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Kinetic Energy

  1. Nov 7, 2005 #1
    Hi Everyone,
    I'm reading in my modern physics book about relativistic kinetic energy. I'm a little confused about what rule of calculus allows the following statement:
    [tex]
    KE = \int_{0}^{s} \frac{d(mv)}{dt} ds = \int_{0}^{mv} v d(mv)
    [/tex]
    I see that they must be saying
    [tex]
    KE = \int_{0}^{s} \frac{d(mv)}{dt} ds = \int_{0}^{mv} \frac{ds}{dt} d(mv)
    [/tex]
    since
    [tex]
    \frac{ds}{dt} = v
    [/tex]
    But I don't understand how you can treat the differentials that way, and how you would know how to change the limits of integration when you made that change of variables. It seems like I must just be forgetting something basic.
     
    Last edited: Nov 7, 2005
  2. jcsd
  3. Nov 7, 2005 #2
    Draw out the mass since it will be constant.

    [tex] KE = m\int_0^x a dx = m\int_0^x \frac{dv}{dt} dx [/tex]

    and [tex] \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v [/tex]

    so the integral becomes [tex] m\int_0^x v\frac{dv}{dx} dx = m\int_0^v v dv [/tex]

    Startin to get the picture? You need to change the integration limits because of the fact that you changed the variables.

    edit: Just noticed we're talking relativistic, in which case m won't be constant, but it goes along the same line if you use momentum instead of just velocity.
     
    Last edited: Nov 7, 2005
  4. Nov 7, 2005 #3
    Thanks!

    Thanks for the reply, Whozum. So you're using the chain rule to rearrange the differentials, with the goal of getting v by itself, and at that point the only differential left standing is d(mv)? I guess I'm wondering what the new limits of integration would be in a more complicated expression. For example, if the upper limit had been something like [itex]sin(s)^2[/itex] or some such crazy thing, what would be the rule for computing the new limit?
     
  5. Nov 7, 2005 #4
    I'm not really sure I understand your question, and am not in a position to answer you completely, but what I can say (unless I'm missing something) is that the limits are dealt with after the integral is solved, so once you solve the integral you would THEN put in the limits at F(b) and F(a).

    I would wait for an authority though.
     
  6. Nov 8, 2005 #5
    What I'm asking is, given that we've rearranged the differentials, what rule of calculus is being used to determine the new limits of integration? When we change from
    [tex]
    \int_{0}^{s} \frac{d(mv)}{dt} ds
    [/tex]
    to
    [tex]
    \int_{0}^{mv} v d(mv)
    [/tex]
    we're changing the limits of integration, which seemed fairly simple, but what if the integral had limits that are more complicated, like
    [tex]
    \int_{s}^{sin^2(s)} \frac{d(mv)}{dt} ds
    [/tex]
    Where could I look in a calculus book to see what to do in a case like that? This isn't like straight substitution (or is it?).
     
    Last edited: Nov 8, 2005
  7. Nov 8, 2005 #6

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    You're using the fact that [itex]mv=0[/itex] when [itex]s=0[/itex] (in other words, you are starting from rest) and that [itex]mv=mv[/itex] when [itex]s=s[/itex].
     
  8. Nov 8, 2005 #7
    Okay, I see that we're starting from rest, but just from a calculus standpoint, I have trouble seeing how to treat these limits -- especially if this hadn't been a physics problem, but just a calculus problem.

    When s=s, mv=mv seems to be true no matter what!

    I sense that I'm just having a conceptual problem here. I'll go sit in a dark room and think about it for a while....

    Thanks for replying!
     
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