from large bodies of mass, are you still moving along geodesics, or is spacetime flat-like (non-curved), aside from the minor spacetime curvature you are creating?

It seems like the idea of "moving along geodesics" is very relative in its nature. The less mass, the less curvy spacetime is. Like I said above, does spacetime, or one's relative motion through spacetime, begin to flatten out the farther one is removed from large bodies of mass?

10. Apr 23, 2013

### Staff: Mentor

If there are no forces acting on you, yes. For example, if you're in a spacecraft with no rocket firing, just floating along.

It will be, but whether or not spacetime is curved is a separate question from whether or not you are moving on a geodesic. See below.

No, it isn't. "Moving along geodesics" is just the mathematical (geometrical) way of describing freely falling, weightless motion. Freely falling, weightless motion is a perfectly objective, observable condition. You can tell whether or not you are in free fall just by testing whether or not you are weightless; you don't have to know anything about the large-scale structure of the spacetime you are in.

What the curvature of spacetime determines is *which* particular paths through spacetime are geodesics, relative to the objects in the vicinity. For example, the curvature of spacetime around the Earth is what determines which paths are geodesic paths, relative to the Earth--the paths of a falling rock and an orbiting spacecraft are geodesics, but the path of someone standing at rest on the Earth's surface is not.

11. Apr 23, 2013

### VeryConfusedP

Yes, Peter, but weight has to be an idea that varies with regard to one's closeness to the body of mass and, equally important, the size of the body of mass.

I don't know which is true, but your weight on earth would be different (more or less) from your weight on the sun. This is telling me that, in a way, everything is in free fall, just varying degrees of it (you're going to roast me on this, but I have to say it, because that's how I'm interpreting it as of this moment). The only way the notion of free fall is dismissed is by the coming together of forms of matter. As you even said, the only reason the aforementioned tennis ball isn't in free fall anymore is because it is falling along--or has fallen along--the curvatures of spacetime being created by the earth's mass (and is on the earth's surface, and by virute of that is experiencing weight). No, I'm starting to get it, and it's all because of you. I may not have it polished off, but a lot of it just clicked. It's also telling me something about the weakness of spacetime curvature. If the sun is so big compared to the earth--and regardless of the earth's "angular momentum--the curvatures of spacetime being created by the sun's mass aren't all that significant.

Question: I am 6'0 and 175 pounds. How much would I weigh on the sun? More or less? (Yes, this is assuming that the sun wouldn't burn me away.)

12. Apr 23, 2013

### Staff: Mentor

Weight is just another name for a felt force. The weight you feel standing on the surface of the Earth is the force of the Earth pushing up on you. Obviously, if you could stand on a tower 1000 miles high, you would feel less weight because whatever was underneath your feet would be pushing up on you with less force, and yes, that's because you would be further away from the center of the Earth. But it's still a felt force, and feeling a force is different from feeling no force at all, and which state you are in depends on your state of motion.

By which you mean, standing on the surface of the Sun, or, equivalently, "hovering" just over its surface in a rocket whose engine was firing just hard enough to keep you from falling in. (And of course the rocket would have to be made of something that could withstand 6000 degree temperatures without melting--but in thought experiments we can help ourselves to these kinds of things. )

Why do you think that? "Free fall" means "weightless"--it means feeling *no* weight. It doesn't mean feeling various degrees of weight depending on where you are. It means feeling *zero* weight. Obviously, the state of motion you have to adopt, relative to other nearby objects, in order to feel no weight will change depending on where you are. But the way you test for that state of motion is still the same: feeling *zero* weight. If you feel nonzero weight--*any* nonzero weight--you are not in free fall.

No, the reason the tennis ball is not in free fall is that it *cannot* "fall along the curvatures of spacetime being created by the Earth's mass"--it can't because the Earth is in the way, pushing on it.

You know this, because the very next thing you say is

But if it's at rest on the Earth's surface, it is *not* "falling"--that's the whole point. It is *not* following a geodesic; it is *not* moving solely in response to the curvature of spacetime in its vicinity. It is being pushed *out* of that path by the Earth. The push is felt as weight.

This is true; spacetime curvature everywhere in the solar system is very small in absolute terms.

Your height doesn't come into it, just your weight. I could cheat by looking up the Sun's surface gravity, but it's more fun, and more instructive, to lay out how one would calculate it. For this problem we don't even need any formulas from GR; we can use the Newtonian formulas, since GR tells us that they are good enough approximations for the Earth and the Sun. The Newtonian formula for an object's weight at the surface of a massive body is:

$$W = \frac{G m M}{R^2}$$

which should look familiar. But we really just want the ratio of your weight at the Sun's surface to your weight at the Earth's surface (since the latter is known), so we have

$$\frac{W_{Sun}}{W_{Earth}} = \frac{M_{Sun} R_{Earth}^2}{M_{Earth} R_{Sun}^2}$$

Notice that $G$ and $m$, the mass of the object, cancel out because we're taking a ratio. You can look up the numbers and plug them in to this formula and see what you get.

