Why two mass attracts each other?

[hansda]

What I know is that GR produces numerical predictions which are in perfect accordance with experiment and observation, so the model works.

My understanding is that GR works on "spacetime curvature", which follows EFEs. This math is fine.

Can you explain "what exactly is space-time", curvature of which generates force or make a mass move? I will give four options.

1) "Space-time" is "space".

2) "Space-time" is "space", which is moving in the "direction of time".

3) "Space-time" is some "field" contained within space and time. Curvature of this field generates force or make a mass move. So, "space-time curvature" means "curvature of this field".

4) "Space-time" is something else other than the above three options.

 

[rpenner]

Using the Schwarzschild coordinates the coordinate speed of light can be practically read off the metric since for any beam of light ds = 0.
Thus we have $$\left(1 - \frac{r_s}{r} \right)^2 c^2 = \left( \frac{dr}{dt} \right)^2 + \left(1 - \frac{r_s}{r} \right) \left(r \frac{d \theta}{d t} \right)^2 + \left(1 - \frac{r_s}{r} \right) \left( r \sin \theta \frac{d \phi}{dt} \right)^2$$

Apparently I went too fast for some.

Starting with
$$ds = 0 \\ ds = - \left(1 - \frac{r_s}{r} \right) c^2 (dt)^2 + \left(1 - \frac{r_s}{r} \right)^{-1} (dr)^2 + r^2 \left( ( d\theta )^2 + \sin^2 \theta ( d \phi)^2 \right)$$,
I substituted in for ds, added $$\left(1 - \frac{r_s}{r} \right) c^2 (dt)^2$$ to both sides, multiplied both sides by $$ 1 - \frac{r_s}{r} $$, divided both sides by $$(dt)^2$$, and expanded the last term to get:
$$\left(1 - \frac{r_s}{r} \right)^2 c^2 = (dr)^2 / (dt)^2 + \left(1 - \frac{r_s}{r} \right) r^2 ( d\theta )^2 / (dt)^2 + \left(1 - \frac{r_s}{r} \right) r^2 ( \sin \theta )^2 ( d \phi)^2 / (dt)^2 $$
Then some gentle application of the exponent laws gives:
$$\left(1 - \frac{r_s}{r} \right)^2 c^2 = \left( \frac{dr}{dt} \right)^2 + \left(1 - \frac{r_s}{r} \right) \left(r \frac{d \theta}{d t} \right)^2 + \left(1 - \frac{r_s}{r} \right) \left( r \sin \theta \frac{d \phi}{dt} \right)^2$$

So for radial light we see that the coordinate speed is a function of position $$\left(1 - \frac{r_s}{r} \right) c$$ and for orthogonal light (dr =0) we see that the coordinate speed is $$\sqrt{1 - \frac{r_s}{r} } c$$ where $$r_s = \frac{2 GM}{c^2}$$.

You sure about the (dr=0) part? You are calculating the radial coordinate speed of light, so what is with the "orthogonal light" bit?
You sure you didn't really want $$d\theta=d\phi=0$$?

$$d\theta=d\phi=0$$ leads to the solution where the light movement is in a pure radial direction, in which case the coordinate speed is $$\left(1 - \frac{r_s}{r} \right) c$$.
$$dr = 0$$, on the other hand, leads to a solution where the light movement is in a direction orthogonal to the radial direction, and the coordinate speed is $$\sqrt{1 - \frac{r_s}{r} } c$$.

If I wished to emphasize the spherical symmetry of this, I would write $$(d\Omega)^2 = (d \theta)^2 + \sin^2 \theta ( d \phi)^2$$ and write the expression obtained by setting ds = 0 as:
$$\left(1 - \frac{r_s}{r} \right)^2 c^2 = \left( \frac{dr}{dt} \right)^2 + \left(1 - \frac{r_s}{r} \right) \left(r \frac{d \Omega}{d t} \right)^2$$
To reduce the number of unknowns, I can assume $$d \Omega = 0$$ and the expression simplifies as follows:
$$\left(1 - \frac{r_s}{r} \right)^2 c^2 = \left( \frac{dr}{dt} \right)^2
\frac{dr}{dt} = \pm \left(1 - \frac{r_s}{r} \right) c
\left| \frac{dr}{dt} \right| = \left(1 - \frac{r_s}{r} \right) c $$
which is an expression for the radial coordinate speed of light for $$r \geq r_s$$.

