Beam-Target Interactions

You've managed to direct the fearsome energies of your death ray beam onto your foe. Great! So now you have an intense irradiated spot on your target. What happens next?

There are several that are useful to know when trying to figure this out. There is the total beam power $P$ incident on your target. There is the beam spot diameter $S$ when it is at your target. When you know both of these, you can find the beam intensity $I=P/(\pi (S/2)^{2})$. There's the length of time your beam stays on that spot $t$. The total energy $E$ delivered by the beam is related to the beam power and duration by $E=P\,t$. And the fluence $F$ is the total energy delivered for a given amount of area, $F=I\,t$.

Okay. Great! Now what happens to our target?

No effect, or Much Ado About Nothing

Well that was disappointing. How will we know if our beam won't do anything? Or more generally, we want to know what are the threshold beam properties we need to cause damage to our target.

We'll start by looking at the amount of energy (and power, intensity, and fluence) actually delivered to our target. There are three things that can happen: the energy can be reflected or scattered out of the target, the energy can be transmitted or pass through the target, or the energy can be absorbed by the target. Only the absorbed energy will actually do anything. The amount of radiation that is absorbed depends on the kind of radiation and the nature of the target.

For most of the commonly encountered laser wavelengths, from all the infrareds through visible light and ultraviolet to the soft x-rays, transmission will be negligible for any reasonable target. Sure, if the target is thinner than paper or made out of glass or something you might have to take transmission into account, but normally we are thinking of shooting things made of steel, aluminum, concrete, exotic carbon allotropes, or skin, meat, gristle, viscera, tendon, and bone. In particular, these will all tend to get absorbed at the surface (although near infrared light going through biological tissue does tend to penetrate deeply and scatter a lot). Metals tend to be initially quite reflective in the infrared part of the spectrum, but reflectivity falls off as wavelengths get shorter and generally metals stop being reflective in the ultraviolet. Non-metals don't tend to reflect much. But for high powered beams, a portion will be absorbed even by metals and this will heat the target, making the metal less reflective so that it absorbs more energy in a runaway process that can end with the metal absorbing nearly all the incident energy once it starts increasing significantly in temperature. From "Laser Machining Processes"

Material	Features
Metals	At room temperature, most metals are highly reflective of infrared energy, the initial absorptivity can be as low as 0.5% to 10%. But the focused laser beam quickly melts the metal surface and the molten metal can have an absorption of laser energy as high as 60~80%. Fusion cutting assisted with gas jet is used.
Non-Metals	Non-metallic materials are good absorbers of infrared energy. They also have lower thermal conductivity and relatively low boiling temperatures. Thus the laser energy can almost totally transmitted into the material at the spot and instantly vaporize the target material. Vaporization cutting is commonly used, nonreactive gas jet is used to prevent charring.

So metallic reflectivity can be important for determining the threshold where the target starts taking damage, but doesn't matter much once it starts taking damage.

For a sufficiently low power flux the thin surface layer will be heated to the fluid state but will stay beyond the evaporation temperature, while the solid-fluid interface will slowly progress into the bulk material by heat conduction. In iron the typical progress rate is about 10^-2 cm ms^-1, and the power flux for this situation may be around 10⁶ W cm^-2. [...]

At a somewhat higher power flux, between 10⁶ to 10⁸ Wcm² [sic], the thin absorbing surface layer is heated up to its evaporation temperature before the solid-fluid interface has progressed appreciably into the material by heat conduction. Thus, with continuing laser power, a less than $\mu$-thick layer of material will continually evaporate, with the material-air interface progressing into the material. Typically a gas jet develops [...]. The gas jet ejects also part of the molten material, thus that the progress rate of the hole is faster than with evaporation of all the material. [...]

At a still higher flux of 10⁹ to 10¹⁰ Wcm-2, after initial evaporation of the surface layer, the gas jet is thermally ionized and absorbs most of the incident radiation, which is such blocked away from the material. The surface layer explodes with an ultrasonic jet, its temperature may rise beyond 10 10⁵ ° K [sic], its pressure beyond 10³ at.

