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## Physical Effects of ball moving at 0.9c

 Quote by pervect It'd be interesting to work out how many atoms there are in the baseball, and how many air atoms are in the path.
These are pretty easy to estimate, at least to order of magnitude:

Atoms in the baseball: average molecular weight around 10 (assuming mostly hydrogen, carbon, nitrogen, and oxygen, but hydrogen atoms will predominate because most atoms of the other types will have at least one and maybe more hydrogens attached). So 145 grams of baseball is about 14 moles, or about 14 times Avogadro's number of atoms. That works out to about 10^25 atoms.

Air atoms in the path: the number we really want here is air atoms per second encountered by the baseball as it flies. A mole of gas at STP takes up 22.4 liters or 0.0224 cubic meters, or about 3 x 10^25 atoms per cubic meter. The standard diameter of a baseball is (on average) 7.4 cm, for a radius of 0.037 meters; so it sweeps out 0.9 x 3 x 10^8 m/s * pi * (3.7 x 10^-2)^2 m^2 = about 6 x 10^4 m^3/s. Multiplying this by 3 x 10^25 atoms per cubic meter gives about 2 x 10^30 atoms per second encountered.

So if the number of baseball atoms disintegrated per air atom encountered is anything close to unity, the baseball will disintegrate in a fraction of a second.

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 Quote by PeterDonis These are pretty easy to estimate, at least to order of magnitude: Atoms in the baseball: average molecular weight around 10 (assuming mostly hydrogen, carbon, nitrogen, and oxygen, but hydrogen atoms will predominate because most atoms of the other types will have at least one and maybe more hydrogens attached). So 145 grams of baseball is about 14 moles, or about 14 times Avogadro's number of atoms. That works out to about 10^25 atoms. Air atoms in the path: the number we really want here is air atoms per second encountered by the baseball as it flies. A mole of gas at STP takes up 22.4 liters or 0.0224 cubic meters, or about 3 x 10^25 atoms per cubic meter. The standard diameter of a baseball is (on average) 7.4 cm, for a radius of 0.037 meters; so it sweeps out 0.9 x 3 x 10^8 m/s * pi * (3.7 x 10^-2)^2 m^2 = about 6 x 10^4 m^3/s. Multiplying this by 3 x 10^25 atoms per cubic meter gives about 2 x 10^30 atoms per second encountered. So if the number of baseball atoms disintegrated per air atom encountered is anything close to unity, the baseball will disintegrate in a fraction of a second.
But given its speed, a second is a long time. However, I get about 10^-7 seconds from pitcher to home plate, which is still order 10^23 atoms encountered. This is still enough, I think, to suggest it is mostly disintegrated by home plate, and that most of the 'magically appearing' KE has been released. With or without fusion, the mushroom cloud image looks not far off.

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 Quote by PAllen I get about 10^-7 seconds from pitcher to home plate, which is still order 10^23 atoms encountered. This is still enough, I think, to suggest it is mostly disintegrated by home plate
If each atom encountered causes approximately 1 baseball atom to disintegrate, then about 1% of the baseball atoms will be disintegrated when it reaches home plate. That's enough to be noticeable, certainly, but not necessarily enough to call the baseball "mostly disintegrated". If the baseball was just traveling through air, it would take about 100 times the distance to home plate to disintegrate every atom at this rate.

Of course, if the baseball encounters any solid objects, that will drastically increase the rate of disintegration, since solid densities are about 1000 times the density of air. So if the event takes place in a stadium, for example, the baseball won't get far through the seats behind home plate. (Even leaving out, of course, the fact that the entire stadium is being obliterated.)

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 Quote by PeterDonis If each atom encountered causes approximately 1 baseball atom to disintegrate, then about 1% of the baseball atoms will be disintegrated when it reaches home plate. That's enough to be noticeable, certainly, but not necessarily enough to call the baseball "mostly disintegrated". If the baseball was just traveling through air, it would take about 100 times the distance to home plate to disintegrate every atom at this rate.
I'm not sure I buy the one for one assumption. The collision will produce to nuclei with high speed (e.g. both well over .1c). These will hit other atoms, etc. It seems to me the ball will be heating up to temps way beyond center of a star, at a time scale comparable to e.g. a .01 c particle takes to cross the baseball. Thus, well before home plate we have a fireball.

