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

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 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. To describe the diffusion of heat we will want a few characteristics of the material

,li> $T_{i}$, 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 $\alpha =K/(\rho \,C_{P})$

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, 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

$T_{1}={\frac {2}{\sqrt {\pi }}}{\frac {I}{K}}{\sqrt {t}}+T_{i}={\frac {8}{\pi \,{\sqrt {\pi }}}}{\frac {P}{S^{2}\,K}}{\sqrt {t}}+T_{i}$

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 $t_{1}={\frac {Z^{2}}{8\,\alpha }}$, whichever is smaller

at which time the surface temperature will be

$T_{1}={\frac {4}{\pi }}{\frac {P}{S\,K}}{\frac {1}{\sqrt {2\,\pi \,\alpha }}}+T_{i}$ or $T_{1}={\frac {4}{\pi }}{\frac {P}{S^{2}\,K}}{\frac {Z}{\sqrt {2\,\pi \,\alpha }}}+T_{i}$, whichever is smaller

Three dimensions

If your beam spot size 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 $S^{2}/(4\,\alpha )$ it reaches a temperature of

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_3 = \frac{2 \, P}{4 \, \pi \, S \, K} + T_i. }

As before, the heated zone will extend to approximately a distance of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): 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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): 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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t < Z^2/(4 \, \alpha)} .

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, 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.

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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): 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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_2 \approx \frac{P}{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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_H = \sqrt{4 \, \alpha \, t}} .

Zero dimensions

When the heated region is large enough to encompass the entire object, the temperature of the object becomes about

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_0 = \frac{P \, t}{M \, C_P} + T_i }

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): 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.