# Power source for a lightsaber

This one has been on my mind for quite some time. What kind of power source would you need to run a lightsaber? I was actually worried recently about this post when I saw the Discovery Channel show “Sci Fi Science“. In that particular episode Michio Kaku talks about how you would actually build a lightsaber. The episode was a little silly, but the science wasn’t too bad. In end Michio decides to build a type of hand held plasma torch. Doing this, he estimated that the lightsaber would need a power source on the order of mega-watts.

He didn’t do what I was thinking. I am thinking about the scene from Phantom Menace where Qui Gon tries to cut through a door at the beginning. It looks like this:

### Comment Notes:

Usually, I save this for the end. However, let me go ahead and pre-emptively address some comments (that are similar to what happened with the flying R2-D2 thing.)

• Yes, I know light sabers are magic. I also know that they run on these cool crystals. This will not stop me from making an estimate anyway.
• Oh, I know I am estimating some quantities. That is ok. At least I can get a ball park figure for this.

Before I get into the calculation, there are a couple of background topics to discuss.

I am not going to get into too much detail, so let me just get to the good stuff. A blackbody is an object that gives off light due to its temperature, not because light is reflecting off of it. When things are hot (even when they are not) they give off electromagnetic radiation. This is true for dense (solidy) objects, not for low density gases. These objects give off a wide range of ‘colors’ of light with the wavelength of the peak of the distribution related to temperature of the object. PhET has a pretty good blackbody simulator. Here is a screen shot:

I added the arrow to point out the “peak” in the spectrum. As the object gets hotter, this peak will get larger and move to the left. More light of a shorter wavelength will also be produced and the color of the blackbody will change. Two excellent examples of a blackbody: the Sun and an incandescent light bulb filament. Here is another example – Built on Facts has a great image of a hot stove element. As this thing heats up, it produces light mostly in the infrared region. You can actually see this with a video camera (since they detect IR even though you don’t want that). As the element gets hotter, this spectrum it produces shifts towards shorter wavelength and starts to look red (don’t touch it). If you got it even hotter, it would look more yellowy.

The point is that you can determine the temperature of a blackbody by the color of light is giving off. Let me leave it at that (although it is possible to make it much more complicated). In the clip from Phantom Menace, I will use the color of the hot door to determine its temperature.

### Thermal Energy

How much energy does it take to increase the temperature of a material? Well, this depends on the change in temperature, the mass of the object and the specific heat of the material. Here is the relationship between these:

Q is the heat (even though I hate that word). Think of Q as the amount of energy you put into the thing. When talking about this thermal energy stuff, I like to talk about a pizza in an oven on some aluminum foil. Suppose you put it in the oven until the temperature is 350 F. Can you touch the aluminum foil? Yes, but don’t touch the pizza. The aluminum foil has a very low mass and thus not very much thermal energy (so it doesn’t really hurt). This is not true for the pizza.

There is another thing to consider with thermal energy. What if the material changes phases – that is to go from a solid to a liquid. This also takes energy that depends on the mass of the stuff and the type of material.

Here Lf is called the latent heat of fusion.

### Temperature

What is the temperature of the metal (I assume it is metal, but it doesn’t really matter) door? Here is a shot of the door being cut (melted) by Qui Gon. I used the PhET simulator to match the color of the heated parts of the door to blackbody temperatures.

This seems pretty rough, but it is a start. One problem is that the screen captures might be a slightly different color. Oh well, it is close enough.

### Mass of stuff

This one is a little bit more difficult. First, the whole door gets hot (you know from thermal conduction). But what is the mass of the stuff that is at 2700 K? What is the mass of the stuff at the 5200 K? Let me start with the hotter stuff. I am going to estimate that the QuiGon makes a line about 2 meters long and that the melted part is the width of a lightsaber blade (7 cm?). How deep is it? Just by looking at the lightsaber poking through the door, I am going to guess 20 cm thick. With these guesses, I can get an estimate for the volume of melted material.

And so the mass would be:

Where rho is the density of the material – which of course I do not know.

What about the mass of the other hot stuff. Instead of making a whole bunch of small estimates, I am just going to say this is 2 times as much as the melted stuff. If you are not happy with this estimation, I am ok with that.

