Four amateur questions about gravity and general relativity

1. Oct 26, 2013

NotEuler

Hi everyone,

I've been reading posts here occasionally, and have been impressed with the amount and quality of knowledge that is being shared.
I've just registered, and as my first thread I thought I would post some questions / thoughts that have puzzled me for a long time, related to gravitation and general relativity.
Now, I'm a complete amateur, so please view the questions as such! They probably seem quite naive to an expert, but to someone like me, they are anything but straightforward. So here goes:

1) My understanding is that we need a unified theory combining general relativity with quantum effects because the two are incompatible at GR singularities. So if string theory is a serious candidate for such a theory, what does string theory predict at the singularities? Same with any other attempts at a unified theory? Do they make some physically sensible statements about what really happens, say, at the centre of a black hole?

2) I've been told that general relativity says there is no force of gravity, but that things tend to 'go straight' in 4D spacetime. The geometry of spacetime itself is warped by mass / energy, which is why it seems like there is a force of gravity affecting trajectories of objects.
But I'm pretty sure I've occasionally seen gravity described as a force in GR texts.
And if it's not a force, then why do we want to find 'gravitons' that supposedly transmit the force of gravity?
So is gravity geometry or force? Or are the two really indistinguishable, and just two ways to look at one thing?

3) With newtonian gravity, we can say that in a particular spot between the earth and the moon (One of the Lagrange points), the gravitational forces from the two objects cancel out. What does this mean in a general relativity context? Does it mean that in this location, spacetime geometry is flat?
And if so, does this mean that if two black holes were orbiting each other at a distance of just slightly over their Schwarzschild radius, would spacetime be flat halfway between them? Could I just float at that point without having to worry too much about being sucked in?

4) This is a bit of a silly one, but it puzzles me!
So Newtonian gravity is just an approximation of general relativity. But how on earth does GR simplify to the beautifully concise and sensible newtonian gravity law? I'm not sure how to explain what I mean, but to me, Newton's law of gravity makes intuitive physical sense: the force of gravity is proportional to both masses, and inversely proportional to the square of their distance. It makes sense!
I'm just amazed that a law that is 'wrong', and as such does not really have any proper physical justification (I assume) can seem so right!

2. Oct 26, 2013

yuiop

Hi and welcome to PF

For question 3 you would be floating on a knife edge in a dynamic environment (because the black hole are orbiting each other) and any stray atom could push you down a slippery slope in either direction. Also the 'flatness' would be very local and if the black holes are small, the extreme tidal forces will probably be spaghettifying your limbs in opposite directions.

3. Oct 26, 2013

WannabeNewton

There is a way to formulate GR wherein gravity is akin to a force (a manifestation of torsion). You get the same experimental predictions so it is really just a matter of viewpoint. The formulation I am referring to is called Teleparallel gravity. Similarly, there is a way to formulate Newtonian gravity in terms of the curvature/geometry of space-time whilst getting the same experimental predictions so again it is just a matter of viewpoint. The latter is called Newton-Cartan theory.

You have to be more careful in GR. Gravitational fields do not simply superpose linearly in GR; the Einstein equation is non-linear due to the self-interactions of the gravitational field hence when you consider two gravitational fields there will be complicated non-linear interactions as opposed to the simple linear superposition in Newtonian gravity.

One writes down a metric tensor solving the Einstein equation in what is called the Newtonian limit. Then one writes down the equations of motion (geodesic equation) for this metric tensor and from there it is easy to show that it reduces to Newton's 2nd law $\vec{a} = -\vec{\nabla} \varphi$. Pretty much every standard GR text will demonstrate this so you can take a look at one of them for the details.

4. Oct 26, 2013

yuiop

I was assuming that if the black holes were identical then there must be a point where due to the symmetry there is no preferred direction to move in, even if it is just an infinitesimal point. For any breakdown in symmetry such as different masses or different rotation rates, then yes in general there is probably no point where you can remain stationary (if that can be defined in the rotating reference frame).

5. Oct 26, 2013

Staff: Mentor

Hi, NotEuler, and welcome to PF!

This is one way of looking at it, yes. Another way would be to say that GR assumes that spacetime is a continuuum, while quantum mechanics says that spacetime ought to be discrete at some small enough length scale (traditionally taken to be the Planck length).

I don't think any of them do at this point, beyond the bare statement that something finite should happen there as opposed to spacetime curvature just increasing without bound.

Yes.

If you look carefully, you should find that they are talking about the Newtonian limit, where gravity is weak and everything is moving slowly compared to the speed of light. In that situation you can view gravity as a force, basically the same way Newtonian mechanics does; the non-Newtonian effects that GR predicts are all too small to matter.

