# How long lasts a second of eternity? [closed]

Given a famous tale, which can be found here:

The king said, the third question is, how many seconds of time are there in eternity. Then said the shepherd boy, in lower pomerania is the diamond mountain, which is two miles high, two miles wide, and two miles deep. Every hundred years a little bird comes and sharpens its beak on it, and when the whole mountain is worn away by this, then the first second of eternity will be over.

Simple question: In similar conditions, how long would such a second be?

I would like reasonable estimates backed up with equations.

Parameters:

• The size of the mountain are given in the story; assume that the mountain is made from a material that the bird can wear down in rubbing its beak as it would against a cuttlebone;
• Assume a bird not at all unlike a budgie, a small seed eater about 7 inches long, keepers of which are known to supply them with cuttlebone;
• Assume that the average cuttlebone is about the same size as the bird in question, about 7 inches long by an inch thick in the middle, and may last about two months of ordinary time:

## closed as off-topic by SPavel, HopelessN00b, Draco18s, JohnWDailey, sphenningsMar 21 '18 at 4:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about worldbuilding, within the scope defined in the help center." – SPavel, HopelessN00b, Draco18s, sphennings
If this question can be reworded to fit the rules in the help center, please edit the question.

• Comments are not for extended discussion; this conversation has been moved to chat. – James Mar 16 '18 at 16:10
• I'm currently seeing reasonable answers between 10^22 and 10^32 years (and one with 10^11 per layer, a few mm at most I'm guessing)- I'm not familiar with any field where an error of that magnitude is acceptable... SK19 - could you maybe provide your definition of "reasonable" parameters for beak hardness etc.? – G0BLiN Mar 18 '18 at 15:02
• Without knowing how large the mountain is and what kind of bird is doing the sharpening we haven't a hope of giving any reasonably accurate answer. – Ash Oct 3 '18 at 15:50
• The main flaw is that this assumes that the mountain will remain unmolested by men with picks for more than 5 minutes once it is known to be made of diamond. – cpcodes Oct 3 '18 at 21:29

The mountain is made of diamond. Diamond on the Mohs scale is a 10.

For the bird beak I have no solid data, so let's assume it has the same value as tooth enamel, which is about 5.

Now let's make the (rather wild) assumption that for every 1 point of difference in the Mohs scale there is a factor 100 in abrasion. This roughly means that while material A with Mohs value X will abrade 100 layers of material B with Mohs value X-1, material B will only abrade 1 layer.

This assumption tells us that it will take $10^{10}$ layers of the beak to take away 1 layer of the mountain spot where the bird sharpens its beak.

If each sharpening takes away 5 layers of the beak, it means that it will take $2*10^9*100$ years to remove 1 layer of the sharpening spot. That is already 200 billion years, way more than the estimated age of the universe (15 billion years).

I am rather confident that weather will be quicker in leveling down that mountain...

• This way, I imagine the diamond will "soon" drown in beak enamel deposit rather than be worn away. ;-) – M.Herzkamp Mar 15 '18 at 10:34
• @M.Herzkamp Obviously, weather removes the broken down beak shavings while also not at all weathering the mountain. Because that's how it works. It makes sense if you don't think about it. – corsiKa Mar 15 '18 at 23:26
• Moh's scale isn't proportional or anything, so using the points difference doesn't make sense – Destructible Lemon Mar 16 '18 at 1:57
• @SK19, for a crystalline material made of a sequence of atoms it can be the first one in the sequence. For the beak is a bit more tricky to give a precise definition, as it is not a crystalline material, – L.Dutch Mar 16 '18 at 9:16
• @SK19 I read it as an infinitidesimal, AKA $dx$ – John Dvorak Mar 16 '18 at 9:48

For purposes of this answer, I am going to assume that erosion of the mountain is not an issue (since I think that is closer to the spirit of this story). I will also have to assume that the bird's beak is going to be able to make any sort of mark on the mountain, or the answer will be "infinite".

For a "small bird", I'm going with a house sparrow, whose beak is 1.5cm long (thanks, Wikipedia!), looks like 0.7cm at the base, and a stubby conical shape. Without going into detailed calculations, I'm going to assume that for sharpening to be meaningful, the sparrow has to abrade about 0.5% of its beak (very generous, I think), so that is $2\times10^{-10}$ $m^3$ of beak.

Let's assume then that 1 unit of beak material abrades 1 unit of mountain material (again, very, very generous here). You are looking at a conical mountain which has approximately $2\times10^{10}$ $m^3$ of material.

That means that it will take $10^{20}$ sharpenings to wither the mountain down to nothing. At the rate of 1 sharpening per 100 years, that's $10^{22}$ years, a.k.a. ten sextillion years, a.k.a. ten billion times the age of the universe, a.k.a. a bloody long time.

• So quite simply, then, we're looking at $k*10^{22}$ where $k$ is the coefficient of beak to diamond abrasion. – corsiKa Mar 15 '18 at 23:25
• @corsiKa: yes, yes we are. Funny how I completely missed that simple correlation. :/ – Xenocacia Mar 16 '18 at 7:34
• Well, you see, if $10^{22}$ years is a bloody long time I'm going to go out on a limb and assume the technical term for $k*10^{22}$ years is a bloodier long time. – corsiKa Mar 16 '18 at 14:37
• @corsiKa: excuse me. I think you mean bloodier longer timer. – Xenocacia Mar 16 '18 at 14:51
• It's only a bloody long time if the bird does not give its beak time to regrow between sharpenings. If it just keeps sharpening it away to nothing, then it will be a bloody long time. – Cort Ammon Mar 19 '18 at 19:01

At the scales we are working on, usually anything except the exponent part is something we can ignore, because error elsewhere is going to be larger than a factor of $10$.

