Markus Hanke

Why make it complicated ? Just model the gravitational field as going radially outward from an isolated point source, and with spherical symmetry ( a physically reasonable model ); this makes the field both irrotational and source-free outside the point source, in other words, it makes it conservative. Mathematically, for a vactor field F, this means that outside the source the following conditions are fulfilled :

[math]\displaystyle{\triangledown \cdot \mathbf{F}=0}[/math]

and

[math]\displaystyle{\triangledown \times \mathbf{F}=0}[/math]

So both the divergence and the curl must vanish. In flat 3-dimensional space this automatically implies an inverse square law, as can easily be shown in spherical coordinates from the above differential equations.

I think you are missing the point here.

The twin paradox has three elements :

1. Both twins start off at rest in the same frame of reference

2. One twin remains at rest throughout the experiment, the other twin undergoes phases of acceleration and uniform relative motion along a closed path which eventually returns him to the first ( stationary ) twin

3. The twins reunite once again at rest in the same frame of reference, and find that their clocks do not agree

Let us take a look at their proper times; in general terms proper time is defined as the arc length of an observer's worldline, which is

[math]\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}}[/math]

where C is the path taken.

Without even needing any specific figures we can immediately evaluate what happens for our twins; the stationary twin travels only through time and does not experience acceleration, so his proper time is simply

[math]\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{\eta _{\mu \nu }dx^{\mu }dx^{\nu }}=\int_{C}dt}[/math]

On the other hand, the travelling twin moves through time and space, and does experience acceleration; his proper time is therefore

[math]\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}=\int_{C}\sqrt{g_{00}dt^2-\sum_{i=1}^{3}g_{ii}(dx^{i})^2}}[/math]

which is always less than the stationary twin. Due to the equivalence principle the metric tensor in the above expression encapsulates the acceleration information. This shows us that stationary observers not subject to acceleration always experience the longest proper time.

It remains to be noted that trying to consider this scenario without acceleration is not physically meaningful, because the travelling observer would not be able to return to his stationary twin, so their clocks could not be compared in any meaningful way. Bear in mind that any deviation from uniform relative motion always involves acceleration.

S.R. or G.R.?

Where there is acceleration, mass, gravitation, what do you use?

It's SR, for Special Relativity. In that framework you don't have acceleration and gravity.

I doubt very much that our friend Mahesh would be able to tackle General Relativity...but if he does I'll be there

As good as this is, crackpots don't tend to be very familiar with mathematical rigor. So while a consistency check is good for those of us who are familiar with relativity and the Poincare group, I doubt it will be very convincing to people who are "completely against relativity."

 

I do appreciate input from competent people, so thank you.

No problem, glad to be able to make a contribution.

Let me attempt to show that within the axioms of SR it is actually impossible to arrive at any paradoxes; SR is an internally self-consistent system, rendering any alleged "paradoxes" physically meaningless.


Fundamental Postulates of Special Relativity

(1) The substance existing at any world point can always be conceived to be at rest, if time and space are interpreted suitably. In other words - locally, all events can be considered to be in an inertial frame
(2) Between all inertial frames the same laws of physics apply, regardless of their states of relative motion
(3) Space-time can be considered isotropic and homogenous
(4) Rulers and clocks function independently of their past history


Self-Consistency Condition

(5) In order for (1)-(4) to hold, the line element measuring the distance between two events in space-time must be the same for all observers, i.e. must not vary if going from one frame to another, regardless of their states of relative motion. Going from one frame to another, and then back to original frame, will yield the same event in space-time.


Mathematical Proof

The distance between two events in space-time can be defined as a line element of the form

[latex]\displaystyle{ds^2=d\mathbf{R}\cdot d\mathbf{R}=g_{\mu \nu }dx^{\mu }dx^{\nu }}[/latex]

wherein [latex]g_{\mu \nu }[/latex] shall be called the metric tensor, and can be thought of as a 4x4 matrix which transforms according to certain rules.
The mathematical description of going from one inertial frame into another is realised by introducing a linear transformation between two vectors x' and x of the form

[latex]\displaystyle{{x}'^{\mu }=L{^{\mu }}_{\nu }x^{\nu }+a^{\mu }}[/latex]

wherein L is a general transformation matrix which represents an as-per-yet unspecified boost and rotation in arbitrary directions, and the 4-vector a represents a shift of origin. We now demand the following restriction to hold :

[latex]\displaystyle{L^{T}gL=g}[/latex]

which corresponds to the simple observation that, when performing a rotation and its inverse, you always arrive at the original vector, i.e. a rotation and its inverse chained together will yield the unity matrix. In tensor language this corresponds to

[latex]\displaystyle{g_{\mu \nu }L{^{\mu }}_{\rho }L{^{\nu }}_{\sigma }=g_{\rho \sigma }}[/latex]

In order to prove (5) one now only needs to show that such a transformation L leaves the space-time line element ds invariant, meaning that the distance between two events in space-time is the same for all inertial observers :

[latex]\displaystyle{g_{\mu \nu }d{x}'^{\mu }d{x}'^{\nu }=g_{\mu \nu }L{^{\mu }}_{\rho }L{^{\nu }}_{\sigma }dx^{\rho }dx^{\sigma }=g_{\mu \nu }dx^{\mu }dx^{\nu }}[/latex]

Quod erad demonstrandum. What this means is that, because above line element is invariant under said transformation, all inertial observers experience the same laws of physics.

