Relativistic Energy: Change of Consts. of Integration

[SamRoss]

In this super short video of the derivation of the relativistic kinetic energy, , I'm just stuck on one thing. Around 1:00 minute in, the constants of integration change from 0 to pv when the integration changes from dx to dv. Where does the pv come from? Thanks!

 

[SiennaTheGr8]

In this super short video of the derivation of the relativistic kinetic energy, , I'm just stuck on one thing. Around 1:00 minute in, the constants of integration change from 0 to pv when the integration changes from dx to dv. Where does the pv come from? Thanks!

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).

 

[SiennaTheGr8]

Also, try this derivation out for size. Note: an overdot indicates a derivative taken with respect to $ct$ (not $t$).

$\vec{\beta} = \vec v / c$
$\gamma = (1 - \beta^2)^{-1/2}$
$\phi = \tanh^{-1}{\beta}$ (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.

 

[SamRoss]

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).

Thanks. That's a load off my mind. And thanks for the alternate derivation as well.