# Wormholes as Weapons of Mass Destruction

A wormhole (or Einstein–Rosen bridge) is a speculative structure linking disparate points in spacetime, and is based on a special solution of the Einstein field equations solved using a Jacobian matrix and determinant. A wormhole can be visualized as a tunnel with two ends, each at separate points in spacetime (i.e., different locations or different points of time). https://en.wikipedia.org/wiki/Wormhole

I want to obliterate my enemy by opening one end of a wormhole at the surface of the Sun and the other end in mid-air above my enemy's city.

Question (edited)

I make a 100 metre radius wormhole at a height of 1km above the Earth's surface connected to the surface of the Sun at the other end. I'd like to incinerate a city of radius 10km. The hole will be open for one second.

What will be the effect on the city - will it

(a) melt into a glass-like substance?

(b) evaporate leaving a clean bowl-shaped crater?

(c) suck the Earth through the wormhole by the force of gravity?

(d) something else?

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– user
Mar 24, 2019 at 12:51
• One thing worth noting: the surface of the Sun isn't that exceptionally hot. It's about 6000 K. The combustion chamber temperature of a jet engine reaches about 2300 K, and even the exhaust gas temperature is on the order of 900 K. Even 2300 K is above the melting point of most metals, but on the other hand the temperature is likely to drop off fast; the second link should be of interest there. If you want hot, try tunnelling to the Sun's corona instead.
– user
Mar 24, 2019 at 13:15
• Wormholes are wespons of mass destruction by definition Mar 24, 2019 at 14:38
• @aCVn The corona is also a lot less dense than the surface (which is itself not very dense), so you are unlikely to get any sort of significant blast at the city over the course of 1 sec. Mar 24, 2019 at 16:51

# c) and d)

Wormholes have mass on their own. A 1-meter wide wormhole would have the same mass as Jupiter. It would already be enough to break the Earth and the Moon apart due to its Roche Limit. A 100-meter wide one will have stellar mass, so the Earth will probably break past spaghettification as well. Within minutes to hours every planet's orbit is modified so fast, the solid ones crack. The sun's orbit around the galactic core tightens a little.

So I'd say the city you wanted to bust is successfully destroyed.

• You might want to mention the reason why they'd have that mass, other than just providing links. It's easy to assume you're just talking about the mass of the sun acting through the wormholes. Mar 24, 2019 at 16:43
• @MishaR good point, I've edited my answer. Mar 24, 2019 at 17:24
• Strictly speaking, and as established by the answer you linked in your answer, the mass of a wormhole is actually the mass of the exotic matter keeping them open. To say, "Wormholes have mass on their own." is slightly misleading. Mar 25, 2019 at 1:40

# Heat - Not much

The Sun is made of several layers, the "surface layer" is the photosphere.

The photosphere temperature is $$5,772 \text{ K}$$, not much. Any element less tungsten and rhenium will evaporate, these two will only melt. But in order to heat materials, you need time, and one second isn't much. Just check it:

Your wormhole has a diameter of $$d= 100 \text{ m}$$ and it's a circle:

$$surface = \pi r^2$$

And $$r = \frac{d}{2}$$. So:

$$surface = \pi \times \frac{100 \text{m}}{2}^2 = 7,853.98 \text{ m}^2$$

Stars are usually close perfect black bodies, so they emissivity is near $$\varepsilon = 1$$.

Using the Stefan-Blotzmann law:

$$P = Aj^\text{*} = A \varepsilon \sigma T^4$$

Where:

• $$P$$ is the total energy emmited in $$J/s$$ or $$W$$.
• $$A$$ is the surface area in $$m^2$$.
• $$j^\text{*}$$ is the radiant emittance with dimensions of energy flux (energy per time per are), like $$J/s/m^2$$ or $$W/m^2$$.
• $$\varepsilon$$ is the emisivity.
• $$\sigma$$ is the Stefan-Botlzmann constant $$\sigma = 5.67036713 \times 10^{-8}W \times m^{-2} \times K^{-4}$$
• $$T$$ is the temperature in $$K$$.

$$P = {7,853.98} \times 1 \times \sigma \times 5,772^4$$ $$P = 494,317,913,890.86 \text{ J/s} = 494 \text{ GJ/s}$$

Your wormhole last one second so $$\text{energy} = P * 1 \text{ s} = 494 \text{ GJ/s}$$.

## Let's melt iron!

It's Enthalpy of vaporization (Also know as heat of vaporization) is $$E_v = 340 \text{ kJ/mol}$$. It's atomic weight is $$A_w = 55.8452$$.

You can boil:

$$m = P_v / E_v / A_w$$

$$m = (494,317,913,890.86 \text{ J/s}) / (340,000 \text{ J/mol}) / (55.4852 g/mol)$$ $$m = 26,202.95 \text{ g} = 26.02 \text{ kg}$$

Or melt $$645.11 \text{ kg}$$ (given the enthalpy of melting is $$E_m = 13.84 \text{kJ/mol}$$).

# Gravity - Deadly but not much

It's surface gravity is $$274 \text{ m/s}^2$$ (28 times Earth's gravity). You wormhole is at 1 km from Earth's surface but that is negigible ($$274.2002 \text{ m/s}^2$$ vs $$274.1994 \text{ m/s}^2$$).

A wormhole of one second will pull all objects to it at $$274 \text{ m/s}$$. In 28 seconds Earth's gravity will fix it. All people will die due to the impact and all cars, bikes and other stuff will also break due to the impact. The whole Earth's movement will slightly change, it may unstabilize the Earth's orbit after a few millions of years, not sure.

I can't say if buildings are enough strong to not break apart. Gravity is a lot, but it's only a second. Maybe all the structure will crack and destroy.

# Particles - Nothing Meaningful

The amount of particles per cubic meter of the photosphere is very low $$~10^{23} m^{-3}$$, about 0.37% of the particle number per volume of Earth's atmosphere. Also, only 3% of the gas is ionized.

These particles carry a negigible amounts of mass and thus force with them. It's density varies from $$1 \times 10^{-3} \text{ kg/m}^3$$ to $$1 \times 10^{-6} \text{ kg/m}^3$$. Even flowing this particles at 7 km/s isn't much (like $$7 \text{ N/m}^3$$ to $$0.007 \text{ N/m}^3$$).

The same with temperature. It's composition is 74.9% H and 23.8% He, which is (using the heat capacity) ($$5,772 \text{ K} \times (0.749 \times 28.836 \text{ J/(mol*K)} / 1.00784) + (0.238 \times 20.78 \text{ J/(mol*K)} / 4.0026022)) \times 10^{-3} \text{ kg/m}^3 = 273.95 \text{ J/m}^3$$ in the worst cases. That amount of energy is very low. It may increase the temperature of air in the surrondings by a few degres, but not much.