# Can I "tack" towards a laser beam from which I am powering my 100% efficient & massless ion engine?

As far as I know, for light and particles taking special relativity into account:

\begin{align} E^2 &= (T + m_0c^2)^2\\ &= p^2 c^2 + m_0^2 c^4 &\text{ (particles)}\\[1.5em] E &= p c & \text{ (photons)}\\[1.5em] F &= \frac{d\,p}{dt\phantom\,} \ne ma.^† \end{align}

If I have a bottle of hydrogen or xenon and 100% efficient and massless ion engine and light to electricity converters, I can accelerate away from a laser beam both by absorbing their momentum and by using their energy to accelerate ions back towards the source of the laser.

I think but am not sure that it is difficult to impossible to accelerate directly into the beam because 1) this comment and 2) a given amount of energy imparts more momentum to a photon than to a particle with nonzero rest mass $$m_0$$.

Questions:

1. Is that right? Even with 100% efficient and massless light to energy converters and ion engines, I can never accelerate directly into a beam of light?

2. If so, for a given particle energy $$T$$ and rest mass $$m_0$$ what is the highest angle at which I can accelerate in the half-space (hemisphere) towards the laser beam, if any? Or can I only accelerate into the half-space away from it?

ref

• Many thanks @MarcusMüller for your MathJax Magic!
– uhoh
Oct 28 '20 at 11:19
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Oct 28 '20 at 20:03
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Oct 28 '20 at 20:07
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– user17550
Oct 28 '20 at 21:33

Because you are making use of fuel (reaction mass), this can only work for a finite time, but it must be possible for a while. We could imagine that we have a "largish" ship so the energy we get from the source imparts a negligible momentum. Then we could use that energy to push off from the bulk of the ship to accelerate toward the energy source. Done.

But it should work well for more realistic setups as well. As an example, I assume the ship is "large" enough that the xenon exhaust is a minor fraction of the total mass.

I receive 1MJ of photons. How hard does it "push" the ship?

$$dp = \frac{E}{c} = 0.0033 \text{kg m/s}$$

An engine can generate a relative velocity with xenon of 20 km/s. Most of the energy will go to the exhaust. How much Xenon can be accelerated from that energy with 100% efficiency?

$$m = \frac{2E}{v^2} = 0.005\text{kg}$$ $$dp = 0.005\text{kg}\ 20\text{km/s} = 100\text{kg m/s}$$

Since this is bigger, we can go in any direction.

The numbers aren't exact even with perfect efficiency (100% of the energy won't go to the exhaust, the engine isn't impulsive so some losses with $$dp$$, etc.) but the difference easily covers those caveats.

• This is beautiful! The incoming photon momentum is absolutely trivial in comparison to what its energy can be used to impart to reaction mass. This result seems to be independent of individual particle mass; protons, xenons or marbles would do if practicalities didn't matter. However if maximum acceleration voltage did matter there would be some preference.
– uhoh
Oct 28 '20 at 6:47
• – uhoh
Oct 28 '20 at 6:48