In the distant future, the Earth Federation decided to turn our Moon into a Death Star which has a powerful laser cannon which produces a high energy laser beam capable of obliterating a terrestrial planet in an instant. Usually high frequency lasers such as gamma and x-rays are invisible to our untrained eyes, and even blue light has more energy than red so why is the world’s most powerful laser beam red?

  • 30
    $\begingroup$ Lasers would not be visible at all as beams. The whole point is that the light travels in one direction for a laser and in space there's nothing for it to hit and illuminate so we can see light scattered from what it hits. When people photograph real lasers they'll introduce smoke to let us see the passage of the laser through the smoke. $\endgroup$ – StephenG Mar 8 at 11:03
  • 41
    $\begingroup$ There's always two kinds of lasers in a space fight: The bad guys have red, the good guys green $\endgroup$ – Hagen von Eitzen Mar 8 at 11:23
  • 29
    $\begingroup$ I thought the laser was green. $\endgroup$ – Daikyu Maryu Mar 8 at 15:13
  • 7
    $\begingroup$ Lasers cannot be practically used to vaporise objects larger than the laser device. The heat losses are significant and the small focus of a laser is only of use if you are trying to penetrate a fuel tank or ablative armour. Using laser on a planet from space would only work to cripple nuclear power stations, arsenals and large dams and volcanic faults. The sun shines more energy onto a planet than any laser ever could and it copes fine. The weapon would need an antimatter beam and then it might glow all sorts of colours for fun. $\endgroup$ – KalleMP Mar 8 at 20:45
  • 20
    $\begingroup$ Why would we build a planet-destroyer in a place where it can only reach the one planet that we really, really don't want to be destroyed? $\endgroup$ – user535733 Mar 8 at 22:40

15 Answers 15


To make a statement.

There is no real requirement for a laser to be red - in fact it is likely 'invisible' unless it shines through a gas or other medium that scatters the light.

So perhaps the makers of the Death Star want to make a statement: they want everyone to know that they mean business. After all, the power of the Death Star is the threat of its use. It is in effect a fleet-in-being.

They would make sure all their weapons have this colour, almost like 'branding'. This would then keep the 'outer systems in line'.

| improve this answer | |
  • 15
    $\begingroup$ The red beam is just for targeting purposes. The beam that actually inflicts the damage is invisible. But since you may sometimes have to fire upon (relatively) pinpoint targets like capital ships, a targeting beam is useful. In addition to being more intimidating to your adversaries. $\endgroup$ – aroth Mar 9 at 5:17
  • 8
    $\begingroup$ So they purposefully add the color (as a holographic projection or something else), for intimidation, just like the Stukas in WW2 used sirens to scare the crap out of their targets during their bombing runs. $\endgroup$ – vsz Mar 9 at 7:11
  • 6
    $\begingroup$ Plot twist - there is a second, much smaller death star, that does the actual shooting with an invisible beam. $\endgroup$ – Sulthan Mar 9 at 12:36

Speak Softly and Carry a Big Stick.

First off, you have to understand that the basic concept behind the Death Star is the same as with the KGS Bismarck and the (never completed) U.S.S. Montana; its a big ship that can blast any other big ship without receiving a single dent, forcing other nations to forever use ships to guard against it instead of attacking the rest of your fleet. These kinds of ships are called "Fleet-in-being" ships.

The one problem with Fleets-in-being is that they are, by definition, overkill. Ships like the Death Star are capable of completely destroying any rebellious planet, but are pretty much useless for anything else. For example, you don't need to be able to destroy a planet if you just want to blow up a small supply base. To compound this, a Death Star constantly attracts undue attention; nobody wants one of those hanging around their planets. As a result, people in general (but especially Rebellions and Resistances) have an annoying tendency to preemptively destroy them.

Regardless of overkill level, a Death Star wouldn't be a very effective patrol ship; it can't be three places at once. While you could conceivably blow up two of those places so that you only need to patrol one, the whole reason why you are patrolling those other two places is so that you can use their resources.

Due to its limited usage-scope, the Earth Federation uses it like the I.J.N. Yamato; they make it as fearsome as possible, but then keep docked most of the time, keeping it as a fleet-in-being.

