Curved space

[ajb]

Everytime I try to wrap my head around space-time illustrations someone drops a heavy ball on a stretched out blanket, and then rolls a smaller ball onto the curve in the blanket made by the big ball. This would be fine if all the universe were flat as a sheet with all the mass floating on top of it. But, isn't everything in the universe completely submerged in the universe? If so, how can the ball on the blanket serve as a valid example of space-time?

 

It is just an analogy. You should not take it too seriously, it is just for illustrative purposes only.

[DrRocket]

It is just an analogy. You should not take it too seriously, it is just for illustrative purposes only.

 

Yes indeed.

 

That particular analogy has done at least as much damage as good.

[ajb]

That particular analogy has done at least as much damage as good.

 

I am going to have to agree with that. This analogy and following misconceptions are quite frequently posted on this forum.

[StringJunky]

The problem with that analogy is it's only representing 2 dimensions (the sheet) but it looks 3D in any illustration. i visualise a cube, as representing a portion of space, with grid lines in it in such that it's made up of lots of smaller cubes. I then visualise an object ebedded in the middle and those grid lines bend all around it; the degree of curvature in the lines diminishing with distance. I make no attempt to imagine what gravity is, only how the mass acts on the grid co-ordinates.

 

Am i far off thinking like this?

[michel123456]

The problem with that analogy is it's only representing 2 dimensions (the sheet) but it looks 3D in any illustration. i visualise a cube, as representing a portion of space, with grid lines in it in such that it's made up of lots of smaller cubes. I then visualise an object ebedded in the middle and those grid lines bend all around it; the degree of curvature in the lines diminishing with distance. I make no attempt to imagine what gravity is, only how the mass acts on the grid co-ordinates.

 

Am i far off thinking like this?

 

All visualizations are off.

The world is not 3D, the world is 4D. There is space, there is time. Spacetime bends. Not space only, not time only.

 

The rubber sheet analogy is very very bad (IMHO). It is a mix of 3D (the round planet) and 2D (the rubber sheet) in a tentative to explain the effect caused by a 3D object translating in a 4D spacetime. The analogy is foul: although you should have for the framework a dimension over the object, the analogy gives a dimension less than the object.

 

It would be better to imagine a coin (a 2D disk) into a 3D framework, where the surface that contains the coin represents space, and the perpendicular represents time.

The disk is then the representation of a sphere at a certain instant in time.

 

-----------------

Take a coin from your pocket, put it on your desk, that's it.

Edited February 6, 2012 by michel123456

[StringJunky]

All visualizations are off.

The world is not 3D, the world is 4D. There is space, there is time. Spacetime bends. Not space only, not time only.

 

The rubber sheet analogy is very very bad (IMHO). It is a mix of 3D (the round planet) and 2D (the rubber sheet) in a tentative to explain the effect caused by a 3D object translating in a 4D spacetime. The analogy is foul: although you should have for the framework a dimension over the object, the analogy gives a dimension less than the object.

 

It would be better to imagine a coin (a 2D disk) into a 3D framework, where the surface that contains the coin represents space, and the perpendicular represents time.

The disk is then the representation of a sphere at a certain instant in time.

 

-----------------

Take a coin from your pocket, put it on your desk, that's it.

 

Yes, I see what what you mean but I'm just considering gravity.

[Jiggerj]

The problem with that analogy is it's only representing 2 dimensions (the sheet) but it looks 3D in any illustration. i visualise a cube, as representing a portion of space, with grid lines in it in such that it's made up of lots of smaller cubes. I then visualise an object ebedded in the middle and those grid lines bend all around it; the degree of curvature in the lines diminishing with distance. I make no attempt to imagine what gravity is, only how the mass acts on the grid co-ordinates.

 

Am i far off thinking like this?

 

 

Well, if you're far off, then I'm off the grid with this: If time is weaved into the fabric of the universe, how come Einstein found no need to insert time into E=MC2 ? Doesn't it take energy to bend space-time? Doesn't Mass bend space-time? Doesn't it take energy for me to make any sense here?

 

All visualizations are off.

It would be better to imagine a coin (a 2D disk) into a 3D framework, where the surface that contains the coin represents space, and the perpendicular represents time.

 

Wouldn't a better illustration be a ball that is weighted and bouyant just enough to float under water?

 

Wait a minute! With some kind of fancy-shmancy camera would we be able to see the water curving around the ball? If not, why not?