- Lasers and their Applications, A. Sona, ed, chapter "Machining with lasers" by Dieter Roess

Da math

Heat Conduction To describe the diffusion of heat we will want a few characteristics of the material $T_{i}$, the temperature of the material before the beam is on $\rho$, the material's mass density $C_{P}$, the material's specific heat, or how much energy it takes to heat up one unit of mass of the material (say, a kg) by one unit of temperature (say, 1 kelvin) $K$, the thermal conductivity of the material. The heat power flowing across an area is the area times the thermal conductivity times the temperature gradient (how quickly temperature changes with distance perpendicular to the area). $\alpha$, the thermal diffusivity. You can find this as $\alpha =K/(\rho \,C_{P})$ $\epsilon$, the emissivity of the object for the beam's radiation. This is the fraction of the beam that gets absorbed. As discussed above, you can usually approximate it as 1, but for metals and beams operating in the infrared, visible, or near ultraviolet you may need to use a value of $\epsilon$ less than one (but more than zero) to see if the beam will heat the target spot to a point where you can ignore the emissivity. The power absorbed by the target will be $P\,\epsilon$. So, can heat diffusion carry the energy you are adding away fast enough to prevent damage? Let's find out! One dimension At very early times for beams that heat only the surface of the target, the heat will diffuse away from the spot into the material as if the spot was simply a flat plate. When the size of the spot is much larger than the distance the heat has diffused, you can treat it as a purely one dimensional problem only looking at the direction into the material and ignoring either of the two directions at the surface. In this case, the material can't wick the heat away fast enough to keep the temperature from rising. The temperature continues to climb for as long as the beam remains on that spot. After a time $t$, the temperature at the surface will be $T_{1}={\frac {I\,\epsilon {\sqrt {4\,\alpha \,t}}}{{\sqrt {\pi }}\,K}}+T_{i}={\frac {4\,P\,\epsilon {\sqrt {4\,\alpha \,t}}}{\pi \,{\sqrt {\pi }}\,K\,S^{2}}}+T_{i}$ The heated region of the material will be about a distance $r_{H}={\sqrt {4\,\alpha \,t}}$. Once this heated region becomes similar in size to the spot size (say the spot radius, $S/2$) or the thickness of the material $Z$, the heat flow is no longer approximately one-dimensional and you can't use this approximation any more. Thus, this approximation only works for a time $t_{1}={\frac {S^{2}}{8\,\alpha }}$ or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_1 = \frac{Z^2}{4 \, \alpha}} , whichever is smaller after which the heat diffuses away as if in three dimensions (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} -limited) or two dimensions (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z} -limited). (The actual temperature field within the heated material is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_1(x,t) = \frac{I\sqrt{4 \, \alpha \, t}}{\sqrt{\pi} K} \, \exp \left[ -\frac{x^2}{4 \, \alpha \, t} \right] - \frac{I \, x}{K} \mbox{erfc} \left[ \frac{x}{\sqrt{4 \, \alpha \, t}} \right] + T_i } as can be verified by showing that the temperature obeys the diffusion equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial t} T(x, t) = \alpha \, \nabla^2 T(x, t) } and the boundary condition that the heat flow into the surface Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q(x,t) = - K \, \mathbf{\nabla} T(x, t) } at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = 0} is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} . ) Three dimensions If your beam spot size (and, if you are using penetrating radiation, the distance the radiation penetrates into the target) is much smaller than the thickness (or any other dimension) of the target material, heat can diffuse away in all directions into the bulk. In this situation, it turns out that the material can reach a steady state, where it is wicking away the heat as fast as it comes in. The spot will gradually rise in temperature until at times much larger than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S^2/(4 \, \alpha)} it reaches a temperature of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_3 = \frac{P \, \epsilon}{2 \, \pi \, S \, K} + T_i. } As before, the heated zone will extend to approximately a distance of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_H = \sqrt{4 \, \alpha \, t}} . If you are beaming a slab of material with thickness Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z} , this three-dimensional approximation will only work as long as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t < Z^2/(4 \, \alpha)} . (The compete temperature profile in the material is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_3(\mathbf{r}, t) = \frac{2 \, P}{K} \frac{1}{4 \, \pi \, r} \, \mbox{erfc} \left[ \frac{r}{\sqrt{4 \, \alpha \, t}} \right] + T_i } which again can be verified by showing that the temperature satisfies the diffusion equation in spherical coordinates and also satisfied the boundary condition that the heat flow into the solid at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \to 0} approaches Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P} . ) Two dimensions If your beam spot size is larger than the thickness of the material, or if the heat has conducted across three dimensions until the heated zone has reached the other edge of the slab you are heating, or if your beam is made of penetrating radiation that heats a cylinder of material deep into the target, you won't be able to conduct heat away in that third dimension. You now get a two-dimensional problem, with the heat only flowing away along the surface or perpendicular to the beam. In three dimensions, you can conduct away heat fast enough to reach a steady state. In one dimension, for as long as the heat is on the temperature keeps rising. So what about two dimensions? It turns out that you can almost conduct the heat away fast enough to reach a steady state, but not quite. The function for the two-dimensional distribution of heat gets complicated and involves strange functions that most people probably never even encounter in college, but at times much larger than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z^2/(4 \alpha)} , the temperature will be approximately Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_2 \approx \frac{P \, \epsilon}{4 \, \pi \, K \, Z} \left( \ln\left[ \frac{Z^2}{4 \, \alpha \, t} \right] - 0.577 \right) + T_i. } As with all such heat diffusion scenarios, the width of the heated region is about Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_H = \sqrt{4 \, \alpha \, t}} . If you have a beam of penetrating radiation and the distance of the heated region exceeds the penetration distance, you switch over to three dimensional heat diffusion. At least until the distance gets larger than the target thickness, in which case you're back to two dimensions again. (The full temperature field in the material will be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_2(\mathbf{r}, t) = \frac{P}{4 \, \pi \, K \, Z} \mbox{E}_1 \left[ \frac{r^2}{4 \, \alpha \, t} \right] + T_i } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} in this case is the cylindrical radial coordinate and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mbox{E}_1} is the exponential integral. It's validity can be verified in the same way as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_3} . ) Zero dimensions When the heated region is large enough to encompass the entire object, the temperature of the object becomes about Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_0 = \frac{P \, \epsilon \, t}{M \, C_P} + T_i } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} is the total mass of the object. The temperature just keeps rising until some other form of heat loss or temperature sink is able to sop up the extra heat that is being pumped in.

