Education and Crank Claims: Special Relativity

Discussion in 'Physics & Math' started by rpenner, Oct 5, 2011.

  1. Motor Daddy Valued Senior Member

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    5,425
    No the student is not correct.

    Coordinate (1,1,0) is the distance of the square root of 2 away from coordinate (0,0,0), because a^2+b^2=c^2.

    1+1=2
    sqrt(2)=1.41

    Did you miss the coordinates of the receivers in my example??

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  3. Motor Daddy Valued Senior Member

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    No it certainly does NOT take the same amount of time to reach the z receiver as the x receiver. The times were measured and the results are in. Are you saying the receivers are faulty or somehow mistaking? The only way to know the time is to measure it from source to receiver, and that is what was measured in the cube, like it or not!
     
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  5. billvon Valued Senior Member

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    BUT 1+1=2! Do you dispute that? How can you claim that is not correct?
     
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  7. billvon Valued Senior Member

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    You actually ran the experiment? How did you get your detector up to .5C?
     
  8. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    19,252
    Woo-powered engines.
    Applied through a crank.
     
  9. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Except in the eyes of relativity it is only a cube in certain frames. In frames where it is moving it is not a cube, it will experience length contraction.

    Saying 'it is a cube' is an assertion, as it's your view it is always a cube but that isn't what relativity says. As such relativity has a different description to the scenario you claim and you're trying to claim there's something wrong with relativity's take on things. However, you cannot disprove relativity if you don't correctly represent it. For instance, if I were to say you're wrong because you think 2+2=5 then you'd say "But I never said that". You're basically doing that with relativity, you are not actually addressing what it says. As such, your calculations (once again done with numbers, do you have some aversion to algebra?) are pointless because any conclusion you reach from them will not have any impact on relativity's validity or relevance.

    No, I am not saying that. If you have to ask what I'm saying then you're just illustrating you don't know what relativity says since I'm telling you what relativity says. You shouldn't need any clarification, you shouldn't need any input from me at all, if you actually knew what relativity says and you were correctly addressing it.

    In your example you don't have any length contraction contributions, nothing related to Lorentz transforms. As such you cannot show relativity has it wrong because you haven't worked out what relativity says. I provided you with the algebra, it was all there. I illustrated the description by relativity was consistent, thus refuting your claim it's description was not. You say relativity can't do something, I demonstrated it can. Everything you've done since then has simply been assertions, either saying relativity is wrong because it says something it doesn't or because you just know how the world works and it isn't how relativity works.

    Do you understand the algebra I provided? Every time you attempt mathematics you always end up posting lots of decimal places, you don't seem to ever work with general algebraic expressions, beyond putting in values to expressions you've been provided with. Are you able to actually do basic algebra?
     
  10. Motor Daddy Valued Senior Member

    Messages:
    5,425
    I never claimed 1+1 doesn't equal 2. If the teacher was worth a damn he would explain to the student that the distance from corner to corner of a square is longer than 1 side but shorter than adding 2 sides together.

    The teacher should explain to the student that if you draw a 1 meter square, and put a circle inside the square so as to touch sides of the square, that the radius of the circle is .5 meters, the diameter of the circle is 1 meter, and the circumference of the circle is 3.14 meters. The circle touches the sides of the square, but doesn't reach the corners of the square. So the distance from the center of the circle to the center of the sides is .5 meters, but the circle falls short of touching the corners of the square. The student would understand when shown a picture.

    You seem to defy that logic, as you've been shown a pic, told how it works, and yet you still act like the student, out in right field, clueless.
     
  11. billvon Valued Senior Member

    Messages:
    21,635
    An excellent explanation. And if your teacher was worth a damn he would have explained to you that neither time nor simultaneity is preserved across frames moving relatively to each other. You are like the student yelling "BUT 1+1=2! 1+1=2!" You don't yet have the math skills to understand why that's not always true in vector math (or, in this case, special relativity.)
     
  12. DonQuixote Registered Senior Member

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    66
    Sorry about all the quoting here, but I thought it important to preserve the context.
    I am of course intersted in how light speed meassurements are actually carried out, but in this theoretical discussion, I don't think it matters much. The point I made is that even conceptually, a speed of something cannot be meassured at a point. I should think everyone would agree to that. I might be wrong, though.
     
  13. Motor Daddy Valued Senior Member

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    5,425
    You don't have the intestinal fortitude to admit when you are wrong.

    Light takes a different amount of time in the cube to reach the different receivers, period. The speed of light is constant and the distance between the source and the receivers is the same. When you understand that you will be on your way to understanding how distance and time are properly measured using light. Until then, all I can say is, ignorance is bliss!
     
  14. OnlyMe Valued Senior Member

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    3,914
    First...

    I think the book your were referring to, (Einstein's Little Book) is, "Relativity: The Special and General Theory" 1920. An English copy is available online in a PDF format at the following link.

     
  15. billvon Valued Senior Member

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    21,635
    You keep yelling "1+1=2!" And again, yes, until you have the math to grasp what's really going on, that's all you will be able to conclude.
     
