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Posts posted by SergUpstart

  1. 1 hour ago, Markus Hanke said:

    See studiot’s comments on intrinsic vs extrinsic to begin with. Furthermore, there is not really any force involved in gravity - when you have initially parallel test particles in free fall, and attach an accelerometer to them, it will always read exactly zero, so no forces; nonetheless in the presence of gravity their geodesics will begin to deviate.

    To do this, the size of the accelerometer must be negligible compared to the curvature of ST. The accelerometer in the form of comet Shoemaker-Levy in July 1994 perfectly "discovered" the gravity of Jupiter.


  2. 6 minutes ago, iNow said:

    How so?

    If you have a debt of $ 1,000 at a rate of 1%, then you will have to spend $ 10 a year on debt servicing, and if the rate increases to 2% , then you will need $ 20 for debt servicing.

    And there is another problem with debts. If the population is heavily credited, then it can no longer take out loans for the purchase of goods, including electronics, cars and real estate. This should inevitably cause a decline in demand, which can develop into a sales crisis.

  3. 1 hour ago, iNow said:

    Again, if debts are being paid, why would inflation and default rates increase? You are conflating them, but they're different things. Debt is not a problem so long as it continues being properly serviced. 

    The financial system is gradually losing its stability. The Fed already does not have the opportunity to significantly raise rates, since this will cause problems with debt servicing.

  4. 19 minutes ago, iNow said:

    And what is the consequence you see coming from this, especially if transactions and money flows are still occurring unimpeded regardless of aggregate debt levels? 

    I don't see anything good in this. Something must eventually happen, either hyperinflation or a parade of defaults. The question is when. How long can an atom be in an excited state? How long can water vapor be in a supercooled state?

  5. 2 minutes ago, Aeromash said:

    You are talking about the speed of the car. And not about the square of the speed or the gravitational potential of the car. You are confusing physical quantities.

    You are confusing this. The gravitational potential of the Earth, not the car. If the car moves so fast that the square of its speed becomes equal to the gravitational potential of the Earth not on its surface, then the car will fly into the Solar System.

  6. 1 hour ago, Markus Hanke said:

    The orbits of planets depend only on the total mass of the central body. If you change only its radius, all other things remaining equal, the planetary orbits will not be affected. This is a direct consequence of Birkhoff’s Theorem (using a simple Schwarzschild model).

    strictly speaking, the motion in a circular orbit is determined by the sum of the masses of the planets and the central body

    The speed (or the magnitude of velocity) relative to the central object is constant:[1]:30

    {\displaystyle v={\sqrt {GM\! \over {r}}}={\sqrt {\mu \over {r}}}}v = \sqrt{ GM\! \over{r}} = \sqrt{\mu\over{r}}


    1 hour ago, inbreeding said:

    So the sun will absorb the planets like mercury and venus?

    Perhaps the Earth too

    36 minutes ago, Markus Hanke said:

    That’s because the sun slowly looses mass and angular momentum through radiation and emission of particles (‘solar wind’), so it’s total mass decreases over time, making planetary orbits larger. The effect is really small though.

    To this effect, you can also add the influence of tides. It is not a solid body. Due to the tides, it gradually loses the angular momentum, which should be compensated by the removal of the planets. In addition, the rotation of the planets around their axis is gradually synchronized with their rotation around the Sun, and this effect also leads to an increase in the radii of the orbits of the planets. But the impact of these effects is also insignificant.

  7. On 9/10/2021 at 4:14 PM, Aeromash said:

    You are absolutely right. Dirac used the classic radius. But he did not pretend to be accurate. He only noticed that the ratio of the classical radius to the radius of the Universe gives a value of 10 ^ 40. The same value is given by the ratio of the roots of the masses of the electron and the Universe. r -radius must be gravitational.

    Well, then it follows simply from the definition of the gravitational radius R=GM/c^2 that the ratio of the gravitational radius of the universe to the gravitational radius of an electron is equal to the ratio of their masses.

    But the mass of the electron is about 10^-30 kg, and the mass of the Universe is about 10^56 kg. So if we use gravitational radii, their ratio will be different.

    1 hour ago, Aeromash said:

    Visitors and astronomers know that the square of speed is the gravitational potential. And the square of the gravitational potential is the force.