[Widdekind]

A coin, or a weight-lifting "plate", would be a better analogy (2D weight, on 2D membrane), than a bowling ball. I offer, that an even better analogy, would be some kind of "flat bag full of lead shot", in lieu of a rigid coin. For, the "rubber sheet" of space-time can, and does, deform, inside of gravitating bodies, i.e. the "flat bag of lead shot" would "slump & sag" towards the middle (until outward pressure forces offset gravity-like compressions).

[IM Egdall]

Well, if you're far off, then I'm off the grid with this: If time is weaved into the fabric of the universe, how come Einstein found no need to insert time into E=MC2 ? Doesn't it take energy to bend space-time? Doesn't Mass bend space-time? Doesn't it take energy for me to make any sense here?

 

 

Yes it takes energy to bend or curve spacetime, And yes mass also bends or curves spacetime. Per E=mc2, mass and energy produce equivalent physical effects. So mass is a source of gravity (spacetime curvature) and space is a source of gravity (spacetime curvature). (In fact, it is difficult to come up with an exact definition of what mass is.)

 

More formally, it is momenergy which is the source of gravity or spacetime curvature, as represented in the stress-energy tensor. And so-called momenergy is a four dimensional spacetime vector with energy as the time component, momentum as the space component, and mass as the magnitude of the vector. See link:

 

http://physics.bu.ed...8_Momenergy.pdf

Edited February 6, 2012 by IM Egdall

[Anilkumar]

I believe that in Euclidian geometry (of flat spaces) then yes, the shortest distance between to points is a straight line.

 

But in non-Euclidian geometry (e.g. of a curved surface such as that of the earth) the shortest distance between to points is effectively an arc rather than a straight line.

 

So 4 dimensional space-time is defined by non-Euclidian geometry I presume, in which case your above assumption is false.

 

In Euclidian geometry; the shortest path between any two points is the straight line.

 

In non-Euclidian geometry; the shortest path between any two points is the Geodesic.

 

In either case, there cannot be more than one shortest distance.

 

 

[DrRocket]

In Euclidian geometry; the shortest path between any two points is the straight line.

 

In non-Euclidian geometry; the shortest path between any two points is the Geodesic.

 

In either case, there cannot be more than one shortest distance.

 

 

 

But there can be more that one path that exhibits that shortest distance. Consider lines of longitude on the globe. They are all geodesics between the poles.

 

The situation is quite different in Euclidean geometry. In non-Euclidean geometry two arbitrary points do not necessarily define a unique geodesic.

[Widdekind]

 

More formally, it is momenergy which is the source of gravity or spacetime curvature, as represented in the stress-energy tensor. And so-called momenergy is a four dimensional spacetime vector with energy as the time component, momentum as the space component, and mass as the magnitude of the vector. See link:

 

http://physics.bu.ed...8_Momenergy.pdf

 

The last equation says "mom-energy = mass x displacement / proper-time". So, some hypothetical CR, impacting earth's atmosphere at high energy, is perceived by "outsiders", e.g. earth observers, to have high mom-energy; whilst perceiving "internally" a small amount of proper-time elapsing ? So, there is a connection, between what the particle perceives (proper-time); and what other observers perceive (mom-energy)? If a high-energy particle is perceiving little proper-time elapsing, then is that why such particles are difficult to accelerate, i.e. you apply what you think is a large force, but the particle perceives the force applied for only a brief moment ??

[Anilkumar]

But there can be more that one path that exhibits that shortest distance. Consider lines of longitude on the globe. They are all geodesics between the poles.

 

The situation is quite different in Euclidean geometry. In non-Euclidean geometry two arbitrary points do not necessarily define a unique geodesic.

In that case there can be infinite possible geodesics.

 

 

[michel123456]

But there can be more that one path that exhibits that shortest distance. Consider lines of longitude on the globe. They are all geodesics between the poles.

(...)

 

Geodesics between the poles is a special case. Only when the 2 points are exactly opposite (antipodes) there are multiple geodesics. Between 2 random points on Earth's surface, there is only one geodesic which is the smallest path.

Edited February 7, 2012 by michel123456

[IM Egdall]

The last equation says "mom-energy = mass x displacement / proper-time". So, some hypothetical CR, impacting earth's atmosphere at high energy, is perceived by "outsiders", e.g. earth observers, to have high mom-energy; whilst perceiving "internally" a small amount of proper-time elapsing ? So, there is a connection, between what the particle perceives (proper-time); and what other observers perceive (mom-energy)? If a high-energy particle is perceiving little proper-time elapsing, then is that why such particles are difficult to accelerate, i.e. you apply what you think is a large force, but the particle perceives the force applied for only a brief moment ??