There are at least two ways of looking at this. One is that classical GR predicts gravitational waves, and gravitons would be the quantum aspects of those waves; i.e., gravitons are just the GR analogue of photons, the quantum aspects of electromagnetic waves, which can do more than just transmit force.

The other way is to think of gravity as a quantum field interaction, the same way we do in the Standard Model of particle physics with the other interactions (electromagnetism, weak, and strong). The graviton is just the quantum particle aspect of the quantum field interaction for gravity.

Both of these viewpoints are of limited usefulness now because, as I noted above, we don't have a good unified theory of GR and quantum mechanics. But they can still have some value within those limits.

I would say they are two different ways of looking at the same thing, but with different domains of validity and levels of difficulty. The geometry way is more general, but also more difficult to apply; the force way is more limited in its domain of validity, but is easier to apply. So it's kind of a tradeoff between generality (and accuracy) and ease of use.

No. First, a clarification: locally, in a small enough neighborhood around any event, we can always approximately view spacetime as flat, just as we can always approximately view a small enough piece of the Earth's surface as a flat Euclidean plane. (How valid this approximation is depends on how accurately we can make measurements.) So when we say the geometry is flat, period, we're talking about more than just being locally flat in the above sense; we're talking about it being "really" flat, as in spacetime curvature is actually zero, not just too small to measure in a small enough neighborhood of a particular event.

That said, the answer is still "no"; there is something that can be viewed as "flat" at a Lagrange point, but it isn't spacetime. In the weak field approximation, we can define a "gravitational potential", the same way we do in Newtonian theory. The gradient of this potential is the Newtonian "gravitational force". At a Lagrange point, the gradient of the potential is zero; that's what the forces "cancelling" means. But the potential itself is still not zero--i.e., it does not have the same value it does at infinity, far away from all gravitating bodies. That means spacetime is not flat; flat spacetime would correspond to zero potential (again, this is in the weak field approximation).

No. The reason is basically the same as for the Lagrange points, as above. But this situation is more complicated than the above, for three reasons:

(1) You can't use the weak field approximation close to a black hole.

(2) We don't actually have an analytical solution to the Einstein Field Equation (the central equation of GR that determines what the spacetime curvature is) for the case of two black holes orbiting each other. GR is nonlinear, so you can't just add together two black hole solutions centered on different points and get a valid solution. The best we can do is numerical simulations.

(3) If the holes are separated by just slightly more than their Schwarzschild radius, they can't orbit each other this way; their separation has to be considerably larger for a stable orbit to be possible. For the case of a single black hole with a "test object" orbiting it of negligible mass, the test particle's orbital radius must be at least three times the hole's Schwarzschild radius to be stable. For the dual black hole case, as I said above, we don't have an exact analytical solution, but AFAIK numerical simulation would still show that the holes' separation had to be something on that order for their mutual orbits to be stable.

All that said, AFAIK, numerical simulation of this case would still show a "Lagrange point" location between the holes, where particle orbits exist that can "hover" between them. (I think that the holes' separation would have to be at least twice the minimum for the holes themselves to be in mutually stable orbits, because the particle would have to be at least three times the holes' Schwarzschild radius away from each hole.) However, as yuiop pointed out, this orbit would be unstable; any small perturbation would end up sending the particle into one of the holes.

The description yuiop gave is valid; i.e., I would not advise trying this experiment yourself.

When viewed properly (at least, properly according to a relativity physicist ), GR is simpler than Newtonian gravity. For one thing, Newtonian gravity requires instantaneous "action at a distance", which we know can't work: causal influences can't propagate faster than light, and physical laws that incorporate that fact actually turn out to be simpler, mathematically.

That said, the derivation of Newton's laws from GR is straightforward:

(1) Solve the Einstein Field Equation for an isolated, spherically symmetric gravitating body. (This is an idealization, of course, but it's the same idealization Newton used.)

(2) The solution turns out to depend on a single parameter, $M$, which can be shown to be the externally measured mass of the body.

(3) Assume that gravity is weak; this turns out to mean assuming that all points of interest are at a radial coordinate $r$ which is much larger than $2 G M / c^2$, the Schwarzschild radius associated with the mass $M$.

(4) Assume that everything is moving slowly compared to the speed of light. (The central body itself is assumed to be at rest.)

