In an astrophysics class I learned about the Poynting-Robertson effect, by which grains of dust orbiting a star slow down and eventually fall into the star. Every source that I have been able to find on this subject explains it by saying that in the star's reference frame the dust mote emits more light in the direction it is moving due to relativistic beaming. In the dust mote's reference frame it emits radiation isotropically, but due to relativistic aberration it absorbs slightly more radiation on the front than the back. I understand this explanation, but it seems a bit unsatisfying.

What about a hot object that is not orbiting a star? If I heated up a cannon ball and launched it out into deep space at a significant fraction of the speed of light, wouldn't it slow down over time due to relativistic beaming? How does this look in the ball's frame of reference if it isn't absorbing radiation from a star?

If there were nothing else but myself and the cannonball in space and I launch the cannonball directly away from myself, what would this look like?

In the ball's frame, I would appear to accelerate away as I absorb radiation emitted by the ball, but otherwise it emits isotropically. The ball and I appear to be accelerating away from each other in this frame.

In my frame, I would expect to see redshifted blackbody radiation from the cannonball. If it is slowing down, the radiation should become bluer over time. I am also absorbing the momentum of that radiation so maybe it would just stay the same color. If I measured its temperature before launching it, I can infer that it must be emitting blueshifted blackbody radiation on the other side, slowing it down. It is not clear to me whether the cannonball is slowing down or accelerating away in this frame.

I must be missing something here, can anyone help me out?

  • $\begingroup$ Slow down relative to what? There has to be another reference frame. $\endgroup$
    – rob
    May 8, 2014 at 21:26
  • $\begingroup$ Relative to an observer. Suppose that I am out in deep space and I throw a ball. Do I see it slow down by emitting thermal radiation? $\endgroup$
    – George G
    May 8, 2014 at 21:37
  • $\begingroup$ Relative to an observer in what reference frame? In the textbook example the observer is in the rest frame of the star. In the absence of a big heavy star generating photons with a preferred direction, some observers would see your hot ball slow down, others would see it speed up. $\endgroup$
    – rob
    May 8, 2014 at 21:40

2 Answers 2


In the case of your cannonball in deep space, you need a reference frame to work with as well as one for the cannonball. So, if the cannonball is all by itself, just radiating heat, then in its own frame it's radiating isotropically and will travel in a straight line, and not decelerate. (Remember that even if the radiation had a substantial momentum it's radiating in all directions so the net force on it is zero).

If the cannonball is absorbing radiation from a source the situation is different. You have a source -- say a star -- and radiation coming in and hitting it which is absorbed and re-emitted according to blackbody/ thermodynamics laws. Now you can think about the frame of the star and see that the cannonball would slow down, in accordance with the Poynting Robertson effect, though it would be very small as anything bigger than a dust grain doesn't experience it.

  • 1
    $\begingroup$ Of course, when all else fails you can use the local frame of the CMB, toward which the cannon ball will tend (but very slowly). $\endgroup$ May 8, 2014 at 23:07
  • $\begingroup$ But the CMB is isotropic, no? So in that sense it could be treated as a sphere surrounding the cannonball, and even treating it relativistically you get radiation from all directions, which once again cancels out. $\endgroup$
    – Jesse
    May 9, 2014 at 15:14
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    $\begingroup$ In it's rest frame it is isotropic, but in any other frame it is (1) red shifted to the back and blue shifted in the front and (2) Lorentz focused to the front. Both effects result in a asymmetric radiation pressure that tends to bring objects closer to the rest frame. The time scale for the effect to matter is cosmological. $\endgroup$ May 9, 2014 at 15:57

If the radiation in ball's rest frame is isotropic, then there is zero net momentum and the ball stays in rest. From a point of an observer that sees the ball moving it may look differently: due to aberation most of the radiation goes in the direction of ball's motion and also photons have higher momentum because of relativistic boots. Therefore if integrated, one may think there IS a net force acting on the ball and slowing it down.

In fact, in the ball's rest frame there is a 4-force acting on it that has only the time component non-zero - meaning the ball is loosing energy. Its 4-velocity also has only its time component non-zero being $v=(1,0,0,0)$. Because 4-force is parallel to 4-velocity, it cannot change 4-velocity direction. It only chnages the balls energy (rest mass).

If boosted to a moving frame, 4-velocity is transformed and will develop some non-zero spatial components. At the same time, however, the 4-force is also transformed, but it still will be parallel to 4-velocity. That is why there is no acceleration/decelaration of the ball due to its own radiation in any reference frame.

Mathematicaly, in rest frame the 4-force is a multiple of 4-velocity and radiative power $W$ $$F^i = W U^i$$ and the equation of motion is $$\frac{d P^i}{ds} = \frac{dm}{ds} U^i + m a^i = W U^i \,,$$ where $a$ is 4-acceleration and $m$ is mass.

Projection to 4-velocity gives $$\frac{dm}{ds} = W$$ and in the perpendicular direction we have $$a^i = 0 \quad {\rm (no\ acceleration)}.$$ The last equation is of course valid in any frame.


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