### Material

This one is just waiting for some discussion. “Hey DOOD! How can you calculate the energy of that stuff. The Trade Federation totally stole some secret transparent aluminum material and it has a really low density!” I agree, this could be something weird. However, I don’t know how to estimate the specific heat or density of weird stuff. Let me make the assumption that this is something like a known material. So, what material is metal and melts around 5000 K? Or at least a melting point greater than 2700 K – because that hot part could be hotter than the melting point. If it was made of an element, probably the best fit would be something like Tungsten or Carbon. Those do not see likely. Titanium melts at 1930 K. If it is an element, I would pick titanium – because it is cool. Another option would be steel. Some steel melts at around 1600 K. What about ceramic? That has a high melting point. Unfortunately, my googling abilities were unable to produce enough data on ceramic.

So, here are my choices along with estimated densities and other stuff. Note that my sources are wikipedia, http://chem.lapeer.org/PhysicsDocs/Goals2000/Laser1.html and http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html.

Notes:

• I couldn’t find the latent heat of fusion for diamond, so I used graphite instead
• Specific heat capacity is tricky, I know this. It really isn’t constant over a wide range of temperatures. The ones listed are most likely for room temperatures.
• For carbon, I picked diamond. They are the Trade Federation. They can afford a diamond door.
• The last material “fake” is my estimate for a material that seems reasonable that will give the lightsaber the lowest power consumption.

### Energy and Power

Now on to the good stuff. First, how much energy did it take to heat up that much stuff and to melt some of it? I will assume that the door started near room temperature (~295 K). Let me break the door into the stuff that just heats up and the stuff that heats up, melts, and then heats up some more. This would require energy on the amount: (I am using ma for the non-melting and mb for the melting stuff. Also, C is the specific heat capacity and L the latent heat of fusion.)

I guess I should be clear about my variables.

• I already declared the masses (above)
• T0 is the room temperature – or starting temp of the door.
• T2 is the final temp of the non-melting door.
• Tmelt is the temperature that the door melts.
• C1 is the specific heat capacity of the solid door.
• C2 is the specific heat capacity of the liquid door.

At this point, I am going to cut the lightsaber a break. Since I have no clue about the specific heat capacity of the door in its liquid state, I am just going to leave that part off.

To calculate the power, I just need the time it takes for this change in thermal energy.

The time is the one thing I feel comfortable determining. You know, because they showed it in the movie. From the movie, Qui Gon took about 9 seconds to make that cut in the door. Bring on the calculator. As I have done before, I will put this in a spreadsheet so you can change the values if you do not like them. (I had a complaint about embedding zoho sheets, so I will try google docs)

This gives a power of 28 kWatts. At least it is not 1.21 gigaWatts – you know what they would need for power like that? Oh, note that in the spreadsheet, I left C2 in the calculation, but its value is 0. If you want to change that to something, go ahead.

### Energy Source

The power needed to cut that door tells me something, but not everything about the energy source for the lightsaber. I want to estimate the energy density of the energy source. To do that I will estimate how long the lightsaber will run without recharging. This is a tough one. Maybe they run forever – I will not assume that because it wouldn’t be as much fun. Really, how long would it have to run for it to be useful? I say at least 2 hours of continuous use. That seems reasonable, doesn’t it? How much energy would that be? Using the same power formula:

How big is the energy source in the lightsaber? Taking an estimate on the high end, I will say it is a cylinder with a radius of 3 cm and a length of 15 cm. I think that is plenty big enough. If this is the case, then the energy density of this power source would be:

What can I compare this to? What about the best evAR battery? Wikipedia has a nice table of energy densities. Just for comparison, 4.7 x 1011 J/m3 is 470 MJ/L. On the table, this puts it somewhere between Octanitrocubane explosive (no idea what that is) and Beryllium + Oxygen (again no idea). As far as batteries go (known Earth-batteries), it seems like the highest energy density is the fluoride ion with 2.8 MJ/L.

Ok, I get it. The jedi have some secret power source.

### Also

Qui Gon also just sticks his lightsaber in all the closed doors of the Trade Federation ship and heats them up. I could have also used this an estimate the power. But, I didn’t.

1. #1 Alex
February 2, 2010

You forgot to take into account the emitter, casing and controls for the lightsaber. So the power source would have to have an even higher energy density.

Also of interest is waste heat; up above you had 28.5kw, if that is at 99% efficiency there is still ~300w to get rid of. If it is 90% then there is about 3kw. They had better be bloody efficient, otherwise Qui Gon is going to cook his hand.

Finally, what happens if this 4.7×10^11 J/m3 power source gets hit?