Alternately, if I chose to assume $$d r = 0$$ by consequence any motion would be in a direction perpendicular to the radial direction, and the expression would simplify this way:
$$\left(1 - \frac{r_s}{r} \right)^2 c^2 = \left(1 - \frac{r_s}{r} \right) \left(r \frac{d \Omega}{d t} \right)^2
\left(1 - \frac{r_s}{r} \right) c^2 = \left(r \frac{d \Omega}{d t} \right)^2
r \frac{d \Omega}{d t} = \pm \sqrt{1 - \frac{r_s}{r} } c
\left| r \frac{d \Omega}{d t} \right| = \sqrt{1 - \frac{r_s}{r} } c$$
which is an expression for the coordinate speed of light for $$r \geq r_s$$ in a direction orthogonal to the radial direction.

Because $$\left(1 - \frac{r_s}{r} \right) c \neq \sqrt{1 - \frac{r_s}{r} } c$$ (for ( $$r \gt r_s$$ ), the Schwarzschild coordinates are anisotropic with respect to the coordinate speed of light. This is a consequence of the choice made to preserve the formula for the circumference of a sphere of radius r.

(This is simple algebra, far simpler than solving for the geodesics of light -- what are straight lines in the geometry need not be straight lines in the coordinate space. And even that is far simpler than solving for the geometry of space-time starting with a distribution of matter.)

I explained all these to Undefined long ago, replete with the $$c_{rcoord}=c\sqrt{1-\frac{r_s}{r}}$$ formula derived from the Schwarzschild solution.

Showing is better than telling. What post, please?

 

[Tach]

$$d\theta=d\phi=0$$ leads to the solution where the light movement is in a pure radial direction, in which case the coordinate speed is $$\left(1 - \frac{r_s}{r} \right) c$$.
$$dr = 0$$, on the other hand, leads to a solution where the light movement is in a direction orthogonal to the radial direction, and the coordinate speed is $$\sqrt{1 - \frac{r_s}{r} } c$$.

Ok, I see, you mixed the radial coordinate speed of light and the tangential coordinate speed of light into the same sentence.
I would do it clearer:

Start with the Schwarzschild solution:

$$
c^2 {d \tau}^{2} =\left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right),
$$

Light follows null geodesics, so $$d\tau=0$$

$$
0 =\left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right),
$$

Now:

1. For $$dr=0$$ AND $$d\phi=0$$ you get the tangential coordinate speed of light (circular photon trajectories)

$$r\frac{d\theta}{dt}=\pm c \sqrt{1-r_s/r}$$

2. For $$d\theta=0$$ and $$d\phi=0$$ you get the radial coordinate speed of light


$$\frac{dr}{dt}=\pm c (1-r_s/r)$$

 

[Markus Hanke]

My understanding is that GR works on "spacetime curvature", which follows EFEs. This math is fine.

Can you explain "what exactly is space-time", curvature of which generates force or make a mass move? I will give four options.

1) "Space-time" is "space".

2) "Space-time" is "space", which is moving in the "direction of time".

3) "Space-time" is some "field" contained within space and time. Curvature of this field generates force or make a mass move. So, "space-time curvature" means "curvature of this field".

4) "Space-time" is something else other than the above three options.

Space-time is modelled as a (3+1)-dimensional pseudo-Riemannian manifold which is endowed with a metric and a connection, more specifically the Levi-Civita connection. The principle invariant of that connection is curvature. Space and time cannot be separated, they are just aspects of the same manifold; that manifold is not embedded in anything else. Gravity then is an intrinsic geometric property of said space-time; in GR that is just curvature. All particles trace out world lines in space-time, so the fact that their trajectories are curved in the vicinity of energy distributions is just simply a consequence of the curvature of the underlying space-time. There are no forces involved. It is all purely geometry.

