laserpropulsionvisible-light
How to calculate propulsion force on a laser pointer in reaction to the emitted light power in watts?
Should I just divide power by speed of light? Since power is energy per unit of time, hence dividing by speed of light gives momentum per unit of time, which is force.
Suppose I have pointer of 20 mW.
Is it's propulsion force $\frac{0.02}{299792458} = 6.67128\times10^{-11}$ Newtons?
There is a limit to how small you can focus an ideal single-mode laser beam. The product of the divergence half-angle $\Theta$ and the radius $w_0$ of the beam at its waist (narrowest point) is constant for any given beam. (This quantity is called the beam parameter product, and is related to the $M^2$ beam quality measure you may have heard of.) For an ideal Gaussian ("diffraction-limited") beam, it is:
$$\Theta w_0 = \lambda/\pi$$
So, to answer what I interpret as your main question:
Let's say that I have a laser beam of some given power that starts with some diameter $D_0$ at the point of emission and increases to $D_f$ at some distance $r$ away. Would this be sufficient information to imply a limit to the power per unit area (W/m^2) that could be obtained through focusing and what would that be?
The answer is no.
The parameters you have given are sufficient for calculating $\Theta$, but only if $r$ is large enough so that the points at which you measure the diameter are in each other's far field.
You would also need to know the beam radius at the waist, so you could calculate the beam parameter product. Then, to get the minimum spot size, you would need to refocus the beam so that it is maximally convergent. The absolute limit is the fictitious divergence half-angle of $\pi/2$, or 90 degrees, although in practice the theory breaks down for half-angles of more than 30 degrees (this number is from Wikipedia) since the paraxial approximation stops being valid. For an ideal beam at this impossible opening half-angle, this gives you a minimum waist radius of $2\lambda/\pi^2$. So yes, it does depend on the wavelength.
What lens characteristics and approaches would someone look for in order to do this with a laser pointer?
You need a lens with a very short focal length. This gives you the largest convergence. Note that the more convergent the beam, and the smaller the waist size, the smaller the Rayleigh range is. That is, the beam radius will get very small, but it won't stay very small, it'll get bigger very quickly as you move away from the focus. (The Rayleigh range is the distance over which the beam radius increases by $\sqrt{2}$.
In addition, thinking of a Gaussian beam as being "straight" is not quite correct. There is always a waist, always a Rayleigh range less than infinity, and always a nonzero divergence angle.
EDIT
Also, it is important to realize that there is no difference between an unfocused and a focused Gaussian beam. Refocusing a Gaussian beam with a lens just moves and resizes the waist.
The aperture size of the laser is not the same as the waist size. If the beam is more or less collimated, then the aperture will still be larger, because the waist radius is usually defined in terms of the radius at which the intensity drops to $1/e^2$ of its peak value. If the beam is cut off by an aperture at that radius, then even if it were close to diffraction-limited, it certainly wouldn't be anymore. So, apertures are always larger.
The waist is the thinnest point of the beam. Usually this point is inside the laser cavity, or outside the laser if there are focusing optics involved, which there often are. So still, the answer to your question is no. You are not missing the definition of $\lambda$; rather, you are comparing your minimum waist radius to the value of $2\lambda/\pi^2$ that I said was "impossible". I called it impossible, because to make a beam converging that strongly, you would need a lens with a focal length of zero!
Let's try a more realistic example with some numbers. Take your red laser pointer with $\lambda$ = 671 nm. Laser pointer beams are often crappy, but not so crappy as you might think, if they are single-mode. Let's assume that this particular laser pointer has an $M^2$ ("beam quality parameter", which is the beam parameter product divided by the ideal beam parameter product of $\lambda/\pi$) of 1.5. A quick Google search didn't give me typical $M^2$s of red laser pointers, but this doesn't seem to me to be too much off the mark.
Note that if you know the $M^2$ and measure the divergence of a beam, then you can calculate the waist radius. We are going to do that now. Suppose the laser pointer beam is nearly collimated: you measure a divergence of 0.3 milliradians, about 0.017 degrees. Then the waist size is
$$ w_0 = \frac{M^2 \lambda} {\pi\Theta} = \frac{1.5 \times 671 \times 10^{-9}} {\pi \times 3 \times 10^{-4}} \approx 1\,\text{mm}. $$
In this case, they probably designed the laser pointer with an aperture radius of 2 or 3 mm.
