View Full Version : What is gravity?

Klausnh

2003-Mar-15, 08:11 PM

I’ve been reading about gravity and I’m not sure what the current accepted theory of gravity is. As I understand it, Einstein “treats the gravitational field as a field of geometrical distortion, a curvature or warping of space-time.” (Superstrings A Theory of Everything? Edited by PCW Davies and J Brown, Pg 16).

But I’ve also read of gravitons, which carry the gravitational force.

Are these 2 ideas from 2 different theories? If gravity the result of a geometrical distortion of space-time, why is there a need for gravitons? Every book and article I’ve read, make it sound like gravity is includes both geometry and particles.

JS Princeton

2003-Mar-15, 08:24 PM

Gravitons are needed if we are ever to marry quantum mechanics and GR. We don't have a consistent theory for this yet. Observation of gravity waves might shed some light on the subject as may the search for the Higgs Boson. As it stands right now, we're not sure exactly how to formulate gravity.

Basically, there are two ways forces have been explained. One was is through Einstein's geodesic formalism, that is that the "force" is really not a force at all but rather the result of the curvature of space time. The other way is to explain the "force" as an exchange of bosons (be the photons, W-bosons, or gluons). So, unification theorists have to deal with both formalisms if they are to make any headway.

Klausnh

2003-Mar-15, 11:23 PM

As I understand it, according to Einstein, space-time is distorted by mass at the speed of light?

I came across a web article saying that was proven in experiment having to do with Jupiter. (Unfortunatley I lost the link and don't remember the details)

JS Princeton

2003-Mar-16, 01:51 AM

Well, the experiment I was pretty enthusiastic about involving Jupiter that I heard about this January has been bogged down by quite a lot of skpetical prodding. We're not sure if they actually measured the speed of gravity or not. The jury's techinically still out.

kilopi

2003-Mar-16, 02:34 AM

On 2003-03-15 15:24, JS Princeton wrote:

Basically, there are two ways forces have been explained. One was is through Einstein's geodesic formalism, that is that the "force" is really not a force at all but rather the result of the curvature of space time. The other way is to explain the "force" as an exchange of bosons (be the photons, W-bosons, or gluons). So, unification theorists have to deal with both formalisms if they are to make any headway.

And, it's entirely possible that what looks like two distinct ways of explaning gravity are actually the same--similar to the situation with quantum mechanics with Schrödinger's wave mechanics and Heisenberg's matrix mechanics, which turned out to be equivalent.

Klausnh

2003-Mar-16, 04:02 AM

On 2003-03-15 21:34 kilopi wrote:

And, it's entirely possible that what looks like two distinct ways of explaning gravity are actually the same

How do gravitons fit into geometric space-time?

Klausnh

2003-Mar-16, 04:20 AM

Theory is that space-time is distorted by mass at the speed of light. I read here in a post, that to a photon, travelling at the speed of light, because of relativity, the trip from a to b is instantaneous. Does it follow that to a person, on a mass (planet), the gravitational attraction is instantaneous because spacetime is distorted at the speed of light?

JS Princeton

2003-Mar-16, 04:41 AM

On 2003-03-15 23:20, Klausnh wrote:

Does it follow that to a person, on a mass (planet), the gravitational attraction is instantaneous because spacetime is distorted at the speed of light?

No, you have mass.

kilopi

2003-Mar-16, 05:16 AM

On 2003-03-15 23:02, Klausnh wrote:

On 2003-03-15 21:34 kilopi wrote:

And, it's entirely possible that what looks like two distinct ways of explaning gravity are actually the same

How do gravitons fit into geometric space-time?

Key word is "possible"--if only I knew, then I wouldn't have to worry about my retirement nest egg.

DStahl

2003-Mar-16, 05:53 AM

We've talked about this before, but it's good stuff to go over again. Just as kilopi says, physicists can and do think of gravity in different ways when they're working on different problems. The relativity specialist Kip Thorne wrote that he finds it very useful to think of gravity as a field in flat spacetime when he's working on gravity waves, for instance, and as a curvature of spacetime when he's working on a black-hole problem.

But as he notes, those two ways of describing gravity--as a field, and as a curvature--turn out to be mathematically exactly the same. They make identical predictions about the behavior of the universe, and so which one is the "real" description of gravity can't be decided scientifically. It's meaningless.

Trouble is, our current descriptions of gravity fail mathematically under certain situations. What's needed, in some opinions, is a theory of gravity that works like the theory of quantum mechanics. Now, it's widely believed that when such a theory is constructed, it will give exactly the same predictions as General Relativity for the scales and regions of the universe in which GR holds true. In a sense it will add another tool to physicists' toolbox, but the mathematics of spacetime curvature, of gravitational fields in flat spacetime, and of "graviton mechanics" will not only overlap they will be mathematically identical for nearly all large-scale situations in the universe. They will all be realistic mathematical models of the mysterious thing philosophers call...<font color=#9900aa>ULTIMATE PURPLE REALITY!</font>...with a /phpBB/images/smiles/icon_wink.gif

<font size=-1>[ This Message was edited by: DStahl on 2003-03-16 00:57 ]</font>

Klausnh

2003-Mar-16, 07:53 PM

On 2003-03-15 23:20, Klausnh wrote:

Does it follow that to a person, on a mass (planet), the gravitational attraction is instantaneous because spacetime is distorted at the speed of light?

On 2003-03-15 23:41 JS Princeton answered:

No, you have mass

I do not understand your answer. Do you mean that any space-time distortion involving mass does not distort at c?

