How to Build a Laser Death Ray
Material Response to Laser Radiation
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. [...]
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).
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
The absorptivity becomes more complicated once the laser intensity is high enough to flash part of the irradiated surface into plasma. The plasma will absorb some or all of the laser light, which then conducts or re-radiates the laser energy to the surface. This is good when the material is highly reflective, since now the laser light is not diectly interacting with the surface. Unfortunately, some of the laser energy can also be re-radiated elsewhere. Further, the plasma can heat the air in front of it to a plasma, and then the air plasma can absorb the beam which will insulate the surface of the material from the beam. You can get interesting phenomena, such as laser-supported detonation or radiation waves rushing as plasma waves through the air away from the target in the direction of the laser source, but this is bad for getting energy to the target. Normally, for laser intensities high enough to cause propagating plasma waves away from the target, it is best to use short pulses where the laser light turns off before the plasma wave can start and then repeat once the plasma has cleared.
You will also want to know some properties of the material being illuminated that deal with heat movement, specifically:
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/tNote 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.
= 2 χ / D.
Conduction from a stationary source in 3D
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(2 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/(2 π κ r) + T0.Conversely, if we know the temperature at some distance r, we know the power being conducted away
P = 2 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/(2 π κ S/2) + T0and if you know the temperature, the conducted power is very approximately
Pc = 2 π κ (T - T0) S/2.
Conduction from a moving source in 1D
Another limiting case will be useful later on. If a moving plane of heat is passing through the material at a uniform velocity of v, adding a constant intensity of heat Ic to the material as it passes, the temperature at a distance x in front of the heat plane is
T(x) = T0 + (Ic/(Cp ρ v)) exp(-v x / χ)This will come in useful when the laser death ray is boring into the surface, drilling a hole at constant velocity while adding enough heat to raise the temperature of the surface high enough to remove the surface.
Conduction from a moving source in 3D
Related to both of the above, a point heat source or spherically symmetric heat source of constant power P passing through a bulk volume of material at velocity v will create a temperature field around it of
T(r,x) = T0 + (P/(4 π κ r)) × exp(-(r+x)v/(2 χ))where r is the distance from the heat source and x is the distance in front of the heat source along the direction of motion. Again, if the laser is drilling into a material this lets us estimate how much heat is leaking out in all directions.
One of the methods by which hot surfaces dispose of their heat is by radiating it as electromagnetic waves. The total radiated intensity (power per unit area) 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.
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 Vmeltto 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/(2 π κ (Tm-T0)).
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 Vvap is
Eevap = Hv ρ Vvap
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.
This isn't a bad approximation when the material is evaporating slowly. However, if the laser intensity is high, you need to take into accound the energy that goes into accelerating the vapor jet away from the surface.
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]or equivalently
p = 1 × 105 Pa exp[Hv Mmol (1/Tb - 1/T) / R]
T = 1/[1/Tref - (R / (Hv Mmol)) ln(p/pref)]where R is the gas constant
T = 1/[1/Tb - (R / (Hv Mmol)) ln(p / 1 × 105 Pa)]
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 vv.
When the vapor jet develops, it will take energy to get the jet going. In addition, the temperature of the jet drops as it expands to the ambient pressure. The jet temperature can be estimated by assuming adiabatic expansion
Tv = T (p/p0)(1-γ)/γwhere γ is the adiabatic index of the gas (in practice, there is an additional drop in temperature at the interface of the melt layer and the vapor when you have rapid evaporation that forms a jet. We will neglect this for now). Many evaporated vapors are made up of single atoms, in which case γ=5/3. If the evaporate is composed of molecules, γ will be smaller but larger than 1 - for molecules made of two atoms γ is approximately 1.4 and for atoms of three or more atoms γ is approximately 1.3 to 1.2, although gamma tends to decrease as the temperature increases.
In addition, by the ideal gas law we can determine the density of the vapor jet
ρv = p0 Mmol/(R Tv).
To estimate the behavior of the vaporizing target material in the presence of a jet, we consider a sample of the target material moving into the beam at a speed vi such that the position of the melt-vapor interface remains at a fixed position x=0. If we consider the material moving into the interface to evaporate, it must be balanced by the material moving away as the fuly developed jet. In a time Δt, a length of target material vi Δt moves into the interface and a portion of the jet of length ve Δt leaves. The mass of material moving into the interface to be vaporized is thus ρ A vi Δt and the mass of material leaving as a jet is ρv A vv Δt. Since these masses must be equal, we find
vi ρ = vv ρv
vi = vv ρv/ρ.