13. Apr 23, 2013

### VeryConfusedP

But, Peter, are you saying your two feet have to be attached to the earth to feel weight? Perhaps this is the last key thing I'm missing in this discussion. If I am floating along a geodesic but I start falling along the spacetime curvatures being created by a large body of mass, won't I start to experience the feeling of weight before my feet actually hit the surface of the body of mass? The closer I get to the surface, the more I start getting familiar with this feeling of weight, right? Or are you saying that your two feet literally have to be planted on the surface of the body of mass before you can feel weight? If we can clear up this last point, I will call this a very, very successful day.

14. Apr 23, 2013

### Staff: Mentor

What do you mean by "attached to"? Do you just mean "in contact with"? Then yes, how else is the force going to be transmitted to your feet? But if you mean do your feet have to be strapped to the Earth for you to feel weight, obviously not, as your own everyday experience will tell you.

Why do you think these are separate things? You are floating along a geodesic as long as there is no force acting on you. That doesn't stop being true just because you get close to a large mass.

No--at least, not if we're ignoring air resistance. Think of the ISS again. The ISS is only a couple of hundred miles above the Earth's surface; the strength of the Earth's gravity there is almost the same as it is on the surface. Yet astronauts in the ISS feel no weight. They are freely falling in orbit about the Earth.

If there is nothing else to exert a force on you, yes. Perhaps the Moon is a better example than the Earth, since the Moon has no atmosphere, so there's no air resistance to complicate things. Suppose you are falling towards the Moon, with no other force acting on you--you are in a spacesuit but you have no rocket pack, nothing to exert a force on you. Then you will feel zero weight until your feet make contact with the Moon's surface. (We're assuming you can land on your feet without damaging your body.)

15. Apr 23, 2013

### Staff: Mentor

Yes, that is exactly what Peter is saying. Consider the weightlessness astronauts experience in space: They are moving freely under the influence of gravity yet feel no weight and it make no difference whether they're in near earth orbit just 100 miles up (force of gravty is a only a few percent less than at earth's surface) or far outside of the moon's orbit.

Imagine you were standing on a spring scale on a trapdoor. The scale reads your weight. We open the trapdoor and you and the scale fall freely. The scale reads zero. It continues to read zero until you touch the floor of the room below and are back to standing on the scale again.

Last edited: Apr 23, 2013
16. Apr 23, 2013

### VeryConfusedP

Got it, got it, got it. Yes, yes, yes. It was the atmosphere of the earth that was throwing me. I'm thinking, I can feel wind. But no, you're saying in a non-atmosphere landing, one has to touch the body of mass to feel the force of weight. Yes, it was the earth's atmosphere that kept throwing me.

Gentlemen, I have learned quite a bit today, and I want to thank Peter (especially) and Nugatory for helping me learn these very counterintuitive things. My brain is already spinning like a ballerina, though. The weakness of spacetime curvature just isn't making any sense. Anyway, that's for another day.

Again, Peter (and Nug), thank you very much for navigating me through these hitherto murky and counterintuitive waters.

Last edited: Apr 23, 2013
17. Apr 24, 2013

### Boy@n

I learned too from this thread :) thanks all.

I hope I don't ruin it now with this 'exotic' question: first, let me sumarise what I understood: in a free fall near a massive object like Earth or Sun we feel no weight and no force, we are falling towards the centre of that object due to space-time curvature which is defined (not created) by that object.

But imagine there are two (average) Black Holes of equal mass very near (say 1km) one another (well, they would have to rotate around one another very, very fast, to not crash together, right? BTW, how fast?), and it happens you (in a spacesuit) get exactly in the middle of them. So, since gravity is huge there, you still feel no pull on you from those two Black Holes, you feel nothing at all since no force is exerted on your body? The space-time there is greatly curved, but that too doesn't affect your body, true? Intuitively I'd think you'd get split in about two parts, but I must be wrong, right?

Last edited: Apr 24, 2013
18. Apr 24, 2013

### Staff: Mentor

If you were a point particle, no problem, you'd free-fall happily along.
But an object of any size will find itself pulled apart because one side will be nearer to one black hole and the other side will be nearer the other, so the two halves would want to free-fall in different directions. This is a tidal effect, something caused by the difference between the gravitational force at two nearby points.