New article shows a fatal math error in SR

Discussion in 'Pseudoscience' started by chinglu, Aug 9, 2013.

  1. Tach Banned Banned

    How do you get

    \(( - \Delta \xi_0 + \frac{v}{c} \left( ( \eta_0 + \Delta) \sqrt{\xi_0^2 + \eta_0^2} - \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2} \right) < 0\)

    it isn't obvious.
  2. Google AdSense Guest Advertisement

    to hide all adverts.
  3. CptBork Valued Senior Member

    I've only looked at Mr. Banks' figures for a few mins, but is he basically making some kind of rookie mistake? Like thinking that if the ray traced by the path of the photon in the unprimed frame hits the left side of the mirror, that means the photon itself struck on the left?
  4. Google AdSense Guest Advertisement

    to hide all adverts.
  5. chinglu Valued Senior Member

    I'll answer this with my post to rpenner.
  6. Google AdSense Guest Advertisement

    to hide all adverts.
  7. chinglu Valued Senior Member

    Show the math.
  8. chinglu Valued Senior Member

    1) Let's start with your misunderstanding of Einstein's thought experiment which motivated LT construction. You continue to claim somehow that Einstein meant x' applies to the "stationary" K system.

    Einstein wrote,

    "From the origin of system k let a ray be emitted at the time \(\tau_0\) along the X-axis to x', and at the time be reflected thence to the origin of the co-ordinates, arriving there at the time \(\tau_2\); we then must have \(\frac{1}{2}(\tau_0 + \tau_2) = \tau_1\)".

    This equation only holds if x' is the location of the mirror and it is in the "stationary" lower case k system, which is moving relative to K.

    Therefore, as the article noted, the mirror is located at (x',0) in the primed system.

    So, you are wrong here as I have explained over and over.

    2) Let's consider you silly rotating mirror hypothesis. That means when we are at rest on the earth, a moving car will rotate. That is complete stupidity. Further, if you are correct, then from system K for Einstein, the mirror at (x',0) in the primed frame must rotate so that the K system believes no light was reflected hence SR is false. Very good.

    3) Finally, we eliminate your crank theory that the mirror runs into the light sphere so it hits the reflective side. But, you claimed the mirror rotates, which is a contradiction since it you are correct, the mirror would have rotated such that the non-reflective side runs into the light sphere. So, you are worse than a crank.

    But, since I have already refuted your mirror rotating theory, I will address and refute your mirror run into the light sphere theory.

    We will use the example in the article in which all units are c units.

    \(y=\frac{5}{3} \), \(v=\frac{3}{5}\), \(x'=-1\), \(\gamma=\frac{5}{4}\)

    First, it is proven that the light sphere strikes \((0,y) = (0,\frac{5}{3})\) before it strikes the mirror.

    Since the mirror is located at \((x>0,\frac{5}{3})\), then that proves the light sphere is behind the mirror in the view of the unprimed frame before it is met by the light.

    Well, since the light sphere in the unprimed frame strikes any coordinate at time \(t=\sqrt{x^2+y^2}\), then \(t_0=\sqrt{y^2}<t_1=\sqrt{x^2+y^2}\) with \(x>0\).

    Now, we must show when the light sphere hit the point \((0,y)\) that the mirror has crossed the y-axis of the unprimed framed.

    Since \(x'=-1\), by length contaction, \(x=-1/\gamma\), so \(x=-\frac{4}{5}\).

    Next, when the light sphere strikes the point \((0,\frac{5}{3})\), \(t=\frac{5}{3}\).

    Now, that means the mirror travels \(vt+x=\frac{3}{5}*\frac{5}{3}-\frac{4}{5}=\frac{1}{5}\).

    Hence, the mirror is located on the positive side of the x-axis when the light sphere strikes the point \((0,\frac{5}{3})\). Therefore, the light sphere has not yet struck the mirror and so the light sphere is behind the mirror..

    Next, we take the partial derivative of the motion of the light sphere along the line \(y=\frac{5}{3}\).

    We have \(x=\sqrt{t^2-y^2}\). Remember we are using c units.

    \(\frac{\partial x}{\partial t}=\frac{t}{\sqrt{t^2-y^2}}=\frac{t}{\sqrt{t^2-\frac{25}{9}\)

    Then, we calculate when \(x=\frac{\sqrt{34}-5}{4}\) as provided by LT in the example in the paper.