Vapor explosion If you are sitting there, watching a whole bunch of fluid moving past you, and you stick your finger into the fluid, your finger will feel a pressure of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = \frac{1}{2} \, \rho \, v^2, } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho} is the fluid density and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} is the speed you see the fluid rushing past. This follows from a relationship known as Bernoulli's principle, and is called the dynamic presure. In this case, the fluid is exerting its dynamic pressure on your finger, and the principle of equal and opposite reaction means your finger is exerting the same dynamic pressure back on the fluid. Now consider if a laser is blasting a hole into the fluid as it goes past you. If the laser can produce a vapor pressure just equal to the dynamic pressure, then, just like your finger, it will seem that the interface where the pressure is being generated is holding steady right in front of you. Now if you look at it from the point of view of someone moving with the fluid, they will see the laser boring a hole into the fluid at a speed of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = \sqrt{\frac{2 \, p_{\rm vapor}}{\rho}}. } Now most of the time the things we think about shooting with a beam are not fluids, but rather solids with at least some internal consistency holding them together – an internal consistency we want to remove in order to violently unmake our target. When dealing with these kinds of penetrating pressures, strong enough to deform solid materials, it has been found that it is useful to add a constant strength term Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_h} to the Bernoulli pressure relation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = \frac{1}{2} \, \rho \, v^2 + K_h. } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_h} is called the cavity strength, and is usually about 3 to 4 times the compressive strength Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_c} of the material (or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_h = \frac{2}{3} K_c \times \left( 1 + \ln \left[ \frac{2 \, G}{K_c} \right] \right)} if you want to get all exact, for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} the shear modulus). So now you can solve for the speed of the laser blasting its way into a solid Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = \sqrt{\frac{2 \, (p_{\rm vapor} - K_h)}{\rho}}. } Multiply this by the duration of the beam Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta t} to find how deep a hole the laser punches into its target Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d = v \, \Delta t . } We can easily find the total volume of the hole or crater left by the beam. Strengths and pressures and such are an amount of energy per unit volume. So if we know the energy of the beam pulse Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} , the volume exploded out of the target is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V = \frac{E}{K_h} } If the beam pulse is very short, it won't have time to burrow in very far before the pulse ends, leading to a nearly spherical crater with radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \left( \frac{3 \, V}{4 \, \pi} \right)^{1/3} } Longer pulses give deeper but narrower holes.

