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So earlier today, Neil Tyson posted an estimate of the mass of Mjölnir, (http://en.wikipedia.org/wiki/Mj%C3%B6lnir)  assuming that it's made out of neutron star matter. And +Cyrus Khan followed up with a calculation of how big a crater it would leave if Thor dropped it. (http://goo.gl/ZYlqS) But I'm afraid this calculation left something out, because while the hammer (if dropped from a height of 10 meters) would only be travelling at 14 meters per second or so when it hit the ground, it turns out that a mass of 300 billion elephants (about 2.1*10^14 kilograms, or roughly the mass of enough dirt to cover Texas a foot deep) hitting the ground at that speed will make a bit of a bang.

Specifically, it will have a kinetic energy of about 2.1*10^16 Joules, or just over 5 megatons. 

(For comparison, that's about the bang you would get from a 42,500 ton meteorite hitting the Earth – one about six times the size of the one that detonated over Chelyabinsk a few days ago – or from a single W-71 thermonuclear bomb)

But OK, just how much is five megatons? What would that do?

We can't compare this to the burst of a nuclear bomb, because bombs (by their nature) create large shells of supersonically expanding gas which fly into the air – that's how they blow things up – whereas this dumps all its energy into the ground. (If you do want to know how large explosions work, then you should look at the Nuclear Weapons FAQ, nuclearweaponarchive.org, perhaps the greatest nonclassified source in the world for understanding things that go "boom") Instead, it's probably best to compare this to an earthquake. Fortunately, the Richter scale is effectively a log scale of energy deposition, so we can make at least a rough guess of the intensity by using that.

The moment magnitude scale for earthquakes (one of the more modern improvements to the Richter scale – http://en.wikipedia.org/wiki/Moment_magnitude_scale) is actually even better, because it directly measures energy dumped into the ground. Using the definition of that, we find that the resulting earthquake would have a moment magnitude of 4.8. Since the Moment and Richter scales were calibrated to match at 5.0, that's about a 4.8 on the Richter scale, too.

But before you go saying "oh, that's not too bad," remember that most earthquakes are a bit more spread out than the size of the average hammer. Rupture areas are normally hundreds of square miles, not less than a square meter. And earthquakes generally go off at least somewhat underground, not right at the surface. So to get a more realistic estimate, we should instead compare this to meteor impacts.

There are a couple of ways to estimate crater sizes, and here's a handy website which runs through them: http://pirlwww.lpl.arizona.edu/~jmelosh/crater_c.cgi . You can just feed in the info; assuming that Mjolnir has a radius of about half a meter, and that it's striking a target of rock (as opposed to loose dirt), you find out that over the next 200 milliseconds, it's going to form a crater with a rim-to-rim diameter of about 21.7 miles.

Boy. It's a good thing he didn't drop it from any higher than that.
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"or roughly the mass of enough dirt to cover Texas a foot deep"

Do it.
 
Frankly, that doesn't sound at all plausible; I don't see how something travelling only 14 m/s is going to transfer enough energy to make a substantial crater, much less one several miles wide, no matter how effectively unstoppable it is.  I'm betting the meteorite models don't scale at all correctly when you extrapolate them that many orders of magnitude out past the regimes they are designed for.

I expect the right model is rather closer to a large bulldozer.  You get a decent localized earthquake and a deep narrow hole as it basically falls through the rock, crushing it out of the way without being slowed up all that much by it.
 
Can someone please convert the Dragon Ball Z powerlevel to how much Kilo calories and jouls are required to have a power level over 9000? How many calories would it take? whats the kinetic output of a kame ha me ha wave? HMMM?
 
+Vincent Litchard... I really love you right now. 

Those are the questions science must answer: not whether there was ever water on Mars or not.
 
Meteors have that kind of Kinetic Energy from their velocity not mass so there's got to be more to the story. What happens when you incorporate the fact that it is not going to disintegrate on impact and that its exerting on the order of 10^15 pascals on whatever surface it touches? 
 
+Brooks Moses I was thinking about the energy transfer issue, actually, but I think it may work properly anyway, at least to order of magnitude. The bolide is travelling well below the speed of sound in the rock, so the standard mechanism of kinetic energy transfer away from the impact should still work. During the very first few milliseconds of impact, until the sound waves have travelled out to the typical sort of radius of a meteorite, the force may be directed more downwards than usual, so you may end up with a deeper initial central depression and less energy going into radial modes; but that downward-thrusting energy is probably going to shoot down and reflect back upwards off the bedrock in a fairly diffuse fashion, so you'll get the energy back into the main crater anyway. It may give the crater a slightly unusual shape, but the total energy transfer into earth movement should be about the same.
 
+Deen Abiola You need to check your physics. Kinetic Energy includes the mass of the item. It's not purely a function of velocity.
 
+Yonatan Zunger, with the way my brain is currently fried right now, you look like Thor to me... A God of Thunder and Math...

For reals.
 
+Yonatan Zunger: The typical bolide stops pretty much instantly.  I seriously doubt the hammer even slows down much when it hits -- and, if that's true, you don't have a sharp compression wave from the impact point.
 
+Deen Abiola I actually ran some numbers for that, too. :) Given a typical meteor velocity for a standard solar system object, this would be the equivalent of a 42,500-ton bolide hitting the Earth. (Compared to about 7,000 tons estimated for the one that disintegrated over Chelyabinsk a few days ago – and NB that that one almost touched surface) Meteors of that size generally don't fully disintegrate before impact, so you get a fairly straightforward impact crater. The insane pressure it generates is exactly what makes that.
 
Good night all. I will check in tomorrow to see if this post makes any more sense after I've had a good nights rest...
 
+Patrick Q No, I made the same point as Yonatan in someone else's reshare (though I misused nukes as an analogy). This object was dropped from 10 meters. It's mass, 10^14 kg is contributing most to the kinetic energy.
 
+Yonatan Zunger Oh okay wait, I wasn't questioning your analogies! I agree with them =)

I was thinking that the hammer would land intact (it would be useless if it broke so easily!) I thought maybe it would sink/drill really deeply into the earth but I don't know what equations would be required to calculate how far. Like say if Thor were to float and drop it lightly, the ground would not give?
 
+Brooks Moses Right, but the way the bolide stops is because the sound waves transfer its energy away.

I guess that the real place you'd see a difference is that degenerate neutron matter is significantly denser than atoms, so you would have to think about the impact cross-section of the hammer against the matter – it might be significantly less than the geometric cross-section, because some of the Mjölnir nuclei are going between the dirt nuclei. There's also the issue that degenerate neutron matter is electrically neutral, so its interaction cross-section with the atomic electrons is only going to be via weak processes of the form ne->pbar nu, with the pbar probably immediately striking a nearby p and annihilating to photons. That's going to have a tiny cross-section compared to the usual ee->ee reaction of matter hitting matter.

Hmm. OK, you may be right: we really have to treat this as a weird sort of particle physics problem to estimate the effective cross-section of the hammer against the Earth. That would let us calculate something like an energy deposition rate per meter of Earth penetrated. That energy would probably exit the impact crater primarily in the form of S-wave oscillations (since the directly emitted photons would thermalize against atomic electrons and gradually convert into ordinary mechanical modes) and produce some kind of crater-like reaction, although at greater depths it might be more like a weird sort of earthquake.

OK. Clearly, some nuclear physics calculations are needed. Or possibly experiments. Anybody got a hammer?
 
Good thing he wasn't in Pisa or Venice.
 
+Deen Abiola Ah, I see what you are saying. I misunderstood and took you more literally.
 
OK, thinking time. Let's see if we can do the calculation that +Brooks Moses' comment implies we're going to need: find the rate at which a lump of degenerate neutron matter deposits energy into ordinary atomic matter, a.k.a. dirt. 