Now, believe it or not, the main qualification for fleet-in-being status is not power (the Yamato was rather underwhelming once it got in an actual battle); instead, the main qualification for being a fleet-in-being is looking incredibly terrifying. As a result, the design team was told the same thing as the designers of the Kill-O-Zap gun in The Hitchiker's Guide to the Galaxy, namely:

"Make it evil... Make it totally clear that this has a right end and a wrong end. Make it totally clear to anyone standing at the wrong end that things are going badly for them. If that means sticking all sort of spikes and prongs and blackened bits all over it then so be it. This is a weapon for going out and making people miserable with."
- Douglas Adams, The Restaurant at the End of the Universe

In your case, the Death Star designers decided to forgo the "spikes and prongs and blackened bits" in favor of a terrifying crimson colored laser.

| improve this answer | |
  • $\begingroup$ "This is not a gun for hanging over the fireplace or sticking in the umbrella stand, it is a gun for going out and making people miserable with." That seems to go against your earlier statement of the purpose of a fleet-in-being here: "Due to its limited usage-scope, the Earth Federation uses it like any other WMD; they make it as fearsome as possible, but then keep it at a starbase most of the time, keeping it as a fleet-in-being." $\endgroup$ – Mast Mar 9 at 10:50
  • 3
    $\begingroup$ @Mast - See where you are coming from but disagree. In the umbrella stand suggests to me either basically forgotten about or only pulled out to surprise attackers. Over the fireplace suggests more ornamental, a trophy piece, something to boast about and show off only to those fortunate enough to be invited into your home. Keeping it in orbit does share the characteristic of not using it like those do but it is putting it front and centre for any that look at Earth for any reason, good or bad. This is Negan's bat, being held menacingly at all times regardless of the situation. $\endgroup$ – RyanfaeScotland Mar 9 at 11:32
  • 3
    $\begingroup$ "Glass cannon" usually refers to something capable of doing a LOT of damage, but easily destroyed/defeated. You seem to use it as "overkill" here? $\endgroup$ – Syndic Mar 9 at 11:46
  • 2
    $\begingroup$ @Syndic A death star would probably be easier to destroy than a fleet, so, it probably fits. $\endgroup$ – Mast Mar 9 at 12:05
  • $\begingroup$ @Syndic Oops, my bad. Its an editing error from an earlier draft. Plus, now that I think about my point in that draft, I tend to agree with Mast: A death star isn't undefeatable; it just takes a huge amount of ammo. $\endgroup$ – The Daleks Mar 9 at 13:52

To advertise that it's on a low setting

Everyone knows a red laser is the least energetic kind of laser. So when they see a planet blown up by a red laser they know their problems can easily get even worse. It advertises to everyone that if they get any cute ideas about shielding their planet you can always switch over to the blue laser mode and blow up their sun instead.

| improve this answer | |
  • $\begingroup$ This is a very good answer. +1 $\endgroup$ – The Daleks Mar 9 at 18:26


Red light goes through an Earth-like atmosphere better than x-rays and other high frequency radiation. In fact that's why sunsets are red 😉

Please notice that being a laser, it will NOT be visible while on its way to the target. We will only know it is red when it hits - and if we are the target, we will only know its color for a very short while.

| improve this answer | |
  • 4
    $\begingroup$ Actually if the weapon is to obliterate the planet in an instant, it needs to carry a ridiculous amount of energy. Even if the atmosphere absorbed every single photon, it would immediately heat up to a temperature that would grill the surface by infrared alone. $\endgroup$ – toolforger Mar 8 at 21:06
  • 1
    $\begingroup$ Does the death star actually shoot a beam of light, or is there more to it than that? $\endgroup$ – Mark Storer Mar 9 at 17:39
  • 3
    $\begingroup$ @MarkStorer in this question it is a red laser, so it is just light. In the Star Wars movies it is something else, because lasers don't work like that. $\endgroup$ – Renan Mar 9 at 19:04

Colour only determines energy per photon, which doesn't matter. For the most part, picking a particular wavelength doesn't sacrifice power or efficiency. The most powerful laser system in the world, the National Ignition Facility, currently generates light in the near infrared at 1053 nm before being converted to ultraviolet in the final stage.

It is true that a single photon at a longer wavelength carries less energy, but a powerful laser beam carries an incredibly high number of photons per second. Two lasers of the same power at different wavelengths will simply transmit a different number of photons.

