1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic Energy Derivation

  1. May 12, 2014 #1
    Whilst reading following a derivation of the Relativistic Energy equation I came across the following:

    d/dt[mu/(1-u2/c2)1/2] = [m/(1-u2/c2)3/2] du/dt.

    I was wondering how that step was done.
     
  2. jcsd
  3. May 12, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    This is just the chain rule and then some simplifications.
     
  4. May 12, 2014 #3
    I see it now, thanks.
     
  5. May 13, 2014 #4
    Okay, so I now have:

    d/dt[mu/(1-u2/c2)-1/2]

    = m((1-u2/c2)-1/2) + mu(-u/c2).(1-u2/c2)-3/2

    I'm not sure how to simplify from here, any hints?
     
  6. May 13, 2014 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I have no clue, how you came to this derivative. First write the whole thing a bit more clearly with LaTeX:
    [tex]E=\frac{m c^2}{\sqrt{1-u^2/c^2}}.[/tex]
    Here, [itex]m[/itex] is the invariant mass of the particle and [itex]u=|\vec{u}|[/itex] the usual (non-covariant) velocity with respect to the computational (inertial) frame.

    Now taking the derivative with respect to time, you indeed just have to use the chain rule:
    [tex]\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{\mathrm{d} \vec{u}}{\mathrm{d} t} \cdot \vec{\nabla}_{u} E.[/tex]
     
  7. May 13, 2014 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Calu is calculating dp/dt using the product rule, but there's a typo, an extra negative sign in the first line. There's another sign error in the second line.
     
  8. May 13, 2014 #7
    Derivation of Relative Energy Equation

    I have no idea how to write in Latex, sorry, I'll type out the entire thing:

    Assume m and c are constants.

    W = ∫x2x1 F . dx = ∫x2x1 dp/dt . dx

    dp/dt = d/dt [mu . (1-u2/c2)-1/2]

    by the product rule:

    dp/dt = m(1-u2/c)-1/2) + mu.(1-u2/c)-3/2 . (-u/c2)

    Then as γ = (1-u2/c)1/2

    dp/dt = m (γ - u/c23))

    I'm not sure how to simplify to get the intended answer of m(1-u2/c)-3/2 . du/dt
     
    Last edited: May 13, 2014
  9. May 13, 2014 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I forgot to point out another error. You've differentiated with respect to u, not with respect to t.
     
  10. May 13, 2014 #9

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Well that is easy to fix as m and c do not depend on t, but it is important to get it right.

    You can combine both fractions to a single one if you expand the first one (the one that gives γ). Then simplify and you get the right result.
     
  11. May 13, 2014 #10
    What do you mean by expand? Sorry I'm very confused.
     
  12. May 13, 2014 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Not quite. You're still have an algebra mistake. The factor of 2 in the second term shouldn't be there.
     
  13. May 13, 2014 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    @vela: Which factor of 2?

    Expand the fraction

    Combining two fractions to a single one is taught several years before special relativity, you should know how to do that.
     
  14. May 13, 2014 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Calu had posted another attempt but edited it out or deleted the post after I replied.
     
  15. May 13, 2014 #14
    Oh I see, we don't call that "expanding" a fraction, sorry for the confusion.

    Also I edited the post because I thought it easier just to continue with what we had.

    EDIT: I'm sorry but could you help me out? I'm not sure what I'm meant to do. And as far as I know we have been taught little about "expanding" fractions as we are usually told to simplify (reduce) them. I can't think of a time I've been asked to do the opposite. Usually to combine 2 fractions I would cross multiply to get a common denominator.

    Let me just step away from this a second, because I have done the above (cross multiplied and found a common denominator) and have arrived at an answer similar to one I arrived at earlier but decided I could go no further.

    The answer I have is as follows:

    dp/dt = m [γ - u2/c23)]

    Is there a way to take out a factor of (1-u2/c2) (i.e. γ2) from this? Or am I completely wrong?

    And I'll post my result from combining the fractions:

    [m(1-u2/c2)3/2) - m(u2/cc)(1-u2/c2)1/2)] / (1-u2/c2)2

    Which when we replace with γ is:

    m(γ3 - (u2/c2)γ)

    which as I said is similar to the answer I got earlier.
     
    Last edited: May 13, 2014
  16. May 13, 2014 #15

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    First take a factor of ##\gamma## outside. Then rewrite the ##\gamma## that remains on the inside in terms of u. Then simplify and think about what the simplified result looks like.

    Edit: I don't think that minus sign is correct.
     
    Last edited: May 13, 2014
  17. May 13, 2014 #16
    Right, so I finally did this after going back and differentiating wrt t as I was supposed to.

    d/dt [mu(1-u2/c2)-1/2]

    = m.du/dt.(1-u2/c2)-1/2 + (-1/2)mu((1-u2/c2)-3/2.2u/c2.du/dt

    =m.du/dt.(1-u2/c2)-1/2 + u2/c2.m.((1-u2/c2)-3/2.du/dt

    = m(1-u2/c2 + u2/c2)(1-u2/c2)-3/2).du/dt

    =m(1-u2/c2)-3/2).du/dt as required.

    Also, how would I solve:

    m ∫uo [u(1-u2/c2)-3/2] . du

    Would it be integration by parts?
     
    Last edited: May 13, 2014
  18. May 13, 2014 #17

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Looks good except that in the second line, 2u/c^2 should be (-2u/c^2), and there's also a left parenthesis too many.

    No need to use integration by parts. You're supposed to write the integrand in a form that enables you to just use the formula ##\int_a^b f'(x)dx=f(b)-f(a)##.

    The work can be expressed as ##\int_0^u mu\gamma^3\mathrm du## or as ##\int_0^t mu\gamma^3\dot u\mathrm dt##. Does either of the integrands look like a derivative?
     
  19. May 14, 2014 #18
    I'm not sure how you would decide whether something "looks like" a derivative.
     
  20. May 14, 2014 #19

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What I meant is that you will have to see that at least one of them is a derivative. Is either of the integrands equal to a derivative that you calculated earlier?
     
  21. May 14, 2014 #20
    I managed to perform the integration by using a substitution of z = 1 - u2/c2.

    I still don't see how either is equivalent to a derivative I calculated earlier though.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted