How to Build a Laser Death Ray

Material Response to Average Intensity
Rough empirical values used for laser machining and welding are described below (Lasers and their Applications, A. Sona, ed, chapter "Machining with lasers" by Dieter Roess)

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 μ-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.

Below is a calculator to estimate the effects of a heat ray incident on a surface of known material properties (or select from a set of pre-defined materials).

Beam Power, P
Beam Diameter at Target, S
Initial (Ambient) Temperature, T0
Ambient Pressure, p0
Surface properties Absorptivity, a
Input custom material properties
Name
Density, ρ
Heat of Fusion, Hf
Heat of Vaporization, Hv
Specific Heat, C
Melting Temperature, Tm
Boiling Temperature, Tb
Thermal Conductivity, κ
Viscosity Coefficient, μ @ T&mu
Surface Tension, λ @ T&lambda
Molar mass
Choose pre-defined material

The physics used in the calculator is described in the following sections. Keep in mind that many of these calculations are very approximate, more accurate treatments of heat flow into a surface generally require complex numerical simulations.

Light contains energy. When that light is absorbed by a material, its energy goes toward heating the material. However, the faster the material gains heat, the faster it sheds heat. If the rate of heat loss gets high enough to equal the rate of heat gain, the death ray can no longer raise the temperature of the surface. Otherwise, the surface continues to gain heat and its temperature rises. The temperature to which a beam of light can bring a material depends on the intensity of the light and the rates and mechanisms by which the material responds to the heat.

There are generally considered to be three mechanisms by which heat moves around: heat conduction, heat convection, and radiation. Heat conduction is what you get when heat moves through a material, transfered from atom to atom by the atoms bumping into each other and exchanging energy. Convection is the movement of heat by moving the material it is stored in. Radiation transfers heat by the emission or absorption of electromagnetic waves. When a surface is exposed to a death ray, it is being heated by radiation.

To estimate the temperature after illumination by a death ray, first note some of the basic parameters of the beam. The section on diffraction allows to to find the diameter of the spot, S, into which you can focus your beam. If you know the time average power, P, of the beam, the total intensity (power divided by area) of the beam is

I = P / (π S2 / 4).

Find the absorptivity, a, of the illuminated material. This is a number between 0 and 1 that determines the fraction of light that is absorbed. The intensity absorbed by the surface is

Ia = I × a.
Experience with laser machining tells us that the absorptivity of materials under intense laser irradiation is typically high. From this reference we find
  • "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%."
  • "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."

You will also want to know some properties of the material being illuminated that deal with heat movement, specifically:
  • ρ, the density. This is the mass of material divided by its volume.
  • κ, the thermal conductivity. This is the power of heat flow (heat energy crossing a surface per unit time) per unit area divided by the temperature gradient (the temperature gradient is the change in temperature ΔT for a given change in distance ΔX). If you know the temperture gradient, the thermal conductivity tells you how much power is flowing P = κ A ΔT/ΔX. Conversely, if you know how much heat power is flowing through a material, you know how fast the temperature changes. Note that this gives all the fundamentals of heat conduction - heat power flows from high temperature to low temperature at a rate proportional to the temperature change per unit distance and proportioanl to the area of the surface across which the heat can flow.
  • Cp, the specific heat. This is the heat energy E a material gains divded by its mass M and divided by its change in temperature ΔT. If you know how much the temperature has changes in a material, you can find out the change in heat energy in it by E = M Cp ΔT. Similarly, if you know how much energy a material has gained, you can determine the change in temperature ΔT = E / (M Cp). Note - if the material reaches its melting point or boiling point, you can't use this equation! Heat will go into melting or vaporizing the material rather than changing its temperature (or, if cooling, the condensing or solidifying material will release heat). Use the specific heat to find how much energy goes into the material up to the melting point, use the heat of fusion (below) to find out how much energy is required to melt the material, and then use the specific heat again to describe the material's rise in temperature once all of it has melted.
  • Hv, the heat of vaporization. This is the heat energy E needed to vaporize a material divided by its mass M, Hv = E/M.
  • Hf, the heat of fusion. This is the heat energy E needed to melt a material divided by its mass M, Hv = E/M.
  • Tm, the melting temperature.
  • χ = κ/(ρ Cp), the thermal diffusivity.
  • λ, the surface tension, at a known temperature Tλ.
  • μ, the coefficient of viscosity, at a known temperature Tμ.
Also note the ambient pressure p0 and ambient temperature T0. If the target of the death ray is of finite thickness, note the thickness X of the slab being heated.

Conduction

One of the first significant means heat is removed from the illuminated spot is via conduction. The heat from the beam immediately begins to diffuse into the surrounding colder material. As described in the definition of κ heat flow is driven by temperature gradients - the faster the temperature changes, the more heat power flows across a given area. As before, the equation is

P = κ A ΔT/ΔX.
After being illuminated for a time t, heat will have penetrated a distance D into the material approximately given by
D = √ 4 χ t .
The front of the heated area will advance at a speed of
vD = √ χ/t 
= 2 χ / D.
Note that the heated area initially advances rapidly, but eventually gets large enough to insulate the illuminated spot from the surrounding cold material. Also note that these equations break down once the material starts melting (or evaporating), since when this happens heat energy is being used to melt or evaporate the material rather than raising the temperature.