Where wavelength does matter is in beam optics and damage thresholds. The diffraction-limited angular beam divergence of a gaussian laser beam (how much it spreads out at far distances) is given by $\theta = \frac{\lambda}{\pi w}$, where $\lambda$ is the wavelength, $w$ is the width of the beam at its narrowest point, and $\pi$ is somewhere around 3. So a laser at half the wavelength will spread out half as much, and more importantly the area of the beam will go up by a factor of 4, decreasing the power density by the same factor. There are also more ways for light to damage materials at shorter wavelengths, and as you go to lower wavelengths eventually everything other than vacuum absorbs a big fraction of your light. A possible workaround is to do exactly what the NIF does, make all your light at a red wavelength and then frequency convert it to a shorter wavelength at the final step before sending it out into space.

As a physicist, if I read a story with a red planet-killing laser, the colour of it is going to be very low on the list of reasons why I would consider it to be unrealistic.

| improve this answer | |

Demolitions protocol. During the demolition of any larger structure a warning needs to be given to ensure people can evacuate in time.

Of course, the builders of this Death Moon weren't too concerned with that. This moon is designed as a statement, and they are going to make damn sure that anyone nearby saw who and what blew up their planet so no one can claim otherwise.

And if that happens to be in line with Galactic Demolitions Protocol... no one is going to complain, right?

| improve this answer | |

Short Answer:

I don't know how to make a visible Death Star beam.

Long Answer:

But I do have few ideas which someone else might be able to calculate the possibility of making work.

Part One of six: Basic Death Star improbability.

Are they going to move the Moon from planet to planet and solar system to solar system during war? If so they have to build very big engines to move it. And the lack of tides on Earth will cause the extinction of numerous species and ecological problems.

Or are they going to leave the Moon in orbit around the Earth and use it to blast distant planets with some faster than light radiation, like the Starkiller Base in The Force Awakens, which is so very, very implausible that it makes the Death Stars seem reasonable and plausible by comparison.

Part Two of six: The energy beam makes interplanetary particles glow.

Do you know how much energy it takes to make a terrestrial planet separate into pieces that will be travelling faster than escape velocity so they won't fall back and recombine? Well, it's more energy than you can imagine. And that incredible amount of energy will cause the planet to slowly expand in a swelling cloud of incandescent gas and plasma. In order for that gas cloud to expand and thin so rapidly that the space where the planet was will seem transparent and empty within a second, countless millions of times as much energy would have to be used.

It seems to me it would be far less wasteful of energy to turn the planet into an expanding sphere of incandescent plasma that will expand much more slowly and take many years to expand to be thin enough to be transparent. That will also leave a sort of visible corpse or skeleton of a planet for visitors to the system to see, and so be a long lasting warning to them.

There are molecules of gas and particles of dust thinly scattered in interstellar space and less thinly, though still extremely thinly spaced, in interplanetary space. So an energy beam, fired at a planet, would strike interplanetary particles and molecules in its path and would heat them up, causing them to glow in infra red, visible, and/or ultraviolet wavelengths of light.

And would that glow be intense enough to be see seen from the side of the energy beam?

I don't know.

But supernovas do explode, and apparently calculations show that they will vaporize any planets which orbit them closely enough. WhatRoughBeast's answer to this question: Can a planet survive a supernova?1 states that the inner planets in the solar system would be vaporized if the Sun turned into a supernova, while the outer giant planets might survive, though not intact.

Note that a supernova explosion is not instantaneous but lasts for many days and would take days to vaporize a planet.

So if the energy flux in the Death Star beam was as strong as in a supernova, it might still take days to vaporize a planet.

Anyway, imagine that the emitter on the Moon is 100 miles in diameter and emits a beam of energy 100 miles in diameter which slowly spreads out in diameter to 200 miles, 400 miles, 800 miles, 1,600 miles, 3,200 miles, 6,400 miles, and so on. Thus the Moon Death Star would have to approach a target planet and reach a distance where its beam will have spread out until the diameter of the beam equals the diameter of the target planet.

Thus the beam of energy from the Moon Death star will be a truncated cone that is 100 miles in diameter at the emitter and spreads out to a diameter of thousands of miles where it hits the target planet at a distance of tens of thousands, hundreds of thousands, or millions of miles.

The interplanetary medium includes interplanetary dust, cosmic rays and hot plasma from the solar wind.

The density of the interplanetary medium is very low, about 5 particles per cubic centimeter in the vicinity of the Earth;


So a square cross section one centimeter by one centimeter through the energy beam at the emitter will contain 16,000,000 cubic centimeters and about 80,000,000 particles. Where the energy beam has expanded enough to reach the diameter of the Earth, about 7,917.5 miles, a square cross section one centimeter by one centimeter through the energy beam will contain about 1,274,194,945 cubic centimeters and about 6,370,974,725 particles.

