Physical Effects of ball moving at 0.9c

[cepheid]

I only just discovered this today:

http://what-if.xkcd.com/1/

I didn't know about it before. My question is simple: what do you think of Randall's analysis of the physics here? Is he "on the ball?"

 

[ghwellsjr]

I only just discovered this today:

http://what-if.xkcd.com/1/

I didn't know about it before. My question is simple: what do you think of Randall's analysis of the physics here? Is he "on the ball?"

It starts off with magic to accelerate the baseball to 0.9c. Take away the magic and you realize you need a tremendous source of energy to get the ball up to that speed. Just think of the effort it takes to get individual particles up to that speed and you'll see that the explosion happens long before the ball even gets started.

But it's a fun read.

 

[cepheid]

Yeah I understand that it takes a tremendous amount of kinetic energy to get a particle to 0.9c, let alone a macroscopic object.

My question is if the *consequences* described are accurate. Nuclear fusion etc

 

[ghwellsjr]

My question is if the *consequences* described are accurate. Nuclear fusion etc

I would think so, since that is one of the things they do with particle accelerators, which is the reason they do it in vacuum. It would be a little fairer if they placed the experiment on the Moon and had astronauts in space suits so that the ball could actually travel some distance undisturbed. The article points out that the hitter would have very little time to react to the sight of the approaching ball. Of course, normal reaction times would make it impossible for him to swing the bat in time but if he anticipated the "pitch" and got the bat in place for the collision, then it would be a similar explosion.

 

[HallsofIvy]

I believe the whole thing was meant humorously.

 

[cepheid]

I believe the whole thing was meant humorously.

The questions being asked may not be very serious, but the point of the site certainly seems to be to use real physics to treat these frivolous topics, as accurately as possible:

 

[pervect]

I wasn't particularly convinced by the claims of "fusion". What's supposed to be fusing with what to yield what product, exactly? And why fusion, rather than fission?

I really don't have much intuition as to what happens if you bombard a baseball with .9c oxygen and nitrogen ions - and I don't think the author of the cartoons really knows either :-(. If he does, he didn't document it convincingly.

In some sense, these are minor quibbles. The major energy input into the system is going to be the magic that accelerates the ball to .9c. And you can pretty much guess that the result is going to be a big nuclear-style fireball when that energy is released. Showers of various particles, ionizing radiation, and fireballs of the sort typically associated with nuclear weapons all seem to me to be reasonable predictions.

 

[PeterDonis]

I wasn't particularly convinced by the claims of "fusion". What's supposed to be fusing with what to yield what product, exactly? And why fusion, rather than fission?

Given the types of atoms in the air and the ball (mostly hydrogen, carbon, nitrogen, and oxygen), it would seem that fusion reactions are overwhelmingly more likely than fission reactions, if nuclear reactions are taking place at all.

As far as whether nuclear reactions would actually take place, the kinetic energy of the baseball at speed 0.9c corresponds to a temperature of about a trillion degrees, far higher than what is required to overcome Coulomb repulsion and allow nuclei to get close enough to each other to fuse. However, xkcd's analysis did leave out one key point: a trillion degrees may actually be too *high* a temperature for many fusion reactions to occur at significant rates (basically, the nuclei fly past each other too fast to stick together).

 

[GarageDweller]

Well the real question is which of the consequences are dominant, for example if i threw a ball at some mundane speed, 4m/s, the electrons inside are moving with respect to the batter, hence he should in principal feel a magnetic force.
But as we know, such forces are tiny and unmeasurable, same concepts are present in this scenario. Now a few things come to mind.

First is that the the space-time around the ball would be severely curved due to the energy-momentum of the ball, thus making everything after that uncertain as we do not understand physics in this scenario. Some very strange things happen when you have energy densities like that, some very very strange things.

 

[DaveC426913]

Well the real question is which of the consequences are dominant, for example if i threw a ball at some mundane speed, 4m/s, the electrons...

... and the protons ...

... inside are moving with respect to the batter, hence he should in principal feel ...

...no net force.

 

[GarageDweller]

i doubt you could find a baseball, or anything macroscopic that is perfectly neutral

 

[OmCheeto]

I saw this on the 10th and posted in https://www.physicsforums.com/showpost.php?p=3990230&postcount=16146.
(I don't have time to start a thread and discuss or analyze such things)

Then I noticed someone had already started another thread earlier in the day.

Interesting problem.

Perhaps we should call up CERN and get them to accelerate a single nitrogen nucleus to .9c and smash it into a baseball and see what happens.