Let me try another approach:

Let me know if any of the following statements are incorrect:

1) c is the upper limit of any object moving within space-time

2) Space-time distortion is not limited by c. (General relativity)

3) Space-time expansion is not limited by c. (That’s why inflation is possible)

4) If gravity is the effect of distorted space-time, an instantaneous interaction between 2 masses is allowed because Space-time distortion is not limited by c and does not take place within space-time.

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JS Princeton

2003-Mar-16, 08:37 PM

On 2003-03-16 14:53, Klausnh wrote:

2) Space-time distortion is not limited by c. (General relativity)

Sure, it's limitted by c. That's why we've got black holes. In Einstein's treatment, gravity (and therefore distortion) is a strictly speed-of-light phenomenon.

kilopi

2003-Mar-16, 08:37 PM

On 2003-03-16 14:53, Klausnh wrote:

Let me know if any of the following statements are incorrect:

1) c is the upper limit of any object moving within space-time

Depends upon your reference frame.

JS Princeton

2003-Mar-16, 08:39 PM

Let's stick to local reference frames. I think they're easier to talk about. The minute we go global, some people's eyes tend to glaze over.

kilopi

2003-Mar-16, 08:47 PM

On 2003-03-16 15:39, JS Princeton wrote:

Let's stick to local reference frames. I think they're easier to talk about. The minute we go global, some people's eyes tend to glaze over.

Still depends upon your reference frame. /phpBB/images/smiles/icon_smile.gif

JS Princeton

2003-Mar-16, 09:10 PM

No, kilopi. I'm putting my foot down here. What I think you're insinuating is not true, and if you aren't insinuating this then you need to post more than a glib one sentence answer. In a local reference frame the speed of light is the fastest speed. That's the definition of a local frame. Anything else is nearly impossible to define.

I assume, of course, you are referring to the old arguments we've had in the past which include the following (Original Source (http://www.math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html):

Stand up in a clear space and spin round. It is not too difficult to turn at one revolution each two seconds. Suppose the moon is on the horizon. How fast is it spinning round your head? It is about 385,000

km away so the answer is 1.21 million km/s, which is more than four times the speed of light! It sounds ridiculous to say that the moon is going round your head when really it is you who is turning, but

according to general relativity all co-ordinate systems are equally valid including revolving ones. So isn't the moon going faster than the speed of light? This is quite difficult to account for.

What it comes down to, is the fact that velocities in different places cannot be directly compared in general relativity. Notice that the moon is not overtaking the light in its own locality. The velocity of the

moon can only be compared to the velocity relative to other objects in its own local inertial frame. Indeed, the concept of velocity is not a very useful one in general relativity and this makes it difficult to

define what "faster than light" means. Even the statement that "the speed of light is constant" is open to interpretation in general relativity. Einstein himself in his book "Relativity: the special and the

general theory" said that the statement cannot claim unlimited validity (pg 76). When there is no absolute definition of time and distance it is not so clear how speeds should be determined.

Nevertheless, the modern interpretation is that the speed of light is constant in general relativity and this statement is a tautology given that standard units of distance and time are related by the speed of

light. The moon is given to be moving slower than light because it remains within the future light cone propagating from its position at any instant.

That's why I was careful to say "local". Once you move non-local, all bets on how to calculate "velocities" are off.

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kilopi

2003-Mar-16, 09:19 PM

On 2003-03-16 16:10, JS Princeton wrote:

No, kilopi. I'm putting my foot down here. What I think you're insinuating is not true, and if you aren't insinuating this then you need to post more than a glib one sentence answer. In a local reference frame the speed of light is the fastest speed. That's the definition of a local frame. Anything else is nearly impossible to define.

Watch where you put that foot down. That might be the definition of a local inertial reference frame. Big difference.

As to whether they are impossible to define, it would seem that engineers might have trouble with it but physicists don't, in the large.

DStahl

2003-Mar-16, 09:31 PM

As I recall, there is a discussion on John Baez's physics website about the "point of view" of a photon, and the upshot is, no such "point of view" is possible. Specifically, time becomes mathematically meaningless at c, if I remember the discussion correctly. And of course nothing with mass can reach c, so in a similar way it's meaningless to talk about what would happen to a ball bearing traveling at c: the situation is undefined.

To put it incoherently, I think you have to be very careful to define your reference frames if you want to describe relativistic effects like mass increase and foreshortening. You have to define the reference frame of the observer and its relationship to the observed thing-a-gummy. But yes: to an observer sitting in the control room of a particle accelerator, the protons which have been boosted to 99% c look more massive than the protons in his fingertip. But from the reference frame of the accelerated protons, they are "normal" but the gold atoms in the target block are more massive than they "should" be and are foreshortened--distinctly pancake-shaped rather than spherical. So I think whether or not spacetime looks distorted depends on where you're sitting.

[Jeez, you guys wrote a bunch of posts while I was writing this one! Now I'll go back and see just how redundant this one is...]

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JS Princeton

2003-Mar-16, 09:49 PM

The local frame is AUTOMATICALLY inertial. Otherwise, how do you define local?

FROM http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec9.html

(Existence of a Local Inertial Frame) If m is any point in the Riemannian manifold M, then there exists a local coordinate system xi at m such that:

(a) g<sub>ij</sub>(m) = ±1

if j = i

; 0

if j i

= ±ij

(b)g<sub>ij,k</sub>) = 0 for every k.

We call such a coordinate system a local inertial frame or a normal frame.

How else would you, kilopi, define a local frame so it's not inertial?