    So, \(t=\frac{5\sqrt{34}-5}{12}-\frac{3}{4}\)

    Therefore, when the light strikes the mirror, \(\frac{\partial x}{\partial t}\approx8.085\)

    Consequently, along the line \(y=\frac{5}{3}\), the light sphere starts out behind the mirror. Then it moves faster than the speed of the mirror such that the instantaneous velocity of the light sphere intersecting the line \(y=\frac{5}{3}\) is greater than the speed of the mirror and is in fact about 8.085c. So, the light sphere strikes the back of the mirror and then moves very fast past it.

    As such, the mirror does not run into the light sphere but the light sphere runs into the back of the mirror.

    So, you have been mathematically refuted on all fronts.
  9. CptBork Valued Senior Member

    Perhaps later, I will.
  10. Fednis48 Registered Senior Member

    The mirror moving faster than the light doesn't contradict the mirror rotating. If we looked at the exact numerics of the reflection, I'm sure both effects would work in tandem to make everything match up. But just qualitatively, the mirror moving faster is enough by itself to show that the light doesn't hit the wrong side of the mirror.

    If we want to know whether the mirror will catch up with the photons or the other way around, we care about the horizontal velocity of the photons themselves, not a constant-y point on the expanding sphere. Therefore, we need a full derivative, not a time derivative. Fortunately, all the photons travel in a straight line, so it's easy to find their horizontal velocity with nothing but trigonometry:

    \(c_x=c \cos(\theta)=\frac{cx}{\sqrt{x^2+y^2}}\)

    Plugging in the numbers from the paper, we get \(c_x\approx 0.124c\), which is less than the mirror velocity of \(\frac{3}{5}c\).
  11. CptBork Valued Senior Member

    I have a hard time believing an allegedly well-reputed editor would approve a paper that ignored basic things like this. The premise of Mr. Banks' argument is based on a falsely-interpreted diagram, nothing more as far as I can tell.
  12. Tach Banned Banned

    It is a junk journal, two people in the back of a room, charging 100$ for printing virtually any garbage. Such "journals" have sprouted in the recent years, an easy way of making money by pushing horse manure.
  13. rpenner Fully Wired Valued Senior Member

    Because this expression is linear in v it is continuous and has exactly one change of sign, which happens at the value of v where the value of this expression goes to zero.
    I assert that since all the variables are greater than zero, this happens at a speed greater than that of light and thus for all physical values of the speed of the detector, this expression is always negative.

    Specifically, I claim this sign-changing value of v is
    \(\tilde{v} = c + c \frac{ \Delta ( \sqrt{\xi_0^2 + \eta_0^2} - \xi_0 ) + \eta_0 \left( ( \sqrt{\xi_0^2 + \eta_0^2} - \xi_0 ) + ( \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2} - \xi_0 ) \right)}{ \xi_0 ( \Delta + 2 \eta_0 ) } = c \frac{ ( \eta_0 + \Delta ) \sqrt{\xi_0^2 + \eta_0^2} + \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2}}{\xi_0 \Delta + 2 \xi_0 \eta_0} \)
    where I have taken the trouble to make clear that under the assumptions \(\Delta, \xi_0, \eta_0 > 0\) this is larger than c.

    Substituting this value into the expression we see:
    \(- \Delta \xi_0 + \frac{\tilde{v}}{c} \left( ( \eta_0 + \Delta) \sqrt{\xi_0^2 + \eta_0^2} - \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2} \right) = - \Delta \xi_0 + \frac{ \left( ( \eta_0 + \Delta) \sqrt{\xi_0^2 + \eta_0^2} + \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2} \right) \left( ( \eta_0 + \Delta) \sqrt{\xi_0^2 + \eta_0^2} - \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2} \right)}{\xi_0 \Delta + 2 \xi_0 \eta_0} = - \Delta \xi_0 + \frac{ ( \eta_0 + \Delta)^2 (\xi_0^2 + \eta_0^2) - \eta_0^2 (\xi_0^2 + (\eta_0 + \Delta)^2) }{\xi_0 \Delta + 2 \xi_0 \eta_0} = - \Delta \xi_0 + \frac{ ( \eta_0^2 + 2 \Delta \eta_0 + \Delta^2 ) (\xi_0^2 + \eta_0^2) - \eta_0^2 (\xi_0^2 + \eta_0^2 + 2 \Delta \eta_0 + \Delta^2 ) }{\xi_0 \Delta + 2 \xi_0 \eta_0} = - \Delta \xi_0 + \frac{ \xi_0^2 \eta_0^2 + 2 \Delta \xi_0^2 \eta_0 + \Delta^2 \xi_0^2 + \eta_0^4 + 2 \Delta \eta_0^3 + \Delta^2 \eta_0^2 - \xi_0^2 \eta_0^2 - \eta_0^4 - 2 \Delta \eta_0^3 - \Delta^2 \eta_0^2 }{\xi_0 \Delta + 2 \xi_0 \eta_0} = - \Delta \xi_0 + \frac{ 2 \Delta \xi_0^2 \eta_0 + \Delta^2 \xi_0^2 }{\xi_0 \Delta + 2 \xi_0 \eta_0} = - \Delta \xi_0 + \frac{ ( \Delta \xi_0 ) (\xi_0 \Delta + 2 \xi_0 \eta_0)}{\xi_0 \Delta + 2 \xi_0 \eta_0} = { \Large 0 }\)
    which completes the demonstration.
  14. Tach Banned Banned


