8. The physical meaning of the Schwarzschild radius
So, black holes do not exist, but does it follow that the gravitational radius or the Schwarzschild radius is devoid of
any physical meaning? After all, it is used in quantum mechanics to determine the maximum
the mass of an elementary particle, if for an object of mass M the Schwarzschild radius is greater than the De Broglie wavelength,
then this object is a macroscopic body, and if the opposite is true, the De Broglie wavelength is greater than the radius
Schwarzschild, then this object is an elementary particle.
To understand the physical meaning of the Schwarzschild radius, remember that it is calculated from classical mechanics
by equating the second cosmic velocity to the speed of light, let's look at this equation by raising
both of its parts are squared.
On the left we see the local gravitational potential created by this body, on the right is the external global < br />
the gravitational potential created by all the bodies in the universe. Hence the physical meaning becomes clear
Schwarzschild spheres. Outside this sphere, the speed of light is determined mainly by the external global gravitational potential,
Inside the Schwarzschild sphere, the speed of light is determined mainly by the local gravitational potential of this body.
But this is in the Galilean coordinate system, and what is in the Einstein coordinate system? A and nay outside the Schwarzschild sphere
the values of dimensional physical constants are determined mainly by the external global gravitational potential,
Inside the Schwarzschild sphere, the values of these dimensional physical constants are determined mainly by the local gravitational potential of this body.