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Thread: A proof for the Stationary Earth, Part 2

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    Troll Magnet Sparko's Avatar
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    Quote Originally Posted by oxmixmudd View Post
    I thought that was kind of obvious by the conclusions he was drawing from them


    Jim
    i generally just skip john's posts like that because I already know he doesn't have a clue what he is talking about, but I assumed he at least knew what the formulas were supposed to represent.

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    42nd Mojave Year DesertBerean's Avatar
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    Even a math dunce like me know you must define the terms you're using so others can evaluate the accuracy of your calculations.


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    Quote Originally Posted by Leonhard View Post
    Quote Originally Posted by JohnMartin View Post
    The value of v does not change at x,y,z,t and x',y'z',t', so v is invariant with respect to both x,y,z,t and x',y'z',t'.
    Again, what are (x,y,z,t) and (x',y',z',t') you leave this completely unexplained. As such basically everything you say is nonsense. Do you mean two different points, at two different times in a euclidian space? Or do you mean the same point in two different inertial reference frames? I assume you mean the latter, because if you mean the former, your argument is incoherent.
    The later. The same point in two different inertial reference frames.

    v = v at both x,y,z,t and x',y'z',t', so v is invariant apparent from any transform.

    v = v is always true, no matter what. Any quantity is always equal to itself. Again, this is slobby notation, as I assume you mean v = v', namely the velocity at the one point, and the velocity at the other. .
    No. Im talking about v =v in both x,y,z,t and x',y'z',t'.

    See this makes me think that you're merely talking about two different spatial points, each with an associated velocity which just happens to be the same. Who cares about that? If I'm running along at 2 meters per second, and a bicyclist somewhere else, is going in the same direction at the same speed... then so what? That's not what an invariant means. That's an equality... the two velocities would be equal to each other.

    Something is invariant, if during a kind of transformation it is unchanged.
    v at x,y,z,t is equal to v at x',y'z',t'. v at both x,y,z,t and x',y'z',t' is not subject to the transform as v is given at both x,y,z,t and x',y'z',t'. Imagine an object at x,y,z,t moving at 100m/s. The same object is then at x',y'z',t' and moves at 100m/s. Both velocities are equal and both are known from the local clocks, or local gauges associated with the local clocks. v at x',y'z',t' is then used to calculate x and t.

    v is a fundamental variable of gamma and is not subject to a transform. If "invariance is something you apply to transformations", as you say, then c is invariant and so is v.

    c is always invariant under a lorentz transformation (going from one inertial frame to another). If v is invariant, then v = c.
    No, v=c is false. An object in SR always has a v<c. Furthermore, v is fundamental to gamma, so v is not transformed. There is no transform of the gamma transform.


    You said this - If something is the same after a lorentz transformation, then its an invarient - this is false.

    Nope this is true.
    Maybe you are using the notion of invariant differently to me. Can you explain further what you mean, with words and an example from SR theory?

    I deny your answer because the transform applies gamma to a variable to transform a value from one frame to another frame. In doing so, the value of the variable changes. This is why SR theory uses equations to calculate x and t.

    A major problem with SR theory is the manner in which v is used within the transform equations. For example in the equations

    Eq 1) x' = (x-vt)/(1-v2/c2)1/2

    Eq 2) t'=(t-vx/c2)/(1-v2/c2)1/2

    Let us call the v in the numerator as vn and v in the denominator as vd. We see in Eq 1), vn in the (x-vt) numerator is used in the manner as a variable multiplied with t, to form a value for a distance, vt. vt, is subtracted from x, to attain a distance from x to x. Then vd in the denominator, (1-v2/c2)1/2 is used to modify vn. vd is used along with c in the denominator to modify vn. As c in the denominator is invariant, then so too, vd must also be invariant. Why? Because c is assumed to be invariant in all frames and is used as the benchmark value to modify all other values. Such use of c, must then be consistently applied to the use of vd. vd must also be assumed as an invariant variable between x,y,z,t, and x,yz,t.