The energy beam will make every particle that it hits super hot, and every particle within the beam will thus emit electromagnetic radiation in all directions, including directions that are more or less sideways relative to the direction of the beam.

Thus it might be possible that the beam might possibly be visible from the side, but someone else will have to try to do the calculations.

And if that is a plausible way of making your energy beam visible from the side, you may need to imitate the purple prose of E.E. Smith to emphasize how incredibly, unbelievably, fantastically intense your energy beam is.

Part Three of six: Vaporized planetary particles.

As I remember, Arthur C. Clarke's novel Earthlight (1955) involved a battle where spaceships used energy weapons against a moon base. The energy beams were invisible in the near vacuum of space, but as they vaporized moon rocks around the base, a local atmosphere of rock particles formed around the base, and the particles in that local atmosphere scattered the light of the energy beams, thus making the parts of the energy beams near the base visible from the sides.

The description of Earthlight in Wikipedia does not mention that, but does mention:

There is also an enigma - the apparent sighting of a 'beam of light', that should not be possible on the airless world. This is explained later in the story as a weapons beam that included metal particulates moving at high velocity.


So what happens when the energy beam from your lunar Death star starts to vaporize the target planet? Plasma will shoot out from the planet in all directions, forming a sort of plasma atmosphere which will be denser near the planet and thinner farther from the planet, and which will be constantly expanding as more and more planetary material is vaporized by the energy beam.

So the planet will be surrounded by an expanding sphere of plasma glowing with heat, and becoming thinner and thinner, and thus dimmer and dimmer, farther from the planet. And where the energy beam passes though the plasma it will will heat up the plasma even more, making it glow brighter than the plasma outside the energy beam.

Thus the energy beams will become visible in the cloud of plasma near the planet, and as the cloud of plasma expands outward the visible part of the energy beam will get longer and longer and extend farther and farther back toward the Death Star.

Part Four of six: Particle beam weapons.

What if your Death Star like weapon shoots subatomic particles or even entire atoms at the target planet? And what if those subatomic particles are highly unstable, so that some of them decay into other particles during the short time between being emitted and hitting the target planet?

A radionuclide is an unstable atom which decays:

A radionuclide (radioactive nuclide, radioisotope or radioactive isotope) is an atom that has excess nuclear energy, making it unstable. This excess energy can be used in one of three ways: emitted from the nucleus as gamma radiation; transferred to one of its electrons to release it as a conversion electron; or used to create and emit a new particle (alpha particle or beta particle) from the nucleus.

So some radionuclides emit gamma rays, highly energetic electromagnetic radiation, which may hit other particles of matter and energize them to emit less energetic radiation, including visible light.

Thus if your weapon is a particle beam shooting radioactive atoms at the target planet it might produced enough light as atoms decay to be see from the side.

If a particle beam consists of subatomic particles, some of them might decay while in transit to the target.

Except for the proton and neutron, all other hadrons are unstable and decay into other particles in microseconds or less.

Most subatomic particles are not stable. All mesons, as well as baryons—except for proton—decay by either strong or weak force. Proton observationally doesn't decay, although whether is it "truly" stable is unknown. Charged leptons mu and tau decay by weak force; the same for their antiparticles. Neutrinos (and antineutrinos) don't decay, but a related phenomenon of neutrino oscillations is thought to exist even in vacuum. Electron and its antiparticle positron are theoretically stable due to charge conservation unless a lighter particle having magnitude of electric charge ≤ e exists (which is unlikely).

Of subatomic particles which don't carry color (and hence can be isolated) only photon, electron, neutrinos with some[7] disclaimers, several atomic nuclei (proton included), and antiparticles thereof can remain in the same state indefinitely.


Part Five of six: MAHAM

The Magneto Hydrodynamic Explosive Munition (MAHEM) is a weapon being developed by DARPA that would utilize molten metal to penetrate enemy armor.1 The molten metal would be propelled by electromagnetic fields from explosions.2 The munition would be delivered to a target as a warhead "packaged into a missile, projectile or other platform." It would penetrate the armor of an enemy vehicle then explode when it gets inside, destroying the vehicle from the inside out. DARPA predicts the weapon will have greater efficiency, control, and precision than conventional explosively formed penetrators.3

It is driven by a compressed magnetic flux generator (CMFG).


The article about Earthlight (1955) is liked to the article about MAHAM, so possibly a MAHAM weapon might possibly be visible from the side.

Part Six of six: Holographic aiming guides.

