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Is The Theory Of Special Relativity Wrong?

  1. Mar 30, 2004 #1
    I think the solution to the following problem proves the theory of special relativity contains an error. I have solved the problem, and can't find any error in my reasoning. Can anyone show me where my error is, if there is one?

    Experimental Setup

    Let us presume that we have two identical digital clocks, clock A, and clock B, at rest with respect to each other, and in an inertial reference frame. Let R1 denote the current reading on clock A, and let R2 denote the current reading on clock B. Let us define D as follows:

    D = difference in clock readings = R1 - R2

    If R1=R2 then the clocks are synchronous.

    Let the clocks currently be synchronous.

    Suppose that when the clocks both read zero, that there is a constant force F applied to clock B, for [tex] \Delta t^\prime [/tex] seconds according to clock B. Let [tex] \Delta t [/tex] denote this amount of time according to clock A. Let V denote the final relative speed of the two clocks. Assuming the theory of special relativity is correct, the times are related by the following formula:

    [tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-V^2/c^2}} [/tex]

    At this point, the clocks are now in relative motion, at a constant speed V.
    As can be seen from the time dilation formula, the clocks are no longer synchronous. In fact, during the time the force was applied, clock B ticked slower than clock A, so that R1-R2 is now positive rather than negative. If the clocks were still synchronous then R1-R2 would equal zero, however we are operating under the assumption that the time dilation formula is correct.

    Consider two ways to bring the clocks to rest with respect to each other again.

    Let T denote the amount of time for which the force was applied to clock B as measured by clock B.

    Method 1: We can apply force F to clock B for time T as measured by clock B, in the direction of clock A.

    Method 2: We can apply force F to clock A for time T as measured by clock A, in the direction of clock B.


    Using either method, the clocks will again be at rest with respect to each other.

    Here is the problem I have detected.

    If we apply the force F to clock A for time T as measured by clock A, then clock A will tick slower than clock B, and after time T has passed according to clock A, the clocks will again be synchronous.

    On the other hand, if we apply the force to clock B for time T as measured by clock B, then clock B will tick slower than clock A, and after time T has passed according to clock B, R1 will be greater than R2.

    Thus, the final readings on the clocks, depends upon whether or not we apply the force to clock A, or clock B.

    Thus, time dilation involves vector dependence on velocity. However, the time dilation formula doesn't involve vector dependence on velocity, v is just relative speed. Thus, there is an error in the theory of special relativity.

    If I have made an error, then where is my error located?

    I don't see any vector term in the time dilation formula, so my analysis ends in the conclusion that the time dilation formula is erroneous.


    Respectfully,

    The Star
     
    Last edited: Mar 30, 2004
  2. jcsd
  3. Mar 30, 2004 #2
    I believe you made a mistake here. If you apply the force to clock B in the direction of clock A, clock B will tick faster than clock A. I suggest reading post #4 in this thread:
    https://www.physicsforums.com/showthread.php?t=16596
     
  4. Mar 30, 2004 #3
    Honestly, sir, you should instead conclude that you misunderstand some point about SR or GR that you have not yet learned. You should not conclude based on this thought experiment that the equation is erroneous.

    Personal note to StarThrower: each and every time you ask for people to prove you wrong you are given just that...what makes you think you'll listen this time?
     
  5. Mar 30, 2004 #4
    I was hoping that someone with incredible mathematical ability would compare R1 to R2, using both methods, and explain to me what happens. I want to see the math. Length contraction is irrelevent in the analysis of this problem.
     
  6. Mar 30, 2004 #5
    Why should I conclude that?

    Why shouldn't I conclude exactly that?


    Personal note to severian596: I just wanted to see a mathematical analysis of the problem, not get into philosophical debate.
     
  7. Mar 30, 2004 #6

    DrChinese

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    I hate to tell you, but you are supposed to do the math. That would be a prerequisite for someone to take your hypothesis seriously. Generally speaking, the burden falls on your shoulders to make sense of things. That is especially true when you are attacking a theory as venerable as SR.
     
  8. Mar 30, 2004 #7
    I have done the math, and my result was that the theory of special relativity self-contradicts. I was looking for someone to independently reach the same conclusion.

    Regards,

    The Star
     
  9. Mar 30, 2004 #8

    Nereid

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    Prove it, mathematically.
     
  10. Mar 30, 2004 #9
    Prove what? That the clocks will again be at rest with respect to each other? Of course thats going to happen.

    Regards,

    The Star
     
  11. Mar 30, 2004 #10

    Nereid

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    Yes, that's what I am asking you to do for us.