References

Minkowski, Hermann (1908/9). "Raum und Zeit". Jahresberichte der Deutschen Mathematiker-Vereinigung: 75–88.

• English translation: Space and Time. In: The Principle of Relativity (1920), Calcutta: University Press, 70-88

That is the one of the reason for which I am completely against relativity.

Address the above, then.

But isn't that idea only valid, in a "Steady State" Universe.

No, it is valid in general for any Lorentzian 4-manifold. You cannot, in general, cover such a manifold with just one coordinate chart; physically that means that there is no universal frame of reference. This would hold both in a "Steady State" as well as an expanding universe. Also, as other posters have quite correctly pointed out, there is no "centre point" to such a manifold.

Two simultaneous points in time separated by vast space is not a possible variable for one would have to know the exact time differential between galaxys as in relative speeds and rotational speeds affecting the immidiate momententum of that specific event ie time this is beyond current sciences. Also the moment we experience now from say 100 light years away is already in the fast distant past . As in viewed from 10000 years in the future if this helps dispel some of your confusion about the nature of time and why its difficult to understand .

Exactly, that is part of a wider problem when talking about simultaneity of two spacially separated events on a general Lorentzian 4-manifold. There just isn't any "universal" frame of reference against which we can decide simultaneity, and attempting to do so in the local coordinate charts will lead to problems.

I've studied the above posts, as best I can. Yet they don't dispel this feeling: That right now - at this exact moment - an event is happening on Earth, and an event is happening in the M.31 Andromeda Galaxy, and the two events are simultaneous.

And how do you decide that they are simultaneous, given the finite speed of light and non-uniform and non-stationary curvature of space-time between here and there ?

Why can't time be constant for everything in the universe?

Because, in the general case, you cannot cover the entirety of a Lorentzian 4-manifold with a single coordinate chart. What that basically means is that both space and time are local phenomena only; observers at different points on the space-time manifold may thus not agree on clock readings and ruler measurements. In fact, even the very notion of comparing such readings in different frames of reference is at best pretty problematic since there is no universal coordinate chart.

Assuming there are no singularities, spacetime can always be considered locally Minkowskian, not just in a vacuum.

That's true, but if you are not in a vacuum then [math]\displaystyle{T_{\mu \nu }\neq 0}[/math], and the Ricci tensor would not vanish.

All I was trying to say is that a vanishing Ricci tensor does not automatically imply a flat space-time; it is a necessary, but not a sufficient condition since, as mentioned earlier, the Riemann curvature tensor is not completely determined by its contractions in 4 dimensions.

In a course, when evaluating the gravitation field by a reasonable mass, Pascal Picard explicitly rejects the Ricci tensor and takes the full Einstein tensor instead. You may ask why, if you wish, but I won't answer that...

In four dimensions the Ricci tensor does not fully determine Riemann curvature, so

[math]\displaystyle{R_{\mu \nu }=0}[/math]

does not necessarily mean that we are deal with a globally flat space-time. All it means that we are considering a vaccum, and that space-time can locally be considered to be Minkowskian. To determine the global geometry one would need additional constraints on the system, as for example the Schwarzschild metric does by demanding that it vanishes at infinity, and that it reduces to Newtonian gravity for weak fields.

Elaborate on the fact that gravity exists due to space time curve.

In the above sentence, swap "due to" with "is" - that will give a more accurate statement.

Consider for a moment elfmotat's avatar, and imagine two observers standing on the equator of that sphere. Now let our observers start walking north, along the lines as drawn in the avatar picture. What happens ? The further north they get, the more they approach each other. At the north pole, they meet. This happens not because of any force between them, but simply because of the geometry of the surface they are moving on.

Likewise GR - two bodies will approach one another not because there are any forces between them, but because of the geometry of space-time itself. Gravity ceases to be a phenomenon external to a body, and becomes a geometric property of space-time.

Binary pulsars are often used to test predictions of GR, namely, the possibility of energy loss due to gravitational waves. But we never see any analysis of SR's predictions, although it is an extremely potential source of empirical data on time dilation due to velocity as well, since during the entire orbit the relative speed between earth-pulsar is variable.

On what grounds do think it is permissible to use SR when analysing a binary pulsar system ? Surely you are aware that SR requires inertial reference frames.

Gravitation by Thorne/Misner/Wheeler is a very in-depth treatment of the subject, and considered the "gold standard" by many. A good understanding of exterior calculus / differential forms is also helpful in understanding GR, but not necessarily required.