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 $a$

  • $\begingroup$ Andrew Palfreyman: "In an inertial frame, the sail acceleration is $A$. [...] $P$ Watts of photon power is beamed towards the sail [...]" -- It seems therefore that you're thinking of a beam source which remains (as good as) stationary in the inertial frame, emitting a constant power beam (i.e. constant as judged by the frame members; not by the reflecting sail). But not all of this power is "converted towards propulsion" of the sail; there's still some power left in the reflected beam. So I get: $$A := 2 \frac{P}{m~c}~(1 - \beta) \times \sqrt{1 - \beta^2}.$$ (Full answer going to follow.) $\endgroup$
    – user12262
    Feb 1, 2015 at 12:29
  • $\begingroup$ @user12262 Thanks for pointing out my unstated assumption that the laser itself is at rest in an inertial frame. Question now edited to reflect [sic] this. $\endgroup$ Feb 1, 2015 at 19:46
  • $\begingroup$ Andrew Palfreyman: "Thanks [...]" -- Sure. Obviously there's a bigger concern still to be resolved: that your question has received two unequal answers; by me (as announced in the comment above), which boils down to "$a = \frac{2~P}{m~c}~\left(\frac{c}{c + v}\right)$" and by @Rob Jeffries "$a = \frac{2~P}{m~c}~\left(\frac{c - v}{c + v}\right)$". So: there still seems something to be learned ... $\endgroup$
    – user12262
    Feb 1, 2015 at 20:31
  • 1
    $\begingroup$ @AndrewPalfreyman in the future, we'd prefer you not make so many small edits. When you make an edit, make sure you go through and change everything you can find that needs to be changed. If you're editing one post more than 4 or 5 times, that's probably too many. $\endgroup$
    – David Z
    Feb 2, 2015 at 6:03
  • $\begingroup$ @Rob Jeffries: "I've checked and agree with your Maths." -- Thx. "I suspect it's a conceptual thing." -- The fun part! I hope my reasoning is airtight to apply "Double-Doppler-ing-down" (uncontroversial when comparing emitted frequency to echoed frequency) to comparing the corresponding power as well, in this case. "My solution has $a$ disappearing as $v$ goes to $c$, whereas yours just halves. To me that seems odd" -- Same for me; but then: it's not a "conventional propulsion" question, either. "I'd be very grateful if you think you could spot a mistake in my argument" -- (See below.) $\endgroup$
    – user12262
    Feb 2, 2015 at 6:46

3 Answers 3


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).

  • $\begingroup$ Rob: Can you justify your choice of this particular factor? $\endgroup$ Feb 1, 2015 at 13:01
  • $\begingroup$ @AndrewPalfreyman See edit. $\endgroup$
    – ProfRob
    Feb 1, 2015 at 14:12
  • $\begingroup$ Rob Jeffries: Given your derivation of "the factor $\gamma^2~(1 - \beta^2)$" I wonder how exactly you went on to the expression of $a$. Does "energy-conservation" obtain? p.s. "Do you have access to Simmons & MacInnes (1993)?" -- Only within weeks, if necessary. $\endgroup$
    – user12262
    Feb 2, 2015 at 6:50
  • $\begingroup$ In the frame of the sail, getting a is just force divided by mass. Energy conservation? Well, when the thing is going extremely fast, the beam Poynting vector tends to zero and it does not accelerate. That seems correct. $\endgroup$
    – ProfRob
    Feb 2, 2015 at 7:37
  • $\begingroup$ Rob Jeffries: "Energy conservation?" -- For instance, I'd like to see a more explicit consideration of the "reflected power"; similar to the "Appendix" treatment, in order to "check the balance". Anyways, this: A. Macchi, arxiv.org/abs/1403.6273 seems to match your answer. "when the thing is going extremely fast, the beam Poynting vector tends to zero and it does not accelerate." -- This approximation/extrapolation seems too crude for a sensible argument: What would happen to "all that power" which is still beamed towards the sail? ... $\endgroup$
    – user12262
    Feb 2, 2015 at 21:13

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:

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


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


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

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

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

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

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

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

Right. Therefore:

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

  • $\begingroup$ @Rob Jeffries: "Just a thought: Your first equation is just the doppler shift of individual photons." -- Yes. "But don't you have to account for the fact that fewer photons per second impact the mirror as it goes faster?" -- Accounting for this gives a (partial) factor $\frac{1}{1 + \beta}$, AFAIU; and together with the "dilation factor $\sqrt{1 - \beta^2}$" obtains the relativistic doppler factor in the first place. So why "double accounting"?? "an extra factor of $(1-\beta)$ [...] would lead to agreement with my result." -- True, but not quite convincing by itself ... $\endgroup$
    – user12262
    Feb 2, 2015 at 21:12
  • $\begingroup$ @Rob Jeffries: "[...] the fact that fewer photons hit the mirror [...] as the mirror recedes with speed v in the same frame." -- I agree: this "kinematics" must be taken into account. I claim: I do take it into account, and together with the "plain dilation factor $\sqrt{1 - \beta^2}$" I derive $\sqrt{\frac{1 - \beta}{1 + \beta}}$ as final "one-way" Doppler result. You, AFAIU, want to apply the "kinematics factor" again, "on top" of that. That's incorrect: Doppler dilation of photon frequency and of "pulse rate" follow from the same calculation. (Has this been asked/solved at PSE?) $\endgroup$
    – user12262
    Feb 3, 2015 at 1:46
  • 1
    $\begingroup$ Yes, I see. Perhaps then it is that the $\beta$ you are using in the doppler formula is not measured at the time the photons were emitted by the laser. This matters because of the acceleration. A (very) brief search suggests that the doppler shift formula is not what you have used in the case of an accelerating receiver. $\endgroup$
    – ProfRob
    Feb 3, 2015 at 12:55
  • $\begingroup$ @Rob Jeffries: "the doppler shift formula is not what you have used in the case of an accelerating receiver." -- True. (I missed that. On, to the drawing board! ;) So, I'd expect some dependence on "$f_{\text{beam}}$", parametrized (at the outset) by $\frac{A}{c~f_{\text{beam}}}$ (in comparison to $\beta$, or to $c$) ... Meanwhile I'm still curious about your derivation of "reflected power in the source system". $\endgroup$
    – user12262
    Feb 3, 2015 at 19:51
  • $\begingroup$ Do you want me to write down the equation of motion (that is what I have shown in the rest frame of the mirror) and the equation of energy conservation in the frame of the laser? i.e. $d\gamma/dt =$ ? I think that is a can of worms and there is another Am.J.Phys. paper that I need to track down to check my result - possibly more tomorrow. In the meantime I removed a couple of my (at least partly incorrect) comments above, to keep the debate concise. $\endgroup$
    – ProfRob
    Feb 3, 2015 at 21:44

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 $$A = \frac{a}{\gamma^3}$$

  • $\begingroup$ Unconvincing. You might be on to something if you deal with a conserved photon number since the $E$ used here should be the energy of the beam, not the energy of a single photon - you multiply by the no. of photons received, which is messed with by the acceleration. But what is $t^{\prime}$ ? The way you have used it to get the transformed power suggests it is a tick of the clock in the moving frame. But in that case $dt^{\prime}/dt = \gamma$, not the factor in the doppler shift formula? $\endgroup$
    – ProfRob
    Feb 7, 2015 at 10:40
  • $\begingroup$ I refer you to p279 of McInnes amazon.com/Solar-Sailing-Technology-Applications-Astronomy/dp/… which I see in free preview. Equation 7.27b. Also physics.ox.ac.uk/users/iontrap/ams/teaching/rel_A.pdf ; p112 is the most apposite to this problem and involves using $\lambda'/\lambda$ and $t'/t$. It also uses N, as you suggest, but this cancels out in the ratio presented. $\endgroup$ Feb 8, 2015 at 4:02
  • $\begingroup$ For some reason I cannot see the same free pages as you, so maybe you should explain/expand the arguments there. I follow Steine's clear argument; they do not use $t^{\prime}$ as you have done, but conserve the number of arriving wavelengths by noting that transformed phase is invariant. Their final formula is indeed how N (the Poynting vector, which they call I) transforms. $\endgroup$
    – ProfRob
    Feb 8, 2015 at 10:04
  • $\begingroup$ Steane uses $N$ to denote the number of wavefronts that enter the bucket $\endgroup$ Feb 9, 2015 at 1:44
  • $\begingroup$ There appears to be a further correction when the beam is not 100% intercepted by the sail, due to divergence. It is as though the solid angle of the beam, due to Doppler, is defocused. Teller & Johnson come up with the square of the factor. $\endgroup$ Feb 10, 2015 at 22:31

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