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Thread: Black Hole force, where are the vectors?

  1. #1 Black Hole force, where are the vectors? 
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    //edited some errors, G is not /~ r^2 sorry. Was thinking of declining of force
    As you might understand, a force needs a vektor, since forces release energy, and for a photon to be released, there must allready be an existing vektor along a coordinate which change direction into the room or, if mechanical force, remains in the room.

    But in a black hole, no vektor can exist since acceleration is as good as total and any vektor would be severely slowed down in a black hole. But it still has gravity which leads to the conclusion that gravity is caused by an outer force, an "empty" spacetime force.

    This is the first idea: The EM force would be caused by that the vector along time has two directions and light speed. The lightspeed would cause the time coordinate to be infinitely thin, and hence, the illusion is that the vektors are not moving away from eachother along this coordinate, and in their comparative system, they are not. Hence positive and negative particles have opposit vektors, and when colliding, they change directions into the room in the form of photons.

    This is the second idea:

    The cosmological constant applies to time distance aswell, and it is attracting along the imaginary time coordinate. Even if time in our RF is infinitely thin, it still has a distance. But the force caused by the distance, only declines from the time vector throughout the room, and it is not only proportional to the time past, it is directly additive and by that I mean that it stacks as if the vektor have not moved but the distance have increased.

    Equations:

    v = everage time speed

    G ~ vt/4pi

    /4pi for the sperical decline

    (1.)But since all energy is equally effected by this force, and all vektors in space slows down caused by force (and just like black holes slows down in timevektor), G is also /~ with vt since force is ~ vektor.

    G ~ 1/4pi



    Also, one must add, that the constant would be proportional to the mass area A, since the bend in spacetime is caused by the tunnel it makes in 4d. For a 2d surface with a tunnel along the third dimension, the force is caused by the tunnel circle as compared to the circle it declines with. Circle against circle, surface against surface.

    G ~ A/4pi

    I don't know if there are any more proportionalities. If you see something more just pm me, until further notice:

    G = A/4pi = (A(m)/m)/4pi

    Note: the energydensity is both ~ and /~ by reasons mentioned (1.)

    Thank you for reading this far.

    Yep, that should definitely describe it:

    F = (A(m)/m)*m(1)*m(2)/(r^2*4pi)

    One kilo mass, 8.38502389 E-10 m^2 surface. Can it be possible?

    surface ~ mass. Look, surface is proportional to mass. Else A(m)/m would not be constant.

    If vectors shortens, so does the force!


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  3. #2  
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    I wouldn't normally even check this forum but since you pmed me about it I thought I'd check it out. To be honest I don't understand this at all, I don't know whether it's because english is not your first language or because your physics is flawed. I'm not a physicist, I only really deal with physics within the earth sciences certainly not cosmology.

    But you lost me in your first sentence when you said "forces release energy". I realize that they are linked but I wouldnt say that, as for "acceleration is as good as total", what does that mean? Sorry, you lost me. I've heard of the total derivative of a force field being linked to acceleration - are you getting at that?

    Could you at least give some refernces (internet links) because i for one need to do some background reading on this before I can respond intelligently.


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  4. #3  
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    when a = max, time does not move ie. relativity. It pass infinitely slow. We move along time.

    Until missaproven I am going to post this pm I made:

    I could prove the proportionalities one by one. First we have the distance the mass moves. A mass moves a distance without us noticing, that's why when pushed, they release photons that moves in the speed of light. The longer they move, the more probable it is that they meet photons and these photons give their vektor to the mass (this caused by mass coordinate being infinitely thin and mass being along the entire distance it travels). The vector entails a force, and this force makes the mass slower, hence it loose some of its time vektor and instead get a room vektor. Over time the roomvektor moves back and fourth and does not add up to the distance the mass moves, so the time vektor lowers. But that applies to all vektors, so the force it creates that draws the photon back is not proportional to the distance the photon travel since it is both ~ and dividedly ~ to timedistance (since the vektors are in amount ~ distance and in itself /~ to distance. Thing is, only the horizontal vektor tells directly of the force. Perhaps your question then is "how many of those are there?" That doesn't matter, since whatever ~ konstant there is, it applies to the horizontal vector aswell, so the force is both /~ and ~ with the same constant to the distance it moves. so gravity won't change with time, only vary slightly since the vektors in space hardly hits at the same time.

    But the force is then not zero but k*t/t = G.

    Now for the constant. We know that force declines spherical in space.
    so G must be /~ to 4pi. It must also be proportional to the sphere that the particle makes since this is the opening of the 4d tunnel that the mass produce, and the force increase with the size of the tunnel, and you will find, if you do experiments, that the size of the opening is ~ to the force the growing tunnel produce, and the momentum ~ to time and force.

    In a tunnel, the opening is circular and if the tunnel would pull an invisible web into the hole, you can imagine this. Ofcourse, the momentum also increase for any invisible particle in the web with the declining distance from the hole, just like any other force. So momentum increase with sphere and growing of tunnel (time) F = momentum/time

    So since the tunnel is 4d, the tunnel opening is spherical and so is the declining of force, sphere to sphere and in previous example, circle to circle.

    so G ~ (A(m)/m)/4pi (G ~ r^2(m))

    A(m)/m is the sphere area per mass.

