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I Force on a Spherically-Uniform Radiator Moving Through Space

  1. Oct 1, 2016 #1
    Let's say I have a spherically-uniform black-body radiator. It is losing energy, and therefore some of its mass, at a particular rate. From the frame of reference of the radiator, it has no momentum, but it has a changing amount of energy. From its frame of reference, the pressure on the radiator yields no net force upon it.

    However, from another frame of reference, and in this case let's choose an independent inertial frame of reference, the radiation would have a net contribution to the momentum due to the relativistic Doppler effect. So there would be a net force on the radiator from this frame of reference. However, according to the radiator's frame of reference, the pressure is uniform so it shouldn't be subject to acceleration. Is then the change of mass of the spherically-uniform radiator as it travels through space at some velocity responsible for producing a "force" on the radiator devoid of any acceleration of the radiator itself?
     
    Last edited: Oct 1, 2016
  2. jcsd
  3. Oct 1, 2016 #2

    Dale

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    Yes. Most of our intuition about force and momentum is strictly for the case of constant mass.
     
  4. Oct 1, 2016 #3

    Ibix

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    Its mass is decreasing but its velocity is constant. ##dp/dt\neq 0##, so ##F\neq 0##, even if the acceleration is zero.
     
  5. Oct 1, 2016 #4
    Is this Newtonian mechanics or special relativity? Doesn't really make sense in either one anyway.:smile:


    A light pulse that leaps off from the back side of a moving platform loses a large part of its energy to the platform.
    A light pulse that leaps off from the front side of a moving platform gets a large part of its final energy from the platform.

    If those light pulses have the same energy in the platform frame, then they leap with the same force in that frame, and also in the frame where the platform is moving forwards.

    The two light pulses do not have a net contribution to the momentum of the platform due to the relativistic Doppler effect, because the relativistic Doppler effect does not make the two forces different. Relativistic Doppler effect only makes the energies and the momentums of the two light pulses different.
     
    Last edited: Oct 1, 2016
  6. Oct 1, 2016 #5

    Dale

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    I am pretty sure that this is incorrect. I will check how forces transform tomorrow.
     
  7. Oct 2, 2016 #6

    Ibix

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    Either. I know it's counter-intuitive but, as Dale says, most of our intuition about force is for constant mass.

    Here's a Newtonian case. Take a platform with a pair of rockets pointing in opposite directions. Each rocket consumes fuel at μ kg/s, and there is no net velocity change of the platform.$$F=\frac {d (mv)}{dt}=\frac {dm}{dt}v+ma $$The second term is clearly zero if velocity is constant, while the first is clearly non-zero (-2μv) unless we are in the rest frame of the platform.

    Consider analysing the system in the case where one of the exhaust plumes is at rest. Then it never carries momentum. The other plume is gaining mass and the platform is losing mass; there must be momentum transfer between the two.

    I don't have pen and paper to hand so will not try analysing your relativistic case right now. I expect the same thing because there will be a term like ##\gamma v dm/dt##.
     
  8. Oct 2, 2016 #7

    Ibix

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    Actually, that's all there is to it, isn't it? The photons carry unequal four-momenta ##(E_+,E_+,0,0)## and ##(E_-,-E_-,0,0)## in any frame except the rest frame of the platform where ##E_+=E_-##. So the momentum of the platform must change, which is what the ##\gamma v dm/dt## term, the only non-zero term from ##F=d (\gamma mv)/dt##, does.
     
  9. Oct 2, 2016 #8
    If somebody says that "the center of mass of those two plumes co-moves with me", then everyone agrees, it is an invariant thing.

    So: The center of mass of an extremely Doppler shifted radiation from a black object co-moves with the object, no matter how much intuition insists that "of course the radiation is beamed to the forwards direction :smile: (the black object will say that the radiation co-moves with it, so that is an invariant fact)

    So: Dumping a ton of fuel causes the same force as burning the fuel in two opposing rocket motors, in all frames. Now the question is: Does dumping fuel cause a force, in frames where the fuel has momentum, just because fuel's momentum is dumped with the fuel?
     
  10. Oct 2, 2016 #9

    vanhees71

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    This is quite a subtle issue, and it's, of course, solved immediately when going from integral laws (in this case the momentum balance of the entire spherical radiating body) to local laws. The reason, why there is apparent trouble is that only the total energy-momentum tensor
    $$\Theta^{\mu \nu}=\Theta_{\text{mech}}^{\mu \nu}+\Theta_{\text{em}}^{\mu \nu}$$
    is conserved in the sense that
    $$\partial_{\mu} \Theta^{\mu \nu}=0.$$
    If and only if that equation is fulfilled the integral
    $$P^{\nu}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \Theta^{0\nu}$$
    is a four-vector.

    So if you take into account all energy, momentum, and stress, i.e., the mechanical stresses holding the radiating sphere together and that of the electromagnetic field (i.e., the radiation) you get a correct description of the energy-momentum balance in terms of a balance equation for energy and momentum.
     
  11. Oct 2, 2016 #10

    Dale

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    So in the rest frame of the platform, and in appropriate units, the pulse off the front will exert a four-force of ##F_f = (-1,-1,0,0)## and the pulse off the back will exert a four-force of ##F_b=(-1,1,0,0)## so there will be a net four-force of ##F_n=(-2,0,0,0)##. In the frame where the platform is moving forward at v=0.6, then the four-forces will be ##F'_f = (-2,-2,0,0)## and ##F'_b=(-0.5,-0.5,0,0)## respectively with a net four-force of ##F'_n=(-2.5,-1.5,0,0)##.

