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Mass must have spatial extent


md65536

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Given that local spacetime is flat, there should be no possible way to measure any curvature at a single point.

With no curvature, there is no gravitation.

If a mass can be a point mass, it would need observations or measurements of spacetime from multiple locations in order to determine any curvature and be affected by it (eg. gravitational attraction). The minimal spatial extent of a mass must include multiple spatial points of observation (as in a particle with size, or multiple point particles somehow combined and sharing information, or a single point particle making observations from different locations).

 

Is this a sound argument?

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What about curvature and extent (duration) in timelike axes as well as spacelike ones?

 

If you could separate the axes, I suppose similar reasoning might imply that there is no way to measure time with a single spatial location, so no passage of time. I think I've argued this before and would try again, but others have pointed out that there is no known relation to spatial extent in some time-related processes like decay of particles.

 

So I would say that a point particle without spatial extent has no mass and does not age.

Then, if something has mass and by my reasoning has spatial extent, then that spatial extent also extends in time, because it makes no sense to speak of multiple locations in a single universal instant. (Unless somehow the locations were time dependent so every observer who sees the mass in a single universal instant would necessarily see the locations as different, compared to what other observers see. I don't see how that could make sense.)

 

What about curvature [...] in timelike axes

 

I think that curvature can only be expressed using multiple axes? And there's nothing that suggests more than one time axis? So any "curvature of time" would only make sense as a curvature of spacetime.

 

I think this is related to the fact that a local clock always ticks at the same rate. If you could measure a local clock deviation I think you'd be able to measure a local spacetime curvature. Similarly if you could measure a local spacetime curvature, you'd be able to specify an absolute direction of motion maybe??? Or at least acceleration. It almost seems like having a local spacetime curvature is like the idea of something moving away from itself/its own location.

 

So does this also mean that proper acceleration cannot be defined for a single point???

 

 

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I've read that certain types of pulsars rival atomic clocks in their accuracy in keeping time. If it were possible to theorize what the period of a particular pulsar would be for an observersitting in a gravity free (flat) region of space, and then one made a measurement of the pulsar's period at the point in space that the observer actually occupied, would it then not be possible to determine the gravitational force acting upon the observer and hence the curvature of space at the observer's location?

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I've read that certain types of pulsars rival atomic clocks in their accuracy in keeping time. If it were possible to theorize what the period of a particular pulsar would be for an observersitting in a gravity free (flat) region of space, and then one made a measurement of the pulsar's period at the point in space that the observer actually occupied, would it then not be possible to determine the gravitational force acting upon the observer and hence the curvature of space at the observer's location?

 

That would involve receiving information from a separate location, rather than detecting a gravitational field locally.

 

 

There is no curvature at the observer's (point) location, at least according to the observer.

 

 

Perhaps I'm confusing curvature and gravitational field. I know that a gravitational field means curved spacetime, but does detecting one mean you can detect the other? Also, a mass in free-fall wouldn't even need to know that it is being pulled toward something; it wouldn't need to know if the spacetime it's moving through is curved or not. So my idea's probably flawed.

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Given that local spacetime is flat, there should be no possible way to measure any curvature at a single point.

With no curvature, there is no gravitation.

If a mass can be a point mass, it would need observations or measurements of spacetime from multiple locations in order to determine any curvature and be affected by it (eg. gravitational attraction). The minimal spatial extent of a mass must include multiple spatial points of observation (as in a particle with size, or multiple point particles somehow combined and sharing information, or a single point particle making observations from different locations).

 

Is this a sound argument?

As it turns out, it isn't. The entire field of Riemannian geometry exists effectively to answer your question. Curvature can be defined entirely by what happens in an arbitrarily small area of a point - there is no minimum distance.

=Uncool-

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As it turns out, it isn't. The entire field of Riemannian geometry exists effectively to answer your question. Curvature can be defined entirely by what happens in an arbitrarily small area of a point - there is no minimum distance.

=Uncool-

I looked it up and came across: http://en.wikipedia....i/Tangent_space

They use an example of a tangent space of a point on a sphere, which is a plane.

 

I think what you're saying is that even though the tangent space is "flat" in this example as it is for a point in spacetime, the curvature of the manifold (the sphere or spacetime) is defined even at that point.

 

By analogy, my argument would be that such a point wouldn't be able to tell it was on a sphere. At a small enough scale, the area around the point becomes indistinguishable from a plane.

