# Robertson-Poynting effect and Relativistic beaming

1. Nov 22, 2014

### universal_101

Statement 1 - Let's assume a charged particle emitting radiation isotropically in it's rest frame, when boosted in another frame we get the relativistic beaming effect. This results in anisotropy of the radiation emitted by the charged particle in the boosted frame.

Statement 2 - But a particle emitting an-isotropically (like beaming) in it's rest frame should have a non-zero back reaction force on it, for the radiated energy comprises a non-zero (i.e net) momentum in the forward direction.

My question is, since an inertial particle emitting isotropically have zero four force vector in it's rest frame, transforming this vector to any boosted frame should again give the zero four force. But according to statement 2 we should have a non-zero back reaction force on the emitting particle.

That is, is there a difference between, a particle described by statement 2 and a particle viewed in a boosted frame in statement 1.

Thanks.

2. Nov 22, 2014

### Staff: Mentor

To summarize, a particle emitting radiation isotropically in its rest frame changes its momentum in other frames.
So what? Where is the contradiction? Note that it will not change its speed! It changes something different, and this does not require a force on it.

3. Nov 22, 2014

### pervect

Staff Emeritus
Wiki http://en.wikipedia.org/w/index.php?title=Poynting–Robertson_effect&oldid=621887101 explains the case where the particle is absorbing external radiation from the sun and re-radiating it, but Wikki's explanation explains the drag by absorption of aberrated radiation in one frame and emission in another. It is not so clear from the wiki explanation what happens if we have a particle or a spaceship generating its own heat to radiate.

I don't have a textbook reference which discusses it, but if we consider a closed system with two parts, a particle and a radiation field, in the particle frame the particle is loosing energy, the radiation field is gaining energy, and there is no momentum transfer from the particle to the field.

As we were careful to keep the system is closed, we should be able to use 4 vectors. However, applying a Lorentz transform to the energy momentum 4 vector, the transform of what appears to be a pure energy in the particles rest frame, (E, $\vec{0}$) should appear as a transfer of both energy AND momentum from the particle to the field in a frame moving with respect to the particle.

4. Nov 23, 2014

### universal_101

Isotopically emitting particle would give rise to relativistic beaming in other frames, and this beaming is anisotropic.

And a particle which is emitting non isotopically in its rest frame experience a back reaction force.

So the problem is, why does a particle which is emitting isotopically in its rest frame when seen in other frames does not experience a back reaction force due to non isotropy of relativistic beaming in these other frames.

5. Nov 23, 2014

### Staff: Mentor

It loses energy and therefore mass, which means it loses momentum even if it maintains a constant speed. I don't think viewing this as a force is useful.

6. Nov 23, 2014

### Staff: Mentor

The four force is non zero in its rest frame. Hint: what is the timelike component of the four momentum and does that give you any clue about the time like component of the four force.

Last edited: Nov 23, 2014
7. Nov 23, 2014

### universal_101

And this transfer of momentum from particle to the field should have a back reaction force on the particle!

Just like a particle experiencing a back reaction force if it emits non-isotropically in its rest frame.

8. Nov 23, 2014

### universal_101

OK, then how is this situation any different from a particle emitting non-isotropically in its rest frame, because such a particle experiences a back reaction force due to net momentum imparted to the radiation field.

9. Nov 23, 2014

### universal_101

Right, the time like component will be dE/dt, and this will give non-zero space like component of the four force in the boosted frame !

Does it mean that the particle will experience a back reaction force due to beaming ?

10. Nov 23, 2014

### Staff: Mentor

Yes, although as mfb mentioned thinking of this as a force in the traditional NewtonIan sense is not necessarily useful. It does not produce an acceleration , just a change in momentum.

This actually is not new to relativity, it is the same thing as in Newtonian physics for systems of variable mass. There are two ways to define force:
1) $F=ma$
or
2) $F=dp/dt$

These two ways of defining force are equivalent only if the mass is constant. If you imagine a Newtonian object which is ejecting material isotropically at a constant rate in its rest frame then in another frame it will be losing momentum but not accelerating. So in Newtonian mechanics F defined by 1) will be 0 but F defined by 2) will be non-zero.

The four-force is the relativistic extension of the second way.

Last edited: Nov 23, 2014
11. Nov 30, 2014

### universal_101

What I'm asking is, how is this scenario any different from a rocket losing its mass(with net momentum in one direction) and the rocket accelerating in other direction. That is, a particle losing its mass in the form of radiation(with net momentum in one direction) should accelerate in other direction.

Thanks.

12. Nov 30, 2014

### Staff: Mentor

It is not any different. It is well known that you can use light just like rocket exhaust.

