On Einstein's explanation of the invariance of c

Discussion in 'Pseudoscience Archive' started by RJBeery, Dec 8, 2010.

  1. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    5,105
    45,220,694.36472 meters.
     
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  3. Neddy Bate Valued Senior Member

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    So the light traveled a little farther than 29,967,254.10 meters, didn't it?

    A^2 + B^2 = C^2

    29,967,254^2 + 45,220,694^2 = C^2

    Haven't we been over this before? :bugeye:
     
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  5. Neddy Bate Valued Senior Member

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    The distance between (.5999,.1508,.1951) and the origin (0000,0000,0000) is easily calculated:


    d = sqrt(.5999^2 + .1508^2 + .1951^2)
    d = sqrt(.3598 + .0227 + .0380)
    d = sqrt(.4205)
    d = .6484

    Which proves that you are claiming light can travels a distance of
    0.6484 lightseconds
    ... during an elapsed time of only 0.6000 seconds.
     
    Last edited: Jan 10, 2011
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  7. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    Wrong again.

    The cube is already traveling a constant velocity and you are going to measure that velocity from within.

    At exactly 12:00:00 the source at the center of the cube is located at (0,0,0) and the source emits a light sphere. At .6 seconds after 12:00:00 the light reaches the receiver at the center of the x wall. At .668 seconds after 12:00:00 the light reaches the receiver at the center of the y wall. At .741 seconds after 12:00:00 the light reaches the receiver at the center of the z wall. I've used the equation v=(ct-l)/t to find the velocity of each component as indicated:

    x-component = .600 seconds (.1666c)
    y-component = .668 seconds (.2514c)
    z-component = .741 seconds (.3252c)


    So at .741 seconds after 12:00:00 all the measurements are complete. I then know the absolute velocity of the center of the cube, which has been traveling a constant velocity since 12:00:00 which was located at (0,0,0).

    In .741 seconds the center of the cube traveled from (0,0,0) to (0.1234506,0.1862874,0.2409732).

    x-component = .741 seconds (.1666c) 0.1234506
    y-component = .741 seconds (.2514c) 0.1862874
    z-component = .741 seconds (.3252c) 0.2409732

    So the center of the cube traveled from (0,0,0) at 12:00:00 to (0.1234506,0.1862874,0.2409732) at .741 seconds after 12:00:00.

    The distance the center of the cube traveled is 98,526,931.187055949068200400463274 meters in .741 seconds. The absolute velocity of the center of the cube from (0,0,0) to (0.1234506,0.1862874,0.2409732) is 132,964,819.41573002573306396823652 m/s.

    Light traveled 222,146,211.378 meters in .741 seconds.

    So since I know the center of the cube's velocity at .741 seconds after 12:00:00 I can then know that at .6 seconds after 12:00:00 the center of the cube had traveled 79,778,891.649438015439838380941914 meters in .6 seconds.

    Since I know the velocities of each component, I know that in .6 seconds, each component traveled:

    x-component = .6 seconds (.1666c) 0.09996 (29,967,254.10168 meters)
    y-component = .6 seconds (.2514c) 0.15084 (45,220,694.36472 meters)
    z-component = .6 seconds (.3252c) 0.19512 (58,495,504.40496 meters)


    That is .6 seconds of travel time at the velocity the center of the cube was traveling.

    That is not saying that light reached the y and z walls in .6 seconds. At .741 seconds after 12:00:00 the light had reached all the walls at the indicated times. At .6 seconds after 12:00:00 the light had not yet reached the y and z walls, and you are trying to say that it had by using the coordinates of the center point of the cube at .6 seconds. You are making the mistake of calculating the light travel as if it had reached all the walls by .6 seconds after 12:00:00. I told you the coordinates of the center of the cube and the distance each had traveled by .6 seconds after 12:00:00 by already knowing the velocities of each component, and I didn't know that until the light had finally reached the z wall at .741 after 12:00:00. After that time I know the velocities of each component and I can then know the distance each traveled in ANY amount of elapsed time prior to .741 seconds after 12:00:00.

