Endercreeper01

If you don't have any math, it won't even be considered. It needs to be proven mathematically and experimentally in order to be a theory. You can't just use logic. Aristotle used logic, but he was incorrect. Much of science is counterintuitive. There is even a rule in speculations about this. It doesn't matter how logical it is to anyone — if it disagrees with actual experiment it's not correct. And if you can't test it, it is not a theory.

World salad. Do you have any mathematics behind this?

It's not a theory, it's word salad. Do you have any mathematics behind your theory?

Relative to the object falling, it would reach the singularity long before the black hole evaporates. With respect to a distant observer, it is different. The observer will see the particle go slower and slower, until it appears to be frozen at the event horizon. However, because the black hole is evaporating, it will be moving very slowly with the event horizon shrinking. It will keep happening like this until the black hole evaporates. This seems to be a paradox, as the object would already have evaporated. The object will most likely evaporate at the event horizon.

Why does the metric require that it changes with time in order to be able to make a black hole?

Yes, but if the radius is changing with time (which is what happens in gravitation), the metric would also change with time.

I mean that it won't reach the event horizon with respect to a distant observer. Relative to the infalling object, it will go beyond the event horizon. But, it will not see anything, as any light in the event horizon will be drawn towards the singularity. Time dilates only with respect to a distant observer. The particle approaching the event horizon will experience no time dilation with respect to itself.

The coordinate time never changes. One second in coordinate time will be one second in coordinate time regardless of proper time.

When [latex]v[/latex] approaches [latex]c[/latex], time dilation does not approach infinity. In special relativity, proper time and coordinate time are related by [latex]\tau = \int_0^t \sqrt{1-\frac{v^2}{c^2}} \ dt[/latex] This does not approach infinity, but 0. As [latex]v[/latex] gets larger, [latex]\sqrt{1-\frac{v^2}{c^2}}[/latex] gets smaller. Therefore, [latex]\tau [/latex] becomes smaller as [latex]v[/latex] approaches c. This is again the same problem. Time dilation for a black hole is given by [latex]\tau = \int_0^t \sqrt{g_{00}} \ dt[/latex] For a Schwarzschild black hole, [latex]g_{00}=1-\frac{R_s}{r}[/latex] Where [latex]R_s=\frac{2GM}{rc^2}[/latex] so [latex]\tau = \int_0^t \sqrt{1-\frac{R_s}{r}} \ dt[/latex] Because [latex]\sqrt{1-\frac{R_s}{r}}[/latex] does not approach infinity as [latex]r[/latex] approaches [latex]R_s[/latex]. This means that [latex]\tau[/latex] does not approach infinity. To answer your question, time would keep dilating more and more as [latex]r[/latex] approaches [latex]R_s[/latex]. An object will never actually reach the event horizon, as time would keep dilating more and more as it approaches the event horizon.

The Schwarzschild metric can vary with time if there is motion, i.e if you have a function r(t). There is certainly motion when the core of a supernova compresses into a singularity.

Yes, time dilates from gravitation. Proper time is related to coordinate time by [latex]d\tau = \sqrt{g_{00}} \ dt[/latex]. In the Schwarzschild solution, [latex]g_{00}=1-\frac{R_s}{r}[/latex], so time is dilated by a factor of [latex]\sqrt{1-\frac{R_s}{r}}[/latex]. Time dilates only with respect to a distant observer. Time freezes with respect to an outside observer at the event horizon, but for the object going into the black hole, it enters the event horizon, so it does not freeze. Time is relative, but it is not a vector. Vectors have both magnitude and direction, but time does not have length.

When [LATEX]v=c[/latex], time stops relative to an outside observer. Relative to the object, it would experience no time dilation. Because everything else would appear to be moving at c relative to the object, time would stop for everything else relative to the object. But this is just using the equation. If something really were to travel at c, we don't know what would happen to the time passed for it relative to an observer. In the equation, the principle square root is used. The principle square root is the non negative value of the square root. This is because the observer will see the object traveling forwards in time if [latex]t[/latex] is positive. They can't see it traveling both forwards and backwards in time.

Speculations. Perpetual motion is impossible in a closed system.