For beams consisting of highly energetic radiation, like hard x-rays or gamma rays or particle beams, the radiation is likely to be far more penetrating. Rather than heating the surface it will go deep into the target and heat a cylinder throughout its volume. If the radiation is penetrating enough, much of it might pass through the target. Radiation formed of energetic forms of light, like x-rays and gamma rays, will deposit more energy near where they are incident than farther in, following the Beer-Lambert law. Charged particles like electrons or ions, tend to deposit a mostly constant but slightly increasing amount of energy as they penetrate deeper, until they reach their maximum depth and dump all the rest of their energy in a localized spot inside the target (if they don't over-penetrate, that is).

Now that we have figured out how much energy has been dumped into our target, we need to figure out how the target gets rid of that energy and how much energy is needed to do something.

Heat conduction

One of the main ways heat energy leaves the hot spot illuminated by the laser is for it to diffuse away. If your beam spot can diffuse away into all three dimensions, eventually you will reach a temperature where the heat is being removed as fast as it is being created. But once the heat is limited so that it can't diffuse away in all three dimensions any more, the temperature will slowly but inevitably rise for as long as the beam is on the spot.

Radiation

All the discussion of heat conduction was assuming that conduction was the only way to get rid of heat. It's not. Hot objects can also radiate heat away. A proper analysis would take into account the radiation as well as the actual geometry of heat conduction. If you need to get this much detail, become an engineer and find an employer who will let you use their FEM software. But for the purpose of understanding how beam weapons heating an object will work, it is easiest to just mention that radiation will put an upper limit on the temperature the target can reach, when the amount of heat being radiated off into the environment is equal to the amount of heat being absorbed by the beam. Emissivity affects radiation away from the target in the same way as absorption of beam energy into the target, so if the beam has the same kind of radiation as is being radiated away, the emissivity will make no difference. Emissivity only comes into play if the beam radiation is significantly different than the heat radiation being shed from the target. Ignoring these differences in emissivity, the maximum temperature the beam-illuminated spot can reach is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{rs} = \left( \frac{P}{(\pi/4) \, S^2 \, \sigma_{SB}} \right)^{1/4} }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_{SB} = 5.67 \times 10^{-8} \mbox{W}/\mbox{m}^2/\mbox{K}^4} is the Stephan-Boltzmann constant. The maximum temperature the body as a whole will reach (which will limit the temperature you can reach for the zero-dimensional heat conduction case) is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{ra} = \left( \frac{P}{A \, \sigma_{SB}} \right)^{1/4} }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is the surface area of the object being heated.

Temperature thresholds

You now should have a rough idea of how hot your beam can get the target in the absence of target damage. Once you pass the threshold where you start actually doing things to the target, the heat you deliver go into the things you are doing rather than increasing the temperature further. Or at least in addition to increasing the temperature further. These other ways for heat to get used mean that you can't use the heat conduction approximations for estimating how hot the target gets. But at least now you know your beam is doing something to your target!

