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Classic and Relativity Kinetic Energy

  1. Nov 15, 2006 #1
    It is said that at low speeds the relativistic kinetic energy formula will give you the same answers as the classical KE formula. I tried this, and it doesn't work. I get different answers.
    I am just wondering what is amiss here.

    How would one derive the Classical KE from the Relativistic KE?

    Thanks for the help,
    Matt
     
  2. jcsd
  3. Nov 15, 2006 #2

    dextercioby

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    Well

    [tex] KE=m_{0}c^{2}\left(\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} -1\right)\simeq m_{0}c^{2}\left(1+\frac{v^{2}}{2c^{2}}-1\right) =m_{0}\frac{v^{2}}{2} [/tex]

    Daniel.
     
  4. Nov 15, 2006 #3

    jtbell

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    Show us an example where you get different results (show your work!) and someone can probably tell you where you went wrong.
     
  5. Nov 15, 2006 #4
    Just to clarify what dextercioby said, the approximation comes from the binomial expansion
    [tex](x+y)^r=\sum_{k=0}^{\infty} {r \choose k}x^ky^{r-k}[/tex]
    [tex](x+y)^r=\sum_{k=0}^{\infty} \frac{r!}{k!\,(r-k)!}x^ky^{r-k}[/tex]
    which we can use to expand
    [tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-0.5}[/tex]

    The first few terms of the expansion are
    [tex]1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\ldots[/tex]

    So we have
    [tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \simeq 1+\frac{v^2}{2c^2}[/tex]
    which can be used, as shown above, to show that the classical expression for kinetic energy approximates the relativistic expression.
     
    Last edited: Nov 15, 2006
  6. Nov 15, 2006 #5

    robphy

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    As written, you'd probably have to explain how to use r= -1/2 in the general formulas you've written.

    It may be conceptually simpler to emphasize the more fundamental Taylor series expansion.
     
  7. Nov 15, 2006 #6
    ah, i forgot about the non-integer r in the 'r choose k'..

    [tex]{r \choose k}={1 \over k!}\prod_{n=0}^{k-1}(r-n)=\frac{r(r-1)(r-2)\cdots(r-(k-1))}{k!}[/tex]

    according to Wikipedia's binomial theorem page
     
    Last edited: Nov 15, 2006
  8. Nov 15, 2006 #7

    rbj

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    you have a small problem assuming an analytic continuation that a formula that is shown valid for integer r is also valid for a non-integer. it is true, you just didn't make the connection. one that doesn't start with the binomial expansion but uses Taylor series is

    [tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \left( {1-\frac{v^2}{c^2}} \right)^{-\frac{1}{2}}

    \quad = e^{-\frac{1}{2} \log\left( 1-\frac{v^2}{c^2} \right)}}

    \quad \approx 1 - \frac{1}{2} \log \left( 1-\frac{v^2}{c^2} \right)

    \quad \approx 1 - \frac{1}{2} \left( -\frac{v^2}{c^2} \right) [/tex]

    for [itex] |v| << |c| [/itex].

    get's the same thing.

    i was also gratified that dextercioby used the symbol [itex]m_0[/itex] which seems to be out of vogue.
     
    Last edited: Nov 15, 2006
  9. Nov 16, 2006 #8

    dextercioby

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    There's an analytic continuation of the binomial formula using the Eulerian gamma function, but yeah, i still prefer Taylor's polynomial.

    Daniel.
     
  10. Nov 16, 2006 #9
    Sorry, I didn't mean to add confusion. I just thought it seemed quite mysterious in the first post that [tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\simeq1+\frac{v^2}{2c^2}[/tex] and wanted to explain where that came from.

    btw in the Taylor series example given above, why use [tex]e^{-\frac{1}{2}}log(1-\frac{v^2}{c^2})\simeq1-\frac{v^2}{2c^2}[/tex]? This seems quite mysterious to me since we can write [tex]e^{-\frac{1}{2}}e^{log(1-\frac{v^2}{c^2})} \equiv e^{-\frac{1}{2}}(1-\frac{v^2}{c^2})[/tex], which is exact.

    There is nothing wrong with doing it this way of course... I just don't like the idea of throwing away precision just to make the answer fit. [tex]e^{-\frac{1}{2}}[/tex] is more like 0.6 than [tex]\frac{1}{2}[/tex]
     
    Last edited: Nov 16, 2006
  11. Nov 16, 2006 #10

    dextercioby

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    You've made a mistake on the e^(ab) expansion. It's not e^a e^b

    Daniel.
     
  12. Nov 16, 2006 #11
    Ah! Thanks! Pretty fundamental mistake I made there, a^b+c = a^b a^c.

    I thought it was odd that someone would consider [tex]0.6\simeq\frac{1}{2}[/tex]. Still, it leaves the question (for me at least) of where that approximation comes from?

    EDIT: Never mind I just worked it out... I think it comes from the power series for [tex]e^x[/tex]
     
  13. Nov 16, 2006 #12

    rbj

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    but
    [tex]e^{-\frac{1}{2}}e^{log(1-\frac{v^2}{c^2})} [/tex]
    is not the same as
    [tex]e^{-\frac{1}{2}log(1-\frac{v^2}{c^2})}[/tex].

    allowing analytical extension from derivatives of integer power functions to non-integer, we know that

    [tex] (1+x)^\alpha \approx 1 + \alpha x [/tex]
    for small |x|.

    so there is a simple expression for what we need. i did the log - exp thing to avoid making that analytical extension since they were defined for noninteger arguments.

    yes, there were two applications of approximation instead of just one. but i didn't need to make an assumption of analytical extension and both approximations are good for small |x|.
     
    Last edited: Nov 16, 2006
  14. Nov 17, 2006 #13
    Sorry about that, I got confused about exponentiation rules, as stated above.
     
  15. Sep 9, 2011 #14
    Is this equation for Relativistic Kinetic Energy really proven?
     
  16. Sep 9, 2011 #15

    clem

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    Yes, millions of times each day at the LHC and other laboratories.
     
  17. Sep 9, 2011 #16

    PAllen

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    Hey, I always just use as every day fact that for small x:

    sqrt(1+x) := 1 + x/2

    1/(1+x) := 1-x

    The former can be derived as trivially as:

    (1+x)^2 = 1 + 2x + x^2

    the latter from sum of geometric series, or even just subtracting and verifying the error is order x^2.
     
  18. Sep 10, 2011 #17
    And it results from Lorentz transformation (which is also proven experimentally).
     
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