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Time derivative of a time-dependent vector and scalar

  1. Jun 29, 2016 #1
    1. The problem statement, all variables and given/known data

    ## \frac{d}{dt}\gamma(t)\vec{u(t)} ##

    2. Relevant equations

    See above

    3. The attempt at a solution

    This comes from trying to verify a claim in Chapter 12 of Griffiths Electrodynamics, 4th. edition (specifically Eq. 12.62 -> Eq. 12.63, if anyone has it on hand).

    I would have expected that the usual product rule applies in this case, and that:

    ## \frac{d}{dt}\gamma(t)\vec{u(t)} = \frac{d\gamma(t)}{dt}\vec{u(t)} + \gamma(t)\frac{d\vec{u(t)}}{dt} ##

    And that all vector operations on the result would proceed as normal (i.e. ## \frac{d}{dt}(\gamma(t)\vec{u(t)}) \cdot \vec{u(t)} = \frac{d\gamma(t)}{dt}\vec{u(t)} \cdot \vec{u(t)} + \gamma(t)\frac{d\vec{u(t)}}{dt} \cdot \vec{u(t)} ##).

    But the text of the book appears to imply that:

    ## \frac{d}{dt}(\gamma(t)\vec{u(t)}) \cdot \vec{u(t)} = \frac{d\gamma(t)}{dt}\frac{d\vec{u(t)}}{dt} \cdot \vec{u(t)} ##

    What am I doing wrong here? Does the product rule not apply when one of the quantities to be differentiated is a vector and the other a scalar? That just doesn't sound right.

    Edit: The exact chain of equations from the textbook is:


    ## \frac{d\vec{p}}{dt} = \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{m\vec{u}}{(1-\frac{u^2}{c^2})^{\frac{3}{2}}} \cdot \frac{d\vec{u}}{dt} = \frac{d}{dt}(\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}) ##
     
  2. jcsd
  3. Jun 29, 2016 #2
    It doesn't; work it out carefully and you'll see that that isn't true. The normal product rule of course applies - all Griffiths did was to show the simplified expression.
     
  4. Jun 30, 2016 #3
    EDIT: poor replay skills

    I followed Griffits and I am stuck in the calculation.

    The scalar field is not time dependent as actually it is the mass of a test particle, so his original γ becomes [tex] \eta \equiv m [/tex]

    In the calculation we want to derive the work-energy theorem thus [tex] W = \int \frac{dp}{dt} dl = \int \frac{dp}{dt} . u dt[/tex]

    Of course p is a vector and is define with the proper velocity η

    [tex] p = m\eta = m\frac{u}{\sqrt{1-u.u/c^2}} [/tex]

    Thus with little work we have to show that

    [tex] W = \int \frac{dp}{dt} . u dt = \int \frac{dE}{dt} dt [/tex]

    Making the calculation on the left leads me to the following expression, where the right thingy in the brackets is missing in the book:

    [tex] \frac{ mu }{ (1-u^2/c^2)^{3/2} } . \frac{du}{dt} \left( (1-u^2/c^2)^3 + u^2/c^2 \right) [/tex]
     
    Last edited: Jun 30, 2016
  5. Jun 30, 2016 #4
    I get a two term expression when attempting this. Namely:

    ## \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) = \frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}} ##

    Take the dot product with u on both terms and it becomes:

    ## (\frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{mu^3\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt} \cdot \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}} ##

    The derivative of relativistic energy, on the other hand, is:

    ## \frac{d}{dt}(\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}) = \frac{mu\frac{du}{dt}}{(1-\frac{u^2}{c^2})^{3/2}} ##

    I'm at a loss for how to simplify the two term derivative into that last line.
     
  6. Jun 30, 2016 #5

    PeroK

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    Why not try a specific pair of simple functions and see whether your solution and/or Griffiths' are correct?

    Sorry, I see now that you are dealing with specific functions. I missed your edit to the OP.
     
