In the case of a black hole, schwarzschild radius exists in a stellar outside.
From inside a Schwarzschild radius, light can not escapes also. It is said that if the mass of the Earth is concentrated in one point, radius will become a black hole of 0.5 cm. It is said that if the mass of the sun is concentrated in one point, radius will become a black hole of 3 km. However, the black hole and Schwarzschild radius does not exist. I'll explain it now. |

In a general way, I'll try finding the Schwarzschild radius in the case of the mass of the sun is concentrated in one point.
Schwarzschild radius, can be find using the formula for the (escape velocity) second cosmic velocity. The radius at which the escape velocity becomes the speed of light is the Schwarzschild radius. U is the potential energy from in the far‐off distance. Speed at which the kinetic energy and the value of U is equal to is the escape velocity. If you substitute a folloing value for the formula, escape velocity of the Earth is the 11.2Km / s. I transform the formula. Put the mass of the sun(1.99 × 10 ^ 30Kg) to M, put the speed of light(3 × 10 ^ 8m / s) in v, then R is the 2.95Km. Indeed, if you concentrate on one point the mass of the sun, black hole of about 3Km radius is created. However, this formula is wrong. |

The formula for finding the escape velocity, you can use the object much slower than light, but it does not work for the light itself or object with a speed close to the light.
This formula, that take no thought of the following. * Speed of light is fixed. * Time delayed by gravity. * Space is shrunk by gravity. I find the radius of the black hole in consideration of these. I invented this method.
At first, I finding the delay time between the ground and the artificial satellite.
Dim GA, hankei, omosa, px, t1, t2, t3, c, p, g, u, v, r As Double Dim bunkatu, i As Long GA = 6.67 * 10 ^ -11 'Gravitational Constant px = 100 'Length of one block which is divided (Unit in meters) u = 20000000 'Distance of artificial satellite and surface of the Earth (Unit in meters) hankei = 6.378 * 10 ^ 6 'Radius of the Earth (Unit in meters) omosa = 5.974 * 10 ^ 24 'Mass of the Earth（Unit in kilogram） bunkatu = u / px 'Number of Partitiors c = 3 * 10 ^ 8 'Speed of light（Units in m/s） t1 = px / c 'Time measured in the static system v = 0 'In order to replace the acceleration and the gravity using the equivalence principle. r = hankei + u 'Distance of artificial satellites and Earth 't1 is the time the observed from the static system at distance which light runs one divided block.(This time is fixed) 't2 is the elapsed time in the rocket. The time will delay as speed of rocket is faster. 't3 is the difference of time the light running through the block which is divided.（Difference of time of static system and rocket system） For i = 0 To bunkatu - 1 g = GA * (omosa / r ^ 2) v = v + g * t1 'In order to replace the acceleration and the gravity using the equivalence principle. Terminal velocity of the rocket. r = r - px Next i t2 = (t1 - v * px / c ^ 2) / Sqr(1 - (v ^ 2 / c ^ 2)) 'Elapsed time in the rocket. t3 = t1 - t2 'The last delay time is answer. Text1.Text = t3 * (1 / t1) 'Delay time between 1 second.
Result
5.26317633122293E-10 The earth's surface time is delayed 5.26E-10 seconds than the artificial satellite. You can be sure this result by an article of GPS satellites.
I will explain the program
To find the answer I used the formula of the special theory of relativity. You can solve a problem of general theory of relativity using formula of special theory of relativity. This is because the special theory of relativity accepted in the small area in bent space-time. |

If the mass of the sun is concentrated in one point, it is said that black hole of 3Km radius is generated.
I will find the delay time of the radius of 2Km. Dim GA, hankei, omosa, px, t1, t2, t3, c, p, g, u, v, r As Double Dim bunkatu, i As Long GA = 6.67 * 10 ^ -11 px = 100 ' omosa = 1.9884 * 10 ^ 30 6.96 * 10 ^ 8 hankei = 2000 u = r - hankei bunkatu = u / px c = 3 * 10 ^ 8 t1 = px / c v = 0 For i = 0 To bunkatu - 1 g = GA * (omosa / r ^ 2) v = v + g * t1 r = r - px Next i t2 = (t1 - v * px / c ^ 2) / Sqr(1 - (v ^ 2 / c ^ 2)) t3 = t1 - t2 Text1.Text = t3 * (1 / t1)
Result
Delay time between the position of 2000m from the center of the sun and the position of the radius of the sun（6.96 * 10^8 m） is about 0.6 seconds between 1 second. Delay time between the position of 3000m from the center of the sun and the position of the radius of the sun（6.96 * 10^8 m） is about 0.4 seconds between 1 second. Delay time between the position of 4000m from the center of the sun and the position of the radius of the sun（6.96 * 10^8 m） is about 0.3 seconds between 1 second.
It is said that the progress of time is stop on the surface of the black hole.
But it is wrong. If you look at the program, you will see that the time stops at the center of the celestial body. In other words, the Schwarzschild radius does not exist. Also in the center of the celestial body, the length of the space is close to 0. Therefore, the light will run at a fixed speed even in a strong gravity. Light from star of strong gravity is dark by red shift, but it will not be completely dark. In other words, the light will come out of the Schwarzschild radius. Mail: Syuta Miyawaki mcs7@mte.biglobe.ne.jp Back |