What about holograms? A hologram reproduces the shape of the object depicted. So it is theoretically possible to create a holographic image of a cone shaped object or a cylindrical object.

Possibly in your world personal laser guns might have a range of about 100 meters. And possibly they come with holograms that emit images of long narrow cones which might be one millimeter wide at the source and one meter wide at the far end. So when someone pulls the trigger of their laser gun the hologram of the cone springs into existence and they can see what objects the beam will hit, and maybe have a second to change their aim before the beam turns on and strikes the target.

And possibly the super giant laser on your lunar Death Star will also have a super giant hologram for aiming purposes.

So there are a few methods you might want to investigate for your giant death ray to be visible from the sides, and thus seen and noticed by surviving people. And possibly an explanation of why the death beam would be red in color instead of another color can be made plausible.

| improve this answer | |

"Because red lasers are cool!"

I asked myself that years ago. But just like who and what decided light saber colors to be that way, there's really no telling except the people who told the story, and as far I know I have not seen an explanation, both in-universe and in interviews. You can, however, choose your own explanation. Beams of light takes on colors depending on their wavelength. Shorter ones take on red, longer ones take on blue, and everything in between is like the color of the rainbow. Everything in the universe exhibit wave properties until measured (then they turn into particles) and that includes light.

| improve this answer | |
  • $\begingroup$ You've got your wavelengths reversed; longer ones are red, short ones are blue. $\endgroup$ – The Daleks Mar 9 at 4:27
  • $\begingroup$ @SEistoopoliticallycorrect Whoops! But yeah, point is that author can use the fact that matter and energy vibrates at different rates and can just use the fact that the Death Star's laser's wavelength is in such a way that it is colored red. $\endgroup$ – Dominic John Maca Mar 10 at 0:38

Helium/Neon laser

This is a helium/neon laser. This is one of the earliest types of laser invented. The tube glows pink, and the colour of the laser beam that comes out one end can be tuned a bit but only in the red or infrared range. Green, blue and UV lasers weren't invented until much later. Neither were the laser diodes that we use to make small, cheap lasers now.

If you don't have a semiconductor factory on the moon, it might be easier to make primitive lasers like this.

| improve this answer | |

Colour-coded for your convenience

This is the name of a trope which is summed up by "indication through coloring". In fiction, colour-coding is usually done for the benefit of the audience, and usually obeys a variety of social conventions. For instance, red is the color of bad guys, but it's also the colour of passion (may it be love or rage), of Communism, of traffic signs that tell you what you can't do, and so on.

Well it's not just fiction. In military strategy, it is usually important to know who is who. NATO standards use four colours to identify four groups: blue (friend), red (hostile), green (neutral), yellow (unknown). That way, when you look at a map and it's chokeful of red, you now you're going to have a bad time. While the choice of colours is strictly a matter of standards, the choice to use colours at all is highly practical when you want to convey simple information quickly.

But wait, aren't lasers invisible?

Well, yes. One way to make a laser visible is to spray a mist of particles, like water, in front of it. Of course, building a giant atomizer to spray water in front of the business end of your Death Star is hardly the most effective use of your budget. But this isn't the only invisible thing in life. You know what else is invisible? The offsides line in football, or the 1st and Ten line in football. This is were you remember the trope:

Colour-coded for your convenience

You probably aren't floating naked in space, looking at your laser. You have to remember that you are always seeing the battle through a screen. Even if it's a window, canopy or other clear glass-like material, you can still paint some augmented reality over it.

In other words, you don't make the laser visible, you just make it look visible. Adding coloured lasers on screen helps the audience understand what's happening, whether that audience is your people at home tuning in for the broadcast, or strategists on the battlefield looking at a real-time rendition of the battle, or pilots of spacefighters looking to avoid crossfire between two capital ships.

From there, you can pick any colour. Red may be the dominant colour of your flag, or it has positive connotations like strength, might, and general betterness. And don't worry too much about the people that will see your laser in its natural, uncoloured state, because those people probably live on a planet that is about to explode.

| improve this answer | |

Red is more energy efficient.

The idea that a laser that is green, blue or even a gamma-ray laser, is more powerful than a red one is wrong. It is true that a blue photon has more energy than a green one, and a green one more than a red one, but that doesn't really matter - you can just use more or fewer of whatever colour of photon you have chosen.

[How many potatoes I can fit in my lorry doesn't depend on whether I have 100 bags of 100 potatoes or 1,000 bags of 10. So how many jouls of energy I can fit in my laser beam...]

