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I haven't found the right search terms for this question, so if it has been answered, references would be welcome.

Suppose we start from experimental station in deep space (interstellar space if need be; ie: very low gravitational interaction), where we launch a photon powered rocket. That is, it sends photons out one end at the speed of light to produce thrust with minimal reaction mass loss, say with a laser. This produces a constant force and a constant acceleration (ignoring any minuscule reduction of the mass of the rocket due to radiated energy).

With constant thrust, the velocity V of the rocket will be a linear function of time, and the accumulating delta momentum mV will also make sense.

But it would seem that the kinetic energy of the rocket (1/2 at^2), relative to the station, is also accumulating at the square of time. The rate of accumulation of kinetic energy (energy per second, or power) is increasing linearly with time.

The same laser "engine" seems to get more and more "efficient" (producing more accumulating kinetic energy with passage of time). The second ten minutes of operation appear to generate 3 times the kinetic energy that the first ten minutes of operation produced (and 5 times as much during the third ten minutes, etc).

This seeming paradox is so obvious that it must be a stock issue resolved long ago, but I'd appreciate being brought up to speed. (No pun intended)

(Obviously there are going to be still other effects if the rocket were to approach relativistic speeds. I'm looking more for the solution within classical mechanics - like hundreds or thousands of meters per second).

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    $\begingroup$ For an ordinary rocket, PE is converted to the KE of the (massive) exhaust and the rocket. So, in the (instantaneous) reference frame in which the exhaust is at rest, the engine is most efficient at converting PE to the KE of the rocket. That is to say, the efficiency of the rocket engine, in this sense, is reference frame dependent. For an engine that converts PE to massless photon energy and to rocket KE, the engine must be more efficient in frames in which the photons emitted have less energy, i.e., are red-shifted. $\endgroup$ – Alfred Centauri Feb 29 '16 at 0:34
  • $\begingroup$ OK, suppose the rocket accelerates to 1 m/s in the first 10 minute,and then to 2 m/s in the second ten minutes. It appears to have gained 3 times as much KE in the second 10 minutes as it did in the first 10 minutes - and it will keep gaining more per 10 minute period as it gains velocity. It's still moving very slowly (compared to the Station), so I don't see how red shifting answers this question. $\endgroup$ – Zeph Feb 29 '16 at 2:34
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    $\begingroup$ You are not including any mass energy equivalence in the calculation. The energy has to come from within the spaceship and the spaceship would lose mass. If the acceleration is to be a material value, the loss in mass due to the release of energy can be significant $\endgroup$ – Peter R Mar 1 '16 at 20:56
  • $\begingroup$ The discrepancy I was concerned about is a constant-power laser engine apparently creating quadratic kinetic energy. This discrepancy of linear vs quadratic starts immediately. I use examples of 1-2 m/s. But this happens at 1-2 mm/sec, or arbitrarily low speeds and short times. Way before the rocket experiences relativistic effects, or we need to worry about mass-energy equivalence. $\endgroup$ – Zeph Mar 15 '16 at 7:56
  • $\begingroup$ Possible duplicate of Why is the work done by a rocket engine greater at higher speeds? $\endgroup$ – knzhou Sep 16 '18 at 17:07
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TL;DR:

As the rocket goes faster, the emitted photons carry less energy in the rest frame. This means more energy is available for the rocket. Conservation of energy and momentum are shown to give exactly the result we expect, even in the classical low-speed limit.


To figure out what is going on, we need to consider what happens to both momentum and energy of rocket and photons.

Without loss of generality we can assume the photons emitted are blue in the frame of reference of the rocket - it makes the description easier.

When the rocket is moving slowly, the photons still look blue in the frame of reference of a stationary observer: they appear to carry all the energy. As the rocket accelerates, the photons will be Doppler shifted, and become "redder" - they carry less energy away after the interaction with the rocket. So for the same number of photons per unit time, the rocket feels the same thrust, but (in the stationary frame) it results in greater increase in energy as the photons have less energy.

