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Math Help - How to express the acceleration of inversion?

  1. #1
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    How to express the acceleration of inversion?

    In inversive geometry, as a point x within a unit circle approaches the center of the unit circle, its inversion point y outside the circle approaches infinity. The distance of the two points to the center of the circle, if I'm not mistaken, can be expressed as follows:

    D_x=\frac{r^2}{D_y}

    (r being the radius of the unit circle)

    Without further preamble, my question is as follows:

    How can we express the acceleration of point y, outside of the circle, with respect to point x?

    In other words, if points x and y are spaceships, what is the acceleration of spaceship y with respect to spaceship x? AND, is this a constant acceleration, or an accerating acceleration (second or higher derivative)?

    Thanks
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    Off the top of my head, I'd say you would have to ask the concept or Relativity whether or not the relative acceleration is constant. Personally, I'd be a bit shocked if it were constant.

    You have D_{x} \cdot D_{y} = r^{2}

    Since the position of Y is dependent on the position of X, we also have: D_{y} = f(D_{x}) (Which you might have stated in a less abbrevaited preamble.)

    This leads nicely to the implicit derivatives:

    D_{x} \cdot D_{y}' + D_{y} = 0

    and

    D_{x} \cdot D_{y}" + D_{y}' + D_{y}' = y

    Then, just a little algebra produces:

    D_{y}' = -\frac{D_{y}}{D_{x}}

    and

    D_{y}'' = -\frac{2 \cdot D_{y}'}{D_{x}} = \frac{2 \cdot D_{y}}{\left( D_{x}\right)^{2}}

    Frankly, this is a bit astounding. As X gets close to the Origin, the crew on Y had better batten down the hatches! If X only flinches, Y will jolt off a very long way... One cannot argue much with that SQUARED Dx in the denomiator. Wow!

    Remember \Delta E = \Delta m \cdot c^{2} from the Special Theory of Relativity? It HAD to be astounding just how much energy differential would be created by a very small amount of converted matter. Same thing. Giant numbers in the numerator or tiny numbers in the denominator are bad enough. SQUARE them and forget it. That's BIG!
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    Quote Originally Posted by TKHunny View Post
    Off the top of my head, I'd say you would have to ask the concept or Relativity whether or not the relative acceleration is constant. Personally, I'd be a bit shocked if it were constant.

    You have D_{x} \cdot D_{y} = r^{2}

    Since the position of Y is dependent on the position of X, we also have: D_{y} = f(D_{x}) (Which you might have stated in a less abbrevaited preamble.)

    This leads nicely to the implicit derivatives:

    D_{x} \cdot D_{y}' + D_{y} = 0

    and

    D_{x} \cdot D_{y}" + D_{y}' + D_{y}' = y

    Then, just a little algebra produces:

    D_{y}' = -\frac{D_{y}}{D_{x}}

    and

    D_{y}'' = -\frac{2 \cdot D_{y}'}{D_{x}} = \frac{2 \cdot D_{y}}{\left( D_{x}\right)^{2}}

    Frankly, this is a bit astounding. As X gets close to the Origin, the crew on Y had better batten down the hatches! If X only flinches, Y will jolt off a very long way... One cannot argue much with that SQUARED Dx in the denomiator. Wow!

    Remember \Delta E = \Delta m \cdot c^{2} from the Special Theory of Relativity? It HAD to be astounding just how much energy differential would be created by a very small amount of converted matter. Same thing. Giant numbers in the numerator or tiny numbers in the denominator are bad enough. SQUARE them and forget it. That's BIG!
    Yeah, I guess you really gotta move fast if you want to make it to infinity.

    D_{y}' = -\frac{D_{y}}{D_{x}}

    Thank you for this. I'm still trying to grasp exactly how you got to this.

    (In particular I don't see how you got from D_{y} = f(D_{x})

    to the implicit derivative: D_{x} \cdot D_{y}' + D_{y} = 0)

    But accepting your result, can we then conclude that the acceleration of spaceship Y A_y is:

    V_{y}' = A_y = -\frac{V_{y}}{V_{x}}= -\frac{1}{V_x^2} ?

    (Since if you substitute V_nT for D_n (where V and T are velocity and time), AND you assume that T= the unit radius r, you get the same equation we started with except in terms of velocity, i.e.:

    V_{x} \cdot V_{y} = 1

    Apply the same steps you applied to D_y and we arrive at this expression of acceleration, don't we?)

    If this is a valid expression of spaceship Y's acceleration with respect to spaceship X, then I guess said acceleration is not constant.
    Last edited by rainer; January 17th 2010 at 10:24 AM.
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    I think we're not there yet. What has been expressed so far is the relationship of the two ships at generally lower speeds. If we are getting anywhere near the speed of light, relative speeds don't quite act the same. Too bad I don't know anything about that sort of thing.
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    Quote Originally Posted by TKHunny View Post
    I think we're not there yet. What has been expressed so far is the relationship of the two ships at generally lower speeds. If we are getting anywhere near the speed of light, relative speeds don't quite act the same. Too bad I don't know anything about that sort of thing.
    If you know nothing about this sort of thing, I know much less than nothing.

    All right, well if you want to go ahead and throw in the towel that's fine, but it seems to me there are ways around the whole "above 10% speed of light" thing. And even if we were forced to take acount of it, Lorentz dilation is just a working up of pythagoras' theorem.
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    FYI:

    I got the following help from "ask Dr. math."

    Apparently, it's not so complicated as I was making it.

    "To work out accelerations, et cetera, begin by first writing down the position of the point
    moving toward the center of the circle in terms of time. For example, if the point moves
    along the x-axis from 1 to 0 taking total time 1 unit, then its position is:
    x(t) = 1 - t.
    y(t) = 0,
    where the circle is the unit circle centered at the origin.
    The position of its inversion is thus:
    X(t) = 1/(1-t)^2
    Y(t) = 0
    The y-coordinates don't enter into it, since they will obviously both be zero always, but
    you can just take first and second derivatives of X(t) to find the velocity and acceleration of
    the inverted point at any time t.
    The nice thing about setting it up this way is that if you want, you could investigate how
    the acceleration of the inverted point changes with different rates of approach of the first
    point to the center of the circle."

    I think he means to write X(t) = 1/(1-t) instead of X(t) = 1/(1-t)^2. The inversion equation is y=r^2/x.
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