    It looks obvious doesn't it?

    Please take us through the (mathematical) proof, step by step.
     
  12. Mar 31, 2004 #11

    russ_watters

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    Because you aren't Einstein, Hawking, Penrose, et al. all rolled into one.
    Because you aren't Einstein, Hawking, Penrose, et al. all rolled into one.

    StarThrower, a little humility would help you a great deal. HUNDREDS if not thousands of the greatest minds the world has ever known have poured millions of hours, dollars, and pages into examination of SR. For you to conclude even that there may be an error without a ready-to-publish paper with hard experimental evidence to back it up is not only arrogant, but it will stand in the way of your ability to learn.
     
  13. Mar 31, 2004 #12

    Integral

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    Not a very interesting problem and not a question about SR.

    As soon as your turn on the Force, you must turn off SR. When you turn off the force you can turn SR on. You have a time difference due to the relative speed of the clocks.

    Turn on the decelerating force, turn off SR, Turn off the force, turn on SR. The difference in time on the clocks will reflect the speed difference and the duration of the difference.

    You cannot apply SR during the periods on Acceleration. If you wish to correctly account for the behavior of the clock in those time intervals you must use GR.
     
  14. Mar 31, 2004 #13

    Hurkyl

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    Differential calculus works just fine in SR.
     
  15. Mar 31, 2004 #14
    Ok Ms Neried no problemo. :)
     
  16. Mar 31, 2004 #15
    I've tried to tell him this before, but he doesn't listen.
     
  17. Mar 31, 2004 #16
    Fallacy of appeal to authority.


    Einstein: Why should I believe Newtonian Mechanics is correct?
    Russ: Because you aren't Galileo, Newton, Leibniz, et al all rolled into one.

    Newton: Why should I believe Relativistic Mechanics is correct?
    Russ: Because you aren't Einstein, Hawking, Penrose, et al. all rolled into one


    My point has been made.


    Perhaps you are right about your humility comment, and then again, perhaps you are wrong about it. I fail to see how you can, inside your mind, figure out whether or not you are correct. I of course know the correct answer. Do you want me to tell you the correct answer?

    Consider the following statement:

    If the fundamental postulate of the special theory of relativity is true, then (X and not X), for at least one statement X.

    Suppose that one such X has been discovered. Then all the experiments performed throughout the whole of time cannot verify the theory of special relativity, since the theory of relativity is falsifiable.

    Please note that in any such proof that SR is wrong, no experimental evidence is needed, rather the evidence is mental.

    Since when is the usage of binary logic to reason a sign of arrgoance. Rather, I believe it would be the sign of intelligence. Please correct me if I am wrong, I'd enjoy seeing your proof.


    Kind regards,

    The Star
     
  18. Mar 31, 2004 #17
    This response is incorrect Mr Integral.

    Clock A is in an inertial reference frame throughout the event.

    The Event

    The acceleration of clock B by application of an outside force to clock B, for an amount of time [tex] \Delta t^\prime [/tex] as measured by clock B, and a corresponding amount of time [tex] \Delta t [/tex] as measured by clock A.

    This event lasts an exact amount of time as measured by clock B, and an exact amount of time as measured by clock A. Since clock A is always in an inertial reference frame, clock A can certainly use the time dilation to relate its measurement to clock B's measurement. If SR is correct, then the formula an observer at clock A must use is:

    [tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]

    You don't need GR at all, because if an observer at clock B wants to figure things out, and he knows that A is always in an inertial refefence frame, and that clock B is being accelerated by a constant force F for time [tex] \Delta t^\prime [/tex] according to clock B, then the observer accelerating along with clock B (in other words an observer at rest with respect to clock B, in other words an observer in this non-inertial reference frame) must use the exact same formula that an observer stationed at clock A would use, since by binary logic, they must reach the same conclusion.

    Regards,

    The Star


    P.S.: If this confuses you, think of an instantaneous impulse instead. Think of clock B as suddenly jumping from a relative speed of zero, to a relative speed of V, over two consecutive moments in time.
     
    Last edited: Mar 31, 2004
  19. Mar 31, 2004 #18
    Hurkyl is correct.
     
  20. Mar 31, 2004 #19
    Here is the proof as promised Ms Nereid.

    The Event

    Two digital clocks A,B are initially at rest with respect to one another, in an inertial reference frame, and synchronous. At the moment both clocks read zero, a constant outside force F is applied to clock B in the exact opposite direction as the vector from clock B to clock A, for time T as measured by clock B. (Hence clock B is moving away from clock A). Once the force on clock B finally vanishes, the relative speed of the clocks is V.