    Unfortunately, I couldn't find how to relate the three force to the spacelike part of the four-force in the case of a variable mass. But clearly the spacelike part of ##F'_f## is not equal to the spacelike part of ##F'_b##.
     
  12. Oct 2, 2016 #11
    Let's consider following "spacecraft": A fuel tank connected by a slack fuel hose to two rocket motors that are bolted together and point to opposite directions.

    Now when the rockets fire, we can agree that the rockets part of this device does not feel a net force in any frame. Right? Can we also agree that the rockets do not exert a force on the tank in any frame?

    (If the last question sounds dumb, let me remind that this is unintuitive stuff, and some kind of momentum exchange maybe could be said to occur between the tank and the motors )
     
    Last edited: Oct 2, 2016
  13. Oct 2, 2016 #12

    Dale

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    No. There is mass and momentum flow both in and out of the rocket motors.

    I generally dislike adding unnecessary complication. The fuel, the tank, the hose, the engine, the exhaust. It is far better to simplify, just the engine and the exhaust, with all the fuel and tubes and pumps etc being included inside the engine.
     
    Last edited: Oct 2, 2016
  14. Oct 2, 2016 #13

    Ibix

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    I suspect that @jartsa's point is that if the fuel tank is floating freely, rather than rigidly attached to the rocket, it's difficult to see how the rockets could be applying a force to the tank. The answer is that they aren't. But if fuel is flowing out of the tank then the tank is a rocket in its own right, albeit with the "exhaust" being constrained to travel in the fuel pipe, and the force comes from the momentum transfer between the emptying tank and the flowing fuel.

    As @Dale says, adding bells and whistles to a thought experiment is generally a mistake. It tends to bury what's going on (in this case: a change of momentum implies a force, whatever the reason for the momentum change) in unnecessary detail.

    I also suspect that we're pushing the edges of where forces are a helpful way to analyse a problem, and we're beginning to see why references to forces in SR are few and far between. They're quite subtle, so everybody just grabs a Lagrangian and extremises it.
     
  15. Oct 2, 2016 #14
    Here's a little quiz relevant to the topic:

    A hyper-relativistic black cube has temperature 3000 degrees Kelvin.

    Photons emitted from the backside of the cube are mostly

    a) very energetic photons
    b) very low energy photons

    a is correct, because the direction of the photons is mostly forwards
     
    Last edited: Oct 2, 2016
  16. Oct 2, 2016 #15

    Ibix

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    Temperature defined in the rest frame of the cube, photons measured in the frame where the cube is moving near the speed of light? Without more detailed analysis I won't go further than (c) On average, lower energy than those emitted from the front face.
     
  17. Oct 2, 2016 #16
    Yes. Temperature defined in the rest frame of the cube, photons measured in the frame where the cube is moving near the speed of light.

    And a low energy photon means a lower energy photon than what is typically emitted from a black object at rest and at 3000 degrees Kelvin temperature.

    And a high energy photon means a higher energy photon than what is typically emitted from a black object at rest and at 3000 degrees Kelvin temperature.
     
    Last edited: Oct 2, 2016
  18. Oct 3, 2016 #17

    vanhees71

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    As I said, everything gets very transparent when you use a covariant formalism. Thermodynamic quantities like temperature and chemical potential, specifying an intrinsic state of the medium are always defined in the (local) rest frame of this medium (fluid cell). Then by definition the phase-space distribution function is a scalar. That said, it's very simple to deduce the black-body spectrum. In the rest frame of the medium, it's given by
    $$f(\vec{p}^*)=\frac{2}{\exp(E_{\vec{p}}^*/T)-1},$$
    where the factor 2 is due to the two polarization states for each photon mode of definite momentum, and ##E_{\vec{p}}=|\vec{p}|## is the "on-shell energy" of the photon.

    Now in the local rest frame the four-velocity is ##u^*=(1,0,0,0)##. Thus you can write
    $$f(\vec{p}^*)=f(p^*)=\frac{2}{\exp(u^* \cdot p^*/T)-1}.$$
    Now ##u## is a four-vector, and thus you have in any frame
    $$f'(p)=f(p^*)=\frac{2}{\exp(u \cdot p/T)-1}, \quad p^0=E_{\vec{p}}=|\vec{p}|$$
    where I used that by definition the temperature ##T'=T## is a scalar.

    The energy-momentum tensor of the radiation is given by
    $$T^{\mu \nu}=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3 E_{\vec{p}}} p^{\mu} p^{\nu} f(p).$$
    This is a Minkowski tensor field, and it's clear how it transforms under Lorentz transformations. So there is no problem at all with the energy-momentum balance in whatever frame.
     
  19. Oct 3, 2016 #18

    Dale

    Staff: Mentor

    I already covered the relevant part of this in post 10.
     
  20. Oct 4, 2016 #19

    Ah now I see. An astronaut jumps off from the back of a forwards moving spacecraft. This causes decrease of momentum of the spacecraft, a force in other words. The more forceful the jump, the smaller the force on the spacecraft.

    No sarcasm. I just understood that thing.
     
    Last edited: Oct 4, 2016
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