What-If 0001: "Relativistic Baseball"

[Max™]

If I'm remembering correctly, the energy of the collision isn't totally transferred to the bouncing object, and since it's not allowed to destroy the bat, it has to go somewhere.

In real baseball, that energy goes back up the bat and into the player as a sort of shockwave.

And some of the energy of that shockwave is released by the bat into the air, i.e., it is emitted as sound. I expect that the sound of an indestructium ball hitting an indestructium bat at .9c would be rather loud. OTOH, this percussive noise will probably be unnoticeable due to the other sounds being produced by air plasmafication and fusion.

http://www.sciencedaily.com/releases/20 ... 132949.htm

"We show that space-time cannot be locally flat at a point where two shock waves collide," said Blake Temple, professor of mathematics at UC Davis. "This is a new kind of singularity in general relativity."

So now we gotta deal with making a singularity too?

Nice job breaking iteverything, hero.

[PM 2Ring]

"We show that space-time cannot be locally flat at a point where two shock waves collide," said Blake Temple, professor of mathematics at UC Davis. "This is a new kind of singularity in general relativity."

Ok. And I guess there's also the possibility of the old kind of singularity, too, since the constructive interference of waves traveling through the bat and ball may be sufficient to induce black hole formation, both by locally compressing matter within its Schwarzschild radius, and through the energy density associated with the stress / pressure. See http://en.wikipedia.org/wiki/Stress-energy_tensor.

OTOH, indestructium has to be strong enough to resist that sort of thing to deserve the name. But it can't be perfectly rigid, since that would require the speed of sound through indestructium to exceed the speed of light. It has to be perfectly elastic, or collisions will deform it. So when you hit it, it will vibrate like a tuning fork for quite a while, but I guess it could radiate some of its energy away as EM (and / or gravity waves) without violating the elasticity condition.

[nate1481]

Would the time dilation (correct term? that time on the ball is moving slower then for the rest of the planet) have any effect on the situation?

[Sockmonkey]

According to baseball rules, what does happen if the pitcher kills the batter with the ball? Do they just have a do-over or does the next guy in the linup take a base?

[ivnja]

According to baseball rules, what does happen if the pitcher kills the batter with the ball? Do they just have a do-over or does the next guy in the linup take a base?

Assuming the game isn't called by the umpire (deaths on the field being a big deal, grieving players and fans, yada yada), a player from the bench will replace the newly deceased player; if the batter was awarded first base for being hit by the pitch (see the earlier discussion of Rule 6.08(b) for reasons he might not be awarded first base), the substitute player would go to first base. The next guy in the lineup would then be up to bat in his own normal spot - nothing is changed for him.

3.03
A player, or players, may be substituted during a game at any time the ball is dead. A substitute player shall bat in the replaced player’s position in the team’s batting order.
<snip>

If the batter, instead of being hit by a pitch, had a heart attack between pitches, the substitute player would come in and assume the same count (balls/strikes) as the injured player. In fact, the batter that's being replaced doesn't even need to be injured - the manager can replace players, even the current batter, at pretty much any point, so long as the ball is not in play. The replaced player can't come back into the game after that, though, and in practice it's quite rare for a batter to be replaced once he's already up at the plate for anything other than injury. It has happened, though:

In his first Major League at-bat, Pittsburgh Pirates rookie Doe Boyland struck out - while sitting on the bench.
In the seventh inning of a home game against the new York Mets, Pittsburgh manager Chuck Tanner sent Boyland in to pinch-hit for pitcher Ed Whitson [note: that's the common type of substituting for a batter - Whitson was taken out of the game before his at-bat, and Boyland came off the bench to take Whitson's spot in the batting order. -ivnja]. The count was 1-and-2 on Boyland when Mets right-handed pitcher Skip Lockwood had to leave the game because he hurt his arm.
New York switched to southpaw Kevin Kobel, so Tanner, going by the book, lifted the left-handed-swinging Boyland and put in right-handed batter Rennie Stennett to pinch-hit for the pinch hitter [Boyland].
While Boyland watched helplessly from the bench, Stennett struck out on Kobel's first pitch. Under the scoring rules, the strikeout was charged to Boyland for his inauspicious debut.

