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I Relativistic photonic rocket

  1. Aug 17, 2016 #1
    For a highly relativistic rocket powered by an external laser such that the thrust is 2*Power/c, the efficiency is dismal at the start when v is low yet approaches 100% as v approaches c. I take efficiency to mean the ability to convert the energy of the beam into usable kinetic energy. Does the momentum kick actually change as the ship gains speed? Is 2P/c just the value when v=0? Conceptually, it's hard to imagine that since most energy is reflected at the start and most energy is converted at the end, that the acceleration is fixed. Yet it is isn't it? Thanks.
     
    Last edited: Aug 17, 2016
  2. jcsd
  3. Aug 17, 2016 #2
    Let us first consider a rocket that absorbs the beam. And let's make use of the law of conservation of momentum.

    When v=0, the laser device emits momentum at rate power/c, and the rocket absorbs momentum at the same rate.

    When v=0.99 c, the laser device emits momentum at rate power/c, and the rocket absorbs momentum at one percent of that rate.

    The longer the beam the more momentum it contains, and the beam is getting longer when v is not 0.

    When rocket speed is 0.99 c, the momentum of the beam is increasing at rate 0.99*power/c, and the momentum of the rocket is increasing at rate 0.01*power/c.
     
  4. Aug 17, 2016 #3
    That is the thrust at any speed, except the P is not the one you think. That P is the one received by the rocket which is a function of v, not the one emitted by the laser which is constant. So the P in your equation decreases with v because of Doppler effect, it is maximum (same as the laser power) at v=0 and goes to 0 as v->c.
     
  5. Aug 17, 2016 #4

    PeterDonis

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    Are you assuming that all of the energy of the beam gets converted into kinetic energy of the rocket? (If not, the rocket will heat up and vaporize.)

    [Edit: removed the rest of the post as it was based on an incorrect understanding of jartsa's post.]
     
    Last edited: Aug 17, 2016
  6. Aug 17, 2016 #5

    pervect

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    To analyze the problem, look first at what happens in the instantaneous rest frame of the mirror. The reflected frequency is equal to the incident frequency, more importantly the reflected power is equal to the incident power. Momentum is transferred to the mirror, but no work is done on the mirror, which may or may not be counter-intuitive depending on one's intuition, but applying the formula power = force * velocity makes it clear that if the velocity is zero, the instantaneous rate of transfer of energy to the mirror via the light beam is zero at v=0. This doesn't mean that the light beam isn't exerting a force, rather it means that the rate at which this force does work is zero, because work = force * distance, and power = work/time = force * velocity.

    Now re-analyze the same situation in a moving frame by applying the Lorentz transform - or better yet, by Bondi's k-calculus, or some equivalent approach. The incoming frequency is multiplied by k, the outgoing frequency is multiplied by 1/k, where k is the relativistic doppler shift factor ##(1-\beta) / \sqrt{1-\beta^2}##. See for instance https://en.wikipedia.org/wiki/Relativistic_Doppler_effect. The ratio of outgoing frequency to incoming frequency is thus k^2. Power, however, is proportional to the square of the frequency. If this isn't obvious (and it probably isn't), we may need to discuss that point more. Assuming for the moment that this is correct, we can see that the ratio of reflected power to incoming power is proportional to k^4.

    Analyzing the problem in the frame of the light source, a frame in which the mirror is moving, the reflected power is still k^4 times the incident power, with k<1. THis implies that the amount of power (energy/time) transfered to the mirror is (1-k^4). So if you have a relativistic doppler shift factor k of 1/2, 1/16 of the power is "wasted" in reflection, meaning that 15/16 of the power is transferred to the mirror.
     
  7. Aug 17, 2016 #6
    You should read jartsa's post again. He explained where the missing momentum gets to.
     
  8. Aug 17, 2016 #7

    PeterDonis

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    Ah, I see. I've edited my post to remove the incorrect statement.
     