    OK, the sign change in the expression happens for

    \(v=c \frac{ \xi_0 \Delta}{(\eta_0+\Delta)\sqrt{\xi_0^2 + \eta_0^2} - \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2}}\)
    Multiplied by the conjugate gives your expression.

    So, you could have done this easier by observing directly that \(\frac{ \xi_0 \Delta}{(\eta_0+\Delta)\sqrt{\xi_0^2 + \eta_0^2} - \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2}}>1\) but I understand your approach. So, the expression does not change sign for any \(v <c\)
    Last edited: Aug 11, 2013
  15. rpenner Fully Wired Valued Senior Member

    It is my assertion, based on ALL of the text in sections 1, 2 and 3 of the translated 1905 paper, that lowercase Latin letters (x,y,z,t) refer to coordinates in the "stationary" K system and that the lowercase Greek letters (\(\xi, \eta, \zeta, \tau\)) refer to coordinates in the system k which has the property that things described as "at rest" in k have constant velocity of v in the X direction in system K. Further, Einstein will use both primes and subscripts to talk about different values of coordinates in the same system. Further, I assert that because you can only add or subtract values in the same coordinate system and get sensible results, Einstein never adds or subtracts x' with any coordinate in system k.

    Einstein introduces in section 3 "point at rest in the system k". What is that point's description in system k?
    It is : \(\xi = \xi(t) = \xi(0), \quad \eta = \eta(t) = \eta(0), \quad \zeta = \zeta(t) = \zeta(0)\)
    And what is that point's description in system K?
    It is : \(x = x(t) = x(0) + v t, \quad y = y(t) = y(0), \quad z = z(t) = z(0)\).
    So when Einstein introduces x', we have "x' = x - v t" with the obvious interpretation:
    \(x' = x - v t = x(t) - v t = x(0) + v t - v t = x(0)\)
    because v only has meaning to the point in system K, the system where the point is not at rest.
    So when Einstein writes: "If we place x'=x-vt, it is clear that a point at rest in the system k must have a system of values x', y, z, independent of time." he is saying ANY point at rest in system k will have a unique description in system K coordinates if take \((x', y, z) = \left( x(0), y(0), z(0) \right)\).

    Only after introducing this \(x' = x(0)\) concept does he begin to explore how we might learn the values of \(\xi, \eta, \zeta, \tau\) beginning with \(\tau\).

    Because you cut off the quote in mid-sentence, I have completed the quote for you.