Combustion

Heck yeah! Let's set stuff on fire! Burn, baby, burn!

If your target is in an oxidizing atmosphere and you bring its temperature above the autoignition temperature, it will ignite and catch on fire. This is also usually a good indication that you are no longer in the "no effect" category. Autoignition temperatures tend to be around 450 K to 600 K for most things of interest, like wood and clothes and gasoline and such. However, rather than trying to compute if you reach the autoignition temperature, it can be more convenient to use rules of thumb on the fluence needed to ignite things. The fluence needed to cause ignition varies with the composition and color of the target and the radiant intensity and duration of the thermal pulse (longer pulses allow more time for the heat to diffuse into the target, thus lowering the surface temperature - on the other hand, a very shallow heated layer may not sustain combustion). The NATO HANDBOOK ON THE MEDICAL ASPECTS OF NBC DEFENSIVE OPERATIONS AMedP-6(B) PART I - NUCLEAR lists the radiant flux for various yields of nuclear explosives needed to ignite various fabrics, and notes that "where the radiant thermal exposure exceeds 125 Joules/sq cm, almost all ignitable materials will flame." So if you can deliver a fluence of 100 - 200 J/cm² within a second or two, you can start things burning.

Unlike all the other damage mechanisms described here that use up the heat of the beam to cause various effects, combustion adds additional heat to the target! We could probably go into coming up with increasingly convoluted models to describe this heat flow, but do we really care? Your target is ON FIRE! Let it burn.

Cooking

The brutal fact is that sometimes people use weapons against things that are still living. Shocking, I know, but true. If you heat a living thing, its tissues will start to cook once they get past a temperature of about 318 to 321 K (42 to 45 °C or so – starting human body temperature is usually 37 °C or 310 K).

Once you get to the threshold for cooking, it is convenient to look at the fluence received within a couple seconds or less. At 10 to 20 J/cm², exposed skin will sustain first degree burns, causing painful, reddened portions of the skin. Between 20 J/cm² and around 35 J/cm² the beam will cause second degree burns. This destroys the top living surface of the skin, causing fluid-filled blisters. And from about 35 to 50 J/cm² or so, the beam can cause third degree burns that destroy the full thickness of the skin. The faster you can deliver these fluences, the less fluence you need to cause the necessary effect. Even higher fluences can cause fourth degree burns, which destroy tissues below the skin, such as muscle. If skin isn't directly exposed, but instead covered by clothing, intense enough heat can still cause burns directly under the clothing (before even considering what happens when the clothes ignite) – about 60 J/cm² can cause second degree burns under army fatigues, and 120 J/cm² will cause second degree burns under chemical protective gear. Second degree or worse burns covering more than 15 to 30% of the body are very serious and will likely result in death if not given medical attention. Incapacitation will be rapid and shock can be expected within minutes. The percentage of skin burned can be estimated with the rule of nines, which assigns 9% of an adult's skin surface area to the head, 9% to each arm, 9% to each upper and lower leg (so 18% for each entire leg), 36% (four 9% sections) to the torso, and the remaining 1% to the genitalia. Because only one side of a person will be facing the beam, the available area to be burned will be halved.

The combination of effects of burning and combustion is that if you are looking for a heat ray that can act like a long-range flame thrower, try to deliver around 120 J/cm² in under a second over as large of an area as possible. You'll ignite anything flammable, like clothes or hair, and cause fourth burns to exposed skin and second or worse degree burns to skin under clothes. And those flaming clothes will further burn the skin that they originally protected from your beam.