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Eta C

2003-Mar-16, 09:51 PM

We're talking about two distinctly different things here. One is the local curvature of space-time caused by mass. In general relativity this is what causes the force we observe as gravity. Quantum gravity theories try to express gravity in the same way the other forces (electro-magnetic, weak nuclear, and color) are, that is by the exchange of an "intermediate vector boson." For EM that's the photon. For quantum gravity it's the still hypothetical graviton. Since gravity has no range limitation, the graviton should be massless and therefore move at c, which is possible only for massless particles.

The other issue is the speed of light in any reference frame. In special relativity this is the same for all intertial frames regardless of the frame's velocity as measured by another observer. Imagine someone turning on a flashlight on a train moving at .95c, both the observer on the train and one on the ground will measure the same speed for the light. I'm pretty sure this still holds true for accelerated or non-inertial frames, which are the realm of general relativity (back to the texts to be sure.)

JS Princeton

2003-Mar-16, 09:57 PM

I think that what may be important to realize is that while one may have a second derivative of position (and therefore a gradient or non-zero curvature) at any given point (Einstein's Equivalence Principle), the local frame will always look inertial from the most pasic differential point of view. Otherwise, there'd be a problem with discontinuities.

Eta C

2003-Mar-16, 10:09 PM

As an attempt to answer a couple of these questions from klausnh

Let me know if any of the following statements are incorrect:

1) c is the upper limit of any object moving within space-time

2) Space-time distortion is not limited by c. (General relativity)

3) Space-time expansion is not limited by c. (That’s why inflation is possible)

4) If gravity is the effect of distorted space-time, an instantaneous interaction between 2 masses is allowed because Space-time distortion is not limited by c and does not take place within space-time.

1) yes. c is the upper limit on the speed of a particle.

2) Space-time distortion really has nothing to do with the speed of an object except to the amount that its speed increases its mass due to relativistic effects ( the effective mass of an object being M0 / sqrt(1-v^2/c^2) ). Note that as v goes to c this expression diverges.

3)Not sure I can answer this one. I know that during the inflationary epoch the apparent speed of expansion exceeded c, but this was for space itself, not of an object moving through space.

4) In a sense, the two objects were always interacting. Each has its effect on space-time. So to speak of an instantaneous interaction is meaningless in this context.

kilopi

2003-Mar-16, 10:16 PM

On 2003-03-16 16:49, JS Princeton wrote:

The local frame is AUTOMATICALLY inertial. Otherwise, how do you define local?

Local usually has a small area of influence, sometimes involves epsilons and deltas.

How else would you, kilopi, define a local frame so it's not inertial?

As long as the local frame doesn't satisfy that definition, it's not inertial. Just make up a few constants and throw 'em in there.

On 2003-03-16 16:51, Eta C wrote:

Imagine someone turning on a flashlight on a train moving at .95c, both the observer on the train and one on the ground will measure the same speed for the light. I'm pretty sure this still holds true for accelerated or non-inertial frames, which are the realm of general relativity (back to the texts to be sure.)

It does not hold true. The local speed of light is a function of the curvature of space. If spacetime is flat, under the local reference frame, the speed of light is c. Of course, you can always find such a reference frame, but it's not necessary and, in some circumstances, could be misleading since that local frame might not match up with the observer's frame.

JS Princeton

2003-Mar-16, 10:40 PM

You can't simply throw in constants to change the definition, because then you aren't in a local frame anymore. If my x<sub>1</sub> dimension is 3 times my x<sub>2</sub> dimension, then I'm not in a well-defined local frame. Yep, epislons and deltas are what I'm dealing with.

The gs are very well-defined things, kilopi, and I don't see how you can treat them so cavalierly.

kilopi

2003-Mar-16, 10:48 PM

On 2003-03-16 17:40, JS Princeton wrote:

The gs are very well-defined things, kilopi, and I don't see how you can treat them so cavalierly.

I'm not. I wasn't arbitrarily taking a given reference frame and throwing in constants--I was using that as an example.

It's possible to transform them to a noninertial reference frame, correct?

JS Princeton

2003-Mar-17, 02:10 AM

I don't know how one would transform a local inertial frame into a non-inertial frame while preserving locality. I don't think it's possible, but I could be wrong.

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kilopi

2003-Mar-17, 02:29 AM

On 2003-03-16 21:10, JS Princeton wrote:

I don't know how one would transform a local inertial frame into a non-inertial frame while preserving locality. I don't think it's possible, but I could be wrong.

Perhaps we differ on the definition of locality. What does it mean to preserve locality?

JS Princeton

2003-Mar-17, 03:06 AM

Well, in as much as locality is defined not discretely but by differentials, I'd say that it means one cannot have a spacelike separation between the two events. However, the point may be that reality could be quantized and then somehow you could invent, on the quantum level, some transformation between an intertial and a non-inertial local frame. I don't have the know-how to do that. I do, however, know how to manipulate a well defined derivative in my g<sub>ij</sub>.

kilopi

2003-Mar-17, 01:05 PM

On 2003-03-16 22:06, JS Princeton wrote:

Well, in as much as locality is defined not discretely but by differentials, I'd say that it means one cannot have a spacelike separation between the two events.

Is that the definition of locality?

JS Princeton

2003-Mar-17, 04:37 PM

Locality, unless we have a dicrete universe, is necessarily defined to be infinitessimal.

kilopi

2003-Mar-17, 04:55 PM

OK, infinitesimal time too, right?

But how does that prevent you from transforming frames?

JS Princeton

2003-Mar-17, 07:02 PM

On 2003-03-17 11:55, kilopi wrote:

OK, infinitesimal time too, right?

Only to make a measurement. But in principle the transformation would require only a non-zero spacelike separation. A timelike separation shouldn't affect the frame.

[/quote]

But how does that prevent you from transforming frames?

