How Does a Relativistic Photon Rocket Calculate Acceleration and Speed?

  • #1
unscientific
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Homework Statement


(a) Show ##a = \frac{a_0}{\gamma^3}##.
(b) Find proper acceleration of rocket
(c) Find speed as a function of time.
(d) Find acceleration of second rocket.

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Homework Equations

The Attempt at a Solution


Part(a)
4-vector acceleration is given by ##\gamma^2 \left[ \frac{\gamma^2}{c}(\vec u \cdot \vec a), \frac{\gamma^2}{c^2}(\vec u \cdot \vec a)\vec u + \vec a \right]##.
For acceleration parallel to velocity, using invariance we have ##a_0^2 = a^2\gamma^6##. Thus we show that
[tex]a = \frac{a_0}{\gamma^3}[/tex]

Part(b)
We know that ##\frac{M(\tau)}{d\tau} = -\alpha M(\tau)##, so solving we have ##M(\tau) = M_0 e^{-\alpha \tau}##. Considering the change in energy of the photon in time ##d\tau##, we have ##\frac{dE}{d\tau} = c \frac{dp}{d\tau} = cm_0 a_0##. Also in time ##d\tau##, the change in mass converted to energy is ##dE=c^2dM = -\alpha M_0 e^{-\alpha \tau} c^2 d\tau##. Thus we have the acceleration as
[tex]a_0 (\tau) = \alpha c e^{-\alpha \tau}[/tex]

Part(c)
I'm not sure how to do this part. I have ##v = c \tanh \left( \frac{a_0 \tau}{c} \right)##.

Part(d)
Same concept as part(b). In time ##dt##, energy of photons produced is ##\alpha M_0 c^2 dt = p c##. Thus upon reflection, the change in momentum in time ##dt## is ##\Delta p = 2\alpha m_0 c dt##. Acceleration is ##a = 2\alpha c##. Finally, we have ##a_0 = \gamma^3 a##:
[tex]a_0 = \gamma^3 (2 \alpha c) = \frac{2\alpha a}{\left( 1 - \frac{v^2}{c^2} \right)^{-\frac{3}{2}}}[/tex]

How do I do part (c)?
 
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  • #2
unscientific said:
Part(b)
We know that ##\frac{M(\tau)}{d\tau} = -\alpha M(\tau)##, so solving we have ##M(\tau) = M_0 e^{-\alpha \tau}##. Considering the change in energy of the photon in time ##d\tau##, we have ##\frac{dE}{d\tau} = c \frac{dp}{d\tau} = cm_0 a_0##.

In the right hand side of the last equation, shouldn't ##m_0## be replaced by ##M(\tau)##?

Part(d)
Same concept as part(b). In time ##dt##, energy of photons produced is ##\alpha M_0 c^2 dt = p c##. Thus upon reflection, the change in momentum in time ##dt## is ##\Delta p = 2\alpha m_0 c dt##.

Due to the motion of the rocket relative to the launch pad, the rate at which photons arrive at the rocket according to launch-pad time is not the same as the rate at which photons are produced according to lauch-pad time. Also in the launch-pad frame, there is a doppler shift in the frequency (or energy) of the photons as they reflect off of the moving rocket.
 
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