    Light traveled 179,875,474.8 meters in .6 seconds and the center of the cube traveled 79,778,891.649438015439838380941914 meters
     
    Last edited: Jan 10, 2011
  8. arfa brane call me arf Valued Senior Member

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    5,305
    Look, NO, yeah?
    I mean like yeah, nah.
     
  9. Neddy Bate Valued Senior Member

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    No, I don't care about the "y" or "z" faces of the cube at 0.6 seconds. All I'm saying is that the light had reached the center of the "x" face at exactly 0.6 seconds. I asked you where that point was located at 0.6 seconds, and you told me that point was located at (0.5999, 0.1508, 0.1951).

    The distance between (0.5999, 0.1508, 0.1951) and the origin (0.0000, 0.0000, 0.0000) is this far:

    d = sqrt(.5999^2 + .1508^2 + .1951^2)
    d = sqrt(.3598 + .0227 + .0380)
    d = sqrt(.4205)
    d = .6484

    Which proves that you are claiming light can travels a distance of
    0.6484 lightseconds
    ... during an elapsed time of only 0.6000 seconds.


    ----------


    If you had calculated the CORRECT velocity, the distance would have been 0.600 light seconds at the time 0.600 seconds. That is because light is supposed to travel at one lightsecond per second. Look, I can do it. Watch:

    These are the components of the CORRECT velocity:
    v(x) = 0.100c
    v(y) = 0.200c
    v(z) = 0.300c

    So, at time 0.600 seconds, the center of the cube is located at (0.060, 0.120, 0.180). Since the center of the "x" face was originally located at (0.500, 0.000, 0.000) we know that at time 0.600 seconds that point would be located at the point (0.560, 0.120, 0.180). Now calculate the distance between that point and the origin:

    d = sqrt(.560^2 + .120^2 + .180^2)
    d = sqrt(.3136 + .0144 + .0324)
    d = sqrt(.3604)
    d = .600

    See? Face it, man, your equation does not work for 3D.
     
  10. phyti Registered Senior Member

    Messages:
    256
    t0=.5 ls/c=.5 s
    vr=1-t0/tr
    frame speed components:
    vx = 1-.5/.6 = .167
    vy = 1-.5/.667 = .251
    vz = 1-.5/.741 = .325
    sum of squares of components:
    (vx)^2 + (vy)^2 + (vz)^2 = .196
    frame speed:
    sqrt(.196) = .443c

    your results:
    79,779,000m = .226 ls
    .226/.6 = .443c

    Your answer is correct (if you ignore time dilation), you just did extra work to get it. It's not however the absolute speed because you can synchronize the clocks without knowing your absolute speed. When you setup your cube/frame, you don't know your absolute speed, therefore you can't know your clocks have absolute synch. If you did know your abs. speed, there would be no point in measuring it!
    The only possibility of determining absolute velocity at present is referring to the cmb or fixed star background,based on the fact that events (light emissions) do not move.
    You're only fooling yourself MD.
     
  11. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    5,105
    Measured times to receivers:

    x time: .761972 seconds
    y time: .761972 seconds
    z time: .92 seconds


    Component velocities:

    v(x) = .2300c
    v(y) = .2300c
    v(z) = .4022c


    Coordinates of source at center of cube at .761972 seconds: (0.17525356,0.17525356,0.3064651384)

    x .761972(.2300c) 0.17525356
    y .761972(.2300c) 0.17525356
    z .761972(.4022c) 0.3064651384


    Coordinates of source at center of cube at .92 seconds: (.2116,.2116,.370)

    x .92(.2300c) .2116
    y .92(.2300c) .2116
    z .92(.4022c) .370


    Coordinates of source at center of cube at 1.0 seconds: (.23,.23,.4022)


    x 1.0(.2300c) .23
    y 1.0(.2300c) .23
    z 1.0(.4022c) .4022

    Distance center of cube traveled from start coordinates (0,0,0) to (.23,.23,.4022) in 1 second is 155,072,655.74 meters, so the absolute velocity of the center of the cube is 155,072,655.74 m/s.
     