One limiting case deserves mention. For a point source of heat on the surface of a half-infinite slab, the temperature distribution at a distance r and time t is

T = [P/(4 π κ r)][1-erf(r/sqrt(4 χ t))] + T0.
Here, erf(X) is known as the error function of X. The main detail we need to know is that for large X, erf(X)→1, and for small X, erf(X)→0. This means that for small distances or large times, the temperature approaches a constant
T = P/(4 π κ r) + T0.
Conversely, if we know the temperature at some distance r, we know the power being conducted away
P = 4 π κ r (T - T0).
These relations only hold if the spot size S is much smaller than the thickness of the slab X, and while the heat diffusion distance D remains less than X. When D is not much larger than S, these relations are only approximate. In particular, for thin slabs, power cannot be conducted away through the bottom of the slab so the temperature will keep rising indefinitely (neglecting radiation). Still, for thick slabs, you can estimate the maximum temperature due to conduction
T = Pa/(4 π κ S/2) + T0
and if you know the temperature, the conducted power is very approximately
Pc = 4 π κ (T - T0) S/2.

Radiation

One of the methods by which hot surfaces dispose of their heat is by radiating it as electromagnetic waves. The total radiated power is given by

Ir = e σ T4,
where T is the temperature in Kelvins, e is the emissivity (a number between 0 and 1 that determines how well the surface can radiate), and σ is a constant of nature
σ = 5.67×10-8 W m-2 K-4.
Note that σ is a constant - the letters after it just tell you want units to use in the equation. If you use temperature in kelvins, for example, it gives you intensity in watts per square meter.

For this rough estimate, you can set the absorptivity equal to the emssivity (this is strictly always true for absorption and emission at the same wavelength, but both a and e vary with wavelength). If we assume that radiation is the only mechanism through which a surface sheds its heat, then the surface will gain heat until the intensity of radiation equals the intensity of beam energy being absorbed. In other words, Ia = Ir. This allows us to solve for the maximum temperature the surface will reach

T = (I/σ)1/4.

This estimate ignores other mechanisms of shedding heat, such as heat conduction from the surface into the bulk material, or transfer of heat to the surrounding air or water which is then carried away by a rising plume of this warmed fluid. When a surface gets very hot, radiation is the dominant method of disposing of heat. However, for cooler temperatures (like "just" the heat of a stove burner on high), the transfer of heat by conduction to cooler matter in contact with the material can be faster than radiation.

Melting

Once a material reaches its melting temperature, any further input of heat goes toward melting the material rather than raising its temperature (although keep in mind that if not all of the material is at the melting temperature, heat will be conducted in to the melting zone from hotter areas and away from the melting zone toward cooler areas). It takes an energy

Emelt = Hf ρ Vmelt
to melt a volume Vmelt of the material. Thus the energy needed to melt a volume V of material initially at temperature T0 will be
E = Emelt + (Tm - T0) Cp ρ V.

If the melt is not removed from the hole, the heat from the laser will need to conduct through the melt to reach the melt front. This will raise the temperature of the surface above the melting temperature. When as much heat is being conducted away from the melt front as is being conducted in to it, the melt front will no longer advance. A rough idea of this condition can be found using the relations for conduction into a thick slab, giving a maximum melt distance

rmelt = Pa/(4 π κ (Tm-T0)).

Vaporization

A material can evaporate at any temperature (although if it is at a temperature less than the melting temperature, it is called sublimation rather than evaporation). If the temperature is low, the rate of evaporation will be very low, but the vaporization rate increases rapidly with temperature, until, at high intensities, most of the energy might be going in to evaporation. The energy needed to evaporate a volume of material is

Eevap = Hv ρ Vmelt

From this, we can see that the energy needed to completely evaporate a volume V of material at temperature T will be

E = Eevap + Emelt + (T - T0) Cp ρ V.
However, a death ray beam may heat the material enough that the evaporate jets off at high speed carrying bits of the melt with it, or at least creating enough pressure that it pushes the melt out of the hole. This is beneficial - it takes much less energy to melt a material than to vaporize it, so the death ray can remove more material by melting a lot of it and only evaporating a little of it.

The amount of material that evaporates depends on the vapor pressure of the material at the temperature to which the laser heats the top layer. The vapor pressure is, as its name implies, the pressure of the vapor emitted at that temperature. If the vapor pressure inside the molten material ever exceeds the pressure on the material at that point, it can push the material aside as a vapor bubble spontaneously grows. When this happens with a pot of water on the stove, we call it boiling. This tells us that the vapor pressure is one atmosphere (1 × 105 Pa) at the boiling temperature Tb. However, at different pressures, the boiling temperature will be different - this is why it is not sufficient to say that the laser has to heat the material to its boiling temperature to vaporize it. Instead, we can use the Clausius-Clapeyron relation to find the vapor pressure p at any temperature T if you know the vapor pressure pref at any other temperature Tref.

p = pref exp[Hv Mmol (1/Tref - 1/T) / R]
p = 1 × 105 Pa exp[Hv Mmol (1/Tb - 1/T) / R]
or equivalently
T = 1/[1/Tref - (R / (Hv Mmol)) ln(p/pref)]
T = 1/[1/Tb - (R / (Hv Mmol)) ln(p / 1 × 105 Pa)]
where R is the gas constant
R = 8.314472 J/(Mol K)
and Mmol is the mass of one mole of the material (be careful - this is usually given in grams, and if you work in SI units, you will need to divide by 1000 to get kg).

