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  1. #31
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    Re: TOE by Brent Harrington

    With all due respect for your knowledge, Dave, your last question seems strange seeing as though you answered it yourself when you said the "ones" represent quantities of qualities.

    Positive charge is "something" that is cancelled by negative charge to render charge as "nothing." The same can apply to mass, energy, space, time, and even you yourself because the particles/waves that you consist of are composites of negating forces.

    My only argument is that fermions, based on velocity (not speed), can't occupy the same place at the same time as well. They are the result of +- decay through time dilation which either has a positive or negative reference frame.

    I'm pretty sure that Brent is referring to keeping universal symmetry, which is promising when we consider that models based on asymmetry can't be correct - including the broken symmetry of matter/antimatter for yet to be discovered reasons.

  2. #32
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    Re: TOE by Brent Harrington

    Something came up. I will be back.

  3. #33
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    Re: TOE by Brent Harrington

    I think this is where I ended my last post, if anybody is still following it.

    Comparing the electrical and the physical circuits.

    The electrical circuit is an inductor made of superconductor, sized to allow one electron at a time to pass. (The circuit could just be a super conductor but being an inductor helps with the comparison). When a voltage is applied to the circuit, charges will begin to move through the circuit. The charges will accelerate, following an exponential curve, until they reach C. At this time an equal and opposite voltage will be produced by the inductor preventing any further increase in current.

    For now the only thing I want to take from this is that the the rise in current when plotted on a graph will follow an exponential curve.

    The comparison physical circuit is a mass in free space. To make it interesting let us make it a series circuit. The circuit is a rocket that applies a given quantity of force regardless of velocity and a series of masses and scales.

    ROCKET-->scale/mass/scale/mass/scale/mass/scale/mass/scale/mass

    This circuit will accelerate from zero to C relative to space.
    The rocket will apply a force on the first scale until it reads 5kg and will maintain this 5kg reading up to C.
    The scales can be considered massless and are only for taking measurements.
    Each of the five masses has a rest mass of 1 kg

    We start the rocket and begin accelerating. The first scale will read 5kg, the second 4kg, the third 3 kg, the forth 2kg and the fifth 1 kg. As the circuit's velocity rises relative to space, the rate of acceleration will decrease. The acceleration will decrease as
    velocity rises until acceleration stops at C. As the velocity rises, the mass of the circuit (resistance to acceleration) increases. The first scale will always read 5kg regardless of velocity because of the design of the rocket control. But as the velocity rises the second scale reading will begin to increase. As will the other scales. Eventually the second scale will read 5 kg, then the third will reach 5 kg, then the third and so on until all the scales
    read 5kg at the C.

    The acceleration of this circuit relative to space, when plotted on a graph should follow an exponential curve just like the electrical circuit.

    Velocity(instantaneous) = C(1-e^-t/T )

    C = speed of light
    t = time in seconds
    T = (tau) = C/a
    a = acceleration
    e = 2.718

    Thats all for today.
    Last edited by Brent; 07-25-2007 at 11:09 PM. Reason: Typos

  4. #34
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    Re: TOE by Brent Harrington

    Hello?

    Velocity(instantaneous) = C(1-e^-t/T)

    Does this equation work?
    Did I re-invent something again?

    I am not sure how to verify this equation works. As far as I can tell, I need to calculate the velocity relative to an observer using the Lorenz transformation.

    S coordinates are t,x,y,z
    S' coordinates are t',x',y',z'

    Lorenz transformation specifies
    t'= Y(t-vx/c^2)
    x'= Y(x-vt)
    y'=y
    z'=z
    Y= 1/square root of 1-v^2/c^2

    I am not sure what to do from here and I wouldn't be sure of my answer anyway.

    Anybody else care to take a stab at it?

    So, if I accelerated at a known rate for a given number of seconds (enough to reach a relativistic velocity), what would be my velocity relative to space by my equation? What would my velocity be relative to an observer considered at rest?

 

 
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