[/quote]

You can transform frames. You'll just always be in an inertial one.

kilopi

2003-Mar-17, 08:51 PM

On 2003-03-17 14:02, JS Princeton wrote:

Only to make a measurement. But in principle the transformation would require only a non-zero spacelike separation. A timelike separation shouldn't affect the frame.

How are you going to avoid a spacelike separation if you have a timelike separation? You're not using an immobile geocentrist frame are you? /phpBB/images/smiles/icon_smile.gif

You can transform frames. You'll just always be in an inertial one.

What's preventing using a noninertial reference frame?

JS Princeton

2003-Mar-17, 09:29 PM

On 2003-03-17 15:51, kilopi wrote:

On 2003-03-17 14:02, JS Princeton wrote:

Only to make a measurement. But in principle the transformation would require only a non-zero spacelike separation. A timelike separation shouldn't affect the frame.

How are you going to avoid a spacelike separation if you have a timelike separation? You're not using an immobile geocentrist frame are you? /phpBB/images/smiles/icon_smile.gif

Nah. The way one can avoid it is to simply make sure one can get there in the future light cone (at a constant velocity). Since we're dealing in a locally flat (inertial) frame, you have to take covariant derivatives all the way, but the transformation is fine, in principle. There is no issue for violating the speed of light because the separation is necessarily TIMELIKE!

You can transform frames. You'll just always be in an inertial one.

What's preventing using a noninertial reference frame?

The fact that I'm staing infinitessimal, for one.

kilopi

2003-Mar-17, 09:36 PM

On 2003-03-17 16:29, JS Princeton wrote:

The fact that I'm staing infinitessimal, for one.

Noninertial reference frames don't have infinitessimals?

JS Princeton

2003-Mar-18, 03:35 AM

When you get to the point of the infinitessimal, the frame becomes inertial. That's why I posted that quote from the GR webpage above.

It's the way GR is connected to SR... at the infintessimal level, GR reduces to SR. Just like on the Earth, the local infinetessimal measurement of the metric is flat.

kilopi

2003-Mar-18, 02:35 PM

On 2003-03-17 22:35, JS Princeton wrote:

When you get to the point of the infinitessimal, the frame becomes inertial. That's why I posted that quote from the GR webpage above.

I think I see what you're getting at. But I disagree.

The frame doesn't become inertial--but you can find a frame that is inertial. Just use a noninertial reference frame, and you have a noninertial frame. The results are the same--that's GR.

It may be simpler (for various reasons) to use an inertial frame, but you don't have to.

JS Princeton

2003-Mar-18, 03:32 PM

The point of what I'm saying, kilopi, is that when you ultimately get to a LOCAL frame, the frame is necessarily inertial. Go back and look at the theorem I posted. It states you can find this inertial frame anywhere, and the way it is found is by simply taking the appropriate derivatives. That's what we consider the local frame. I don't know of any other frame that can be considered local in the same sense of the word. Therefore, I don't buy the argument that you can simply create a local frame that's noninertial. Noninertial frames exist, but they are never the local type as the inertial frames I defined on the last page are.

JS Princeton

2003-Mar-18, 03:34 PM

Let me be more clear. I don't think you can define a local frame by means of simply taking appropriate derivatives and get off-diagonal terms in you g<sub>i,j</sub> matrix. If you can prove otherwise, I'd be glad to see that proof.

kilopi

2003-Mar-18, 04:44 PM

Local reference frames seem to be defined by how they are applied, rather than by their components. You can transform any inertial reference frame into a non-inertial reference frame, just use an accelerated reference.

JS Princeton

2003-Mar-18, 06:39 PM

kilopi, "local" indeed can be a confusing term. However, I'm referring to it in the strictest sense as I described on the previous page and in my last post. Indeed, you can describe metrics that do not conform to diagonal matrices. That's not the issue. The issue is at the FUNDAMENTAL level (that is, taking the differential at any point) you will see that it all reduces to flat. This is simply a statement of calculus if you like: you can always find a tangent to a curve, and the most local describtion of the curve will be identical to the tangent (in the limit).

The "law of local inertial frames" implies directly that space is described in a continuous and differentiable fashion. I understand that one can transform into a noninertial frame if you are dealing with other points of view... but there is a neccessary spacelike separation involved then (you need to determine where your other point of view is from). This was expressly forbidden by any transformation of the local inertial frame we initially described.

There is one caveat, this technique only works in the limit and we don't know how far we can take limits. However, any theory that claims that we can't take the limits arbitrarily will end up probably explaining quantum gravity and so will be farther along than we are now.

Charlie in Dayton

2003-Mar-18, 08:12 PM

Gravity is that force that demonstrates itself when you loosen your grip on the building brick held a meter or so directly above your toes.../phpBB/images/smiles/icon_smile.gif

kilopi

2003-Mar-18, 09:23 PM

On 2003-03-18 13:39, JS Princeton wrote:

I understand that one can transform into a noninertial frame if you are dealing with other points of view... but there is a neccessary spacelike separation involved then (you need to determine where your other point of view is from).

I disagree. The noninertial POV can be from the same POV as the inertial--if POV means Point of View, and not Reference Frame. So, you can have local noninertial reference frames. That is, noninertial reference frames that are applied locally.

JS Princeton

2003-Mar-18, 09:47 PM

On 2003-03-18 16:23, kilopi wrote:

I disagree. The noninertial POV can be from the same POV as the inertial--if POV means Point of View, and not Reference Frame.

Technically correct, but then you aren't as local as you used to be. Do you understand why?

Think of it this way. Take a sphere and look at the tangent plane to the sphere. It only touches the sphere at a single point. The sphere itself is non-inertial but the plane is flat.

