Velocity of an object accelerating towards a mass

[PWiz]

I attempted to find a formula relating the velocity of an object released from infinity moving towards a mass $M$to the distance $r$ from $M$. According to Newton's law of gravity, $a=\frac{GM}{r^2}$. Since $a=v \frac{dv}{dr}$ , $\int a dr = \int v dv = \frac{v^2}{2} + k$. Since initial velocity equaled 0, k=0 . So $v=\sqrt{2 \int a dr}= \sqrt{\frac{2GM}{r}}$ (ignoring the -ve in the integral).
I've checked my maths but I'm still not sure if this formula is correct, because I immediately noticed that the formula appears to violate the second postulate of the special theory of relativity in a particular case. If $M$ is considered to be spherical and have a radius of $x$ where $x<\frac{2GM}{c^2}$(by substituting v with c) with most of it's mass concentrated at the center, then any object moving towards this mass from infinity will exceed the speed of light when $r$ becomes less than $\frac{2GM}{c^2}$(i.e. when the object moving towards $M$ enters the region between $x$ and $\frac{2GM}{c^2}$). This also means that the escape velocity for a distance less than $\frac{2GM}{c^2}$ from $M$ will be unachievable, which doesn't make any sense. Where have I gone wrong?

 

[HomogenousCow]

The inverse squared potential is solvable in one dimension.
Write out the full DE, and substitute r''=(dr'/dr)(dr/dt)=(r' dr')/dr
Multiply the dr over and integrate, and take it from there.

 

[HomogenousCow]

The integrals are quite nasty though, and you end with an implicit solution.

 

[PWiz]

Um I've done that. a=r'', so r"= r' (r'/r) (where r'=dr/dt), which is the same as what I've written (a= v [dv/dr], and the dr s cancel out)

 

[SteamKing]

I attempted to find a formula relating the velocity of an object released from infinity moving towards a mass $M$to the distance $r$ from $M$. According to Newton's law of gravity, $a=\frac{GM}{r^2}$. Since $a=v \frac{dv}{dr}$ , $\int a dr = \int v dv = \frac{v^2}{2} + k$. Since initial velocity equaled 0, k=0 . So $v=\sqrt{2 \int a dr}= \sqrt{\frac{2GM}{r}}$ (ignoring the -ve in the integral).
I've checked my maths but I'm still not sure if this formula is correct, because I immediately noticed that the formula appears to violate the second postulate of the special theory of relativity in a particular case. If $M$ is considered to be spherical and have a radius of $x$ where $x<\frac{2GM}{c^2}$(by substituting v with c) with most of it's mass concentrated at the center, then any object moving towards this mass from infinity will exceed the speed of light when $r$ becomes less than $\frac{2GM}{c^2}$(i.e. when the object moving towards $M$ enters the region between $x$ and $\frac{2GM}{c^2}$). This also means that the escape velocity for a distance less than $\frac{2GM}{c^2}$ from $M$ will be unachievable, which doesn't make any sense. Where have I gone wrong?

You haven't necessarily gone wrong, you've just achieved a result first obtained by scientist John Michell more than 200 years ago, when he was studying Newton's theory of gravitation:

http://en.wikipedia.org/wiki/John_Michell

In his work, Michell showed that for certain massive stars, the escape velocity of those objects would exceed the speed of light, and thus would not be visible to an external observer. In effect, Michell was the first scientist known to have described what are now called black holes.

Many years after Michell, German physicist Karl Schwarzschild used parts of Einstein's theory of general relativity to derive a similar result as Michell had for the escape velocity of a spherical, non-rotating object of mass M.

http://en.wikipedia.org/wiki/Schwarzschild_radius

The Schwarzschild radius r_s of an object with mass M is r_s = 2GM/c².

