# Would a pin head heated to 15 million degrees Celsius kill everyone in a 1000 mile radius?

The YouTube video How Hot Can it Get? contains, at the 2:33 mark, the following claim:

A pin head heated to 15 million degrees will kill everyone in a 1000 miles radius.

On what basis can this claim be true? Some of the things I can think of:

1. Radiation of the metal as it cools down

2. Energy released in fusion (not sure if this works for an iron pin)

Would the damage be only to organic matter or will it destroy other structures within that radius?

• At that high temperature, it wouldn't be a pin anymore, rather a lump of superheated metal plasma. May 16, 2017 at 14:57
• The conventional way to kill everyone in a 1000 mile radius May 17, 2017 at 5:55
• If you're interested in that kind of stuff, you probably want to check out What If by Randall Munroe.
– Mast
May 20, 2017 at 15:13

In this occasion Vsauce rather dropped the ball, I should think. As the other answers show, the claim as stated doesn't make much sense when you put in the numbers, and if you chase the source to its origin there's some crucial context that got dropped along the chain.

The video description attributes the quote to the book The Universe and the Teacup: The Mathematics of Truth and Beauty, by KC Cole, which contains the quote, attributed to James Jeans (but without an actual reference), in the second page of chapter 2,

A pinhead heated to the temperature of the center of the Sun, writes Jeans, "would emit enough heat to kill anyone who ventured within a thousand miles of it."

The quote itself comes from The universe around us (Cambridge University Press, 1930), p. 289, and it reads

The calculated central temperature of 30 to 60 million degrees so far transcends our experience that it is difficult to realise what it means. Let us, in imagination, keep a cubic millimetre of ordinary matter ─ a piece the size of an ordinary pin-head ─ at a temperature of 50,000,000 degrees, the approximate temperature at the centre of the sun. Incredible though it may seem, merely to maintain this pin-head of matter at such a temperature ─ i.e. to replenish the energy it loses by radiation from its six faces ─ will need all the energy generated by an engine of three thousand million million horse-power; the pin-head of matter would emit enough heat to kill anyone who ventured within a thousand miles of it.

OK, so having filled in the references, let's pick apart the calculation and see what the claim actually is. What Cole and Vsauce missed in the quoting is a crucial qualifier:

merely to maintain this pin-head of matter at such a temperature ...

The claim is therefore that an object at that high a temperature, were it to radiate away as a blackbody, while also having a magical energy pump to keep it at that temperature, would be as deadly as claimed.

To see whether this is true, let's put in some numbers. A blackbody at temperature $T$ radiates away a power determined by the Stefan-Boltzmann law, which reads $P=\sigma AT^4$, where $A=6\:\mathrm{mm}^3$ is the area of the cubic pinhead of the paragraph and $\sigma = 5.67\times 10^{-8}\:\mathrm{W\:m^{-2}\:K^{-4}}$ is the Stefan-Boltzmann constant, and this power then gets distributed equally over a sphere of radius $R=1000\:\mathrm{mi}$, so this gives a power density at that 1000-mile radius of $$j = \frac{\sigma \, A \, T^4}{ 4\pi R^2} \approx 0.0104 \left(T/\mathrm{MK}\right)^4 \:\mathrm{W/m^2}.$$ Now, here we get to one of the sticky points: the claim in Jeans' book has over-estimated the temperature of the core of the sun by about $50/15\approx 3.33$, but Vsauce has missed that discrepancy and he's repeated the claim for the more modern value of $15\:\mathrm{MK}$. Normally, this would not be a problem, because factors of $3$ are pretty ignorable in Fermi analyses, but the Stefan-Boltzmann law has a quartic dependence in $T^4$ and this can mount up quickly, giving a discrepancy of $(50/15)^4\approx 120$ between the Vsauce claim and its Jeans source.

In this case, the difference does matter, probably because Jeans has chosen his numbers so they're roughly at the edges of what they can give. If we put in the numbers for the modern value of the core temperature, we get $$j_\mathrm{Vsauce} \approx 0.0104 \times 15^4 \:\mathrm{W/m^2} \approx 530\:\mathrm{W/m^2},$$ which curiously enough is just under half of the solar constant, i.e. the energy flux density from the actual Sun at the surface of the Earth. Thus, from a pure heat-flow perspective, if you were exposed to this for an extended period of time, you might get slightly sunburned, but it's very far from deadly.

