# What is the energy requirement needed to make a Visser wormhole and what kind of civilisation on the Kardashev Scale could do this?

So, after discussing how to make a wormhole in chat (after I started with how long Earth could last in the middle of an invasion of Earth in the mid 1970s - early 1980s), I was curious:

How much energy is necessary to make a wormhole and what kind of civilisation could make a network of those things?

NOTE: I am specifically referring to Matt Visser's model of a polyhedral wormhole (in simpler terms, one that looks like a giant floating space cube).

Basically, I am basing my wormholes on this report here: Traversable Wormholes: Some Simple Examples

So, for each model, what energy requirements can I expect and where on the Kardashev Scale would such a civilisation be?

• – HDE 226868 May 10 '17 at 20:13

# I have no idea if this is the right answer...

...but I'd like to help. In your linked report, I find the quote

The stress energy tensor for the wormhole is then concentrated entirely on the edges of the cube where $\mu = T=−\frac{1}{8G}=−1.52×10^{43}$ Joules/metre

Using non-relativity units, you have to divide this number by $c^4$ to get $-1.88\times10^{9} \text{ J/m}$ as $T$, a Cauchy Stress tensor. If I'm interpreting this paper right (which I probably am not), you need this much energy applied to the quarter cylinders which represent the edges of a cube with rounded edges in order to create the wormhole. I suppose the stress must be applied to spacetime itself? That isn't clear to me either.

Its also not obvious how big the rounded edges must be with respect to the overall size of the proposed wormhole cube. I imagine the wormhole cube must be large enough to encompass whatever vessel you want transported, so it has to be hundreds to thousands of meters in length per edge.

The number of meters barely matters, since you are multiplying by something raised to the 439th power for energy, so an order of magnitude or two of distance doesn't make that much difference. The real problem is that the energy value is negative, not positive. How you will get so much negative energy in one place, and apply it to a large enough area to 'wormhole' as starship, I don't know.

If that were positive energy, you'd be looking at about a tank of gas' worth of energy per meter, so the total amount of energy needed isn't great, assuming that negative energy can be stored and applied as easily as positive energy, which is not an assumption I would make.

• In this context, G is Newton's gravitational constant. – KareemElashmawy May 10 '17 at 19:56
• I think it's the choice of units that caused the confusion. In relativity, natural units are often used, where $c=1$. If you plug in $c^4$ on the top of that fraction, in SI units, you get the correct result - with $G$ being the gravitational constant, as @KareemElashmawy said. – HDE 226868 May 10 '17 at 20:09
• So, a K 1.2/K 1.6 is stuck with STL technology? Oh....great. – Future Historian May 10 '17 at 20:26
• @FutureHistorian With my errors pointed out, and now corrected, I have edited the question. I have no idea how hard it is to get negative energy together, so this answer still isn't very useful. – kingledion May 11 '17 at 2:55
• @Draco The correct way of looking at energy is as the square of energy. The square of a negative number is positive. See the full form of the Einstein equation in special relativity. Negative energy doesn't make the Kardashev scale negative. LOL. – a4android May 11 '17 at 6:03