    Here Einstein is talking about leaving one object (the origin of k) hitting another object (our point on the X-axis which is coincident in both systems) and returning to the first object. Both objects are at rest in k, but we only have descriptions of them in system K. Thus we have for the origin:
    \(x_O(t) = 0 + v t, \quad y_O(t) = 0, \quad \z_O(t) = 0\)
    and for the point:
    \(x_P(t) = x_P(0) + v t, \quad y_P(t) = 0, \quad \z_P(t) = 0\).
    Einstein then posits a linear functional relationship between between descriptions in K of points at rest in k and time in K with time in k
    \(\tau\left( x(t) - vt, y(t), z(t), t \right) = \tau\left( x(0), y(0), z(0), t\right) = \tau(x', y, z, t)\)
    because Einstein established that \(x' = x(0) = x(t) - v t\) was a time-independent description of the point in system K.
    \(\begin{array}{c|cccc|cccc|c} Event & x & y & z & t & x' & y & z & t & \tau \hline 0 & x_O(t) = v t & y_O(t) = 0 & z_O(t) = 0 & t & vt - vt = 0 & 0 & 0 & t & \tau(0,0,0,t) 1 & x_P(t + \Delta_1) = x_P(0) + v t + v \Delta_1 & y_P(t + \Delta_1) = 0 & z_P(t + \Delta_1) = 0 & t + \Delta_1 & x_P(0) + v t + v \Delta_1 - v ( t + v \Delta_1 ) = x_P(0) & 0 & 0 & t + \Delta_1 & \tau(x_P(0), 0,0,t + \Delta_1 ) 2 & x_O(t + \Delta_1 + \Delta_2 ) = v t + v \Delta_1 + v \Delta_2 & y_O(t + \Delta_1 + \Delta_2 ) = 0 & z_O(t + \Delta_1 + \Delta_2 ) = 0 & t + \Delta_1 + \Delta_2 & v t + v \Delta_1 + v \Delta_2 - v ( t + \Delta_1 + \Delta_2) = 0 & 0 & 0 & t + \Delta_1 + \Delta_2 & \tau(0,0,0,t + \Delta_1 + \Delta_2 ) \end{array}\)
    Because \(c \Delta_1 = x_P(t + \Delta_1) - x_O(t) = x_P(0) + v \Delta_1\) it follows that \(\Delta_1 = \frac{x_P(0)}{c - v}\) (which follows only from coordinates in system K and the speed of light in system K. Likewise \(\Delta_2 = \frac{x_P(0)}{c + v}\), and so we have:
    \(\tau_0 = \tau(0,0,0,t), \quad \tau_1 = \tau(x_P(0), 0, 0, t + \frac{x_P(0)}{c-v}), \quad \tau_2 = \tau(0, 0, 0, t + \frac{x_P(0)}{c-v} + \frac{x_P(0)}{c + v})\) which allows Einstein to finish his sentence as a coherent inference from assumption. So not only does Einstein's textual use of subscripts and primes imply that he is using the same coordinate system (K) as the unprimed Latin letters, but his math requires it.

    This was not the conclusion of Einstein's section 3.

    You assert over and over, but your arguments are based on misconceptions and fail to hit the target.

    It appears you still haven't grasped what I mean by "effective".
    Another way to show this is by using a parametrized circle instead of a point-like detector, and the math shows the light comes in at the same parametrized point on the circle in both frames, thus the light must hit on the same "side" as defined in a way that makes physical sense.

    I see that this Andrew Banks/chinglu thread has properly been moved to pseudoscience -- the category of those that only ape the form and not the substance of science, so I believe that constitutes a formal evaluation of those worthless ideas.
    Last edited: Aug 12, 2013
  16. chinglu Valued Senior Member

    You are leaving out the y component speed which is about \(c_y\approx 0.9923c\). When the point mirror moves 3/5 x units, the mirror moves in the y direction \(\approx 0.9923c\) units. So, in order for the photon to strike to point mirror on the reflective side, it is not simply the x-component speed that you need to consider.

    Therefore, by using x-component speed only, you leave out an important part of the problem. Once the y-component speed is considered, there is no point (x,5/3) where the light sphere is ahead of the mirror in order for the mirror to run into it.
  17. AlphaNumeric Fully ionized Registered Senior Member

    Chinglu, still peddling nonsense by Andrew Banks? I guess you still haven't bothered to learn relativity properly then. Funny how you're stuck whining on forums, can't find a reputable journal who'll publish all these supposed disproofs? Unsurprisingly...

    Until such time as you can demonstrate the soundness of your claims and not resort to repeating already retorted claims put your whining in the fringe section, not the main forum. You've shown you are not capable of discussions that forum is supposed to be for. That's why we have this fringe section. Next time you get a holiday... again!
  18. chinglu Valued Senior Member

    Why exactly is this fringe.
  19. origin Heading towards oblivion Valued Senior Member

    Your inabiltiy to understand why this is in the fringe section is not surprising in the least. Have you missed the responses to the article you cited? Apparently your head is buried too deeply in the sand to see them.
  20. brucep Valued Senior Member

    It means you're so consistently intellectually dishonest that you are only being allowed to comment in the space reserved for intellectually dishonest cranks. The moderator leaves an opening for you to reform yourself and escape 'the rings of Hades'. What's the liklihood of that coming to fruition?
  21. chinglu Valued Senior Member

    If I hold a GPS unit over an mmx experiment, the GSP unit claims sagnac exists and mmx does not.

    How does that work?
  22. chinglu Valued Senior Member

    If I hold a GPS unit over an mmx experiment, the GSP unit claims sagnac exists and mmx does not.

    How does that work?
  23. chinglu Valued Senior Member

    I was wondering since you claim superiority,

    If I hold a GPS unit over an mmx experiment, the GSP unit claims sagnac exists and mmx does not.

    How does that work? Both are in ECEF.

    Can you explain this?

Share This Page