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I Partial integration vs total integration and time-dependent force

  1. Jul 13, 2016 #1
    Integration as antiderivative

    Question: Why isn't a distinction made between anti-total-derivatives and anti-partial-derivatives in common usage of integration?

    Consider the functions ##G_1(x, t)## and ##F(x, t)## such that ##F(x, t)=\frac{d}{dx}G_1(x, t)=\frac{\partial G}{\partial x}+\frac{dt}{dx}\frac{\partial G}{\partial t}=\frac{\partial G}{\partial x}+\frac{1}{v}\frac{\partial G}{\partial t}##, where ##v=\frac{dx}{dt}## is the velocity.

    Then we define ##G_1(x, t)## as the anti-total-derivative (also known as the total-integration) of ##F(x, t)## with respect to ##x##. ##F(x, t)=\frac{d}{dx}G_1(x, t)\iff G_1(x, t)=\int_a^xF(x', t(x'))d_{full}x'+c##, where ##a## and ##c## are constants.

    Next, consider the functions ##G_2(x, t)## and ##F(x, t)## such that ##F(x, t)=\frac{\partial}{\partial x}G_2(x, t)##.

    Then we define ##G_2(x, t)## as the anti-partial-derivative (also known as the partial-integration) of ##F(x, t)## with respect to ##x##. ##F(x, t)=\frac{\partial}{\partial x}G_2(x, t)\iff G_2(x, t)=\int_{a(y)}^xF(x', t(x'))d_{partial}x'+c(y)##.

    Example: Partial integration

    Question: Is equation (2) below correct?

    Consider a force ##F## that varies with displacement ##x## and time ##t##. In general, the acceleration ##a## and velocity ##v## of a particle subjected to such a force also vary with ##x## and ##t##.

    ##a=\frac{\partial v}{\partial x}\frac{dx}{dt}+\frac{\partial v}{\partial t}=v\frac{\partial v}{\partial x}+\frac{\partial v}{\partial t}##

    Since ##F=ma=mv\frac{\partial v}{\partial x}+m\frac{\partial v}{\partial t}##,

    ##mv\frac{\partial v}{\partial x}=F-m\frac{\partial v}{\partial t}.\,\,\,\,\,## -------- (1)

    Integrating (partial-integration) both sides wrt ##x##,

    ##\int mv\frac{\partial v}{\partial x} d_{partial}x=\int(F(x, t)-m\frac{\partial v}{\partial t})d_{partial}x##

    ##\frac{1}{2}mv^2=g(x, t)+h(t)\,\,\,\,\,## -------- (2) Is this correct?

    where ##\frac{\partial}{\partial x}g(x, t)=F(x, t)-m\frac{\partial v(x, t)}{\partial t}##.

    Example: Total differential

    Question: I obtain below ##\frac{\partial v}{\partial t}=0##, which is inconsistent with ##F(x, t)##. What's wrong?

    Multiplying ##dx## to both sides of (1),

    ##mv\frac{\partial v}{\partial x}dx=(F-m\frac{\partial v}{\partial t})dx##.

    Adding ##mv\frac{\partial v}{\partial t}dt## to both sides,

    ##mv(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt)=(F-m\frac{\partial v}{\partial t})dx+mv\frac{\partial v}{\partial t}dt.\,\,\,\,\,## -------- (3)

    Since ##dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt## and ##vdt=dx##,

    ##mvdv=(F-m\frac{\partial v}{\partial t})dx+m\frac{\partial v}{\partial t}dx=Fdx##.

    Integrating both sides,

    ##\frac{1}{2}mv^2=\int F(x, t)dx##.

    I believe the integration on the RHS should be a partial-integration. (But why?) Then

    ##\frac{1}{2}mv^2=G(x, t)+H(t)## (This is consistent with equation (2).)

    where ##\frac{\partial}{\partial x}G(x, t)=F(x, t)##.

    But simplifying the RHS of (3), we have

    ##mv\frac{\partial v}{\partial x}dx+mv\frac{\partial v}{\partial t}dt=Fdx+0dt##

    By comparing terms with ##dt##, we deduce

    ##\frac{\partial v}{\partial t}=0##.

    But this is inconsistent with a time-dependent force ##F(x, t)##. What's wrong?
     