As a baseline, when normal matter hits normal matter, the main reaction is just the repulsion of electrons in the valence shells of the respective atoms. These electrons are basically smeared out over the entire impact surface, so the effective cross-sectional area is equal to the geometric area of the impact. That forms nice vibrational waves which cart the energy off in the form of sound waves moving through the earth.

For degenerate neutron matter, though, the reactions are instead going to be

nN -> nN (elastic collisions of neutrons from Mjölnir against nucleons from the dirt)
ne -> pbar nu (electron capture by the neutrons from Mjölnir against the electrons from the dirt)

But the second reaction is weak-mediated, so it's going to be tiny compared to the electromagnetic/strong nucleon scattering. (NB that this scattering gets a big EM component, even though neutrons are electrically neutral, because their component quarks aren't) So we can treat this as a straight-up nuclear scattering problem. 

The Earth's crust is about 47% Oxygen, 28% Silicon, and the rest miscellaneous other stuff. The O is almost entirely 16O and the Si 28Si. Their neutron scattering cross-sections are 4.29 and 2.12 barns, respectively, so we can estimate that the average cross-section of a terrestrial nucleon is going to be about 3.1 barns.

These are thermal nucleon cross-sections, which is good because the neutrons are coming in at very nonrelativistic speeds, but realistically degenerate neutron matter behaves pretty much nothing like free neutrons. (We're assuming here, like Tyson, that Mjölnir is made out of the free neutron drip part of a neutron star, despite the obvious questions of what would hold that together against a rather spectacularly rapid weak decay. I'm going to assume that it was bound that way by some particularly crafty Svártalfar technology, or maybe it's been wrapped in the hide of the world-serpent from a previous incarnation of Midgard by Odin. That sounds like the sort of thing he would do.) I am not a neutron star astrophysicist so I don't have a good guess off the top of my head about how this would interact. But since it is a degenerate gas, I would expect that the effective radius of each neutron would grow to be roughly the size of the hammer as a whole, basically a continuum of nuclear matter rather than individual nucleons. Fortunately, the density of the stuff roughly accounts for that, so I'm going to wave my hands in the air and assume this doesn't matter.

OK, so we have two intersecting columns of nucleons, in (as previously supposed) a circle of diameter 0.5m, one at a density of roughly 1.2*10^3 kg/m^3 (the density of dirt, which if we guess that the average nucleon has a mass of about 30u translates to 2.5*10^28 nucleons/m^3) and one at a density of about 2*10^14 kg/m^3 (or 1.2*10^41 neutrons/m^3). Each matter nucleon has an effective cross-sectional area of 3.1 barns.

The number of scattering events per unit length of dirt is therefore sigma*rho(dirt)*rho(neutrons)*volume(Mjölnir), where the rho's are number densities of nucleons, or 6.1*10^40 / m. (Using standard formulae for the definition of cross-section; I yoinked this out of Peskin & Schroeder, eqn 4.59, because I couldn't remember it. Too many years doing CS instead of physics.) 

Since this is a nonrelativistic elastic collision, we can easily estimate the momentum transfer per collision as 

delta p = 2mMv/(m+M),

where m and M are the masses of the neutron and the nucleus, and v is the impact velocity. Over a period of time (dt), the total number of collisions will be X*v*dt, where X is the number of scattering events per unit length computed above, and so the total momentum transfer will be

dp = 2mMv^2Xdt / (M+m) 

and so the force is simply dp/dt = 2mMXv^2/(m+M). Plugging in those numbers, that translates to 

F = (1.96*10^14 kg/m) * v^2

At the moment of impact, that translates to 3.9*10^16N, which in turn should translate to a deceleration of the bolide at 187 m/s^2, enough to bring it to a halt within 75msec.

This is comparable to the stopping time of ordinary matter against ordinary matter, so I conclude that the degenerate nature of the neutron matter does not, in fact, affect the original calculation.
 
Oops: Slight correction. Since the force is proportional to v^2, I can just solve the differential equation for its motion. Dividing the force equation by the mass of Mjölnir, we get the equation

dv/dt = - v^2 / x0
v(0) = v0

where x0 = 1.07m (The mass of Mjölnir / 1.96*10^14 kg/m, the characteristic depth of the equation) and v0 = 14.14m/s is the initial velocity. The solution is

v(t) = v0/(1 + v0 t / x0) == v0 / (1 + t/t0)

where t0 = x0 / v0 = 75.7msec is the characteristic time. This never goes all the way to zero, but to a reasonable approximation it comes to a stop when t = 5 t0 = about 380msec. 

The position can be derived from integrating v(t); it's simply

x(t) = x0 log (v0 t + x0) - x0 log x0 
      = x0 log (1 + t/t0)

(all logs being natural, of course) so at this time it would have penetrated a depth of roughly x0 * log 6 = 2 meters or so into the ground. 

Basically, Mjölnir will stay on the surface, while the resulting forces spread out in an expanding circular ring and form a really damned big crater. 
 
But momentum must be conserved. If the hammer stops in 75msec, then its entire momentum is transferred to the mass it displaces. In 75msec it did not travel very far (it was slow moving to begin with) so it did not displace a lot of mass. The ground is much less dense than the hammer. To conserve momentum the displaced mass would have to be ejected at relativistic speeds, that's an explosion. 
 
+Fedor Pikus I think this should be OK. Energy is being transferred away from the hammer via sound waves. The speed of sound in dirt is roughly 350m/s, much greater than any of the velocities involved in this problem; so no shock wave should form, and instead we should see a smooth wave deforming the ground, like a pebble dropped in a pond.

A really big pebble, maybe.
 
+Deen Abiola See my long comments below. Looks like it would only burrow about two meters into the earth. Apparently the incredible density of neutrons in a neutron star is more than enough to compensate for the fact that neutrons aren't stopped by matter particularly easily.
 
The conclusion is that Thor can carry 300-billion elephants :)
 
One big missing piece is the force of gravity. Another is accounting for the transformation of rock into effectively sand.
 
+Oleg Mihailik Gravity shows up here in two places – one in the speed of Mjölnir when it hits the ground, and second in the model of crater formation, since gravity determines how high the berms of the crater go. 

The transformation of the rock into sand is also accounted for in the standard crater formation models.
 
I estimate a crater of ~200m diameter at first 30 minutes, and a horrible volcano in about 45 minutes.

The hammer will hit the iron of the Earth core.

The force that set the hammer in motion doesn't cease when it hits the ground.
 
+Luke Youngman That's nothing. He managed to lower the water in a drinking-horn that was attached to the ocean, lift one paw of a kitten which was actually the Midgard serpent, and wrestle an old woman to nearly a draw who was actually old age herself. 
 
Suddenly xkcd what-if seems tame. 
 
You are treating it as a projectile, but it is more like a hydraulic press.
 
+Oleg Mihailik Why? It should be a simple, nonrelativistic, parabolic drop onto a rigid medium. (With the approximate rigidity confirmed with that nuclear physics calculation above)
 
Why does hydraulic press not create an explosion? There's a lot of energy to dissipate too.

It's not rigid. The speed is slow enough to start treating the rock/sand as a liquid. The kinetic energy gets wasted on viscosity and transferred into heat.
 
+Oleg Mihailik Why do you say the speed is so slow? We're looking at an impact velocity of 14 m/s, the sort of speed at which one normally thinks of the ground as being fairly hard. For the dirt to be effectively liquid, the impact velocity would have to be orders of magnitude smaller.
 
+Yonatan Zunger  You can transfer energy this way, sure, but sound waves won't carry enough momentum to stop a hammer that is 10^10 times denser than the rock after displacing the volume of rock about equal to its own. It's more like dropping a pebble into a foam, the pebble will just keep sinking until it compresses the foam enough to offer resistance (and if you have a hammer that is 10^10 times denser than the rock, the earth crust is very compressible from your point of view).
 