(5) Find that all of the effects of gravity can be described by the Newtonian potential $\phi = - GM / r$. (Notice that $\phi / c^2 = - r_s / 2 r$, where $r_s$ is the Schwarzschild radius; so the potential is just a measure of how strong gravity is at radius $r$.) The (negative of the) gradient of the potential describes the Newtonian force of gravity (more precisely, the force per unit mass; the potential itself is the potential energy per unit mass), which is $f = - G M / r^2$. Notice that the $1 / r^2$ dependence of the force pops out naturally; it's just a reflection of the fact that the potential is proportional to $1 / r$, which makes sense since, as noted above, it's just the ratio of $r_s$ to $r$.

The Newtonian law is not "wrong", and it does have physical justification: it's the correct weak field, slow-motion limit of the general GR solution for an isolated gravitating body. See above.

6. Oct 26, 2013

TumblingDice

When I read WannabeNewton's perspective, it triggered the recollection that gravity begets gravity. Furthermore, gravity 'here' creates a little gravity 'over there'. I'm wondering if that is what was being referred to? I was imagining kind of a rolling plains effect with two BH gravity sources, and the great speed they would have to be circling at. So even a center of symmetry so to speak, could be undulating like crazy.

Have I lost my way?

7. Oct 27, 2013

NotEuler

Wow, thanks for the replies everyone. The quality of the responses is pretty amazing!
That being the case, I'll throw in a couple more naive questions!

5) So gravity is geometry. I know these things are hard to visualize and get your head around, but one thing I'm having particular trouble with is tidal forces in the geometrical interpretation. It's easy enough to understand in Newtonian gravity that your feet are pulled with a stronger force than your head (or vice-versa) if you're falling into a black hole. But what does that mean in curved spacetime? Aren't you just floating (going straight, inertially) in that curved spacetime? So if there's no force, what is it that causes the 'stretching' effect, since both your head and feet are just freely floating objects? Is it that in that particular small region of spacetime, your feet want to cover a longer spatial distance in a shorter time than your head does? And even though your head is just floating too, it wants to float slower, so a pull is felt between the two? Or am I totally off the tracks here?

6) Again on spacetime curvature.. Another thing that puzzles me (yes, I'm easily puzzled): Curvature is supposedly pretty low in the solar system, or at least when we're far outside the orbit of Mercury. Yet planets stay in orbits around the sun. It feels difficult for me to reconcile that with the idea of low curvature! It seems to me that SOMETHING must be pretty curved to make an object with the mass of the earth orbit around the sun, and keep coming back to more or less the same spot! So there must be something I'm misunderstanding here. Perhaps I'm thrown off by the time component of spacetime? It seems that space can't be too warped here where I sit typing this post, so is time somehow more warped, causing the orbits? Or is it that we're 'moving' at different speeds through the space and time components, and therefore somehow seeing their curvatures differently?

TumblingDice, do you mean that the mutually orbiting black holes are causing gravitational waves? If so, would that mean that you couldn't even theoretically float still at the midway point between two identical black holes? Then again, it seems that just because it's symmetrical, you should be able to! I'll surely let you know if I get a chance to try it.

8. Oct 27, 2013

dvf

Nah. Gravity can be modeled with geometry. The former is physical and the latter is mathematical.

The Sun has a LOT of mass.

9. Oct 27, 2013

Staff: Mentor

Suppose you are falling vertically towards a gravitating body, with your feet below your head. If your head and feet were not attached, so that they could fall freely, they would separate, because your feet are slightly closer to the gravitating body than your head and so they will accelerate downward slightly more than your head (relative to the gravitating body--we are basically taking the Newtonian viewpoint here, but for this case that works well enough). So if your head and feet were not attached, the distance between them would gradually increase as they fell: that is a direct manifestation of tidal gravity.

But since your head and feet *are* attached, they are *not* falling freely; there are internal forces between them that try to keep them the same distance apart. In other words, even if the center of mass of your body is in free fall (i.e., weightless, feeling zero force), your head and your feet will not be: your head will feel a small downward force and your feet will feel a small upward force, from the rest of your body trying to keep your head and feet the same distance apart--i.e., your body is trying to keep itself from stretching. That force is what is usually called "tidal force", but it's important to note that it's not tidal gravity that's actually exerting the force: it's the various internal parts of your body exerting forces on each other. The only role tidal gravity (i.e., spacetime curvature) plays here is in determining what force an individual part of your body will experience if you impose the constraint that your body remains the same length as it falls.

It's actually low *everywhere* in the solar system, including inside the Sun. See below.

Remember that we're talking about *spacetime* curvature, not just *space* curvature, so you have to include time as well as space when you're figuring out the curvature of planetary orbits, or the curvature inside the Sun.