We can interpolate from there.

This isn't just an interesting problem, this is a fun problem.

 

[PeterDonis]

First is that the the space-time around the ball would be severely curved due to the energy-momentum of the ball

No, it wouldn't. Even if we leave out various subtleties in attributing "extra" spacetime curvature due to the ball's motion (briefly, curvature is not frame-dependent, but the ball's kinetic energy is), the relativistic gamma factor at 0.9c is a little more than 2, so the ball's total energy is a little more than twice the energy equivalent of its rest mass. The curvature due to the ball's rest mass (given the ball's volume) is negligible; a little more than twice negligible is still negligible.

 

[pervect]

I saw this on the 10th and posted in https://www.physicsforums.com/showpost.php?p=3990230&postcount=16146.
(I don't have time to start a thread and discuss or analyze such things)

Then I noticed someone had already started another thread earlier in the day.

Interesting problem.

Perhaps we should call up CERN and get them to accelerate a single nitrogen nucleus to .9c and smash it into a baseball and see what happens.

We can interpolate from there.

This isn't just an interesting problem, this is a fun problem.

I'm sure collisions at much higher speeds have been done, for instance in the relativistic heavy ion collider (RHIC). No idea of what the results are though.

 

[OmCheeto]

I'm sure collisions at much higher speeds have been done, for instance in the relativistic heavy ion collider (RHIC). No idea of what the results are though.

I was trying to comprehend a paper last night, but it's been so long since I've studied particle physics, I don't remember how to convert Gev's to velocity.

Secondary charged particles emitted at 13° from
a platinum target bombarded by 4 GeV nitrogen ions
have been observed as a function of momentum and
time-of-flight. Preliminary analysis indicate a
rather surprisingly copious production of α-particles,
³He, ³H and deuterons relative to protons.

hmmm... Can the breaking up of atomic nuclei with a particle accelerator be referred to as "fission"?

ps. I just found some equations on an old thread, but when I did the math, v came out to be zero. Naprawde, ja nic nie wiem...
Wait! I think I've figured out where I went wrong.

 

[PAllen]

I wasn't particularly convinced by the claims of "fusion". What's supposed to be fusing with what to yield what product, exactly? And why fusion, rather than fission?

I really don't have much intuition as to what happens if you bombard a baseball with .9c oxygen and nitrogen ions - and I don't think the author of the cartoons really knows either :-(. If he does, he didn't document it convincingly.

In some sense, these are minor quibbles. The major energy input into the system is going to be the magic that accelerates the ball to .9c. And you can pretty much guess that the result is going to be a big nuclear-style fireball when that energy is released. Showers of various particles, ionizing radiation, and fireballs of the sort typically associated with nuclear weapons all seem to me to be reasonable predictions.

To underscore that this is the main point: in some, fairly short, distance, the KE of the ball will be dissipated. The amount of that KE (at gamma a little over 2 for .9c) is (assuming standard weight of baseball, 145+/-5 grams) is about 150 times the energy released by the Nagasiki nuclear bomb, or about the same as 3 megaton H bomb. This is without even considering possible fusion.

 

[pervect]

I was thinking about this some more, and my best guess goes something like this. It seems pretty obvious that the incoming atoms would quickly ionize. Then you'd probably have some sort of rutherford scattering pinball going on as the heavy nitrogen nucleus (you'd have the usual assortment of elements, the majority would be nitrogen though) bounces around. I don't know if you'd need relatistic corrections to the Rutherford formula or not, you might. Exactly how much stuff gets knocked out of the baseball per incoming ion isn't at all clear, but I would guess you'd get more than one atom knocked out per incoming atom. There'd be a lot of ionizing radiation - ions recombining with electrons, Bhremsstrauling from various disturbed electrons, etc. I could imagine the occasional nuclear event happening, but I'm not sure how often.

It'd be interesting to work out how many atoms there are in the baseball, and how many air atoms are in the path. If we could guess how many atoms got knocked out by the assumed Rutherford pinball per incomming atom, we could get a handle on how long before the baseball was totally disintegrated by that process.

 

[PeterDonis]

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.

 

[PAllen]

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.

 

[PeterDonis]

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.)

 

[PAllen]

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.

 

[OmCheeto]

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=r_oA^1/3
where r_o = 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.

 

[PeterDonis]

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.)

 

[PAllen]

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?

 

[OmCheeto]

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.

 

[PAllen]

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.

 

[PAllen]

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 vacuum 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.]