# Using light as a barrier to block people without grilling them

We know that light exerts pressure upon matter.

I'm imagining a hallway where an intense light source on one end shines towards the other end. The goal is to have the beam of light exert so much pressure that a human being cannot enter (or reach the end of) the hallway. Of course, we only want to deter the human, not to burn them to a crisp.

Would this be realistic?

• Naw, fam. I'm not sure what the math looks like (someone will draft that and put it into an answer, I'm sure), but that would definitely kill the person you are trying to repel. – user49466 Jan 15 at 4:14
• I have a different idea which might be useful to you. The light is just a distraction, a laser pointer of sorts which shows that defenses are working. Strong enough to blind unwary person and hiding the true defenses: giant fans blowing air in constant stream and doing pressure that way. – jo1storm Jan 15 at 9:53
• Randall Munroe of XKCD fame has answered what’s basically the same question: what-if.xkcd.com/87 – Mike Scott Jan 15 at 10:45
• The closest you are going to get is the microwave area denial system, which has low risk of deep tissue injury instead burning only the skin. Most people flee before they are injured. en.wikipedia.org/wiki/Active_Denial_System – John Jan 15 at 14:27
• Keep a lid on and make sure there is enough steam, this should make sure you steam them instead of grilling them... oh wait, this is wb not cooking.... – rackandboneman Jan 15 at 16:52

No, it won't be realistic.

The radiation pressure produced by an electromagnetic radiation of intensity, or better irradiance, $$I_f$$ impinging at an angle $$\alpha$$ on a surface can be calculated according to

$$P_{Inc} = \frac{I_f}{c} \cdot \left(\cos \alpha\right)^{2}$$

where c is the speed of light.

You immediately see that, due to the c factor, you need huge irradiance to get meaningful pressures at human scale. Let's say you want to achieve $$1 \ \frac{N}{m^{2}}$$, you would need an irradiance of about $$3 \cdot 10^8 \ \frac{W}{m^{2}}$$. That would char any human on which it impinges.

In case of a person wearing a perfect reflecting suit,

if the wave is specularly reflected, then the recoil due to the reflected wave will further contribute to the radiation pressure. In the case of a perfect reflector, this pressure will be identical to the pressure caused by the incident wave

$$P_{Inc}= 2 \frac{I_f}{c}$$

It just halves the required radiance to achieve the same pressure, thus still leaving an extremely high value.

• Unless, of course, that human is very, very shiny. ;-) – Joe Bloggs Jan 15 at 7:43
• – Fabian Röling Jan 15 at 7:45
• @L.Dutch: The very shiny comment was referring to the human charring/not charring, not the energy density required. I’m right with you on the daft numbers required. – Joe Bloggs Jan 15 at 8:50
• @Renan just to put it into further perspective: 1 N/m^2 is equivalent to around 2 m/s wind speed. In order to get to storm like levels of pressure(20 m/s) we would need around 120 N/m^2. We're definitely vapour before light stops us. – Christoph Jan 15 at 12:34
• You might want to clarify: The human in the perfectly reflecting suit won't get hot. The human in the 99.9999% reflecting suit will char. – Yakk Jan 15 at 14:15

Not with light pressure. You need 300 megawatts of radiation flux for each Newton of force on the object (equivalent to the force exerted by a ~100 gram weight).

However, you can use microwaves as non-lethal deterrent. A moderate microwave flux is extremely painful on skin, well before it becomes damaging. This is the basis of several experimental crowd control weapons in real life, such as https://en.wikipedia.org/wiki/Active_Denial_System

• You could combine this with visible light such that subjects would believe the visible light was the problem. There is a protection "Reflective materials such as aluminum cooking foil should reflect this radiation and could be used to make clothing that would be protective against this radiation." (from the same source). Also a knight in shining armour would be protected. – chasly from UK Jan 15 at 9:06
• @chaslyfromUK luckily, we haven't used steel armor in well over a hundred years, I believe - and Kevlar and ceramics let microwaves just fine. – John Dvorak Jan 15 at 17:10

No

It's true that light exerts a pressure, but photons have no rest mass and very little else (thanks AlexP). You're dealing with pressure created by a distribution of energy, which is miniscule.

If you think about it, your skin burns on a beach — but you don't feel even the slightest pressure from the light that's burning you.

Now increase the light such that you could feel the pressure. You'd flash into a fine ash before you could comprehend that you were feeling pressure.

But!

What if you change your goal just a bit? What if the original intent of the light was to flash-burn anyone trying to get down the hall? Your protagonish knows this and dresses accordingly in attire that protects he/she from the burning properties of the light. And yet, as he/she walks down the hall, feels the pressure! pressure that's great enough to hinder progress! That would be a cool twist to the story.

• Photons have no rest mass. Photons most definitely do have mass, namely $\frac {h\nu}{c^2}$. – AlexP Jan 15 at 9:01
• I really like this answer in that it potentially adds another plot hook in getting said armor - which in turn can be used to reveal or explore some other important aspect of the protagonist or the world. – MrSpudtastic Jan 15 at 14:47
• If the armor is ablative, the force doesn't need to come only from the light pressure, but also reaction from the material being ablated. – azurefrog Jan 15 at 21:34
• @azurefrog, that's true, but it's not the point. I was trying to create a means of using the OP's idea. Injecting another force or pressure other than the light is counter to the OP's expectations. – JBH Jan 15 at 21:37

Literally , no. Other answers have dealt with this.

However, if you allow that the possible intruders will not be armored, you can do something like it.

Produce the light by an array of high-powered emitters. Each separate emitter produces a beam which focusses down to a very, very small spot about 1/4 to 1/10th the length of the hallway. The beams are all aimed at different spots within the hallway, and none of them is aimed directly down the corridor. As a result, virtually all of the emitted light will be absorbed by the walls.

The result will be hazardous to look at from the end of the hall, due to unavoidable (hopefully small) reflections. However, if you attempt to traverse the hall, you'll eventually encounter a wall of "hot spots", where the power density is so great as to burn holes in you. Before you get to that point, the perceived temperature will rise to unbearable levels, and this should keep intruders out.

Of course, this won't work in the presence of suitable armor - if the intruder simply rushes the corridor, she will pass almost instantly through the "death zone", and ought to be able to make it.

Maybe not visible light, but possibly an electrostatic wall