Okay, so at that point you might say you'd want to switch to a non-inertial frame. Specifically, switch to the frame of the sphere.

However, the sphere and the plane are NECESSARILY the same thing in the limit of that point. You cannot determine the difference between the two of them. In particular, in the limit the only information you have is that the surface is flat. This is a simple result of the fact that you are in the limit and not describing the feature in another manner.

In order to describe this differently, you'd have to switch reference frames to some other system that had off-diagonal terms in your metric. This is all fine and dandy, but you simply don't have the information to do it. In particular, your metric at that local point will always read the same thing if you sit at the limit. So I have no idea how one could transform to a metric that wouldn't be diagonalized. If you have an idea, please share it.

What I know CAN be done is transferring to a metric that takes into account the curvature of the sphere (going back to our analogy). BUT... BUT.... in order to do this one has to have a second point because the only way to have a consistent frame of reference is to measure the metric difference (measure the curvature). Otherwise you aren't describing reality.

There's something very fundamental about this question, in a sense. Probably in the most technical sense, you're right since one could have any function at any given point. After all, it's two points that define a line, not one point, and the only way you can measure is if you are given two points. But I'm maintaining that we're staying local, so you aren't allowed off that point. The only information you have then is due to the time-like separation between events, which will necessarily always march forward with the diagonalized metric since you aren't changing reference frames. This is the issue: the localization of the point is so precise as to be trivial mathematically. So we are left with flat.

This, of course, makes no sense physically, but it is the way the math must work out, I'm fairly certain.

So, you can have local noninertial reference frames. That is, noninertial reference frames that are applied locally.

I agreee, only when you take the derivative (get closer and closer to having your metric being taken in the same spot) the answer is you ALWAYS get it to be flat. Always. That's just the rules of limits in calculus.

JS Princeton

2003-Mar-18, 10:09 PM

Let me put it a completely differnt way:

Let's say you are in a freefall rocketship... You are in a noninertial frame of constant acceleration.

Now prove that you are in a noninertial frame. What can you do? Well, you have to make a measurement from the outside. You can determine that the rest of the universe is behaving oddly (non-inertially). Of course, you think that you are moving in the flat universe. Now you can switch reference frames, fine, from the inertial to the noninertial since you've been given information about the rest of the universe.

But inasmuch as the Equivalence Principle applies, there is no way to deviate from flat without being given an observation outside of your own reference frame (or more techinically, within ds = sqrt[dx<sub>i</sub><sup>2</sup> - dt<sup>2</sup>] from your reference frame). You will be in a flat frame. Now you can PRETEND that you are seeing other things and SWITCH to that reference frame that is non-inertial, but that's an arbitrary switch for which you can make no claim. In effect, without having the observation you have no way of determining whether that's a meaningful reference frame, and as it only describes things outside of the ds<sup>2</sup> area, you are out of luck.

So, this gets back to the original question of whether the speed of light is constant at any given point in space and whether it depends on your reference frame. Here I am, falling in my reference frame and I measure the speed of light. The only thing I can do is use the metric I'm given to do it since I can only get as far away as ds from it. Actually, I can actually get a bit farther than that, as long as I keep my dx<sub>i</sub> terms in the limit, I can have, basically, delta(s)=delta(t) (modulo i, I suppose). The metric I'm given is necessarily flat by Einstein's Equivalence Principle. I calculate that light is the fastest thing there is between me and any event I'm allowed to measure in order to maintain locality.

The physical problem is when I go to do the experiment, I can't get any data. Since I'm sitting at the local point in the limit, the light beam will leave me and I'll never see it again. The measurement that occurs ds away is a different measurement mathematically, but it's not clear whether it's a different one physically.

Where does that leave us? With no information?

I'd say, "no" because mathematically the reference frame is well defined by a flat metric in the limit. Remember that ds does STILL represent a change... it may be infintessimally small, but it's still a change. If there was a noninertial term, our ds would look different. Macroscopically, at a different point in space and then comparing notes I might be able to determine that I'm in a gravitational field and that I'm in freefall, but if I compare notes with someone ds away from me I will get "flat" for an answer and only "flat".

<font size=-1>[ This Message was edited by: JS Princeton on 2003-03-18 17:12 ]</font>

AgoraBasta

2003-Mar-18, 11:24 PM

On 2003-03-18 17:09, JS Princeton wrote:

I'd say, "no" because mathematically the reference frame is well defined by a flat metric in the limit.JS,

You've just defined a non-inertial frame - it's the one where the metric is non-diagonizable.

Regarding the equivalence principle, I must say that effects of gravity always differ from effects of accelerations exactly by the tidal forces, which are just one order higher a derivative of the metric and exactly the components of the field equations. So one must always remember that GR tells how exactly there's no "equivalent of equivalence" for the higher order derivatives.

JS Princeton

2003-Mar-18, 11:31 PM

On 2003-03-18 18:24, AgoraBasta wrote:

JS,

You've just defined a non-inertial frame - it's the one where the metric is non-diagonizable.

Right. In the limit of the differential, the frame is inertial, though.

Regarding the equivalence principle, I must say that effects of gravity always differ from effects of accelerations exactly by the tidal forces,

Incorrect. Tidal forces are also present in regimes where accelerations are applied in the "same way" the gravitation occurs. In particular, you can have accelerations which produce tidal effects, no problem. You just need to have higher order derivatives which is fine if your acceleration isn't constant.

which are just one order higher a derivative of the metric and exactly the components of the field equations.

Yep, but also perfectly describably dynamicaly.

So one must always remember that GR tells how exactly there's no "equivalent of equivalence" for the higher order derivatives.