In addition to balancing mass between the melt and the jet, we can balance momentum in the same way. Momentum is mass times speed, so the momentum of material entering the interface is M vi = ρ A vi2 Δt, where we used our previous result for the mass of target material entering the interface. In addition, forces can transfer momentum between the target and the jet. In a time Δt the momentum transfer is the force times Δt. Since force is the pressure times the area, we have ad additional momentum transfer of p A Δt from the melt across the interface. Applying the same reasoning to the vapor jet, we find
ρ vi2 + p = ρv vv2 + p0.If we now use the relationship between vi and vv found by the mass balance
vv2 = (p - p0)/(ρv (1 - ρv/ρ))
vv = √[(p - p0)/(ρv (1 - ρv/ρ))]
At this point, we can describe all relevant properties of the melt and the jet in terms of the temperature T of the interface between the melt and the vapor, but we do not yet know the interface temperature. To determine this, we now balance the energy flow between the interface and the jet, similar to the momentum and mass flow. The energy leaving the interface as a jet is the heat energy of the jet plus the kinetic energy of the jet
Ev = ρv A vv Δt (γ R Tv / Mmol + vv2/2)where we used ρv vv2/2 as the kinetic energy per unit volume and ρv γ R Tv / Mmol as the heat energy per unit volume. Since the evaporation of a given mass M of material adds an energy Hv M, and since the resulting vapor, if at rest and at the same temperature, has a heat energy of γ (M/Mmol) R T, we determine that the heat energy in the liquid at melting temperature must be γ (M/Mmol) R T - Hv M. Therefore, the energy carried into the evaporating interface is
El = ρ A vi Δt (γ R T / Mmol - Hv + vi2/2)There is also energy being conducted away from the interface into the melt.
= ρ A vi Δt (vi2/2 - Hv) + ρv A vv Δt γ R T / Mmol.
Ec = κ A Δt ΔT/ΔX.At steady state, this must equal the energy needed to raise the solid at ambient temperature to a liquid at the vaporization temperature
Ec = ρ A vi Δt (Hf + (T-T0) Cp).Finally, there is energy being added by the laser and energy being radiated away as heat from the high temperature interface
Ea = Ia A ΔtWhen we require the energy in to equal the energy out, we find
Er = e σ T2 A Δt.
El + Ea = Er + Ec + Evwhich evaluates to
Ia - e σ T2 - κ ΔT/ΔX + ρ vi (vi2/2 - Hv) + ρv vv (γ R (T - Tv) / Mmol - vv2/2) = 0.For steady state drilling, this becomes
Ia - e σ T2 - ρ vi (Hf + (T-T0) Cp) + ρ vi (vi2/2 - Hv) + ρv vv (γ R (T - Tv) / Mmol - vv2/2) = 0.All of the variables are either determined by the material properties, the incident laser radiation, or the temperature of the evaporation front. Thus, we can use this to solve for the temperature at which evaporation occurs, T. This equation has no analytic solution for T, but it can be solved numerically using a computational root finding algorithm or with pencil and paper by graphing the function and finding where it crosses 0. Once T is determined, we will have found along the way the speed at which the hole is being drilled vi, along with other useful properties such as the pressure p and jet velocity vv.
The high pressure of the hot evaporated vapor jetting from the molten surface can act like a piston to squirt the melt layer off to the side, where it is then broken up into droplets and blown out of the hole by the high speed jet. This removes the molten material without evaporating it. Since melting takes much less energy than vaporization, the more melt ejection that occurs, the faster the laser can drill at a given energy.
Melt ejection can be the dominant effect in laser machining when the spot sizes are small and powers are in the multi-kilowatt range. For larger spot sizes or higher powers, melt ejection becomes less important compared to vaporization. For lower power lasers, the vapor pressure is insufficient to overcome the surface tension and melt ejection does not work.
To determine the rate of melt ejection, it is first necessary to estimate the thickness of the liquid layer. This can be determined from the equation for heat conduction in front of a moving heat source. The intensity being conducted into the surface has to be enough to raise its temperature from the initial temperature T0 to the temperature of the surface T when advancing at a speed of v
Ic = ρ v (Hf + (T-T0) Cp).The thickness of the melt layer is then approximately
Dm = (χ/v) ln[Ic/((Tm-T0) Cp ρ v)]
= (χ/v) ln[(T-T0 + Hf/Cp)/(Tm-T0)]
Melt ejection is driven by the pressure ramming down onto the molten layer. It is resisted by the viscosity of the melt, which retards shear motion of the liquid - in this case, of the melt layer squirting out to the sides due to the pressure pushing down on it. Further, the central pressure of the beam is resisted by the ambient pressure and the surface tension from the liquid at the edge of the spot. The surface tension gives an effective pressure
pst = 4 λ / Swhere λ is the surface tension.
It has been found that the rate at which the melt squirts out the bottom of the hole and up onto the sides can be reasonably approximated by assuming a rate of mass loss of
dM/dt = π Dm3 (p-p0-pst) / (3 μ)where μ is the coefficient of viscosity. This adds a speed of vm to the interface speed
v = vm + viwhere vi is the interface speed from vaporization alone and
vm = 4 Dm3 (p-p0-pst) / (3 ρ μ S2).This requires the interface speed and melt layer to be determined self-consistently
[4 (p-p0-pst) / (3 ρ μ S2)] Dm3 + vi Dm - χ ln[(T-T0 + Hf/Cp)/(Tm-T0)] = 0.This gives a quartic equation, which can be solved analytically.
