I Effect of Radiation Pressure on a Moving Object

[wnvl2]

A 1 kg mass slides over a frictionless floor at a speed of 10 m/s. The mass has a top surface area of 1 m². The floor is uniformly irradiated with an intensity of 1 W/m², and the radiation is perpendicular to the floor in the floor’s rest frame. All incident radiation is absorbed by the mass—there is no re-emission.

My question is: Does the mass slow down in the rest frame of the floor?

 

[PeroK]

If the radiation is absorbed by the object, then its mass must increase. And, by conservation of momemtum, it must slow down.

 

[JimWhoKnew]

In addition to @PeroK 's answer in #2:
In the reference frames where the body is instantaneously at rest, the radiation is not perpendicular to the floor, but has a component that pushes the body backward (aberration).

Edit: It is analogous to the Newtonian problem of a bucket sliding under a perpendicular rain.

 

[HansH]

same question but then for the red mass not receiving radiation but only emitting radiation, so effectively loosing mass. Does the mass then speed up in the rest frame of the floor? (I think it will keep its speed)

 

[Dale]

Does the mass then speed up in the rest frame of the floor?

It depends on the details of how it emits the radiation. If it emits isotropically in its own frame, then no. If it emits more light to the left in its own frame then yes it will accelerate to the right.

 

[jbriggs444]

same question but then for the red mass not receiving radiation but only emitting radiation, so effectively loosing mass. Does the mass then speed up in the rest frame of the floor? (I think it will keep its speed)

It is the same situation. One has to pay attention to the direction of the emitted radiation. If the radiation is vertical in the rest frame of the floor then the object must lose mass and gain speed. If we shift to the [instantaneous] rest frame of the object we see that the emission direction must be angled slightly rearward.

 

[HansH]

the direction is equally in all directions, at least equal in the direction of the motion. (because we (at the dutch forum)) want to understand the basic idea of what radiation does without unnecessary complicating factors of non uniform directional.

 

[Dale]

the direction is equally in all directions

In which frame?

 

[jbriggs444]

the direction is equally in all directions, at least eqaual in the direction of the motion. (because we (at the dutch forum)) want to understand the basic idea of what radiation does without unnecessary complicating factors of non uniform direcitional,

If the direction is uniform in the frame of the emitter then it will be non-uniform in the frame of the floor.

Edit: Once again, simultaneous to a response from @Dale. Sorry.

 

[HansH]

It is the same situation. One has to pay attention to the direction of the emitted radiation. If the radiation is vertical in the rest frame of the floor then the object must lose mass and gain speed. If we shift to the [instantaneous] rest frame of the object we see that the emission direction must be angled slightly rearward.

but from the perspective of the restframe of the moving mass; how can it know to sent its emission be angled slightly rearward because I would expect in that restframe it radiates simply in both directions in the same way as it is at rest.

 

[wnvl2]

HansH’s question is inspired by the following scenario:

A cannonball is fired from reference frame 1 (the cannon). After a brief acceleration, it continues to move at a constant velocity relative to reference frame 1. Reference frame 2 is the frame moving along with the cannonball. Due to the firing, the cannonball is heated and at that moment has a (rest) mass m. Naturally, the cannonball will cool down, and its mass m will therefore decrease.

What happens to the velocity of the cannonball?

If we assume that, in the reference frame of the cannonball, the radiation is isotropic, then I can understand why the cannonball would not accelerate. But when we look at it from the reference frame of the cannon, is the radiation emitted by the cannonball still isotropic? If the radiation is not isotropic in the cannon’s frame, how can we then explain that the cannonball does not accelerate in the cannon’s frame?

 

[HansH]

so from that my conclusion would be that because it looses mass, it looses momentum so that momentum can only be transferred to the radiation. from the restframe of the moving mass you will see symmetrical radiation in both directions, but from the still standing refernece frame you will see red shift and blue shift, so different momentum in both directions, so I think the momentum lost from the radiated away mass transfers to momentum in the radiation.