Take planetary orbits first, and let's use the Earth as an example. (Note: all of this is just heuristic, to give an idea of how to visualize the curvature of a path in spacetime, and the numbers I will calculate are just to give an idea of orders of magnitude. The actual math to calculate curvature components is more complicated.) The Earth has an orbital period of one year, and an orbital radius of 150 million kilometers. To compare these quantities, we need to put them into the same units; we can either multiply the time by the speed of light, or divide the distance by the speed of light. The former is easier: $3.1 \times 10^7 s$ times $3 \times 10^8 m/s$ gives about $10^{16} m$ for the orbital period, as compared to $1.5 \times 10^{11}$ meters for the orbital radius in space, or 2 pi times that, about $10^{12}$ meters, for the orbital circumference.

So the spatial distance traveled per orbit is *much* smaller (by a factor of $10^4$) than the "time distance" per orbit. That means that thinking of the Earth as traveling in a circle with a radius of $1.5 \times 10^{11}$ meters is wrong; it is really traveling along a helix in spacetime, which if you drew it on a spacetime diagram, would look like a very, very gently curved helix, stretched out by a factor of 10,000 along the time dimension as compared to its spatial radius.

Visualizing the curvature of a worldline inside the Sun is a bit more difficult, because a given small piece of matter inside the Sun is not in a free-fall orbit; to a first approximation, it's just sitting at rest in the overall gravitational field of the Sun as a whole, so there's no "orbital curvature" to look at the way we did for the Earth above (its worldline is just a vertical line in a spacetime diagram). But we can generalize what we did for the Earth above to something that we *can* apply to a small piece of matter inside the Sun. Consider Kepler's Third Law formula for an orbit: $\omega^2 R^3 = GM$. If we write $\omega = 1 / T$ and divide through by $R$, we get $R^2 / T^2 = GM / R$. In other words, the ratio we computed above, of the spatial radius of the orbit to the spacetime arc length, is just the square root of the Newtonian potential $\phi$, which as we have seen is a good measure of the "strength of gravity". Our computation above basically told us that in the vicinity of the Earth's orbit, $\phi \approx 10^{-8}$. (Note that $\phi$ here is dimensionless; basically we are using units where $G = 1$ and mass has the same units as length and time.)

But we can easily compute the Newtonian potential $\phi$ inside the Sun. At the surface, it's just $G M_{sun} / R_{sun}$. We have $M_{sun} \approx 10^{30} kg$ and $R_{sun} \approx 7 \times 10^8 m$. Converting $M$ to "length units" means multiplying by $G \ c^3$, so we have $M_{sun} \approx 1.5 \times 10^3 m$. This gives $\phi = M / R \approx 2 \times 10^{-6}$.

As we go further inside the Sun, $\phi$ will increase (actually we should say "decrease" because $\phi$ is negative, indicating a bound system; we are actually calculating $- \phi$, but we'll gloss over that). We don't know exactly how much it will increase as we go to the center of the Sun, but we can easily compute an upper bound to the increase: the rate of change of $\phi$ with radius at the Sun's surface is just the acceleration due to gravity there, which is $M / R^2$ in our units, i.e., it's just $\phi / R \approx 10^{-13} m^{-1}$. The rate of increase will grow smaller inside the Sun, until it goes to zero at the center (again, the same result as for Newtonian gravity, although the exact shape of the curve will be slightly different), so the value of $\phi$ at the center will certainly be less than a straight linear extrapolation, which would give $\phi_{center} = 2 \phi_{surface} \approx 4 \times 10^{-6}$. So again, even at the center of the Sun gravity is weak.

(Note that $\phi$ is not exactly the same as spacetime curvature, but all of the curvature components will have more factors of $R$ in the denominator than $\phi$ does, so if $\phi$ is small, everything else will be small squared or small cubed--as we saw with the acceleration just now. So spacetime curvature even at the center of the Sun is also weak.)

They would emit gravitational radiation, yes.

I think you still could, because I think that point would be a "null point" in the gravitational waves--i.e., a point where they destructively interfere and so cancel each other out. But I don't know if anyone has actually done the math to verify this.

10. Oct 28, 2013

dauto

According to quantum mechanics all waves have quanta (particles) associated with them even if those waves have nothing to do with force transmission. Look at phonons which are particles associated with sound waves.

11. Oct 28, 2013

Bill_K

Even if the gravitational potential is constant but nonzero, the spacetime is still flat. As it is, for example, in the interior of a spherical shell of matter.

The midpoint between two black holes rotating about one another is an equilibrium point, in the same sense as the Langrange points. This means that a test particle placed at the origin will remain there, and in the context of GR this means it's the location of a geodesic. This is true simply because of symmetry, even if you take the gravitational waves into account.

