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SergUpstart

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Everything posted by SergUpstart

  1. Matter and energy arise and disappear constantly in accordance with the Heisenberg uncertainty principle. So the origin of the universe is most likely a quantum effect. God is not needed for this.
  2. Yes, and after BB, the universe rises in a gravitational well. This means that the red shift in the spectra of galaxies has not only a Doppler component, but also a gravitational one.
  3. As Hawking said, God is not needed to explain the origin of the Universe, God is superfluous.
  4. Sure. If we take the above formula g=C^2/r, then it gives only the module g, but g is a vector.
  5. Although I see an error in the above quote, near a more massive body, r will be smaller, not larger.
  6. 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, . Near a very massive object such as a black hole, this value is considerably greater. https://demonstrations.wolfram.com/SpacetimeCurvatureForAFallingObjectNearTheEarthsSurface/
  7. @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.
  8. There are not one equivalence principle, but two, strong and weak. It is necessary to distinguish between "weak equivalence principle" and "strong equivalence principle"[13]. A strong equivalence principle can be formulated as follows: at each point of space-time in an arbitrary gravitational field, one can choose a "locally inertial coordinate system", such that in a sufficiently small neighborhood of the point under consideration, the laws of nature will have the same form as in non-accelerated Cartesian coordinate systems of SRT, where "laws of nature" mean all laws of nature[14]. The weak principle differs in that the words "laws of nature" are replaced in it by the words "laws of motion of freely falling particles"[13]. The weak principle is nothing but another formulation of the observed equality of gravitational and inert masses, while the strong principle is a generalization of observations of the influence of gravity on any physical objects. https://ru.wikipedia.org/wiki/Принцип_эквивалентности_сил_гравитации_и_инерции I believe that these equivalence principles should be applied depending on the situation. If the same system is observed by several observers in different places of the gravity well, they should all observe the same picture only at different time scales. There is a strong equivalence principle at work here. If one observer observes several identical systems but located in different places of the gravity well, then only a weak equivalence principle should work here.
  9. 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
  10. 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.
  11. If you want to write your own program for reading QR codes, for example in C#, you can use the ZXing library
  12. Of course, the solutions will be different and the event horizon and singularity will not appear in them.
  13. This is not about refuting the GRT, but about correcting it. G in the GRT equation must be a variable whose value is inversely proportional to the time dilation. The fact that gravity is a curvature of space-time by the momentum energy tensor, the fact that it propagates in the form of gravitational waves at the speed of light, which is constant, all this is beyond doubt.
  14. And what about the reconstruction of pyramid construction technology? Elon Musk the year before last generally said that the pyramids were built by aliens Or reconstruction of bronze melting technology. The problem here is that there are no forests in Egypt, and charcoal is needed for metallurgy.
  15. No, wrong. The consequence of this post is that the event horizon should not arise at all. If we plot the time dilation function for the mass of a material point The graph of the function T (r)/To looks like this The graph shows that at r=2GM/c^2, the time dilation becomes infinitely large, i.e. the so-called event horizon takes place. Now we take into account the change in the gravitational constant inversely proportional to the time dilation, for which we solve the following system of equations. The solution of this system of equations Graph of the received function It can be seen from this graph that the time dilation for an external observer becomes infinite only at r=0. Thus, there is no event horizon. There are many massive and supermassive bodies in the universe that are asymptotically similar to black holes, but they should not have an event horizon.
  16. If we express the height of the tree in the number of wavelengths of a laser, then it will be the same for all observers. But for a remote observer, this wavelength will be longer due to the gravitational redshift. This means the laser, which is located next to the tree.
  17. I'm just saying the opposite. Quote from the beginning of the topic "From this it can be seen that in the reference frame of a remote observer, the acceleration of gravity on the surface of planet X should be k times less."
  18. Calculating the acceleration of free fall on the surface of the planet is only an intermediate task here. The final task is to calculate the gravitational constant. And in the condition of the problem there are two independent distances, the height of the fall of the apple H and the radius of the planet R. Do you assume that there is such a way to recalculate two independent distances that G will be left behind by a constant? And it should also be borne in mind that the gravitational well is formed not only (or even not so much) by the gravity of the planet, but also by the gravity of the star, galaxy....
  19. Here we consider acceleration in the non-relativistic case. The apple is falling at a speed much less than the speed of light. Let's look at the second part of the topic related to the oscillatory circuit. There it is proved that in the reference frame of the remote observer the dielectric constant decreases in proportion to the decrease in the standard length. It is logical to assume that G should change in the same way ( taking into account the fact that the dielectric constant is included in the denominator in Coulomb's law, and G is included in the numerator in Newton's law of gravity )
  20. I assume that in order for the speed of light to remain constant, the distance traveled by light and the time interval for which it is traveled would change the same factor.
  21. By time slowed you mean it takes longer according to that clock, right? The distant observer's clock runs faster than the one next to the tree. Also, it's not the weaker gravity, it's being higher up in the gravity well that causes time dilation. g could be constant and you would still have this effect. Yes, that's right, by weaker gravity, it meant that the remote observer is higher in the gravity well. But if we substitute time into the equation of motion according to the readings of the remote observer's clock, then the magnitude of the acceleration of gravity g will be different.