  16. OnlyMe Valued Senior Member

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    3,914
    I am not quite sure what you mean and are asking here. It could go several ways. If you can restate it, perhaps I can get a better understanding of what you mean.
     
  17. AlexG Like nailing Jello to a tree Valued Senior Member

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    4,304
    States Motor Daddy. A bare assertion, backed by nothing than his own belief. As he's repeated countless times before on many other threads.
     
  18. DonQuixote Registered Senior Member

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    66
    Yeah, that's it. Thanks for the link.

    Although it's a popularised explanation, the SR part should be allowed as a background for discussion in this thread, I hope.

    Thanks a lot.
     
  19. OnlyMe Valued Senior Member

    Messages:
    3,914
    You are correct you cannot measure the velocity of anything at a single point.

    Measuring the speed of light has gone through some significant changes as our timing capabilities have improved. I have read through some of the methods but would not present myself as an expert on any of the methodology.
     
  20. DonQuixote Registered Senior Member

    Messages:
    66
    Yes. As I said, I will work on making the question precise. Just ignore this for the time beeing. Sorry.
     
  21. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    This false claim of yours has been discussed at length in other threads. If you insist on repeating it ad nauseam in unrelated threads I may have to start banning you for spamming.

    I explained to you in the other threads that the light travel time is different in different frames of reference. Obviously you have either forgotten the previous discussions or you didn't understand them or you are insisting that your fantasy version of reality is correct despite having zero experimental or observational evidence for that and a stack of experimental and observational evidence against it.
     
  22. chinglu Valued Senior Member

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    1,637
    I am incrediblized by your intellect.

    Can you explain where any frame will claim the light sphere will appear from the coordinate (3ls,15ls,0) along the line y = 15 after further propagation of the light wave?
     
  23. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    Education: Assuming that two observers agree on a particular point in space and time as having the same Cartesian coordinates and that the directions of their axes are parallel, what can the Lorentz transform tell us about their absolute (Poincaré) transformation between coordinate systems?

    One of the useful properties of the Lorentz transformation is that it is linear. This makes it easy to compute the Lorentz transformation of a zero vector or the sum or difference of vectors.

    Let us call the event with the same coordinates in both systems O, \(t_O = t'_O, \, x_O = x'_O, \, \cdots\).
    The for any point, E, in space and time, we use the difference in coordinates between that point and O. \(\Delta t = t_E - t_O, \, \Delta x = x_E - x_O, \, \cdots \; \Delta t' = t'_E - t'_O, \, \cdots\). Then we have

    \(\begin{eqnarray} \begin{pmatrix} x'_E \\ y'_E \\ z'_E \\ t'_E \end{pmatrix} & = & \begin{pmatrix} x'_O \\ y'_O \\ z'_O \\ t'_O \end{pmatrix} + \begin{pmatrix} \Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} \\ & = & \begin{pmatrix} x'_O \\ y'_O \\ z'_O \\ t'_O \end{pmatrix} + \begin{pmatrix} \cosh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & - c \, \sinh \, \tanh^{-1} \, \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \frac{1}{c} \, \sinh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & \cosh \, \tanh^{-1} \, \frac{v}{c} \end{pmatrix} \left( \begin{pmatrix} x_E \\ y_E \\ z_E \\ t_E \end{pmatrix} - \begin{pmatrix} x_O \\ y_O \\ z_O \\ t_O \end{pmatrix} \right) \\ & = & \begin{pmatrix} \cosh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & - c \, \sinh \, \tanh^{-1} \, \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \frac{1}{c} \, \sinh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & \cosh \, \tanh^{-1} \, \frac{v}{c} \end{pmatrix} \begin{pmatrix} x_E \\ y_E \\ z_E \\ t_E \end{pmatrix} + \begin{pmatrix} x'_O \\ y'_O \\ z'_O \\ t'_O \end{pmatrix} - \begin{pmatrix} \cosh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & - c \, \sinh \, \tanh^{-1} \, \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \frac{1}{c} \, \sinh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & \cosh \, \tanh^{-1} \, \frac{v}{c} \end{pmatrix} \begin{pmatrix} x_O \\ y_O \\ z_O \\ t_O \end{pmatrix} \end{eqnarray} \)

    Naturally this is simplest when \(t_O = t'_O = 0, \, x_O = x'_O = 0, \, y_O = y'_O = 0, \, z_O = z'_O = 0\), a situation which is sometimes called the standard configuration.