[gmalivuk]

Would the time dilation (correct term? that time on the ball is moving slower then for the rest of the planet) have any effect on the situation?

It could affect the rate at which different kinds of energy are released, but I don't think it would much impact the total amount of energy released or the fact that it happens over a relatively short time (fraction of a second).

[Coyne]

Okay, I'll buy the part about the batter "being hit" and advancing to first base. But what if the bases are loaded? Is this relativistic throw then counted as an unearned run?

[TheGrammarBolshevik]

No, but it would be an earned run (assuming the pitcher was responsible for those baserunners in the first place).

[ehque]

From LHC FAQ:

"In the LHC, under nominal operating conditions, each proton beam has 2808 "bunches", with each bunch containing about 100 billion protons. Bunches of particles measure a few centimetres long and a millimetre wide when they are far from a collision point. However, as they approach the collision points, they are squeezed to about 16 micrometres (a human hair is about 50 micrometres thick) to allow for a greater chance of proton-proton collisions. When the bunches cross, there will be a maximum of about 20 collisions between 200 billion particles. Bunches will cross on average about 30 million times per second, so the LHC will generate up to 600 million particle collisions per second."

100 billion protons in a 16 micrometer thick beam hitting another 100 billion protons generates a "maximum of about 20 collisions".

The baseball has roughly 11 moles of carbon (6.6*10^13 more stuff) in a 7.5cm diameter (21 million times bigger). Air is roughly 1000 times less dense than a baseball.

Rough calculations means for each 7.5cm of air the baseball passes through, we have about 6000 nuclei-nuclei collisions. A concrete backstop would raise that to 12 million collisions per 7.5 cm.

For a baseball that contains 10^25 nuclei, that's not a lot.

If the baseball doesn't hit a backstop, I'm fairly confident to say most of the baseball will make it to space intact. Even if it did hit a backstop, the baseball would go through a significant chunk of it before a significant number of baseball-molecules hit backstop-molecules. Basically, the scenario would be less like mushroom cloud (single point) and more like radiant beam of death.

[UnstableMongoose]

According to baseball rules, what does happen if the pitcher kills the batter with the ball? Do they just have a do-over or does the next guy in the linup take a base?

Well, the original analysis of the problem was posted in an extremely finite timescale, so I'll try to do the same with the ruling on the field.

We're going to make two quick assumptions. First, we're going to assume the pitch is on-target and whatever remnants of the ball remain will cross home plate in the dead center of the strike zone. We are going to, as this is an important distinction in the way the question was asked, assume that the ball remains whole so that it can be hit. Second, the batter is actually going to hit this sucker (again, part of the way the question was originally asked).

Zero time: Ball races outwards at .9c

t = 1ns (approximate): Ball creating plasma field, pitcher begins to disintegrate. Halfway through this disintegration, there would theoretically be a possibility for an injury timeout after the play. After the pitcher is disintegrated, the pitch would still be played, but play would halt to replace him.

t = 67ns (approxmiate): Ball reaches home plate and is struck by the bat. Our erstwhile batter is now ineligible for a hit-by-pitch under 608(b) because he has made contact with his bat. The ball is now considered to be in-play.

t = 67ns (approximate, but slightly later than the time the ball hits the bat): Batter disintegrated by nuclear fire. But because he hit the ball first, it will still be played. As the batter apparently can swing at and hit lightspeed projectiles, he may be super-human and immune to this firestorm. However, unless his bat is as well (and it can't be, it's a regulation MLB bat, made of wood), this timeline is not bothered by his survival.

t = 68ns (approximate): Catcher fails to catch ball as it whizzes through him at near-lightspeed. He disintegrates too. Play will definitely stop to deal with fatalities after the pitch is played out.

t = 69-73ns (approximate): Home plate umpire disintegrates, thus meaning that the pitch cannot be called a ball or strike. Thankfully, in this scenario, the bat made first contact. The failure of the catcher to contain the ball and the backwards motion of the ball make the current call foul ball (which is a call that, in the event that the home plate umpire doesn't have a good view, can be made by the other umpires)

t = 69-79ns (approximate, depends on depth infield playing at and umpire position): The other two umpires (and several infielders) disintegrate. With the departure of the final umpire from the field before any are able to react and actually make a call, this incident is now ruled a no-pitch. Play resumes at the discretion of the league.

t = 134ns (approximate, depends on the ballpark): Ball shoots through home plate backstop. If there was anywhere still alive to call it, the ball has clearly left the playing field and is out-of-play and foul. The play is now dead so that we can sort out all of this mess.