Mathematically, if we emit $N$ photons of wavelength $\lambda_0$ in a short time interval, then for a rocket of mass $m$ with velocity $v$ the total change in momentum is

$$m\Delta v = N\frac{h}{\lambda}$$

The energy increase of the rocket is

$$\begin{align} \Delta E &= \frac12 m (v_1^2 - v_0^2) \\ &= \frac12 m (v_1+v_0)(v_1-v_0) \\ &= v\Delta p\\ &= \frac{Nvh}{\lambda_0}\end{align}$$

The energy of the (Doppler shifted) photons is (in the non-relativistic approximation)

$$\lambda_1 = (1+\frac{v}{c})\lambda_0$$

And the energy is

$$\begin{align} E &= N\frac{hc}{\lambda_1}\\ &= N\frac{hc}{\left(1+\frac{v}{c}\right)\lambda_0}\\ &\approx N\frac{hc}{\lambda_0}\left(1-\frac{v}{c}\right)\end{align}$$

As you can see, the sum of this energy plus the energy gained by the rocket is constant - as the rocket goes faster, the photons carry away $N\frac{hv}{\lambda_0}$ less energy - and that is precisely the energy gained by the rocket.

If the rocket reaches relativistic velocities, the calculation is more complex but the conclusion is the same. A proper calculation would also not ignore the mass loss of the rocket due to the energy emitted, but again that doesn't detract from the fundamental principle at play.

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  • $\begingroup$ Let me see if I understand. The postulated photon drive is adding energy to the overall system at a constant rate. That total includes the KE of the rocket + the energy of the photons emitted so far. The KE of the rocket is indeed increasing quadratically with time, but the energy going into the photons is decreasing to exactly balance the rocket's increase in KE. However the decreasing energy going into the photons manifests as a non-relativistic doppler red shift. Is that the gist of your answer? $\endgroup$ – Zeph Mar 4 '16 at 20:33
  • $\begingroup$ Yes that sums it up nicely $\endgroup$ – Floris Mar 4 '16 at 21:48
  • $\begingroup$ I'd be careful about saying the drive is "adding energy" to the system. All the energy is already in the rocket+photons system, but the photons are stored in the rocket initially then released. The total energy doesn't change. $\endgroup$ – Asher Mar 15 '16 at 18:45
  • $\begingroup$ @Asher OK - but I think the intention is clear, especially since it is specified as "energy of emitted photons plus KE of rocket". $\endgroup$ – Floris Mar 15 '16 at 19:01
  • $\begingroup$ Indeed. I know that you know, but given that the purpose of the site is to provide answers which are available to everyone, I wanted to clarify that point, since it is a fairly common misconception that engines generate energy, when they don't. $\endgroup$ – Asher Mar 15 '16 at 20:42
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I apologize in advance, but I'm going to make the problem look worse first so that I can explain it:

Imagine that instead of you measuring the velocity from the station where the rocket is launched - let's call it station A - you measure instead from station B, which is moving away from station A at some high speed. And imagine that station A launches two identical rockets, one toward you and one away from you.

Since we're operating under the well-tested current theory of special relativity, we understand that we can either say "B is moving away from A" or with equal correctness "A is moving away from B" since we can declare either one as "stationary" for purposes of measuring velocities with respect to our chosen coordinate system. So let's start over and say that Station A is moving at speed away from Station B and launches two rockets.

Of course, before they launch from A, both rockets will have a certain kinetic energy based on their motion, and that energy will be equal for the two identical rockets. When we launch them, however, that begins to change: one of them is starting to move away from us even faster, while the other begins moving away from us slower. After some time, in fact, one will be moving away from us at twice the speed it originally had - that is, four times the kinetic energy - while at the same time the other is moving away from us at zero relative velocity: no kinetic energy whatsoever! Now we're really mixed up, because both rockets did the exact same thing but one ended up with a very high kinetic energy and one with no energy whatsoever.

The issue I hope to highlight with that example is that kinetic energy is frame-dependent and not an intrinsic energy for an object. That's why you can safely ride along in a car moving at 100km/h along a road but not be safely struck by a car moving at 100km/h along a road; if you travel with the car, it has zero kinetic energy in your frame, but if you're standing on the road it has a very high kinetic energy in your frame. Kinetic energy is relative.