Try this visualisation for clarity :

http://demonstrations.wolfram.com/GravitationVersusCurvedSpacetime/

 

[quantum_wave]

Space-time is modelled as a (3+1)-dimensional pseudo-Riemannian manifold which is endowed with a metric and a connection, more specifically the Levi-Civita connection. The principle invariant of that connection is curvature. Space and time cannot be separated, they are just aspects of the same manifold; that manifold is not embedded in anything else. Gravity then is an intrinsic geometric property of said space-time; in GR that is just curvature. All particles trace out world lines in space-time, so the fact that their trajectories are curved in the vicinity of energy distributions is just simply a consequence of the curvature of the underlying space-time. There are no forces involved. It is all purely geometry.

It isn't often that I am on a talking basis with someone who knows GR so well. Your reponse brings up a question. In GR there are three "curvatures" or "shapes" of the universe and the real shape is determined by the cosmological constant, I think. We are now thought to be in a very nearly flat "shape" with slight "open" curvature, if I'm not too wrong. My question takes the title of this thread to the extreme "closed" curvature option which has as its fate a final big crunch where all the matter and energy accumlate at the end of the game. What is the current thinking as to the size, in light years, of the diameter of such a final crunch, or have you never thought about it?

 

[hansda]

Space-time is modelled as a (3+1)-dimensional pseudo-Riemannian manifold which is endowed with a metric and a connection, more specifically the Levi-Civita connection. The principle invariant of that connection is curvature. Space and time cannot be separated, they are just aspects of the same manifold; that manifold is not embedded in anything else. Gravity then is an intrinsic geometric property of said space-time; in GR that is just curvature. All particles trace out world lines in space-time, so the fact that their trajectories are curved in the vicinity of energy distributions is just simply a consequence of the curvature of the underlying space-time. There are no forces involved. It is all purely geometry.

So, according to you none of the four options i mentioned is correct.

Now, consider a case there is only space and our Earth. The Earth will follow world-line.

Consider another case, there is only space and the Sun. The Sun will follow world-line.

Consider another case, there is only space and The Sun and the Earth. Here the Sun will follow world-line but the Earth will orbit the Sun. Here the Earth will not follow world-line. So, dont you think there is some force between the Sun and the Earth?

 

[quantum_wave]

It isn't often that I am on a talking basis with someone who knows GR so well. Your reponse brings up a question. In GR there are three "curvatures" or "shapes" of the universe and the real shape is determined by the cosmological constant, I think. We are now thought to be in a very nearly flat "shape" with slight "open" curvature, if I'm not too wrong. My question takes the title of this thread to the extreme "closed" curvature option which has as its fate a final big crunch where all the matter and energy accumlate at the end of the game. What is the current thinking as to the size, in light years, of the diameter of such a final crunch, or have you never thought about it?

I Googled around and found this whimsical answer:
http://science.howstuffworks.com/dictionary/astronomy-terms/question221.htm

If you were to move all of the matter in the universe into one corner, how much space would it take up?

It's hard to answer this question exactly because there are some unknowns. But if you are willing to accept three assumptions, we can come up with an answer that sounds reasonable...

The first question is, "How big is the universe?" No one knows, but this Question of the Day assumes that the universe is a cube that is 30 billion light years on each side. That means that the whole universe contains about 2.7E+31 cubic light years.

The next question is, "How much matter does the universe contain?" The mass of the universe is a source of debate right now because there is no easy way to put the universe on a scale. This NASA page and this " Extension, Age and Mass of the Universe" article discuss different techniques that scientists use to estimate the mass. The latter article also includes an estimate of about 1.6E+60 kilograms for the mass of the universe. Other estimates give other numbers, but all are in that ballpark.

The next question is, "What density do you want to assume the mass will have once you push all of it into one corner?" Now, if you were really to do this -- if you actually did move all of the mass of the universe into one corner -- it would condense instantly into a black hole and potentially ignite another big bang. But let's say that you could keep it from doing that, and you were somehow able to keep all of the mass evenly distributed at the density of the sun. According to "Magnitudes of Physics", the density of the sun is about 1,410 kilograms per cubic meter. (For comparison, the density of water is 1,000 kilograms per cubic meter.)