    Last edited: Jul 13, 2016
  2. jcsd
  3. Jul 13, 2016 #2

    Charles Link

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    Just a quick comment or two: Your force F you can write as a field that is a function of x and t, but the velocity is simply a function of t or a function of x where x=x(t). You can not separate the velocity with v=v(x,t). a=a(x,t)=F(x,t)/m seems to be ok, but a(x,t) is a two parameter function and really should not be confused with the actual acceleration a(t)=a(x(t),t) that occurs... editing... Your "force" F=F(x,t) is really a field function and not an actual force. The actual force experienced is F(t)=F(x(t),t). It would be good to even subscript the ## F_p(t) ## and ## a_p(t) ## so that it doesn't get confused with F(x,t) and a(x,t). I would also subscript the particle position ## x_p(t) ## and write e.g. ## F_p(t)=F(x_p(t),t) ## so that it doesn't get confused with the parameter x that is used in the field function ## F(x,t) ##.
     
    Last edited: Jul 13, 2016
  4. Jul 14, 2016 #3

    Charles Link

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    Besides my response in post #2, you may find it of interest that a ## \vec{v}(\vec{x},t) ## formalism is used in plasma physics where the particles (e.g. electrons) have a density ## n(\vec{x},t) ## and a velocity ## \vec{v}(\vec{x},t) ## in a field ## E(\vec{x},t) ##. Derivatives (vector derivatives, etc.) are done on the ## n(\vec{x},t) ## and ## \vec{v}(\vec{x},t) ## in some applications. Sometimes, the density is written ## f(\vec{x},\vec{v},t) ## where the density function includes both position and velocity. Equations then can be written for ## f(\vec{x},\vec{v},t) ## in an electromagnetic field and look similar to those of a charged particle in an electromagnetic field, but are somewhat different including the sign of the terms. Anyway, you may find this application of interest...editing... I just looked this up in one of my plasma physics textbooks (author Ichimaru). The equation that describes the behavior of ## f(\vec{x},\vec{v},t) ## in an electromagnetic field is called the Vlasov equation.
     
    Last edited: Jul 14, 2016
  5. Jul 14, 2016 #4
    So I guess equation (2) is correct?

    ##\frac{1}{2}mv^2=g(x(t), t)+h(t)##

    That can be interpreted as the kinetic energy (or rather, the change in kinetic energy) depends on ##x(t)##, which is the path the particle takes.

    But I'm still confused with why ##\frac{\partial v}{\partial t}=0## and how do we know if the integration ##\int dx## in the steps following equation (3) is a partial-integration. What if we express ##F(x, t(x))## as a function ##J(x)## of ##x## and then perform the integration ##\int dx##? That is,

    ##\frac{1}{2}mv^2=\int J(x)dx=K(x)+c##.
     
  6. Jul 14, 2016 #5

    Charles Link

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    The difficulty I think is your use of v(x,t). In the plasma physics, (and also in fluid flow), v(x,t) describes the velocity of the flow at (x,t). In the case of a single particle, it is incorrect to use this formalism. It seems you may be eager to use this mathematics in some form=it really is an interesting formalism. e.g. The derivative ## dn(\vec{x},t)/dt=\vec{v} \cdot \nabla n(\vec{x},t)+\frac{\partial{n(\vec{x},t)}}{\partial{t}} ## describes the rate of change of ## n ## with time when moving along with the fluid. If you want to research it further, you can find it being applied in some form in most introductory plasma physics textbooks.
     
    Last edited: Jul 14, 2016
  7. Jul 14, 2016 #6
    nonsense
     
  8. Jul 14, 2016 #7

    Charles Link

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    It is not uncommon for an electric field to be described by ## \vec{E}=\vec{E}(\vec{x},t) ##. The force ## \vec{F} ## for a particle of charge ## Q ## is given by ## \vec{F}=Q \vec{E}(\vec{x},t) ##. The OP's mistake is to use equations that are designed for a fluid flow, such as ## \vec{v}=\vec{v}(\vec{x},t) ## to describe a single particle in which case it must be written ## \vec{v}=\vec{v}(t) ##.
     
  9. Jul 14, 2016 #8
    what exactly is not uncommon? to claim that the velocity of the particle depends upon its position in the space?
     
    Last edited: Jul 14, 2016
  10. Jul 14, 2016 #9

    Charles Link

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    I've been trying to explain to the OP how using v=v(x,t) is incorrect for a single particle, but it can be used in applications such as fluid flow It appears the OP is still on a learning curve with some of these concepts... v=v(t) can also be written as v=v(x) since x=x(t) for the particle (where x is the location of the particle at time t), but it is incorrect to write the velocity v=v(x,t) as a field equation except in the case of a fluid. I do think the OP, with a little further study, will be able to see why what he attempted to do simply is incorrect.
     
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