 

But I don't think that matters, because though the tangent space is flat, if the point is moved in any direction on that tangent space, the point moves on the sphere and not on the tangent space.

 

So if there was a point mass, it would follow the curvature of spacetime without having to measure the curvature at that point.

 

 

 

It's kinda over my head so I don't know if what I wrote makes sense. If so, does this imply that a point mass wouldn't "know" which direction to accelerate toward, but any movement at all (even random) would cause it to behave properly (ie. accelerate toward gravitational masses)?

 

If location is subject to Heisenburg uncertainty but curvature isn't, then perhaps a (point) mass could not be defined in a way that it can "remain at rest in a fixed location, experiencing local flat spacetime" and thus avoid gravity, so it wouldn't require any specific movement at all in order to behave as a mass does. --- Sorry if I'm quickly jumping into nonsense.

Edited by md65536
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I looked it up and came across: http://en.wikipedia.org/wiki/Tangent_space

They use an example of a tangent space of a point on a sphere, which is a plane.

 

I think what you're saying is that even though the tangent space is "flat" in this example as it is for a point in spacetime, the curvature of the manifold (the sphere or spacetime) is defined even at that point.

 

By analogy, my argument would be that such a point wouldn't be able to tell it was on a sphere. At a small enough scale, the area around the point becomes indistinguishable from a plane.[/QUOTe]

And that's false. It is possible to tell that you are on a sphere no matter how small an area you get.

 

One way to do it for a surface is the following: mark a circle of sufficiently small radius R. Find the circumference of the circle. When R is sufficiently small, the Gaussian curvature at that point will be approximately proportional to C/R - 2\pi.

 

This is one of the reasons that it is impossible to make a map without distorting distances somewhere, even if you do break up the sphere into arbitrarily many pieces - you will always have curvature.

=Uncool-

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One way to do it for a surface is the following: mark a circle of sufficiently small radius R. Find the circumference of the circle. When R is sufficiently small, the Gaussian curvature at that point will be approximately proportional to C/R - 2\pi.

So the Gaussian curvature is defined for a point, and wouldn't matter how small R was chosen?

The limit of the Gaussian curvature would be a non-zero number, as R approaches 0?

 

Still, your example and the ones on wikipedia (ants on a surface drawing a triangle and measuring the angles) requires making measurements from more than one location, which is exactly my point in this thread. Is there any way to measure the Gaussian curvature at a point on a surface without using measurements from more than one location? (Nor using some measure over a spatial extent other than a single point.)

Edited by md65536
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Now you are getting somewhere.

 

Curvature is only defined at a point, as the limit you describe.

It makes no sense to talk of the curvature at more than one point at a time since it may differe from point to point.

 

However in order for curvature to exist you need at least one extra dimension for it to exist in. Is this what you are trying to describe?

 

 

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Now you are getting somewhere.

 

Curvature is only defined at a point, as the limit you describe.

It makes no sense to talk of the curvature at more than one point at a time since it may differe from point to point.

 

However in order for curvature to exist you need at least one extra dimension for it to exist in. Is this what you are trying to describe?

That might be an important part of the answer but if so I'm not seeing it.

 

What I'm trying to describe is the notion that "local spacetime is flat".

So a point observer should not be able to measure any curvature using only local measurements.

If the spacetime curvature is defined and non-zero at that point, it can only be measured by other observers???, or by incorporating information from other observers.

 

What I don't get: Does spacetime have an intrinsic curvature (as in a Gaussian curvature) that is independent of how or from where it is measured? If so then what is the meaning of "local spacetime is flat"? Are both measures speaking of the same property, the same meaning of "curvature of spacetime"?

 

 

 

So as an example, if you have a function like "y=x", with a slope of 1, that slope can be measured at any x using only local measurements. If you sample the function at y1=x1, and at y2=x1+epsilon, you can determine the slope no matter how small epsilon is (as long as it's non-zero, so maybe my point is lost anyway because you couldn't measure the slope at a point even if the function isn't "flat" locally). The limit of the slope as epsilon approaches 0, is 1.

 

If local spacetime is flat, then the only curvature that should be measurable in the space around a point P + epsilon, should be 0 as epsilon approaches 0. The only local curvature that should be measurable is "completely flat".

 

I'm not sure now where my reasoning goes wrong.