If you are willing to accept these three assumptions, then:

1.1E+57 cubic meters of matter in the universe
A cubic light year contains about 1E+48 cubic meters. So all of the matter in the universe would fit into about 1 billion cubic light years, or a cube that's approximately 1,000 light years on each side. That means that only about 0.0000000000000000000042 percent of the universe contains any matter. The universe is a pretty empty place!
For more information on this topic, check out the links on the next page.

 

[Undefined]

Reading back over this, I agree with you that "apparent" was an unfortunate choice of words, brucep. My apologies. You are right to stress that all frames are equally valid, it is just that they don't necessarily agree; that is fine though, because in GR there is no requirement for them to do so.

Now you know why naive understandings like mine get confused by jargon that is loosely used and not clearly explained as to what it means and what it does not mean. My naive undestandings are that something is "apparent" from a remote frame until that "apparent something" is Lorentz Trasformed into an "invariant something" using a theoretical algorithm to match with an object's "proper" something.

A proper frame measurement is that without any transforms needed to make it in that proper frame. But a remote frame "measurement" is not a measurement until it is transformed into the invariant measurement of the proper frame. I would say that until that is done, the object's remotely observed "properties" are just raw data which has no meaning (and actually only "apparent" therefore) unless and until it is made sense of by some algorithm treatment to put that raw data into a form or value which agree with the object's proper frame "invariant property" being remotely studied? In my naive opinion, it is people using terms like "measurement" confusingly like that which causes all the problems for my naive understandings when listening to a discussion like this.

 

[Undefined]

I think there is a misunderstanding here somewhere. Light always traces out null geodesics; it cannot do anything else. Therefore the local measurement of its speed will always yield exactly c. In the vicinity of a black hole these null geodesics are not straight lines pointing outwards from the event horizon at right angles.

If light (a photon) is not leaving the bh horizon or dropping into it, then it is going nowhere in fact despite any theoretical coordinate manipulations to make it "virtually" appear as if it is. The local measurement would also be zero for 'c' in that situation no matter what a theory interpretation is put on it by "abstract coordinate" theories and "arbitrary choices" we mentioned before.

I also asked where is the null geodesic for that photon if it is not moving in fact but only in some "chosen coordinate theoretical interpretation" which hasn't explained what is actually happening to that photon as it fades away in situ? Also my naive understanding asks about that rock sitting on the Earth's surface? It is going nowhere, so what "null geodesic" is it tracing out when it does not move down in "freefall"?

 

[Undefined]

I have given you a very detailed explanation about how light traces out null geodesics in space-time. In the vicinity of black holes these null geodesics are curved trajectories in curved space-time, and locally you will always measure exactly c.
If you decide not to acknowledge answers given to you, then that is your own problem. Don't try to blame those who make the effort to explain things.

Can I suggest this : http://www.adamtoons.de/physics/gravitation.swf.
These aren't null geodesics ( so not light ), but it does illustrate what happens to trajectories in the vicinity of massive bodies.

I should point out to you also that photons/light do not undergo acceleration - not surprising, since their speed is always constant.

I naively understand already what theory says. That is not my problem. I asked what real "null geodesics" is a trapped photon "following" as it fades away in situ and not "free falling" into the bh horizon or leaving its vicinity?

All such "animations" and "simulations" are not explanations of what happens, but only theoretical interpretations of what may be happening.

The Epicycles "explanation" was an "theoretical interpretation" of the sun, moon and stars "going around a supposedly geocentric Earth". That can also be animated and simulated, but it is not what happens in fact but in interpretation of theory put on those observations?

I do not want just more interpretations or simulations, but actual explanations of where is the theoretical "space-time" null geodesic for a photon trapped and fading away to nothing in situ at that location above event horizon? I have the theory "explanations", I just want to naively understand the actual physics in fact and not some "epicyclic" or other "theory" interpretations of it. Can anyone help me on that?

