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.

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