Incorrect. One can have a "freely falling" reference frame that has a change in acceleration or an acceleration that has higher order derivatives. It's all equivalent-like. Of course, "freely falling" must be in quotes since we now have higher order derivatives to contend with!

<font size=-1>[ This Message was edited by: JS Princeton on 2003-03-18 18:32 ]</font>

AgoraBasta

2003-Mar-18, 11:54 PM

On 2003-03-18 18:31, JS Princeton wrote:

Incorrect. Tidal forces are also present in regimes where accelerations are applied in the "same way" the gravitation occurs. In particular, you can have accelerations which produce tidal effects, no problem. You just need to have higher order derivatives which is fine if your acceleration isn't constant.For your objection to stand, you need a specifically non-uniform field of accelerations acting on your test body within it's finite size, in such conditions no equivalence can stand anyway. That's not possible if you simply have your ref frame accelerated.

Incorrect. One can have a "freely falling" reference frame that has a change in acceleration or an acceleration that has higher order derivatives. It's all equivalent-like. Of course, "freely falling" must be in quotes since we now have higher order derivatives to contend with! That's not right. Attach a ref frame to a chunk of matter, and you can measure the real tidal forces in it in case of gravity and you get no such forces if you simply jerk ref frame around. And you should know better that those forces may get very strong in some exotic cases...

<font size=-1>[ This Message was edited by: AgoraBasta on 2003-03-18 18:59 ]</font>

JS Princeton

2003-Mar-19, 01:56 AM

On 2003-03-18 18:54, AgoraBasta wrote:

For your objection to stand, you need a specifically non-uniform field of accelerations acting on your test body within it's finite size, in such conditions no equivalence can stand anyway. That's not possible if you simply have your ref frame accelerated.

Yes, equivalence stands because you can specify a such a field. In other words, If one tied ones' rocketship to a spring, you wouldn't be able to tell the difference. There is an equivalence between accelerations and gravity, even in the strong field limit.

That's not right. Attach a ref frame to a chunk of matter, and you can measure the real tidal forces in it in case of gravity and you get no such forces if you simply jerk ref frame around.

Sure you do. All you need is non-uniform accelerations in the radial direction. If your feet are accelerated at a different rate from your head, that's a tidal force. How do you know that it's due to gravity or due to different springs that have different spring constants attached to you at every point (an acceleration field?) The answer is, you don't.

Just because it's hard to visualize such a field doesn't mean that equivalence doesn't apply.

And you should know better that those forces may get very strong in some exotic cases...

Doesn't matter. It may seem strange, but the equivalence principle still applies. It's just an equivalence to a situation that is abnormal, but the gravitational situation is abnormal too.

AgoraBasta

2003-Mar-19, 09:06 AM

JS,

You are arguing against the very basics here. Tidal forces constitute a tensor, and for this very reason cannot be transformed away by coordinates choice. Sure you can, in principle, construct a tensor accelerations field "by hand", but that's not a free fall!

JS Princeton

2003-Mar-19, 02:28 PM

I know it's not free-fall, and I never said it was. The fact of the matter is, it is not constant acceleration. How could it be?

The fact that you can put in the accelerations "by hand" proves that it is possible, in principle, to make an equivalent non gravitational system that would mimic the strong limit. This is, of course, all incidental to the main point of the argument, but it's important to note that because there is equivalence and every given differentiated point (even with tidal forces), it is possible to invent an equivalent system.

AgoraBasta

2003-Mar-19, 04:20 PM

On 2003-03-19 09:28, JS Princeton wrote:

The fact that you can put in the accelerations "by hand" proves that it is possible, in principle, to make an equivalent non gravitational system that would mimic the strong limit.That's not much more than a tautology - tidal forces/accelerations are equivalent to accelerations. So einsteinian gravity is reduced to coordinate transform plus tidal forces, but newtonian gravity can be reduced in the very same way, so nothing's really different here. The only real difference is that all forces/fields other than gravity use the gravitationally modified metric as a background. Yet tidal forces always betray the presence of gravity even in free fall conditions.

JS Princeton

2003-Mar-19, 04:39 PM

You're contradicting yourself, Agora.

As you rightly stated previously, tidal forces cannot exist in "free fall" because you need uniform acceleration of the reference frame.

So if tidal forces are "giving away" the presence of the graviational field in "free fall", then we aren't in free fall, are we?

If your reference frame is not in the limit of uniform acceleration then you will experience tidal forces. But then it is not in the limit where the criteria for free fall are met. Therefore the Equivalence Principle looks different.

It's just like Newton's Second Law when taken macroscopically can be disproven. A good proof of this is looking at a rocket's acceleration. Since there is a mass and accleration gradient, you end up with tensor cross terms. Does this mean that Newton's Second Law isn't valid? No! It just means you aren't looking at it in the appropriate conditions. In fact, Newton's Second Law is behaving just as it is predicted it would for every given differential. It's just the macroscopic reference frame that leads to confusion (and the appropriate conditions).

The same thing holds true for Einstein's Gravity, which, for example, explains the precession of the perhelion of Mercury and the proper effect of gravitational lensing while Newtonian gravity does not.

One can also experience tidal forces due to a whole host of other phenomena (as you well know), so stating that it must be due to gravity is a misnomer. You could have some bound currents in you, for example, that would give you magnetic tidal effects that could mimic gravitational tidal effects in the right conditions.

There is no difference between the geodesic effects and gravity. That's the gist of GR.

<font size=-1>[ This Message was edited by: JS Princeton on 2003-03-19 11:45 ]</font>

AgoraBasta

2003-Mar-19, 06:01 PM

On 2003-03-19 11:39, JS Princeton wrote:

You're contradicting yourself, Agora.No, I'm not. You may wish to read about it here (http://math.ucr.edu/home/baez/PUB/efe).