On another note, I said before that the two rockets are doing "the same thing," but in any given single frame of reference that's not quite true. Obviously, in my example above, from the frame of station B one rocket is losing kinetic energy and the other is gaining kinetic energy, for example. But that's only considering the rockets themselves: those are being propelled by launching photons the other direction, and those photons are expelled with a certain energy themselves. If you were to measure the energy of the photons themselves, (attaching a small mirror to the back of the rocket headed toward you so that you can see the "exhaust") you would notice that over time the exhaust of the rocket moving away from you seems less energetic and the exhaust of the rocket moving toward you seems more energetic. That is to say, photons launched toward you are less energetic than photons launched away from you... which was the same conclusion we came to with the rockets themselves launching from Station A, and again is a consequence of the relativity of kinetic energy. But this time we can more directly say that one is "red-shifted" and the other "blue-shifted."

EDIT: It may also be simpler to imagine the solution if you use a normal reaction drive with massive exhaust. The exhaust is low mass but leaves at high speed; the rocket is higher mass and thus gains less velocity from the reaction. The kinetic energy increases by the same amount in opposite directions for each body of mass (taking all the exhaust as a "body"), otherwise we're getting energy for free from somewhere. Also, since massive exhaust has subliminal expulsion speed, you can have a case where the rocket is accelerating away from you but the exhaust is also moving away from you, which isn't possible with photon drives. But in all cases, photons or not, the total energy of the system is conserved: measured from an inertial coordinate system, the rocket "gains" the same amount of energy that the exhaust "loses," and the measured exchange rate does increase quadratically.

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  • $\begingroup$ First off - thanks for making the seeming paradox "worse". Or to be more accurate, making such a strong case that KE is relative to the frame by giving an even more perplexing scenario. Alas, I'm still no closer to understanding it, however. We still have a constant-power rocket engine which appears to be increasing the accumulated kinetic energy of the rocket quadratically, from the initial reference frame. And this effect shows up at arbitrarily small speeds, far from relativistic. The photons from a rocket moving away at 1 vs 2 m/s is going to be differently red shifted, but barely... $\endgroup$ – Zeph Feb 29 '16 at 18:44
  • $\begingroup$ To focus the question a little more: suppose the rocket starts some distance from Station A, at rest relative to each other. After T seconds of constant acceleration, the rocket impacts station A and it's kinetic energy is released, so it's not just theoretical. The amount of KE at the point of impact seems to increase quadratically with acceleration time, tho it's produced by a constant-power photon engine whose energy input/output increases linearly with acceleration time. All this at tiny velocities with negligible relativistic effects. $\endgroup$ – Zeph Feb 29 '16 at 18:54
  • $\begingroup$ Doppler shift, in other words non-relativistic redshift, is is 1% at 0.01 c, 2% at 0.02 c ... and so on. $\endgroup$ – stuffu Feb 29 '16 at 20:13
  • $\begingroup$ @Zeph When you say "The photons from a rocket moving away at 1 vs 2 m/s is going to be differently red shifted, but barely" it's important to note that this redshift is not trivial. To cause a significant change in velocity you have to either emit photons with huge energies or a huge number of photons, and those "barely" redshift adjustments add up to a large difference of energy. I'll add another edit after I eat this lunch and get my coffee, hang on. $\endgroup$ – Asher Mar 1 '16 at 19:42
  • $\begingroup$ The discrepancy is between a constant power laser engine (energy used = linearly proportional to time) and the accumulated kinetic energy (kinetic energy = proportional to time squared). This happens at arbitrarily small velocities. Consider the rocket accelerating to 1 mm/sec, 2 mm/sec and later to 1 cm/sec. Versus C, the doppler shift is going to be tiny - how does this make the linear/quadratic inconsistency work out? Also, the doppler shift is going to increase linearly with time (as velocity does); how does that account for the quadratic increase in Kinetic Energy of the rocket? $\endgroup$ – Zeph Mar 15 '16 at 8:02
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Your analysis ignores Special Relativity; this is fine until the speed attains any considerable fraction of the speed of light. The relevant measure is the Lorentz factor, which is hardly measurable at ordinary speeds, but is $\gamma =1.001$ at $v=0.05 c$; increases to $\gamma =1.033$ at $v=0.25 c$, and becomes significant with $\gamma =1.250$ by $v=0.6 c$.

This means that an ever-increasing amount of power must be provided in order to add to the energy, and this power must be provided by the laser. Hence the linear model which you have proposed cannot be extended past a certain level due to relativity.