+Fedor Pikus The displacement of target material doesn't have anything to do with how effectively it's stopped. Newton's rule for approximating impact depth is meant only for high-velocity collisions, where you can assume that there's no cohesion in the target material. In the far subsonic range I don't think that assumption holds; sound waves can cart the energy off quite effectively. 
 
About the speed of a car cruising populated streets. Car collisions like that don't explode their kinetic energy, they deform and crush in more or less liquid/viscosity physics fashion.
 
+Yonatan Zunger Awesome answer and excellently explained. If every story problem were in this form Physics would be the most popular subject in school.

Yet I'm still uncomfortable with the final answer. It describes what happens during contact by looking at momentum transfer at the subatomic level (so cool) but I'm interested in what effect that much pressure has on the dirt as it rests there. 

It seems you're assuming uniform density of the material it's resting on but intuitively it seems there should be a failure from stress and that it should sink until some kind of equilibrium is reached - say some point where density is effectively uniform. Am I making sense? I don't know enough physics to tell, but an answer to that is worth looking silly for!
 
The physics is partly liquid/pebble partly solid body collision. To which proportion is hard to judge, but simply considering insane density intuitively it should be much more liquid than solid physics.

So, initial impact crushes the rock, sends splinters and boulders flying, while the hammer digs into the ground, slows down and it converts into more and more liquid model.

Eventually the displacement will create a crack, the hot material from underneath the Earth crust will shoot up and expand that crack and the hammer will freely sink through the magma. 
 
+Yonatan Zunger: It's not even close to rigid.  You're thinking like a nuclear physicist, and completely forgetting the macro-scale effects -- and, as you noted, this is all very subsonic so those matter a great deal!

Concrete -- a reasonable stand-in for rock -- has a modulus of elasticity around 50 GPa (http://en.wikipedia.org/wiki/Properties_of_concrete).  Which is to say that a hammer pressing down on it with a pressure of 10^15 Pa -- assuming +Deen Abiola's number is correct -- is more than capable of squishing it well into the "much smaller than it was" nonlinear range.  The compressive strength of rock is likewise many orders of magnitude smaller than the pressures involved, which means that the rock might as well be fluid as far as this is concerned.

I maintain that the rock will barely slow it down -- not because of atomic effects, but because the force required to simply push the rock out of the way is trivial compared to the scales involved here.  Even if you set it down gently it will start falling through the ground like a heavy pebble falls through a lake.  Or, more accurately, like a heavy pebble falls through the atmosphere.

The energy-deposition calculation you want is going to be more one of "supposing the hammer continues moving essentially unimpeded, how much energy is required to get the rock out of the way by a combination of accelerating it, fracturing it, and squeezing it?"
 
+Deen Abiola That's a fair point; I had been focusing my attention on the initial impact, but once the hammer comes to a stop its static pressure on the ground should lead to very different dynamics; essentially the "hydraulic press" dynamics that +Oleg Mihailik is raising.

So OK, modified prediction: The hammer will hit the ground in a fairly traditional bolide collision, coming to a stop within ~400msec as it transfers its initial energy and momentum into sound waves, and penetrating roughly 2m below the original surface. This will trigger an expanding circular wave which will leave an impact crater about 22mi across. Once it slows down enough that its initial velocity is basically dissipated, its mass will start to compress the soil beneath, and it will sink down as though through a viscous fluid. It should, in fact, keep sinking all the way down to the center of the Earth, because it's much denser than anything beneath it, enough so that the hydrostatic approximation should keep making sense even in the inner core. So at some point (I'd need to think about the fluid dynamics to figure out how long this would take) the hole it would start sinking into would penetrate into some pressurized region of the mantle, and lava would start to flow upwards.

Exactly what happens next would depend on just how much pressure the region beneath was under, but I suspect that most regions aren't under explosive amounts of pressure normally, so we'd get more of an oozing eruption than a big bang. I'm also not sure just how much lava would come up – whether it would make it all the way to the surface at all, whether it would form a mount inside the crater, etc. 

Is there a vulcanologist in the house?
 
Did you know that if you took those 300 billion elephants and laid them end to end into space, they'd all die? ;-)

Anyhoo... You have got to love that man. He's easily the most legendary astrophysicist since Carl Sagan... maybe even greater all round.
 
+Brooks Moses Hmm. Is the force required to push the rock out of the way really negligible during the initial impact? 

Maybe a way to calculate this would be to think about the time between successive impacts on each nucleon compared to the relaxation time of the molecular structure of the rock, since the latter is what carts away energy into vibration modes. 

But perhaps you're right and we should simply assume unimpeded forward motion. In that case, we could use Newton's formula and the penetrating depth would be Lrho(M)/rho(Earth), where L is the diameter of the hammer, which is effectively infinity. In effect, it wouldn't manage to deposit any of its energy into radial modes, no matter how low the speed was, so it would form a bullet hole.

That seems wrong to me OTTOMH, but I really need to get some sleep right now so I may need to ponder this more in the morning. :)
 
+Yonatan Zunger: Why does the hammer stop?  What force stops it?

Your typical 5m bolide hitting the earth at 30 km/s stops because if it did not slow down it would sweep out its volume in solid rock in a small fraction of a millisecond, which means it would need to accelerate that rock to a comparable velocity either downwards or sideways in order to make room.  And, given that the densities are comparable, it's clear that the bolide cannot do that for many fractions of a millisecond without losing all of its energy.

The hammer, being 0.5m in diameter and moving at only 14 m/s, has no such large mass of rock to accelerate in order to keep moving.
 
Who says its not hollow?

Wouldn't this bring it down to a more manageable mass?
 
Disgree. Like a pile driver the energy would be released at the surface would be low.  It would however, increase progressively down to the core of the earth. If it went through soft sand you would barely notice it. It would accelerate at 32ft/sec^2 and after a km through rock would be really expending energy, but bear in mind the 5megatons of TNT would be dissipated over the 6000km to the centre of the earth.  in fact probably more because it would overshoot and oscillate for a while before settling to form a new center of gravity for the earth.
You would feel some rumbling but not a massive earthquake.
Sorry, no crater.  This stuff is going right to the bottom.
 
Wouldn't neutron star matter explosively revert to normal matter if it wasn't in a neutron star?
 
+Christina Talbott-Clark magic enough to not explode with the force of a thousand suns, but not magic enough not to fall down into the core of the earth. Check.
 
I see a mistake in your equation. Dr Sheldon Cooper.
 
Dude ask Dr.Sheldon Cooper if your gonna post these
 
Side note: When is someone going to convince Dr. Tyson to join G+? I can't imagine him not enjoying threads like this.
 
+Greg Slocum the earth's crust isn't nearly that cohesive. It doesn't matter what you poke at it (even a moon), it's still not going to drag down more than a few kilometers without breaking apart.
 
+Jasper Janssen The idea here is suspending all but the disbeliefs which make the problem most interesting =)
 
Lesson: this stuff is harder to reason about than it might first appear. If this thread reaches a reasonable consensus, it should absolutely be a guest "What If?" post on XKCD.

Also, is anyone aware of software that would allow you to stimulate this interaction? Seems a lot easier if, as Yonaton suggests, the nuclear physics aspects can be largely ignored.
 
One of the few instances that actually allow the use of the word "epic." So there, +Yonatan Zunger THAT is an epic thread! 
 
Ha, another big elephant missed!

The hammer will explode regardless of the Earth. Neutron star is held together by gravity, in small quantities that weird matter is unstable.

So really, it will be a majestic explosion like none of our nuclear bombs put together would approximate.