The Jeans claim, on the other hand, is somewhat different because of that factor of a hundred, giving $$j_\mathrm{Jeans} \approx 0.0104 \times 50^4 \:\mathrm{W/m^2} \approx 65\:\mathrm{kW/m^2} = 6.5 \:\mathrm{W/cm^2},$$ and that's a lot closer to the damage thresholds. Going by Safety with Lasers and Other Optical Sources: A Comprehensive Handbook (Sliney and Mellerio, Springer, 1980, p. 162), the threshold for flash burns is at around $12\:\mathrm{W/cm^2}$, while second-degree burns start at $24\:\mathrm{W/cm^2}$ - for a flash exposure under half a second in duration. Stick around for more than a minute and it sounds about right that you'll very quickly develop some very severe burns, and succumb to them not long after that.

However, as pointed out in the comments, the bulk of the radiation that carries this energy will be in the form of high-energy photons, peaking at around $1.3\:\mathrm{keV}$ (for $T=15\:\mathrm{MK}$; it's $4.3\:\mathrm{keV}$ at $T=50\:\mathrm{MK}$), and that's at the beginning of the ionizing-radiation regime (more specifically grenz rays), which means that the effects are somewhat harder to model, and the detailed radiometry of what would happen is maybe an interesting exercise for an xkcd What if? episode.

As a rough estimate, if you assume that all of the radiation is absorbed (reasonable given this plot of attenuation lengths in water), and taking a surface area of $1\:\mathrm{m}^2$ and a body mass of $75\:\mathrm{kg}$, the Jeans energy flux, when seen as ionizing radiation, is equivalent to an absorbed dose of about $870\:\mathrm{Gy/s}$, which immediately gets out of hand. The lower-temperature source at $15\:\mathrm{MK}$ delivers an equivalent dose of about $7\:\mathrm{Gy/s}$, which I guess is a bit over the ballpark of a Chernobyl liquidator after a couple of seconds. It seems, then, that at both temperatures you're likely to die through radiation sickness, though the details will be messy to work out - but then again it's not really what either of the original sources imply.

To emphasize, though ─ this calculation assumes that you have a magical source of energy that can supply the ${\sim}2\times 10^{18}\:\mathrm{W}$ (!) required to keep that pinhead at $50\:\mathrm{MK}$. It's a reasonable thing to assume if you're already in hypothetics land, but it's a completely different question to the energy that's actually stored in that tiny bit of highly ionized iron plasma, and it's important to state that up front.

I'll keep this around for the next time I need to scare someone into checking their sources ─ it's a monument to academic carelessness, if you will ─ because it's such a good example of how things fall apart if you don't look carefully enough. The claim, in its original context, is roughly reasonable - but the Vsauce claim falls flat under even mild scrutiny.

• @DavidHammen The problem with that calculation is that you'd need a believable estimate for the heat capacity of the highly ionized plasma, which is probably doable, but I think that's where the calculation stops being worth the hassle. Also: is the ball of plasma confined? Is it set loose at $t=0$? How does the radiative cooling compare with the adiabatic expansion? How much energy is released by that expansion itself? But yeah, there's no way this thing remains dangerous after a second. May 16, 2017 at 19:29
• @Nathaniel I think the part you quoted refers to what being exposed to 530W/m2 would do to you.
– Seth
May 17, 2017 at 8:33
• @Nathaniel neither source even begins to invoke radiation sickness - they both unambiguously call out the heat as the mechanism. May 17, 2017 at 8:58
• @Seth sure, but it assumes the 530W/m2 is in the form of thermal radiation at around 6000K (like sunlight), whereas it wouldn't be. Emilio: that's true of course, but it doesn't stop the statement that you would get sunburn from being false. (You would die.) May 17, 2017 at 10:37
• @NickT As I said below Floris' answer, the heating caused by that x-ray absorption is precisely what causes the fireball in a nuclear explosion, and that is what causes the most physical devastation. If you want to see it that way, then you want to start working out what happens in a nuclear blast if you change the bomb for one that pumps in a megaton of TNT every two milliseconds. Personally, I think that the air absorption is where it stops being worthwhile, but whatever rocks your boat. May 18, 2017 at 22:22

A pin head is maybe equivalent to a spherical piece of iron with a diameter of 2 mm. That gives it a volume of about 4 mm$^3$ and a mass of $3.2 \times 10^{-6}~\rm{kg}$; computing the heat capacity of matter at these kinds of temperatures is hard, but whatever method you use, the energy content of the pin that you calculate would be insufficient to kill all living things at that distance.

But what if you went to the extreme limit - the matter somehow converted entirely to energy? In that case, the energy would be

$$E = mc^2 = 3\cdot 10^{11} \:\mathrm{J}\, .$$

That's a lot more energy - but if you spread that over a sphere of 1000 mile radius, you would have 9 mJ (milli Joules) of energy per square meter: this is clearly not enough to kill "everything" at that distance.

On the other hand, if you could heat a pin head to that temperature, and keep it that hot, that would require (and release) a very significant amount of energy.