So if the energy-per-photon makes no difference to your choice of colour, what does? Well two things: "which kinds of photons can I make most energy efficiently?" and "Which kinds of photons can I keep control of most easily".

It probably matters to your finance department a lot whether the laser is 10% efficient and powered by a hundred million nuclear reactors or instead 1% efficient and powered by a thousand million. It probably matters to your crew that they don't get radiation sickness from your gamma ray death-laser.

I have no idea which lasers are going to be the most efficient in the year 4,500 AD, and if I did I wouldn't tell you. But their is no good reason to assume they are NOT going to be red, so lets roll with that.

As a final point:

A lot of people here are saying you can't see laser beams. But you can if they are going through a gas: spray an aerosol and you can make them appear very clearly. The tiny amount of gas in interplanetary space will scatter a tiny proportion of any laser beam, but your planet-killer is so gigantically powerful that this negligible proportion lost to scatter may actually be visible.

| improve this answer | |

The Beam isn't Red

What you're seeing is a remnant of the fact that the local system used had a vibrant asteroid mining trade. The asteroids in the region happen to be rich in strontium, which is the element they use to make red flares or fireworks.

The beam is highly energetic across most of the electromagnetic spectrum, but especially gamma. The atomic nuclei in its path decay into energetic radiation (stray protons and alpha particles, mostly). This happens so quickly that very little of the radiation has time to scatter, which means that you don't see its light. The electrons, too are scattered by the photoelectric effect.

As the energy dissipates radially off the beam, it continues to do the same thing to the matter it interacts with, until a critical threshold is met. Once the energy drops below the blast-apart-a-nucleus threshold, it is 'caught' by the electron orbitals of whatever matter happens to be around, and emits light according to their emission spectra. If you shot it into a part of space right in sodium, for instance, you'd see yellow (more about this here).

The beam is actually much thinner than you see. You're seeing the cylindrical region of space where it stops destroying matter directly and instead starts interacting with it chemically.

| improve this answer | |

The color of the laser is irrelevant. Once the planet is destroyed the Death Moon will have no well defined centre of mass to orbit. It will quickly become impossible to aim the laser as the orbit becomes irregular. After that targeting other planets will become essentially impossible.

| improve this answer | |

Red means dead.
(I'm guessing "red means death" would gramatically be more correct, but that doesn't rhyme).

Red is a warning color which most sentient beings recognize immediately. Why do you think stop signs or red traffic lights are red (not blue, which would be visible much further). Remember Ethan Hunt's wrist-mounted decibel meter in the Mission Impossible computer room heist (the "One - Two - Toast - Toast" scene). The scale has two shades of green, and red for "Toast". Red is bad. You do not want red because when you see red, you're dead.

Thus, when bubbling light strings form a growing, stationary red blob in front of your superlaser which stands still for an instant before bursting into a massive beam, and it's green (like in Star Wars 1977), then that's only moderately impressive. I mean, seriously, what did the Empire think! The ability to destroy a planet is insignificant next to the message delivered by a red beam.

When you see a red beam forming (just a second or two before the laser beam starts to, uh, actually move at considerably less than light speed), you know what's coming at you. That's extremely important because where's the point in killing a few million people if they don't fear you?

It's also much more impressive for bystanders who aren't going to die. When they see the red beam forming, they'll know that those poor bastards who are going to get blown to pieces in a few seconds see it coming. And they know they know.

Besides, doing things in a deliberately difficult way and still succeeding at it (such as blowing a planet apart using a red laser) yields a tremendous amount of coolness and will guarantee that your enemies respect you.
Such as the kid with the sword in "Picard". I mean, fighting people who have ray guns with nothing but a silly little sword means that you are either fucking stupid, or insane (or both). But it also means that if you actually win the fight, then your enemies will definitively fear you.

| improve this answer | |

The power of the Death Star comes from the Kyber Crystals, as seen being transported in Rogue One.

As with light sabers, when a sith uses the crystal, it 'bleeds'..and the light saber/Death star 'lazer' turns red.

Also, because the bad guys use red, it's an evil colour in Star Wars.

| improve this answer | |
  • 1
    $\begingroup$ The OP is not asking about an "in SW universe" explanation. $\endgroup$ – L.Dutch - Reinstate Monica Mar 10 at 11:59
  • $\begingroup$ surely, outside of the SW universe there isn't a Death Star....but point taken :) $\endgroup$ – Mr_DW_Brighton Mar 10 at 12:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.