It will pulverise and melt the area size of Siberia and immediately kill all life on Earth.

But if we can rely on magic to keep the hammer together, we get a medium size boom at the impact, then sinking and crackling through the crust, slowing down on the way. A brief acceleration as the hammer enters the liquid matter and smooth slowing down as it travels through the mantle and fuses with the core.
 
OK, so we're proving a "comic book fact"? What's next - shall we prove that a coyote can survive riding an explosive rocket that plummets 2,000 feet from a cliff top?
 
So basically there was no way that Helicarrier was gonna stay in the sky once Thor came on board.  ;)
 
All hail Thor, Odin son. The one true God. 
 
+Phil Gowers More a "Prose Edda Fact" than a "Comic Book Fact."

+Oleg Mihailik I'm assuming that the neutron star matter was bound in place with a strap cut from the hide of the Midgard serpent, so it doesn't dissociate. Those Svártalf engineers know what they're doing.

(I did think through more of the physics overnight, btw, and will update shortly)
 
Guess what we talked about in the car on the way to work? Yeah, this. 

The husband's take on it: since, as +Zachery Jensen pointed out, Mjolnir wasn't forged from a neutron star, but in a neutron star, it doesn't actually have that much mass. Rather, through sympathetic magic, the hammer took on attributes of the neutron star as it was forged. It could have variable mass, but that isn't as good an explanation for, for example, Thor being able to pin Loki down under Mjolnir without crushing him; so, instead, it must have selective inertia, tuned to Thor (and, ultimately, to Odin). This explains why Thor can swing it around so easily at times, yet at other times it cannot be moved; and why it doesn't either sink through the earth, or yoink the Helicarrier out of the sky. 

He also says that clearly Mjolnir must have been made from a component of a starship drive's inertial dampers, and the the shipyard for those starships logically would be in a Dyson sphere built around a neutron star. (Is it any wonder I married him?)
 
I think the "slowly sinks through" scenario is right (more like a hydraulic splitting wedge than like a bullet). The other scenario, "stops after short distance and dissipates all energy into sound" has a self-contradiction: we assume that the hammer will travel the distance about equal to its own size and the energy is dissipated into sound waves and all involved speeds are slow. Ok, where did the momentum go? To the initial displacement of the rock, and then re-dissipated into sound waves. How fast was the initial displacement? The hammer is 10^10 times denser, the displaced volume is about that of the hammer, so the initial speed was 10^10 times faster (before relativity stops this nonsense :) We started with the assumption of subsonic speeds and got a relativistic shock wave. So the scenario of the quick stop cannot satisfy both energy and momentum conservation. The other scenario is, the hammer slowly sinks deeper, losing energy and momentum to viscosity and sound waves (of course, gravity keeps accelerating it).
 
if it was neutron star matter, wouldn't stuff orbit around it?
 
+Bryce Li Be attracted to it, at the least, yeah. I was just thinking that myself: it's not that Mjolnir sinks into the earth so much that it pulls the earth around itself. 

ETA: Although... how much mass do 300 billion elephants have relative to the mass of the planet? Anybody?
 
Stuff orbits around mass, regardless of the material of that mass. Hammer is heavy but no t heavy enough for the Earth to even minutely shift its position, nevermind rotate around the hammer.
 
+Yonatan Zunger Thank you! It would have taken me way too long and too much scratch paper to try to work that out myself. 

Not even remotely a problem, then. :) I guess even being made of magically bonded atoms can't do everything. 
 
Another little detail. The hammer will collect dust, dirt and gravel like a greedy tortoise.

It's like magnets collecting little specks of iron, except gravitational and thus omnivorous. With the mass of a mountain placed a meter away (radius of the hammer) the gravity is strong enough to be apparent.

Won't affect our carnage story, just a cute side note.
 
If you dropped the hammer, I'm thinking it would fall straight through solid rock without enough resistance to actually release the energy.  Probably fall all the way to the core either compacting all the mass it hit along the way. 
 
Picture the ancient aliens guy saying "nerds"
 
Just skimmed some of the entries in this long and cleverly argued thread, so forgive me if I missed something. BUT... Carl Sagan actually played out a very similar thought experiment in the original Cosmos series. He discussed what would happen if he had a small ball of neutronium and "dropped" it from a few feet above the earth's surface. (He illustrated with a small, dark grey ball about the size of an American handball or racquetball.) We say "dropped" here to indicate that it was at rest (somehow) just a few feet above the earth before being released and allowed to fall.

The crucial property, according to Sagan's telling, is that the density of this material so far surpasses any Earth-bound matter that we have ever encountered that many normal rules do not apply. He said something to the effect that its density, as compared to solid rock, or even the molten iron of the Earth's core, is like the density of solid rock is to smoke. So it would cut through the Earth itself like a bullet cuts through smoke, essentially not even noticing it was there. It would accelerate rapidly, pass through the center of the Earth and come out the other side, stopping at the same altitude, roughly, that it was dropped from, not having lost very much energy to friction. It continues to oscillate thus for some time. As the Earth turns, the object cuts holes through slightly different paths through the Earth, eventually giving it a some what Swiss-cheesy consistency. (This does not get into the details of what happens when magma flows into the tunnels being cut, etc.)

I'd bet Sagan's discussion has be clipped and uploaded to YouTube, I'll go look...
 
It seems to me the scenarii where the hammer sinks through the soil, should account for the work of gravity further than just the initial 10 meters drop.
 
I did some calculations with some online earth impact simulator, here's results:
Your Inputs:
Distance from Impact: 14.00 meters ( = 45.90 feet )
Projectile diameter: 20.00 cm ( = 7.87 inches )
Projectile Density: 1100000000000000000 kg/m3
Impact Velocity: 14.00 meters per second ( = 45.90 feet per second )
Impact Angle: 90 degrees
Target Density: 2500 kg/m3
Target Type: Sedimentary Rock
Energy:
Energy before atmospheric entry: 4.52 x 1017 Joules = 1.08 x 102 MegaTons TNT
The average interval between impacts of this size somewhere on Earth during the last 4 billion years is 1.5 x 107years
Major Global Changes:

The Earth is completely disrupted by the impact and its debris forms a new asteroid belt orbiting the sun between Venus and Mars.

Good, he didn't dropped it... Or something is wrong here?
 
Really doubt merely a mountain worth of mass would retain any energy to oscillate about. I don't think it is heavy enough to accelerate through rock even, it will merely sustain speed as do normal objects falling long way through air or water.
 
+Yonatan Zunger Feature request: Permalinks to individual comments. Your math is too beautiful to leave buried behind a top-level post.
 
Theia has hit the Earth 4 billion years ago and only made enough damage to create the Moon out of splinters. This little hammer is not heavy enough to break the planet like Earth to pieces.
 
+Oleg Mihailik, the crucial property is density, I think, and the mass itself is not exactly important. It oscillates because it is so dense that is just pushes through normal matter without even taking note of it. Like I said, a bullet through smoke. I could be mistaken, but this is how I recall Sagan's discussion going.
 
Normal object falling through water reaches terminal velocity. The velocity is determined by the ratio of mass to friction, i.e. for a spherical object by density. The hammer is 10^10 times denser than water, its terminal velocity is 10^5 times that of a normal matter (it's proportional to the square root of density).
 
What will be the energy output rate at that terminal velocity?
 
(OK, so the clip from Sagan does not contain the comparison "neutronium is to iron as iron is to smoke," I must have picked up that analogy from somewhere else. It might have been one of my college professors. At any rate, there is a several orders of magnitude difference in density, and I expect they would interact in a similar manner. I'm inclined to think Sagan's analysis is about right. I don't predict an impact crater, but rather a bunch of tunnels about the size of Thor's hammer, being cut right through the Earth.)
 
