why do x Schwarzschild radii equal time dilation effects of speed of light going y times faster than an object^2?

Question

let me walk you through the math.

First you start with the gravitational time dilation formula where:

$$ T_1=T\sqrt{1-\frac{2GM}{rc^2}} $$

and rather than entering $r$ for the radius we replace $r$ with the Schwarzschild radius formula $(2GM/c^2)x$ with an $x$ at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:

$$ T_1=T\sqrt{1-\frac{2GM}{\frac{2GM}{c^2}xc^2)}} $$

Which when simplified breaks down to:

$$ T_1=T\sqrt{1-\frac{1}{x}} $$

and if you make $T=1$ then you just get

$$ T_1=\sqrt{1-\frac{1}{x}} $$

This is very similar to the one in many physics books $=\sqrt{1-r_0/r}$, where $r_0$ is equal to the Schwarzschild radius and then $r$ equals the radius from the center. The formula above it just makes it slightly simpler due to making $r_0$ equal to 1 and $x$ equal to how many radii a point you are observing is from the center of the mass.

That is the gravitational time dilation side portion of this relationship. Now for the velocity time dilation side we use a similar methodology and start with:

$$ T_0=T\sqrt{1-\frac{v^2}{c^2}} $$

Now we make $T$ equal to 1, $v$ equal to one, and $c$ to $y$ because now we are going to make $c$ a variable.

$$ T_0=\sqrt{1-\frac{1}{y^2}} $$

What you see now "$1/y^2$" is showing the velocity as a constant 1 and $y$ represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as $1/5^2$ then this would be the same as saying an object is going at a velocity 1/5th the velocity of light. So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:

$$ \sqrt{1-\frac{1}{x}}=\sqrt{1-\frac{1}{y^2}} $$

We can simplify this to

$$ x=y^2 $$

What does this mean?

Answer 1

It means that if move from far away from a black hole to a distance of $x$ Schwartzschild radii away from a black hole, the relative time dilation you experience is the same as if you instead accelerate from rest to a speed of $\frac{1}{\sqrt{x}}$ times the speed of light.

For example, you would have the same time dilation relative to a stationary observer far from a black hole if you were at a radius of 3 Schwartzschild radii, or if you were at a velocity of $\frac{1}{\sqrt{3}}c$

Those two actions, of moving close to a black hole, and speeding up, both cause time dilations, and you have found how to achieve the same effect in two different ways.

Answer 2

Just because you can "derive" the same physical phenomenon in two different settings does not necessarily make them related, or allow you to just set things equal. In particular, the time dilation formula familiar from special relativity (i.e. $T = \frac{T_0}{\sqrt{1-v^2/c^2}}$) is only valid for uniformly moving observers in flat space. The spacetime around a black hole is not flat, so you cannot use that time dilation formula there, and your comparison is meaningless.

However, there is a unifying framework that encompasses both the time dilation of special relativity as well as the time dilation in a gravitational field (e.g. near a black hole); while the comparison between $v$ and $r$ is still meaningless, in this sense the two time dilation effects are due to the same physics. The idea is just that time dilation measures the relative difference between two clocks held by different observers. This "time measured on a clock" is what is referred to as the proper time of an observer, and it can be calculated as the length* of the path of an observer in four-dimensional spacetime, where this length* is calculated by using the spacetime metric.

In flat space, the time dilation effect happens when two observers A and B are moving relative to one another, and therefore as seen by B, the proper time elapsed along the trajectory followed by A is different than that elapsed along B's trajectory. This causes B to think their clocks to disagree. (Likewise, A also thinks their clocks disagree)

Near a black hole, instead, the time dilation effect happens because the spacetime itself is curved, and this curvature is different at different values of the coordinate $r$. Therefore, when you calculate the proper time elapsed along the trajectories of observers A and B sitting at different $r$ values, you get different results, and once again the two observers' clocks disagree.

The key here is that there is only one time dilation effect: calculate the proper time of the spacetime trajectory of two different observers and compare. The two apparently different effects you quote are just different manifestations of this same unified approach.

*Here I'm using "length" in a loose sense, since the invariant interval $ds^2$ along the world line of an observer is negative. What I specifically mean by "length" here is the integral $\int \sqrt{-ds^2}$ along an observer's world line.

Answer 3

$$ T_1=T\sqrt{1-\frac{2GM}{rc^2}} $$

Besides of being the gravitational time dilation formula, the above formula is also the kinetic time dilation formula for a clock that has been free falling from infinity to radius r.

A reason for that is that kinetic time dilation is proportional to the ratio of kinetic energy and rest energy, while gravitational time dilation is proportional to ratio of potential energy and potential energy at infinity.

When calculating kinetic time dilation at radius r, we might first calculate how much kinetic energy a clock that fell from infinity has at that point, and that thing we might calculate by calculating how much potential energy the clock has lost at that point.

When calculating gravitational time dilation we might first calculate potential energy at that point.

There must be some simple way to say all that. Maybe this way: Gravitational time dilation is proportional to potential. Kinetic time dilation is proportional to how large potential well can be climbed by the kinetic energy.