 

[jbriggs444]

but from the perspective of the restframe of the moving mass; how can it know to sent its emission be angled slightly rearward because I would expect in that restframe it radiates simply in both directions in the same way as it is at rest.

If the radiation is uniform in the frame of the ground, it must be angled rearward in the frame of the emitter. If the radiation is uniform in the frame of the emitter it must be angled forward in the frame of the ground.

Your new hypothetical is that the radiation is uniform in the frame of the emitter. So the latter applies. It must be angled forward in the frame of the ground.

 

[wnvl2]

If the radiation is uniform in the frame of the ground, it must be angled rearward in the frame of the emitter. If the radiation is uniform in the frame of the emitter it must be angled forward in the frame of the ground.

Your new hypothetical is that the radiation is uniform in the frame of the emitter. So the latter applies. It must be angled forward in the frame of the ground.

Assume the radiation is uniform in the emitter's frame. In that case, there is no acceleration in that frame. However, from the perspective of the ground, the radiation is not uniform. How can you explain that there is no acceleration in the ground frame?

 

[HansH]

that reasoning (from

jbriggs444
) I do not understand. if i put a radiating spherical mass in free space then the only thing that can happen would be radiating in all directions at the same intensity. so if it looses mass it does not loose momentum seen from its own reference frame as there is no speed.

 

[HansH]

but it looses momentun seen from a moving reference frame, so in that moving feference frame you must conclude that the lost momentum is treansferered to the radiation.

 

[Ibix]

In that case, there is no acceleration in that frame.

No, in that case the light carries away mass but no net momentum in the ground frame. So the object must speed up to maintain a constant momentum.

Sorry, got my frames mixed up. See @jbriggs444"s reply below for the correct analysis.

Time for bed...

 

[jbriggs444]

Assume the radiation is uniform in the emitter's frame. In that case, there is no acceleration in that frame. However, from the perspective of the ground, the radiation is not uniform. How can you explain that there is no acceleration in the ground frame?

Indeed, in this situation, there is no acceleration of the emitter in either frame.

We have an emitter with constant velocity and decreasing mass. Its forward momentum is declining.

We have a forward-biased cloud of radiation with constant average velocity equal to that of the emitter and increasing mass. Its forward momentum is increasing.

Note that a cloud of radiation can have a positive total mass despite being comprised of massless individual bits.

 

[Dale]

but it looses momentun seen from a moving reference frame, so in that moving feference frame you must conclude that the lost momentum is treansferered to the radiation.

Yes. If the radiation is isotropic in the frame of the emitter then it is not isotropic in the ground frame. In the ground frame the emitter loses momentum and the radiation gains momentum.

The emitter does not accelerate in any frame. It loses momentum in the ground frame by decreasing mass while maintaining the same velocity.

 

[HansH]

now 1 next step: suppose you have this radiating sphere being at rest in its restframe called frame1. at t=0 you apply a force to it F during an interval t1 so adding a well defined momentum F*t1. then you let the mass radiate for some time so reducing its mass. and then you apply the same force F during same t1 but then in the opposite direction so having the sum of both momentum =0. Now the question is: will there be a residual speed of the moving mass in the original frame frame1 (still at the same speed as it was at the beginning)

 

[Dale]

will there be a residual speed of the moving mass in the original frame frame1 (still at the same speed as it was at the beginning)

Yes. The second force was applied to a smaller mass. Don’t produced a larger acceleration.

 

[HansH]

so then is this cycle an example of propulsion by transferring mass into energy and transferring momentum of radiation into momentum of mass I suppose because finally you remain with a residual speed while you started with no speed.

 

[jbriggs444]

so then is this cycle an example of propulsion by transferring mass into energy and transferring momentum of radiation into momentum of mass I suppose because finally you remain with a residual speed while you started with no speed.

It is a hugely inefficient method of propulsion, yes.