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Given that local spacetime is flat, there should be no possible way to measure any curvature at a single point.

With no curvature, there is no gravitation.

If a mass can be a point mass, it would need observations or measurements of spacetime from multiple locations in order to determine any curvature and be affected by it (eg. gravitational attraction). The minimal spatial extent of a mass must include multiple spatial points of observation (as in a particle with size, or multiple point particles somehow combined and sharing information, or a single point particle making observations from different locations).

 

Is this a sound argument?

 

First off -

Given that local spacetime is flat, there should be no possible way to measure any curvature at a single point.

Your subject matter is differential geometry and it'sapplication to generl relativity. When I see someone say local spacetime is flat, I wince. :P The curvatue of a manifold is a local property. A local property such as this is primarily geometry in the small or local geometry.

 

The smaller the spacetime you confine yourself to the more precisely you can measure the curvature. There are articles about this in the American Journal of Physics as well as in a few GR texts. E.g

What is the principle of equivalence?, Hans C. Ohanian, Am. J. Phys. 45(10)), October 1977

Abstract - The strong principle of equivalence is usually formulated as an assertion that in a sufficiently small, freely falling laboratory the gravitational fields surrounding the laboratory cannot be detected. We show that this is false by presenting several simple examples of phenomena which may be used to detect the gravitational field through its tidal effects: we show that these effects are, in fact, local (observable in an arbitrarily small region). Alternative formulations of the strong principle are discussed and a new formulation of strong equivalence (the "Einstein principle") as an assertion about the field equations of physics, rather than an assertion about all laws or all experiments, is proposed. We also discuss the weak principle of equivalence and its two complimentary aspects: the uniqueness of free fall of a test particles in arbitrary gravitational fields ("Galileo principle") and the the uniqueness of free fall of arbitrary systems in weak gravitational fields ("Newton's principle").

I have the article in a PDF file if you want to read it.

 

Second --

With no curvature, there is no gravitation.

Not only do I disagree with this I know as fact that Einstein disagreed with it to. See my article http://xxx.lanl.gov/...physics/0204044 for details and Einstein's written references.

 

The important part of the paper regarding this point is where I quote Einstein who said

... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of the affine connection], not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.

Basically what all this says is that the observations you make in a free-fall enviroment when objects are accelerating towards/away from each other are observations of tidal forces. The equivalent GR name for such tidal gradients is spacetime curvature.

 

Third ---

If a mass can be a point mass, it would need observations or measurements of spacetime from multiple locations in order to determine any curvature and be affected by it (eg. gravitational attraction). The minimal spatial extent of a mass must include multiple spatial points of observation (as in a particle with size, or multiple point particles somehow combined and sharing information, or a single point particle making observations from different locations).

Sorry but I can't make sense out of this.

 

I assume by If a mass can be a point mass you are referring to a test particle. If so then I can't make out what the following means regarding it

(test particle) ... would need observations ... of spacetime from multiple locations ....

Are you observing the test particle or are you obseving spacetime?

 

The rest I can't understand at all

 

Is this a sound argument?

Sorry, but no, it isn't. :(

 

Please read my article from front page to back page and if you read it close enough and ask me questions when you think I'm unclear, then I'm sure we can come to s place in the topic in which we both agree. In the meantime I'm going to write a quote from a differential geometry text which defines some terminology so we have the same basis of terminology there too. That will have to wait a bit. My legs are giving out (due to spinal cord damage and sitting too long).

Edited by pmb
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I have the article in a PDF file if you want to read it.

Yes please! I obviously don't know the answer to the title of the article, and I can't imagine how tidal forces could be measured locally.

 

Thanks... I've only begun to look at your article, and I doubt I'll understand it all, but I guess I've had it all wrong. I thought spacetime curvature corresponded to the magnitude of gravitational force, but it only corresponds to the change in gravitational force?, or as you quoted Einstein: "The equivalent GR name for such tidal gradients is spacetime curvature."

 

So a uniform gravitational field would have no spacetime curvature? The paths of objects in freefall in such a field would be curved, but spacetime itself would be flat? And would geodesics or the path of a light signal would be straight according to any relatively stationary observer in the uniform field?

 

I assume by If a mass can be a point mass you are referring to a test particle. If so then I can't make out what the following means regarding it

 

Are you observing the test particle or are you obseving spacetime?