Note: The speed of the ball from the pitcher all the way to the batter is reckoned at .9c, using 3E8 m/s as the approximation for c. Under the circumstances asked in the question, I assume that the ball holds this speed from the pitcher's mound all the way to home plate (and that, paradoxically, any related events also propagate at .9c rather than c). Times are rounded to the nearest full nanosecond. I messed up a decimal point and screwed up my times, but I think they're fixed now.

[Max™]

But there is no joy in Mudville, mighty Casey and half of the town has been vaporized.

[Manabu]

Cosmic rays hit earth at close to light speed, yet we don't see fusion. Would this ball really trigger fusion? Do we have the density and time necessary for that? Ehque comment is interesting.

[Max™]

http://en.wikipedia.org/wiki/Cosmic_ray_spallation

http://en.wikipedia.org/wiki/Cosmogenic_nuclide

http://en.wikipedia.org/wiki/Ultra-high ... cosmic_ray

Cosmic ray particles with even higher energies have since been observed. Among them was the Oh-My-God particle observed on the evening of 15 October 1991 over Dugway Proving Ground, Utah. Its observation was a shock to astrophysicists, who estimated its energy to be approximately 3×10²⁰ eV (50 J)[3]—in other words, a subatomic particle with kinetic energy equal to that of a baseball (5 ounces or 142 grams) traveling at about 100 kilometers per hour (60 mph). It was most probably a proton traveling very close to the speed of light, slower by only about 1.5 femtometers (quadrillionths of a meter) per second, or about 0.9999999999999999999999951c, based on its observed energy. At that speed, in a year-long race between light and the particle, the particle would fall behind only 46 nanometers, or 0.15 femtoseconds (1.5×10⁻¹⁶ s).[4]

[Hafting]

Rough calculations means for each 7.5cm of air the baseball passes through, we have about 6000 nuclei-nuclei collisions. A concrete backstop would raise that to 12 million collisions per 7.5 cm.

For a baseball that contains 10^25 nuclei, that's not a lot.

If the baseball doesn't hit a backstop, I'm fairly confident to say most of the baseball will make it to space intact. Even if it did hit a backstop, the baseball would go through a significant chunk of it before a significant number of baseball-molecules hit backstop-molecules. Basically, the scenario would be less like mushroom cloud (single point) and more like radiant beam of death.

So the baseball mostly escapes at 0.9c, with only a few nuclear reactions. (A few million collisions in the backstop won't be too unpleasant.) Most of the atoms along the trajectory will remain. But molecules will be destroyed along the path and atoms ionized, for electron orbits are much bigger than nuclei. (They normally prevent baseballs from going through stuff.)

So the noticeable effects on earth will mostly be what happens to the ionized plasma along the trajectory. Air will glow as ions and electrons recombine, it'll look like a horizontal bolt of lightning - from the pitcher, along the surface end eventually into the upper atmosphere as the earth curves away. Spectacular, but not necessarily too dangerous to be around. Fully ionized holes through solid matter (the backstop, possibly the bat or batter, buildings) might be more unpleasant. Ionization energy may not be enough to slow the ball, but should be on the scale of chemical explosions as all those disturbed nuclei and electrons recombine and start forming molecules again.

Even if the batter isn't hit, the game will likely be stopped due to the explosion in the backstop. Possible terrorist activity and all that.

[WIMP]

The baseball has roughly 11 moles of carbon (6.6*10^13 more stuff) in a 7.5cm diameter (21 million times bigger). Air is roughly 1000 times less dense than a baseball.

Rough calculations means for each 7.5cm of air the baseball passes through, we have about 6000 nuclei-nuclei collisions. A concrete backstop would raise that to 12 million collisions per 7.5 cm.

For a baseball that contains 10^25 nuclei, that's not a lot.

