Relativistic beamed photon propulsion

Question

I am analysing the flight profiles possible with a lightsail powered by a photon beam, up into the relativistic regime. In an inertial frame, the sail acceleration is $A$. The onboard acceleration experienced by the sail+payload is $a$. $P$ Watts of photon power is beamed towards the sail from a laser maintained at rest in an inertial frame. The sail intercepts 100% of it, the sail has 100% reflectivity, and the sail+payload all-up mass is $m$. Then for force we have $F = \gamma m A = 2 P/c$ and thus $A = 2 P / (\gamma m c)$ and so, since $a = \gamma^3 A$, $a = 2 \gamma^2 P / (m c)$

Are these expressions for accelerations $A$, $a$ correct? Clearly they are not correct - witness the $\gamma^2$ in the numerator for $a$

Answer 1

I think I understand your problem, but I think there is at least one mistake.

If I work in the frame of reference of the spacecraft, its mass is constant. The Poynting vector intercepted from the photon beam and hence force exerted is diminished by a factor $\gamma^2 (1-v/c)^2$ (see below). Thus I think that $$ a = \frac{2P \gamma^2 (1 - v/c)^2}{mc} = \frac{2P}{mc}\frac{(c-v)}{(c+v)},$$ which tends to zero as $v$ tends towards $c$. $A$ then follows as $a/\gamma^3$.

This calculation is done differently to, but the result agrees with, that presented for a relativistic light sail by Simmons & MacInnes (1993).

Appendix: Where the factor comes from.

The transformation of E- and B-fields from frame $S$ to frame $S^{\prime}$ moving with velocity ${\bf v}$ are

$$ {\bf E^{\prime}} = {\bf E}_{\parallel} + \gamma [ {\bf E}_{\perp} + {\bf v} \times {\bf B} ] $$ $$ {\bf B^{\prime}} = {\bf B}_{\parallel} + \gamma [ {\bf B}_{\perp} - \frac{1}{c^2}{\bf v} \times {\bf E} ]\, , $$ where $\gamma = (1-\beta^2)^{-1/2}$, $\beta = v/c$ and the parallel and perpendicular subscripts indicate fields parallel or perpendicular to ${\bf v}$.

Let the laser emit waves travelling in the z-direction (e.g. $ {\bf E} = E_0 \sin(kz - \omega t){\bf \hat{i}}$ and ${\bf B} = (E_0/c)\sin(kz-\omega t){\bf \hat{j}}$) and let ${\bf v} = v{\bf \hat{k}}$. In this case, the E- and B-fields of the electromagnetic waves are perpendicular to the z-axis and so ${\bf E}_{\parallel} = {\bf B}_{\parallel}=0$.

The E- and B-fields in the frame $S^{\prime}$ of the sail are then $$ {\bf E^{\prime}} = \gamma [ {\bf E} + {\bf v} \times {\bf B} ] = \gamma E_0 (1-\beta)\sin(kz-\omega t) {\bf \hat{i}} $$ $$ {\bf B^{\prime}} = \gamma [ {\bf B} - \frac{1}{c^2}{\bf v} \times {\bf E} ] = \gamma (E_{0}/c) (1 - \beta)\sin(kz-\omega t){\bf \hat{j}}\, . $$ The Poynting vectors (${\bf N} = {\bf E} \times {\bf B}/\mu_0$) in the rest and primed frames are related by $$ {\bf N^{\prime}} = {\bf E^{\prime}} \times {\bf B^{\prime}}/\mu_0 = \gamma^2 (1-\beta)^2\ {\bf E} \times {\bf B}/\mu_0 = \gamma^2 (1-\beta)^2\ {\bf N}$$

EDIT: As an interesting aside: Velocities larger than about $\gamma =100$ are unobtainable due to the increasing opposite force due to the doppler-boosted cosmic microwave background - McInnes & Brown (1990).

Answer 2

In an inertial frame, the sail acceleration is $A$. [...] P Watts of photon power

... let's say "photon power, or wattage" of the source: $P_{\text{source}}$ ...

is beamed towards the sail

It seems therefore that you're thinking of a beam source which remains a member of an inertial frame, emitting a constant power beam (i.e. constant as judged by the frame members; not by the reflecting, accelerating sail).

the sail intercepts 100% of it, the sail has 100% reflectivity,

But not all of this power (or energy, received by the sail during some period) is "converted towards propulsion" of the sail and payload, increasing their kinetic energy with respect to the (members of the) inertial frame. Instead, the reflected beam shines with some finite (but varying) power, too, namely (again, as determined by members of the inertial frame):

$$P_{\text{refl}} := \left(\frac{1 - \beta}{1 + \beta}\right)~P_{\text{source}},$$

where the factor $\frac{1 - \beta}{1 + \beta}$ can be thought of as the product of two equal Doppler factors $\sqrt{\frac{1 - \beta}{1 + \beta}}$, and
$v = \beta~c$ is of course the speed of the sail as determined by the (members of the) inertial frame.