 

The rest I can't understand at all

I'll try to explain what I meant, with the realization that it is a flawed argument and I'm using the notion of spacetime curvature incorrectly.

 

Yes, I mean a point particle with mass, and a size of zero (not negligible size or infinitesimal, but actually zero).

 

I am speaking of the test particle observing or measuring local spacetime.

 

I'm basing my argument on the idea that if an object behaves a certain way due to a given environment, then the properties of the environment that cause the behavior must be measurable or observable by the object. If it is impossible to measure a property, then it is impossible to be "properly" affected by that property and behave expectedly.

 

However as I realized, this is not the case in my argument because the behavior of the test particle is relative, or different according to different observers. The "proper" behavior of a test mass is to behave as if it is in free fall. It would not have to (or even be able to) know if it was in freefall in the absence of gravity or in the presence of a uniform gravitational field (with an absence of tidal effects in either case). It would not need to be able to measure a gravitational field in order to behave expectedly.

 

Or the other way to see it, which might be related, is that the test particle may "know" its tangent space, and know how it can move, and if it moves it will follow the curvature of spacetime without ever needing to know what that curvature is.

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From Relativity, Thermodynamics and Cosmology by Richard C. Tolman, Dover Pub. (1987 version of 1934 text), page 29

 

 

In this language it is important to guard against the fallacy of assuming all directions in the hyper-space are equivalent, and of assuming that extension in time is of the same nature of extension in space merely because it may be convenient to think of them as plotted along perpendicular axes. …. That there must be a difference between the spatial and temporal axes in our hyper-space is made evident, by contrasting the physical possibility or rotating a meter stick from an orientation where it measures distance in eh y-direction, where the impossibility of rotating it into a direction where it would measure time-intervals-in other words the impossibility of rotating a stick into a clock.

This is an excellant explanation about what the cautions are regarding the nature of spacetime.

 

I recall countless times people speaking about the speed at which a photon travels through spacetime.

 

You can just as well talk about the speed at which light moves through spacetime as you can talk about rotating a clock into a rod. :P

 

 

Edited by pmb
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Thanks for all the information.

 

I'll try to revise the speculation. Sorry if I go off the rails again...

 

 

The proposition of this thread is wrong. Gravitational acceleration is not a local effect (a test mass accelerating in free-fall feels no proper acceleration) so a test mass need not measure a gravitational force in order to be affected by it.

 

Gravitational acceleration is therefore a relative effect, such as can be observed as a difference in acceleration between two spatially separate masses. (So there we have the main idea, that gravitational acceleration may be meaningless without multiple points of observation.)

 

Further speculation: All gravitation does require non-zero spacetime curvature, just not locally. A uniform gravitational field that is not bounded by some space with curvature is the same as having the entire universe experiencing a constant acceleration in free-fall, which is meaningless without something to accelerate relative to, and can't be measured (no proper acceleration).

 

 

Does that mean that gravity wouldn't need to be a field effect??? Locally, gravity is just inertial movement. Only the spacetime curvature is a field. Only a gravitational gradient could be measured locally.

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This is an excellant explanation about what the cautions are regarding the nature of spacetime.

 

Yes indeed your quote is excellent.

 

It hints at further deductions that may be made about the common misconception of "time travel" in general.

 

An object has an extent (duration) along the time axis, just as along the space axis.

 

So why all the talk of time travel as moving one point of that extent when the same proponents would never dream of moving an isolated section in space.

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Yes indeed your quote is excellent.

 

It hints at further deductions that may be made about the common misconception of "time travel" in general.

 

An object has an extent (duration) along the time axis, just as along the space axis.

 

So why all the talk of time travel as moving one point of that extent when the same proponents would never dream of moving an isolated section in space.

When people talk of time travel they are either speaking of the distant future or the past. The first one requires waiting, the second requires faster than light travel. This is because such travel requires moving on a closed timelike curve which requires faster than light travel.

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When people talk of time travel they are either speaking of the distant future or the past.

 

Time travel is change of temporal coordinates. I see no reason to limit (Pun intended) this to large changes.

 

The first one requires waiting,

 

Waiting is no more a statement of time travel than the statement that Africa extends from 10 degrees to forty degrees longitude at the equator is a statement that Africa is travelling round the equator 30 degrees.

 

the second requires faster than light travel. This is because such travel requires moving on a closed timelike curve which requires faster than light travel.