JS Princeton

2003-Mar-19, 06:22 PM

From the site you mentioned (which is a good discussion, I'd say):

Another way to understand the relationship between Newtonian gravity and

gtr is to consider "tidal forces" near an isolated massive object such as

the Earth. In Newtonian gravity, our space capsule is actually very

slightly stressed by the differential between the "gravitational

attraction" on parts of the capsule which are a few meters closer to the

Earth and the farther parts. That is, the parts of the space capsule

which are a few meters closer to the Earth are attracted a bit more than

those which are further away, so the matter of the capsule experiences a

tiny "radial tension". In addition, if you compare the direction from the

front of the capsule to the center of mass of the Earth and the direction

from the rear of the capsule to the center of mass of the Earth, there is

a tiny difference between the "direction of gravitational attraction"

which cause a tiny -compression- between the front and rear of the

capsule. These tiny tensile and compressive forces are the tidal forces

acting on the space capsule.

If the distance from the center of mass of the capsule to the center of

mass of the Earth is R, the tidal forces scale like M/R^3, where M is the

mass of the Earth. Very similar tidal forces arise in gtr, and they scale

just the same way! The difference is that in Newtonian gravity, we think

of a "gravitational force" scaling like M/R^2, and say that -differences-

in these forces give rise to the tidal forces scaling like M/R^3, but in

gtr, there is no "gravitational force"; rather, there are only "tidal

forces", which turn out to be direct consequences of the "curvature" of

spacetime itself.

This is exactly right, but notice we're not talking about the equivalence principle here in terms of a free-falling coordinate system. That's because your macroscopic object is not in the "free fall limit"... that is, the tidal forces represent a deviation from the free fall. It is all due to the curvature, but, as we've been discussing, spacetime is locally flat no matter what funky metric you're sitting in. That's why Einstein's Equivalence Principle works for the limit, and by extension to the rest of a macroscopic system as long as you take appropriate integrals. ANY GIVEN POINT will be accelerating in free-fall, true, but the macroscopic system is not in free-fall. That's because a curved metric isn't flat.

Here's the meat of the matter:

in gtr there are always infinitely many

ways of slicing up (four dimensional) spacetime into (three dimensional)

surfaces in this way, and each slicing gives an equally valid description

of the spacetime! Furthermore, whereas in Newtonian physics "space" is

ordinary euclidean three dimensional space, in gtr, our three dimensional

"spaces at a time" will generally be -curved- three dimensional spaces,

and moreover, their geometry will in general "evolve over time".

If you take a point in this curved spacetime, you can define a tangent plane. It's these points that follow Einstein's Equivalence principle most closely, but, as with all discussions, we cannot envision points. So we switch to the weak-field limit (at least with respect to the size of the coordinate system) and try to ignor the calculus of curvature as well as we can. Since differentials are perfectly well defined, this excersize is perfectly descriptive. There's no extensions necessary. Notice, Baez doesn't mention anywhere that you need to "add" tidal effects to get GR to work. Rather, tidal effects are a direct consequence of GR's curvature relations.

<font size=-1>[ This Message was edited by: JS Princeton on 2003-03-19 13:23 ]</font>

AgoraBasta

2003-Mar-19, 06:55 PM

On 2003-03-19 13:22, JS Princeton wrote:

Notice, Baez doesn't mention anywhere that you need to "add" tidal effects to get GR to work. Rather, tidal effects are a direct consequence of GR's curvature relations.Here's another quote from the same page:

The central "law of physics" in gtr is an equation which says how concentrations of matter at each even, or more generally, concentrations of any form of (non-gravitational) mass-energy such as an electromagnetic field-- which carries energy, according to Maxwell's theory, and therefore gravitates--- curve the spacetime near that event, resulting in tidal forces. This law is called "Einstein's equation".

That's more or less the thing I was trying to tell you in our discussion here.

JS Princeton

2003-Mar-19, 07:36 PM

Oh, well, that doesn't matter in the Equivalence Principle discussion because we aren't dealing with the sources of gravity but rather with the effects. The reference frames are merely subject to the conditions of curvature. They are, for the purposes of a theoretical discussion, empty of energy and mass.

AgoraBasta

2003-Mar-19, 08:12 PM

On 2003-03-19 14:36, JS Princeton wrote:

Oh, well, that doesn't matter in the Equivalence Principle discussion because we aren't dealing with the sources of gravity but rather with the effects. The reference frames are merely subject to the conditions of curvature. They are, for the purposes of a theoretical discussion, empty of energy and mass.A few points - first, to measure the field one needs a test mass or something like that; second, an equation like G=0 (pseudotensor neglected) is mostly uninformative without boundary conditions. Once we get a test mass, we immediately get a tidal force tensor which is exactly the higher-order deviation from equivalence, and that deviation is quantified by the Einstein field equation. In fact, such quantification is one of the meanings of the field equation itself, the rest is free field connecting whatever massenergies are out there.

JS Princeton

2003-Mar-19, 08:32 PM

On 2003-03-19 15:12, AgoraBasta wrote:

A few points - first, to measure the field one needs a test mass or something like that;

Nobody's talking about measuring the field. Rather we're talking strictly theory here. The question is, is the most local frame you can define necessarily inertial? The answer is, yes.

second, an equation like G=0 (pseudotensor neglected) is mostly uninformative without boundary conditions. Once we get a test mass, we immediately get a tidal force tensor which is exactly the higher-order deviation from equivalence, and that deviation is quantified by the Einstein field equation.

Well, it's more like equations (there are dozens of those bad boys), but the fact remains that we are defining coordinate systems and not physical measurement.