    If vd is applied consistently with c in the denominator, then vd is an invariant velocity, modifying vn. vn is then a variant velocity at x,yz,t in the numerator, being modified by an invariant velocity vd. The inconsistent application of velocity in the numerator and denominator means the Eq 1) is used to determine x as a value which is meaningless.

    To show this further, we can use the simple comparison of two equations -

    Eq 3) x = x-vinvarianttinvariant

    Eq 4) x = (x-vvarianttvariant) . gamma

    Eq 3) is the non controversial formula for x, whereby v and t are standard, invariant variables. v is the velocity at x, which does not change by shrinking to another variant velocity. Likewise, t does not dilate and become t'. What then happens to v when multiplied by another variable in gamma? v changes from vinvariant to vvariant in Eq 4). As SR assumes such equations can be used, v must be assumed to be both invariant in the numerator of Eq 1) and 2), and variant in the denominator. There is simply no way out of this problem.

    The manner in which x is calculated is analogous to saying I went to the shop 1km away, in 10 minutes, by travelling at vd at x, and vn /(1-vd2/c2)1/2 at x. Such a statement is jibberish.

    A similar critique can be made of Eq 2) to calculate t based upon an invariant vn and a variant vd.

    Furthermore, if x and t are calculated by Eq 1) and 2), why then not also calculate v as x-x/t-t and then have two values of velocity at x,y,z,t? If v determines x and t, why not have x and t determine v? Such a calculation would show SR to be the illogical double speak that it is. To have v and v at x,y,z,t would mean a body was moving at two velocities, v and v.

    Also, if v and t are known directly at x,y,z,t and v is known directly at x,y,z,t, why then cannot t' also be known directly at x,y,z,t? And then x' deduced from x'=v/t'?

    You should have said this - If something is the same after a lorentz transformation, then there is a contradiction. For the same would be the same both before and after the transform, but the transform is a variable other than 1. Hence, the same cannot be the same both before and after a transform. The implied contradiction within your statement is another indication that SR theory is false.

    The identity transformation, basically going from the inertial reference frame... to the same reference frame, is one special case of the lorentz transformation. And yes, c is invariant under this transformation. So are all quantities. Its impossible for any number to not be invariant under an identity transformation.
    The identity transformation is a meaningless term. A transform changes the value of a variable within an equation. A transformed variable means the variable can change value, an hence is not an invariant variable.

    However c, and only c, is invariant under a lorentz transformation of any kind, and any rotation. Its tedious to prove it for the general case from the lorentz transformation alone, however I don't mind showing the proof for one of the simpler cases.
    SR wants us to believe only c is invariant, but the SR equations assume vd is invariant when calculating x and t.

    Lets take two inertial frames of reference, moving in the same direction. One at v, the other at v'. In that case we can calculate what any velocity would be like after the reference frame shift (lorentz transformation).

    In this case its given by the simple velocity addition formula from the Special Theory of Relativity.

    In the special case that u is equal to c, then we have for any difference of velocities (v - v').

    So if u = c, then u' = c after a lorentz transformation.

    However there is no u, less than c, which will be the same after a lorentz transformation. Only c is invariant under all lorentz transformations.

    Same problems as posed above. v is both variant in the numerator and invariant in the denominator.

    SR is sophistry.

    JM
    Last edited by JohnMartin; 05-21-2016 at 06:14 PM.

  4. #634
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    Quote Originally Posted by Sparko View Post
    i generally just skip john's posts like that because I already know he doesn't have a clue what he is talking about, but I assumed he at least knew what the formulas were supposed to represent.
    I generally skip your posts, because I know you are an inventor. Example, see your recent ramble about the SR formulas.

    JM

  5. #635
    Troll Magnet Sparko's Avatar
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    Quote Originally Posted by JohnMartin View Post
    I generally skip your posts, because I know you are an inventor. Example, see your recent ramble about the SR formulas.

    JM
    which one is that?

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