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.(adsbygoogle = window.adsbygoogle || []).push({});

I've been working through the [itex]E=mc^2[/itex] derivation at theShady Crypt Observatorysite, 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].

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# Dot product of 4-force and 4-velocity

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