Assuming a perfect black body radiator with a 1 mm radius at 15 MK, the power emitted per unit time would be

$$P = A\sigma T^4 = 4\pi 10^{-6} \cdot 5.67\cdot 10^{-8} (15\cdot10^6)^4 = 4\times 10^{16} \:\mathrm{W} \, .$$

That is some serious power, but when we distribute that power over a sphere with 1000 miles radius, the power density is about 1.2 kW/m$^2$, which is roughly the intensity of sunlight.

Now it's worth noting (as pointed out by David Hammen) that the wavelength distribution of this power is "far beyond the visible". In fact, Wien's displacement law tells us that the peak is at $\lambda = \frac{b}{T}$ where $b=2.9\cdot 10^{-3} m K$. At a temperature of 15 MK, that puts the peak wavelength at 0.2 nm - the realm of X rays. In fact, the handy conversion from wavelength to eV is E = (1240 eV nm) / $\lambda$, so 0.2 nm has an energy of about 6 keV. Lucky for you, that is an energy that is well absorbed by air - according to this table the attenuation coefficient at 6 keV is about 23 cm$^2$ / g. With the density of air at about 1.2 kg / m$^3$ or 1.2 mg/cm$^3$, the attenuation length in air is 0.027 cm$^{-1}$. That means none of that radiation would reach very far. The local air would be massively ionized, but would then re-emit energy at progressively longer wavelengths; at 1000 miles you would be quite well shielded.

Clearly, getting very close to such a hot body would kill you, but at 1000 miles you get "just" the power of the sun - which you should be able to survive. Just don't look directly at the pin - you will probably go blind.

And given the power needed - no, you can't make (and keep) a small blob of matter that hot.

• @DanielSank, $A$ = area, $4\pi r^2$, $\sigma$ = Stefan-Boltmann constant (=$5.67\cdot10^{-8} W m^{-2} K^{-4}$) May 16, 2017 at 17:54
• One key problem with your answer (the same goes for Emilio Pisanty's answer) is that almost all of that 15 million kelvin pinhead's radiation will be in the form of gammas, X-rays, and extreme UV. It's not enough to look at the energy produced by the pinhead. You also need to look at the frequency and the radiation dosage that would result. May 16, 2017 at 18:41
• @DavidHammen I did mention in a comment that the peak wavelength is around 0.2 nm. I agree that ionizing radiation can be more deadly but the peak wavelength corresponds to 6 keV - that's a rather soft X-ray that is easily attenuated by the atmosphere. I updated my answer to reflect this. May 16, 2017 at 19:50
• The absorption by air is an interesting point - but then again, this rapid heating of air by absorption of radiation is precisely what causes the fireball in a nuclear bomb, so the resulting expansion would carry all of that energy regardless. If you buy that concept for Jeans' constant-power source, though, you're pumping about 500 kilotons per millisecond into that explosion, which is pretty sizeable. On the other hand, though, it'd take you a day to get to Chicxulub-level energies, so there's that. May 16, 2017 at 20:03
• @Floris: I have a hard enough time imagining one impossible thing before breakfast, let alone six. This blob of impossibly hot iron is more than enough for my imagination. May 17, 2017 at 1:14

A small atomic bomb converts about 1g of its mass into energy.

So even in the best (worst?) case of a pinhead being perfectly turned into energy this is only about the amount of energy needed to severely redevelop a small city's downtown. It certainly didn't kill everyone within 1000mi

• You're assuming that the superheated pin head could have total energy (with which to radiate) no greater than the rest mass energy of a cold pin head. This is not so. May 16, 2017 at 15:32
• @gj255 Well, 15 MK is only about 1300 eV per degree of freedom, which is a non-relativistic temperature for a gas of nuclei.
– rob
May 16, 2017 at 15:47

It depends how quickly you heated it. If you did not do it instantly like within a fraction of a second, but lets say took an hour to do it. the real small mass of steel that you speak of would gradually go through the following phase changes as it was heated up.

solid to liquid

if somehow confined to the pinhead volume:

liquid to gas

finally gas to plasma

Since according to Boyle's law, the volume of a quantity of gas increases with temperature, you would have to confine the original pinhead mass somehow to stay in that pinhead volume, and even more so as the mass changes form gas to plasma. In any sense, a gradual heating will not cause the thermal shock wave that a rapid instantaneous heating will cause.

Pinhead is made out of steel which is around 80% iron and 20%carbon. While you may get some carbon nuclei to fuse through quantum tunneling at 15 million degrees, this temperature is still insufficient for iron to iron fusion or carbon to iron fusion.

Even if you were able to cause fusion of all the carbon nuclei in this minute sample, you would not have enough energy to accomplish said devastation; not even a fraction of it.