The Light Speed Postulate and its Interpretation in Derivations of the Lorentz Transformation

Discussion in 'Pseudoscience' started by tsmid, Apr 24, 2016.

  1. Neddy Bate Valued Senior Member

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    You say that now, but in the OP you said this:

    You claimed to have "derived" the equation x1'=x2 as as a "transformation". You are now admitting that it is a transformation of two different events. Yet the Galillei and Lorentz transformations pertain to transforming one single event between two frames. What is the use in using two different events, and then pointing out that it is different to SR???
     
    Last edited: May 18, 2016
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  3. Neddy Bate Valued Senior Member

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    Yes, just as in the convention, you have the origins of both systems adjacent to each other at time t1=0=t2=0, and that is the time the light is emitted from the origin. Assuming x1=x2 and no length contraction, that means the only time when coordinate x1 is adjacent to coordinate x2 is at t=0. But that is the time the light is emitted, and due to the finite speed of light, it cannot also be the time the light is detected at the other end of the rod. Yet, you somehow require x2 and x1 to be in the same place at the time the light is detected, which is what you use to justify your claim of x1'=x2=x1.

    That is a contradiction of your own conventions. Surely you admit that t1>0 and t2>0 right? So you have to use v to figure out how the x1 and x2 axes are aligned at those times. You cannot assume it will be the same as if v=0 in which case nothing has changed since t1=t2=0.
     
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  5. tsmid Registered Senior Member

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    x1'=x2 is not a transformation as x1' and x2 relate to different events. The transformation would be x1'=a*(x1-b*t).

    Let me recapitulate: in frame 1 (the frame co-moving with the light source) we have stationary detector(s) located at the distance x1 from the origin, in frame 2 (the frame moving relatively to the light source) we have (different) stationary detector(s) at the distance x2 from the origin.
    If the light signal reaches x1 after the system time t in frame 1, this means

    (1) x1=ct

    Now assuming that the speed of light does not depend on the state of motion of the light source (as postulated by Einstein), then in frame 2 the light signal (also emitted from the origin) must after the same time t reach the same distance

    (2) x2=ct

    (assuming that both systems use the same length and time units, as discussed earlier).

    Now obviously, the detection events relating to the the frame 1 and frame 2 detectors are different events. Let's call them class 1 and class 2 events respectively. According to Eqs(1) and (2), the speed of light in frame 1 is defined by class 1 events (involving only detectors stationary in frame 1), and the speed of light in frame 2 is (independently) defined by class 2 events (involving only detectors stationary in frame 2).
    Eqs(1) and (2) must necessarily hold in this sense if the speed of light is independent of the speed of the light source. However, Einstein defines the speed in frame 2 differently: he uses also class 1 one events here, i.e. he does not consider x2 but x1' (the location of x1 in frame 2). Now if the class 1 events are supposed to obey the light speed postulate in frame 2 as well , this means that after time t, x1' must also have reached the location x2=ct, so this would require

    (3) x1'=x2

    (I have illustrated this below in an animated graphic where blue means "frame 1/class 1" and red means "frame 2/class 2". The bottom half of the section corresponds to Einsteins procedure of using class 1 events (blue dots) for the definition of the speed of light both in frame 1 and frame 2. In either case, the dots progress in both frames with the same speed).

    But if you combine equations (1),(2),(3) you get

    (4) x1'=x1

    which is not a possible transformation unless the two frames are at rest relatively to each other.
    From this we can conclude that it is not permissible to use class 1 events to define the speed of light in frame 2.


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    Last edited: May 21, 2016
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  7. Neddy Bate Valued Senior Member

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    You are contradicting yourself. In the very paragraph where you finally introduce x1' you clearly want your equation to be a transformation. There, you claim you want to do the same thing that Einstein does by considering x1' instead of x2, and you admit it is not two different classes of events, here:

    Yet, even there, you still arbitrarily require x1'=x2 even though you know that was obtained by keeping x2 and x1' in different "classes" of events? Why? Is it something you derive from your equations 1 and 2 which have a simple t in them, rather than t1 and t2? Your equations should be like this:

    (1) x1 = ct1
    (2) x2 = ct2
    (3) x1' = ct1'
    (4) x2' = ct2'
    Can you still derive x1'=x2 from those equations? If so, how?