Melting

This is one of the big ones. If you bring a material up to its melting temperature, it will – no surprise – start to melt. The act of melting absorbs heat. The amount of heat it takes to melt a given mass (a kg, say) of a material is its specific heat of fusion (which, confusingly, has nothing to do with nuclear fusion). If your beam's heat has time to diffuse through the body, then once melting starts the temperature will never get any higher than the melting temperature until the entire target is melted. In practice, though, you are adding heat much faster than this. So you have a hot surface with a temperature higher than the melting temperature that conducts heat to the melt-solid interface, At this interface, heat is used to turn the material from solid to liquid rather than increasing the temperature. And then beyond that, you have heat being conducted from the interface into the rest of the material.

Vaporization

If you heat something enough, it will begin to turn into vapor at a high enough rate to cause damage. All solid or liquid surfaces have an equilibrium vapor pressure in which if in contact with vapor made of that material at that partial pressure (pressure when considering only the material of interest, not other gases which may be present like ordinary air), the pressure of the vapor neither increases nor decreases. If the vapor has a lower partial pressure than this, net material will evaporate and escape from the solid or liquid surface. This is what we want to do with our beam. The opposite case, where the partial pressure of the vapor is higher than the equilibrium vapor pressure, makes the vapor condense on the surface and is much less interesting when we're trying to blast things.

The equilibrium vapor pressure depends on temperature – the higher the temperature, the higher the vapor pressure. As long as this vapor pressure is less than atmospheric pressure, it will diffuse away slowly – the vapor layer is impeded from leaving by the surrounding air and, hanging around in the vicinity of the surface, is likely to simply be re-asborbed rather than escaping. But once the vapor pressure exceeds the ambient pressure, that pressure actively pushes away the surrounding air to give bulk transport of the material away from the surface. You then get rapid removal of material as the bulk vapor flows away as a jet. Unless you are very close to the threshold, this vapor jet will be shooting out at high speeds.

Note that the temperature where the vapor pressure exceeds the ambient pressure will allow bubbles of vapor to spontaneously form in liquid because it can push the liquid aside to make more vapor – a phenomenon known as boiling. Hence the temperature where the vapor pressure equals the ambient pressure is often also called the boiling temperature. Also note that the boiling temperature depends on the ambient pressure – if you go where the pressure is lower the boiling temperature will decrease, and if you go where the pressure is higher the boiling temperature will increase. Actual boiling of molten material can occur when the melt is heated from below, like when you heat water in a pot on a stove. Heat added at the base can create the vapor bubbles that pushes the molten stuff aside. If the material is heated from its free surface where the vapor is able to escape, bubbles won't actually form because the evaporation is occurring at the surface rather than inside the material – the pressure of the vapor may push the molten surface out of the way but it won't create bubbles deeper in the material and you don't actually get boiling. This later situation is the usual case when a beam is heating a surface, so most beam heating will not actually cause boiling.

As with melting, heat that goes into vaporization doesn't go into raising the temperature. Perhaps not surprisingly, the amount of heat necessary to vaporize a given mass of material (a kg, for instance) is called the specific heat of vaporization. Unlike melting, your beam will tend to go through the vapor to directly impinge on the liquid-vapor interface. This raises the temperature of the surface of the melt; for high radiant intensities, this can raise the temperature well above the boiling temperature. You then have a bunch of things happening to the heat. The heat delivered by your beam goes partially into vaporizing the material at the surface, partially into the kinetic energy of the blazing hot jet of evaporate blasting away from the surface, and partially into conducting through the melt layer to the melt-solid interface (which is held at a fixed temperature of the melting temperature). Then some of the heat goes into melting the solid into a liquid, and then you finally get diffusion of heat from the melt interface into the bulk.

Now if you are crazy enough to try to actually estimate the heat flux into the material from this combination of effects, you'll have to deal with an energy balance equation requiring a solution of the temperature and speed of the vapor jet along with all the effects mentioned above. On top of that, note the the formulas worked out above for heat conduction were for a stationary heat source, not a spot of heat burrowing its way in to a chunk of solid. But if you are really this dedicated, the math is worked out in more detail here , and it also includes a handy calculator for implementing the calculations.

