# Impossible to graph position vs time?

• leok
In summary: This is correct - the differential equation I gave should yield an explicit position versus time function. It just won't work all the way to the singularity.
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

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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:
$$F=ma$$
$$\frac{Gm_{1}}{r^{2}}=a$$

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

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

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 wouldn't that make relativity no longer a theory but rather a law?

i do however think it is possible to graph without relativity. i just haven't been able to incorporate time into my equation where time is my independent variable. LOLwhat 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 don't think its relativity that we need)

Last edited:
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.

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

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

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

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

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

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

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

## 1. What does "impossible to graph position vs time" mean?

It means that it is not possible to create a graph that shows the relationship between an object's position and time. This could be due to various factors such as the data being incomplete or inaccurate.

## 2. Why is it important to be able to graph position vs time?

Graphing position vs time allows us to visually understand the movement and behavior of an object. It is a useful tool in analyzing data and making predictions about future movements.

## 3. Can any type of movement be graphed as position vs time?

No, not all types of movement can be graphed as position vs time. For example, circular motion cannot be accurately represented on a position vs time graph.

## 4. What are some limitations of using a position vs time graph?

One limitation is that it only shows the overall trend of an object's movement and does not provide detailed information about its velocity or acceleration. Additionally, it may not be able to accurately represent complex or non-linear movements.

## 5. How can we overcome the challenge of graphing "impossible to graph position vs time"?

There are a few strategies that can be used to overcome this challenge. One approach is to collect more precise and accurate data. Another option is to use advanced mathematical and analytical techniques to interpret the data and create a visual representation. Additionally, it may be helpful to consider other factors that could affect the object's movement, such as air resistance or friction.

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