- #1

SamRoss

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- #1

SamRoss

Gold Member

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- #2

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Looks like a mistake to me. Should be v (he annotates a correction to that effect at the 3:00 mark, but as far as I can tell it should come sooner).

- #3

- 469

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##\vec{\beta} = \vec v / c##

##\gamma = (1 - \beta^2)^{-1/2}##

##\phi = \tanh^{-1}{\beta}## (giving ##\cosh{\phi} = \gamma## and ##\sinh{\phi} = \gamma \beta## by hyperbolic identities)

##\vec p = \gamma m \vec v##

##\vec f = \dot{\vec p} c = \dfrac{d\vec p c}{d(ct)} = mc^2 \dfrac{d}{d(ct)} (\gamma \vec{\beta}) = mc^2 \dfrac{d}{d(ct)} (\hat{\beta} \sinh{\phi})##.

Now do the work–energy thing:

##\begin{split}

\int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec r_f}_{\vec r_i} \dfrac{d}{d(ct)} (\hat{\beta} \sinh{\phi}) \cdot d \vec r \\[3pt]

&= mc^2 \int^{\vec r_f}_{\vec r_i} \left( \hat{\beta} \dot{\phi} \cosh{\phi} + \dot{\hat{\beta}} \sinh{\phi} \right) \cdot d \vec r \\[3pt]

&= mc^2 \int^{\vec r_f}_{\vec r_i} \left( \hat{\beta} \cdot d \vec r \right) \cosh{\phi} \, \dfrac{d \phi}{d (ct)}

\end{split}##

(because ##d \vec r## and ##\dot{\hat{\beta}}## are orthogonal). Then change variable using ##d \vec r / d(ct) = \vec \beta = \hat{\beta} \tanh{\phi}##:

##\begin{split}

\int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\phi_f}_{\phi_i} \left( \hat{\beta} \cdot \hat{\beta} \right) \tanh{\phi} \, \cosh{\phi} \; d \phi \\[3pt]

&=mc^2 \int^{\phi_f}_{\phi_i} \sinh{\phi} \; d \phi \\[3pt]

&= mc^2 \, \Delta \cosh{\phi} \\[3pt]

&= mc^2 \Delta \gamma ,

\end{split}##

etc.

- #4

SamRoss

Gold Member

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Thanks. That's a load off my mind. And thanks for the alternate derivation as well.Looks like a mistake to me. Should be v (he annotates a correction to that effect at the 3:00 mark, but as far as I can tell it should come sooner).

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