As far as, aside of the beam, there is no other "power source" supposed to contribute towards driving, or breaking, or otherwise affecting the spacecraft, therefore the "effective driving power" is

$$P_{\text{eff}} := P_{\text{source}} - P_{\text{refl}} = \left(\frac{2~\beta}{1 + \beta}\right)~ P_{\text{source}}.$$

Further, as far as the spacecraft mass $m$ remains constant throughout:

$$P_{\text{eff}} := \frac{d}{dt}[~\frac{m~c^2}{\sqrt{1 - \beta^2}}~] = m~c^2~\frac{d}{dt}[~\frac{1}{\sqrt{1 - \beta^2}}~] = m~c^2~\left(\frac{\beta}{(1 - \beta^2)^{(3/2)}}\right)~\frac{d}{dt}[~\beta~] \equiv m~c^2~\left(\frac{\beta}{(1 - \beta^2)^{(3/2)}}\right)~\frac{A}{c} = m~c~\left(\frac{\beta}{(1 - \beta^2)^{(3/2)}}\right)~A.$$

Consequently,

$$A = \frac{P_{\text{eff}}}{m~c}~\left(\frac{(1 - \beta^2)^{(3/2)}}{\beta}\right) = \frac{P_{\text{source}}}{m~c}~\left(\frac{2~\beta}{1 + \beta}\right)~\left(\frac{(1 - \beta^2)^{(3/2)}}{\beta}\right) = \frac{2~P_{\text{source}}}{m~c}~(1 - \beta)~\sqrt{1 - \beta^2}.$$

Finally:

$$\int dt = \frac{m~c^2}{2~P_{\text{source}}}~\int \frac{d\beta}{(1 - \beta)~\sqrt{1 - \beta^2}},$$

$$t = \frac{m~c^2}{2~P_{\text{source}}}~\left(\frac{1 + \beta}{\sqrt{1 - \beta^2}} - 1 \right),$$

$$\beta = 1 - \left(\frac{\left(\frac{m~c^2}{P_{\text{source}}}\right)^2}{\left(\frac{m~c^2}{2~P_{\text{source}}}\right)^2 + \left(\frac{m~c^2}{2~P_{\text{source}}} + t\right)^2}\right),$$

$$\int dx = c~\int dt \left(1 - \left(\frac{\left(\frac{m~c^2}{P_{\text{source}}}\right)^2}{\left(\frac{m~c^2}{2~P_{\text{source}}}\right)^2 + \left(\frac{m~c^2}{2~P_{\text{source}}} + t\right)^2}\right)\right),$$

$$x = c~t - \frac{m~c^3}{P_{\text{source}}}~\left(\text{ArcTan}[~\frac{m~c^2}{2~P_{\text{source}}} + \frac{t}{\frac{m~c^2}{2~P_{\text{source}}}}~] - \text{ArcTan}[~\frac{m~c^2}{2~P_{\text{source}}}~]\right).$$

The onboard acceleration experienced by the sail+payload is $a$. [...] since $a=\gamma^3~A$ [...]

Right. Therefore:

$$a = \frac{A}{(1 - \beta^2)^{(3/2)}} = \frac{2~P_{\text{source}}}{m~c}~\frac{(1 - \beta)~\sqrt{1 - \beta^2}}{(1 - \beta^2)^{(3/2)}} = \frac{2~P_{\text{source}}}{m~c}~\frac{1}{1 + \beta}.$$

Answer 3

Thank you both for working this problem. Rob Jeffries has it right, but I'll present a derivation that's purely SR, now that I better understand the problem. I take the relativistic Doppler effect as a given: $$\lambda'= \sqrt\frac{1+\beta}{1-\beta} \lambda$$ Since $t=\lambda/c$ , this is also how time transforms; i.e. $$t'= \sqrt\frac{1+\beta}{1-\beta} t$$ Because $E = h c/\lambda$, energy transforms as $$E'= \sqrt\frac{1-\beta}{1+\beta} E$$ so power transforms like this: $$P'=\frac{dE'}{dt'} = \frac{dE}{dt} \frac{dE'}{dE} \frac{dt}{dt'}$$ or $$P' = \frac{1-\beta}{1+\beta} P$$

This allows the accelerations to be calculated. In the sail frame we have $$a = \frac{F}{m} = \frac{2 P'}{m c} = \frac{2 P}{m c} \frac{1-\beta}{1+\beta}$$ and in the laser frame $$A = \frac{a}{\gamma^3}$$