2. #2 David
February 2, 2010

I always thought that Lucas missed a great opportunity. I would have made it so that the Jedi focuss the force into the lightsaber. Then only a powerful jedi could get a fully potent weapon.

3. #3 Moopheus
February 2, 2010

A light-saber runs on 3 D-cell batteries.

(Or at least, the original prop, which was made out of a 3-cell Graflex flash tube.)

4. #4 Matt Springer
February 3, 2010

We do know the Star Wars universe has access to some truly ludicrous power generation capabilities. The Death Star packs several times the gravitational binding energy of an entire planet into a power core that’s at best a few cubic kilometers in size.

5. #5 Dunc
February 3, 2010

“They had better be bloody efficient, otherwise Qui Gon is going to cook his hand.”

That close to all that molten whatever, he’s going to cook his hand just off the radiant heat.

6. #6 IanW
February 3, 2010

David is on to it. You didn’t include “The Force” in your calculations, Rhett.

7. #7 Dave
February 3, 2010

Don’t forget, when figuring the energy density that the light saber may not be required to produce the full power if nothing is in the beam. Thus, without anything interacting with the beam, the power consumption may only be a few Watts (or even less).

Dave

8. #8 Rhett Allain
February 3, 2010

@Dave,

Very good point. I thought of that, but did not take it into account because I had no way of doing so. But you are correct.

9. #9 Mu
February 3, 2010

Cubane is a hydrocarbon arranged like a cube, you have 8 C atoms at the corners, bonding to 3 other C atoms. You have one bond sticking out into space. Since this gives you a C-C-C bond angle of 90 degrees, there’s a lot of strain energy already in the unsubstituted molecule. in ONC, the eight hydrogens are replaced with nitro groups, for a C8N8O16 total molecular formula. Which likes to decompose to 8 CO2 and 4 N2 and a lot of energy. When I was in grad school that was our geek-LOL molecule, but surprisingly, that thing got synthesized and is stable to 200 C or so.

10. #10 Michael Varney
February 3, 2010

Awesome! I am glad I stumbled on this blog! I just love this sort of back of the envelope type problem!
http://scientificilliteracy.blogspot.com/2010/02/fun-back-of-envelope.html

11. #11 rob
February 4, 2010

I used to bullseye womp rats in my T-16 back home, it’s energy capacity is not much bigger than 470 MJ/L.

12. #12 Rhett Allain
February 4, 2010

@Rob,

good one.

13. #13 Rick61
April 10, 2011

What about a microscopic black hole. That would explain why only Jedi or the Seth could construct such weapons. By focusing the force on a single atom at the heart of the power cell, the power of a singularity is what powers a high energy plasma stream shaped into a tube by a containment field forming a saber. This is also why construction of this weapon is the last test of a Jedi or Seth. The fail safe of such a power source is that if containment of the singularity is lost, say during battle, the microscopic blackhole simply evaporates and becomes inert.

14. #14 Rick61
April 10, 2011

What about a microscopic black hole. That would explain why only Jedi or the Seth could construct such weapons. By focusing the force on a single atom at the heart of the power cell, the power of a singularity is what powers a high energy plasma stream shaped into a tube by a containment field forming a saber. This is also why construction of this weapon is the last test of a Jedi or Seth. The fail safe of such a power source is that if containment of the singularity is lost, say during battle, the microscopic blackhole simply evaporates and becomes inert.

15. #15 DarthDrax
April 25, 2011

According to what sources we have for lightsabers, while the Force is necessary in the construction of your most basic lightsaber, a generic “energy cell” is all that is required to power it, which is basically the Star Wars equivilant of a battery. However, through some sort of psuedo science, a lightsaber recycles its energy while the blade is not in contact with anything, therefor, it can run indefinately, and only loses energy when it strikes another saber or object. However, there are some Jedi and Sith who do construct more complex sabers, and some of these can be powered directly by the Force, plus there’s nothing stopping a Jedi or Sith from focusing the Force through their saber to intensify it, probably why in all of the tabletop RPGs to date, a lightsaber does more damage when wielded by a more powerful Jedi/Sith. I’d have to say there are way too many variables to determine how much power the “energy cell” produces, but kudos for actually attempting the calculation. However, I must say, i disagree with your assumption that a lightsaber’s blade is 7cm wide, that would put it roughly 3 times wider than the emitter shroud of my custom-built saber prop, and a 7cm blade width is roughly double the diameter of most replica sabers, and saber wide enough to produce a blade that wide would be nearly impossible to grip comfortably one-handed by your average sized Jedi.