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C+V and c-V

  1. Mar 2, 2007 #1
    Consider a scenario which involves a stationary machine gun (MG) and a moving target T. MG emits successive bullets at constant time intervals t(e) which hit the moving target at constant time intervals t(r). The bullets move with velocity U and the target with velocity V. The first bullet hits the target at t=0. Equating the distnce traveled by the second bullet with the distance traveled by the target between the reception of the first and of the second bullet we obtain
    U(t(r)-t(e))=Vt(r) (1)
    where from
    t(r)=Ut(e)/(U-V) (2)
    Replacing the MG with a source of light we obtain
    We could follow scenarios which lead to an expression containing the terms U+V and c+V. Are U+V, U-V, c+V and c-V the result of a velocity addition?
  2. jcsd
  3. Mar 2, 2007 #2


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    I would think the answer to your question lies in the Minkowskian interpretation of the problem.
    What is your answer, in light of http://arxiv.org/abs/physics/0703002...
    [which, in my humble opinion, is lacking many references and attributions of credit]?
  4. Mar 2, 2007 #3


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    Please TEX your formulae.
  5. Mar 2, 2007 #4
    c+V, c-V

    Thanks. What is the meaning of attributions of credit? The paper is under revision and criticism is highly appreciated. I think that it would be better to use a relativistic space-time diagram which displays the space-time coordinates in true magnitude.
  6. Mar 3, 2007 #5

    Meir Achuz

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    No, no, no.
  7. Mar 3, 2007 #6
    c+V, c-V

    Thanks. But what would be a YES, YES, YES answer. Of course sine ira et studio.
  8. Mar 4, 2007 #7

    Meir Achuz

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    "Are U+V, U-V, c+V and c-V the result of a velocity addition?"
    This is a relativity forum. Yes would be the answer to a correct statement.
    I had to repeat the nos because the forum does not accept "no" for an answer. It would not even accept "yes".
    Gedanken experiments are thoughtful, but they don't measure anything.
    One like a machine gun just shows how to get to an inference you know is not correct.
  9. Mar 4, 2007 #8


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    This step is just nonsensical. Surely the correct expression for light is
    t(r)=ct(e)/c = t(e).

    What point are you trying to make with this post ?
    Last edited: Mar 4, 2007
  10. Mar 4, 2007 #9
    nonsensical for sure?

    1. Please take into account that t(r)=t(r)-0 and t(e)=t(e) are time intervals of emission by a stationary source and of reception by an uniformly receding receiver, respectively. In special relativity both are proper time intervals. As you know such time intervals are related by the Doppler shift formula (acoustic in the case of the bullets and optic in the case of light signal. In the limits of your free time have a look please at
    "Teaching the Doppler shift using a machine gun analogy" The Physics Teacher Vol.39, 468-469 (2001). I would never use such hard statements
    as (nonsensical) and avoid them if you continue the discussion with me.
    2. I find in some of the derivations of the formula which accounts for length contraction the expression
    giving the time interval during which a light signal propagates forth and back along a moving a rod considering I thinck a composition of velocities. Are they correct? Do you understand my humble point?
    Are you a teacher, user or learner of special relativity? I am an eternal learner!
  11. Mar 4, 2007 #10

    My problem, as an eternal learner is

    I find in some of the derivations of the formula which accounts for length contraction the expression
    giving the time interval during which a light signal propagates forth and back along a moving a rod considering I thinck a composition of velocities. Are they correct? I think the problem is of intertest for many participants on the Forum.
  12. Mar 5, 2007 #11
    Of course is 2L/c(1-VV/cc).
    My question is: can you show one of those derivations? (It doesn't seem to me that those two terms L/(c-V) and L/(c+V) can have a real physical meaning).
    Thank you.
  13. Mar 5, 2007 #12

    Meir Achuz

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    I think that this line demonstrates that there are transparent derivations and confusing derivations. I am happy with a transparent derivation using simple algebra and the LT. I try not to waste my time on confusing derivations.
    This is not an insult, because I think you are just trying to say that those derivations are confusing.
  14. Mar 5, 2007 #13


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    I reserve the right to use any terms I think appropriate.
    If you start by using Gallilean velocity addition and then plug in relativistic velocities, it will not make sense.

    As LightArrow points out expressions like u-c and u+c do not have any meaning. Use the Lorentz formula to add velocities.

    I still don't understand what point your original post is making. Are you alerting out to us some errors in some text ? Maybe this is just a language problem and it's all a misunderstanding, so I bow out of this discussion.
  15. Mar 5, 2007 #14
    c+V, c-V

    I just remembered that in
    N. David Mermin, It's about time Princeton University Press 2005
    in which the Lorentz transformation are not mentioned you will find in many of its pages such terms.
  16. Mar 17, 2007 #15
    I don't have that book and I haven't seen that derivation, but it's simple to make it.

    The equation you wrote:
    deltat = L/(c-V) + L/(c+V) = 2L/c*(1-VV/cc)
    is correct.

    You obtain it from: L = (c-v)deltat_1 = (c+v)delta2
    where :deltat_1 is the interval of time from when the photon is sent towards the mirror, which is fixed in a rod moving along the same line and versus with speed v, and when it hits the mirror; deltat_2 is the i. of time in the reverse path, from when the photon hits that mirror to when hits the photon's source, which is solidal to the rod; L is the distance between source and mirror, as seen from a stationary observer, that is, the rod's lenght in the stationary observer ref. frame; deltat = deltat_1 + deltat_2 is the int. of time of the total path (forth and back).