 

I do not think you are correct here.

 

Proper Time

Proper time is the spacetime interval in time units. It is absolute -- the same value for all uniformly moving frames of reference. Proper time is not the "perceived" time, as you call it.

 

Co-ordinate Time

 

The perceived time is relative or different in different reference frames. It is called co-ordinate time -- the time interval in the reference frame chosen.

 

Momenergy

 

In the momenergy vector, the momentum and energy are relative, but the mass is absolute. (Similar to the relative space and time intervals and the absolute spacetime interval.) They are related by:

 

mass ^2 = energy^2 - momentum^2

 

The momentum and energy change with relative motion, but the mass is invariant. Say in my reference frame, a particle is at rest. So it's momentum is zero. Thus its mass equals its energy. (This is E=mc^2 with c in units like light-years per year so c = 1).

 

Now say the same particle is moving at velocity v in your reference frame (you are moving at v with respect to the particle). Here the momentum is not zero. The particle's energy, however, as you measure it, is greater than the energy I measure. Why? Because your measurement includes the particle's rest energy plus its energy of motion. So you measure greater momentum and greater energy for the particle than I do. But when you take the difference of the squares of the energy you measure and the momentum you measure, you get the same value I get for the mass.

 

This mass is the magnitude of the momenergy vector. And we both measure the same mass, the same momenergy magnitude for the particle -- regardless of how we are (uniformly) moving with respect to that particle.

Edited February 7, 2012 by IM Egdall

[Widdekind]

I understand, that the "mom-energy" 4-vector is calculated, from quantities, derived from two different reference frames. First, the observer measures a 4-displacement, [math]\Delta x^{\mu}[/math], between two events, on the trajectory, of some particle, propagating through their laboratory rest frame. Then, that 4-displacement is divided by the proper time, [math]\Delta \tau[/math], elapsing between those two events, in the particle's own rest-frame. I.e. "the observer measures the displacement, but the particle reports the time".

 

The "mom-energy" is the velocity 4-vector, multiplied by the particle's mass. What about 4-acceleration ? I understand that

 

If the 4-velocity "points" in the direction of the particle's time-axis, i.e. "is the tangent vector to the particle's world-line"; then must the orthogonal 4-acceleration "point" in the direction of the particle's space-axis ?? I observe, that, indeed, in a particle's own rest-frame, the 4-acceleration has no time component, and so is wholly space-like. More generally, in (2+1)D, I suspect, that the 4-acceleration must lie in the "tilted (x'y') plane" in the particle's reference frame.

[DrRocket]

Geodesics between the poles is a special case. Only when the 2 points are exactly opposite (antipodes) there are multiple geodesics. Between 2 random points on Earth's surface, there is only one geodesic which is the smallest path.

 

Nope.

 

On a sphere there are at least two geodesics between any two points -- each a segment of a great circle passing through the two points. Geodesics, even in the truly Riemannian case, do not necessarily minimize the arc length between two points, they only do so for points that are sufficiently close to one another. In the case of the Lorentzian metric used in general relativity, geodesics actually maximize local arc length.

[Widdekind]

http://physics.bu.ed...8_Momenergy.pdf

 

Please ponder "mom-energy", in curved space-time. Particles residing near massive bodies experience gravitational time dilation. Accordingly, for a distant observer, dividing by the particle's proper time, induces a "gravitational gamma factor", [math]m \rightarrow \frac{m}{1-\frac{2GM}{c^2 r}}[/math]. Note, this relation can also be derived, from [math]g_{\mu \nu}p^{\mu}p^{\nu}=\left(mc^2\right)^2[/math], using the Schwarzschild metric, on a stationary test charge [math]p^{\mu} = \left( \begin{array}{cccc} E & 0 & 0 & 0 \end{array} \right)[/math], which "picks out" the time-time component, of the metric tensor, [math]g_{00} = 1 - \frac{2GM}{c^2r}[/math]. If so, then particles residing near massive bodies, are perceived, by remote observers, to have a gravity time dilation; and a gravity gamma-like factor, which increases the effective mass, i.e. "time slows & mass increases, for fast-moving particles; and for particles in gravity 'wells'".