    Or is it something to do with your animation which shows what appears to be two rods of equal length which are STATIONARY with respect to one another? Please revise your animation so that there is some relative motion apparent between the two frames, and then perhaps you will see how v would have to be part of the transformation equations. You cannot possibly deny that the transformation would be x1'=gamma*(x1-(v*t1)) under SR, so what is x1'=x2 supposed to be?
     
    Last edited: May 24, 2016
  8. Neddy Bate Valued Senior Member

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    2,548
    Can you please explain your animation, tsmid? Here is the first slide:

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    What is the reason for having an upper half and a lower half which appear to be identical? Are the two halves supposed to represent the two different reference frames? If so then you should not have the red linear rod marked with velocity v in its own frame, because the red rod would consider itself stationary with respect to its own reference frame.

    So, if the two halves do not represent the two reference frames, then what are they supposed to represent?

    Here is the second slide:

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    I assume the dots represent the wavefront of the light moving to the right. But what is the reason for having two blue dots in the lower half, while the upper half has one blue and one red? What concept were you hoping to convey with this color coding, if anything?

    And finally, why does there not appear to be any relative motion between the red and blue linear rods? There are not even any marks depicting x1 or x2 on the horizontal axes, so all we see are two stationary rods.
     
    Last edited: May 24, 2016
  9. tsmid Registered Senior Member

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    368
    To repeat (again): I do not require a transformation. I would perfectly happy to define the light speed postulate just through

    (1) x1=c*t1
    (2) x2=c*t2

    that is, the speed of light in frame 1 is defined by class 1 events, and in frame 2 by class 2 events. But that obviously wouldn't get Einstein anywhere. So he considers class 1 events to define the speed of light in frame 1 and frame 2, thus requiring a coordinate transformation between the two frames. And I am merely addressing the implications and consequences of Einstein's requirement.


    If we have

    (1) x1=c*t1
    (2) x2=c*t2

    and apply additionally the convention that the light signal is emitted in both frames at the system time 0, i.e. t1=0 <=> t2=0, and also the convention that the clocks are physically identical in both systems, we have therefore t1=t2 throughout and can instead replace the two independent variables t1 and t2 by one global independent variable t, so

    (1) x1=c*t
    (2) x2=c*t

    Now (1) still describes class 1 events and (2) class 2 events. This is useless for Einstein. In order to introduce a transformation equation, he needs class 1 events to define the speed of light in frame 2 as well. Taking the variable x1' for these events and applying the light speed postulate to these as well (i.e. x1'/t =c) we have therefore

    (2a) x1'=c*t

    so from the above follows x1'=x2 (in other words the class 1 and class 2 events must coincide in frame 2 IF one requires that the class 1 events propagate with the speed of light as well in frame 2 (which is what Einstein does)).
     
  10. tsmid Registered Senior Member

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    368
    The two halves represent the two sets of equations

    (1) x1=c*t (frame 1)
    (2) x2=c*t (frame 2)
    ------------
    (1) x1=c*t (frame 1)
    (2a) x1'=c*t (frame 2)


    The events are seen from the respective rest frames, so there is no relative motion of the rods visible here (the diagram does not represent a particular reference frame).


    Yes, the dots represent the motion of the light front in the two frames (blue=frame 1; red=frame 2). And as per above we have

    (1) x1=c*t (class 1 (blue))
    (2) x2=c*t (class 2 (red))
    ----------------------------
    (1) x1=c*t (class 1 (blue))
    (2a) x1'=c*t (class 1 (blue))
     
  11. Neddy Bate Valued Senior Member

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    2,548
    Just because the clocks are physically identical does not mean t1=t2. Remember, you defined x1 as a rest length measured in system 1, which could be referred to as the proper length of rod 1. And you defined x2 as a rest length measured in system 2, which could be referred to as the proper length of rod 2. At this point in your "derivation" you have not required the rods to have equal proper lengths. If x1 and x2 are unequal, then t1 and t2 will also be unequal.

