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Dot product of 4-force and 4-velocity

  1. Feb 15, 2010 #1
    In what follows, I'll use bold for 4-vectors and an arrow over a letter to denote a 3-vector. I'll use a dot both for the Minkowski dot product and the Euclidean one, and multiplication of real numbers; the meaning in each case should be clear from the symbols on either side. [itex]\vec{v}[/itex] and [itex]\vec{a}[/itex] are 3-veclocity and 3-acceleration. The same letters without the arrow are their magnitudes.

    I've been working through the [itex]E=mc^2[/itex] derivation at the Shady Crypt Observatory site, which someone posted in a recent thread. I've got up to the detour establishing that [itex]\mathbf{F} \cdot \mathbf{U} = 0[/itex], equation 18. I'm with them for the first three lines of the derivation that follows this equation, but I don't yet see how

    [tex]\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \frac{\gamma^3}{c^2} \, \vec{a} \cdot \vec{v}.[/tex]

    I get

    [tex]\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \frac{\mathrm{d} }{\mathrm{d} t}\left ( \left [ 1-\left ( \frac{v}{c} \right )^2 \right ]^{1/2} \right )[/tex]

    [tex]=-\left [ 1-\left ( \frac{v}{c} \right )^2 \right ]^{-3/2}\cdot\left ( \frac{-2v}{c} \right )\cdot\frac{1}{c}\cdot a[/tex]

    [tex]=\frac{2av}{c^2\left [ 1-\left ( \frac{v}{c} \right )^2 \right ]^{3/2}} = \frac{2av \gamma^3}{c^2}=\frac{2av\gamma}{c^2-v^2}.[/tex]

    But [itex]\vec{a} \cdot \vec{v} = a \, v\, \cos{\theta}[/itex] and, whatever the angle, [itex]\cos{\theta} \neq 2[/itex].
     
  2. jcsd
  3. Feb 15, 2010 #2
    You forgot to put a 2 in the denominator of the derivative of [tex]\gamma[/tex].

    AB
     
  4. Feb 15, 2010 #3
    Ah, yes, so I did. Thanks, Altabeh. But how do we know that acceleration due to this arbitrary force will be in the same direction as the velocity of the object experiencing the force?
     
  5. Feb 15, 2010 #4

    DrGreg

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    In your argument you have taken

    [tex]a = \frac{d|\vec{v}|}{dt}[/tex]​

    when in fact it is

    [tex]a = \left|\frac{d\vec{v}}{dt}\right|[/tex]​

    To get the result you want, differentiate

    [tex]v^2 = \vec{v} \cdot \vec{v}[/tex]​
     
  6. Feb 15, 2010 #5
    We don't need to know how a and v vectors are directed if there is a force acting on an object! Why are you asking this?

    AB
     
  7. Feb 15, 2010 #6
    Got it! Thanks, DrGreg.

    I see that now. I asked because I thought they'd just substituted [itex]\vec{a}\cdot \vec{v}[/itex] for [itex]av[/itex]. But as DrGreg pointed out,

    [tex]\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\vec{v}\cdot \vec{v}\right)^{1/2}=\frac{1}{2v} \left(\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \cdot \vec{v} + \vec{v} \cdot \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \right)[/tex]

    is not the same thing as

    [tex]\vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}.[/tex]

    So

    [tex]\frac{\mathrm{d}\gamma}{\mathrm{d}t}=\frac{\vec{a}\cdot \vec{v}}{c^2}\gamma^3.[/tex]
     
    Last edited: Feb 15, 2010
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