    Not so, the field strength is the gradient from the potential with a minus sign. And the force is the field strength multiplied by the mass of the test body ( if we are talking about the gravitational force in the Newton paradigm) or by the test charge (if we are talking about the electrostatic force).

  8. 9 minutes ago, Aeromash said:

    Dirac meant the electron. But as it turned out, Dirac's formula works on any scale.

    The electron does not have a certain radius and certain coordinates. An electron is not a ball. This follows from the Heisenberg uncertainty principle. Maybe you meant the radius of the orbit of an electron in a hydrogen atom??? But the electron does not have a clear orbit in the atom, at present physicists are talking not about orbits, but about orbitals.

  9. 14 minutes ago, Aeromash said:

    Dirac a little earlier than I noticed that the ratio of the mass of the smallest particle in the Universe to the mass of the Universe gives the same order of magnitude as the ratio of the squares of their radii. To be precise, the roots of the mass ratios give the same order as the ratio of the radii. Those.:

    What do you think is the smallest particle and what is its radius???

  10. On 9/8/2021 at 3:41 AM, iNow said:

    One of the best ways to do so is to enforce tax laws and prosecute tax evasion. 

    I remembered how in 1998, a couple of weeks before Russia's default, Sergey Kiriyenko (then he was the prime minister of Russia) came to the Gazprom office and demanded to pay all taxes.

  11. On this forum, they communicate in English, take the trouble to make a translation. With the help of Yandex-Translator, it will take a couple of minutes https://translate.yandex.ru

    ( На этом форуме общаются на английском языке, потрудитесь сделать перевод. С помощью Яндекс-Переводчика это займет пару минут https://translate.yandex.ru)

  12. 10 hours ago, Markus Hanke said:

    There is no such thing as “gravitational potential” in general curved spacetimes - the concept only makes sense under some very special conditions, and certainly not for large regions of the universe. I seem to remember that this has been pointed out numerous times in past threads on here.

    The gravitational redshift can be expressed as

    {\displaystyle z={\frac {\Delta \nu }{\nu _{1}}}=(1+\alpha ){\frac {\Delta U}{c^{2}}}}

    where {\displaystyle \Delta \nu =\nu _{2}-\nu _{1}}is the gravitational redshift, {\displaystyle \nu _{1}} is the optical clock transition frequency, {\displaystyle \Delta U=\Delta U_{2}-\Delta U_{1}} is the difference in gravitational potential, and }\alpha denotes the violation from general relativity.


    It is impossible to do without the gravitational potential.

    If you look at the definition of the gravitational potential, then this is the energy that must be communicated to a resting body of a unit mass so that it flies to infinity. Probably, the fact that this energy is difficult to calculate from the GRT equation does not mean that this energy does not exist.

    In addition, it follows from the definition of the gravitational potential that it is equal to the square of the escape velocity from a given point in space. Does the escape velocity exist?

    10 hours ago, Markus Hanke said:

    I’m pretty sure he didn’t actually say this, since metric expansion is itself a gravitational phenomenon. Can you provide an exact reference, so we can see the context?

    Metric expansion means simply that measurements of distances depend on when you make them - the results get bigger (on large enough scales) as you age into the future.


    I can't give you a link. Hawking's books do not contain formulas, so it is convenient to listen to them in audio format, which I did. But here I found the link http://kosmos-x.net.ru/news/kuda_rasshirjaetsja_vselennaja/2018-08-09-5403 It is in Russian, but now this is not a problem, since many browsers automatically translate the text. Here is a quote from there, Space expands only where the gravity of matter and energy are limited. Therefore, space does not expand inside galaxies or complex galactic groups, but only between galactic clusters and superclusters.


  13. 53 minutes ago, swansont said:

    Good. Now please explain what you mean by the fifth force, and why you phrased it as if this were mainstream physics.


    Here's what you wrote "Frankly, this does not seem to be a difficult concept; the concepts of motion and force it evokes are Newtonian. If gravity in a region is strong enough to prevent expansion, that’s what happens. " 

    I just asked a counter question, the meaning of which is "Won't we need the fifth interaction to explain the nature of this Newtonian force ?"

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