In fact, such quantification is one of the meanings of the field equation itself, the rest is free field connecting whatever massenergies are out there.

That's right. So, if you ask, "What is the field at a given point?" then you give me an answer. In the limit where the test-particle goes to the vacuum, that's where we deal with our situation.

AgoraBasta

2003-Mar-19, 09:04 PM

On 2003-03-19 15:32, JS Princeton wrote:

Well, it's more like equations (there are dozens of those bad boys),...Sure, unless your tensor is a scalar /phpBB/images/smiles/icon_smile.gif

That's right. So, if you ask, "What is the field at a given point?" then you give me an answer. In the limit where the test-particle goes to the vacuum, that's where we deal with our situation.That's not a trivial question, actually. The stress-energy and the derivative of tidal tensor go to zero at the same speed; consider the simplest newtonian case - stress-energy reduces to mass and tidal forces go as mass*size/r^3, so no matter how small the test thingy - the ratio of tidal tensor derivative to test mass trends to a specific non-zero limit and thus is rather a property of spacetime than of that test thingy. Surely that's a higher order deviation, yet it has a profound physical meaning distinguishing gravity from accelerations at any given point.

JS Princeton

2003-Mar-19, 10:08 PM

On 2003-03-19 16:04, AgoraBasta wrote:

That's not a trivial question, actually. The stress-energy and the derivative of tidal tensor go to zero at the same speed;

If you keep density constant for your test mass, that's true. But I don't keep density (or the off-diagonal pressures either) constant. I consider the metric to be more important so all you have to do is choose your stress-energy tensor in such a way that the stress-energy tensor for your test-particle is empty. That's perfectly fine because the solutions to Einstein's Equations work in an empty metric.

consider the simplest newtonian case - stress-energy reduces to mass and tidal forces go as mass*size/r^3,

For any real object, sure. But I'm dealing with theory now. That's what we really care about anyway.

so no matter how small the test thingy - the ratio of tidal tensor derivative to test mass trends to a specific non-zero limit

Unless the stress-energy tensor goes to zero. Which I have determined it must do in order to stay consistent. After all, if you don't do that you're going to end up with a singularity, and I don't think you want us playing that game.

and thus is rather a property of spacetime than of that test thingy.

Only because there is a relationship of the stress-energy tensor of the "test thingy" with the curvature. However, if you demand all your densities and pressures go to zero the problem is no longer there.

Surely that's a higher order deviation, yet it has a profound physical meaning distinguishing gravity from accelerations at any given point.

Again, this is not true because you can invent any field of accelerations you choose. There is absolutely no way to distinguish between one kind of acceleration and another if you aren't allowed to look out the window. They're all the same.

AgoraBasta

2003-Mar-19, 11:07 PM

On 2003-03-19 17:08, JS Princeton wrote:

Only because there is a relationship of the stress-energy tensor of the "test thingy" with the curvature.But that construct as I described it carries no properties of the test body. It really is a local property of spacetime. The other aspect is that we know of nothing that would couple to that property value in a point-like interaction, but are we guaranteed against inventing such one?

Again, this is not true because you can invent any field of accelerations you choose.If I need to invent, I also necessarily need the means to transport that field at destination points/area in space, so I need some massenergy there to couple to my invention. It's impossible to invent an unphysical field in reality. It seems like a technological limitation, but it really is a fundamental one.

There is absolutely no way to distinguish between one kind of acceleration and another if you aren't allowed to look out the window.AFAIK, so far it was technologically possible to do using accelerometers/gravimeters in rather practical/practicable setups. Although you can always require the use of smaller test areas that go beyond the sensitivity of existing tools, you can as well demand development of higher sensitivity tools to perform measurements at those smaller scales. This particular task is purely technological and has no fundamental limitations (within classical physics).

JS Princeton

2003-Mar-20, 02:17 AM

Now, Agora, you're stearing the discussion towards an area I've admitted we don't know anything about. Either we accept the derivative limit or we don't.

You are not right in saying that your limit doesn't contain any information about the test particle, because you necessarily have a stress-energy tensory that is not null for the given test particle. It's a similar (and equally valid) criticism you give for the lack of "physicality" to a field of accelerations. How are you going to build such a thing without affecting the metric? Well, the answer is you cannot. However, we're dealing with theory. If you think it is impossible to set up a system that isn't gravitational and has tidal effects you are wrong. All I need to do is get enough ideal springs. There's nothing wrong with an ideal spring, is there?

This is why we deal with empty spaces first and build up from there.

In principle, the measurement of local inertial frames will have to be done at the quantum level to see if it's "true", but since we don't have a consistent theory to describe that at this time, I'm afraid there's nowhere left to go with this discussion.

AgoraBasta

2003-Mar-20, 02:41 PM

Well, JS,

I think we really are pretty close to consensus over the issues discussed in this thread. We sure could continue arguing over tinier yet details ad infinitum, I'm just not sure if it's worth our efforts /phpBB/images/smiles/icon_wink.gif

kilopi

2003-Mar-20, 04:12 PM

On 2003-03-20 09:41, AgoraBasta wrote:

I think we really are pretty close to consensus over the issues discussed in this thread.

Before you lock it, let me finish reading it. /phpBB/images/smiles/icon_smile.gif

kilopi

2003-May-02, 04:58 AM

Nobody's talking about measuring the field. Rather we're talking strictly theory here. The question is, is the most local frame you can define necessarily inertial? The answer is, yes.

What the h*ck, we've hashed this out before, but I still want to know, what in the world does it mean to be the "most" local frame? I don't see that as being a unique frame. And not necessarily inertial.

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