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 Quote by PAllen But given its speed, a second is a long time. However, I get about 10^-7 seconds from pitcher to home plate, which is still order 10^23 atoms encountered. This is still enough, I think, to suggest it is mostly disintegrated by home plate, and that most of the 'magically appearing' KE has been released. With or without fusion, the mushroom cloud image looks not far off.
Yes. I came up with around 100 nanoseconds also.
Another thing to consider is the cross sections of the atomic nuclei interacting.
The electron shells I would imagine would be stripped away almost immediately, leaving a plasma baseball.

According to wiki, the radius of the average atomic nucleus can be estimated by:
R=roA1/3
where ro = 1.25 femtometers = 1.25e-15 meters
and A is the atomic mass number
Which for nitrogen, yields a radius of 3 femtometres, and a cross sectional area of ~3e-29 meters
Multiplying by 10^25 particles, per PeterDonis, yields a solid cross sectional area of 3e-4 meters for the baseball. Which is somewhat less than the actual cross section of the ball, which is 4.3e-3 meters. Roughly 1/10 the actual size.

Multiplying that by the distance to the plate, 18 meters, yields a volume of 0.005 cubic meters, then multiplying by 3e25 nitrogen atoms per cubic meter yields 1.5e23 atoms on the way.

Which is 65 times fewer atoms than in the baseball. So I think there will be something left once it hit the bat.

Poof!

ps. please don't shoot me if I got anything wrong. I'm only practicing my nano's and fempto's. But if I got those wrong, well, yes, then you can shoot me.

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 Quote by PAllen I'm not sure I buy the one for one assumption. The collision will produce to nuclei with high speed (e.g. both well over .1c). These will hit other atoms, etc. It seems to me the ball will be heating up to temps way beyond center of a star, at a time scale comparable to e.g. a .01 c particle takes to cross the baseball. Thus, well before home plate we have a fireball.
Even if only 1% of the baseball atoms are disintegrated, that's still enough kinetic energy released to make a fireball--the equivalent of a 30 kiloton nuclear explosion (based on the entire baseball being equivalent to 3 megatons).

I agree that each initial collision between a baseball atom and an air atom will cause secondary collisions. But it's also quite possible that many baseball atoms will miss an air atom altogether, at least over a short distance (compared to the ball's speed) like the distance to home plate. More precisely, I can see many *nuclei* missing each other; the atoms themselves will be ionized, and I would expect the electrons to interact via Coulomb repulsion (basically forming an electron plasma), but nuclei are much smaller (in diameter, not mass) than electrons under these conditions (the electron wavefunctions spread over atom-size distances, but the nuclei wavefunctions are confined to a spatial dimension five orders of magnitude smaller or so), so a significant fraction of the nuclei could just pass right through the air between the mound and home plate without being significantly deflected.

So there are two effects that work in opposite directions: secondary collisions, vs. nuclei missing each other altogether. That could average out to an effective one for one disintegration rate. (Or it could not, of course; I haven't tried to estimate the order of magnitude of either effect.)

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 Quote by PeterDonis Even if only 1% of the baseball atoms are disintegrated, that's still enough kinetic energy released to make a fireball--the equivalent of a 30 kiloton nuclear explosion (based on the entire baseball being equivalent to 3 megatons). I agree that each initial collision between a baseball atom and an air atom will cause secondary collisions. But it's also quite possible that many baseball atoms will miss an air atom altogether, at least over a short distance (compared to the ball's speed) like the distance to home plate. More precisely, I can see many *nuclei* missing each other; the atoms themselves will be ionized, and I would expect the electrons to interact via Coulomb repulsion (basically forming an electron plasma), but nuclei are much smaller (in diameter, not mass) than electrons under these conditions (the electron wavefunctions spread over atom-size distances, but the nuclei wavefunctions are confined to a spatial dimension five orders of magnitude smaller or so), so a significant fraction of the nuclei could just pass right through the air between the mound and home plate without being significantly deflected. So there are two effects that work in opposite directions: secondary collisions, vs. nuclei missing each other altogether. That could average out to an effective one for one disintegration rate. (Or it could not, of course; I haven't tried to estimate the order of magnitude of either effect.)
Ok, to get to the next step, we need an estimate of the mean free path for a nitrogen nucleus in a baseball. Model the problem as a beam of .9c nitrogen atoms of density of air hitting a baseball. I believe this is about 1 Gev / nucleon for the incoming atoms.