Spacetime at the origin is not flat, however. The curvature tensor depends on the second derivative of the Newtonian potential, and that is nonzero.

12. Oct 28, 2013

A.T.

Yes. Slow objects advance mostly through time, so the time distortion has a larger effect. For photons the effects of space and time distortion are comparable.

13. Oct 28, 2013

Staff: Mentor

Yes, good point; the relationship between potential and curvature is in general more complicated than I implied in my post. In the particular case of a Lagrange point, the derivative of the potential only vanishes at a point, not in an extended region.

14. Oct 28, 2013

NotEuler

Thanks again everyone! It's all starting to make more sense, but I'll have to put some more time into re-reading and thinking about all your posts.

A question to those of you who have been thinking about these things for years: is it possible, even in principle, to 'visualize' 4D spacetime geometry? I wonder if it's possible to intuitively understand everything about gravitation as geometry, or is it best to use geometry and force as alternative ways of thinking about it (to prevent eventual insanity from brain overload)?

15. Oct 28, 2013

Staff: Mentor

If you take "visualize" literally, then no, at least not with a standard issue human brain.

I think yes, because there are other ways of intuitively understanding it besides trying to visualize all 4 dimensions at once.

16. Oct 29, 2013

NotEuler

I'm guessing you mean using symmetry, for example? I suppose you could, in principle, then visualize what's going on around a non-spinning black hole using just 2 dimensions... Everything would have to be symmetric if you're at a distance r from the center, and you'd just be left with one spatial dimension (distance from the center) and one time dimension. Am I on the right track here?
Not that I actually understand what it would look like, even in that 2d-representation! I always had trouble understanding infinite curvature, or 'spacetime curling around itself' or whatever description is used for black holes.

Speaking of confusing descriptions... what does it mean when it is stated that in general relativity, coordinates don't have a direct physical meaning, whereas in special relativity they do?

17. Oct 29, 2013

A.T.

Yes, you have to go down to 2D, which you can embedd in 3D to visualize the non-Euclidean geometry.

Check this out.
http://relativitet.se/spacetime1.html

18. Oct 29, 2013

Staff: Mentor

That's one way of doing it, yes.

Another way is by looking at the motion of test particles, i.e., objects that are so small that their mass does not measurably affect the spacetime curvature. Start two such objects near each other and at mutual rest, and watch what happens to them: by picking different initial conditions, you can explore different aspects of spacetime curvature. Two quick examples for the spacetime around a nonspinning black hole:

(1) Start the two objects along the same radial line, but at slightly different altitudes; they will gradually separate as they fall towards the black hole.

(2) Start the two objects at the same altitude, but on slightly different radial lines (i.e., with a small tangential separation). They will gradually get closer together as they fall towards the hole.

What these examples illustrate is that spacetime curvature is really the same thing as tidal gravity. So anything that helps you visualize tidal gravity also helps you visualize spacetime curvature.

Each of my two examples of tidal gravity gives a visualization of the curvature in two dimensions (but a different two in each case):

(1) gives a visualization of curvature in the t-r plane (i.e., the time and radial dimensions); the curvature is negative, like a saddle (imagine the r dimension running along the length of the saddle, and the t dimension running sideways--the initial time at which both particles are at rest corresponds to the centerline of the saddle).

(2) gives a visualization of curvature in the t-phi plane (i.e., the time and one of the tangential dimensions); this curvature is positive, like a sphere (imagine the r dimension running along lines of latitude and the t dimension running along lines of longitude--the initial time at which both particles are at rest corresponds to the equator).

Notice that both of these examples illustrate that it is *spacetime* curvature, not just *space* curvature, that's important.

I've never seen it stated in quite that way; do you have a source?

The way I would state it is that coordinates do not *have* to have a direct physical meaning in either GR or SR; but if we pick the coordinates carefully, they *can* have a direct physical meaning in both GR and SR. The difference is that in SR (more precisely, in flat Minkowski spacetime), it is possible to find coordinates that are globally inertial; in curved spacetime, it's not. Since globally inertial coordinates match our intuitions about coordinates having a "direct physical meaning", that makes SR easier to intuitively grasp using coordinates.

19. Oct 29, 2013

NotEuler

Thanks again for the replies!

I can look up a source later on today. Could be that I've also slightly misquoted, what I wrote was my interpretation of it. Anyway, I'll get back to this soon.

20. Oct 29, 2013

NotEuler

I still don't have the book I was referring to here, but a google search came up with this:
http://arxiv.org/abs/1308.0394

First sentence of the abstract: "Every general relativity textbook emphasizes that coordinates have no physical meaning".