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

I Special relativity

  1. Dec 12, 2017 #1
    When expanding Kinetic energy as
    T=mc2-m0c2
    where m0=rest mass and
    m=m0/sqrt(1-β2)
    the first term correction is coming out to be -p4/8m03c2
    but taking T=E-m0c2
    where E=sqrt(c^2p^2+m02c4) the correction is coming to be 3p^4/8m03c2
    is there any contradiction?
     
  2. jcsd
  3. Dec 12, 2017 #2

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    You would need to show your work. Are you using the relativistic expression for ##p##?
     
    Last edited: Dec 12, 2017
  4. Dec 12, 2017 #3

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you trying to get ##\frac12 m_0v^2##?
     
  5. Dec 12, 2017 #4
    this term is coming from both but the second term is coming different
     
  6. Dec 12, 2017 #5

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    I think he's looking at the next term in the series.
     
  7. Dec 12, 2017 #6
    for the second case p remains as it is but for the first case I have taken p=m0v
     
  8. Dec 12, 2017 #7
    yes
     
  9. Dec 12, 2017 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You must have made a mistake with the binomial expansion in the first case.

    I don't see how you can take ##p = m_0 v##.
     
  10. Dec 12, 2017 #9

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    Well, that's likely the reason for the discrepancy in your two results!

    In the relation ##E=\sqrt{(pc)^2+(m_oc^2)^2}##, ##p## is the relativistic momentum given by ##\frac{m_o v}{\sqrt{1-\beta^2}}##, not the newtonian ##m_o v##.
     
  11. Dec 12, 2017 #10
    ok thanks, I have mislooked that thing
     
  12. Dec 12, 2017 #11
    but that doesn't match the coefficient of -1/8 and 3/8
     
  13. Dec 12, 2017 #12

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not sure how you apply the binomial expansion to that, given that neither term need be larger than the other.
     
  14. Dec 12, 2017 #13
    the term is coming as
    m0c2(-.5C2(-v2/c2)2)
     
  15. Dec 12, 2017 #14

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's a mess in any case.

    If you assume ##p## is small, then you can get an expansion in terms ##p/c##. But, ##p## has a gamma factor, so you cannot compare this expansion directly with the expansion involving ##v/c##.

    The first terms are ##\frac12 m_0v^2## and ##\frac{p^2}{2m_0}## respectively. But, these are not equal; and neither are the second terms, which in fact have different signs.
     
  16. Dec 12, 2017 #15

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    You should get ##T \approx m_oc^2\left(\frac{1}{2}\beta^2 + \frac{3}{8}\beta^4\right)##

    Mine is ##\frac{3m_ov^4}{8c^2}##.

    Replace ##m_o v## with ##p## and you get ##\frac{3p^4}{8m_o^3c^2}##.

    I'm not sure how you did that, but it matches what I got above!

    By the way, you may find this thread interesting:
    https://www.physicsforums.com/threads/kinetic-energy-and-momentum-of-a-relativistic-particle.895897/
    There the usual notation is used, where ##m## is the ordinary mass, what you are calling the rest mass.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted