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Classical mechanics question

  1. Aug 14, 2008 #1
    1. The problem statement, all variables and given/known data
    I am self studying Chow's Classical Mechanics. I have realized that I am at my best if I leave no stone unturned

    2. Relevant equations
    [tex] \hat e_t . \hat e_t = 1 [/tex]

    [tex] d(\hat e_t. \hat e_t)/dt =0[/tex]
    Also understood

    But [tex] 2\hat e_t .d\hat e_t/dt=0[/tex]
    Not understood (tex was first time too)
  2. jcsd
  3. Aug 14, 2008 #2


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    Hi sirius0! :smile:

    (hmm … tex not bad :smile: … but you could use use \frac{}{} and \cdot :wink:)

    [tex] \frac{d}{dt}(\hat e_t. \hat e_t)\ =\ 0[/tex]

    Use the chain rule, then a.b = b.a:

    [tex] \hat e_t \cdot\frac{d\hat e_t}{dt}\ +\ \frac{d\hat e_t}{dt} \cdot\hat e_t\ =\ 0[/tex] :smile:
  4. Aug 14, 2008 #3
    Oh thanks a good clue, explains where the two comes from! Pathetically I have to have a quick re-visit to the chain rule but at least I remember what it looked like and I think I can mop it up from here. Good tip on the tex too.
  5. Aug 27, 2008 #4
    I am now up to pp 20 of Chow's book. I am puzzled as the text states for cylindrical coordinates that [tex]
    \hat r =p\hat e_p +z\hat e_z
    [/tex]. But I thought that any 3D coordinate system needed three dimensions. Should the above have been [tex]
    \hat r =p\hat e_p +z\hat e_z+\phi e_
  6. Aug 27, 2008 #5


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    Hi sirius0! :smile:

    I don't have the same book, but I assume that [itex]\hat{\bold{e}}_p[/itex] is two-dimensional and variable (just as [itex]\hat{\bold{r}}[/itex] is three-dimensional :wink:), and that only [itex]\hat{\bold{e}}_z[/itex] is fixed. :smile:
  7. Aug 29, 2008 #6
    Understood progressing well for now....
    Thank you.
  8. Nov 25, 2008 #7
    Re: Classical mechanics question With a Relative twist

    The whole reason for my self study is in order to relearn SR and learn GR in an initial manner. The notation conventions and clear thinking were an issue so I turned back to classical mechanics. I have gotten distracted from this agenda on another forum and have been looking again at SR. I have a long path ahead WRT GR, tensors Riemann groups etc.
    But as a result of the distraction I made an assumption regarding time dilation and gravity.
    This is what I came up with using SR. Is it familiar or even right I wonder? Am I bordering on to GR via SR?

    [tex] \Delta t \sqrt{\frac{GM}{rC^2}+1} = \Delta t' [/tex]
  9. Nov 26, 2008 #8
    Just had a look http://en.wikipedia.org/wiki/Gravitational_time_dilation" [Broken] . There is something of a resemblance but I don't appear to be right. However there must be something to be learned from this.
    Last edited by a moderator: May 3, 2017
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