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I have been fascinated by a very intriguing question - Can lasers push objects up?

Let's say we have a $1000~\text{mW}$ laser and we would like to lift an object of weight $100~\text{g}$.

By definition: $1~\text{W} = 1 \frac{~\text{J}}{~\text{s}}$

That means the laser is emitting $1~\text{J}$ of energy per second.

On the other hand, the energy required to lift an object off the ground is by a height $h$ is given by $m \cdot g \cdot h$.

Putting in the numbers and let's say we want to solve for

$0.1~\text{kg} \cdot 9.8 \frac{~\text{m}}{~\text{s}^{2}} \cdot h = 1~\text{J}$

So, $h \approx 1~\text{m}$.

You see, if we had a $1000~\text{mW}$ laser we could lift an object of $100~\text{g}$ weight up to 1 meter in one second.

I can't see anything wrong with the above math. If this is correct, can anyone tell me then why on Earth we use heavy rockets to send objects into space?

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    $\begingroup$ see also optical levitation, directly related to optical tweezers $\endgroup$
    – scrx2
    Commented Apr 10, 2016 at 14:22
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    $\begingroup$ Are you asking why we're using heavy rockets to lift objects with an exclamation or is your question specific to this planet? $\endgroup$
    – Sebb
    Commented Apr 10, 2016 at 19:46
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    $\begingroup$ someone should forward this to Randall Munroe. $\endgroup$
    – Federico
    Commented Apr 11, 2016 at 8:38
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    $\begingroup$ @Federico Yes indeed. This is the question of "What if we tried more power?" $\endgroup$
    – Tomáš Zato
    Commented Apr 11, 2016 at 8:49
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    $\begingroup$ The basic fallacy here is you're computing "how much energy does it take to lift a 100 g weight one meter?" and then jump, without any argument or reason, to the conclusion that this amount of energy packaged up as laser light in particular would constitute a workable lifting apparatus. That's like saying that just because the price of a Big Mac is \$4, one ought to be able to construct a machine that produces a Big Mac when supplied with \$4 worth of crude oil. $\endgroup$
    – hmakholm left over Monica
    Commented Apr 11, 2016 at 10:41

4 Answers 4

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Your approach is incorrect. You cannot do this calculation by considering that the energy absorbed by the object is converted into a change in gravitational potential energy. For one thing the object would just get hot and radiate away most of the energy and for another this is a dynamical problem, you have to be able to accelerate the object upwards.

What is important is the product of the power per unit area of the laser and the area over which it is incident. More precisely, to "levitate" an object by shining a laser onto its underside requires that the force exerted upwards by the laser is equal to the force $mg$ acting downwards. A general expression one could use for a perfect absorber is $$\frac{1}{c}\int \vec{S} \cdot d\vec{A} \geq mg,$$ where $\vec{S}$ is the time-averaged Poynting vector of the laser, with a magnitude equal to the power per unit area in the beam, and the component of this normal to the surface is integrated over the surface area of the object to be levitated. If instead the object was perfectly reflective then the upward force could be as much as double that (for normal incidence).

Hence assuming I had a completely black cube of surface area $A$ oriented so that a surface was perpendicular to a laser beam with Poynting vector $S$: $$ \frac{SA}{c} \geq m g$$ $$ m \leq \frac{SA}{cg}$$ and if $SA = 1$ W, then $m \leq 3.4 \times 10^{-10}$ kg is the mass which it could accelerate upwards against gravity.

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  • $\begingroup$ With perfect mirror, the laser would lift twice as much, because the reflected light also exerts the force too (the leaving photons also have momentum, that must be taken from the lifted object). $\endgroup$
    – Jan Hudec
    Commented Apr 11, 2016 at 18:50
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    $\begingroup$ Rob, this is conspicuously missing the term Radiation Pressure somewhere. Is it intentional or just an oversight? $\frac{S}{c}$ is the Radiation Pressure on an ideal black surface. $\endgroup$
    – Tamoghna Chowdhury
    Commented Apr 12, 2016 at 4:14
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    $\begingroup$ @TamoghnaChowdhury that concept is not helpful if the surface is not black or if the surface vector is not parallel to the laser beam. I.e. in all practical cases. $\endgroup$
    – ProfRob
    Commented Apr 12, 2016 at 5:51
  • $\begingroup$ @RobJeffries Very well $\endgroup$
    – Tamoghna Chowdhury
    Commented Apr 12, 2016 at 5:52
  • $\begingroup$ And if SA=1e11W, F will be ~3e2 N, and acceleration of a 1g probe consequently ~3e5 m/s^2 (!). An acceleration phase of 100s will bring it to 0.1c, non-relativistically (which is a good enough estimate for that speed). Actually, because the probe must be reflective (who wants to absorb 1e11 W?), it's 0.2c. $\endgroup$
    – Peter - Reinstate Monica
    Commented Apr 13, 2016 at 12:47
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Your consideration implies you have a device that can convert laser light with 100% efficiency and convert it to mechanical power. This is theoretically possible, but there is not much of physics. You still need some ladder to climb along, and building a space elevator requires much tougher materials than we currently know.

In practice, you would have to use some nonideal conversion mechanism, e.g. a photovoltaic panel. With the state-of-the-art technology, you can get to roughly 40% efficiency; the mechanical part of your device can be made fairly efficient, so we can expect your laser could provide power for 1/3 m/s climbing speed with a 100 gram weight of the device. Seems possible.

Without any ladder, you can use a very powerful laser to exert direct optical pressure. If you divide the momentum of any photon by its energy, you arrive to the fact that 1000 mW laser exerts a 1/c=3.3 nN pressure on any absorbing surface. Provided the light would be fully reflected, you would need a continuous 150 megawatt laser to lift a 100-gram mirror. This is three orders of magnitude more power than what we have built so far, although the total power requirements could be significantly reduced if a mirror was added on the launchpad to form a giant optical resonator.

Another option would be to use a less powerful laser to heat up some evaporative bottom of the rocket, so that it does not have to bring all its energy in the form of relatively inefficient fuel. This appears as one of plausible ways for rocket propulsion of the future.

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Lookup Optical Tweezers.

The limitations of the technique are due to the damage thresholds; see laser ablation.

This idea has been used extensively in science fiction, especially when implemented as solar sails. It's even practical for some applications, as noted in the article. But laser propulsion from the ground suffers losses due to the atmosphere.

The photonic laser thruster is the hot new technology.

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Laser is stimulated emission of highly energetic photons. Fundamental use of laser is heating, propulsion is very distant aim which lasers can achieve.

Few kW rating lasers can actually lift the mass (very small values though) because incident energy beam has momentum associated with it. Your assumption is not correct as you are comparing heat energy with potential energy (with 100% conversion efficiency). One simply can't achieve mechanical power of scale you specified using lasers.

The simple fact that process efficiency is very low in case of laser limits our aim to use lasers for rocket propulsion. Typical laser efficiency is nearly 5-10 % of input energy (typically electricity). Converting heat energy to mechanical energy will also have many losses, extra systems will also be needed.

Video for reference: https://www.youtube.com/watch?v=3F1FDwg4XRc

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