Sublimation

Some materials never melt, but rather transition directly from a solid into a vapor. This process is called sublimation. Dry ice and graphite are probably the most commonly known materials that sublimate. There will always be some (usually very small) amount of sublimation from any solid, but materials that are well known for sublimating start to lose material at a substantial rate when the temperature gets high enough before they are able to melt. For graphite, the temperature at which the vapor pressure exceeds the ambient pressure of Earth's atmosphere at sea level is about 3150 K, so if you can't heat graphite armor above 3150 K you won't do much to it.

Sublimation is vaporization, but where you don't have the extra complication of a liquid melt layer between the escaping vapor and the solid material.

Melt ejection

Now we're really starting to get serious. Melt ejection is where the intense vapor pressure of the evaporated material is so crushing that it literally squishes the molten layer out of the hole that is being drilled like squeezing a tube of toothpaste. This sends sparks of molten material flying, for a pretty light show along with your generous helping of beam-caused destruction. Most industrial laser cutting and drilling occurs with the help of melt ejection. Melt ejection helps you blast bigger and deeper holes into your target because you don't have to waste as much energy vaporizing your target. Just melt it and then use a bit of extra energy to make the vapor to blast that molten junk out of the way.

Vapor explosion

if you thought melt ejection was getting serious, wait until you get to vapor explosions! If the vapor pressure exceeds the mechanical strength of the material being zapped by the beam, the material will experience mechanical failure and be blasted out of the way. This will form cavities and craters from the blast. This is not the beam "burning" its way through the material, this is the same kind of mechanical deformation you get from jamming your finger through a soft stick of butter, or a tack into a wall. A sustained beam (and by this we mean maybe a millisecond long) will continue to hit the back of the cavity it is making, producing a moving source of vapor of sufficient pressure to push the material out of the way and making a deep hole. A short pulse (on the order of a few nanoseconds or less) will just vaporize a thin chunk of the surface and blast out a spherical crater.

Beams that are made of highly penetrating radiation, like particle beams or x-ray lasers, have a slightly different dynamic. Because they are not stopped at the surface they don't have to tunnel in like their lower frequency laser brethren. Instead, they can simultaneously heat an entire column of material to sufficiently high pressure that it all explodes outward. This line explosion is something like what would happen if you drilled a hole in the target, threaded the hole with det cord, and set it off.

One nice feature of using a beam to make the target explode is that the explosions are causing mechanical damage rather than thermal damage. It is one to two orders of magnitude (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 10 \times} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 100 \times} ) more efficient at causing damage than via thermal means. So a beam that uses high power pulses can be more effective for the same energy than one that relies on evaporation, melt ejection, or heat ray effects.

The threshold for causing these steam explosions in flesh appears to be around 1 MW/cm². For stronger and more refractory materials, the threshold is significantly higher, to around 1 GW/cm² for steel.

Pulse trains

If a rapid pulse just blows out a spherical crater, how do you drill a deep hole in someone without expending all the energy needed to blow them entirely to bits? One way is to emit a rapid train of pulses so closely spaced that they land on top of one another. The first pulse explodes out a crater. The second pulse explodes a crater in the back of the first pulse, making a hole that is twice as deep. Each subsequent pulse continues this progression, digging the hole deeper.

Decomposition

Some materials decompose if they reach a high enough temperature. This is common of most organic materials, speaking in the chemistry sense here so organic also includes things like plastic and benzene and other carbon-containing substances whether or not they were ever alive. If heated in the absence of oxygen, they start to break apart into simpler molecules. While the temperature at which this happens obviously depends on the substance, you might estimate that for a "typical" organic molecule you can get thermal decomposition at somewhere between 420 and 700 K. If the material contains water, this will usually be driven out at about boiling temperature – 373 K at the pressure of Earth's atmosphere at sea level, or more generally when the vapor pressure of the heated water exceeds the ambient pressure. It also happens to diamond, which decomposes into graphite at temperatures of about 2,000 K.