It is impossible to have backwards time in any part of the universe (according to our current understanding of the universe). In special relativity, time dilates by the equation [latex]\tau = t \sqrt{1-\frac{v^2}{c^2}}[/latex]. In order for [latex]\tau [/latex] to be negative, [latex]\sqrt{1-\frac{v^2}{c^2}}[/latex] must be negative also. Because the square root refers to the positive value in the equation, it can't be negative. This also applies to general relativity. Time also dilates as a result of gravity. Time is dilated by the equation [latex]\tau = t \sqrt{g_{00}}[/latex]. In the Schwarzschild solution, [latex]g_{00}=1-\frac{2GM}{rc^2}[/latex], so it becomes [latex]\tau = t \sqrt{1-\frac{2GM}{rc^2}}[/latex] for a spherical, non-rotating, and uncharged mass. Even if [latex]R_s=r[/latex] or [latex]r>R_s[/latex], there is no way to make the equation have negative time.

Some claims involving black holes may be pseudoscience, but black holes themselves are not. Black holes have a mathematical framework surrounding them, such as the Schwarzschild metric: [latex]ds^2=\left(1-\frac{2GM}{rc^2}\right)dt^2-\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2-r^2d\theta ^2 - r^2 sin^2\theta \ d\phi ^2[/latex] and the event horizon radius for a Schwarzschild black hole (Schwarzschild radius) : [latex]R_s=\frac{2GM}{c^2}[/latex] Pseudoscience has no mathematics (or no scientific evidence for mathematical claims), unlike black holes. There is lots of evidence for black holes, such as supernovas. If a star is massive enough, it will collapse into a supernova. The matter left behind will have to compress into a black hole, and that is what happens.

[latex]\partial V[/latex] is just an infinitesimal volume. It can mean different things because [latex]V[/latex] is a function of multiple variables. Let's say that [latex]V[/latex] is a function of some variable [latex]n[/latex]. To find [latex]\partial V[/latex], we can imagine [latex]\frac{\partial V}{\partial n}[/latex] is a ratio between two infinitesimals, [latex]\partial V[/latex] and [latex]\partial n[/latex]. Because of this, we just multiply the partial derivative by [latex]\partial n[/latex] to get [latex]\partial V[/latex], i.e [latex]\partial V=\frac{\partial V}{\partial n}\partial n[/latex] This can be applied to the different variables V depends on, such as x, y, and z. For example, let's say that [latex]V(x, y, z)= x^2 + y^2 +z - xyz^2[/latex] In this case, we can take the partial derivative with respect to 3 variables: x, y, and z. We will find [latex]\partial V[/latex] with respect to x. We find that [latex]\frac{\partial V}{\partial x}=2x+y^2-yz^2[/latex] and so [latex]\frac{\partial V}{\partial x} \partial x=(2x+y^2-yz^2) \ dx[/latex] For cylinders, [latex]V[/latex] is a function of [latex]z[/latex] and [latex]r[/latex]. The volume of a cylinder is given by [latex]V=\pi r^2 z[/latex] And so we have 2 values of [latex]\partial V[/latex], [latex]\partial V=\frac{\partial V}{\partial r} \partial r=2\pi r z dr[/latex] and [latex]\partial V= \frac{\partial V}{\partial z}\partial z=\pi r^2 dz[/latex] In a cylinder with uniform density, [latex]dV[/latex] is just [latex]dV=\frac{1}{\rho }dM[/latex]

Oh, just wondering.

Relative to the particle, the black hole is accelerating towards it with the same acceleration as the particle. All accelerating reference frames are equivalent, so the black hole appears to experience time dilation relative to the particle. When [latex]R_s=r[/latex], time freezes for the particle with respect to a distant observer. When the particle reaches the event horizon, the black hole appears to have the same acceleration as the particle with respect to the particle. So, shouldn't the black hole experience time dilation with respect to the particle?

What about people who are good at multiple things?

This is a little late, but I am new to this forum.

I still don't understand what you are trying to say.

It is an argument that nothing can fall into a black hole with respect to a distant observer and with respect to the object falling into the black hole.

I don't see how it is the hidden code of pi.

I mixed up [latex]t[/latex] and [latex]\tau[/latex]. They should have been switched. But, it would still apply. The black hole would still be accelerating relative to the infalling object.

The event horizon is the place where the escape velocity is the speed of light. The distance from the singularity at the event horizon is represented by the Schwarzschild radius, Rs. It is given by [latex]R_s = \frac{2GM}{c^2}[/latex]. Nothing is ever able to go past the event horizon, even relative to the object accelerating towards the black hole. This is because the black hole is accelerating relative to something accelerating towards a black hole. Because of the equivalence principle, the black hole will experience time dilation. The time dilation is the same as that relative to an outside observer, i.e [latex]t=\tau \sqrt{1-{\frac{R_s}{r}}}[/latex]. When Rs=r, time freezes for the black hole relative to the object accelerating, so an object cannot get past the event horizon.