jbriggs444 said:
As long as you are willing to shrink your payload indefinitely (i.e. ignoring any quantum complications and assuming a continuum model), you can thrust forever, becoming smaller and smaller as you go.
I had objected to this originally, but based on PM conversation, I now understand the type of model you are proposing here and I agree that it is consistent. (It does raise an additional question about an alternate possibility which I'll save for a separate post.)
Using the relativistic rocket equations in the article
@Dale linked to in post #2, we have the following (I am using units in which ##c = 1##):
$$
v = \frac{at}{\sqrt{ 1 + a^2 t^2}}
$$
$$
\gamma = \frac{1}{\sqrt{1 - v^2}} = \sqrt{ 1 + a^2 t^2 }
$$
$$
\frac{M + m}{m} = \gamma \left( 1 + v \right)
$$
The last equation is slightly rewritten from the FAQ article. What it gives is the ratio of the total mass present at the start (when the rocket is sitting at rest on the launch pad) to the rest mass of the remaining portion of the rocket, at the time when the rocket has achieved speed ##v##. Note that the FAQ article interprets this point as the
end of the burn, so that ##m## is the rest mass of the "payload", i.e., rest mass that never gets converted into rocket exhaust. However, we can just as easily apply the same equation to a burn that goes on indefinitely, i.e., for the case ##t \to \infty##, ##v \to 1##.
To facilitate that interpretation, we will change notation slightly; instead of ##M + m##, we will write ##E##, since this quantity is the total energy present in the scenario. (Since at the start, everything is at rest on the launch pad, ##E## is also the total rest mass that is sitting on the launch pad at the start.). Then our third equation becomes:
$$
\frac{E}{m} = \gamma \left( 1 + v \right)
$$
This now tells us that, since ##\gamma \left( 1 + v \right) \to \infty## as ##v \to 1##, we must have ##m \to 0## as ##v \to 1## (since ##E## is a fixed finite number). That is what "shrinking the payload indefinitely" refers to. Now we simply need to work out the consequences.
The above equation obviously implies
$$
m = \frac{E}{\gamma \left( 1 + v \right)}
$$
We know that we must have energy conservation and momentum conservation, so we can write down two more equations as follows:
$$
E = E_r + E_e = \gamma m + E_e
$$
$$
P = 0 = P_r + P_e = \gamma m v + P_e
$$
where ##E_r## and ##P_r## are the total energy and momentum of the rocket, ##E_e## and ##P_e## are the total energy and momentum of the rocket exhaust. The above equations give us the following equations for the energies and momenta:
$$
E_r = \frac{E}{1 + v}
$$
$$
P_r = \frac{E v}{1 + v}
$$
$$
E_e = \frac{E v}{1 + v}
$$
$$
P_e = - \frac{Ev}{1 + v}
$$
These equations tell us several things. First, since we have ##|E_e| = |P_e|##, the rocket exhaust is indeed photons (as it should be, since our equations are all based on the assumption of a photon rocket). Second, in the limit ##v \to 1##, we have ##E_r = E_e##, i.e., half of the original energy ends up in the rocket and half ends up in the exhaust. And third, in the limit ##v \to 1##, we have ##E_r = P_r##, which makes sense mathematically since the hyperbolic worldline of the rocket asymptotes to a null line, i.e., to the worldline of a photon.
An interesting point about this, though, is that, even though the payload "shrinks indefinitely", to zero rest mass in the limit, it does
not "vanish"--the payload has nonzero energy and momentum even in the limit. This observation raises a follow-up question that I'll put in a separate post.