    That is why you invoked your special case of x1/x2=t1/t2=q=1 in the OP. When you did that, you were able to make t1=t2, but only because you were then requiring the special case where x1=x2, or rods of equal proper length.


    Which means nothing more than x1=x2, rods of identical proper length, as explained above.


    And there you fail to consider the movement of rod 1 relative to system 2. If the light was emitted at time t=0 and rod 2 has a non-zero length, then the light arrives at the end of rod 2 at time t2>0. During the time interval from 0 to t2>0, (while the light is in transit along rod 2), system 2 would measure that rod 1 moves a distance equal to v*t2. That is the meaning of the velocity v term. Therefore one must consider that change in distance, but you do not do so.

    Note that there is a v term in the Lorentz transformations, so this is your mistake, not Lorentz or Einstein's.
     
  12. Neddy Bate Valued Senior Member

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    2,548
    Below is an animation I made which assumes there is no such thing as length contraction:

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    Since there is no length contraction in the above animation, the only time that x1 and x2 coincide is at the moment the light is emitted from the origins. Therefore tsmid's expectation that x1 and x2 should coincide when the light reaches detectors at either x1 or x2 is not even possible in this arrangement.

    ---------------

    And below is another animation I made, this time assuming length contraction in accordance with SR:

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    Since there is length contraction in the above animation, there is no time after the light is emmitted that x1 and x2 coincide. Therefore tsmid's expectation that x1 and x2 should coincide when the light reaches detectors at either x1 or x2 is once again shown to be impossible in this arrangement.
     
  13. tsmid Registered Senior Member

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    368
    Yes, x1,x2 and t1,t2 are the proper distances and times in the two systems respectively. But you have to bear in mind which are the independent and which the dependent variables. In this case I took t1 and t2 as the independent variables. And if the speed of light is invariant, this means that within the same time t1=t2 the signal travels the same distance x1=x2 (as defined by stationary detectors at these points in the respective system). We can therefore write this also as in terms of the new independent variable t (assuming, as mentioned, that the clocks in the two systems are physically identical)

    (1) x1=ct
    (2) x2=ct

    But Einstein also demands that the transformed distance x1' travels the same distance ct in the same time t in frame 2, so additionally he assumes

    (3) x1'=ct

    There is nothing to prevent you from choosing rods of equal length, but what the above equations express is that the light signal travels the same distance in the same time along either rod (you should be seeing it more as a continuous array of distance markers defining the reference frames here)

    Yes, there should be a change in distance for the class 1 events as observed in frame 2 depending on the relative velocity, but that would directly contradict the light speed postulate applied to these events because the above equations (1),(2),(3) would require x1'=x1. So we can conclude that Eq.(3) does not hold (the class 1 events are not permissible to define the speed of light in frame 2).
     
  14. tsmid Registered Senior Member

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    368
    First of all, you seem to have forgotten the prime on x1 in your comments (it should be about x1' and x2 coinciding).
    Secondly, I am not expecting that x1' and x2 coincide. It is what Einstein's assumption of the class 1 events defining the invariance of the speed of light in frame 2 implies (because we know that these events start at the origin of frame 2 as well, and in time t they would travel a distance ct, which is co-located with x2).
     
  15. Neddy Bate Valued Senior Member

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    2,548
    Not so. Your equation (3) should be

    x1' = ct1'

    Where there is no requirement that t1' be equal to your t. As you concede below, the distance traveled x1' would not be equal to either x1 or x2 because of the relative velocity v. Therefore the travel time t1' could not be equal to your t either.

    Yes, thank you. So the change in distance due to relative velocity MUST be considered when calculating x1' because that is the location of the detector x1 as recorded by system 2 (which finds that detector to be moving at velocity v). Now that you have conceded this, you must stop writing x1'=x1 or x1'=x2 which do not consider the relative motion between the systems.

    No it wouldn't. Even if x1' does not equal x1 or x2, the light postulate could still be upheld if x1'/t1'=c.