You cannot extrapolate from LHC energies to low energy QCD. The strong force coupling runs with energy, and is larger at low energy. Therefore, the probability of interaction for any two nucleons is significantly larger for the baseball than for LHC protons.

http://en.wikipedia.org/wiki/Asymptotic_freedom

A little back of the envelope math tells me the probability of interaction (which goes as the square of the coupling) is around 30-40 times higher for the baseball. The basic conclusion stands, of course!

[bmonk]

could the curve ball of light, be doing this?

whenever I see the EM wave depicted, it is always like this.

Notice that there should also be another E and another H, as the EM wave disappears.



It seems unbalanced, and i wonder how it maintains a straight direction, being lop-sided, as depicted.

i would think the EM wave would actually make a spiral, not a straight line, under those conditions.

has experimental evidence proved that the EM wave is not as depiction below?



i do realize the unlikeliness of this, and would like to rule it out as a possibility.

Are there any quantifiers/mathematics as to the 1/2 oval shapes themselves.

Who should i research to find the derivation of this two-lobe manifestation...Hertz? Maxwell? Others?

Thanks for any information or urls regarding the experimental derivation of the EM two-lobe wave format.

Wave/particle duality does a *much* better job of explaining observations than weird things like "maybe it's a triangle".

You mixed up two concepts, I believe.

i suggest that the double-slit experiment has never been tried with as a triple-slit.

I conject, that in place of the wave/duality conundrum, that a "photon" is a cluster of equal spheres. These spheres are way way way smaller then any gluon or Higgs-Boson. That is, this assertion is that photons have different masses and manifest their finger-print spectral lines, due exclusively/mathematically from the external parameter of the clusters. Indeed, sphere clusters have two distinct fields types externally. One, centered the x,y,z axis directions is triangularal/hexagonal format. The other field locations are at the 8 x=y=z axis, where now the packing is done is formatted into square matrix, [ regardless of cluster quanta ( number of spheres in the sphere cluster).]

I think the equations you want are Maxwell's equations: he recognized light as a combined electric field that produced a magnetic field which in turn produced the electric field. All propagating at the speed of light, of course.

The reason the fields are offset is that's how it works: the electric field is changing at a maximum when the magnetic is a minimum and vice versa. (It's similar to why cosine and sine are offset in graphs of the imaginary powers of e.)

As for light making a spiral, well, it can. There are polarized forms of light, and I believe some of them are spirally polarized. Or at least certain organic solutions that can turn light in spirals--this is the principle that allows LCD displays to function.

[ehque]

The baseball has roughly 11 moles of carbon (6.6*10^13 more stuff) in a 7.5cm diameter (21 million times bigger). Air is roughly 1000 times less dense than a baseball.

Rough calculations means for each 7.5cm of air the baseball passes through, we have about 6000 nuclei-nuclei collisions. A concrete backstop would raise that to 12 million collisions per 7.5 cm.

For a baseball that contains 10^25 nuclei, that's not a lot.

You cannot extrapolate from LHC energies to low energy QCD. The strong force coupling runs with energy, and is larger at low energy. Therefore, the probability of interaction for any two nucleons is significantly larger for the baseball than for LHC protons.

http://en.wikipedia.org/wiki/Asymptotic_freedom

A little back of the envelope math tells me the probability of interaction (which goes as the square of the coupling) is around 30-40 times higher for the baseball. The basic conclusion stands, of course!

Thank you. I am not a physicist. My knowledge ends at general relativity.

Of course, as you say, a 30-40 fold decrease in 6*10¹⁶m is not going to be significant. You'll also need to tow a few concrete planets into place to use as backstops.

Thinking about it, these calculations would probably show that the pitcher could chuck that baseball straight through the center of the earth.

[formivore]

Fascinating thread guys .

Why is it OK to ignore electromagnetic interaction? An alpha particle at .05c is stopped by a few centimeters of air. Won't there at least be a huge transfer of KE from this? More than just the "lightning bolt" someone described.

[eran_rathan]

Fascinating thread guys .

Why is it OK to ignore electromagnetic interaction? An alpha particle at .05c is stopped by a few centimeters of air. Won't there at least be a huge transfer of KE from this? More than just the "lightning bolt" someone described.

Mostly because the baseball is travelling at significantly higher speeds (0.9c).

