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The fall of the apple and the LC oscillatory circuit in the reference frame of the remote observer.


SergUpstart
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Mathematics aside ( because we know GR is incomplete, and all its predictions are not consistant with scaling effects ), while I agree G cannot be variable locally, or wrt time dilation ( ? ), there is the possibility that G is a global variable, and in a Machian manner, could depend on the size, and/or energy distribution, of the universe.
You would then, have G evolve with the evolution of the universe such that earlier eras of the universe would have differing values of G.
Examining far-away ( and therefore, much earlier ) parts of the universe has yielded no variation in the value of G.
And observational evidence, not pretty mathematics, is the final arbiter.

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8 hours ago, MigL said:

there is the possibility that G is a global variable, and in a Machian manner, could depend on the size, and/or energy distribution, of the universe.

I don’t think it is very meaningful to try and define G across a larger region of curved spacetime; it’s numerical value would then depend on the distribution of matter-energy in that region, and also on the observer whose clocks and rulers are used. It wouldn’t be an invariant.

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12 hours ago, MigL said:

You would then, have G evolve with the evolution of the universe such that earlier eras of the universe would have differing values of G.
Examining far-away ( and therefore, much earlier ) parts of the universe has yielded no variation in the value of G.
And observational evidence, not pretty mathematics, is the final arbiter.

A very difficult, I would say practically unsolvable task. If we observe two objects that revolve around each other, then in order to measure G we must measure their masses and the distances between them. Any deviation in the period of rotation can be explained by both the deviation of G and the deviation in the masses of objects. How to distinguish a change in the masses of objects from a change in G??

In addition, in the first post of this topic, I showed on a thought experiment with an oscillatory circuit that in the reference frame of a remote observer, ε0 and µ0, and therefore h, should change. 

And here is the news on this topic. An international team of astronomers has discovered a giant dead galaxy that existed 12 billion years ago, when the age of the universe was 1.8 billion years. This is reported in a press release on Phys.org.

The team conducted spectroscopic observations using the MOSFIRE spectrograph (Multi-Object Spectrograph for Infrared Exploration) located at the Keck Observatory (Hawaii).

The galaxy, which was designated as XMM-2599, was characterized by a high rate of star formation (one thousand solar masses per year), but then this process completely stopped. Scientists do not yet know what the cause of death of XMM-2599 is.

This may mean that in the early universe h was much smaller, which is why the stars did not light up.

 

 

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1 hour ago, SergUpstart said:

This is reported in a press release on Phys.org.

Interesting link, thanks!

1 hour ago, SergUpstart said:

This may mean that in the early universe h was much smaller, which is why the stars did not light up.

Is the above a speculation of yours or by the authors of the article? I did a quick read of the Astrophysical Journal Letters: 

Quote

While gas-rich major mergers are important in building up the stellar mass at early times, a reduction in the number of these events would limit the amount of gas available for star formation. Virial shocks and increased feedback from active galactic nuclei could provide the energy necessary to keep any remaining gas heated, thus prevent the cooling and collapse necessary for forming stars

from https://iopscience.iop.org/article/10.3847/2041-8213/ab5b9f, linked from the phys.org source.

 

Edited by Ghideon
clarified the source link
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1 hour ago, SergUpstart said:

Of mine

!

Moderator Note

We have requirements in speculations, demanding a certain level of rigor. We also require discussions stay on topic, and this thread was about G, not h

If you have more than conjecture(i.e. more than what you have presented here), open up a new thread

 
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12 minutes ago, swansont said:

We have requirements in speculations, demanding a certain level of rigor. We also require discussions stay on topic, and this thread was about G, not h

If you have more than conjecture(i.e. more than what you have presented here), open up a new thread

 

12 minutes ago, swansont said:
!