    Here is an example Lorentz transformation based on the same data I used in my second animation (the one with length contraction included):

    c = 1.00
    v = 0.60c
    gamma = γ = 1 / √(1 - (v²/c²)) = 1.25
    x2 = 1.00
    t2 = 1.00
    x2' = γ(x2 - (v*t2)) = 0.50
    t2' = γ(t2 - ((v*x2) / c²)) = 0.50

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    Note that x2/t2=c and x2'/t2'=c even though x2' does not equal x2, and even though t2' does not equal t2.
     
    Last edited: May 26, 2016
  16. Neddy Bate Valued Senior Member

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    You seem to have forgotten that x1' is defined as the location of x1 as recorded by system 2. That means x1' and x1 must coincide. So if you think x1' and x2 coincide, then you must think x1 coincides also.

    That's good.

    So you know x1' does not coincide with x2 at time t, but you claim x1'=x2 anyway, and then you conclude that t1'=t. Or else you assume t1'=t first, and use it to justify your absurd notion that x1'=x2. Meanwhile, if you had ever used the actual Lorentz transformations, you would know that x1' does not equal x2, and t1' does not equal t. Yet x1/t1=c and x1'/t1'=c.

    You are tilting at windmills -- fighting perceived demons which do not even exist. What you are doing is not SR by any stretch of the imagination.
     
    Last edited: May 26, 2016
  17. tsmid Registered Senior Member

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    But from the first two equation we know already that t1'=t, because after the proper time t in frame 2 the signal travels the distance x2=ct, and after the same proper time t in frame 1 it reaches the identical distance x1=ct. So when the signal reaches x2 in frame 2, it must also reach x1 in frame 1. But if x1' is supposed to change with the speed of light as well, we must also have x1'=ct after the same time t, which means we would have x1'=x2 and with this x1'=x1. And the latter, as you pointed out yourself, is obviously not possible if the frames are moving relatively to each other. From which we can conclude that the light speed postulate can not be applied to x1'.
     
  18. tsmid Registered Senior Member

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    That doesn't seem to make any sense: if you locate an object at coordinate x1 in an earth based coordinate system, and another object at the same coordinate x2=x1 in some coordinate system in a different galaxy, then certainly you could not say that x1 and x2 coincide, even though they have the same value. These two objects would physically only coincide if x1'=x (two objects coincide if they have the same coordinates in the same reference frame).

    It depends what you mean by 'doing SR'. The LT is not introduced as a postulate, so you can not take it for granted here. The postulate is the invariance of the speed of light, and we are discussing here merely whether this is compatible with a velocity dependent transformation (of any form) for the coordinates of a light signal.
     
  19. Neddy Bate Valued Senior Member

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    Alright, now I think I finally understand your argument completely. If so, then I would suggest that you should have made your claim of t1'=t prior to claiming x1'=x2. Since x1'=x2 does not appear to be justifiable on its own, (due to the relative motion between the two systems), I had trouble getting past it.

    But the only problem with t1'=t is that it assumes that system 2 would record the clocks of system 1 to be synchronised at all times with its own clocks. That is the same assumption that Galilean relativity made, and everyone already knew that the light postulate was not compatible with Galilean relativity. Once the Lorentz transformations were derived, and Einstein realised their meaning, the idea of universal time would have to be abandoned in order for the light postulate to be upheld.

    Surely you have heard of time dilation in SR. The clocks which are relatively moving tick at a slower rate than the clocks which are relatively stationary. That alone prevents t1'=t from being a safe assumption, but there is even more reason in a thing called "relativity of simultaneity". But let's not get ahead of ourselves.

    If you honestly want to examine the veracity of the light postulate, you will have to do so in terms of SR and the Lorentz transformations, which are not t1'=t or x1'=x2. They are t1'=γ*(t1-((v*x1)/c²)) and x1'=γ*(x1-(v*t1)) where gamma=γ=1/√(1-(v²/c²)) and you will not be able to derive them if you assume universal time at the outset.
     