    Last edited: Jan 12, 2011
  12. Neddy Bate Valued Senior Member

    Messages:
    1,403
    Time = 0.7619 seconds
    (0.1752, 0.1752, 0.3064)
    (0.6752, 0.1752, 0.3064)
    d = sqrt(0.6752^2 + 0.1752^2 + 0.3064^2)
    d = sqrt(0.4559 + 0.0307 + 0.0939)
    d = sqrt(0.5805)
    d = 0.7619 lightseconds

    Time = 0.92 seconds
    (0.2116, 0.2116, 0.370)
    (0.2116, 0.2116, 0.870)
    d = sqrt(0.2116^2 + 0.2116^2 + 0.870^2)
    d = sqrt(0.0448 + 0.0448 + 0.7569)
    d = sqrt(0.8465)
    d = 0.9200 lightseconds

    YES!! YOU DID IT!!
    What equation did you use to find the velocities?
     
  13. phyti Registered Senior Member

    Messages:
    256
    MD;

    v(x) = .2300c
    v(y) = .2300c
    v(z) = .4022c

    Why didn't you just square the speed components, add them together and take the square root of the sum?
     
  14. Neddy Bate Valued Senior Member

    Messages:
    1,403
    You can only do that if you know the three components in the first place. For this exercise, the only known variables are the size of the cube, and the travel times for the light signals. In the above case, the travel times were:
    t(x) = 0.7619 seconds
    t(y) = 0.7619 seconds
    t(z) = 0.9200 seconds

    From there, we are supposed to figure out the velocity. So MD was able to calculate these:
    v(x) = .2300c
    v(y) = .2300c
    v(z) = .4022c

    We don't have a a simple equation for this yet, so we have been doing it with interpolation, or whatever works. The idea is to come up with velocity components which do not violate the idea that the speed of light should be c for all the light signals, when viewed from MD's "absolute rest frame".
     
  15. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Neddy Bate, I need help with making an equation.

    I have a partial equation that finds the velocity of x, and the time to the receiver at x, if the y time is known, and the y and z component velocities are zero, which goes like this:


    v(x)=sqrt(t(y)^2-l(y)^2)/t(y)
    t(x)=l(x)/(c-v(x))




    Using the same cube with sides of length 1 light second, and a source at the center of the cube with receivers in the center of the x,y, and z walls, given a y and z time of .65 seconds, I know the x component velocity and the x component time to receiver.



    x time= 1.384930 seconds
    y time= .65 seconds
    z time= .65 seconds

    x component velocity = 0.638971 c
    y component velocity = 0 c
    z component velocity = 0 c

    The location of the y receiver at .65 seconds is (0.41533, .5, 0)

    x= .65(.63897c)=.41533
    y= .5


    (0.41533, 0.5, 0)
    d = sqrt(0.41533^2 + 0.5^2 )
    d = sqrt(0.1724990089 + 0.25)
    d = sqrt(0.4224990089)
    d = 0.65 light seconds



    Does that look right for a zero y and z component velocity?

    BTW...look what v=(ct-l)/t says the x component velocity is:

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    v(x)=(1.384930-.5)/1.384930

    v(x)=.63897 c
     
    Last edited: Jan 14, 2011
  16. phyti Registered Senior Member

    Messages:
    256
    In the absolute frame, light speed in the direction of motion will be <c.
    This is MD's argument from the beginning, Einstein substitutes constant c for closing speeds in the clock synch definition. If you calculate the distance as total transit time * c, the longer time is obscured by lengthening the distance. This appears to preserve constant light speed. This is in conjunction with 'length contraction' which 'adjusts' the distance along with the time.
    Use just the x-axis frame moving along x at some speed v. Calculate the times as usual. Then calculate light speed from the origin point of view.
     