Also, I'm not sure how significant interactions between nuclei and electrons would be for energy transfer. Is there a good argument that these can be ignored?

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 Quote by PAllen Also, I'm not sure how significant interactions between nuclei and electrons would be for energy transfer. Is there a good argument that these can be ignored?
That's a good point. But being around 2000 times less massive that the nucleons, I would ignore them.
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Using 10^25 atoms in the baseball, that gives about 5*10^28 atoms/cubic meter (7.5 cm diameter for the ball). Using a figure I found for nuclear radius for nitrogen or carbon of 4*10^-15 m (pretty similar to OmCheeto's figure), and assuming a billiard ball analogy, this gives a mean free path of 40 cm, which is about 10 times the radius of baseball. This simple model suggests the typical nitrogen atom will be stripped of electrons but pass through the ball otherwise 'unscathed'. Even if this is true, besides the significant energy transfer from electron interaction, we have the ball becoming charged to an extreme degree for a massive object. It is clear that even if passed home plate is a vaccuum, the ball's temperature and charge ensure it will disintegrate with atomic bomb energy. Much less clear is how much happens within the 100 nanoseconds reaching home plate. I did find papers on experiments with several Mev/nucleon ion beams heating up less than micron thick foils to thousands of degrees in 10s of nanoseconds (with ion densities far less than air). Here we have 1 Gev/nucleon atoms hitting the ball. Thus, there really is no doubt that the air will 'end up' violently disintegrating the ball (even if it enters a vaccuum) - the open question is how much of this will happen between pitcher's mound and home plate.
 Blog Entries: 1 Recognitions: Gold Member Science Advisor I guess an additional comment is (assuming 40 cm is ballpark correct for mean free path), is that the 'ball' in whatever state it is in, will still have most of its momentum on reaching home plate - assuming only a nucleus-nucleus collision will effectively transfer momentum. This all suggests a model (contrary to the cartoon in the OP) that the ball is still 'mostly a ball' on reaching home plate (primarily due to lack of time to fully disintegrate), but is heated to inconceivable temperatures, it's electron structure significantly shredded, and it is negatively charged to a degree never seen by humans for a macroscopic object. Even if it then entered a vaccuum tube, the ball would effectively explode in a rapidly moving fireball, and those it passed on the way to home plate would 'soon' be killed by the side effects of its passage, but the passage time is too small for most effects to reach the bleachers or even the outfield before the ball reaches home plate. The energy transferred from ball by home plate would be much larger than the Nagasaki nuclear bomb (20 kilotons) but much smaller than a modern H-bomb (1 megaton). Up shot: other than that 'you wouldn't want to be there', I think most of what is written in the OP cartoon is not correct. [Edit: Using Peter's figures combined with flight time of 100 nanoseconds, the ball 'hits' 2*10^23 atoms. If we assume 90% of these nuclei pass thru the ball without major deflection (as implied by the mean free path), then we have 2*10^22 nuclear collisions, each releasing about 14 Gev of energy as a wide mixture radiation type (fusion energy, whether it occurs or not, is insignificant; that is a matter of a few Mev per nucleon, which is de minimus compared to the KE). This works out to about 15 kilotons of energy carried off as intense radiation - a bit less, actually, than the Nagasiki atom bomb. If the ball magically vanishes at home plate, this is all that will be released. It would appear that over 1000 times the distance to home plate would be needed for the ball to give up most of its KE in air.]