Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bondi k calculus :why the k?

  1. Apr 13, 2006 #1
    i have this question.we are being taught sr using bondi k calculus.it is said here that 2 observers A and B are moving with some relative velocity with respect to each other. observer A sends a signal at his time t and B recieves it at B time t' .we say that t'=kt.my question is how do we know that such a linear relationship exists?is it an assumption or can it be justified on the basis of the postulates of SR?
     
  2. jcsd
  3. Apr 13, 2006 #2

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The answer to first question is think "similar triangles"... a scaling argument.
    The postulates of SR will suggest that t"=kt', reversing the roles of the emitter and the receiver. In this latter case, it's also "similar triangles [with a flip]".. but similarity in the Minkowski geometry sense: the "Minkowski-angle [rapidity] between a timelike ray and a lightlike ray is infinite".

    Check out my posts in
    https://www.physicsforums.com/showthread.php?t=113915

    By the way, what text are you using?

    (If you think about Lorentz Transformations from a matrix point of view, you'll find that its eigenvectors are the lightlike vectors and k and k-1 are eigenvalues.)
     
    Last edited: Apr 13, 2006
  4. Apr 13, 2006 #3
    can u elaborate on thesimilar triangles argument.
     
  5. Apr 13, 2006 #4
    wont this linear relation hold if i use any other wave instead of light?
     
  6. Apr 13, 2006 #5

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In the first case:
    draw a triangle with two future-pointing timelike-vectors from an event O, with a lightlike vector connecting the tips. Suppose that the elapsed time is t on source watch and the reception time is t' on the receiver's watch. Certainly, one can write: t'=kt, for that pair of t and t'. We want to show that we get the same k for any corresponding pair of T and T' on the respective watches.

    Well... by drawing a magnified triangle from O with longer timelike legs, you can see that this triangle is similar to the original triangle. All corresponding [Minkowski-]angles are congruent. So, the legs are in proportion. That is, T=at and T'=at'... so, since t'=kt, we have (T'/a)=k(T/a)... or T'=kT.... for any corresponding pair of T and T' on the respective watches.

    In the second case, we are going to eventually exchange the role of source and observer. We can physically argue via SR that the same relation with the same k must occur. Geometrically [which is probably not done to introduce the idea... but to verify its consistency later], we first Minkowski-"rotate" ("boost") the original triangle so that the receiver is vertical [which changes no "Minkowski"-angles or -lengths]. Under a Minkowski-"rotation", events slide along Minkowski-"circles", which are hyperbolas on a spacetime diagram. If it's hard to visualize, play around with Twin Paradox Applet (select "equal intervals from O" and "Light Cone of Q").

    Then reflect about the vertical axis... then scale upward so that the vertex off the vertical axis coincides with the vertex in the original triangle. This triangle is similar to the original triangle... so, t'' (on the original source-watch)=k t'(on the original receiver-watch).



    [tex]
    \]

    \unitlength 1mm
    \begin{picture}(55,90)(0,0)
    \linethickness{0.3mm}
    \put(20,10){\line(0,1){80}}
    \linethickness{0.3mm}
    \multiput(20,90)(0.12,-0.12){250}{\line(1,0){0.12}}
    \linethickness{0.3mm}
    \multiput(20,30)(0.12,0.12){250}{\line(1,0){0.12}}
    \linethickness{0.3mm}
    \multiput(20,10)(0.12,0.2){250}{\line(0,1){0.2}}
    \put(15,30){\makebox(0,0)[cc]{t}}
    \put(15,60){\makebox(0,0)[cc]{\gamma t}}
    \put(15,90){\makebox(0,0)[cc]{t''=k(t')=k^2t}}

    \put(55,60){\makebox(0,0)[cc]{t'=kt}}

    \end{picture}
    \[
    [/tex]
     
  7. Apr 13, 2006 #6

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The linear relation should hold with any other wave.
    However, I think you'll run into computational problems when you try to use the principle of relativity to get the important relation t''=k t', then try to relate k to the relative velocity. (Note that in the non-light case, the signal-speed of emission according to the emitter is different from the signal-speed of reception according to the receiver.)
     
    Last edited: Apr 13, 2006
  8. Apr 14, 2006 #7
    thanks robphy that was really helpful.
     
  9. Apr 14, 2006 #8

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're welcome.

    But you never answered what text you are using...
     