  7. Jun 30, 2016 #6
    Pull out the common factor
    [tex] \frac{m \vec {u} \cdot \frac{d \vec{u}}{dt}}{(1-u^2/c^2)^{3/2}}[/tex]
     
  8. Jun 30, 2016 #7
    The fact that the one of the time derivatives is ## \frac{du}{dt} ##, and the other is ## \frac{d\vec{u}}{dt} ## doesn't make a difference with regard to pulling them out? I'd think the unit vector in the u direction wouldn't be time independent in general (so ## \frac{d\vec{u}}{dt} = \frac{du}{dt}\hat{u} + u\frac{d\hat{u}}{dt} ##), so shouldn't that add another product rule expansion?

    I see that this does, in fact, get the right answer if you assume that ## \frac{d\hat{u}}{dt} = \vec{0} ##, but is that assertion always valid?
     
  9. Jun 30, 2016 #8
    You end up with this conundrum because you are not differentiating correctly.
    [tex]\frac{d}{dt} u^2 = \frac{d}{dt} \left(\vec{u} \cdot \vec{u} \right) = 2 \vec{u} \cdot \frac{d \vec{u}}{dt}[/tex]
     
  10. Jun 30, 2016 #9
    Be careful of the [tex] u^3 [/tex] as despite not explicitly labelled with a arrow it is a vector, thus always respect the dot and write it. In this case do it as [tex] u.u u [/tex]

    The easiest way to think about it is by writing it by components using Einstein summation convection with upper and lower indexes. Thus: [tex] p_i = \left( \frac{d}{dt}\left( \frac{mu}{\sqrt{...}} \right) . u \right)_i = \left( \frac{d}{dt} \frac{mu}{\sqrt{...}} \right)_i u^i = \frac{m}{\sqrt{...}} \frac{du^i}{dt} u_i + m u_i u^i \frac{d}{dt} \frac{1}{\sqrt{....}} = \frac{m}{\sqrt{...}}u.\frac{du}{dt} + \frac{m}{(...)^{3/2}}u.\frac{du}{dt} \frac{u.u}{c^2} [/tex]

    Unfortunately no idea how the get rid of the extra term in brackets, maybe some series....
     
  11. Jun 30, 2016 #10
  12. Jun 30, 2016 #11

    PeroK

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    You can break this down logically as follows. First, you need to show (by differentiating carefully) that:

    ##\frac{d\gamma}{dt} = \frac{\gamma^3}{c^2} \vec{u} \cdot \frac{d\vec{u}}{dt}##

    Hence, with ##E = \gamma mc^2## you have:

    ##\frac{dE}{dt}= \gamma^3 m \vec{u} \cdot \frac{d\vec{u}}{dt}##

    And, with ##\vec{p} = \gamma m \vec{u}## and using the product rule, you have:

    ##\frac{d\vec{p}}{dt} \cdot \vec{u} = \frac{d\gamma}{dt} mu^2 + \gamma m \vec{u} \cdot \frac{d\vec{u}}{dt} = \gamma^3 m (\frac{u^2}{c^2} + \frac{1}{\gamma^2}) \vec{u} \cdot \frac{d\vec{u}}{dt}##

    Finally, as ##\frac{u^2}{c^2} + \frac{1}{\gamma^2} = 1## we have:

    ##\frac{dE}{dt} = \frac{d\vec{p}}{dt} \cdot \vec{u}##
     
    Last edited: Jul 8, 2016
  13. Jul 6, 2016 #12
    I don't see where the ## \frac{d}{dt} u^2 ## term is in the initial expression. The initial one as written is:

    ## \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = (\frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} ##

    And only one of the vectors is inside the time derivative on the left (in your expression, both u vectors are inside the time derivative: ## \frac{d}{dt} \left(\vec{u} \cdot \vec{u} \right) ##; I know that expression is correct, I just don't see how it applies here). Am I supposed to bring the dot and vector on the right inside the time derivative before applying it, so that it looks like this?

    ## \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \vec{u}) ##
     
  14. Jul 6, 2016 #13
    It comes from the derivative of the denominator.
     
  15. Jul 7, 2016 #14
    Of course; I should have noticed that sooner, I was a bit too fixated on the parentheses there. Thanks for the assistance.
     
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