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# Experiment verification of General relativity

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3 hours ago, SergUpstart said:

He also showed that the variation of the elementary charge violates the law of conservation of energy.

So does the idea that c is a variable. Consider the local conservation of the energy-momentum tensor in the presence of gravity:

$T{^{\mu}}{_{\nu ||\mu}}=0$

Since the covariant derivative depends on the metric, which explicitly contains c, and because in your idea c varies in a way that is not covariant, the above relationship ends up being no longer valid. This whole idea puts you in a situation where there is no longer any conservation of energy-momentum, not even locally. This is clearly in direct contradiction to experiment and observation.

Edited by Markus Hanke
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3 hours ago, SergUpstart said:

Yanchilin assumed that the splitting of spectral lines in distant quasars will not change, so the constant of the fine structure does not change. He also showed that the variation of the elementary charge violates the law of conservation of energy.

How about addressing the clock experiment I have brought up twice.

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6 hours ago, Markus Hanke said:

So does the idea that c is a variable. Consider the local conservation of the energy-momentum tensor in the presence of gravity:

Tμν||μ=0

Since the covariant derivative depends on the metric, which explicitly contains c, and because in your idea c varies in a way that is not covariant, the above relationship ends up being no longer valid. This whole idea puts you in a situation where there is no longer any conservation of energy-momentum, not even locally. This is clearly in direct contradiction to experiment and observation.

The law of conservation of energy will not be violated, the rest masses of all bodies are also variable and are determined by the gravitational potential. Here I found a link to this theory in English https://vixra.org/pdf/1603.0398v1.pdf

6 hours ago, swansont said:

How about addressing the clock experiment I have brought up twice.

It is useless to accumulate impulses, this only narrows the Delta f. With a gravitational shift all frequencies of the spectrum will change by the same relative value

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1 hour ago, SergUpstart said:

It is useless to accumulate impulses, this only narrows the Delta f. With a gravitational shift all frequencies of the spectrum will change by the same relative value

I’m talking about an atomic clock. What are you talking about? What relevance do those drawings have?

If I have a clock running at a frequency F, and I count the number of “ticks”, I measure time. Now we take a clock, running at F, to some new height, H, above the reference clock. According to your earlier post, the frequency will be lower by a relative value 2gH/c^2. I can count the number of ticks at this new height, return it to the original height and compare it to a reference clock.

GR predicts a higher frequency at H, and you predict a lower one. This is easily checked and trivially falsified.

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2 minutes ago, swansont said:

I’m talking about an atomic clock. What are you talking about?

Let's say we count 1000 pulses from the Geiger counter. In fact, we form a rectangular pulse, whose front front coincides with the first pulse, and the back - with the 1000th. when transmitting it to an external observer, we will get the same rectangular pulse, the duration of which will be formed by a proportional change in all the frequencies of its spectrum.

Nothing will work, we check the gravitational frequency shift.

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22 minutes ago, SergUpstart said:

Let's say we count 1000 pulses from the Geiger counter. In fact, we form a rectangular pulse, whose front front coincides with the first pulse, and the back - with the 1000th. when transmitting it to an external observer, we will get the same rectangular pulse, the duration of which will be formed by a proportional change in all the frequencies of its spectrum.

Nothing will work, we check the gravitational frequency shift.

That’s not how clocks work. That’s not even how these measurements work - we don’t care about this “rectangular pulse” We just count the decays. Discrete values.

We have a radioactive sample. We measure 1000 dps at the reference system. Now we move the clock up such that the frequency changes, according to your theory, to 998 dps. GR predicts 1001 dps. Let the system sit there for an hour. Then we compare to the reference. From what you’ve said, your prediction is 120 decays fewer, what GR predicts is 60 more.

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52 minutes ago, swansont said:
52 minutes ago, swansont said:

We measure 1000 dps at the reference system

Yes, you are right, such an experiment will help you find the truth

Edited by SergUpstart
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21 minutes ago, SergUpstart said:

Yes, you are right, such an experiment will help you find the truth

We’ve done the equivalent experiment with clocks, several times.

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11 hours ago, SergUpstart said:

The law of conservation of energy will not be violated

Yes it will be. If you look at the above equation, if c is variable, the covariant derivative will contain extra terms including derivatives of c. These terms don’t cancel out, so there is no way to not violate the relation.

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10 hours ago, swansont said:

We’ve done the equivalent experiment with clocks, several times.