Moderator Note

We have requirements in speculations, demanding a certain level of rigor. We also require discussions stay on topic, and this thread was about G, not h

If you have more than conjecture(i.e. more than what you have presented here), open up a new thread

 
Initially, the topic was broader. Here is a quote from the first post of the topic
 
Thus, this thought experiment with an oscillatory circuit shows that in the reference frame of a remote observer, the dielectric and magnetic constants must change in different directions in accordance with the change in the time/distance scale. Then it is not difficult to show
that the Planck constant will also change in the reference frame of the remote observer, since it is equal to
hhhh.jpg
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25 minutes ago, SergUpstart said:

 

Initially, the topic was broader. Here is a quote from the first post of the topic
 
Thus, this thought experiment with an oscillatory circuit shows that in the reference frame of a remote observer, the dielectric and magnetic constants must change in different directions in accordance with the change in the time/distance scale. Then it is not difficult to show
that the Planck constant will also change in the reference frame of the remote observer, since it is equal to
hhhh.jpg

 

And thereby lies a mathematical contradiction.

Mu-nought and epsilon-nought in your formula are scalar constants.

That means they are independent of direction.

The condition for them to become direction dependent is that they are no longer scalars but in fact tensors (not even vectors will do).

This is in fact achieved under non isotropic conditions.

However the rub is that there is no such thing as the square root of the ratio of two tensors.

Edited by studiot
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1 hour ago, SergUpstart said:

 

Initially, the topic was broader. Here is a quote from the first post of the topic
!

Moderator Note

You were working on establishing the premise of this little exercise, which you haven’t done. As I suggested earlier, you would/could be making the same mistake all over again.

Regardless, you haven’t made any connection between the article and h changing other than wishful thinking, and that’s woefully insufficient 

 
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@Marcus Hanke said "So visual appearance in curved spacetime does depend on the observer, because different events are linked by different sets of null geodesics."

Exactly. The acceleration of gravity is related to the radius of curvature of spacetime by the approximate formula

r = c^2/g or g=c^2/r

In the reference frame of a remote observer, the radius of curvature of spacetime increases by k times, as do all distances, and, accordingly, the acceleration of gravity decreases by k times, as mentioned above.

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6 hours ago, SergUpstart said:

The acceleration of gravity is related to the radius of curvature of spacetime

What do you mean exactly by ‘radius of curvature of spacetime’?

7 hours ago, SergUpstart said:

the radius of curvature of spacetime increases by k times, as do all distances, and, accordingly, the acceleration of gravity decreases by k times, as mentioned above.

Again, I don’t know what you mean by this, since the tidal curvature of spacetime certainly does not follow a linear relationship like this. For a radial infall such as the one I spoke about the relevant quantity is the (t,r,t,r) component of the Riemann tensor - which is an inverse cube law.

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17 hours ago, Markus Hanke said:

What do you mean exactly by ‘radius of curvature of spacetime’?

The radius of curvature of spacetime is a quantity inversely proportional to the scalar curvature of spacetime. It has the dimension of length. For weak gravity there is an approximate formula r=c^2/g

 

The graphic shows a projection of the path followed by a falling object in four-dimensional spacetime into three-dimensional space. Objects fall because they follow geodesics in this spacetime. They are of maximal length, maximizing proper time. For all trajectories, the initial angle or speed does not matter and the curvature is the same, desc1.png. Near a very massive object such as a black hole, this value is considerably greater. https://demonstrations.wolfram.com/SpacetimeCurvatureForAFallingObjectNearTheEarthsSurface/

 

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1 hour ago, SergUpstart said:

Although I see an error in the above quote, near a more massive body, r will be smaller, not larger.

Yes.

Also, remember previous discussions we had about degrees of freedom. Scalar curvature (I presume you mean the Ricci scalar) only measures one particular aspect of curvature; it can’t account for all necessary degrees of freedom.

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19 hours ago, Markus Hanke said:

Scalar curvature (I presume you mean the Ricci scalar) only measures one particular aspect of curvature; it can’t account for all necessary degrees of freedom.

Sure. If we take the above formula g=C^2/r, then it gives only the module g, but g is a vector.

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