When did spacetime form?

[MigL]

Textbook geometry might be fine with singularities, but I don't think space-time geometry is.
My opinion; take it for what it's worth.

[Genady]

1 minute ago, MigL said:

Textbook geometry might be fine with singularities, but I don't think space-time geometry is.
My opinion; take it for what it's worth.

Space-time geometry is fine with singularities, too. Take a triangle. It is a perfect geometric shape in spite of having three singularities.

[joigus]

7 hours ago, Genady said:

This is incorrect. GR requires geometry with certain smoothness. It fails if the geometry is not sufficiently smooth.

 

7 hours ago, MigL said:

Exactly.
As happens when geometric curvature goes to infinite at a singularity.

The argument that always creeps in is that,

looks (and is) smooth enough, while,

doesn't so much, although it's equally smooth. It's just that the curvature at the corners is enormous as compared to scale 1.

These are called squircles, btw, and they're fascinating.

 

[Genady]

22 minutes ago, joigus said:

 

The argument that always creeps in is that,

looks (and is) smooth enough, while,

doesn't so much, although it's equally smooth. It's just that the curvature at the corners is enormous as compared to scale 1.

These are called squircles, btw, and they're fascinating.

 

In this example, a smooth curve appears not smooth as one zooms out. The opposite is also possible, i.e., a rugged curve appears smooth as one zooms out.

[J.C.MacSwell]

11 hours ago, iNow said:

On a Tuesday

The second day?

...yep, you're right!...

Our new Trump Bibles sure come in Handy!

[MigL]

Thanks for that Joigus.
It does relate to scale, as my original assertion was that at  Planck time, or at Planck scale, physical geometry might vanish.
And I've been unsuccessfully trying to make my position clearer to Genady that, while 'textbook' geometry does allow for singular points, the physical geometry of space-time cannot support such structures.
IOW infinities are useful, and fine, in mathematics, not so much in the Physics of the real world.

[joigus]

9 hours ago, Genady said:

In this example, a smooth curve appears not smooth as one zooms out. The opposite is also possible, i.e., a rugged curve appears smooth as one zooms out.

Yes. Scale can dilute our delusions.

6 hours ago, MigL said:

It does relate to scale, as my original assertion was that at  Planck time, or at Planck scale, physical geometry might vanish.
And I've been unsuccessfully trying to make my position clearer to Genady that, while 'textbook' geometry does allow for singular points, the physical geometry of space-time cannot support such structures.

It shouldn't should it? Einstein seems to have been of the same opinion.

I find it very difficult to disagree with that. The most likely candidate to smooth things out is, of course, quantum mechanics.

And Genady is very hard to win over, probably because he lives on an island.

[Markus Hanke]

6 hours ago, joigus said:

It shouldn't should it? Einstein seems to have been of the same opinion.

This got me thinking - setting aside singularities for the moment, are there (in a purely mathematical sense) metrics that are not valid solutions to the EFE? IOW, could one write down a metric for which the equations cannot be worked backwards to obtain a corresponding energy-momentum tensor, given we are on a semi-Riemannian manifold?

Again, I’m talking purely mathematically, never mind physical realisability.

I suspect the answer is no.

Edited June 8 by Markus Hanke

[joigus]

5 hours ago, Markus Hanke said:

This got me thinking - setting aside singularities for the moment, are there (in a purely mathematical sense) metrics that are not valid solutions to the EFE? IOW, could one write down a metric for which the equations cannot be worked backwards to obtain a corresponding energy-momentum tensor, given we are on a semi-Riemannian manifold?

Again, I’m talking purely mathematically, never mind physical realisability.

I suspect the answer is no.

That's an outstandingly good question IMO. I know inverse problems in physics are notoriously difficult. Eg, the inverse-scattering problem: Given the scattering properties, find the scatterer. What you're proposing, in terms of scattering problems, would be: Given any scattering, can I always find a scatterer that does it?

My guess would be no too. That you can always find a T that does it, even though you might have to make it un-physical. But I cannot go beyond the guess at this point.

[Genady]

5 hours ago, Markus Hanke said:

This got me thinking - setting aside singularities for the moment, are there (in a purely mathematical sense) metrics that are not valid solutions to the EFE? IOW, could one write down a metric for which the equations cannot be worked backwards to obtain a corresponding energy-momentum tensor, given we are on a semi-Riemannian manifold?

Again, I’m talking purely mathematically, never mind physical realisability.

I suspect the answer is no.

If the metric is twice differentiable everywhere, then its Einstein tensor is everywhere defined, and you can just take this Einstein tensor as the energy-momentum tensor of your equation. Then, this metric is a solution of this equation.

P.S. Of course, the metric has to be locally Lorentz to start with.

[Genady]

Developing on my previous post, I think that the metric \[diag(-1,1,1,1+H(z)z^2+H(-z)z^4)\] is not a valid solution to the EFE.

*H() is Heaviside step function.

[Markus Hanke]

17 hours ago, Genady said:

If the metric is twice differentiable everywhere, then its Einstein tensor is everywhere defined

Very good point!