    For standard configuration, the descrition of inertial motion gives us:
    \(\begin{eqnarray} \begin{pmatrix}x' \\ y' \\ z' \\ t' \end{pmatrix} & = & \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} u_x (t-t_0) + x_0 \\ u_y (t-t_0) + y_0 \\ u_z (t-t_0) + z_0 \\ (t-t_0) + t_0 \end{pmatrix} = \begin{pmatrix} \frac{ (u_x - v)(t - t_0) }{\sqrt{1 - \frac{v^2}{c^2}}} \\ u_y (t-t_0) \\ u_z (t-t_0) \\ \frac{1 - \frac{u_x v}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} (t-t_0) \end{pmatrix} + \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} x_0 \\ y_0 \\ z_0 \\ t_0 \end{pmatrix} = \begin{pmatrix} \frac{ (u_x - v)(t - t_0)}{\sqrt{1 - \frac{v^2}{c^2}}} + \frac{x_0 - v t_0}{\sqrt{1 - \frac{v^2}{c^2}}} \\ u_y (t-t_0) + y_0 \\ u_z (t-t_0) + z_0 \\ \frac{1 - \frac{u_x v}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} (t-t_0) + \frac{t_0 - \frac{v x_0}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \\ & = & \end{eqnarray} \begin{pmatrix} u'_x (t'-t'_0) + x'_0 \\ u'_y (t'-t'_0) + y'_0 \\ u'_z (t'-t'_0) + z'_0 \\ (t'-t'_0) + t'_0 \end{pmatrix} \)
    So inertial motion is preserved by the Lorentz transform.

    Education: What does the Lorentz transform have to say about about the length of moving objects?

    Let's measure the distance between world-lines in the x-direction. Clearly the length is only well-defined if the left and the right edges have the same state of motion. So we have the left world line: \( L = \begin{pmatrix} u_x (t-t_{\textrm{left}}) + x_{\textrm{left}} \\ u_y (t-t_{\textrm{left}}) + y_{\textrm{left}} \\ u_z (t-t_{\textrm{left}}) + z_{\textrm{left}} \\ (t-t_{\textrm{left}}) + t_{\textrm{left}} \end{pmatrix} \) and the right world-line: \( R = \begin{pmatrix} u_x (t-t_{\textrm{right}}) + x_{\textrm{right}} \\ u_y (t-t_{\textrm{right}}) + y_{\textrm{right}} \\ u_z (t-t_{\textrm{right}}) + z_{\textrm{right}} \\ (t-t_{\textrm{right}}) + t_{\textrm{right}} \end{pmatrix} \) so the difference in x-coordinate at any time is \(\left( R - L \right)_{x} = u_x (t-t_{\textrm{right}}) + x_{\textrm{right}} - \left( u_x (t-t_{\textrm{left}}) + x_{\textrm{left}} \right) = x_{\textrm{right}} - x_{\textrm{left}} - u_x \left( t_{\textrm{right}} - t_{\textrm{left}} \right) \).

    For another observer, we need \(u'_x = \frac{u_x - v}{1 - \frac{u_x v}{c^2}}, \; \begin{pmatrix} x'_{\textrm{left}} \\ t'_{\textrm{left}} \end{pmatrix} = \begin{pmatrix} \gamma & - \gamma v \\ - \gamma c^{-2} v & \gamma \end{pmatrix} \begin{pmatrix} x_{\textrm{left}} \\ t_{\textrm{left}}\end{pmatrix} , \; \begin{pmatrix} x'_{\textrm{right}} \\ t'_{\textrm{right}} \end{pmatrix} = \begin{pmatrix} \gamma & - \gamma v \\ - \gamma c^{-2} v & \gamma \end{pmatrix} \begin{pmatrix} x_{\textrm{right}} \\ t_{\textrm{right}}\end{pmatrix} \) to compute
    \(\begin{eqnarray} \left( R - L \right)_{x'} & = & x'_{\textrm{right}} - x'_{\textrm{left}} - u'_x \left( t'_{\textrm{right}} - t'_{\textrm{left}} \right) \\ & = & \left( \gamma x_{\textrm{right}} - \gamma v t_{\textrm{right}} \right) - \left( \gamma x_{\textrm{left}} - \gamma v t_{\textrm{left}} \right) - \frac{u_x - v}{1 - \frac{u_x v}{c^2}} \left( \left( \gamma t_{\textrm{right}} - \gamma \frac{v}{c^2} x_{\textrm{right}} \right) - \left( \gamma t_{\textrm{left}} - \gamma \frac{v}{c^2} x_{\textrm{left}} \right) \right) \\ & = & \left( \gamma + \gamma \frac{v}{c^2} \frac{u_x - v}{1 - \frac{u_x v}{c^2}} \right) \left( x_{\textrm{right}} - x_{\textrm{left}} \right) - \left( \gamma v + \gamma \frac{u_x - v}{1 - \frac{u_x v}{c^2}} \right) \left( t_{\textrm{right}} - t_{\textrm{left}} \right) \\ &= & \gamma \frac{c^2 - v^2}{c^2 - u_x v} \left( R - L \right)_{x} \\ & = & \frac{1}{\left( 1 - \frac{u_x v}{c^2} \right) \gamma} \left( R - L \right)_{x} \end{eqnarray}\).

    When \(u_x = 0\), \(u'_x = -v\) and this reduces to \(\left( R - L \right)_{x'} = \sqrt{1 - \frac{v^2}{c^2}} \left( R - L \right)_{x}\) and so the moving object is length-contracted.
    When \(u_x = v\), \(u'_x = 0\) and this reduces to \(\left( R - L \right)_{x'} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( R - L \right)_{x}\) and so the more-moving object (i.e. x as opposed to x') is length-contracted.
     

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