  9. Aug 17, 2016 #8


    One alternative is to emit the absorbed energy back, which makes the situation similar to reflection. In the rocket frame we can easily see that that would double the acceleration. Interestingly in the laser device frame the light emitted back has just one percent of the momentum and energy of the light that was absorbed by the rocket, when speed of rocket is 0.99 c.
     
  10. Aug 18, 2016 #9
    I think that was my understanding, that if in your example 15/16 th of the power is transferred to the mirror, or ship, then that ends up as kinetic energy.
    I don't quite understand what you mean by the momentum of the beam. That is fixed. There should not be an accumulated momentum of the beam at the point of interaction with the ship.
    But in the observer frame, the power of the beam remains the same yet my understanding is that most of it gets converted to the kinetic energy of the ship. I thought that extreme red shifting means that most of the energy is converted to motion as seen in the observer frame. One paper described light hitting a ship near c is like light hitting a black hole.

    So if I were watching the ship pass by me at .99c and the beam hit it, I think I would see most of that beam energy convert to a change in ship kinetic energy as compared to watching it just start and most energy does not get translated to ship kinetic energy but gets reflected. In my frame then, does the effective momentum transfer change?

    Not to end discussion, but thanks to all for the answers so far!
     
    Last edited: Aug 18, 2016
  11. Aug 18, 2016 #10
    Not sure what observer you mean, but I still think you are confusing the emitted power in the rest frame of the laser with the power received by the rocket. They're not the same thing.
    If you actually calculate a numeric example I think you'll find out that most of the energy gets reflected back and only a very small part transferred to KE of the rocket. It would be very red-shifted indeed but you are ignoring the time component, the reflected beam is very low power but over a very long period of time. It integrates to almost the same amount as the incident energy. I am assuming a perfect mirror at the rocket so energy is not lost as heat.
     
  12. Aug 18, 2016 #11

    We should be able to account for all things in any frame if we are consistent. I choose the frame where the laser is located, say the earth frame where the ship starts out. I agree most energy gets reflected at the start. I've seen charts for photon rockets which say the conversion is near 100% as the rocket approaches c. They say the efficiency of the rocket nears 1. What I am trying to clear up is if that means nothing, if the power goes to zero, 100% of zero is zero. Or if it means a lot, the mass grows forever as the speed asymptotically nears c.
     
  13. Aug 18, 2016 #12
    The momentum of the beam grows with it's length.
     
  14. Aug 18, 2016 #13
    By momentum of the beam I mean the momentum of the beam. :)

    In order to not get distracted, let's discuss a "momentum pulse" which is a pulse of momentum emitted by a laser device that emits a pulse of light.

    In the laser device frame: The pulse has momentum p. When the rocket absorbs the pulse, the momentum of the rocket increases by p.

    In the rocket frame: The pulse has momentum p * relativistic redshift factor. And that's how much the pulse increases the momentum of the rocket, when absorbed.

    At speed 0.99 c the relativistic redshift factor is: 0.01 / gamma = 0.01 / 7.1 = 0.0015

    (in the rocket frame the laser device is a photon rocket with 99% efficiency, that's where the 0.01 comes from, and then we must also divide the energy by gamma, thats where the 7.1 comes from)
     
  15. Aug 19, 2016 #14
    Thanks, so your answer is that if I shine a laser pulse of 10000J at a ship going 0.99c 15J will actually go into changing the kinetic energy of the ship.

    I've been thinking a lot about this and currently I think the following. Let's look at energy. The ship is going at v. In my frame the input energy is E and the reflected energy is downshifted by a relativistic red shift which goes as E(1-beta)/(1+beta). The difference goes into the ship and adds to its energy. The ratio of that difference over the input energy is simply R= 2 beta/(1+beta). At v=0, no energy is transferred. At v=1 m/s, R is 6.67E-9. At v=0.1c, R=0.18 and at v=0.99c, R=0.995. In my version, 9950J goes into the ships energy. If gamma is appropriate, and I'm not sure it is since I am looking at the problem from the earth frame and not the ship frame, then it's still about 1400J. I say that because the pulse is going at c wrt to the ship and it's energy is E just as it hits it. Sure, in the ships frame (or a co-moving frame) it's red shifted but I am not in that frame. I may be wrong but there it is, at least for now. I don't mind being corrected. Thanks.
     