Let's go back to the clock experiment. The new theory States that as the gravitational potential increases, Planck's constant decreases, which should REDUCE the rate of radioactive decay. We need an experiment not with one, but with a couple of hours, where one will work on the principle of radioactive decay, and the other as a quantum frequency standard. This pair of hours is set at the top and synced, then we lower the pair of hours down and they will have to be out of sync.

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3 hours ago, SergUpstart said:

Let's go back to the clock experiment. The new theory States that as the gravitational potential increases, Planck's constant decreases, which should REDUCE the rate of radioactive decay. We need an experiment not with one, but with a couple of hours, where one will work on the principle of radioactive decay, and the other as a quantum frequency standard. This pair of hours is set at the top and synced, then we lower the pair of hours down and they will have to be out of sync.

How does the rate of radioactive decay depend on Planck’s constant?

Why not address the experiment I described, instead of trying to come up with a more complicated experiment?

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On 6/28/2020 at 8:42 AM, SergUpstart said:

The new theory States that as the gravitational potential increases, Planck's constant decreases

Again, gravitational potential - if it can be meaningfully defined at all - is a gauge field with a gauge freedom to choose a zero point, whereas Planck’s constant obviously isn’t. It is not physically meaningful to relate the two in this manner.

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On 6/28/2020 at 8:23 AM, Markus Hanke said:

Yes it will be. If you look at the above equation, if c is variable, the covariant derivative will contain extra terms including derivatives of c. These terms don’t cancel out, so there is no way to not violate the relation.

Yanchilin's equations are slightly different from Newton's

Newton's Phi (r)=Gm/r g (r) = Gm/r^2

Yanchilin'sPhi(r)=2Gm/r g(r)=(dPhi/dr)/2=Gm/r^2

This reflects the law of conservation of energy. When an Apple accelerates when it falls from a height H, the energy of the Apple changes not by the amount of mgH, but by the amount of 2mgH. Half of that energy to change the kinetic energy mv^2/2, and the other half goes to the change in internal energiei that of Apple, which is Einstein's E=mc^2, and Yanchilin's respectively E=-mphi.

A photon has a rest mass of 0, so all this energy  is used to change the kinetic energy of the photon. 2mgH and not mgH, as in Newton. This explains and corrects the error Of Newton's theory, which predicts the deflection of a ray of light in a gravitational field exactly 2 times less than GRT.

On 6/28/2020 at 2:13 PM, swansont said:

How does the rate of radioactive decay depend on Planck’s constant?

Why not address the experiment I described, instead of trying to come up with a more complicated experiment?

How are you going to measure local time and determine when to stop counting decay events? There's no way without a second watch. Have there been experiments with two different clocks?

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4 hours ago, SergUpstart said:

How are you going to measure local time and determine when to stop counting decay events? There's no way without a second watch. Have there been experiments with two different clocks?

You don’t have to stop counting until you bring them back together. No timing required, though that’s not a big issue (if you do a measurement that lasts a day, who cares if you have a few nanoseconds of dilation. It has a negligible effect on the answer, if you do the experiment properly). You start and stop with the two systems next to each other.

Yes, this has been done with clocks. Tom van Baak’s version was to bring some clocks up on Mt Rainier for few days while the family went camping, and compare to the clocks left at home.

(I think someone linked to this recently, but I don’t recall which thread)

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On 6/29/2020 at 2:09 PM, SergUpstart said:

Yanchilin's equations are slightly different from Newton's

I did not make any reference to Newtonian gravity or any particular form of potential, I am only using the fact that the energy-momentum tensor has to be locally conserved. The relation I gave follows from Noether’s theorem, and not any particular theory of physics. The point was simply that, if you allow c to vary, this conservation law no longer holds, because the underlying symmetry that gives rise to the conserved quantity is no longer there. If c is not constant, energy-momentum cannot be conserved, irrespective of what else you attempt to change.

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• 3 weeks later...
On 6/24/2020 at 11:16 PM, swansont said:

How do you arrive at that interpretation? They measured the magnetic field near a black hole binary, not the permeability of free space.

But in Our galaxy, you can probably find more than one pair of main-sequence stars, so that they have approximately the same mass, temperature, and speed of rotation around their axis, but that one would be close to the center of the galaxy, and the other close to the edge of the galaxy. That would be a rough measure and compare the strength of their magnetic fields.

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• 1 year later...

I found an explanation in the framework of V. Yanchilin's quantum theory of gravity that when approaching a massive body, the course of the atomic clock on caesium slows down.