    Last edited: May 28, 2016
  20. tsmid Registered Senior Member

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    368
    t1' is actually a variable that is not defined here (and in fact can't be used in the way you want it here). The initial equations were

    (1) x1=c*t1 (class 1 events define speed of light for frame 1)
    (2) x2=c*t2 (class 2 events define speed of light for frame 2)

    where t1 and t2 are both independent variables (designating the system (proper) time in each reference frame.)

    Einstein now requires that the class 1 events also obey the light speed postulate in frame 2, i,e,

    (3) x1'=c*t2

    So for instance, if we have a row of photon detectors at rest in frame 1, and at each of those detectors have a blue lamp lighting up when it detects the photon, then according to Eq.(3) these 'blue lamp' events should, like the 'red lamp' events described by Eq.(2), progress with speed c as a function of the system time t2 in frame 2. The system time t1 does not come into Eq.(3) at all, neither directly nor indirectly as a transformed variable.
    And if you further assume that the light signal is emitted at t1=t2=0 in each reference frame and that all clocks are physically identical, then you can set t1=t2=t, which gives the equations I stated earlier.

    As I mentioned earlier already, it is not legitimate to use the results of SR here, as they are derived only based on a subset of all possible events (the class 1 events). As explained above, they contradict the speed of light postulate if you include class 2 events as well.
    ---------------------------
    P.S.: I'll be away now for a couple of weeks, so you can take your time to consider my argument and your reply.
     
    Last edited: May 30, 2016
  21. Neddy Bate Valued Senior Member

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    2,548
    If you would write the full set of equations, you would see that t1' is a defined variable:

    (1) x1=c*t1 (class 1 events define speed of light for frame 1)
    (2) x2=c*t2 (class 2 events define speed of light for frame 2)
    (3) x1'=c*t1' (class 1 events define speed of light for frame 2)
    (4) x2'=c*t2' (class 2 events define speed of light for frame 1)

    No. The light speed postulate is satisfied if all of the following hold true:

    x1/t1=c
    x2/t2=c
    x1'/t1'=c
    x2'/t2'=c

    Even in the special case where x1=x2 and thus t1=t2, there is no requirement that t1' must be equal to t1, nor is there any requirement that t1' must be equal to t2. I have already given you the reasons -- time dilation and relativity of simultaneity. Just because your approach precludes you from deriving the Lorentz transformations does not mean that they cannot be derived. Lorentz did it, and so did Einstein.
     
    Last edited: May 31, 2016
  22. tsmid Registered Senior Member

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    368
    The speed of light in frame 2 must be measured in terms of the system time in frame 2 (whether it is for class 2 events or for class 1 events), which however is already given by the (independent) variable t2. You can not put a dependent variable t1' on the right hand side of the equation and claim it would define the speed of light in frame 2.

    You are arguing again in circles here, taking results for granted that are yet to be proved. And as I explained in detail in my previous post, Einstein's derivation is based on excluding class 2 events (events related to detectors at rest in the stationary frame) from his definition of the speed of light in the stationary frame.
     
  23. tsmid Registered Senior Member

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    368
    Just to summarize the line of argument again::

    assuming the lightspeed postulate holds, we have the following equations for the propagation of the light signal as registered by detectors at rest in frame 1 (class 1 events) and by detectors at rest in frame 2 (class 2 events)

    (1) x1=c*t1
    (2) x2=c*t2

    where t1 and t2 are the respective system times. They are independent variables here (after all, the clocks in system 2 are completely independent of those of system 2). Any events are timed in terms of t1 in system 1 and in i n terms of t2 in system 2. So even if you ask how the location of the class 1 events propagates in system 2, this is done in terms of t2. So if those should propagate with the speed of light as well, this would mean

    (3) x1'=c*t2

    A time t1' does come into it at all, and as I said before, it is indeed completely undefined. We only have the two independent system times t1 and t2 here. And since Eq.(3) must hold true, your suggestion x1'=c*t1' would imply t1'=t2, which would not only be pointless, but incorrect, as t2 is by definition an independent variable, which can not be identical to an obviously dependent variable (whatever you mean by a dependent time variable).
     

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