+Fedor Pikus the hammer is not falling through water. Also you need effective viscosity of rock to calculate the terminal velocity.
 
+Ronald Gainey in the ideal case of falling through the hammer is acting like hydraulic press, crushing its way on. Which implies similar damage as the press would have done, or compare it to a bulldozer taking down a house.
 
Most of the earth is magma, and its viscosity is actually quite high, from 10^3 to 10^12 Pa*sec, compared to 10^-3 for water. Terminal velocity is proportional to the square root of density over viscosity, so if the hammer is 10^10 times denser than the rock and magma is 10^10 times more viscous, it comes out the same. Interesting, seems like terminal velocity of the hammer (assuming it breaks through the thin earth crust) is about the same as that of a brick in water. At that speed it won't oscillate much when it reaches the center, it'll just settle in.
 
+Oleg Mihailik, re: hydraulic press, I don't think so. Normal matter is mostly empty space, including atoms themselves. Almost all the mass is in the nucleus and almost all of the size is in the electron cloud. The atomic radius is 10^5 times the size of the nuclear radius. Neutronium acts like a giant atomic nucleus. There are also no electrons around a chunk of neutronium (no protons to attract them), so there are no Van der Waals forces, etc.

I guess it's not clear how such a mass would actually interact with normal atoms, but I believe the strong force is repulsive beyond a certain radius, and so I tend to stick with the explanation that the normal matter would just get pushed aside.
 
+Ronald Gainey We have to assume that the hammer somehow repels the normal matter that comes into immediate contact with it (i.e. it pushes aside the atoms that touch it, by interacting with their nuclei or whatever, and that sets off the normal Van der Waals forces in the rest of the matter and everything gets pushed away). This is the point where the assumption that neutronium somehow just holds together comes to bite us: whatever forces maintain the pressure high enough to form degenerate neutron matter somehow stop working at the boundaries of the hammer and don't let the neutrons of the normal matter to join the mass of the hammer (and don't smash electrons into protons and then absorb them as neutrons into the hammer). So we basically have to assume that the hammer is surrounded by some impenetrable barrier.
 
The real question is, what does a "normal" atom, composed of a nucleus surrounded by a much larger electron cloud, do when it collides with a chunk of neutron star? I'm thinking it just bounces off, but I don't know for sure.

I wonder if someone like +Alexander Natale has an opinion?

(Edit: Cross posted with +Fedor Pikus' immediately preceding post, not a response to it.)
 
A regular normal atom and a real neutron star? The atom gets absorbed, the protons and electrons are squished together and the neutrons join the neutron Fermi liquid of the star.
 
Right, right, but yes, in our scenario there's some sort of magic involved, giving us neutronium without the normally requisite amount of mass, and your previous comment gets at the (essentially unresolvable) messiness of deciding what happens in the "magic zone."

I'll consider the horse dead, to at least a first order approximation, and stop beating it now.
 
Actually, let me spell it out more clearly, perhaps merely for my own benefit, but perhaps for someone else who may be following the discussion. As +Fedor Pikus and others correctly point out, there must actually be a zone of magic surrounding Thor's hammer, if it is to be composed of neutron star material (neutronium), in order for it to hold together as such. In the real universe, you need a stellar-sized mass to make a neutron star, you just won't get neutronium otherwise.

How such an entirely magical object will interact with normal matter depends essentially on the properties of the "magic zone." If normal matter bounces off the magic zone, having a very tiny coefficient of friction (approaching zero) then the hammer just slices through the Earth for a long time, oscillating around the center of mass of the Earth, as described by Carl Sagan. If, OTOH, the magic zone is very sticky to normal matter (very high coefficient of friction) then the hammer rapidly accumulates a ball of material around it, and it does, in fact, burrow into the Earth some modest distance, forming an impact crater of some size, before it is ground to a halt by this accretion of normal matter.

My intuition was shaped by Sagan's description, which I saw as an adolescent, and he implicitly posited a slick, very low friction magic zone. Others, including +Yonatan Zunger, imagine a sticky magic zone, making this resemble a meteor impact. I don't think there's any answer because anyone's intuition/imagination is as good as anyone else's, when it comes to magic. :-)
 
Slow stuff will actually stick to that hammer, in the same fashion iron pecks stick to a magnet.

But gas and liquid is not slow, their molecules fly and wobble quite fast. Solids can be slow, so dust and pieces of grass and dead beetles will stick to it.
 
+Oleg Mihailik, I don't think we can make statements like "slow stuff will stick, fast stuff will not." There is no electron cloud surrounding a chunk of neutron star, and the electron cloud determines how atoms and molecules interact in normal matter. On a neutron star, a captured atom gets absorbed and converted into more neutron material, as +Fedor Pikus described. On Thor's hammer, it's up to the magical properties of the magical force that makes the hammer exist. Maybe slow stuff sticks, maybe not.
 
Actually, viscous friction of an object moving through a fluid does not require "sticky" surface: the friction is generated by the liquid itself. A perfectly slick object still has to displace liquid in front of it and make it flow around it, and liquid has friction even if the object is slick. I don't know what will happen to the outer earth crust (viscosity of the solid rock is huge, about 10^24 Pa*s, but it may simply break and let the hammer through) but if the hammer reaches the magma, its viscosity is known (varies widely depending on composition and temperature) and it seems that the hammer will drift down at about the same speed as a rock sinking in water (give or take an order of magnitude :)
 
We have to assume it an ideal solid body of a given mass. The molecules will bounce off it, as long as the gravity can be overcome (which it can for such a medium-sized hammer).

It being slick and glossy doesn't change the fact that Earth's material will have to crush bend and get out of the way for the hammer to move down. That moving out is essentially the same dance that a house and a bulldozer do: one is going through another unstoppable and generally slow.
 
It's actually more complex than that, because that small a mass of neutron star material is not stable;  it does not have enough self-gravitation to overcome degeneracy pressure, and as a consequence it will undergo beta decay, producing protons, electrons, neutrinoes, gamma rays and a bunch of energy.  It will do this with a halflife of a few minutes, so given how dense it is, the thermal power is likely to be pretty enormous.  Which is another way of saying that it will explode, violently, with a total energetic payload of several thousand megatons.
 
Sure, but from the start we are assuming that there is a magic force field around the hammer and this field can exert arbitrary high force on the matter it contains (at least high enough to equal the gravitational pressure inside a neutron star). The outside matter actually interacts with that force field and not the neutron Fermi liquid it contains, so we should probably postulate that the barrier is impenetrable from outside also, and beyond that we can't say anything.
 
We discussed that earlier: there must be magic holding it together (+Yonatan Zunger have even described some implementation of such magic). The best approximation therefore is an ideal solid body.
 
Right, but does that apply when Thor's not controlling it?
 
The hammer does not blow up when left alone, does it? So it must still apply. Or Thor always controls it, even at a distance. :)
 
There are two possible answers: yes it does or no it does not.

If it does not we have a fantastic explosion leaving a sea of lava the size of a small continent, a ring of debris at low orbit slowly coalescing into a small moon over fifty million years and total life extermination.

If it does hold, we get to our current line of discussion.
 
+Yonatan Zunger I may have found an error in your calcuations (disclaimer: I am no physicist). This equation:

sigma*rho(dirt)*rho(neutrons)*volume(Mjölnir)

seems to be a small-density approximation.  I'd imagine that most of the dirt-neutronium interactions happen at the leading edge of the neutronium since it is so dense.  Once a dirt particle has interacted with the neutronium (absorbing some energy/momentum) it is not available to interact with the other neutronium particles along the vertical axis of Mjollnir (at least, not in a way that absorbs energy).  If this equation is normally used to model particle beam interactions in an accelerator, you wouldn't see the same effect there because the beams are much more diffuse.  The cross terms probably disappear.