I see it as a Rube Goldberg variant of a photon rocket. In a normal photon rocket, we would shine a flashlight out the back. In this variant we would accelerate an incandescent bulb rearward, turn it on, allow it to radiate [isotropically in its own frame, anisotropically in ours], turn it off and bring it to a relative stop.

 

[Dale]

so then is this cycle an example of propulsion by transferring mass into energy and transferring momentum of radiation into momentum of mass I suppose because finally you remain with a residual speed while you started with no speed.

If you wanted propulsion then why would you do the second force? Just do the first force and skip the light and the second force.

Or skip the forces and just use a mirror to make anisotropic radiation, like a flashlight.

 

[HansH]

The second force was only applied because that was the way the idea came into my mind. I could use that to support the thought experiment to show the principle that you can harvest impulse from a radiating mass by describing the process split up in cycles similar to a carnot diagram. Of course you could focus the radiation into one direction to make a better propulsion. but that is obvious how that works.

 

[A.T.]

I could use that to support the thought experiment to show the principle that you can harvest impulse from a radiating mass

See also:
https://en.wikipedia.org/wiki/Pioneer_anomaly

 

[Dale]

Of course you could focus the radiation into one direction to make a better propulsion. but that is obvious how that works.

Note that the cycle mechanism works the same way. It is just a very poor way to focus the radiation. In the cycle approach (in the initial rest frame), the radiation has a net momentum in the opposite direction of the final motion. Momentum is still conserved and the complicated approach only works by the same mechanism as the obvious approach works.

There is no avoiding the conservation of momentum.

 

[HansH]

at the Dutch forum we summarised it as follows: in the moving mass frame there is no momentum (speed=0) so the mass without momentum is converted into radiation without momentum (symmetrical radiation)
in the lab frame there is momentum (speed>0) so the mass with its momentum is converted into radiation with the same momentum. In the lab frame you see that as asymmetric radiation as result of the difference in speed between lab frame and moving mass frame.

 

[Ibix]

in the moving mass frame there is no momentum (speed=0) so the mass without momentum is converted into radiation without momentum (symmetrical radiation)

I would say "radiation without net momentum" or "radiation with zero total momentum" because the radiation does carry momentum, but taken as a whole it does so in equal and opposite amounts. A similar comment applies to the lab frame analysis. Otherwise, I agree.

 

[HansH]

yes, I had already assumed adding it as total momentum

 

[wnvl2]

On a Dutch forum, there is a discussion about the following problem:

A $1 \mathrm{kg}$ mass slides across a frictionless floor with an initial velocity of $10\, \mathrm{m/s}$. The top surface of the mass has an area of $1\, \mathrm{m^2}$. The floor is perpendicularly irradiated with light of intensity $1\, \mathrm{W/m^2}$. How long does it take for the speed of the mass to decrease to $1\, \mathrm{m/s}$?

This can be seen as an interaction between the mass, the floor, and the incoming radiation.
Can I assume that the horizontal momentum of the mass remains constant when viewed from the frame of the floor?

More precisely: does the relativistic horizontal momentum of the mass remain constant from the initial state (1), throughout the interaction, until the final state (2)? That is,

$$
\gamma_1 m_1 v_1 = \gamma(t) m(t) v(t) = \gamma_2 m_2 v_2
$$

Is it valid to calculate the increase in mass via the energy absorbed from the incoming radiation, and then determine the change in velocity by assuming that the relativistic horizontal momentum remains constant?

ps No idea what is wrong with the LaTeX rendering.
[Mentors' note: You needed double hash delimiters not single dollar signs for the inline delimters. We've fixed this post for you, but next time check out our Latex help guide]

 

[Ibix]

Didn't we discuss this already?

Yes, the point is that the horizontal momentum is conserved and the mass is the only thing with horizontal momentum. As long as you're allowed to assume there's no reflection and the radiation is perpendicular to the floor in the floor rest frame, anyway. And that it doesn't lose any energy by radiation during whatever period it's absorbing radiation.