OK guys... here's an exact calculation.
A laser, and a target, each have rest mass M. They approach each other at velocity v, with a gamma factor γ. From the perspective of either laser or target, the other one is moving, and has a total energy of γ.M.c
2. The momentum p of the moving mass satisfies the following:
\begin{equation*}\begin{split}<br />
(\gamma M c^2)^2 & = (pc)^2 + (M c^2)^2 \\<br />
(\gamma^2 - 1) (M c^2)^2 & = (pc )^2 \\<br />
\frac{(v/c)^2}{1-(v/c)^2} (M c^2)^2 & = (pc )^2 \\<br />
\gamma M v & = p<br />
\end{split}\end{equation*}
In general, for a particle of energy E and momentum p, you can find the rest mass as
m = \sqrt{(E/c^2)^2 - (p/c)^2}
Now the laser fires a single photon at the target, with frequency f in the frame where the laser is initially at rest. The laser recoils, and has a reduced rest mass. The photon goes on to collide with the target, which is slowed by the impact, but gains rest mass.
Case 1. Analysis in the frame where the laser is initially at rest
The momentum of the photon is hf/c, and the energy of the photon is hf. By conservation of momentum, the laser recoils, with momentum -hf/c. It also has a total energy of M.c
2-hf, in this frame.
The new rest mass of the laser is
\begin{equation*}\begin{split}<br />
m_l & = \sqrt{(M - hfc^{-2})^2 - (hfc^{-2})^2} \\<br />
&= \sqrt{M^2 - 2Mhfc^{-2}} \\<br />
& \approx M - hfc^{-2}<br />
\end{split}\end{equation*}
I've given the exact new rest mass; but as you can see it is about what you expect to give up the energy required to fire the photon. The energy of the recoiling laser is negligible by comparison.
Meanwhile the photon collides with the target. The target now has momentum of magnitude γ.M.v - hf/c, and energy of γ.M.c
2 + hf.
(Note that v and γ are treated as constant throughout, reflecting the velocity before the photon exchange!)
Hence the new rest mass of the target is
\begin{equation*}\begin{split}<br />
m_t & = \sqrt{(\gamma M + hfc^{-2})^2 - (\gamma M v/c - hfc^{-2})^2} \\<br />
&= \sqrt{(\gamma^2 - (v/c)^2 \gamma^2 )M^2 + 2(1+v/c)\gamma Mhfc^{-2}} \\<br />
&= \sqrt{M^2 + 2(1+v/c)\gamma Mhfc^{-2}} \\<br />
& \approx M + (1+v/c)\gamma hfc^{-2}<br />
\end{split}\end{equation*}
The factor (1+v/c)γ turns out to be precisely the factor of the Doppler blue shift. That is, the target has gained in rest mass as if colliding with a blueshifted photon. In the frame of the laser, most of this energy comes NOT from the laser, or from the any firing of engines, but from kinetic energy lost by the target as it ploughs into the photon!
Case 2. In the frame where the target is initially at rest
The photon is blueshifted, by a factor
\sqrt{\frac{c+v}{c-v}} = (1+v/c)\gamma
The momentum of the photon is (1+v/c)γhf/c, and the energy of the photon is (1+v/c)γhf. By conservation of momentum, the laser recoils, and now has momentum
p = \gamma M v - (1+v/c) \gamma hf/c
The new energy of the laser, in this frame is
E = \gamma M c^2 - (1+v/c) \gamma hf
Hence the new rest mass of the laser after delivering this very energetic photon is
\begin{equation*}\begin{split}<br />
m_l & = \sqrt{(\gamma M - (1+v/c)\gamma hfc^{-2})^2 - (\gamma M v/c - (1+v/c) \gamma hfc^{-2})^2} \\<br />
&= \sqrt{\gamma^2(1-(v/c)^2)M^2 - 2\gamma^2M(1+v/c)hfc^{-2}(1-v/c)} \\<br />
&= \sqrt{ M^2 - 2 M hfc^{-2}} \\<br />
& \approx M - hfc^{-2}<br />
\end{split}\end{equation*}
... as before. Most of the energy in the incoming photon is actually taken up from energy lost by the recoil of the laser, in this frame!
The new rest mass of the target, from the energetic photon, is
\begin{equation*}\begin{split}<br />
m_t &= \sqrt{ ( M + (1+v/c) \gamma hfc^{-2})^2 - ( (1+v/c) \gamma hfc^{-2})^2} \\<br />
&= \sqrt{ M^2 + 2 \gamma M (1+v/c) hfc^{-2} } \\<br />
&\approx M + \gamma (1+v/c) hfc^{-2}<br />
\end{split}\end{equation*}
... as before.
Cheers -- sylas
PS. Once you understand what happens for a single photon, generalizing is easy. The total number of photons delivered is invariant. Hence the total energy is scaled by the factor
\sqrt{\frac{c+v}{c-v}}
The time dilation means that this energy is also received over a smaller interval of time than it took to deliver it.
For example. You fire your 5mW laser for one minute, at a target approaching at 80% light speed. The blueshift is a factor of 3. You are expending 0.3 Joule, to deliver 0.9 Joules of damage.
The target gets energy at 45 mW, over a span of 20 seconds. It is receiving photons three times as rapidly, for one third as long, but each one packing three times the punch.