Now suppose you focus your collimated beam with a 1 cm focal length positive lens, which is quite a strong lens. The beam's new waist will be at the lens's focal length. That means you can calculate the divergence half-angle: it is the smaller acute angle of a right triangle with legs 1 mm and 10 mm. So,
$$\tan\Theta = 1/10,$$
or $\Theta\approx$ 6 degrees. Applying the formula once more to calculate the waist yields a waist radius of 3.2 microns, which is quite small indeed.
A "safe" laser pointer might have a power of 1 mW. The peak intensity is equal to $2P/\pi w_0^2$, so before the lens the peak intensity is about 600 W/m^2. After the lens it is about 100000 times larger.
So, to summarize:
This is a fun, high-quality qualifying exam question. The algebra is not hard; the physical insight takes some real thought; there are many ways to be partly right. Here's my take on it.
From Einstein's equation $E^2 = p^2 + m^2$ we have for each photon $E = p = hf_0$ (in the reference frame of the laser). We can use the power of the laser $P_0$ to find the rate at which individual photons are emitted: $$P_0 = \dot N_0 hf_0.$$ (Pardon me for writing $\dot N$ rather than $dN/dt$.) The total force exerted by the laser is the time derivative of its momentum, which is the time derivative of its energy, which is just the power, $$F = \dot p = \dot E = P_0,$$ and since the laser's momentum must be coming from the rocket we have its acceleration is $a = P_0/m_0$. You'll have to stick some $c$s back in to make the dimensions come out right.
If the rocket is moving away from Earth at constant speed $\beta = v/c$, with corresponding relativistic factor $\gamma = 1/\sqrt{1-\beta^2}$, there are three factors that affect the power received at Earth:
Combining these we have a power recieved at Earth of $$P' = \dot N' hf = \frac{P_0}{\gamma(1+\beta)}\sqrt\frac{1-\beta}{1+\beta} % = \frac{P_0}{\gamma^2(1+\beta)^2} = P_0 \frac{1-\beta}{1+\beta} .$$
It's always possible in relativity problems to get identical results using classical EM fields for light instead of photons, with Poynting vectors carrying momentum, etc. I would not know how to go about that in this case.
This part was not immediately obvious to me. The messy option is to try and integrate the expression from the previous section; that probably requires assumptions about the time profile of the acceleration. Usually when you only know the initial and final conditions for a problem, conservation of energy is a good strategy. I wasted some time before I remembered to use conservation of momentum, too.
We know that the final momentum of the rocket is $p_f = \gamma_f v_f m_f$, and that the combined momentum of the rocket and its laser exhaust, in the initial rest frame, is zero. Then, using Einstein's equation again, we have a bunch of backwards-going photons with total energy $$E_{\text{exhaust}} = \left|-p_f\right|$$ and a rocket with total energy $$E_{\text{rocket}}^2 = p_f^2 + m_f ^2.$$ If we assume that none of the laser power was wasted as heat, these components must add up to the initial rest mass of the rocket,
\begin{align} p_f + \sqrt{p_f^2 + m_f^2} &= m_0 \\ m_f \left( \gamma_f v_f + \sqrt{\gamma_f^2 v_f^2 + 1} \right) &= m_0 \\ m_f \left( \gamma_f v_f + \gamma_f \right) &= m_0 \\ m_f &= m_0 \sqrt{\frac{1-v_f}{1+v_f}} \end{align}
Actually this result also holds if the laser power supply is inefficient, so long as the rocket is thermally insulated such that the thermal waste photons are all emitted in the same direction as the exhaust --- tail of the rocket hot, head of the rocket cold. If there's forward-going thermal radiation, it will be focused in the forward direction in a complicated way, and the problem becomes much more difficult.
Best Answer
The Poynting vector $\vec{N}$ is the power per unit area of your beam.
If the beam is perfectly absorbed, then the force is given by $$ F = \frac{1}{c} \int \vec{N} \cdot d\vec{A}$$
So, providing you have the beam incident normally upon something, the force on it will just be the power of the laser divided by the speed of light.
Of course, if the light is reflected, then you have more force. In the limit of a perfect, specular reflection, the force is doubled.
If what you mean is, suppose I have a laser pointer suspended in space and I turn it on; What is the apparent accelerating force on the laser pointer? The answer is exactly the same as above. Power divided by the speed of light.