 

[Undefined]

Can you map the surface of the Earth on a flat two-dimensional map without distortion? No.

Please excuse my need for naive understandings, but why would we want to attempt that and expect no distortion? Does the universe want to do that before it can do what it does naturally before our philosophy, science and mathematics allowed humans to "want" to do these things for "exercises" in our abstract understandings and speculations? If the universe does not need to do that "mapping" like a mathematician wants to, and the universe works without such "mapping exercises", why does that "prove" anything about the questions I asked about to improve my naive understandings of what the universe (despite what mathematicians do) does physically in fact? Can you tell me how such failures in attempted "mathematical mapping without distortion" in such arbitrary situations causes and explains the real physical phenomena? I don't have a position either way, I just want to naively understand what is real and physical and what is theoretical plays about things which nature doesn't care about one way or the other to function as it has been doing before man and mathematics came to think and philosophize about it and wanting to "map" things and such "mind" exercises.

 

[Tach]

I naively understand already what theory says. That is not my problem. I asked what real "null geodesics" is a trapped photon "following" as it fades away in situ and not "free falling" into the bh horizon or leaving its vicinity?

All such "animations" and "simulations" are not explanations of what happens, but only theoretical interpretations of what may be happening.

The Epicycles "explanation" was an "theoretical interpretation" of the sun, moon and stars "going around a supposedly geocentric Earth". That can also be animated and simulated, but it is not what happens in fact but in interpretation of theory put on those observations?

I do not want just more interpretations or simulations, but actual explanations of where is the theoretical "space-time" null geodesic for a photon trapped and fading away to nothing in situ at that location above event horizon? I have the theory "explanations", I just want to naively understand the actual physics in fact and not some "epicyclic" or other "theory" interpretations of it. Can anyone help me on that?

No one has had any interaction with a BH, so you will need to understand that the explanations given to you are all purely theoretical, they are the GR predictions for photon trajectories in the vicinity of a BH. Some of these trajectories (outside the EH, $$r>r_s$$) are testable, others (inside the EH, $$r<r_s$$) are not because no one can send any information from behind the EH. Since GR is one of the most tested theories (alongside with QM and SR), and since , to date, no experiment has falsified GR, it is safe to assume that the explanations given to you, though only theoretical, are correct. For the time being, until experimentalists can interact with a BH, this is all that we know.

 

[Undefined]

No one has had any interaction with a BH, so you will need to understand that the explanations given to you are all purely theoretical, they are the GR predictions for photon trajectories in the vicinity of a BH. Some of these trajectories (outside the EH, $$r>r_s$$) are testable, others (inside the EH, $$r<r_s$$) are not because no one can send any information from behind the EH. Since GR is one of the most tested theories (alongside with QM and SR), and since , to date, no experiment has falsified GR, it is safe to assume that the explanations given to you, though only theoretical, are correct. For the time being, until experimentalists can interact with a BH, this is all that we know.

I already naively understood that position regarding "proof". I just would like to know what the "theory" about "null geodesics" has to say (clearly) about what the photon is doing if it is trapped there as described, going nowhere and not freefalling into or propagating away from the bh horizon at that position? Can you help me with that, Tach? Anybody else?

 

[Tach]

I already naively understood that position regarding "proof". I just would like to know what the "theory" about "null geodesics" has to say (clearly) about what the photon is doing if it is trapped there as described, going nowhere and not freefalling into or propagating away from the bh horizon at that position? Can you help me with that, Tach? Anybody else?

No one can help you with that since the above is not a well formed question, it is gibberish. Markus, rpenner and I gave you answers to what is going on and I have a good guess as to what you are trying to ask, but your question is so malformed that I am concerned that I answer it and you claim that this is not what you were asking. So, you will need to try harder to come up with a question that isn't gibberish. Sorry.