For more on the Lorentz factor, see https://en.wikipedia.org/wiki/Lorentz_factor

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    $\begingroup$ I don't think that quite answers the question; I think the question is more along the lines of "In time interval $(t_1, t_2)$, the kinetic energy change is $1/2 m a (t_2^2 - t_1^2)$ and as such clearly not merely a function of $t_2 - t_1$. And that would imply that the energy output of our "motor" needs to increase with time. $\endgroup$ – Lagerbaer Feb 29 '16 at 0:46
  • $\begingroup$ Let's only look at a rocket accelerating by 1 m/s per 10 minutes, as an example. And we can demonstrate the effect in question in the first minutes or hours. It seems to gain 3 times as much KE (relative to the launch station) in the second 10 minute as it did in the first, and this quadratic increase in kinetic energy based on a fixed power thrust engine (the laser) is my question. But it's still moving very slowly indeed compared to light (a few meters/sec) so I don't yet see how Special or General Relativity makes any difference. $\endgroup$ – Zeph Feb 29 '16 at 2:44
  • $\begingroup$ @Zeph: using obsolete terminology for a moment, it's because of the relatavistic mass increase, so the kinetic energy produced by the constant force doesn't keep up with the Newtonian model. If the system runs long enough, it asymptotically approaches the speed of light. Of course by that time the laser would have consumed its "fuel". $\endgroup$ – Peter Diehr Feb 29 '16 at 2:53
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    $\begingroup$ @PeterDiehr The discrepancy between a constant-power rocket engine, and apparent quadratic increase in KE, occurs at arbitrarily small velocities. We can terminate the experiment at speeds of mm/hour (hardly approaching the speed of light) and there is still a quadratic/linear energy discrepancy. So I'm not yet seeing how relatavistic mass increase plays into this. $\endgroup$ – Zeph Feb 29 '16 at 18:58
  • $\begingroup$ @Zeph: Apparently I've lost track of what your question is. Please restate just the absolute minimum, perhaps as a new question. $\endgroup$ – Peter Diehr Feb 29 '16 at 23:50
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The same laser "engine" seems to get more and more "efficient" (producing more accumulating kinetic energy with passage of time).

This is not specific to laser based propulsion; increasing rate of change of kinetic energy is true also for chemically propelled rockets.

The reason it seems strange, or even "paradox", is that one forgets that part of the released energy is taken by the exhaust gas. In fact, just after the rocket began to move, most of the released energy in unit time is taken away by the exhaust gas, the energy of the rocket changes by amount that is only small part of total energy released. So while the mass loss has negligible impact on the mass of the rocket, in the beginning it takes almost all energy away.

Later on, as the rocket acquires higher and higher speed, the speed of the exhaust gas in the Earth frame gets lower and lower, thus the exhaust gas takes away less and less energy per unit time. This implies the balance changes in favor of the rocket, which gets larger and larger share of the energy released per unit time.

When both energy increases per unit time, that of the rocket and that of the exhaust gas, are summed, this sum equals the energy released from the fuel per unit time.

Assuming the rocket motor works the same way in the beginning as well as 10mins later, the energy released per second is constant; let us denote it by $P$. Conservation of energy implies

$$ P\Delta t = \Delta [ \frac{1}{2}mv^2 ] + \Delta E_w $$

where the first term gives change of kinetic energy of the rocket of speed $v$ during time $\Delta t$ and the second term $\Delta E_w$ is change of energy of the exhausts expelled in time $\Delta t$.

Since the thrust force and rocket mass are also assumed constant (valid in the early stage of flight when only small percentage of fuel has been burnt), the speed of the rocket is linear function of time and its kinetic energy is quadratic function of time, as you wrote, so change of kinetic energy increase in time $\Delta t$ increases.

Since left-hand side of the above equation is constant in time and the kinetic energy term increases in time, there comes a point when $\Delta E_w$ becomes negative. What does that mean? It means the fuel that becomes the exhaust gas decreases its kinetic energy. This is possible because the fuel is moving with high speed at that point and burning and exhausting it out in the direction opposite to the velocity of the rocket slows it down. This happens roughly when the speed of the rocket gets higher than half of the speed of exhaust gas with respect to the nozzle.

