Radius of curvature of Spacetime

[studiot]

Would I be right in calculating that Eddington's value for the grazing deflection of light past the Sun of 1.6 seconds of arc implies a value of around 9.6 x 10¹⁷ metres for the curvature of spacetime in the vicinity of the Sun?

Edited April 7, 2018 by studiot

[geordief]

 

Am I right to be reminded of an old thread I started 18 months ago?

If so can I  also ask what might be the value of pi for  one of the neutron stars in the binary black hole merger recorded not so long ago.

Is Studiot's measure of curvature essentially the same thing as this divergence from pi in a massive object such as the neutron star or a black hole?

 

 

Edited April 7, 2018 by geordief

[studiot]

Thank you for finding that thread geordief, I half remembered it, but  I seem to have trouble finding old threads again.

I note I should have proposed the 'effective radius' of curvature to be of the order of 1 x 10¹⁸ metres.

This is from a simple geometrical calculation of actual measurement, which can be and has beeen made.

[studiot]

On ‎4‎/‎7‎/‎2018 at 3:05 AM, geordief said:

Is Studiot's measure of curvature essentially the same thing as this divergence from pi in a massive object such as the neutron star or a black hole?

 

No it is calculated from the deviation of starlight passing the Sun, as observed by Eddington.

[mistermack]

Am I right in saying that curvature is four dimensional curvature, a combination of speed and mass and distance?

For instance, if your speed relative to the Sun is zero, then your free fall is a straight line towards the centre of the sun. The faster you move tangentially, the greater your deflection from that line in free fall. And the speed of light gives the greatest deflection. 

So it's wrong to picture a fixed curvature of space ?

[Strange]

1 minute ago, mistermack said:

Am I right in saying that curvature is four dimensional curvature

Yes, it is space-time curvature.

1 minute ago, mistermack said:

, a combination of speed and mass and distance?

No. Space and time.

1 minute ago, mistermack said:

For instance, if your speed relative to the Sun is zero, then your free fall is a straight line towards the centre of the sun. The faster you move tangentially, the greater your deflection from that line in free fall. And the speed of light gives the greatest deflection. 

That doesn't change the curvature of space-time. (Except, of course, there will be some curvature associated with your own mass and that will move as you do.)

2 minutes ago, mistermack said:

So it's wrong to picture a fixed curvature of space ?

In a given distribution of mass, the curvature is fixed. 

[mistermack]

So what does curvature actually mean in words? How does it relate to the free fall of an object?

[Strange]

11 minutes ago, mistermack said:

So what does curvature actually mean in words? How does it relate to the free fall of an object?

It describes the way that the relative paths of particles moving forwards in time change in the spatial direction(s). 

As an analogy, imagine two people walking forwards, side by side, on a flat plane. Their paths will remain parallel over time. In fact, we can consider the direction they are walking as the "time" dimension (they are moving steadily into the future) and the distance between them as a space dimension.

On the the flat plane, the distance between them doesn't change over time. Now put them on the surface of the Earth and have them both walk towards the North Pole. As they move forwards (in time) they get closer together. No force is acting on them, it is just a consequence of the curved geometry they are travelling in. You can consider them falling towards one another because of the gravity of the curved space-time they are in.

More here: http://math.ucr.edu/home/baez/einstein/einstein.html

[koti]

26 minutes ago, mistermack said:

So what does curvature actually mean in words? How does it relate to the free fall of an object?

An easy way of picturing this for me is this statement: Gravity is spacetime curvature.

[Mordred]

Another hint the curvature is followed by the null geodesic light path as the primary metric. The primary reason for this is the curvature affects how we see and measure the cosmos. This includes distortions of CMB which is one of the primary measuring tools to confirm what curvature constant "k" is the best fit ie Planck dataresults.

 So for example here is the 4d spacetime null geodesic of the FRW metric

[latex]d{s^2}=-c^2d{t^2}+a(t^2)[dr^2+S,k(r)^2d\Omega^2][/latex]

[latex]S\kappa(r)= \begin{cases} R sinr/R &k=+1\\ r &k=0\\ R sinr/R &k=-1 \end {cases}[/latex]

On the first k=+1 light paths converge positive curvature. In the second k=0 light paths remain parallel (flat) when k=-1 light paths diverge (negative curvature)

Edited April 24, 2018 by Mordred

[studiot]

11 hours ago, mistermack said:

So it's wrong to picture a fixed curvature of space ?

 

The curvature is a local phenomenon, which means it varies from point to point in spacetime and can be determined with reference to nearby points in spacetime.

However remember the fourth axis of spacetime is ct not time. This has the dimension of length, and units of metres.

So, unlike Strange's example where the radius arm points to a common point in 3D spacetime, there is no such common centre in 5D spacetime.
The radius arm at every point in our 4D manifold extends in a different direction in the mathematical 5D spactime.

 

 

11 hours ago, Strange said:

It describes the way that the relative paths of particles moving forwards in time change in the spatial direction(s). 

So long as this doesn't give the impression that the radius is somehow in 'time'.

Which brings me back to my OP calculation (in metres).

 

Still no takers?

 

 

[Strange]

1 minute ago, studiot said:

So long as this doesn't give the impression that the radius is somehow in 'time'.

Well, as far gravity is concerned, it is the curvature in the time dimension that is more significant.

It is not clear to me that intrinsic curvature of a 4D manifold can be represented as a radius (or even four radiuses).

4 minutes ago, studiot said:

So, unlike Strange's example where the radius arm points to a common point in 3D spacetime, there is no such common centre in 5D spacetime.

The "common point" in 4D space-time would be the centre of a massive object (more obvious in the case of black hole, where that becomes a point of infinite curvature).

And what is 5D space-time? 

[studiot]

2 hours ago, Strange said:

Well, as far gravity is concerned, it is the curvature in the time dimension that is more significant.

Really?

Why do you say that?

 

And how does that pertain to the OP?

Or, if you like, how does any bending in the ct direction allow light to go round a solid obstacle?

 

[Strange]

4 minutes ago, studiot said:

Really?

Why do you say that?

Because in the analogy, it is curvature in the time dimension that brings the two walkers together.

5 minutes ago, studiot said:

And how does that pertain to the OP?

It doesn't. I was answering the question from mistermack. He seems to have taken the thread of track slightly - apologies for following him!

What is 5d space-time?

[studiot]

3 minutes ago, Strange said:

It doesn't. I was answering the question from mistermack. He seems to have taken the thread of track slightly - apologies for following him!

What is 5d space-time?

And yet spatial displacement was the originally proposed test of GR and the actual one which provided the first practical verification.

I am aware of your longitude analogy, and am hoping to offer a fuller answer to that and your other questions, but in turn, I would also be grateful for some discussion of the OP.

[mistermack]

On 4/7/2018 at 1:12 AM, studiot said:

Would I be right in calculating that Eddington's value for the grazing deflection of light past the Sun of 1.6 seconds of arc implies a value of around 9.6 x 10¹⁷ metres for the curvature of spacetime in the vicinity of the Sun?

Is the curvature of spacetime directly related to the strength of the gravitational field at a point? I was viewing the one as a way of modelling the other for mathematical purposes.

[studiot]

18 hours ago, mistermack said:

Is the curvature of spacetime directly related to the strength of the gravitational field at a point? I was viewing the one as a way of modelling the other for mathematical purposes.

Basically yes.

That is why the Sun's gravity was used for the test.