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KE in reletivity? Strange integration

  1. Mar 14, 2005 #1
    I'm reading through a modern physics text (on my own) and it did some very strange manipulation when deriving the relativistic form of the kinetic energy equation. It starts off deriving relativistic force which I understand. Basically, the implicit derivative of (d/dt)(ymv). Y is gamma and is equal to the Lorentz ratio 1/sqr(1-v^2/c^2). (I do not know how to write the notation.) Work or kinetic energy(if no work goes to potential energy) is equal to the integral from 0 to s of Fds (Must know how to write notation of this site.) With substitution we get the integral from 0 to s of (d/dt)(ymv)ds. This is where the book does some strange things. The author claims that the integral for 0 to s of (d/dt)(ymv)ds is equal to the integral from 0 to mv of v*d(ymv), and once y is substituted in, the integral is equal to the integral from 0 to v of
    v*d(mv/sqr(1-v^2/c^2)). He has changed the bounds of his integration twice in a way that I do not understand! Does anyone know how he changed the bounds accordingly as he substituted variables in? I fallowed the rest of his work. He does integration by parts simplifying things greatly.
     
  2. jcsd
  3. Mar 15, 2005 #2
    Here you got Relativistic Force which equals the Change in Momentum with respect to time. Where Relativistic momentum is

    [tex]mv\gamma \equiv \frac{mv}{\sqrt{1 - v^2/c^2}}[/tex]

    We thus take the Derivative of Relativistic momentum to get relativistic force, whereas we then apply the integral of relativistic force from x to 0. Where [tex] x = vdt [/tex]

    [tex]{\int_{0}^{x}} \frac{d(p)}{dt}dx}[/tex]

    = [tex]\frac{d(p)}{dt}{\int_{0}^{x}}dx}[/tex] (1)

    = [tex] Fx [/tex] (2)

    = [tex] Fvdt [/tex] (3)

    = [tex] \frac{d(p)}{dt}vdt [/tex] (4)

    = [tex] vd(p) [/tex] (5)

    = [tex] vd[mv\gamma] [/tex] (6)

    where [tex]mv\gamma \equiv \frac{mv}{\sqrt{1 - v^2/c^2}}[/tex]

    Once dx is integrated, you can still claim x = vdt. Thus dt's divide and cancel out on (4)

    Thats the best I can do on that for you. It's too late at night for me to truly dive into the idea of integrating from 0 to mv on v*d(mvy).

    An admin here should come in here and correct everything done wrong. I do not think I am right, since I didn't get into any integration by parts with the approach. Tomorrow when its not late I shall take the approach using the other imformation you have given.
     
  4. Mar 15, 2005 #3
    I wish the author had shown all that work. It is no wounder I'm confused. I thought there was some integration property I didn't know. What I'm still not understanding is how he changed the bounds. I hope the rest of this book isn't going to be so obscure. Thank you for the help. That is way farthure than I would have gotten. I will look forward to tomarrow or whenever you get back to me.
     
  5. Mar 16, 2005 #4
    Clarifying My Question

    I finally figured out notation on this site, so I'm going to rewrite my question cleanly. Force in reletivity is given by
    [tex]
    \begin{displaymath}
    F = \frac{d}{d(t)}({\gamma}mv) \qquad {\gamma} = {\frac{1}{\sqrt{1-v^2/c^2}}
    \end{displaymath}
    [/tex]
    The formula for kenetic energy in classical physics can be given by
    [tex] KE = \int_{0}^{s}Fd(s) [/tex]
    if all all the work goes into motion. Using substitution, we obtain the equation for reletivistic kenetic energy.
    [tex] KE = \int_{0}^{s} \frac{d({\gamma}mv)}{d(t)}ds [/tex]
    I hope everything up to this point makes since because this next part is where I get confused. The author claims that
    [tex]
    \begin{displaymath}
    KE = \int_{0}^{s} \frac{d({\gamma}mv)}{d(t)}ds = \int_{0}^{mv} vd({\gamma}mv) = \int_{0}^{v} vd(\frac{mv}{\sqrt{1-v^2/c^2}})
    \end{displaymath}
    [/tex]
    How did the author change his bounds from s to mv to v? How did he convert d(s)/d(t) to v. By definition, velocity is change in distance over time, but the derivitive of [tex]\frac{d}{d(t)}({\gamma}mv}) [/tex] is attatched to the function [tex] ({\gamma}mv) [/tex], so mathematically d(s) cannot be devided by d(t) even though the notation makes it seem like it can be. In other words, I do not think he devided d(s) by d(t) to get v. I hope all this clarifies my question.
    I was thrilled to get a response. Unfortunatly, the answer given above also abuses some mathematical rules. You cannot pull a derivitive out of an integral like that. Thank you for at least trying to answer my question though. Any more ideas? Thanks.
     