    Last edited: Aug 19, 2016
  16. Aug 19, 2016 #15
    No that is not my answer. My answer is about the same as your answer.

    This is correct, as far as the redshift factor is correct. If anyone doubted this, I would refer to a law called "conservation of energy".:smile:

    What was the problem again? According to post #1 changing acceleration is a problem. Well, the acceleration changes, and that is not a problem, that is how I solve this problem. :smile:

    Self-powered photon rockets have a more constant acceleration than externally powered photon rockets, isn't that understandable, as there is a changing relative velocity between different parts of the propulsion system in the latter case.
     
    Last edited: Aug 19, 2016
  17. Aug 20, 2016 #16
    That went wrong. Wikipedia says the redshift is much smaller. So let me try again:

    When a source of waves moves away from us at speed 0.99 c, a wavefront moving towards us gains distance to the source at rate: 1.0 c + 0.99 c = 1.99 c. This makes the waves produced by the source 1.99 times longer. I mean longer compared to the situation when the speed of the source is 0.

    As the wave source is time dilated by a factor of 7.1, it takes 7.1 times longer time to produce a wave crest than "normally". This makes the waves produced by the source 7.1 times longer than "normally".

    The correct way to combine these two effects is to multiply them: 1.99 * 7.1 = 14.13

    So the waves are 14.13 times longer than "normally". So the frequency of waves is 1/14.13 of the "normal" frequency.

    This result agrees with a formula on the Wikipedia, so this time I made no error, probably.


    So, what is the efficiency of this wave source, if we decide that it is a photon rocket?
     
  18. Aug 20, 2016 #17

    pervect

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    I think I need to revise my answer - I'll do it by means of an example. Suppose we have a gigawatt laser accelerating our object via light pressure, and that in the frame of the laser, it shines for 1000 seconds. Let's assume the object being shined on is moving at 3/5 c, so that the doppler shift factor (which I'll call the k factor) of 2:1.

    If an observer on the object measures the power of the pulse, it will go as k^2, i.e. it will be a 250 megawatt pulse. But it will last for 2000 seconds, not 1000 - the same redshift that reduces the frequency doubles the duration of the pulse. So from the point of view of the object, the pulse has a power of 1000/4 = 250 megawatts, but a total energy of 500 gigajoules.

    Applying the same symmetrical formula, the power of the reflected pulse will be 1/4 of 250 megwatts, i.e. 62.5 megawatts. But it will be 4000 seconds long so the total energy in the pulse will be 250 gigajoules.

    So at 3/5 c, the object will reflect 250 gigajouls of the incoming 1000 gigajoules as measured in the laser frame. So the overall energy effeciency is 75 percent.
     
    Last edited: Aug 20, 2016
  19. Aug 20, 2016 #18

    Thanks. I made one error in that I forgot a square root for the Doppler formula but as it turns out the ratio I got was correct for the instantaneous efficiency according to Marx [1] and the total efficiency is 1- SQRT[(1-beta)/(1+beta)].



    I think you mean 500 gigajoules but I get your point. I agree that for beta of 0.6, the instantaneous efficiency is 75% according to Marx [1].

    [1] Marx, G., "Interstellar Vehicle Propelled by Terrestrial Laser Beam," Nature, Vol. 211, July 1966, pp. 22-23.
     
    Last edited: Aug 20, 2016
  20. Aug 20, 2016 #19

    pervect

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    You're right about the gigajoules, I edited the original post. Thanks for the reference, it's reassuring that it matches.
     
  21. Aug 22, 2016 #20
    Thanks all for the answers. I was playing with some numbers and noticed that even though near 100% of the energy of a beam gets converted to rocket kinetic energy, something seems funny. It doesn't give greater acceleration than at the start, all it does is account for the huge delta of energy required in the observer frame to add incremental velocity. I had wondered that perhaps this phenomenon held a clue to the behavior of the controversial EmDrive and Cannea drive devices.
     
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