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Kenneth Krane's Modern Physics muon example (SR)

  1. Oct 17, 2012 #1
    This is the frirst SR problem in his book. I think the question is flawed.

    The solution assumes a speed of 3.00 E8 m/s to calculate Δt=333us since it traveled 100km.

    Next, 333us and 2.2us is used to solve for velocity using the time dilation formula. Velocity is reported to be .999978c.


    This does not make any sense. If you calculate Δt using C as the speed, then C should be the answer at the end, after you solve for velocity using the time dilation formula.

    Shouldnt .999978c be reported as 3.0 E8 m/s due to the significant figures?


    The answer you get at the end just ends up being whatever value you assumed for "very high speeds."

    Isn't this question impossible to answer unless you are given the actual velocity for the muon????


    Thanks :-)




    PS.

    The next example in the book is the lengh contraction example for the muons reference frame. It uses 0.999978c as the velocity.
     
  2. jcsd
  3. Oct 17, 2012 #2

    mfb

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    Let's assume the initial assumption "speed close to c" would be wrong. What happens? You get a larger travel time, which requires a larger gamma factor - this corresponds to an even higher speed (compared to .999978c). That is a contradiction, and the assumption "speed close to c" has to be right.

    You can do it properly and introduce a parameter for the velocity, but then you get (at least) quadratic equations. And you don't gain much, as the speed is really close to c anyway.


    However, the question itself has a big flaw:
    There is no minimal speed. 2.2µs is the average decay time - but some muons survive longer. In particular, about 0.1% of the muons will live longer than 15µs (10*half-life), and 0.0001% longer than 30µs (20*half-life).
     
  4. Oct 17, 2012 #3

    Nugatory

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    No, it can be answered. A fair paraphrase of the problem would be:

     
  5. Oct 17, 2012 #4
    Perhaps you mean "reported as 3.00 E8 m/s"?
    No. Here is a similar example, but simpler: suppose that you know that a door opening is 1.00 m wide (w=1.00 m), and you also know (don't ask how) that a table is 0.9990w m wide.
    How much tolerance do you have if you move the table horizontally on wheels through the door? The answer is, at best 0.5 mm on each side - no matter if the door is 995 mm wide or 1005 mm wide. Check out what happens if you round the table size to 1.00 m for your calculation. :wink:

    Note however the very good remark by mfb: the question is wrong, but for a different reason than what you thought!
     
  6. Oct 17, 2012 #5
    I meant 3.0e^8 because final equation involves using 2.2us.

    Doesnt that require the answer to be two sig figs?

    0.999978c = 2.999934 E8 m/s
    Which should be rounded to 3.0 E8m/s (due to the 2.2us)


    Actually, I knew the problem was simplified in the half life aspect. :-)


    Thanks for the help everyone. I'm just getting back into physics and I'm rusty with it all.
     
    Last edited: Oct 17, 2012
  7. Oct 17, 2012 #6
    No, it is not.

    It tells you that seen from an observer at rest on earth, the distance to the myon is 100km.

    In the frame the myon is at rest in however, this distance is not 100km,

    but L = Lo * sqrt(1-(v^2/c^2))

    You have Lo, you have c -

    You do not have L or v directly, but you know how long the (unrealistic as mfb pointed out) myon survives in the frame it is at rest in. 2.2us

    Therefore, since the myon in the frame it is at rest in stays alive only for 2.2us, earth traveling towards the myon (remember the myon considers itself at rest) at vrel will shorten the distance by vrel * 2.2us. If vrel was too small to shorten the distance of L the myon sees between itself and earth, then the myon will be dead before earth reaches it. So vrel*2.2us = L if you want the myon to survive that long.
    Luckily, vrel is the same for both and you can solve the equation now.

    vrel * 2.2us = Lo * sqrt(1-(vrel^2/c^2))
     
    Last edited: Oct 17, 2012
  8. Oct 17, 2012 #7
    Once more, no. Did you not understand my example? It's just the basics of correctly working with inaccuracies.

    1. What will happen if you do that with the table? Try it!
    2. And with the time dilation factor?

    Answers below. :wink:


    1. You should then find, uselessly, around zero tolerance (because you removed the significant figure)
    2. You should then find, uselessly, up to infinite time dilation
     
    Last edited: Oct 17, 2012
  9. Oct 17, 2012 #8

    mfb

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    You cannot use the concept of significant figures here, it does not work. You can assume that the lifetime has an uncertainty of 0.1µs and use proper error propagation, and you will see that the relative uncertainty of v-c is similar to the relative uncertainty of the lifetime (maybe with a factor of 2).

    In other words, "0.000022c" has an uncertainty of about 0.000001c, and your result (0.999978c) has the same uncertainty. Its uncertainty is about 1 at the last digit.
     
  10. Oct 18, 2012 #9
    He/she can use the concept of significant figures here, and I started to explain how. Anyway, you indicate the same already, as I want to show rear naked that the relevant term is c-v (after he/she does a basic subtraction exercise!).
     
    Last edited: Oct 18, 2012
  11. Oct 18, 2012 #10
    from earth's frame, the time taken is given by
    [tex]t=\frac{d}{v}[/tex] [tex] [/tex]
    where [tex]d[/tex] is 100km and [tex]v[/tex] is the relative velocity of the muon and earth.



    the relativistic time interval of the muon is [tex]\frac{t}{γ}=\frac{d}{γv}[/tex]



    in order for the muon to reach earth [tex]\frac{d}{γv}≤τ[/tex]

    where [tex]τ[/tex] is the muon's lifetime.




    solve the inequality
    [tex]\sqrt{1-\frac{v^{2}}{c^{2}}}×\frac{d}{v}≤τ[/tex]

    to find the minimum speed.
     
    Last edited: Oct 18, 2012
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