  10. Apr 14, 2006 #9
    im not using any text we are being taught using the k calculus in college and we havent been told to follow any particular book.
     
  11. Apr 14, 2006 #10

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    A more detailed graphic and argument as to why k is constant. (If I ever get the latex to work :-( )


    Light travels in straight lines. So we can use geometry.

    Trinagle (O,t,kt) and (O,2t, 2kt) are similar. Therefore we know that if light is emitted at the source at time t and arrives at the destination at time time kt, that light emitted at time 2t must arive at time 2kt - by geometry, and the fact that light travels in straight lines.

    How do we know the triangles are similar? The included angle is obviously the same, and the two light rays (t,kt) and (2t,2kt) must be parallel. The light rays must be parallel because they must cover equal distances (x-coordinate distance) in equal times (t-coordinate times), meaning that their slope, dx/dt, must be the same.

    The argument that light emitted from the destionationat time kt arrives back at the source at time k^2 t is based on the physical argument that the receiver and the transmitter can be interchanged, as there is no "absolute" motion, only relative motion.

    We can use similar triangles again to show that

    (O, kt, k^2t) and (O, 2kt, 2k^2t) are similar, completing the argument.

    [tex]
    \[

    \unitlength 1mm
    \begin{picture}(60.00,95.00)(0,0)

    \linethickness{0.15mm}

    \multiput(30.00,40.00)(0.12,0.24){83}{\line(0,1){0.24}}
    \put(40.00,60.00){\vector(1,2){0.12}}

    \linethickness{0.15mm}

    \put(20.00,20.00){\line(0,1){20.00}}
    \put(20.00,40.00){\line(0,1){20.00}}

    \linethickness{0.15mm}

    \multiput(20.00,20.00)(0.12,0.24){83}{\line(0,1){0.24}}
    \put(30.00,40.00){\vector(1,2){0.12}}

    \linethickness{0.15mm}

    \multiput(20.00,50.00)(0.12,-0.12){83}{\line(1,0){0.12}}
    \put(20.00,50.00){\vector(-1,1){0.12}}

    \linethickness{0.15mm}

    \multiput(20.00,40.00)(0.12,0.12){167}{\line(1,0){0.12}}


    \linethickness{0.15mm}

    \multiput(20.00,40.00)(0.12,0.12){167}{\line(1,0){0.12}}
    \put(40.00,60.00){\vector(1,1){0.12}}

    \linethickness{0.15mm}

    \multiput(20.00,80.00)(0.12,-0.12){167}{\line(1,0){0.12}}
    \put(20.00,80.00){\vector(-1,1){0.12}}

    \linethickness{0.15mm}

    \put(20.00,40.00){\line(0,1){40.00}}
    \put(20.00,80.00){\vector(0,1){0.12}}

    \put(0.00,60.00){\makebox(0,0)[cc]{}}

    \linethickness{0.15mm}

    \multiput(20.00,30.00)(0.12,0.12){83}{\line(1,0){0.12}}
    \put(30.00,40.00){\vector(1,1){0.12}}

    \linethickness{0.15mm}

    \put(10.00,30.00){\makebox(0,0)[cc]{}}

    \put(18.00,30.00){\makebox(0,0)[cr]{T=t}}

    \put(18.00,40.00){\makebox(0,0)[cr]{T=2t}}

    \put(18.00,50.00){\makebox(0,0)[cr]{T=k^2 t}}

    \put(32.00,40.00){\makebox(0,0)[cl]{T'=kt}}

    \put(42.00,60.00){\makebox(0,0)[cl]{T'=2kt}}

    \put(18.00,80.00){\makebox(0,0)[cr]{T=2k^2t}}

    \put(60.00,60.00){\makebox(0,0)[cc]{}}

    \put(30.00,40.00){\makebox(0,0)[cc]{}}

    \linethickness{0.15mm}

    \put(20.00,80.00){\line(0,1){10.00}}
    \put(20.00,90.00){\vector(0,1){0.12}}

    \put(20.00,92.00){\makebox(0,0)[bc]{T axis}}

    \put(20.00,90.00){\makebox(0,0)[cc]{}}

    \linethickness{0.15mm}

    \multiput(40.00,60.00)(0.12,0.24){125}{\line(0,1){0.24}}
    \put(55.00,90.00){\vector(1,2){0.12}}