  17. Neddy Bate Valued Senior Member

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    1,403
    These are the equations I use:

    \( (ct_x)^2 = (t_xv_x+L)^2 + (t_xv_y)^2 + (t_xv_z)^2 \)

    and

    \( (ct_y)^2 = (t_yv_x)^2 + (t_yv_y+L)^2 + (t_yv_z)^2 \)

    and

    \( (ct_z)^2 = (t_zv_x)^2 + (t_zv_y)^2 + (t_zv_z+L)^2 \)


    I'm not sure how to simultaneously solve those equations for velocity or time, so I just use interpolation. Basically I use trial and error until I get the right answer.

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  18. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    As you can see in my calculations, the speed of light is measured to be .361c in the x direction, as it took 1.384930 seconds for light to travel from the center of the cube to the x receiver. .5/1.384930=.361.

    In the y and z directions, the speed of light is measured to be .5/.65=.7692c, as it took .65 seconds for light to travel from the source to the y and z receivers.

    The ONLY time the speed of light will ever be measured to be 1.0c is when the cube has an absolute zero velocity in all directions.

    Did you happen to notice that in my calculations, the y and z component velocities are zero, and yet it took .65 seconds for light to reach the receivers?

    Once again, I prove Einstein to be wrong using actual numbers of distance and time that can't be disputed!

    There is no way light can be measured to be 1.0c unless the cube is at an absolute zero velocity!
     
  19. phyti Registered Senior Member

    Messages:
    256
    This equation should work, using (1,2,3) for (x,y,z).

    d is distance from origin to each clock

    time for light to reach clock with v=0 is:

    t0=d/c= d light seconds, with c=1

    time (ls) for light to reach each clock from the origin is:

    t1, t2, t3

    speed of origin is:

    v=sqrt[{(t1-t0)/t1}^2+{(t2-t0)/t2}^2+{(t3-t0)/t3}^2]

    It doesn't assure that v will be less than c. You need an additional relation for that.
     
  20. phyti Registered Senior Member

    Messages:
    256
    Her is a space-time drawing showing MD's issue with speed measurements.
    It's quite obvious, if A is moving at .8c, and his times result from closing speeds of c-v and c+v, those are what A would measure (if possible).

    Please Register or Log in to view the hidden image!

     
  21. Neddy Bate Valued Senior Member

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    1,403

    OK so you have given me these times:
    \(t_x = 1.38493\)
    \(t_y = 0.65000\)
    \(t_z = 0.65000\)

    By my calculations, the velocity components look like this:
    \(v_x = 0.63897c\)
    \(v_y = 0.00000c\)
    \(v_z = 0.00000c\)

    At time 1.38493, the center of the "x-face" is located at point (1.38493, 0.000, 0.000) and the distance from that point to the origin is:
    \(d = sqrt(1.38493^2 + 0.00000^2 + 0.00000^2)\)
    \(d = 1.38493\)

    At time 0.65000, the center of the "y-face" is located at point (0.41533, 0.50000, 0.00000) and the distance from that point to the origin is:
    \(d = sqrt(0.41533^2 + 0.50000^2 + 0.00000^2)\)
    \(d = 0.65000\)

    Yes, everything matches your calculations. Good job!


    Yes, that is why we needed to look at all three dimensions. If you had only looked at the 0.65000 seconds, you might have thought there was a "y" or "z" component to the velocity. (For example, using your old equations).


    No, all you have done is come up with a self-consistent model of how you think the universe should work. In actual practice, the speed of light is always measured to be 1.0c, and there is no absolute rest frame required.
     
  22. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    5,105
    Measuring the speed of light in the box proves the speed of light is not always measured to be c. If the box has a velocity the light can not be measured to be c. That is simply impossible!
     
  23. Neddy Bate Valued Senior Member

    Messages:
    1,403

    In that case, I guess my microwave oven must be at absolute rest:


    http://en.wikipedia.org/wiki/Speed_of_light#Cavity_resonance

     

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