V. Yanchilin explained the increase in the frequency of photons emitted by an atom when approaching a massive body, relying on the Niels Bohr model of the atom. According to this model, an electron in an atom can only be at strictly defined energy levels, each level corresponds to its own
the radius of the orbit.

According to the quantum theory of gravity, when approaching a massive body, the square of Planck's constant decreases inversely
according to the absolute value of the gravitational potential, this means that all the radii of the orbits in the Bohr atom model also decrease
inversely proportional to the absolute value of the gravitational potential. In the formula for the energies corresponding to the orbits of
the radius of the corresponding orbit is in the denominator, which means that the electron energy for each energy level
increases, and therefore increases the energy that is transferred to the photon that is emitted during the electron transition
from a higher energy level to a lower one. The frequency of the photon increases accordingly.

And now let's look at how the cesium atomic clock works. In them, the frequency is set by the electron transition with non-medzhu energy
the levels of the model of the Bohr atom, and the transition of an electron between sublevels, which are formed by the splitting of one of the levels due to spin-orbit interaction of the electron's magnetic moments. For the energy of such a transition, there is the following formula

In this formula, the radius of the orbit is not in the denominator and in the numerator, which means that when approaching a massive object
the energy of the photons emitted in the transition between sublevels, not increases and decreases, thus decreasing their frequency and

therefore,  the course of a cesium atomic clock.

It follows from the above that the deceleration of the current of the atomic clock does not occur in spite of, but in full accordance with the quantum theory of gravity.
To finally decide which theory is correct, the quantum theory of gravity or the general theory of relativity,
we need an experiment with a clock whose frequency is set by the transition of electrons between the allowed energy levels of the Bohr model.
But, unfortunately, there are no such watches yet.

In addition to the above, it should be added that another effect can be detected here. For an external observer, the photons are located at the top,
the photons emitted during the transition between such sublevels should experience a stronger redshift, 4gH/C^2, instead of gH/c^2 for photons emitted at
normal transitions.
From this point of view, you should pay attention to the wavelength of 21 cm, because it is emitted by hydrogen during the transition between sublevels.
If it is possible to measure the red shift at a wavelength of 21 cm for a distant galaxy, and it turns out to be relatively large,
than the red shift of the spectral lines of other chemical elements, this will testify in favor of the quantum theory of gravity.

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53 minutes ago, SergUpstart said:

I found an explanation in the framework of V. Yanchilin's quantum theory of gravity that when approaching a massive body, the course of the atomic clock on caesium slows down.

V. Yanchilin explained the increase in the frequency of photons emitted by an atom when approaching a massive body, relying on the Niels Bohr model of the atom. According to this model, an electron in an atom can only be at strictly defined energy levels, each level corresponds to its own
the radius of the orbit.

Strictly speaking, we know the Bohr model to be wrong, but the energy solutions for hydrogen-like systems is correct.

Quote

According to the quantum theory of gravity,

There is no quantum theory of gravity

Quote

when approaching a massive body, the square of Planck's constant decreases inversely
according to the absolute value of the gravitational potential, this means that all the radii of the orbits in the Bohr atom model also decrease

This is the speculation?

Quote

inversely proportional to the absolute value of the gravitational potential. In the formula for the energies corresponding to the orbits of
the radius of the corresponding orbit is in the denominator, which means that the electron energy for each energy level
increases, and therefore increases the energy that is transferred to the photon that is emitted during the electron transition
from a higher energy level to a lower one. The frequency of the photon increases accordingly.

Why do clocks that don't rely on photons (e.g. quartz oscillators) feel the same effects?

Quote

And now let's look at how the cesium atomic clock works. In them, the frequency is set by the electron transition with non-medzhu energy

non-medzhu? What is that?

Quote

the levels of the model of the Bohr atom, and the transition of an electron between sublevels, which are formed by the splitting of one of the levels due to spin-orbit interaction of the electron's magnetic moments. For the energy of such a transition, there is the following formula

Spin-orbit is the fine structure. The clock transition is in the hyperfine structure.

e.g. in hydrogen, it's the fine structure is splitting of the excited P state. The hyperfine structure is seen in the ground state

The spin-orbit interaction is not involved in the atomic clock operation.

Quote

In this formula, the radius of the orbit is not in the denominator and in the numerator, which means that when approaching a massive object
the energy of the photons emitted in the transition between sublevels, not increases and decreases, thus decreasing their frequency and

therefore,  the course of a cesium atomic clock.

It follows from the above that the deceleration of the current of the atomic clock does not occur in spite of, but in full accordance with the quantum theory of gravity.