This may be incorrect, but may help to reconcile your calculation with the comments from others in this thread (and Carl Sagan apparently).
 
+Martijn van Schaardenburg There is something related to that -- I think I figured out the mistake in my calculations last night, but haven't had time to post some refined calculations yet. I think that I'm in agreement with the rest of this thread, though, modulo some possible details about what the entrance and exit wounds might look like.
 
Oh my, well I'm glad I issued my disclaimer at the very beginning, but it does look like there's a lot going on, and the answer(s) are rather unclear (at least to me). I had not actually read Yonatan's treatment of the scenario as a nuclear scattering problem, which seemed pretty compelling. I was prepared to withdraw all my previous comments (and provisionally concede that Sagan was missing some details), but everything just got murky again. Fascinating discussion, Yonatan!
 
It's worse than that, +Martijn van Schaardenburg : the atoms cannot enter the hammer at all, it's in a degenerate neutron liquid meaning all energy states below the Fermi level are occupied, the Pauli exclusion principle says that any new neutrons entering the hammer must go above the Fermi level. This is a large energy barrier for neutrons of the atom nuclei (and again we have to wave our hands about the magic force field vs protons and electrons and assume that the hammer does not rip the matter apart at the nuclear level). So basically the barrier which contains the neutron liquid in the hammer manifests itself as a high energy barrier to the outside world (at least the neutron part of the world).
 
+Fedor Pikus If the atoms stayed relatively fixed (e.g. because the atomic restorative forces could push them back into place) then they could still interact with the full stack of neutrons. But as has been pointed out, they won't, so instead Mjolnir would act as a sort of giant reamer.
 
+Yonatan Zunger They would interact with a full stack of occupied energy states, sure, but it won't be an individual neutron cross-section. The calculation of the scattering cross-section  assumes that all initial and final states are available. In our case only high-energy states are available on the "inside" semispace.
 
+Yonatan Zunger  Holy deja vu Batman!!! According to my calculations it's the bloody brilliant pot roast... in slow motion...  This is a problem for shock physics believe it or not...  The pressure of the hammer merely placed on the surface of the earth would exceed 4x10^15 Pa.  Placing this hammer on the surface of the earth with zero velocity and then releasing it would launch a very strong shock wave (really? yeah really.  It's a version of the classic piston driven shock problem in an admittedly unusual corner of thermodynamic phase space).  This is the same principal at work in shock tubes for gases or liquids by a piston driven shock.  By exceeding the bulk and shear moduli of rock diamond... the core of jupiter... by >>10x you get this type of response (keep in mind that this is about 1/6 the pressure at the center of the sun!!! Very extreme conditions).   (Math warning, I didn't spend as much time with the #'s as I should have so I encourage ppl to research their own solution to the problem if they wish to verify my conclusions).  Assuming the #'s are good though here's some words.  

So what would happen in our 0 m/s spherical cow model?  First a large vertical column of the crust would break into an outline resembling bugs bunny running through a door (this is the brittle/compressible/plastic flow phase of the process very messy with unimportant/confusing details).  This rapidly compressed column underneath the hammer would heat adiabatically as the hammer accelerates to its terminal velocity.  At this point the bucking and shearing and squishing processes would grind to a halt since the column of rock underneath the hammer literally can't get out of the way fast enough.  A density/pressure discontinuity would form and propagate forward at the speed of sound when the critical pressure was reached.  We now have launched a shock just like a normal piston driven shock in a semi/incompressible fluid (I assumed incompressible for easier math/EOS).  Keep in mind that in this problem we don't need to stop a really heavy 14 m/s hammer, we need the material underneath it to absorb all its gravitational potential energy like quadrillions of tiny springs and then act on the hammer with an equal and opposite force...long story short... our earth springs are puny so they cant.  This is much harder to do than one would suppose due to all sorts of funny instabilities that occur when a medium of very high density impinges on a medium of very low density at very high pressure...  At these pressures/densities the effects of solids vs liquids or gasses just don't add much to the general picture.  The technical term is "splat" followed by viscous flow.

As for beginning this process at 14 m/s it would probably make a satisfying noise on impact but static or moving it's kind of like trying to support an anvil with a single piece of paper only in this case the paper tears, melts, then flows as the shock front hits...yeah I know that's friggin crazy!  In a scenario like this what matters is the pressure and the potential energy of the object... which in technical terms is "a metric fuck ton" in your favorite system of units.    :) cheer's 

I hope this means I passed my quals... numbers are hard.   

Related paper demonstrates the shock tube phenomena with good Schlerin images of shock propagation out of a tube.
http://www.currentscience.ac.in/Volumes/104/02/0172.pdf
 
+rob morgan I think you're right, and that model actually captures what's going to happen. (Together with the nuke physics part to make sure that it actually does push on the ground and doesn't just tunnel straight through it like Mjolnir's less-successful neutrino-based prototype)

One question I wanted to do some quick calcs on was whether the initial 14m/s impact would result in some of that energy being laterally transferred along the crust via sound waves -- it might not be much, but compared to the total KE of the hammer it could still create an appreciable crater. 

The next question is what the shock dynamics would look like, since the piston shock would form long before the hammer hit the core, so it's going to be travelling through a fairly interestingly nonuniform medium. Since the density is increasing as it goes, there's a lot for the shock wave to reflect off of, in particular the mantle/core boundary. Once Mjolnir approaches that boundary, the shock strength should increase (as it gets sandwiched between the two) and it's probable that some kind of shock wave would detach as Mjolnir passed that boundary. (Not enough to cause serious damage anywhere, but it should make the seismology of tracking this interesting)

The other thing to work out would be how much energy Mjolnir would lose due to friction, etc., during its pass through the Earth; would it make it out of the opposite end? (Probably not, given that it only had 10m of dropping potential energy to start with) Would some of the propelled column make it out the other end?

My guess is that Mjolnir itself would lose enough energy during this plough through the earth that it would stop partway past the core and then gradually sink to the bottom, but the resulting shock column might make it to the surface on the other side and leave an interesting-looking exit wound full of lava. But I'd have to figure out some good ways to simulate those fluid flows to work out those details...
 
The radius of the Earth being only some 10^7 the linear dimension of the hammer, at a density of 10^10 it could absorb all the matter it traverses without growing significantly (together with all the energy coming out of the process for good measure). If we assume magic to entail the meta-property of picking properties so as to minimize the ripples of its known paradoxical properties, this is a winner.

The contender is that the hammer has a frictionless surface, is perfectly immutable in its local coordinate system(s), has mass, inertia, and weight, and can crush what it collides with pressure at least as high as that which stabilizes neutronium.

So the first question to solve is perhaps, could possibly the stated properties entail for the hammer to induce in front of its trajectory dynamical pressure high enough that the rock will by itself collapse to neutronium? If not, how far do we get from that possibility? 
 
Well, we know the energy loss in a viscous flow (which is what it looks like to Mjolnir), we know the viscosity of the liquid layers inside Earth. the terminal velocity is not that high, so I agree, it would just sink to the center of the planet.
I did not think of the initial shock wave though, that's pretty cool.
 
+Yonatan Zunger If this is right I feel like RBL deserves a nod in that he was criticized for posing a problem with no practical real life application...  I mean after all we are participating in a global dialog on almost the exact same subject which was touched off by Neil deGrasse Tyson... :)  seems pretty realistic AND useful from here.  I think his error was framing the problem around frozen meat instead of magic hammers.    