    \put(55.00,92.00){\makebox(0,0)[bc]{T' axis}}

    \put(55.00,70.00){\makebox(0,0)[cc]{}}

    \put(60.00,60.00){\makebox(0,0)[cc]{}}

    \put(85.00,75.00){\makebox(0,0)[cc]{}}

    \put(80.00,85.00){\makebox(0,0)[cc]{}}

    \put(18.00,20.00){\makebox(0,0)[cr]{O}}

    \end{picture}

    \]

    [/tex]
     
    Last edited: Apr 14, 2006
  12. Apr 15, 2006 #11

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Here is a simpler diagram with the same argument as above. Thanks to robphy for helping me learn how to use the jpicedt tool and fix-up the latex so that it will display properly on the forum.

    [tex]

    \[

    \unitlength 1mm
    \begin{picture}(85.00,95.00)(0,0)

    \linethickness{0.15mm}
    \put(20.00,0.00){\line(0,1){90.00}}


    \linethickness{0.15mm}
    \put(20.00,90.00){\line(0,1){0.12}}

    \linethickness{0.15mm}
    \put(20.00,90.12){\line(0,1){0.12}}
    \put(20.00,90.24){\vector(0,1){0.12}}

    \linethickness{0.15mm}
    \multiput(65.12,90.24)(0.12,0.24){1}{\line(0,1){0.24}}
    \put(65.24,90.48){\vector(1,2){0.12}}

    \linethickness{0.15mm}
    \multiput(40.12,40.12)(0.12,0.12){1}{\line(1,0){0.12}}
    \put(40.24,40.24){\vector(1,1){0.12}}

    \linethickness{0.15mm}
    \multiput(60.12,80.12)(0.12,0.12){1}{\line(0,1){0.12}}
    \put(60.24,80.24){\vector(1,1){0.12}}


    \linethickness{0.15mm}
    \multiput(20.00,0.00)(0.12,0.24){375}{\line(0,1){0.24}}
    \put(65.00,90.00){\vector(1,2){0.12}}


    \linethickness{0.15mm}
    \multiput(20.00,20.00)(0.12,0.12){167}{\line(1,0){0.12}}
    \put(40.00,40.00){\vector(1,1){0.12}}

    \linethickness{0.15mm}
    \multiput(20.00,40.00)(0.12,0.12){333}{\line(1,0){0.12}}
    \put(60.00,80.00){\vector(1,1){0.12}}

    \put(20.00,0.00){\makebox(0,0)[cr]{O\,}}

    \put(20.00,20.00){\makebox(0,0)[cr]{A\,}}

    \put(40.00,40.00){\makebox(0,0)[cl]{\,B}}

    \put(20.00,40.00){\makebox(0,0)[cr]{C\,}}

    \put(60.00,80.00){\makebox(0,0)[cl]{\,D}}

    \put(20.00,92.00){\makebox(0,0)[bc]{T axis}}

    \put(65.00,92.00){\makebox(0,0)[bc]{T' axis}}

    \put(20.00,20.00){\makebox(0,0)[tl]{\beta}}

    \put(20.00,40.00){\makebox(0,0)[tl]{\beta}}

    \put(20.00,7.00){\makebox(0,0)[bl]{$\alpha$}}

    \end{picture}

    \]

    [/tex]

    In the above diagram, we have a stationary observer, represented by the T axis, who sends out light signals at events A and C.

    We also have a moving observer, represented by the T' axis, who recieves the emitted light signals at events B and D, respectively.

    The time interval between events O and A according to the stationary observer is represented by [itex]\overline{OA}[/itex], similarly the time interval between events O and C is represented by [itex]\overline{OC}[/itex].

    Similarly, the time interval between events O and B according to the moving obserer is represented by [itex]\overline{OB}[/itex], and the time interval between events O and D is represented by [itex]\overline{OD}[/itex].

    We wish to prove that the ratio of the time intervals between received and transmitted events is a constant, i.e. we wish to show that

    [tex]
    \overline{OB}/\overline{OA} = \overline{OD}/\overline{OC} = k
    [/tex]

    where k is some constant. The fact that light travels in straight lines, and that the speed of light relative to the stationary observer is constant is sufficient to prove the above statement.

    We do this by noting that the triangles AOB and COD are similar. We note that the angle marked [itex]\alpha[/itex] is common to both triangles. The two angles marked [itex]\beta[/itex] on the space-time diagram must be equal, as they are formed by the angle between a light beam and the T axis. Because the speed of light relative to the stationary observer on the T axis does not change with time, the slope of the light beam cannot change with time, hence the angle that a light beam forms with the T axis on the space-time diagram must not change with time.