Much like melting, thermal decomposition will absorb heat, and heat going into the decomposition won't go into raising the temperature. Unlike melting, you don't always get a sharp interface between composed and decomposed material, but if you consider an interface of finite thickness the dynamics should be somewhat similar.

Warping and cracking

As a material heats, it expands. Differential expansion between parts that are at different temperatures will cause stresses on the material, which can cause permanent deformation or stress relief via crack propagation. Estimating these mechanical effects can get quite involved, and will probably require expensive engineering analysis software to get estimates of when it will happen. But do note that if your beam heats up part of an object and causes large thermal gradients, it can make it bend, deform, or crack.

Dazzling and blinding

If the laser is not bright enough to structurally damage the target, it can interfere with its in-band sensors. In-band, in this case, means sensors that can detect the beam. So the beam might produce so much glare that your enemy’s targeting sensor cannot see you, and thus your enemy can’t shoot you. The beam might even be blinding - while on its own it can’t do things to your enemy, the enemy’s optics on its sensors collect enough light to concentrate the beam enough to burn the sensor elements.

Irradiation

Beams of deeply penetrating ionizing radiation can cause damage even if they are not tightly focused just by virtue of their radiation getting inside the target and doing stuff. If you shine a beam of this kind on living organisms, they can develop acute radiation poisoning that will eventually sicken and possibly kill them. Ionizing radiation can also mess up microcircuitry and make it not work. This is generally described by the dose, in absorbed energy per unit of mass. For example, if a person absorbs a Joule (1 J) of energy for every kilogram (kg) of body mass, they will take a dose of one Grey (1 Gy). Actual calculations of received dose are rather involved, and generally require running simulations that throw millions of virtual particles at a virtual person with virtual organs and things and tracking how the radiation is absorbed and scattered. But just the scattered radiation from the nearby hit of a weapons-grade x-ray laser or particle beam will probably give a person a really bad week.

Ultraviolet light can cause sunburns. People with very light skin can get sunburns from a fluence of approximately 100 J/cm². Darker skin can take an order of magnitude larger fluence to cause sunburn. Ironically, this means that it takes less fluence to cause thermal burns than to cause sunburn. The difference is that for thermal burns, the fluence needs to be delivered within a second or so while for sunburn it can be delivered over the course of several hours. Because you usually won't be shining your death ray on a living target for more than a few seconds at a time, sunburn will not be a problem until after thermal burns are.

Getting the beam into the target

Particle beams and lasers of ionizing radiation are pretty easy to get their energy to the target - as long as they are not too penetrating and just mostly go through, there's not much that will block them or reflect them.

Infrared, visible light, and near ultraviolet light lasers, on the other hand, have bunch of tricky things that go on at the interface between the beam and the target. As was mentioned, at low intensities reflection of the beam can be significant. As the beam raises the temperature, the reflectivity drops and most to nearly all of the beam gets absorbed. But as the beam heats the surface further, some of the evaporating material will become ionized. This creates a plasma, and the plasma will absorb the laser beam. Now, you have the laser beam heating the plasma and the plasma heating the target material. This is where things can get really complex. The plasma can make it so the target absorbs more of the laser energy, by absorbing the laser and then conducting or radiating that energy into the target. However, the plasma can also shield the target. The main way it does this is by heating the air in front of it until that air becomes a plasma as well. Now the air-plasma is absorbing the beam, and this is further away from the surface. This air-plasma can then heat the air next to it, making the plasma progress even farther from the target material. Depending on how much intensity the laser is pumping in to the plasma wave propagating away from the target – and into the laser beam – you can get a laser-supported combustion wave, laser-supported detonation wave, or laser-supported radiation wave. A combustion wave can actually enhance the laser-target coupling in some circumstances (but not all). Detonation and radiation waves just shield the target from the laser so you want to avoid them.

For short intense pulses intended to cause vapor explosions, you will practically always get a plasma. But the pulse will be so short that this plasma won't matter much. It won't have time to expand while the laser is on, so you dump all your energy into it and let it explode when you are done.

Credit

Author: Luke Campbell