[joigus]

On 6/8/2024 at 7:28 PM, Genady said:

Developing on my previous post, I think that the metric

diag(−1,1,1,1+H(z)z2+H(−z)z4)

is not a valid solution to the EFE.

 

*H() is Heaviside step function.

Ah, yes! But this introduces discontinuity in the metric by hand. So it's not a surprise that PDE's (never mind their being non-linear or having a good deal of geometric meaning) cannot encompass them.

[Genady]

1 hour ago, joigus said:

Ah, yes! But this introduces discontinuity in the metric by hand. So it's not a surprise that PDE's (never mind their being non-linear or having a good deal of geometric meaning) cannot encompass them.

To be precise, this metric is continuous, differentiable, and its first derivative is continuous. It is not twice differentiable though.

Of course, it was introduced by hand, as defined in the question: "could one write down a metric for which ..."

[joigus]

33 minutes ago, Genady said:

To be precise, this metric is continuous, differentiable, and its first derivative is continuous. It is not twice differentiable though.

Of course, it was introduced by hand, as defined in the question: "could one write down a metric for which ..."

Sorry, Genady. That's what I should have said. Should  have meant? In the general theory of PDE's, as you probably know, the eqs. can be proven to have unique solutions for any complete set of initial/boundary data iff the highest order derivatives can be solved in terms of the rest by Lipschitz-continuous functions (simply continuous is a stronger condition that of course suffices).

g''=F(g',g,T,x)

where (schematically) g is the "function to be solved for", F is a function, T is a source, and x is the point.

Heaviside's step function is not continuous, but you're right that the "damping factors" z² and z⁴ seem to patch up the zeroth and first derivatives, but not the second. I'm next to sure that the moment you second-differentiate, you'd run into problems though.

49 minutes ago, Genady said:

Of course, it was introduced by hand, as defined in the question: "could one write down a metric for which ..."

I wasn't critizising your example. I understood the motivation, and it's very nicely devised.

[KJW]

13 hours ago, joigus said:

Heaviside's step function is not continuous, but you're right that the "damping factors" z² and z⁴ seem to patch up the zeroth and first derivatives, but not the second. I'm next to sure that the moment you second-differentiate, you'd run into problems though.

Why would there be problems? So what if the second derivative has a jump discontinuity. If the source term of the EFE was describing a ball of matter, then there would be a jump discontinuity at the surface of the ball, and the solution metric would be a cookie-cut of the metric describing the ball of matter into the Schwarzschild metric, with matched zeroth and first derivatives at the surface of the ball.

However, the metric:

[math]\text{diag}(−1,1,1,1+H(z)z^2+H(−z)z^4)[/math]

does describe a flat spacetime, so all the derivatives will cancel to zero in the Riemann tensor field.

 

[joigus]

1 hour ago, KJW said:

However, the metric:

diag(−1,1,1,1+H(z)z2+H(−z)z4)

does describe a flat spacetime, so all the derivatives will cancel to zero in the Riemann tensor field.

 

Are you sure? What is H'(z) and H''(z)?

You might be right, and I wrong.

You might be wrong, and I right.

Or we both right or wrong, and not know.

In any case, it's interesting.

[KJW]

23 hours ago, joigus said:

Are you sure?

Am I sure that the metric describes flat spacetime? Yes, I am. Any metric of the form:

(ds)² = T(t)² c² (dt)² – X(x)² (dx)² – Y(y)² (dy)² – Z(z)² (dz)²

describes flat spacetime. The following coordinate transformation exists between this metric and the Minkowskian metric:

t' = t'(t)    ;    x' = x'(x)    ;    y' = y'(y)    ;    z' = z'(z)

where t'(t), x'(x), y'(y), and z'(z) are solutions to the differential equations:

dt'(t)/dt = T(t)
dx'(x)/dx = X(x)
dy'(y)/dy = Y(y)
dz'(z)/dz = Z(z)

Ignoring your use of the (–,+,+,+) signature, your metric is of the above form:

T(t) = X(x) = Y(y) = 1
Z(z) = H(z)(1 + z²)^½ + H(–z)(1 + z⁴)^½

 

Edited June 12 by KJW

[joigus]

When you construct the Riemann tensor, second derivatives of the metric coefficients are involved, as you well know.

As you didn't address my question (only half of it), I'll repeat:

23 hours ago, joigus said:

What is H'(z) and H''(z)?

Those are certainly involved in the calculation of the Riemann tensor, aren't they?

[KJW]

2 hours ago, joigus said:

When you construct the Riemann tensor, second derivatives of the metric coefficients are involved, as you well know.

As you didn't address my question (only half of it), I'll repeat:

On 6/11/2024 at 11:00 PM, joigus said:

What is H'(z) and H''(z)?

Those are certainly involved in the calculation of the Riemann tensor, aren't they?

I didn't calculate the Riemann tensor because I had an alternative method of proving the metric is flat. And because the metric is flat regardless of the function [math]Z(z)[/math], in the Riemann tensor, the cancellation of derivatives must be exact regardless of any pathologies in the function [math]Z(z)[/math]. However, I will attempt to answer your question:

[math]H'(z) = \delta(z)[/math]

[math]H''(z) = \dfrac{d}{dz} \delta(z) =\ ...[/math] I'll have to refresh my memory of distributions before answering this.