My objections aside, I see no analysis of the actual amount of frequency shift you would expect. You have only hand-waved the answer. It needs to be quantified.

Quote

To finally decide which theory is correct, the quantum theory of gravity or the general theory of relativity,
we need an experiment with a clock whose frequency is set by the transition of electrons between the allowed energy levels of the Bohr model.
But, unfortunately, there are no such watches yet.

There are none because the Bohr model is wrong. But if you mean by transitions between levels where the principle quantum number (n) changes, they do exist. They are called optical clocks, because these transitions are often in the visible part of the spectrum. And they have measured frequency shifts perfectly consistent with general relativity.

https://www.nist.gov/publications/relativity-and-optical-clocks (this is from > 10 years ago)

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In addition to the above, it should be added that another effect can be detected here. For an external observer, the photons are located at the top,

The top of what?

Quote

the photons emitted during the transition between such sublevels should experience a stronger redshift, 4gH/C^2, instead of gH/c^2 for photons emitted at
normal transitions.

You mean the redshift that you have yet to calculate for the conjecture you are championing? Why is there no calculation?

How do you get from your formulas to 4gh/c^2?

Quote

From this point of view, you should pay attention to the wavelength of 21 cm, because it is emitted by hydrogen during the transition between sublevels.
If it is possible to measure the red shift at a wavelength of 21 cm for a distant galaxy, and it turns out to be relatively large,
than the red shift of the spectral lines of other chemical elements, this will testify in favor of the quantum theory of gravity.

The 21 cm transition in H is the hyperfine transition I mentioned before, not the fine-structure splitting.

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23 minutes ago, swansont said:

Why do clocks that don't rely on photons (e.g. quartz oscillators) feel the same effects?

Have they been detected? Does the accuracy of a quartz watch allow you to detect these effects?

24 minutes ago, swansont said:

non-medzhu? What is that?

A typo, it was meant that a transition is used that does not change the main quantum number

28 minutes ago, swansont said:

You mean the redshift that you have yet to calculate for the conjecture you are championing? Why is there no calculation?

How do you get from your formulas to 4gh/c^2?

I wrote above that According to Yanchilin's theory, the atom at the bottom emits a photon at a frequency with a violet shift of 2gh/c^2, but going up it experiences a shift to the red side by 3gh/c^2. If, during the transition without changing n at the bottom, a photon is emitted with a shift to the red side gh/c^2 and there is another shift by 3gh/c^2 during the ascent, then the total shift will be 4gh/c^2.

35 minutes ago, swansont said:

Spin-orbit is the fine structure. The clock transition is in the hyperfine structure.

e.g. in hydrogen, it's the fine structure is splitting of the excited P state. The hyperfine structure is seen in the ground state

The spin-orbit interaction is not involved in the atomic clock operation.

Ok, I will deal with an ultra-thin structure.

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2 hours ago, SergUpstart said:

Have they been detected? Does the accuracy of a quartz watch allow you to detect these effects?

I didn’t say it was a watch. But the oscillators used in GPS are adjusted to run slower (on the ground) so that the relativistic effects in orbit make them have the same frequency as a ground-based clock.

2 hours ago, SergUpstart said:

A typo, it was meant that a transition is used that does not change the main quantum number

I wrote above that According to Yanchilin's theory, the atom at the bottom emits a photon at a frequency with a violet shift of 2gh/c^2, but going up it experiences a shift to the red side by 3gh/c^2. If, during the transition without changing n at the bottom, a photon is emitted with a shift to the red side gh/c^2 and there is another shift by 3gh/c^2 during the ascent, then the total shift will be 4gh/c^2.

Bottom of what? Ascent of what? You are saying things without giving context.

2 hours ago, SergUpstart said:

Ok, I will deal with an ultra-thin structure.

I don’t see how this means anything. We’re talking about atoms. Fine structure and hyperfine structure refers to energy level splitting. There is no such thing as ultra-thin structure.

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15 minutes ago, swansont said:

Bottom of what? Ascent of what? You are saying things without giving context.

Bottom - the lower point from which the photon rises to the height h

17 minutes ago, swansont said:

I don’t see how this means anything. We’re talking about atoms. Fine structure and hyperfine structure refers to energy level splitting. There is no such thing as ultra-thin structure.

I'm sorry, naturally an hyperfine structure. I use an online translator for communication and such incidents happen.

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Quote

the total shift will be 4gh/c^2.

We have experimental evidence that GR is correct and this is wrong. Why are you wasting time with a model that’s contradicted by experimental evidence? The Pound-Rebka experiment.

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