More seriously I think you make some very good points about the wave detaching as the density increases (reflection at boundaries should be observable and if it's strong enough it might cause spall on the way out which could generate your "exit wound") it just depends on how much friction matters here.  If the hammer accelerates to a super sonic velocity (5-10km/s?) there's a good chance for things heading out the other side otherwise the earth will just crack at the seams.  But because we lack good data we really don't know the EOS for these conditions other than the rough borders/constraints of the param space. :)  I can probably get you rough estimate but that would take more time than I currently have.  Any hydro/shock guys watching?
 
The shock wave model does not marry with the facts.

The mass of the hammer has little effect on the rock beneath it, it is the front edge of the hammer that makes the effect.

Moving a solid object at the terminal velocity of rock/rubble doesn't create that much of a shock. The resulting effect is pretty much the same as in a hydraulic press or a bulldozer case. No weird sonic waves, no columns of matter, just unstoppable movement of a solid object.
 
+Oleg Mihailik I think the issue here is the boundary conditions at the interface between the hammer and the fluid. The normal viscous flow assumes that the pressure exerted by the solid is much smaller than the molecular cohesion of the liquid, so the pressure from the moving solid reaches equilibrium "infinitely fast" (much faster than the liquid moves) by Van der Waals forces. This is not the case here, the pressure is too high even if the speed is slow. The liquid cannot convert this pressure into macroscopic flow fast enough.
 
As the hammer reaches terminal velocity and pushes on, all of its potential energy is as good as lost. That is the point of terminal velocity in the first place.

Now, as an object passes through the layers of Earth the gravity forces subside: outer spherical layers balance themselves out and the hammer is only pulled by the amount of Earth matter within the sphere below the hammer.

So the terminal velocity within the crust is slooow.

The terminal velocity at the entry into the liquid magma is moderate, but slowing down as the hammer sinks. The core is solid and the gravity is much smaller in there, so the hammer slows down to a stop at the edge of the core, sinking further very slowly, could be in thousands of years.
 
+Fedor Pikus If you look at the microscopic level, the speed of the hammer is tiny comparing to the normal heat speed of molecules.

At the macroscopic level the pressure is not created by the weight of the hammer, it is created by the resistance of the rock. Which is mediocre, bland and predictable.
 
+Oleg Mihailik The key here is ~4x10^15 Pa.  A pressure discontinuity this extreme in as much excess of the shear and young's moduli of a solid will compress that rubble significantly.  However since the density of the hammer is still significantly greater than the rock it will accelerate.  It will continue to do this until the rocks can push back as hard as it can push forward.  The rock will continue to be crushed between the rocks in front and the hammer behind until it has the consistency of a granular medium at this point it will flow and friction will add heat until it melts at which point a thin boundary layer of fluid will form and start growing.  When this boundary layer gets thick enough dense enough and hot enough a shock will detach.  This is literally an astronomical pile driver the conditions are very far removed from those a bulldozer would encounter.  I definitely understand your reasoning.  However when I looked at the numbers more carefully and calculated the approximate pressure I found a thermodynamic regime that was consistent with processes in the interior of small stars and really big gas giants...
 
+Fedor Pikus: The pressure is not that high.  The only way you get a pressure that high is by assuming a static-equilibrium solution and asking what pressure is required to sustain it.  But obviously such a solution would require an impossible pressure, so it cannot be sustained, and the hammer will accelerate downward -- with the acceleration limited primarily by its own inertia rather than pressure from outside.

Or, in other words, you need to solve your fluid equations with a fixed-motion boundary condition, not a fixed-pressure boundary condition.
 
You've calculated the pressure wrong.

You are assuming that the rock resisting the movement completely and calculating the weight it needs to support. Whereas the rock is giving in and therefore the force is simply the residual force of the rock resisting the movement. It does not equal the weight of the hammer.

We can easily imagine how much rock can resist the movement, from the hydraulic press model. Not too much.
 
+rob morgan: How far does it have to fall before you get into that regime?  Even neglecting drag entirely, it would only get up to about 200 m/s after falling through 2km of rock.  You won't start getting a shock wave at least until it gets well into the mantle, if not in the core itself.
 
If only Thor was real then we would know for sure.
 
+Brooks Moses Merely applying a driven load of 4x10^15 Pa to any solid for even a fraction of a second will launch a very strong shock wave.  The issue here is the strain rate.  While it's true that the strain rate starts off low it will steepen as the hammer acts to continuously compress the medium.  Unfortunately for our medium the hammer can maintain this pressure and near constant acceleration for a very long time relative to the thermo parameters of the medium... This is the principle behind Inertial Confinement Fusion (ICF).  In the lab a laser acts like a bullet of light compressing the gas for fusion.  Assuming the earth is solid iron and has a density of 10000 kg/m^3 and the cross sectional area of our hammer is 1m^2 and it is 1 meter long we have a hammer with a mass that is ~1x10^4 greater than the material directly below it (i.e. all the way to the core and out the other side of the planet).  The momentum transfer between earth and hammer is negligible especially since we are at pressures so much greater than those the material can sustain.  It will accelerate and it will hit a significant velocity until the force on the hammer is balanced by that of the planet which occurs on its way back out toward the surface.  This is not a gentle process like pushing rocks with a bulldozer its really violent.  It just starts off gracefully.  Keep in mind this thing is really small it's drag would be negligible given its surface area.  It would achieve between 1 km/s and 2 km/s (which is probably close to the speed of sound at that depth)@ ~5500km delta RadiusEarth <~20% or falling 1/5 of the way in toward the core (very high inertia very high pressure, very low surface area, very low drag). However a shock wave should launch and detach long before this occurs since we are exceeding the shear and youngs moduli by orders of magnitude.  If this seems too crazy its basically why rockets are tall and skinny.  They need the mass to beat the drag.  Make em too short and fat and they loose too much energy fighting drag.  We don't have that problem here.  If those #'s are correct this is outside the realm of normal experience and difficult to imagine without the math to serve as a guide.  I encourage you to check the #'s if they seem way off as they were hastily calculated.
 
+Brooks Moses If it's 200 m/s after 2 km, if we assume gravity roughly constant across the first 1% we multiply the height by about 32 and the speed to about 1.12 km/s.
 
+Brooks Moses pressure = force * surface area in this case 9.8m/s * 2*10^14 kg/(.5m^2) ~ 4*10^15 Pa.   
 
+rob morgan: My disagreement starts at your first sentence.  By what means are you "applying a driven load of 4x10^15 Pa to [a] solid for ... a fraction of a second"?

I agree that this is violent.  I dispute that it is violent at impact in the sort of way that makes miles-wide craters.

+Boris Borcic: Ok, but at that point you're well below anything that constitutes "solid rock", and magma ought to be viscous enough to dissipate the shock wave before it does much to the surface.
 
+Boris Borcic the earth is 6000km in diameter after falling a few hundred km you really start moving now you are in the regime where you start dragging a lot more mass in your wake maybe even enough to stop accelerating as the pull of gravity starts decreasing significantly!  I'm assuming a very generous amount of drag so I don't accidentally draw the wrong conclusions (drag typically scales like the cube of the velocity but there's a whole lot of inertia here). the easiest way to do this is to use Newtons formula for gravitational potential energy and then divide out the mass of the test object and take the square root you can then look at the wiki's to guide you intuition for how this object will behave.  This way you don't over estimate the velocity.  
 
This is somewhat tangential to the whole discussion, but the biologist in me would like to know if you're using Asian or African Elephants? The size difference between the two (Asian being the smaller species) would probably influence the overall weight of Mjolnir once you'd rounded up all 300 billion. Or is there some special SI unit of elephant that I'm unaware of? 
Thank you for your help, you may return to your physics :) 
 
+rob morgan I was just continuing the sentence of +Brooks Moses who spoke of neglecting drag entirely - because I did not see what he was meaning to be compelling in his sentence. I do dimly remember timing iron balls falling in glycerine to establish relevant coefficients in another life.