 

 

Let [math]T(u)[/math] be an arbitrary test function of [math]u[/math]. Then:

[math]\int_{-\infty}^{\infty} T(u)\ \delta(u-x)\ du = T(x)[/math]

[math]\int_{-\infty}^{\infty} T(u)\ \dfrac{d}{du} \delta(u-x)\ du = -\int_{-\infty}^{\infty} \dfrac{d}{du} T(u)\ \delta(u-x)\ du = -T'(x)[/math]

 

 

Edited June 12 by KJW

[joigus]

Yes, it's flat space-time, almost everywhere. It seems obvious, doesn't it? A simple re-scaling of the variables takes you back to Minkowski. When I said "are you sure?" I meant that all the derivatives cancel. That there is total continuity. That "there is no problem", as you implied. This requires a calculation, and perhaps a little thought too. I'm getting,

\[ g''_{zz}\left(z\right)=2H\left(z\right)+2z\delta\left(z\right)+12z²H\left(-z\right)-12z³\delta\left(-z\right) \]

This is discontinuous because of the step function, not because of the delta terms, which have damping factors as I mentioned before. I had to do the calculation to actually see the step discontinuity. Has this discontinuity come from a sloppy parametrisation of empty space, or does it correspond to a weird distribution of matter at z=0? That's another question.

I mention all this because, eg, a cone is flat space everywhere, except at a point where it has infinite curvature. That's why you must be careful and look at the topology, global properties, and so on. It seemed to me you were just blissfully saying it was flat space because a purely formal re-scaling does it.

Does Genady's example answer Marcus' question then?

 

Edited June 13 by joigus
Latex editing

[Markus Hanke]

13 hours ago, joigus said:

Does Genady's example answer Marcus' question then?

I think he answered the question, and the answer was in fact so obvious that I couldn’t even see it myself 🤨

As for step functions, I don’t see how these would arise in a real-world physical metric, though I’m aware of exact solutions containing delta functions, eg the Aichlburg-Sexl ultraboost.

[KJW]

20 hours ago, joigus said:

It seemed to me you were just blissfully saying it was flat space because a purely formal re-scaling does it.

I don't think you fully grasped the logic behind this sentence from my previous post:

"And because the metric is flat regardless of the function [math]Z(z)[/math], in the Riemann tensor, the cancellation of derivatives must be exact regardless of any pathologies in the function [math]Z(z)[/math]."

That is, the cancellation of derivatives in the Riemann tensor must occur for reasons that have nothing to do with the properties of [math]Z(z)[/math]. To be specific, of the 64 first-order partial derivatives of the above metric tensor components, the only one that is not exactly zero is [math]g_{zz,z}[/math], and of the 256 second-order partial derivatives of the above metric tensor components, the only one that is not exactly zero is [math]g_{zz,zz}[/math]. Therefore, the only component of the corresponding Riemann tensor that could possibly be non-zero is [math]R_{zzzz}[/math]. But the Riemann tensor is antisymmetric on the first two indices, [math]R_{ijkl} = -R_{jikl}[/math]. Therefore, [math]R_{zzzz} = -R_{zzzz} = 0[/math]. Thus, all components of the Riemann tensor for the above metric are exactly zero (everywhere).

 

 

Edited June 14 by KJW

[joigus]

3 hours ago, KJW said:

I don't think you fully grasped the logic behind this sentence from my previous post:

Oh, believe me, I did.

One would think it's you who doesn't want to grasp the logic behind mine: Are you sure?

Symmetry of partial derivatives is a consequence of continuity:

https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Schwarz's_theorem
Somewhere on the back of my mind I was trying to remember an argument that validates it. It might be something like this:

https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Distribution_theory_formulation

But this is becoming farther and farther away from the OP with every post --and less and less interesting for everybody else, I suspect.

9 hours ago, Markus Hanke said:

I think he answered the question, and the answer was in fact so obvious that I couldn’t even see it myself 🤨

As for step functions, I don’t see how these would arise in a real-world physical metric, though I’m aware of exact solutions containing delta functions, eg the Aichlburg-Sexl ultraboost.

Mm... Yes, you're right. Genady actually said it (if I understood correctly). That, as \( G_{\mu\nu}=8\pi G T_{\mu \nu} \), any \( G_{\mu \nu} \) obtained from the metric connection by means of the \( g_{\mu\nu} \) you give yourself can be used as the \( T_{\mu \nu} \) just after dividing by \( 8 \pi G \).

Edited June 14 by joigus
Latex editing

[KJW]

1 hour ago, joigus said:

Symmetry of partial derivatives is a consequence of continuity

But, we are not dealing with mixed partial derivatives. So, even if [math]R_{ijkl} = -R_{jikl}[/math] doesn't hold in general due to discontinuity, [math]R_{zzzz} = -R_{zzzz} = 0[/math] must still hold.