Genady

3 minutes ago, joigus said:

The mechanisms of denial. Procrastination being but one aspect perhaps. In the end it's all one big shrugging gesture.

Sour grapes.

2 hours ago, Dhamnekar Win,odd said:

Do you agree with my derivation for the expression of Pn=n⋅rn2⋅R−rn,R2=R−rn . If not, I shall show you my working over these derivation.

Look at the picture for case 2:

You can count, \(n=13\). You can measure, \(R/r_n \approx 5\). If your formula, \(P_n = \displaystyle\frac{n \cdot r_n}{2\cdot R -r_n}\) were correct, then \(P_{13}=\displaystyle\frac{13 \cdot 1}{2\cdot 5 -1} \gt 1\), which cannot be correct because it has to be \(\lt 1\) for any \(n\).

So, your result is wrong.

3 minutes ago, Dhamnekar Win,odd said:

Do you agree with my derivation for the expression of Pn=n⋅rn2⋅R−rn,R2=R−rn . If not, I shall show you my working over these derivation.

Please show it.

1 hour ago, Genady said:

I'd need to see your step-by-step derivation.

Surely your result, \(P_n=2.0944\) is wrong because \(P_n\) has to be \(\lt 1\) by its definition.

42 minutes ago, Dhamnekar Win,odd said:

If limn→∞n⋅rn→2⋅π⋅R2

 

We got R−rn=R2⇒R=R2+R2=2R2  

 

Putting these values in our final expression for Pn we get 2⋅π⋅R24⋅R2−R2=23⋅π=Pn=2.0944 approx.

 

But author's answer is Pn=0.9069 approx. 

How is that?

 

 

 

I'd need to see your step-by-step derivation.

Here is the explanation of how time is defined, from the Gravitation by Misner, Thorne, and Wheeler: 

1 hour ago, chron44 said:

How can science at all deal with universal physical issues such as Newtonian and GR/ SR/ QM physics without the true scientifical "formula" for time?

There are many scientific formulas for time. For example, \(t= \frac d v\), \(t'=\gamma(t- \frac {vx}{c^2})\), \(d \tau^2=dt^2-ds^2\), etc.

1 hour ago, grayson said:

you could have at least explained why we do not use kelvin

You did not ask.

2 hours ago, Dhamnekar Win,odd said:

In that case Pn would be 

limn→∞,rn→0Pn=n⋅rn2R−rn=0

 

 

 

No, not necessarily.

For example, if \(r_n= \frac c n\) then \[\lim_{n \to \infty} n \cdot r_n=c\]

"Why do we use kelvin to measure heat?"

We do not.

9 minutes ago, Dhamnekar Win,odd said:

But in this case limn→∞Pn=∞ Isn't it?

Not necessarily. As \(n \rightarrow \infty\), \(r_n \rightarrow 0\).

2 hours ago, Airbrush said:

Most of this is beyond my understanding.

"Time and energy are Fourier conjugates (or more generally, spacetime and energy-momentum) and cannot exist in the physical reality without each other. In other words, GR states that spacetime is the field produced by matter just like the electromagnetic field is produced by charges. Vacuum solutions are unphysical, they don’t exist in reality. Their flaw is that the equations are solved without realistic physical initial conditions. This approach and resulting solutions are physically meaningless."

But then there is also this from the same source:

"No, general relativity doesn't make any claim as to whether matter must exist or not. In fact, the simplest of the solutions to the Einstein equations are vacuum solutions. For example, the Kerr-Newman blackholes and their special cases such as the Schwarzschild blackholes and Kerr blackholes. The dimensionality of spacetime is still 4D in these solutions with one dimension being time-like." 

general relativity - Does Time Require Matter to Exist? - Physics Stack Exchange

The central postulate of general relativity is that local physics is physics of special relativity. And special relativity does not make any claims about matter.

3 minutes ago, chron44 said:

cognitive ability

Cognitive abilities are not a topic in physics. This post belongs to another forum.

