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Mass growing with speed

  1. May 30, 2017 #1
    1. The problem statement, all variables and given/known data
    Fast electrons in synchrotron approach the speed of light:
    Snap2.jpg
    I don't get to the "one part in 8,000,000"

    2. Relevant equations
    Momentum: ##~P=mv=\frac{m_0v}{\sqrt{1-v^2/c^2}}##

    3. The attempt at a solution
    $$\frac{v^2}{c^2}=1-\frac{1}{4E6}~\rightarrow~\frac{v}{c}=\frac{1}{1.000,000,5}$$
    $$\rightarrow~v=\frac{c}{1.000,000,5}$$
    $$c-v=c\left( 1-\frac{1}{1.000,000,5} \right) \neq c\cdot \frac{1}{8,000,000}$$
     
  2. jcsd
  3. May 30, 2017 #2

    scottdave

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    1 part in 8 million is 1/8000000. But since it differs from 1 by this much, you have 7,999,999 / 8,000,000
     
  4. May 30, 2017 #3

    haruspex

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    How do you get that? What is the Taylor expansion of (1-¼10-6)?
     
  5. May 31, 2017 #4
    I executed ##~1-\frac{1}{4E6}##
    What do you mean by (1-¼1016)? maybe you made a mistake, do you mean the Taylor expansion for ##~\sqrt{1-\frac{1}{4E6}}~##?
    Taylor expansion:
    $$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+...$$
    It has a function f(x) in it and i don't, i just have numbers, so how will i take derivative?
     
  6. May 31, 2017 #5

    haruspex

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    Yes, sorry, the - in the exponent was a typo. Should have been (1-¼10-6)½. And I meant binomial expansion. Other than that....
     
  7. May 31, 2017 #6

    scottdave

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    I am not sure why you need Taylor Series to approximate it. You could do this: let r = (v/c). Then we have (1 - r^2) = (1 + r)(1 - r).
    Since v differs from c by 1 part in 8 million, then (1 + r) is approx 2 {since r is approx 1}, and (1 - r) is 1 divided by 8 million. So you have: (1 - r^2) is approx 2/(8 million) = 1/(4 million).
    The square root of that is 1/2000. You could run this in reverse, if you know the 2000 and want to determine the 1/(8 million).
     
    Last edited: May 31, 2017
  8. May 31, 2017 #7

    hilbert2

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    I've seen the Taylor polynomial approximation used in quantum mechanics if there's a need to include the relativistic "mass increase" in the form of a lowest order perturbation in the Schrödinger equation, but I'm not sure why that is done here.
     
  9. May 31, 2017 #8

    PeroK

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    This is all based on, for small ##\epsilon##:

    ##\frac{1}{1 \pm \epsilon} = 1 \mp \epsilon##

    And

    ##(1 + \epsilon)^2 = 1 + 2\epsilon##

    In particular, if ##\frac{v}{c} = 1- \epsilon##, then ##\frac{v^2}{c^2} = 1- 2\epsilon## and ##1 - \frac{v^2}{c^2} = 2\epsilon##
     
  10. May 31, 2017 #9

    Ray Vickson

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    The positive solution of ##1/\sqrt{1-\beta^2} = 2000## is ##\beta = \sqrt{3,999,999}/2000##, so ##1 - \beta = 0.1250 \times 10^{-6} = 1/8,000,000.##

    Actually, that last should be ##1/7,999,999.500000031250##, approximately
     
    Last edited: May 31, 2017
  11. May 31, 2017 #10
    But in binomial expansion the exponents are positive integers
    Why?
    $$c-v=\frac{1}{8E6},~1-r=1-\frac{v}{c}=\frac{c-v}{c}=\frac{1/8E6}{c}=\frac{1}{c\cdot 8E6}$$
    But you don't use ##\frac{1}{1 \pm \epsilon} = 1 \mp \epsilon##, you logically and correctly say ##\frac{v}{c} = 1- \epsilon##
    $$\frac{v}{c} = 1- \epsilon~\rightarrow~\frac{v^2}{c^2}=(1-\epsilon)^2=1-2\epsilon+\epsilon^2 \approx 1-2\epsilon$$
    $$1-\frac{v^2}{c^2}=\frac{1}{4E6}=2\epsilon~\rightarrow~\epsilon=\frac{1}{8E6}$$
    $$\frac{v}{c}=1-\frac{1}{8E6},~~c-v=\frac{c}{8E8}$$
     
  12. May 31, 2017 #11

    PeroK

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    That looks right to me.
     
  13. May 31, 2017 #12

    scottdave

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    You should notice something that when you see: c - v = 1/(8 E6). On the left side, you have speed (Length/Time), and the right side is dimensionless. So something didn't get put in, correctly. Now with this: c-v = c/(8 E6), both sides have the same dimensions.
     
  14. May 31, 2017 #13
    c-v = c/(8 E6) has correct dimensions and, as PeroK said, is probably the correct answer but how do you explain "v differs from c by one part in 8,000,000"?
    I expected to see c-v=1/8,000,000
     
  15. May 31, 2017 #14

    PeroK

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    Well, you expected wrong. That equation is dimensionally nonsense.
     
  16. May 31, 2017 #15
    How did you get that?
    $$\frac{1}{\sqrt{1-\beta^2}}=2000~\rightarrow~\sqrt{1-\beta^2}=0.0005~\rightarrow~\beta=0.999,999,875$$
     
  17. May 31, 2017 #16

    haruspex

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    Which differs from 1 by one part in 8 million.
     
  18. May 31, 2017 #17

    haruspex

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    Please see my correction in post #5.
     
  19. May 31, 2017 #18

    Ray Vickson

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    I am lazy in my old age, so I just let Maple solve the problem for me, and then evaluated the solution using 20-digit numerical computations.

    Of course, the solution of ##1/\sqrt{1-\beta^2} = \gamma## is ##\beta = \sqrt{1-\gamma^{-2}},## which expands out as
    $$\beta = 1 -\frac{1}{2 \gamma^2} -\frac{1}{8 \gamma^4} - \cdots,$$
    by applying the series expansion for ##\sqrt{1-x}## for small ##x = 1/\gamma^2.## If you keep only the ##1/\gamma^2## term you get exactly 1 part in 8,000,000 because we are using ##\gamma = 2000.##
     
  20. May 31, 2017 #19

    scottdave

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    What if it said something like v differed from c by 1%, would it be clear that v = 0.99c ?
    Instead of 1% they could say v differs from c by 1 part in 100. It means the same thing.

    So instead of 1 part in 8million difference, they could say differ by 0.0000125%
     
  21. Jun 1, 2017 #20
    Thank you very much scottdave, PeroK, Ray, Hllbert2 and Haruspex
     
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