  6. Mar 16, 2005 #5
    Well, the Integral

    [tex] \begin{displaymath}KE = \int_{0}^{mv} vd({\gamma}mv) \end{displaymath}[/tex]

    Does make sense,, for it has [tex] d({\gamma}mv) [/tex] in the integral.. However.. why isn't it


    [tex] \begin{displaymath}KE = \int_{0}^{{\gamma}mv} vd({\gamma}mv) \end{displaymath}[/tex] ?

    Anyways, I took the d(p)/dt out of integal due that the integral was integating with respect to distance x.

    In Textbooks, they usually define Work as

    [tex] W = \int_{0}^{x} Fdx = F \int_{0}^{x} dx = Fx [/tex]

    Where Force is ofcourse the derivative of momentum with respect to time.
     
    Last edited: Mar 16, 2005
  7. Mar 17, 2005 #6
    Because the integral is now [tex] d({\gamma}mv) [/tex], the integral should have matching bounds, but the author did not arrange that. Also, if you pull F out of the integral and integrate, you would not get the right answer. There would still be an s left over. See what I mean? I think you are going the right way though.
     
  8. Mar 17, 2005 #7
    You want Relativitistic form of Kinetic Energy.

    Given that

    [tex] P = mv\gamma = \frac{mv}{\sqrt{1 - v^2/c^2}} [/tex]


    and [tex] KE = \int_{0}^{v} Pdv = [/tex]

    We will integrate relativistic momentum from c to 0. First I'm going to rewrite relativistic momentum as

    [tex] P = mv\gamma = \frac {mvc}{\sqrt{c^2 - v^2}} [/tex]

    So we have

    [tex] E = \int_{0}^{c} \frac {mvc}{\sqrt{c^2 - v^2}}dv [/tex]

    Now I am going to say [tex] u = c^2 - v^2 [/tex] and [tex] du = -2v dv [/tex]

    I will rewrite integral as

    [tex] E = -\frac{mc}{2} \int_{0}^{c} \frac {du}{\sqrt{u}} [/tex]

    which then equals

    [tex] E = -\frac{mc}{2} [\frac{\sqrt{u}}{\frac{1}{2}}]_{0}^{c} [/tex]

    given that [tex] u = c^2 - v^2 [/tex]

    then

    [tex] E = -mc[\sqrt{c^2 - v^2}]_{0}^{c} [/tex]

    = [tex] E = -mc[\sqrt{c^2 - c^2} - {\sqrt{c^2 - 0}] = -mc[\sqrt{0} - \sqrt{c^2)] [/tex]

    = [tex] E = -mc[0 - c] = -mc[-c] = mc^2 [/tex]

    Therefore

    [tex] E = mc^2 [/tex]

    Wow,, didn't know [tex] E = mc^2 [/tex] could be so beautiful.
     
    Last edited: Mar 17, 2005
  9. Mar 23, 2005 #8
    Cool but this does not give me the relitivistic form of KE just mc^2. No one I have asked can figure out what the book did, so if anyone out there has another book and would like to share the derivation with me and the rest of the world please feel free to post here.
     
  10. Mar 23, 2005 #9

    krab

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    Don't worry about the integration limits. You have an initial state where s=0, v=0, and at a final state s=s_f, v=v_f. Integrate through. If at some stage the integration variable is momentum, then it is from 0 to mv\gamma, and so on. Just do the integration manipulation and when you finally end up with a formula between two limits, take the difference between the endlimit, where s=s_f, v=v_f, p=p_f=mv_f/sqr(1-v_f^2/c^2), etc., and the initial state, where all these are zero.
     
  11. Mar 23, 2005 #10
    Can you expand on what you mean be finding the limits and doing the integral minipulation? The point is I do not know what he did or even how he minipulated those bounds. One thing I am sure of is that the way he did it is one of the few ways of solving the problem. Look at what happens if I solve the momentum derivative.
    [tex]
    \begin{displaymath}
    F = \frac{d}{dt}({\gamma}mv) =
    \frac{d}{dt}(\frac{mv}{\sqrt{1-v^2/c^2}}) =
    m{\frac{d(v)}{dt}}\bigg(\frac{(1-v^2/c^2)^{\frac{1}{2}}+\frac{v^2}{c^2}(1-v^2/c^2)^{\frac{-1}{2}}}{(1-v^2/c^2)}\bigg) =
    \end{displaymath}
    \begin{displaymath}
    m{\frac{d(v)}{dt}{\bigg[\frac{1}{(1-v^2/c^2)^\frac{1}{2}}+{\frac{v^2/c^2}{(1-v^2/c^2)^{\frac{3}{2}}}\bigg]} =
    m{\frac{d(v)}{dt}}{\bigg[\frac{1}{(1-v^2/c^2)^{\frac{3}{2}}}\bigg]}
    \end{displaymath}
    [/tex]
    Since d(v)/dt = a or acceleration, this formula becomes:
    [tex]
    F = \frac{ma}{(1-v^2/c^2)^{\frac{3}{2}}}
    [/tex]
    Now, if I try and subsitute this value back into the equation for KE, I get:
    [tex]
    KE = \int_{0}^{s}{\frac{ma}{(1-v^2/c^2)^{\frac{3}{2}}}}{ds}
    [/tex]
    We can see immediatly that all this does is add another variable, s into our equation. The way the integral bounds and variables change has to be the key to getting the formula found in most text books for KE. Also thank you zeronem for finding a very nifty way of getting E=mc^2 although multiplying the momentum by the change in velocity is sort of strange and does not return the whole formula for KE. Thanks for all the help.
     