If you had a rocket which was ejecting exhaust isotropically in its rest frame, then in another frame you would have anisotropic exhaust and the rocket would lose momentum without any acceleration.

It makes no difference if the rocket exhaust moves at c or at some v<c.

13. Nov 30, 2014

### universal_101

I think you are missing the point, I'm trying to make. Let's consider it the other way around, the point is if a rocket emits non-isotropically(as in relativistic beaming) in it's rest frame, then i can go to a frame where it emits isotropically, Right ?

And since the rocket is emitting isotropically in this frame, I should conclude that there is no acceleration for the rocket! But that is not the case, because it was accelerating in it's rest frame then it should accelerate in all other frames.

14. Nov 30, 2014

### Staff: Mentor

Sure. You'll see the rocket accelerating and losing mass in a rate that exactly cancels the momentum changes. The rocket accelerates but keeps its momentum.
No, you should conclude there is no change in its momentum.

15. Nov 30, 2014

### Staff: Mentor

I don't think that every non-isotropic emission pattern has an isotropic frame, but let's posit that we have an emission such that it is isotropic in some frame, let's call that the unprimed frame, and for simplicity let's consider just time and one dimension of space (t,x) so that "isotropic" means 2 exhaust particles moving in opposite directions. Let's say that the exhaust particles have an energy $E$, and a velocity $v \le 1$ in the unprimed frame, and that the rocket has velocity $u < 1$ in the unprimed frame and an invariant mass $m>2E$.

Let's actually work through the math and see if this logic is valid. In the unprimed frame the four-momentum of the exhaust particles are:
$e1=(E,Ev)$
$e2=(E,-Ev)$
Clearly the exhaust is isotropic in this frame.

In this frame the four-momentum of the rocket before exhaust is:
$$r0 = \left( \frac{m}{\sqrt{1-u^2}},\frac{m u}{\sqrt{1-u^2}} \right)$$
and the velocity before exhaust is $u$

and the four-momentum of the rocket after exhaust is:
$$r1=r0-(e1+e2)=\left( \frac{m}{\sqrt{1-u^2}}-2E, \frac{m u}{\sqrt{1-u^2}} \right)$$
and the velocity after exhaust is
$$\frac{m u}{m-2E\sqrt{1-u^2}} \ne u$$

So, even though the spatial component of the four-velocity is unchanged in the unprimed frame, the rocket does in fact accelerate in the unprimed frame.

It does accelerate in all frames, including the frame where the exhaust is isotropic. The reasoning that you pose sounds very nice at first glance, but if you actually work through the math you find that it is simply wrong.

EDIT: I see mfb put it much more succinctly. The rocket is maintaining its momentum in this frame, but just as you can lose momentum without acceleration, so you can maintain momentum while accelerating. Again, this is a fact of variable mass systems. The same effect occurs in pre-relativistic physics also.

Last edited: Nov 30, 2014
16. Nov 30, 2014

### Staff: Mentor

I thought that it might be worth posting the math for the non-relativistic case also. For convenience and later comparison with the above results, we will stick to units where c=1 as before. Now, let's say that the exhaust particles have a mass $dm$, and a velocity $v$ in the unprimed frame, and that the rocket has velocity $u$ in the unprimed frame and a mass $m>2dm$.

In the unprimed frame the momentum of the exhaust particles are:
$e1 = dm ~ v$
$e2 = -dm ~ v$
Clearly the exhaust is isotropic in this frame.

In this frame the momentum of the rocket before exhaust is:
$r0= m u$
and the velocity before exhaust is $mu/m=u$.

And the momentum of the rocket after exhaust is:
$r0 = r1 -(e1+e2) = m u$
and the velocity after exhaust is $mu/(m-2dm) \ne u$.

How is this possible with no change in momentum? Remember, we are using $F=dp/dt$ which, since $p=mv$, expands to, $F=m~dv/dt+v~dm/dt$. In this frame $F=dp/dt=0$, and since $v\ne 0$ and $dm/dt \ne 0$ then $dv/dt \ne 0$ also.

You can clearly see that the relativistic results reduce to the non relativistic results in the limit $u<<1$ for $dm=E$. So your confusion here is not a result of relativity at all, it is simply a result of not carefully dealing with a variable mass system.

Last edited: Nov 30, 2014
17. Nov 30, 2014

### universal_101

I get the point ,that is, the system in which particle is emitting isotropically in its rest frame is different from a system in which we can go to a moving frame(w.r.t the emitting particle) in which the particle is emitting isotropically.

Thanks all.

18. Nov 30, 2014

### Staff: Mentor

Yes. Those are physically different systems.