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I Laser momentum question

  1. Apr 5, 2016 #1
    Hi, a general question involving heat, light, kinetic energy, and momentum.

    A man has a 100W laser. He fires the laser at a black body which absorbs the photons of the laser and gains heat energy from it.
    In practice the conversion efficiency from the 100W laser power to the black body molecules should be high, 90% should be attainable.
    Heat is molecular kinetic energy, so the molecules in the black body are given momentum by the laser photons, at a rate of 90J/sec.

    Q : By conservation of momentum, if the photons are giving 90J/sec of KE to the BB molecules, why isn't the man experiencing a kick back from laser with a power of 100W, similar to the kickback from a 100W water-hosepipe ?

    My Answer : Guess : Since light waves exert EM forces perpendicular to their line of travel...
    a.) kickback and knock on momentum they exert is mostly perpendicular to their line of travel.
    b.) as a horizontal traveling wave oscillates it will exert a EM force up then down...
    so for large objects the energy given by one half of the EM wave is often taken back in the next half. Only very small objects, typically ions or electrons have so little mass they can be knocked clear by the 1st half of the wave with enough velocity to so as to escape the 2nd half of the wave that would return them to their original position and rest velocity.
    c.) any kickback is also thermalized instantly, just like the BB molecules, i.e. the laser device heats up a bit.

    Is this about right?
    At 1st glance I couldn't understand why since a 100W laser, or a light torch shouldn't have a 100W kick back.
     
  2. jcsd
  3. Apr 5, 2016 #2

    jbriggs444

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    Watts are not a unit of force. How much kickback (in Newtons) is there from a 100W water-hosepipe? How would you go about calculating that?
     
  4. Apr 5, 2016 #3
    >Watts are not a unit of force. How much kickback (in Newtons) is there from a 100W water-hosepipe? How would you go about calculating that?

    I know watts are not a unit of force. But I also know that to give KE to an object there needs to be power. So there is always a (M/E) MP associated with induced change of motion. The point was not to calculate the force for a specific hose. But we can do that.

    For a 100W hosepipe, with a thruput of 100kg mass emitted at 1m/s, the stopping force per sec per kg would be 1 N.
    For a 100W hosepipe, with a thruput of 10kg mass emitted at 4.5m/s, the stopping force per sec per kg would be 10 N.

    Generally, the kickback force or stopping force for a set power and mass depends on how quickly the subject goes from stop to motion , or motion to stop.
    I don't see that its helps with my question though. There should be an force, end energy, and power associated with a projectile beam, regardless of the value of force.

    I wasn't thinking of kickback force. Unless the objects involved are all static, there is always a power associated with a kickback force or stopping force. So its often legit and useful to associate kickbacks with power, without having to consider force, e.g rockety. The same is true for kickback energy if the force is over a short time, e.g.guns.
     
  5. Apr 5, 2016 #4

    ZapperZ

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    Why are you using power and energy? Why can't you just use the wavelength or frequency of the light source, calculate the momentum carried by each photon, and figure out (from the power), how many photons per second hits the person? Then use basic mechanics to get the rate of change of momentum, and voila! You get the force acting on that person due to the impact of the photons.

    {scratching head}

    Zz.
     
  6. Apr 5, 2016 #5

    jbriggs444

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    Please show your work. What I get is...

    100 kg accelerated to 1 m/sec at a rate of one such mass per second is 100 kg m/sec2. That's 100 N.
    100 kg at 1 meter/sec is ##\frac{1}{2}mv^2## = 50 Joules of energy. At a rate of one such mass per second, that's 50 W.

    10 kg accelerated to 4.5 m/sec at a rate of one such mass per second is 45 kg m/sec2. That's 45 N.
    10 kg at 4.5 meter/sec is ##\frac{1}{2}mv^2## = 202.5 Joules of energy. At a rate of one such mass per second that's 200 W.

    There is a pattern that can be found for the thrust to power ratio as a function of exhaust velocity.
     
  7. Apr 5, 2016 #6
    I'll have a go.
    1.Over 1 sec, a 100W laser emits Et=100J of Energy as photons

    get number of photons
    2.For n photons Et=nhf
    3. choose a frequency, -> blue light has f = 6.8E14
    4. so blue photon current is n = 2.2E20 / sec

    now get momentum per photon with debroglie
    5. pv = h, where p is momentum , v is wavelength,
    6. for blue light v = 4.4E-7 so p = 1.5 E-27

    get result
    7. So photons convey np = 3.3E-7 kgm/s of momentum per sec, or 3.3E-7 newtons force. Thats the mechanical kick back from a 100W blue laser.