    4pi caused by the outer sphere with variable radius r.

    You can see now, that any other ~ would be caused by the timespan the universe was created on, hence the length past difference of the different masses along the time coordinate? My postulate are that all distances are equal, hence:

    G = r(m)^2/m
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  5. #4  
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    Earth to leaving quietly.......... that doesn't help. in fact i didn't even bother reading past the first paragraph!

    You need to cater for your audience and it seems that audience is me, try talking in layman's terms, it might make things a little clearer.
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    Ok, this is how I would explain it. An electron has a force field, a force cause a particle in the force field to move, a movement has energy. If the electrons charge is caused by a particle moving through time, then this particle must, for time to become infinitely thin, move in lightspeed. It's called length contraction. it must also has a vector that causes the energy, unless EM is caused by an outer vektor, which is not probable.

    As for acceleration, it has a maximum value, and when it reach this value, then time dilation, and dilation of movement in general, occur.

    In a black hole this dilation is considerable, so the gravity force, which remains, locks or partially locks any inner force, like EM. But it does not lock itself, and that implies that it is not dependent of inner vektors.

    Every force needs a vektor to release energy, which i have had confirmed earlier. Along a coordinate, let it be time. If we would travel through time in lightspeed, it would become infinitely thin. As I understand it, the vektor would then exist for the energy release of the force. The strong and the weak force would be caused by the vector changing roomcoordinates. But since time as mentioned, was infinitely thin, the vector remains were the particle existed until the energy has been released. In the case that time was really thin in our reference frame. That is my theory.
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    Not sure I understand what you're saying but it kind of reminds of a theory I read somewhere that all "particles" are actually just one particle moving through time.
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    The beginning in the universe, must have been a funktion of itself. The way I see it, forces entails that they were, if independent of outer factors, that is, ~ and /~ to energy density, a funktion of the properties of a non existent object it is clear that the force is /~ to a n-dimensional sphere that is zero in value, and the force is proportional to the sphere of the object itself, that is zero in value, still generates a force that is 1 in value, and needs a vektor that is proportional to the force, so the universe was forced to exist. Makes sence? so what I am trying to say is, such a vektor created by this force is the smallest photon, and that photon is undevideable, since it has got the size 1. But photons can unite or come to exist right next to eachother, why there is a smallest photon and the universe can come to exist by itself.

    If you want me to explain it more clearly, I will do so after dinner.

    Ok, let's try my gravity theory on the beginning of the universe.

    F ~ E, F = /~ E, F /~ 4pi*r^2, F ~ 4*pi*r(m)^2/m

    E = 0, r = 0

    F(1) = 1, F(2) = -1

    F(1) = 1, F(2) = -1 --> To be continued, after renting movie. It leads to energy and vector in opposit direction, not yet determined. Just like positive and negative particles.

    Sorry, these equations would take to long time to solve
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  9. #8  
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    Hello:

    There are many things wrong with these speculations.

    The first, and easiest things to correct is the spelling. I always take things I will be posting, even this note, out into a program that knows how to spell things. If you don't put in the effort, you cannot expect anyone else to put in effort. Misspellings say to me, your effort was small. Note: that is independent of the actual time invested, it just is what is communicated. A vector is not spelled with a K, average starts with an a.

    Newton's law of gravity and Coulomb's law of electrostatics are fine for simple classical laws. Both laws do not get along with special relativity. The way to be consistent with special relativity is dramatically different. In EM, one uses the Lorentz 4-force expression:
    F^u= gamma e (V/c . E}, E + V X B/c)
    where gamma = 1/(1 - V^2/c^2)^(1/2)
    There was no mention of the Lorentz 4-force, and I suspect the author is unfamiliar with it. This is the law that must be confronted, not Coulomb's law.

    Several very big names in physics has shown that the way to make Newton's law consistent with special relativity will in the end derive Einstein's equations for gravity, a rank 2 tensor expression. Since the author again makes no references to general relativity, I suspect he does not know the relevant details, a serious shortcoming.

    Suggesting that surface ~ mass runs into a basic problem: the units are wrong.

    I am sorry to report I see no value in this work.

    doug
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  10. #9  
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    I will update soon, I also saw some errors. But not today. Mañana, mañana.

    F(G) = ct*(M*m*4pi*r(p)^2/m(p)/(4pi*r^2))/t = c*r(p)^2*M*m/(m(p)r^2)

    G = c*r(p)^2/m(p)

    Given that ct*4pi*r(p)^2/m(p) is the drain, just like in a tunnel with four dimensions. The flow would cause an attractive force.

    How to verify:

    c*r(p)^2 = |E

    distance = r(p) gives |E = c*r(p)^2/r(p)^2 = c

    Since this force is based on flow, The force will be almost constant.


    I don't think this radius should be confused with the schwarzschild radius, since if:

    2*|E*r^2/c^2 = r(s)

    if r(s) = r, then |E = c^2/(2r)

    Clearly they are speaking of that any object within that radius that has mass, will be traped within. Not photons.
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