    Given that two angles of the triangles are equal, and that the sum of the angles of a triangle is 180 degrees, all three angles of the triangles must be equal, and the triangles must be similar.

    Because the triangle are simlar, the ratios of the sides are constant. This proves the desired result that

    [tex]
    \overline{OB}/\overline{OA} = \overline{OD}/\overline{OC} = k
    [/tex]
     
    Last edited: Apr 15, 2006
  13. Apr 16, 2006 #12
    i think that Bondi's k calculus has its origins in the radar detection and in the Doppler effect. its linearity consists in the fact that to one time in one reference frame corresponds a single time in another inertial frame
     
  14. Apr 16, 2006 #13

    Garth

    User Avatar
    Science Advisor
    Gold Member

    Actually K-Calculus was the invention of E.A.Milne and found in his books "Relativity, Gravitation and World-Structure", 1935 (before the invention of radar), and its sequel "Kinematic Relativity", 1948, both Oxford at the Clarendon Press.

    Milne's cosmology, kinematic relativity, was so-called because of its tight links to the kinematics of Einstein's Special Theory of Relativity.

    Kinematics, in Einstein's Special Theory, is especially concerned with observers trying to make measurements of the behavior of objects moving in systems relative to themselves. According to Milne's understanding, these measurements could only be made by the observers signalling among themselves using light or other electromagnetic radiation, using clocks to time the signals.

    Milne recognised that cosmology is only about observed photons that have arrived from the depths of space and asked of any cosmological concept, "How do we measure it?"

    Given the hypothesis of the invariance of the speed of light Milne allowed equivalent observers only to send and receive light signals and measure time intervals using their own clocks. Distances were thus 'radar' distances, even before its invention.

    As Bondi later remarked, “Milne's idea led straightaway to the radar speed gun!”

    Garth
     
    Last edited: Apr 16, 2006
  15. Apr 16, 2006 #14
    k calculus

    the diagram is presented by Asher Peres in Am.J.Phys.
     
  16. Apr 16, 2006 #15

    Garth

    User Avatar
    Science Advisor
    Gold Member

    Yes - Milne uses a more primitive form of the diagram on page 48 of "Relativity, Gravitation and World-Structure", (1935).

    Garth
     
  17. Apr 16, 2006 #16

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you referring to "Relativistic telemetry" by Asher Peres, Am. J. Phys. 55, 516 (1987) [keyword search for "telemetry" in http://scitation.aip.org/ajp/ ]? That paper doesn't show the full spacetime diagram [with the Minkowski-similar triangles with timelike legs] in my earlier post above... although it could be deduced from it.


    I only have the 1948 edition on my shelf. Thanks to your comment on Milne's work predating Bondi's, I've opened it up and looked at it in more detail. Indeed, the k-calculus calculations are there for special relativity [in the 1948 edition]... and Milne's approach can handle some additional cases.

    However, in the 1948, "Fig 3 on p. 40" with "A's diagram and B's diagram" appears to be a pair of spatial diagrams, rather than spacetime diagrams. Is this true in the 1935 edition as well? If so, then Milne's diagram doesn't reveal the geometrical structure [i.e. the similarity of the triangles] in Minkowski spacetime... although it can be deduced from his algebraic equations.

    (digression from this thread:
    Milne is apparently interested in pursuing a modified theory of general relativity in which "galaxies constitute the natural frames of reference". Here are some webpages with some discussion:
    http://world.std.com/~mmcirvin/milne.html
    http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1996QJRAS..37..365W
    http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1936ApJ....83...61R
    http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1941Obs....64...11.
    However, this digression is really off-topic.)
     
  18. Apr 16, 2006 #17

    Garth

    User Avatar
    Science Advisor
    Gold Member

    You are correct, Milne's motivation was to reproduce the results of SR and generalise it to cosmology without conceeding space-time as a continuum. Therefore his diagrams only reflect 3D space with time as a parameter. In that sense his version of 'K-Calculus' is not geometric, only algebraic.

    Garth
     
  19. Apr 20, 2006 #18
    so would you say that k is constant because of homogenity of time?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Bondi k calculus :why the k?
  1. K=-1 closed space (Replies: 3)

Loading...