  12. Mar 24, 2005 #11
    [tex] KE = \int{F}{dx} = \int{\frac{m\frac{d^2x}{dt^2}}{(1-\frac{dx}{dt}/c^2)^{3/2}}{dx} [/tex]
     
    Last edited: Mar 24, 2005
  13. Mar 24, 2005 #12

    krab

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    You want
    [tex]\int_0^{s_f} {d(\gamma mv)\over dt}\,ds[/tex]
    Now forget about the limits. Just integrate, and when you are done, apply what you know about the end and subtract from the start condition. The reason you do this is because the limits should always be written in terms of the dummy variable. But you are changing dummy variables like you change underwear...
    So first to notice is that you can put the dt under the ds and that is basically your v. So
    [tex]\int v\,d(\gamma mv)[/tex]
    m is constant so take it outside. We get:
    [tex]m\int v\,d\left({v\over\sqrt{1-v^2/c^2}}\right)[/tex]
    Integrate by parts:
    [tex]m\left({v^2\over\sqrt{1-v^2/c^2}}-\int {v\,dv\over\sqrt{1-v^2/c^2}\right)[/tex]
    Now substitute [itex]X=v^2/c^2[/itex]:
    [tex]mc^2\left({X\over\sqrt{1-X}}-{1\over 2}\int{dX\over\sqrt{1-X}}\right)=mc^2\left({X\over\sqrt{1-X}}+\sqrt{1-X}\right)={mc^2\over\sqrt{1-X}}[/tex]
    Now remember X is v^2/c^2, which is zero at the start and [itex]v_f^2/c^2[/itex] at the end, so the answer is:
    [tex]\left.{mc^2\over\sqrt{1-X}}\right|_0^{v_f^2/c^2}=mc^2\left({1\over\sqrt{1-v_f^2/c^2}}-1\right)[/tex]
     
    Last edited: Mar 24, 2005
  14. Mar 24, 2005 #13
    That has to be the sexiest derivation/equation ever.
     
  15. Mar 24, 2005 #14
    No problem.

    Notice though, that we have original Kinetic Energy as [tex] \frac {mv^2}{2} [/tex]

    Now lets look at the simple integral [tex] \int_{0}^{v} P dv [/tex]

    We know P = mv so we replace the P with mv in the Integral

    [tex] \int_{0}^{v} mv dv = m \int_{0}^{v} v dv = \frac {mv^2}{2} [/tex]

    So Instead of taking the integral of momentum with respect to dv, On my last post I simply took the integral of relativistic momentum with respect to dv, from c to 0 since the velocity goes from 0 to c.

    The class I am in will be going over Special Relativity as the last lesson in my class. In my book they actually show a different derivation of the "Relativistic kinetic energy" that you are trying to derive.
     
  16. Mar 24, 2005 #15
    Wow, everything is starting to come together!! I am very excited. I have learnt my lesson. Forget everything my math class every told me about integration and just integrate. Physicists (not that I am complaining) do not necessarily fallow the strict rules of pure math. On the other hand, I am not qualified to speak on this matter. If I could learn or figure out the mathematical reason of why ignoring limits work and how you can turn ds/dt into v, I would be even happier. For now, I will simply except that I am freely able to do those tricks at will. Thanks bunches.
     
  17. Mar 25, 2005 #16
    Hey, I just glanced through the post quickly, and please forgive me if I'm wrong...but isn't the derivative of position (s) velocity? Therefore, d(s)/dt is equal to v.
     
  18. Mar 25, 2005 #17

    dextercioby

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    Yes,it is,i'm sure Krab's post used this "detail"...So what...?

    Daniel.
     
  19. Mar 28, 2005 #18
    I know that ds/dt is v; however, in my math class I was taught that reducing variables in such a manner can lead to wrong results. How do you know that ds/dt create a linear function v? What formula would we use for ds/dt^2? Would we use a for acceleration or a = adf/dfa^(adf/dfaj)? If ds/dt^2 where replaced like that, is it a constant or a variable? Where did these seemingly meaningless limits of integration (mv to v) come from? I'm pointing out that intuitivily placing v for ds/dt seems right but mathematically there needs to be stronger grounds to allow such things.
    One final question. Does anyone find it odd that when deriving KE from momentum we get a different equation than when deriving KE from work? What does that mean?
     
  20. Mar 29, 2005 #19
    Did you read the rest of the posts? Specifically the part I quoted in my post?
     
  21. Mar 29, 2005 #20

    dextercioby

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    I read,so what?Your post,though correct,brought nothing new.

    Daniel.
     
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