If the vapor pressure is less than the ambient pressure p0, the vapor cannot push the surrounding air out of the way. At most, it slowly diffuses away, remaining close to equilibrium and re-absorbing almost as much of the material as evaporates. As soon as the vapor pressure exceeds p0, the evaporate will be blown away by its excess pressure at speed ve. The rate of mass flow dM/dt (mass evaporated divided by the time to evaporate it) will greatly increase.

dM/dt = (π S2/4) ρvapor ve
where ρvapor is the density of the vapor
ρvapor = p Mmol/(R T)
The force F produced by the escaping gas on the surface is
F = ve dM/dt + p0 (π S2/4) Newton's first law plus force balance
= p (π S2/4) definition of pressure as force divided by the area over which the force is applied
and by Newton's third law, the surface produces the same force on the gas (the surrounding atmosphere produces a force in the opposite direction of p0 (π S2/4), so the net force on the gas is just ve dM/dt). Therefore
dM/dt = (π S2/4) (p - p0) / ve.
Relating this expression to the original formula for the rate of mass flow, we immediately find the jet velocity of the evaporate
ve = ((p - p0) / ρ)1/2
= (R T (1 - p0/p) / Mmol)1/2

Estimating how much material splatters out due to the gas jet pressure is rather involved. If Dm is the depth of the melted zone, then it will become unstable and break up into droplets when a parameter known as the Weber number Weg is greater than 1

Weg = ρvapor Dm ve2 / λ > 1.
An analogous condition relating the wind shear breaking droplets off of a liquid dominated by viscosity rather than surface tension would be
ρvapor Dm ve / &mu > 1;
Thus, the maximum thickness of melted region we can have is
Dm = min(&mu/(ρvapor ve), λ/(ρvapor ve2))
Let us denote by dMf/dt the rate at which molten material is blown out of the hole. The rate at which material is melted is dMf/dt + dM/dt. The power going in to melting Pm, vaporization Pv, conductive heating of the solid Pc,s, and conductive heating of the liquid Pc,l are thus
Pv = dM/dt Hv
Pc,l = (dMf/dt + dM/dt) (T - Tm) Cp
Pm = (dMf/dt + dM/dt) Hf
Pc,s = (dMf/dt + dM/dt) (Tm - T0) Cp
and we have the obvious relation
Pa = Pv + Pc,l + Pm + Pc,s.
The power Pm + Pc,s must be conducted across the melt zone to the melt front
Pm + Pc,s = κ (&pi S2/4) (T - Tm)/Dm
= Pa - Pv - Pc,l.
If we know dM/dt, we can use
P = dM/dt Hv + dM/dt (1 + (dMf/dt)/(dM/dt)) [Hf + (T - T0) Cp]
to solve for
dMf/dt = (P - dM/dt Hv)/[Hf + (T - T0) Cp] - dM/dt
Unfortunately, solving backwords from the power and beam parameters gives no algebraic solution. You can, however, solve forwards from a trial value of the pressure to get Dm and dMf/dt. This allows you to solve these non-linear equations numerically. You can do this by plotting the lines
Pm + Pc,s = κ (&pi S2/4) (T - Tm)/Dm
and
Pm + Pc,s = (dMf/dt + dM/dt) Hf + (dMf/dt + dM/dt) (Tm - T0) Cp
on a graph and finding where they cross. Alternately, you can program these equations into a root-finding algorithm and solve them on a computer. It may be that the self consistent solution has a negative value for dMf/dt. Since this is impossible, it means that the material is not vaporizing fast enough to break the surface into droplets. Set dMf/dt = 0, solve Dm = κ (T - Tm) / (Pm + Pc,s), and use this value to self consistently find the solution.

Using the pressure and melt depth from these equations, you can find the rate of melting, from which you can determine the speed at which the death ray drills into its target

vdrill = (dMf/dt + dM/dt) / (&rho &pi S2/4).

One complication is that the diffusion coefficient μ and surface tension λ depend on temperature. Since you have to solve for the rate of drilling numerically anyway, this isn't too much more of a problem. The viscosity for many materials is approximately given by

μ = μ0 exp[Hv Mmol (1/T - 1/Tμ) / (3 R)]
and the surface tension by
λ = λ0 + k (Mmol/ρ)2/3 [Tλ - T]
where k = 2.1×10-7 J / (K mol2/3).

It is interesting to note that laser-heated surfaces will never boil. Since the heat enters the surface, the surface is the hottest part of the material. Thus, the surface has the highest vapor pressure. The escaping evaporate pressurizes the surface to its vapor pressure, preventing any boiling anywhere in the material.

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