10 minutes ago, Dhamnekar Win,odd said:

I am sorry. I wrongly computed Pn My new rectified Pn=n⋅rn2R−rn

You need to express it as a function of \(n\).

19 minutes ago, geordief said:

Not sure how it affects the point I was trying to make.

It removes the consideration of superluminal expansion from the point you were trying to make.

1 hour ago, geordief said:

it is receding from us faster than the speed of light and so has become invisible to us.

This is a common misconception. In fact, we can and do see light coming to us from sources which recede faster than \(c\). It is so because as the emitted light moves away from its source along the line connecting us to the source, it gets to parts which recede from us slower than the source. Eventually, it gets to parts which recede slower than \(c\).

Here is a more detailed description:

([astro-ph/0310808] Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe (arxiv.org))

7 hours ago, Dhamnekar Win,odd said:

Replacing n ... by 2

Anyway, n cannot be 2 in these configurations if n is:

On 1/5/2024 at 6:35 AM, Dhamnekar Win,odd said:

the number of the small circle in the outer layer.

1 hour ago, Dhamnekar Win,odd said:

Would you explain your answer in details?

Never mind, I got it.

1 hour ago, Dhamnekar Win,odd said:

How is that?

Looking at it ...

1 hour ago, Dhamnekar Win,odd said:

Replacing n in Rn by 2

How can you replace \(n\) by \(2\) in \(R_n\), but leave it \(n\) everywhere else in the expression?

22 minutes ago, geordief said:

does the light ray  with which we see the boat disappear exist  in  an embedding 3rd dimension making the surface extrinsic after all?

Yes, this is it. If the light were somehow tied to the 2D surface and went strictly along it, the ship would not disappear. A straight line on the surface of Earth is a great circle. But the light ray does not follow it - it rather follows a straight line in the 3D space.

Consider the example of a cylinder again. Imagine that the planet has a cylindrical shape. Unless the ship goes in the direction of the cylinder's axis, it will be disappearing behind the horizon as usual. But we know that the intrinsic curvature of the cylinder is 0. So, the disappearance of the ship over the horizon is a consequence of the extrinsic curvature, i.e., of how the surface is curved in the embedding space.

1 hour ago, geordief said:

Less and it is a positive curvature ;more and it is negative (like a saddle)

A minor correction here: one measures Δ=(sum-of-the-angles-of-a-triangle)−π .

If Δ>0 , i.e., sum of the angles is more then 180o , then the curvature is positive (like a sphere).

If Δ<0 , i.e., sum of the angles is less then 180o , then the curvature is negative (like a saddle).

1 hour ago, geordief said:

Not sure how you would do it in  4d spacetime.

You would not. It works only for space, more precisely, for a metric signature +++... .

The spacetime metric signature is -+++ (or +---, depending on convention). One calculates Riemann curvature tensor in a general case.

2 hours ago, TheVat said:

if I'm a flat bug on the shore of a calm lagoon I can measure a curvature of the "flat" water as flat bug-ships drop over the horizon

No, that would be an extrinsic curvature.

5 minutes ago, geordief said:

are we talking about spacetime curvature or the topology

The former.

48 minutes ago, geordief said:

very distant= still local ?

That wouldn't  be evidence of a global topology, would it?

The observable universe is still  "local" isn't it?

No. "Local" is space-time volume where the metric can be approximated by Minkowski's one with vanishing first derivatives.

9 minutes ago, geordief said:

How do we know that a global topology exists at all and that we don't just have a patchwork of local topologies?

We see the light coming from very distant sources in the observable universe. It appears to be very homogenous and isotropic.

1 hour ago, Dhamnekar Win,odd said:

Then in both cases PnR2π=nr2nπ+PnR2nπ

I see how \( P_n R^2\pi = n r^2_n \pi + P_n R^2_n \pi \) is correct in the first case, but not in the second one because in the second case the radius \( R_n \) partially covers the circles in the outer layer.

Exactly. This is the solution:

1 minute ago, MigL said:

Mathematics has no such limitations.