    Over at the black body, it also takes a 3.3E-7 N mechanical force from the incoming photons. But this can't possibly be the way the photons get 100W of heat energy into the black body, hence my electrodynamic analysis in my 1st post.
     
  8. Apr 5, 2016 #7
    there's no point. I was doing it in my head, hence i kept forgetting the 1/2 in the Ek formula. Your calcs are correct.
     
  9. Apr 5, 2016 #8
    let me rephrase the original question : Why, when we know that a photon laser will convey 100W of Ek to a black body that absorbs the laser and converts it into heat, is there not a similar Ek given to the object holding the laser?

    My rephased answer : because the laser energy is conveyed in the electromagnetic wave that is light, not as mechanical momentum.

    Hence I don't think looking at de broglie or the planck relation will give the answer, it will just prompt the question again : since they give the miniscule mechanical momentum of photons which isn't sufficient to heat the black body.

    To find the energy that heats the black body at 100W, a good question is 'What is the energy stored in the E M field of a light wave ? ", hence my electrodynamic analysis in my 1st post.
     
  10. Apr 5, 2016 #9

    ZapperZ

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    But isn't this just an example of inelastic collision?

    When a ball collides into a squishy object and sticks to it, why doesn't all the ball's initial KE transfers to the ball+squishy object?

    All of the light's energy doesn't get converted into KE of the BB simply because most of it are absorbed and converted into heat. This is no different than any other inelastic collision problem.

    Zz.
     
  11. Apr 5, 2016 #10
    >But isn't this just an example of inelastic collision?
    I don't think so. I think is an eletrodynamic collision.

    >When a ball collides into a squishy object and sticks to it, why doesn't all the ball's initial KE transfers to the ball+squishy object?
    I think it does, as long as the ball comes to rest within the squishy object.

    >All of the light's energy doesn't get converted into KE of the BB simply because most of it are absorbed and converted into heat.
    But heat is mostly molecular and sub atomic KE. So nearly all of the photons energy is converted to KE.


    If you think this situation only requires light's mechanical properties to solve, I'd be interested to know if there are any situations which you think require light's electrodynamics to solve.
    Seeing photons for example. Does the eye react to light's mechanical interaction, or its electrodynamic interaction? I've heard it said that a good eye can detect a single photon.
     
  12. Apr 5, 2016 #11

    ZapperZ

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    Wait, did I read this correctly? Are you insisting that KE is conserved here? I'm not going to address the rest of your post till this is sorted out.

    Zz.
     
  13. Apr 5, 2016 #12
    Yes. For our purpose its practically all conserved. A negligible fraction exists as a photon gas within the coalesced object, an even smaller amount get lost as radiant heat at the event of impact, but practically all KE ends up within the coalesced object, as either molecular thermalised KE or net KE.
     
  14. Apr 5, 2016 #13

    ZapperZ

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    So when you did your undergraduate intro physics lab, and did your experiment on collisions or HW problems on this topic, how did it go when you argued this with your instructor or when you did your HW/exams? I can show you a lot of collision experiments where KE isn't conserved in the mechanics of the collision. How did you get away with this world view when you when through your education?

    Here's the OTHER thing that you are missing. The energy that has been transferred into molecular vibrations DOES NOT translate into a uniform and "directional" KE. It is an increase in the statistical, RANDOM motion, and averages out to ZERO in terms of net velocity. It means that the macroscopic velocity of the object is not affected by the increase in thermal energy of the object.

    Zz.
     
  15. Apr 5, 2016 #14
    >implying you've stated an initial thing I'm missing.
    But you haven't

    Glad that finally sunk in. I agree. Now go back and read all my posts again. I've been telling you from the 1st post....

    If you've been stuck on the idea that only macro things can have mechanics, you won't have been able to follow any discussion involving micro mechanics.
     
  16. Apr 5, 2016 #15

    ZapperZ

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    Sorry, but I did not miss it nor did it finally "sunk in". I deal with the physics of photoemission, and the issue of energy and momentum transfer from photons to (i) the photoelectron and (ii) the lattice ions has been carefully treated. They are NOT one of the same!

    You are trying to compare an "ordered" energy and thinking that this translates into a statistical distribution, and somehow, this is due to a transfer of momentum. It will not match! Otherwise, you won't be asking this here.

    Zz.
     
  17. Apr 5, 2016 #16

    berkeman

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    Thread is done.
     
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