On the whole thread, as soon as the idea came about that the hammer would sink, I just got impressed that you guys made such a story of dissipating the energy from a minuscule 10m drop in the gravity field but felt no compulsion to discuss the energy from the further drop.
 
+Nich Walker: African elephants, to go with the swallows, clearly.  (This is how we know that they are not European swallows, as there are no European elephants.)

+rob morgan: I got a comment notify for a comment you posted answering my question about the force, but G+ seems to have eaten it it was earlier in the thread than I thought.  You correctly compute a force (and an area), but it's an irrelevant force: mass*g is the force that gravity is applying to the hammer.  That is not the force that the hammer applies to the rock.
 
+Boris Borcic: Let me clarify what I was talking about with neglecting drag entirely.  The maximum amount of pressure that the rock can provide on the hammer is roughly on the order of the material strengths of 10^9 Pa, at least at subsonic speeds.  (That's a handwave, but it's certainly not more than a couple of orders of magnitude of handwave.)  The maximum amount of force that the hammer can apply to counteract this pressure is on the order of 10^15 Pa, because that's what it would apply if it were not accelerating.

Thus, the drag force is trivial compared to the force required to slow down the hammer, and so can be neglected in the computation of the hammer's motion -- the hammer will accelerate at nearly the same rate it would in vacuum.  This is not dropping an iron ball in glycerine; it's dropping an iron ball in air a typical "medium" vacuum.

(The drag force is of course entirely non-negligible in estimating what happens to the rock!)
 
As a Thor fan who read every single Thor comic every made, this gave me the biggest nerdgasm reading.
 
Sounds plausible, +Brooks Moses.

Force is not measured in Pa though, so it's still not clear how negligible the rock's effect on the hammer would be.
 
+Brooks Moses until the failure process occurs no matter how infinitesimally brief all the atoms in the rock are at rest and all the atoms in the hammer are at rest and gravity is bearing down with its full force and though its very brief the brittle medium that has yet to fracture.  The material experiences the full pressure of the hammer and then they both start to move.  The pressure on the hammer decreases by several orders of magnitude very abrubtly.  Once the crust fractures since we are well out of the elastic regime it doesn't crack or crater its like an anvil being supported by a big piece of paper or a big truck on thin ice and that fracture process races forward into the medium at the speed of sound.  That initial wave dissipates in a conventional interaction and the falling object is stopped.  In this case material piles up in front of the hammer and the density increases on a ramp as the hammer continuously drives this process into the ground.  

As more material accumulates on the front surface of the ram the highly compressed region grows at an accelerating rate eventually catching up with and overtaking the normally propagating sound wave (this is how a driven shock wave is created).  This is also known as steepening and occurs in closed constant velocity or accelerating systems like shock tubes or blast waves.  The enormous density difference and low surface area means drag will likely be negligible.  The inertia of the material driven by the ram is also negligible compared to the mass of the ram itself which places a nice bound on the degree of the equal and opposite reaction.  If it ever did reach a point where drag is significant enough to appreciably alter the terminal velocity of the hammer say 10% restmass@1g, the pressure on the leading surface of the hammer will have to be a sustained ~0.5-1x10^15Pa to appreciably decelerate this object (conservation of energy via Fdx) this is a significant degree of over pressure for the earth's interrior and should therefore behave similarly to a blast wave up to some geometry constraints.  Otherwise it will act as a freely falling object accelerate to ~8km/s which will definitely create a conventional shock wave creating Yonatan's ejecta crater.  

Thus either the pressure is too extreme to support normal subsonic flow around the object traveling through earth's interior which will create a piston driven shock effect due to the overpressure in the medium (shock tube effect), or the hammer itself will break the sound barrier in the conventional sense.  The physics of shocks is indeed strange stuff but it's more thermo than kinematics, the energy has to go somewhere and this object is very small and contains way more energy than our largest nuclear bombs.  However, its good to root that stuff in this kind of conventional analysis to see that it is in fact consistent with the equations of motion.  
 
You keep claiming that superpressure. Where does the astronomic pressure come from? There's no force to create such pressure.
 
It seems to me that the gravitational attraction cannot be neglected. It does more than vacuum up dust - when the hammer is 1m above the ground, I'm coming up with an upward force at the surface of several hundred times earth gravity. It seems likely that this would disrupt the surface all on its own?
 
Good point, but have you got the numbers? 
 
Yeah, actually, it looks like it's irrelevant until the hammer starts breaking up the rock, at which point it's just building up a trail of rubble that's following it to the center of the earth. When it's at the surface, the gravity is just barely strong enough to break the nearest 1m^3 of rock free, but there's no time for it to accelerate before the hammer hits.

Of course, when Thor's walking around, Mjölnir will constantly strip everything above the bedrock and form it into a tidy sphere, but the bedrock and concrete appear to be safe until he finds a weak spot.
 
Hmm, that's a good point... 2.1*10^14 kg in a nice, tidy, 0.25m-radius sphere is actually a bit of a hazard. :)
 
Oops. I misremembered the mass when I was doing the numbers. I had 10^13, not 10^14. So yes, I think that would be able to break up the bedrock before it reached it. I think it happens about 4m out, and starts accelerating at 1000m/s^2. By the time it hits the surface (assuming .25m radius) it's traveling at 250m/s. Most importantly, I'm thinking this means that (assuming the "travel straight through the earth's core" model holds) the tunnel will be at least 4m diameter through the bedrock. That's about the limit of my imagination and math, though.
 
+Boris Borcic Sorry about the delayed response to your comment I just got a few minutes to write a quick reply.  

The difference in density of an iron ball and glycerin is ~10:1.  the proposed difference in density between this hammer and solid iron @30010^9 Pa (aka the pressure at the earth's core) is 10^10:1.  The ten orders of magnitude separating these examples makes a dramatic difference in the possible interaction dynamics.  Assuming a 3km/s-5km/s terminal velocity approximately 1*10^15-2*10^15 N would be distributed over a surface of !0.25m^2 is ~2*10^15-8*10^15 Pa.  This is ~25,000 x the pressure at the earth's core.  

Thus, as this hammer initially begins its descent, it should accelerate freely, pushing and compressing the earth in front it in a manner similar to a boat's wake.  Because this fluid is a viscous compressible medium when the hammer reaches a critical velocity the fluid will no longer be able to move around it, the fluid will be torn by the shear induced by the passing hammer causing cavitation (i.e. the hammer will cut/tear through the rocky fluid).  This happens at sub sonic velocities.  At this point the hammer is now acting like a ram and the critical shear front will grow in thickness (i.e. lengthening the ram).  As long as the hammer is accelerating (or moving at a constant velocity), the ram's length (aka critical shear front) will continue growing at a rate faster than the ram itself is moving (since this critical cavitation/ram pressure characterized by the shear and young's moduli is much less by a factor of ~100-10000 than the max pressure that the hammer is capable of exerting).  As the hammer's velocity approaches the speed of sound (~5km/s) after falling 1/4 of the way to the center of the earth, the ram front will exceed the velocity of sound in the medium and a shock front (aka overpressure blast wave) will continue to grow until the hammer's velocity falls below 3-5kms.  At this point the blast front will detach from the hammer's surface and freely propagate the remaining 1000km to the earth's surface.  Once it's no longer a ram "driven" blast wave, if the pressure of the shock wave's reflection off the different density discontinuities on the way out of the earth plus the rarefaction pressure exceeds the maximum strain of the medium, a phenomenon called spall will occur.  If this spallation event is energetic enough material will be ejected out the other side creating an eruption.  

Here's an interesting reference from LLNL that will shed some light on what to expect from this type of physics.
https://e-reports-ext.llnl.gov/pdf/305598.pdf
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