 

[HomogenousCow]

I attempted to find a formula relating the velocity of an object released from infinity moving towards a mass $M$to the distance $r$ from $M$. According to Newton's law of gravity, $a=\frac{GM}{r^2}$. Since $a=v \frac{dv}{dr}$ , $\int a dr = \int v dv = \frac{v^2}{2} + k$. Since initial velocity equaled 0, k=0 . So $v=\sqrt{2 \int a dr}= \sqrt{\frac{2GM}{r}}$ (ignoring the -ve in the integral).
I've checked my maths but I'm still not sure if this formula is correct, because I immediately noticed that the formula appears to violate the second postulate of the special theory of relativity in a particular case. If $M$ is considered to be spherical and have a radius of $x$ where $x<\frac{2GM}{c^2}$(by substituting v with c) with most of it's mass concentrated at the center, then any object moving towards this mass from infinity will exceed the speed of light when $r$ becomes less than $\frac{2GM}{c^2}$(i.e. when the object moving towards $M$ enters the region between $x$ and $\frac{2GM}{c^2}$). This also means that the escape velocity for a distance less than $\frac{2GM}{c^2}$ from $M$ will be unachievable, which doesn't make any sense. Where have I gone wrong?

Well this is expected, you're not working with a relativistic theory.

 

[PWiz]

Wow, I must admit I'm very surprised that my result was correct. But then what happens to the object going towards $M$ (black hole)? Does it's velocity exceed light speed after crossing the Schwarzschild radius?
@HomogeneousCow Assume that $M$ is stationary. Which relativistic equation should I then use to eliminate this problem?

 

[HomogenousCow]

Wow, I must admit I'm very surprised that my result was correct. But then what happens to the object going towards $M$ (black hole)? Does it's velocity exceed light speed after crossing the Schwarzschild radius?
@HomogeneousCow Assume that $M$ is stationary. Which relativistic equation should I then use to eliminate this problem?

The geodesic equations in GR. Although it's far more than just using an equation, the gravitational field (the metric) becomes a dynamical entity like the EM field in GR.
For a constant spherically symmetric mass, you would have the schwarzschild metric in the geodesic equation.

 

[SteamKing]

Wow, I must admit I'm very surprised that my result was correct. But then what happens to the object going towards $M$ (black hole)? Does it's velocity exceed light speed after crossing the Schwarzschild radius?
@HomogeneousCow Assume that $M$ is stationary. Which relativistic equation should I then use to eliminate this problem?

The laws of physics break down inside the Schwarzschild radius, also known as the event horizon:

http://en.wikipedia.org/wiki/Event_horizon

For the most part, what happens inside the event horizon stays inside the event horizon.

 

[PWiz]

@SteamKing Alright, thanks!
@HomegenousCow Ah, then I better leave it. I can tackle SR problems mathematically, but I haven't touched GR yet, and it's safe to say that its out of my league for now (I haven't started with topology yet)

 

[Dale]

I immediately noticed that the formula appears to violate the second postulate of the special theory of relativity

Of course it does. You derived it from non relativistic physics. You shouldn't expect it to be relativistically correct. For that you would need to derive it from general relativity instead of Newtonian gravity.

Edit: I see others already mentioned that. I found this paper helpful http://arxiv.org/abs/gr-qc/0311038

 

[stevebd1]

In 'Exploring Black Holes' by Wheeler & Taylor, $v_{shell}=\sqrt(2M/r)$ is the velocity of an object that has fallen from rest at infinity (rain frame) relative to a shell frame (i.e. a specific radius) and becomes undefined when $r\leq 2M$ as there is no stable r (or shell) within the event horizon. For the velocity of the object as observed from infinity, multiply by $(1-2M/r)$.

The above applies for a static black hole, I looked for a similar equation for a rotating black hole and found the following-

$$V^{\hat{r}}_{\text{ff,BL}}=-\left(\frac{2M}{r}\right)^{1/2}\ \ \frac{r\ (r^2+a^2)^{1/2}}{\Sigma}$$

where $M=Gm/c^2,\ a=J/mc,\ \Sigma=\left[(r^2+a^2)^2-a^2\Delta\sin^2\theta\right]^{1/2}$ and $\Delta=r^2+a^2-2Mr$

and to be expected, v=1 at the event horizons $r_\pm=M\pm\sqrt(M^2-a^2)$

source- Black Hole Astrophysics: The Engine Paradigm page 240 eq. 7.56I'm also came across another equation for v from 'Painleve-Gullstrand Coordinates for the Kerr Solution' by J Natario -

$$v=-\frac{\sqrt{2Mr(r^2+a^2)}}{\rho^2}$$
where $\rho^2=r^2+a^2\cos^2\theta$

This is referred to as 'radial proper velocity of a zero angular momentum observer dropped from infinity'. With this equation, there doesn't seem to be any significance to where v=1, which occurs outside the event horizon and ergosphere. Does someone have an idea of the significance of this equation?