In a time Δt, this mass removal carries away a heat energy of
Ee = (π S2/4) vm Δt ρ (T-Tm) Cpand thus an intensity (energy per time per area) of
Ie = vm ρ (T - Tm) Cp.This can be directly added to the energy balance equation
Ia - e σ T2 - ρ vi (Hf + (T-T0) Cp) + ρ vi (vi2/2 - Hv) + ρv vv (γ R (T - Tv) / Mmol - vv2/2) - vm ρ (T - Tm) Cp = 0.
One complication is that the viscosity 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).
These results are for steady state beams. In practice, it is found that melt ejection can be significantly although not overwhelmingly enhanced by illuminating the target surface with a relatively long but low intensity pulse to make a sizeable melt pool, and then deliver a short but very high intensity driver pulse to blast the melt pool away. This is not yet accounted for in the calculator.
If the vapor pressure exceeds the material strength of the target, the pressure of the vapor can drive an expanding cavity in the target material. We can approximate the effects of this pressure driven penetration by the beam by modifying the theory of long-rod tank penetrators. This starts with Bernoulli's equation for the dynamic pressure of a moving fluid
p = ½ ρ v2.Consider a purely fluid target moving the target into the beam at a speed v chosen so that the point of deepest penetration remains the same. Then we have to balance the dynamic pressure of the fluid with the pressure of the vapor in order to keep the interface stationary. This corresponds to the laser boring a hole into the fluid at a speed of v.
Of course, the targets we are interested in are not fluids, but rather solids. It is found that a reasonable approximation is to add a constant to the dynamic pressure
p = ½ ρ v2 + Kcrater.This approximation performs reasonably well for estimating the penetration of shape-sharge explosive jets, anti-tank long-rod kinetic penetrators, hypervelocity impacts with debris in orbit by satellites, and bullets from high powered smokeless powder centerfire rifles. In all cases, the penetration is largely due to the dynamic pressure of the projectile balancing the dynamic pressure of the target across the interface, with dynamic pressures in excess of the target's strength pushing aside the target material. Here, we replace the dynamic pressure of the projectile with the vapor pressure generated by the beam. This means that the beam will bore into the target at a speed of
v = √(2 (p - Kcrater) / ρ).
The material constant Kcrater has a simple physical interpretation. It is the energy needed to gouge a cavity in the material divided by the volume of the cavity. Kcrater is approximately 3 to 4 times the compressive strength Kcomp, or can be found as
Kcrater = (2/3) Kcomp × (1+ln(2 G/Kcomp))where G is the shear modulus. From this definition, we know the volume of the crater excavated by the beam
V = E / Kcrater.For a given cross-sectional area of the hole A and depth x of the hole, the volume is V = A x. In a time t we can then write V/t = A x/t = A v, where v is the speed of the deformation-driven hole into the target material. Since E = P t, we therefore have P/Kcrater = A v. Since the hole will be more or less cylindrical, the radius r of the hole will be approximately r = √(A/π). Thus
r = √(P/(π Kcrater v)).
More generally for pulses of finite duration, consider the depth of the hole made while the laser is on
h = v t.This describes a cylinder directly illuminated by the laser of depth h and diameter S (where, again, S is the spot size). If the pressure is sufficient to expand the cavity by a distance rc, the volume of the final cavity is
E / Kcrater = π [ (4/3) rc3 + (π S / 2 + h) rc2 + (S2/2 + S h) rc + h S2/4 ]This requires solving a cubic equation, which has analytical solutions. The total depth D is then
D = h + rcand the width of the hole is
r = S + 2 rc.This latter version has the advantage of smoothly transitioning from the steady state case for long pulses to the circular cratering limit of very short pulses.
Using deformation to drive a hole into your target has an advantage in that it doesn't matter if most of the beam gets absorbed by the plasma and little of that energy goes into heating the target. The high pressure of the plasma itself can act to deform the target just as well as evaporated vapors. However, you do have to avoid laser supported plasma waves rushing away from the target where the front of the wave is too far away from the target for its pressure to support the target. Consequently a beam composed of a series of pulses, where each pulse is too short to initiate a plasma wave, would be preferred. Each pulse is incident on the back of the previous crater, making a deeper and deeper hole.
If subsequent pulses are incident on the bottom of the hole before the crater from the previous pulse has had time to fully expand, the pressure can pile up, leading to a shallower but wider hole. This is accounted for in the calculatr by using the time average power over the entire pulse train to compute the depth of the hole if this results in a wider hole than adding together the individual pulses. Otherwise, the depth is the sum of the depths of each individual pulse.
Since these relations are approximate, at edge cases we might end up with inconsistent results, such as a hole that is narrower than the beam. In this case we enforce consistency by expanding the hole to the beam diameter and reducing the rate at which the hole propagates into the target so that E = V Kcrater.
Thanks to Jez Weston and Anthony Jackson for stimulating discussions and useful information.