# Impossible to graph position vs time?

1. Jan 26, 2009

### leok

picture a body falling toward a black hole.
at the center of the black hole you have an infinite amount of force pulling you toward the center: F=Gm1m2/x^2 while x=0. F=infinity
so at a certain distance away from the black hole you have a certain force pulling you in.
but as you approach the center your acceleration increases.
but as you accelerate your velocity increases
so you were once not moving at you initial position. after a few seconds you start moving due to the force pulling you in.
but what happens is you gain velocity so now your acceleration is increasing.
but since you are accelerating your velocity starts to jerk.
the jerk effects the acceleration and you get a cycle of one derivative influencing the next.
so the position vs time graph becomes infinitely differentiable.

if you look at it in terms of energy:
work=force x distance
so Kinetic energy would be the integral of f=Gm1m2/x^2
your velocity is the square root of 2 times the integral

if change in momentum=impulse

m x dv= f x t

we have m2 x our velocity = f x t

we have force for a given X but not a given T so we cannot integrate Gm1m2/x^2
so to be able to graph x vs t we need f vs t

2. Jan 26, 2009

### Nabeshin

The "infinite differentiability", as you call it, isn't a problem in and of itself. Take the function $$e^{x}$$ for example. This is a similar case, and yet clearly it is graphabe.

Your problem is not really unique to black holes, however. Any body will produce this effect. The fact that the force of gravity goes to infinity as the distance goes to zero is mooted by the fact that our laws of physics (classical laws, at least) do not work after a certain point so we cannot describe that.

I think the energy equation you were talking about is this:
$$\int\frac{Gm_{1}m_{2}}{r^{2}}dr=\frac{m_{2}v^{2}}{2}$$

But this equation is rather useless. If we assume a singularity, a force will be felt by the infalling object will increase without bound as r approaches zero, which it can. Since the singularity has no size, there is no boundary to stop the gravitation. Therefore, the gravitational potential energy would be infinite.

I would write something more along the lines of:
$$F=ma$$
$$\frac{Gm_{1}}{r^{2}}=a$$

$$\frac{Gm_{1}}{r^{2}}=\frac{d^{2}r}{dt^{2}}$$

3. Jan 26, 2009

### Chronos

Classical metrics do not apply, and, yield invalid results compared to general relativity - as Nabeshin noted.

4. Jan 27, 2009

### leok

so the graph x vs t impossible to graph with a function?
and yes i realize this is not unique to black holes i was just trying to get ppl's attn

is it necessary to use relativity?
if so wouldnt that make relativity no longer a theory but rather a law?

i do however think it is possible to graph without relativity. i just havent been able to incorporate time into my equation where time is my independent variable. LOL

what i am really trying to get at is if i have a body at a certain distance from a planet there is no function that will tell me the time it takes to get there. HA! such a simple question to ask. (and i refuse to use euler's method- that %error is gross)

(Also if it is solvable using relativity then the same scenario applies to forces and test charges-bam! i said it first XD.... just kidding i dont think its relativity that we need)

Last edited: Jan 27, 2009
5. Jan 27, 2009

### Nabeshin

The differential equation I gave should yield an explicit position versus time function. It just won't work all the way to the singularity.

Relativity won't help, because although it may predict a singularity, it doesn't help to describe it. Relativity would, however, likely yield a better answer to the problem, simply because Newtonian mechanics are merely an approximation which will become more obvious as the gravitational field of the black hole (or other celestial body) increases.

6. Jan 28, 2009

### leok

well your differential equation does not work the way i did it.
it yeilds t=abs(x) which i know isnt right

im still working on it so ill get back to this post soon

7. Feb 3, 2009

### leok

yeah i got nothing.
seems it is impossible to graph position vs time without the euler method

8. Feb 3, 2009

### Nabeshin

$$\frac{Gm_{1}}{r^{2}}=\frac{d^{2}r}{dt^{2}}$$

$$Gm_{1}dt^{2}=r^{2}d^{2}r$$

$$\int \int Gm_{1}dt^{2}=\int \int r^{2}d^{2}r$$