 

principle of equivalence:

 

Whether in "flat" space-time, or "curved" space-time, particles "prefer" to propagate along "straightest possible paths", i.e. geodesics, through space-time. Per Newton's first law, each & every deviation, from the "desired" geodesic "free fall" path, must result from forces; and are all experienced equally as accelerations. Is this accurate ?

 

I understand, that, ultimately, all force-ful interactions arise, from the three fundamental forces, i.e. Strong, Weak, EM, via boson exchange. E.g. standing on earth is, ultimately, an EM interaction, between electrons, via virtual photons; whereas standing on a neutron-star would, hypothetically, be a Strong interaction, via virtual pions. Is that accurate ?

[imatfaal]

IME - that's not the definitions I would have used - take a look at this page on Gravitational Time Dilation

 

I know wiki can sometimes play a little fast and loose but here is one section

 

 

 

A common equation used to determine gravitational time dilation is derived from the Schwarzschild metric, which describes spacetime in the vicinity of a non-rotating massive spherically-symmetric object. The equation is:

 

, where

• t₀ is the proper time between events A and B for a slow-ticking observer within the gravitational field,
• t_f is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate),

 

 

The way I think of it is that proper time is the time between two events as experienced by the clock that passes through both these events. Coordinate time is more complicated - but in a situation like this is that of a distant theoretical observer

Edited February 8, 2012 by imatfaal

[IM Egdall]

IME - that's not the definitions I would have used - take a look at this page on Gravitational Time Dilation

 

 

I gave the definition of coordinate time in special relativity (flat spacetime -- zero gravity). You gave the definition in general relativity (curved spacetime).

 

I understand, that the "mom-energy" 4-vector is calculated, from quantities, derived from two different reference frames. First, the observer measures a 4-displacement, [math]\Delta x^{\mu}[/math], between two events, on the trajectory, of some particle, propagating through their laboratory rest frame. Then, that 4-displacement is divided by the proper time, [math]\Delta \tau[/math], elapsing between those two events, in the particle's own rest-frame. I.e. "the observer measures the displacement, but the particle reports the time".

 

 

In special relativity, momenergy is calculated in a single reference frame. The 4-displacement is measured in that frame, as you indicate. But the proper time is invariant -- it is the same for all (uniformly moving) frames of reference.

Edited February 8, 2012 by IM Egdall

[michel123456]

Nope.

 

On a sphere there are at least two geodesics between any two points -- each a segment of a great circle passing through the two points. Geodesics, even in the truly Riemannian case, do not necessarily minimize the arc length between two points, they only do so for points that are sufficiently close to one another. In the case of the Lorentzian metric used in general relativity, geodesics actually maximize local arc length.

 

Why nope?

I suspect a language issue.

I maintain: on a sphere between 2 random points there is only one geodesic which is the smallest path, meaning that any other geodesic has a longer path. When the longest path is equal to the smallest, then the 2 points are at the antipodes, and only in this case there are multiple path of the same length.

That was in response to your statement that

But there can be more that one path that exhibits that shortest distance. Consider lines of longitude on the globe. They are all geodesics between the poles.

I don't think such a configuration can be found easily in outer space: the Universe has no poles and random points in space cannot be found at the antipodes of the universe.

Edited February 8, 2012 by michel123456

[StringJunky]

Why nope?

I suspect a language issue.

 

 

It's just a more emphatic version of "No"....I suspect he's confident it's wrong.

[imatfaal]

My school level maths agrees with Michel. Assuming a perfect sphere and not the squidged shape of earth (and possibly with that too), then the only two points with more than one geodesic - ie that there are multiple shortest routes between are antipodes. Any point that are not antipodes will have one and only one shortest route.

 

reasoning:

The shortest route is a section of a great circle.

A great circle is the intersection of the surface of the sphere and a plane which passes through the centre of a sphere.

There can be only one plane which passes through three points as long as those points are not in a line.

The points cannot be in a line (start of journey, end of journey, centre of earth) unless the start and finish are antipodal

[Santalum]

In Euclidian geometry; the shortest path between any two points is the straight line.

 

In non-Euclidian geometry; the shortest path between any two points is the Geodesic.

 

In either case, there cannot be more than one shortest distance.

 

 

 

 

A geodesic is an arc according to this: http://en.wikipedia.org/wiki/Geodesic

 

And the shortest distance between two points in 4D spacetime could be a wormhole........equivalent to boring through the earth from one point on the surface to another.

 

Correct?

[michel123456]

My school level maths (...)

 

You are too modest.