  22. Then he should also use a tape measure to measure the distance not from his laboratory. This means that it will switch from a relative coordinate system to an absolute one. In the absolute coordinate system, G is of course a constant.
  23. Let's do a thought experiment. A tree grows on planet X, an apple falls from it from a height H. There is an observer next to the tree, and in his frame of reference, the apple will fall in time Let another observer from outer space also observe the fall of an apple through a telescope. Due to the fact that the second observer is < br /> in a weaker gravity, in its frame of reference, the time of the apple's fall will be slowed down by a factor of k, and the height of H will also be increased by a factor of k. From this it can be seen that in the reference frame of a remote observer, the acceleration of gravity on the surface of planet X should be k times less. This change also corresponds to the dimension of the acceleration of free fall, m/s^2. But on the other hand where M and R are the mass and radius of planet X, respectively. In the reference frame of a remote observer, the radius of planet X will also increase by a factor k, and the acceleration of gravity on the surface of planet X g' will be This shows that in order for the acceleration of free fall on the surface of planet X in sesame starting a remote observer g' would be k times less than it is in the system of reference of the observer insider, requires that the gravitational constant G in the reference frame of an external observer would also k times greater than in the reference system of observer-insider. Otherwise, in the reference frame of the remote observer, the acceleration of gravity g' will not be in k, but in k^2 times less than it is in the frame of reference of the insider observer. This simple thought experiment proves that in the reference frame of a remote observer, the gravitational constant must change back proportional to the change in the distance/time scale. Next, we consider the LC oscillatory circuit. The capacity of the capacitor in the frame of reference of the insider observer is calculated by the formula. First, let's see how the capacitance of the capacitor will change in the reference frame of the remote observer, if we assume that the dielectric constant on changes. In the reference frame of an external observer, the area of the capacitor plates will increase by k^2 times, and the distance between the plates will also increase by k times, thus, the capacitance of the capacitor will increase by k times. Next, let's see how the inductance of the coil will change in the reference system of the remote observer, which is in the reference system of the insider observer it is calculated by the formula. First, let's see how the capacitance of the capacitor will change in the reference frame of the remote observer, if we assume that the dielectric constant on changes. In the reference frame of an external observer, the area of the capacitor plates will increase by k^2 times, and the distance between the plates will also increase by k times, thus, the capacitance of the capacitor will increase by k times. Next, let's see how the inductance of the coil will change in the reference system of the remote observer, which is in the reference system of the insider observer it is calculated by the formula. We see that the oscillation period increases by k times, as it should be. But the dimension of the electric constant F/m, so that in the reference frame of a remote observer, it should decrease by k times, then, in order for the speed of light to remain constant, the magnetic constant must increase by k times. Then, in the reference frame of the remote observer, the capacitance of the capacitor will not change, but the inductance of the coil will increase by k^2 times. Thus, the oscillation period of the oscillating circuit will also increase by k times in this case. To find out which of these two options is correct, we will introduce an active resistance into the oscillatory circuit so that the fluctuations in it become attenuated. Let's see how the resistance value changes when switching to the reference system of a remote observer. Resistance is the ratio the difference of electrical potentials to the current strength in the conductor. The dimension of the electric potential is Joule/Coulomb. The joule does not change when the time/distance scale changes synchronously, and the Сoulomb is not related to the time/distance scales at all, therefore, the difference of electrical potentials in the reference system of the remote observer will be the same as in the reference system an insider's observer. The strength of the electric current in the conductor is equal to the ratio of the charge that has passed through the cross-section of the conductor to the time for which this charge passed. The charge in the reference frame of the external observer will be the same as in the reference frame of the insider observer but the time will increase by k times. Thus, the resistance in the reference frame of the remote observer will be k times greater than in the reference frame of the insider-observer. In order for both the insider observer and the remote observer to see the same picture of damped oscillations on different time scales, it is required that the Q-factor of the oscillatory circuit in the reference systems of both observers would be the same. (Q-factor is a parameter of the oscillatory system that determines the resonance width and characterizes, how many times are the energy reserves in the system greater than the energy losses during the phase change by 1 radian.) In the first variant, when the dielectric and magnetic constants are considered unchanged, in the reference frame of the remote observer, the Q-factor the oscillating circuit will be And in the second variant, when the dielectric and magnetic constants change in different directions by k times, in the reference frame of the remote observer the Q-factor the oscillating circuit will be 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 For those who still do not believe that the numerical value of the gravitational constant in the reference frame of a remote observer changes back in proportion to the change in the time/distance scale, I propose to solve the following problem. On a planet with a mass of 5.97*10 ^ 24 kg, a radius of 6317000 m, an apple falls from a tree from a height of 5 m in 1 second, calculate the value of the gravitational constant G. And then calculate the value of the gravitational constant G' in the reference frame of the remote observer, in which the apple falls for 1.1 seconds from a height of 5.5 m and the radius of the planet is 6948700, respectively, and compare the obtained values of G and G'.
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