 

[rpenner]

Briefly, that what is invariant proceeds from the space-time geometry of a situation:

• The objects collide / do not collide,
• The segment/polygon is/is not a subset of the line/plane,
• the objects are co-moving/have a relative velocity which is not-zero,
• the line is time-like/space-like/light-like,
• the object moves inertially/does not move inertially,
• the object moves inertially and has proper length of x meters,
• the object experiences a proper time of x seconds between space-time points A and B, etc, etc.

What is not invariant proceeds from coordinates in a way that does not give rise to invariants.

• The object is motionless/The object moves with speed v (where $$v \neq c$$,
• the object has length L,
• gravity was first described by physicists of the Andromeda galaxy/Milky Way galaxy/simultaneously
• Susan met me here, last week . etc

Coordinates give algebraic meaning to the words "here", "now" and "speed" but only in a way that depends on a standard of rest. But "here" in the dining car of a train in mid-trip has different meanings than "here" at a Chicago nightclub, so we need to be careful and specify which coordinates we are using when we describe space-time because another person could equally validly make a different choice.

What was perplexing enough in special relativity becomes even more confusing in general relativity because now the relation between coordinate systems may vary widely from place to place (and time to time) even though both describe the same events/world-lines in the same space-time geometry.

Examples of space-time metrics for the same space-time geometry:
$$
\begin{array}{l|cl}
\textrm{Coordinates} & \quad & \textrm{Metric}
\\ \hline
\\ (t,r,\theta,\phi) & \quad & ds = - \left( 1 - \frac{2 G M}{c^2 r} \right) c^2 (dt)^2 + \left( 1 - \frac{2 G M}{c^2 r} \right)^{-1} (dr)^2 + r^2 (d \theta)^2 + r^2 \sin^2 \theta (d \phi)^2
\\ (\mathbb{T},\mathbb{X},\mathbb{Y},\mathbb{Z}) & \quad & ds = \left( \frac{ 8 c^2 G M \sqrt{\mathbb{X}^2 + \mathbb{Y}^2 + \mathbb{Z}^2}}{\left( GM + 2 c^2 \sqrt{\mathbb{X}^2 + \mathbb{Y}^2 + \mathbb{Z}^2} \right)^2} - 1 \right) c^2 (d\mathbb{T})^2 + \frac{ \left( GM + 2 c \sqrt{\mathbb{X}^2 + \mathbb{Y}^2 + \mathbb{Z}^2} \right)^4}{16 c^8 \left(\mathbb{X}^2 + \mathbb{Y}^2 + \mathbb{Z}^2 \right)^2 } \left( (d\mathbb{X})^2 + (d\mathbb{Y})^2 + (d\mathbb{Z})^2 \right)
\\ (\mathcal{T}, \mathcal{R}, \Theta, \Phi) & \quad & ds = -c^2 (d\mathcal{T})^2 + \left( \frac{4GM}{3c^2 \left(\mathcal{R} - c \mathcal{T} \right) } \right)^{\tiny \frac{2}{3}} (d\mathcal{R})^2 + \left( \frac{9 GM \left( \mathcal{R} - c \mathcal{T} \right)^2}{2c^2} \right)^{\tiny \frac{2}{3}} \left( (d \Theta)^2 + \sin^2 \Theta (d \Phi)^2 \right)
\end{array}
$$

By contrast, the Cartesian coordinates of two inertial coordinate systems in flat space-time look like this:
$$
\begin{array}{l|cl}
\textrm{Coordinates} & \quad & \textrm{Metric}
\\ \hline
\\ (t,x,y,z) & \quad & ds = - c^2 (dt)^2 + (dx)^2+ (dy)^2+ (dz)^2
\\ (t',x',y',z') & \quad & ds =- c^2 (dt')^2 + (dx')^2+ (dy')^2+ (dz')^2
\end{array}
$$
and so special relativity is going to be a special case of general relativity -- one that is easier to study.

 

[Undefined]

No one can help you with that since the above is not a well formed question, it is gibberish. Markus, rpenner and I gave you answers to what is going on and I have a good guess as to what you are trying to ask, but your question is so malformed that I am concerned that I answer it and you claim that this is not what you were asking. So, you will need to try harder to come up with a question that isn't gibberish. Sorry.