    Mathematical Analysis Of The Event

    Let [tex] \Delta t^\prime [/tex] denote the amount of time of this event as measured by clock B.

    Let [tex] \Delta t [/tex] denote the amount of time of this event as measured by clock A.

    Assumption 1: The time dilation formula of SR is correct.

    By assumption 1, the measurements of both clocks are related by the following formula:

    [tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]

    where [tex] T = \Delta t^\prime [/tex]

    Thus, by the time dilation formula, once the event is over, the clocks are no longer synchronous. Instead, the reading of clock A is greater than the reading of clock B, which means that clock B ticked slower than clock A during the event. Let R1 denote the reading of clock A at any moment in time, and let R2 denote the reading of clock B at any moment in time.

    Conditions before force F was applied to clock B for amount of time T as measured by clock B:

    Before the force F is applied, the relative speed of the two clocks was zero, and so by assumption 1 the following equation was true:

    [tex] \Delta t = \Delta t^\prime [/tex]

    Where delta t is an arbitrary amount of time as measured by clock A.

    Since additionally the clocks are identical, the equation above implies that they ticked at the same rate. Since the clocks were synchronous at the beginning of the event, they were synchronous before the event began. Thus, prior to the beginning of the event, R1=R2.

    Initial Clock Readings

    By stipulation, the event begins at a moment in time at which both clocks read zero. Thus, the initial conditions of the event were:

    R1=R2=0

    Final Clock Readings

    Now, as clock B moved further away from clock A, by assumption 1, the two clocks fell out of sync, and clock B ticked slower. Thus, at the end of the event we have R1>R2, and the relative speed of the clocks is V. At the end of the event, clock B is again in an inertial reference frame.

    Additionally, clock B now ticks slower than clock A.

    Now to answer Ms Nereids question.

    Suppose that a constant force F is applied to clock A for an amount of time T as measured by clock A, in the direction of clock B. Prove the clocks are again at rest with respect to each other.

    This case is obvious. Here, A will undergo changes which are exactly analagous to any changes that clock B underwent. Hence, at the end of the application of the force, clock A will have a final velocity V relative to a stationary observer fixed at the initial location of clock A. And V is also the velocity of clock B with respect to a stationary observer fixed at the initial location of clock A. Hence, the difference in the velocity vectors of clocks A and B, from the initial position of clock A is:

    V-V=0.

    Hence, if a constant force F is applied to clock A, for an amount of time T as measured by clock A in the direction of clock B, then after the force vanishes, the relative speed of clock A to clock B is zero, in any and all reference frames.

    Additionally, the tick rate of clock A will approach the tick rate of clock B, so that once the force on clock A vanishes, the clocks will again tick at the same rate.

    Now, suppose that a constant force F is applied to clock B for an amount of time delta t as measured by clock A, in the direction of clock A, where delta t is related to T through the following equation:

    delta t = T/ sqrt(1-v^2/c^2)

    Prove the clocks are again at rest with respect to each other.

    In this case, use vectors which are defined in a coordinate system where the position of clock A is fixed. At the moment when force F is re-applied to clock B (only this time in the opposite direction), the initial velocity of clock B relative to clock A, is V. We must have the final velocity vector of clock B relative to clock A come out to be the zero vector.

    Let vi denote the intial velocity of clock B relative to clock A.
    Let vf denote the final velocity of clock B relative to clock A.

    We know that vi=V.
    We want vf=0.

    Thus,

    vf-vi = 0 - V = -V

    Here is the definition of acceleration:

    a = dv/dt

    Whilst the force F is applied, the acceleration of clock B relative to clock A is constant, hence we have to use the kinematic equations for constant acceleration.

    a dt = dv

    Then we integrate both sides to get:

    a2(delta t) = vf - vi = -V

    Now, the first time the force was applied to clock B using the same math we would have to get:

    a1(delta t)= vf - vi = V

    Thus, the acceleration vectors must satisfy the following equation (in order to have the final relative speed of the two clocks come out to be zero):

    a1(delta t) = -a2 (delta t)

    From which it follows that

    a1= -a2

    Thus, as long as vector a1 and vector a2 have the same magnitude, but opposite directions, the two clocks will have a final relative speed of zero, and that was stipulated. Keep in mind, that all vectors in this case need to be defined in one frame only, and I chose a frame at rest with respect to clock A. The same result would have been obtained in any other reference frame whatsoever.

    QED
     
    Last edited: Mar 31, 2004
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