The bird pecks at the mountain. It sharpens off $0.01 \mathrm{mm}$ of beak. The beak has a density of about $1 \frac{\mathrm g}{\mathrm{cm}^3}$ and a thickness of $0.1 \mathrm{mm}$, so this involves abraiding $0.0001 \mathrm{mm}^3$ or $10^{-12} \mathrm{kg}$.

The mountain is made of diamond. It has a density of about $3.5 \frac{\mathrm g}{\mathrm{cm}^3}$. It has a volume of $3\cdot 10^{16} \mathrm{cm}^3$, so a weight of about $10^{14} \mathrm{kg}$.

The mountain is $10^{26}$ times larger than the amount of beak sharpened off, give or take.

It takes about $2 \frac{\mathrm{kJ}}{\mathrm{mol}}$ of diamond to convert it to graphite -- basically, peel off layers of atoms. To sheer off a layer of graphite takes $0.2 \frac{\mathrm J}{\mathrm m^2}$. If we model the bird's beak as graphite (The bird's beak is going to stick together better than graphite does, so we'll use this as a lower bound): every $0.22\mathrm{nm}$ requires $0.2 \frac{\mathrm J}{\mathrm m^2}$ to sheer off, or every $4\cdot 10^{-8} \ell$ requires $1 \mathrm J$, or about $44$ micrograms per Joule or $4\cdot 10^{-5} \frac{\mathrm g}{\mathrm J}$.

$1 \mathrm{mol}$ of diamond is $12$ grams (assuming pure carbon-12), so $\frac{12\mathrm g}{2000 \mathrm{kJ}}$ is $6\cdot 10^{-3} \frac{\mathrm g}{\mathrm J}$ of diamond to sheer off carbon.

So, the diamond takes about $10^2$ more energy to break carbon off than it would take to sharpen matter off the beak.

If $1\%$ of the energy used to sheer off the beak also nicks off atoms of diamond, then $10^4$ more beak is worn off than diamond is in the sharpening process.

$10^{26}\cdot 10^4$ is $10^{30}$ beak sharpenings to wear the mountain down.

One sharpening every 100 years (aka $10^2$), so $10^{32}$ years, aka 1 followed by 32 zeros.

• Even if the question is off topic, I like this answer for having dived into just how much energy it takes to wear away the mountain and makes some assumptions about how much of the beak sharpening energy goes into that process. That is, it approximates the k in Xenocacia's answer (which is not an insignificant value). – Draco18s Mar 16 '18 at 17:57

As pointed out by others, the bird is definitely the underdog here. Its beak just isn't hard enough to really make much progress. However, you were specific:

"...only a literal diamond mountain is in the spirit of my question." -SK19

Well let's use real diamond. While De Beers might want you to think otherwise, a Diamond is Not Forever. I'll give 'em credit. It's pretty close. But it's not actually stable. It's metastable. Diamond decays into graphite over time... a lot of time. At high temperatures, we can make this happen fast, but at room temperature, we've never actually observed diamonds decaying into graphite..

Fortunately, chemistry is amazing. We can find that the activation energy of the transition is -540kJ/mol, which is utterly gigantic. But by recognizing that this is an activation energy problem, we can leverage the Arrhenius equation. This is an equation which has some decently strong theory behind it, but more importantly, it has a history of being empirically effective for many sorts of reactions, ranging from gas reactions to crystallization.

One of the consequences of the Arrhenius equation is that the pace of reactions tends to double roughly ever time you raise the temperature by 10C. It's not an exact rule, but we are talking about a bird wearing away an eternal mountain here, so I think some scientific license is permitted.

From this video, we can see that in the 1400-1500 range, it takes just a few seconds to convert diamond to graphite. For simplicity, let's say it takes 1s at 1500C. That's roughly 150 10C steps away from room temperature. Each one of those doubles how long the reaction takes, so $2^{150}s$ is not unreasonable. That's roughly $10^{45} seconds$, or $10^{38}$ years.

That number happens to be close to the lower bound on the half-life of a proton. The upper bound is quite a lot higher.

I'd say $10^{33}$ years to $10^{38}$ years to destroy the mountain, depending on whether proton decay occurs first, or if the diamond all turns to graphite first. If diamond to graphite takes longer (I've seen estimates of $10^{80}$ years for a half life), the upper bound of $10^{45}$ for proton decay will take over.

Oh, and the bird? Well, given the timeline of 1,000,000,000,000,000,000,000,000,000,000,000 years for the fundamental structure of the mountain to give way, I don't think it will have much trouble shaving off the graphite in a nominal amount of time after that occurs. Or maybe its protons will decay, in at which point the question really starts to become one of those abstract philosophical ones, doesn't it?

Short answer: A geological amount of time. Over 32 million years, probably longer, if tectonic activity raises the mountain.

Long answer: I am gonna assume the mountain is just called diamond mountain and not actually made of diamond.

Your mountain is 2 miles high, which is about 3.2 km (using metric, as I am doing science). This website states, that Everest erodes at a rate of 0.1 mm/yr.

Following the calculation 3,200,000 mm / 0.1 mm/yr = 32,000,000 yr or 32 million years.

This does not account for any growth of the mountain, as I don't have any information on the tectonic activity of the area (Everest grows 5 mm every year for example), but just keep in mind that it could take much longer.

The bird does not make any difference though.