    Infact, from the stationary observer:
    in the forth path the photon travels for c*deltat_1 while the source travels for v*deltat_1. So, the distance between the mirror and the source is:
    c*deltat_1 - v*deltat_1 = (c-v)*deltat_1 = L.

    In the back, the photon covers a path c*deltat_2 and the source a path v*deltat_2 and their sum must equals L: L = c*deltat_2 + v*deltat_2 = (c+v)delta2.

    So, you find:
    deltat = L/(c-V) + L/(c+V) = 2L/c(1-VV/cc) = (2L/c)/(1-beta^2)

    In the ref. frame of the moving rod:
    c*deltat' = 2L'.

    Now, since deltat' = deltat*sqrt(1 - beta^2) (time contraction):
    2L' = c*deltat' =
    c*deltat*sqrt(1 - beta^2) = c*[(2L/c)/(1-beta^2)]*sqrt(1 - beta^2) =
    = 2L/sqrt(1 - beta^2) --> L = L'*sqrt(1 - beta^2)
    which is the lenght contraction.
    (Very complicated way to obtain it).

    The time contraction law: deltat' = deltat*sqrt(1 - beta^2) can be obtained in a simpler way using Pitagora's theorem in the case of the rod moving with a velocity v transverse to its lenght. In that case you equals L in the two ref frames (we can consider the forth travel only):
    for a stationary observer L = sqrt[(c*deltat)^2 - (v*deltat)^2] =
    = deltat*sqrt(c^2 - v^2)
    in the rod's ref. frame: L = c*deltat'.
    So: c*deltat' = deltat*sqrt(c^2 - v^2)
    --> deltat' = deltat*sqrt(c^2 - v^2)/c = deltat*sqrt(1 - beta^2)

    However, the quantity L/(c-V) (for example) which represents the int. of time deltat_1, cannot be interpreted as the time in which a body travels the distance L at the speed c-V, because the photon travels a different distance during that int. of time, and the rod travels that distance but in a different int. of time; so that quantity is a physical time, but not travelled from a physical object.

    Similarly for L/(c+V).
    Last edited: Mar 17, 2007
  17. Apr 10, 2007 #16
    "c+v" and "c-v" are the foundations of the subject

    I would like to point out that the whole of special relativity is based upon just such additions as 1/(c+v) + 1/(c-v) - as exemplified in section 3 of Einstein's 1905 paper. It is precisely these formulae for relatively moving observers, where a clock that is Einstein synchronised at half the time between emission & reception of a to-&-fro signal, appears to lag according to an observer who sees a longer 1/(c-v) "to" time than the 1/(c+v) "fro" time, that gives rise to both relativity of simultaneity and reciprocal time dilation. This is essential knowledge, because anyone who only knows SR from the "Lorentz" transformations cannot be said to have understood special relativity at all, since they will be unable to distinguish the manifestly different kinematical nature of Einstein's SR from Lorentz's aether theory with its ad hoc hypothesis of length contraction.

    Incidentally, I find it strange that nobody ever seems to query the fact that although t and x appear perfectly symmetrically in the "Lorentz" transformations, one is treated as time "dilation" while the other is regarded as length "contraction".
  18. Apr 10, 2007 #17


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    Horse feathers.
  19. Apr 11, 2007 #18
    Unfortunately Boustrophedon, that is the terminology that was picked in the early days (Zeitdilatation and Längekontraktion) and it appears we are stuck with it.

    Is it confusing? You bet!
    Is it consistent? Not at all!

    The contraction in "length contraction" relates to the contraction of a spatial interval as measured by an observer in relative motion to it.
    The dilation in "time dilation" is actually a misnomer, it would be far more consistent to call it "time contraction" since clearly the temporal interval would contract as well under the same conditions.

    And while we are at it, the term "contraction" is not best term in the world either. Nothing physically contracts.

    But again, these terms are standard terms and taught widely.

    We just have to live with them. :smile:
    Last edited: Apr 11, 2007
  20. Apr 11, 2007 #19
    Bravo ! A penetrating example of how the meaning of the 'Lorentz' transformations differs markedly between the two theories. However, I think your "misnomer" is misplaced.....

    The inconsistency between time 'dilation' and length 'contraction' stems from the contrary way they are derived from the Lorentz tranformations. For time, two clock readings at the same position in S' ( i.e. the same clock in S' ) are compared to clock readings at two different places in S, whereas for length, the ends of a rod in S' at the same time in S are compared to their positions at two different times in S'. Now surely the first procedure is reasonable and shows that a time interval in S' appears longer from the point of view of S, and is supported experimentally by high speed particle decays.

    However, the procedure for length means that the front end is recorded before the rear end, from the point of view of the rod's frame and so cannot represent the true length. To be consistent with the first procedure the ends must be recorded at the same time in S'. A common example of how this might be done is a simultaneous flash or explosion at each end of the rod which leave marks in S. Thus I would argue that "length contraction" is the misnomer and that a consistent approach reveals length dilation.

    Another confirmatory way of showing this is for S' to measure the length of the rod by timing a light pulse sent along it to and from a mirror at one end. Since S finds that the clocks in S' run slow, S will find a greater time for the traverse and therefore deduce a longer length.
  21. Apr 11, 2007 #20
    You got it exactly upside down Boustrophedon.

    Take for instance a twin experiment. A stays home and B travels. The time interval for B is obviously shorter than for A.

    It has nothing to do with some kind of a dilation effect that makes clocks slow down. We are simply comparing lengths of intervals, and their ratios are related to the Lorentz factor.

    Clocks that really slow down, or measuring rods that really shrink are obviously not good measuring instruments, they belong in the wastebasket. :smile:
    Last edited: Apr 11, 2007
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