# Till infinity and beyond!

In this post we will consider the velocity function of a space ship, taking relativistic mass and time dilation into account. So suppose we are given such a space ship. While stationary, let its mass be $m_0$ and its power output $P_0$. Suppose also, that it launches at $t = 0$. Now, its relativistic mass is

$\displaystyle m(t) = \frac{m_0}{\sqrt{1 - \frac{v^2(t)}{c^2}}},$

and its power output, due to time dilation, is

\displaystyle \begin{aligned} P(t) &= \frac{P_0}{\frac{1}{\sqrt{1 + \frac{v^2(t)}{c^2}}}} \\ &= P_0 \sqrt{1 - \frac{v^2(t)}{c^2}}. \end{aligned}

Since

$\displaystyle P = \frac{W}{t} = \frac{Fs}{t} = Fv = mav,$

we have that

\displaystyle \begin{aligned} P(t) &= \frac{\mathrm{d} v(t)}{\mathrm{d}t} v(t) m(t) & \Longleftrightarrow \\ P_0 \sqrt{1 - \frac{v^2 (t)}{c^2}} &= \frac{\mathrm{d} v(t)}{\mathrm{d} t} v(t) \frac{m_0}{\sqrt{1 - \frac{v^2(t)}{c^2}}} & \Longleftrightarrow \\ \frac{P_0}{m_0} \Bigg( 1 - \frac{v^2(t)}{c^2} \Bigg) &= \frac{\mathrm{d} v(t)}{\mathrm{d}t} v(t) & \Longleftrightarrow \\ \frac{\mathrm{d} v(t)}{\mathrm{d} t} &= \frac{P_0}{m_0}\Bigg( \frac{1}{v(t)} - \frac{v(t)}{c^2} \Bigg). & \end{aligned}

Rewriting to conventional differential equation notation we obtain

$\displaystyle y'(t) = \frac{P_0}{m_0} \Bigg( \frac{1}{y(t)} - \frac{y(t)}{c^2} \Bigg),$

which has solution

\displaystyle \begin{aligned} v(t) &= y(t) = \sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0}}} \Longleftrightarrow \\ \frac{\mathrm{d} v(t)}{\mathrm{d} t} &= \frac{ \frac{\mathrm{d}}{\mathrm{d}t} \Bigg(c^2 - e^{C - \frac{2P_0t}{c^2m_0}} \Bigg) }{ 2\sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0}}}} \\ &= \frac{ \frac{\mathrm{d}}{\mathrm{d}t} \Bigg( - e^{C - \frac{2P_0t}{c^2m_0}} \Bigg) }{ 2\sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0}}}} \\ &= \frac{ - e^{C - \frac{2P_0t}{c^2m_0}} \frac{\mathrm{d} }{\mathrm{d}t}\Bigg( C - \frac{2P_0t}{c^2m_0} \Bigg) }{ 2\sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0}}}} \\ &= \frac{ e^{C - \frac{2P_0t}{c^2m_0}} \frac{2P_0}{c^2m_0} }{ 2\sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0}}}} \\ &= \frac{P_0 e^{C - \frac{2P_0t}{c^2m_0} } }{ c^2m_0\sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0} }} }. \end{aligned}

Now let us verify the validity of the solution to the differential equation:

\displaystyle \begin{aligned} \frac{P_0}{m_0}\Bigg( \frac{1}{v(t)} - \frac{v(t)}{c^2} \Bigg) &= \frac{P_0}{m_0} \Bigg( \frac{1}{ \sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0} }} } - \frac{ \sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0} }} }{c^2} \Bigg) \\ &= \frac{P_0}{m_0} \Bigg( \frac{ c^2 - \Big( c^2 - e^{C - \frac{2P_0t}{c^2m_0} } \Big) }{c^2 \sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0} }} } \Bigg) \\ &= \frac{P_0}{c^2m_0} \Bigg( \frac{e^{C- \frac{2P_0t}{c^2m_0} }}{\sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0} }}} \Bigg) \\ &= \frac{\mathrm{d} v(t)}{\mathrm{d}t}, \end{aligned}

as expected. Now let us calculate the integration constant $C$. Since $v(0) = 0$, we have

\displaystyle \begin{aligned} \sqrt{c^2 - e^{C - \frac{2P_0t}{c^2 m_0}}} &= 0 &\Longleftrightarrow \\ e^{C - \frac{2P_0t}{c^2m_0}} &= c^2 &\Longleftrightarrow \\ e^C &\overset{t = 0}{=} c^2 &\Longleftrightarrow \\ \ln e^C &= \ln c^2 &\Longleftrightarrow \\ C &= 2\ln c, & \end{aligned}

and we have that

\displaystyle \begin{aligned} v(t) &= \sqrt{c^2 - e^{C - \frac{2P_0t}{c^2m_0}}} \\ &= \sqrt{c^2 - e^{2 \ln c - \frac{2P_0t}{c^2m_0}}} \\ &= \sqrt{c^2 - c^2 e^{-\frac{2P_0t}{c^2 m_0}}} \\ &= c\sqrt{1 - e^{- \frac{2P_0t}{c^2 m_0}}}. \end{aligned}

Now, let us check that all pieces fit together. We have that

\displaystyle \begin{aligned} v^2(t) &= c^2 \Bigg( 1 - e^{-\frac{2P_0t}{c^2m_0} } \Bigg) , \\ \frac{v^2(t)}{c^2} &= 1 - e^{-\frac{2P_0t}{c^2 m_0}} , \\ e^{-\frac{2P_0t}{c^2m_0}} &= 1 - \frac{v^2(t)}{c^2}. \qquad \qquad \qquad (1) \end{aligned}

Also,

\displaystyle \begin{aligned} \frac{\mathrm{d}v(t)}{\mathrm{d}t}v(t)m(t) &= \frac{m_0}{\sqrt{1 - \frac{v^2(t)}{c^2}}} c \sqrt{1 - e^{-\frac{2P_0t}{c^2m_0}}} \frac{P_0 e^{-\frac{2P_0t}{c^2m_0}}}{c m_0 \sqrt{1 - e^{-\frac{2P_0t}{c^2m_0}}}} \\ &= \frac{ P_0e^{ -\frac{ 2P_0t }{ c^2m_0 } } }{ \sqrt{ 1 - \frac{v^2(t)}{c^2} } } \\ &\overset{(1)}{=} P_0 \frac{\Big( 1 - \frac{v^2(t)}{c^2} \Big)}{ \sqrt{ 1 - \frac{v^2(t)}{c^2} } } \\ &= P_0 \sqrt{1 - \frac{v^2(t)}{c^2}} \\ &= P(t), \end{aligned}

as expected.