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  • $\begingroup$ I like this explanation of a conventional (mass thrusting) rocket. So suppose a rocket engine develops 1 MW of power (in terms of the particles it sends backwards from the rocket). Initially almost all of that power goes into the thrust particles, and almost 0% of the 1 MW is seen by rocket (and almost 100% is seen by the thrust particles). But with time, a higher and higher proportion goes to the rocket - linearly with time. So we have 1% going to the rocket, then 2% etc. Integrating that over time, the accumulated energy is quadratic - matching the KE of the rocket. But... $\endgroup$ – Zeph Mar 15 '16 at 8:20
  • $\begingroup$ It gets more tricky with the laser engine. Another answer makes the case that non-relativistic doppler red shift takes the place of your particles. I'm still absorbing that. $\endgroup$ – Zeph Mar 15 '16 at 8:22
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I don't think laser propulsion is going to work as far as creating linear increases in velocity with time for any sustained period of time. I think it's impossible - energy conservation won't allow for it. I believe that the rate of velocity increases will progressively slow.

It's not that a constant thrust laser couldn't potentially make this happen, it's just that it would require quadratically increasing power consumption to do it. This is analogous to automobile acceleration and how fuel consumption is mostly straightlined with kinetic energy, as force gets applied to increasingly greater distances. Velocity is the result.

Kinetic energy is inherently incompatible with classical mechanics in that it exists hand-in-hand with relative time, and without relative time, kinetic energy doesn't happen. It is a failing of 19th century physics that it went ahead and constructed a model in which kinetic energy existed alongside absolute time rather than fully consider that time may be relative.

But we need not repeat the failures of previous times today. Any discusion of kinetic energy, classical kinetic energy, that doesn't mention relative time is, at best, incomplete.

Time dilates mostly at the quadratic at low velocities. Time literally passes slower for an object travelling at classical velocities by an amount that progresses quadratically with velocity increases.

That quadratic is this quadratic.

See https://www.quora.com/Since-time-is-relative-how-much-time-has-passed-for-Voyager-1-in-comparison-to-Earth for a great example of this. The symmetries of special relativity are such that this very real dilation effect can often be lost in the minutiae of the details of reference frame specific math. But it is what connects various reference frames that most interests me, and the model in my head is that it is the relative rate of the passing of time that does the connecting.

I believe that time is real and that it is real in a way that means it actually passes. And the motion of electromagnetic radiation is itself the very essence of time passing. This is an unorthodox viewpoint - probably one worthy of elevating me to crank status - but I am convinced of it, and it is very obvious to me that the quadratic of kinetic energy is in fact the very small amounts of time dilation occurring to this passing of time.

It has to be this way since the infinitesimal of acceleration is light speed invariant. Only by a changing of the relative rate that time is passing can an accelerating object's invariant perspectives of the speed of light both before and after acceleration be made sense of.

Kinetic energy's quadratic is not explained by Doppler redshift. That effect is a linear one with respect to velocity, and it vanishes when the relative velocity of an object returns to zero, with no quadratic energy transfer to account for. This effect is a natural partner of the second term in the Lorentz transformation of time.

Conversely, the quadratic energy increases that correspond to those same linear velocity increases associated with Doppler are in fact "real" in a way where the energy needs to transfer, or be accounted for, when the attained relative velocity is again negated. The quadratic is "squarely" rooted in the Lorentz transformation for time's first term - gamma.

See specifically this comment to another answer by the asker of this very question for context on what we are really pursuing here:

To focus the question a little more: suppose the rocket starts some distance from Station A, at rest relative to each other. After T seconds of constant acceleration, the rocket impacts station A and it's kinetic energy is released, so it's not just theoretical. The amount of KE at the point of impact seems to increase quadratically with acceleration time, tho it's produced by a constant-power photon engine whose energy input/output increases linearly with acceleration time. All this at tiny velocities with negligible relativistic effects. – Zeph

The quadratic isn't explained by Doppler, it's explained by time dilation. Time dilation that is empirically proven to exist in a real way, to the best of my knowledge.

This will work only if quadratically increasing amounts of fuel are consumed to apply force over the ever-increasing distances that the acceleration is occurring over, corresponding to an ever-increasing number of points in space over which infinitesimal amounts of time dilation accrue.

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protected by Qmechanic Mar 2 '16 at 7:44

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