 

No one has evidence of the validity of this statement.

Edited by studiot
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From Relativity, Thermodynamics and Cosmology by Richard C. Tolman, Dover Pub. (1987 version of 1934 text), page 29

 

 

 

In this language it is important to guard against the fallacy of assuming all directions in the hyper-space are equivalent, and of assuming that extension in time is of the same nature of extension in space merely because it may be convenient to think of them as plotted along perpendicular axes. …. That there must be a difference between the spatial and temporal axes in our hyper-space is made evident, by contrasting the physical possibility or rotating a meter stick from an orientation where it measures distance in eh y-direction, where the impossibility of rotating it into a direction where it would measure time-intervals-in other words the impossibility of rotating a stick into a clock.

 

 

This is an excellant explanation about what the cautions are regarding the nature of spacetime.

 

I recall countless times people speaking about the speed at which a photon travels through spacetime.

 

You can just as well talk about the speed at which light moves through spacetime as you can talk about rotating a clock into a rod. :P [/font]

 

So that's the answer to the question I asked a millenium ago:

Q. Is time rotatable?

A. No.

(if i understand clearly your quote)

But then: how is it possible that time & space are interchangeable, that space for an observer changes in time for another?

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So that's the answer to the question I asked a millenium ago:

Q. Is time rotatable?

A. No.

(if i understand clearly your quote)

But then: how is it possible that time & space are interchangeable, that space for an observer changes in time for another?

Well it's off topic, and I don't know for sure, but I'd say...

 

The first question's a bit of a nonsense question, because time is a scalar value, not a vector (or object, or something else rotatable). It doesn't make sense to ask if something is rotatable when there is no rotation operator defined for it.

 

The answer to the second question is that a rotatable vector with a time component can be rotated.

 

For example, the 4-vector representing the length and orientation of a ruler can be rotated onto the time dimension. Then it represents the time it takes light to cover the length of the ruler, or perhaps the ruler traveling at the speed of light. Just because you can do this rotation mathematically doesn't mean there is any physically possible equivalent. A material ruler is not the same as a 4-vector representing its length and orientation, and a 4-vector of (t,0,0,0) is not the same as a physical clock.

 

 

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Fair comment, I apologise for introducing time to your thread.

No worries! It's fair to be asked to deal with time when bringing up the topic of spacetime.

 

I didn't mean to try to own or police the thread, I just didn't want to go off topic myself, however the time-related stuff is interesting.

 

 

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What is the principle of equivalence?, Hans C. Ohanian, Am. J. Phys. 45(10)), October 1977

Abstract - The strong principle of equivalence is usually formulated as an assertion that in a sufficiently small, freely falling laboratory the gravitational fields surrounding the laboratory cannot be detected. We show that this is false by presenting several simple examples of phenomena which may be used to detect the gravitational field through its tidal effects: we show that these effects are, in fact, local (observable in an arbitrarily small region). [...]

 

 

In the pdf you sent me, the author describes a physical process of measuring curvature using measurements taken at two spatially separated points, and then shows that the space could be reduced arbitrarily. But then the author admits that there are practical limits to reducing the size of the apparatus due to "quantum properties" of materials.

 

This means it is not physically possible to practically measure curvature in an arbitrarily small space due to limitations of the instruments, but wouldn't any theoretically ideal measurements themselves also have "quantum properties" that make them undefined at arbitrarily small distances?

 

What is the point of using a definition of curvature that is well-defined at a point, and so claim that that mathematical definition corresponds to physical reality even at that point, when quantum mechanics suggests that such a definition breaks down at sufficiently small distances? We don't have an accepted theory of quantum gravity, so why would anyone talk about GR as valid "locally" even at small enough distances that it doesn't work at, where QM is needed to describe reality at such distances? Isn't that simply a case of ignoring the incompatibility of GR and QM at all scales, and pretending that one is valid anyway at all scales?

 

 

 

 

Edit: I think I understand that the limit of measured curvature approaches the curvature defined at the point, as the distance between measurements approaches 0... but that this does not correspond to reality due to "quantum properties".

 

To revise my main speculation in this thread, I would say: "There are no local effects of spacetime curvature." Except perhaps in case of singularities? Also excepting any results that might be predicted by a hypothetical quantum theory of gravity that incorporates spacetime curvature.

Edited by md65536
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