Which is the "gibberish", the question or the "explanations" so far attempted in response? You are entrenched in a particular "camp" and use other people's work to try and convince that you "know". I am not entrenched either way, I just want to get some sense out of what is "out there" from anyone that can help improve my naive understandings without the need or confusion of the jargon and theoretical assumptions being presented as "fact" or "proof" of something when it isn't when you look deeper and more naively ask for clarification. Thank you for trying, Tach. I hope you have better luck another time.

 

[Tach]

Which is the "gibberish", the question

Your question, definitely, I made that quite clear.

 

[Undefined]

Your question, I made that quite clear.

In your opinion. No real explanations or real proof as usual. ok thankyou.

 

[rr6]

If It Did Not We Would Not Exist

If mass did not attract we would not exist.

Pulling-in/contractive is the essential primary/apriori/inherent/fundamental force( boson? )--- phenomena --- of our finite Universe.

Barring like charges repeling for the moment, all pushing-out or pushing-in phenomena of Universe, are a resultant of pulling-in/contractive.

Pulling-inward( contravtive ) is the path-of-least resistance.

On TV I have seen strong men pulling a box car or the locomotive engine but never ever have I seen a strong man push a box car or locomotive engine.

In Japan they make these light weight hand saws--- and I presume use them ---that are designed to cut only on the pull motion.

Action attracts attention. If we see something move our attention is drawn to it, so motion attracts attention.

Motion attracts other motion. In some wave dynamics two independent, circular waves, having opposite spins attract each other.

I offer a simple scenario for gravitational spacetime, where two opposite spinning motions attract each other and at same time are part of a greater integral whole.


Imagine a initiating source 'A' that emits two tubes, as each were one half of a torus ( ).


One half--- ( ---of the torus spins clockwise to the axis of its trajectory/path and the other half of the torus--- ) ---spins counter clockwise to the axis of its trajectory/path.


.......'A'........
.......( )........
.......'B'........


Here above we se that the two spinning halfs, come back together at the destination 'B'.

So even tho there initial path was away from each other, there spin attracts them back together.

So, where, or is there an integral relationship between the two halfs attributed to the spin attraction?

....(<--><-->).....i.e. is there a line-of-relationship between the two spinning tubes?


Or is it not that that the spins attract each other but only that the initial circumstances at source 'A' initiate the opposite spins the illusion of a line-of-interrelationship between the two spinning tubes?

So we may be also asking ourselves does the torus have a central axis--- between 'A' and 'B' ---that exists because of a constant line-of-relationship between the two spinning halfs.

H,mm suddenly I'm not sure which scenario above I like better. A constant line-of-relationship due to spinning motion or initiating circumstances at source leading us to false conclusions.

r6

 

[przyk]

pryzk: see this from another thread.

...So what you really mean is something moves through space-time along some worldline...

I rest my case.

Um, what case? You just posted a knee-jerk reaction to the particular language AlphaNumeric used without saying anything about the substance of his post. This is exactly what I was criticising you for.

You did the same with me in [POST=3067812]this[/POST] post:

First, it merely states that "empty space" is inhomogeneous, which is not controversial (e.g. the geometry or curvature of spacetime within the solar system is inhomogeneous).

No! He said spacetime in the previous sentence. Now he says space. Not spacetime. Spot the difference!

In this particular instance, what I said about spacetime also applies to space. In fact, I might be wrong but my impression is that "inhomogeneity" is normally a reference just to the spatial part (e.g. the FLRW metric is typically said to be "homogeneous" despite the fact it is time-dependent). So an "inhomogeneous spacetime" would mean a spacetime in which the spatial slices were inhomogeneous, which makes "inhomogeneous spacetime" pretty much synonymous with "inhomogeneous space". The distinction between "space" and "spacetime" was not relevant to the point I was making. Yet you jumped on it anyway and left the actual point I was making unaddressed: there is an alternative and more literal interpretation of what Einstein said that is perfectly in accord with the mainstream understanding of general relativity.

(Incidentally, if you can't tell the difference between something Einstein said and something I'm saying in response, then we're really in trouble.)