[Yakk]

There is someone who worked out the heat deposited by E-M interaction if the baseball's electrons and protons where unassociated. It worked out to be significant.

However, is it justified to treat them as unassociated? I can think of two reasons why that might be reasonable: first, they are pretty far apart given the velocity of the ball. Second, energy deposit from various other processes would heat the baseball up quickly, turning it into plasma, and disassociating the particles. However, I'm insufficiently competent to know if either of these are good justifications...

[Tass]

There is someone who worked out the heat deposited by E-M interaction if the baseball's electrons and protons where unassociated. It worked out to be significant.

However, is it justified to treat them as unassociated? I can think of two reasons why that might be reasonable: first, they are pretty far apart given the velocity of the ball. Second, energy deposit from various other processes would heat the baseball up quickly, turning it into plasma, and disassociating the particles. However, I'm insufficiently competent to know if either of these are good justifications...

I think it is justified to think of the electrons and nuclei as disassociated. The de Broglie wavelength of the nuclei is very little compared to an atomic orbital. Besides the electron would also interact with each other, slowing each other down in each others reference frame. The EM are just as strong for them, but they are ligther. This means that the front parts of the ball would quickly consist of only nuclei as they outrun the electrons.

There is no way it is going to shoot out of the atmosphere like a lightning bolt. You can't ignore EM. We know that single relativistic particles doesn't make it that far through the atmosphere. The chances of hitting nuclei are too low to matter, thus EM must work to stop even higly relativistic particles. A bunch of particles, like the ball, is only going to survive longer if the front parts scoop up everything in its way and thus protecting the rear parts, but this contradicts the assumption that it will go straight through. And if it does scoop, like in Randalls drawings, then it hits equivalent amounts of mass in tens of meters. This is therfore the lengthscale over which it will slow drastically down and have much energy converted into sideways motion. The plasma then hits more and more air, slowing more down, spreading more and hitting yet more air. I think Randall is right that the Bragg peak of the ball, and thus the ground zero of the explosion, will be some way behind the former backstop.

I think Randall is wrong in that there will be significant fusion, there will likely be more fission. This is not going to matter much to the final conclusion though.

[rmsgrey]

I don't think you can think of the electrons and nucleii as dissociated - at least not unless they get stripped from each other by the passage through the air - in the ball's rest-frame just after launch, it's got regular thermal motion, plus a high-intensity wide relativistic particle beam firing at it for a modest fraction of a second.


Whatever happens to the ball, you have about 9*10⁷kg m s^-1 of linear momentum to share between the ball and any collision products (145g ball moving at .9c gives 0.145*0.9*3*10⁸ classical momentum times relativistic scale factor of about 2.3)

To get the average velocity of the particles sharing that momentum down to Mach-1 (roughly 300 m s^-1) you're looking at about 3 kilotonnes of matter - roughly a trillionth of the whole atmosphere - or the amount contained in a vertical cylinder about 40 metres in diameter, or a tangential cylinder about 5 metres in diameter (assuming my quick calculations are correct).

If you assume the ball's momentum is averaged out over all the particles it interacts with, getting it to the point where it's interacting with the air as a gas rather than as independent particles requires it to have spread into a cloud about the size of a house before it leaves the atmosphere. Getting it to "stop" closer to the launch point requires it to spread out faster...

[Yakk]

I think it is justified to think of the electrons and nuclei as disassociated.

I think so too, but I'm not certain. The electron and atom are close together and electrically neutral. With a charged alpha particle, it is "naked" and charged from a long distance away, and as such it should be pulling at electrons from far away -- even far away relative to the speed of the alpha particle!

The neutral atom will only interact with electrons and protons that get relatively close, and being relatively close (close enough to see the electron and protons as relatively disassociated), they will blink away really quickly. You still run the risk of ionization of the neutral atom, but if you presume ionization because alpha particles interact strongly, you are presuming that we can treat them as disassociated in order to treat them as disassociated!

The thing is, what is being described is very strange. We don't, as far as I know, have the ability to generate electrically neutral yet fast moving atoms. Most of our methods involve exploiting the charge of the atomic nucleus, don't they? And I don't doubt that over interstellar distances, the electrons and nuclei would be ionized.

There is no way it is going to shoot out of the atmosphere like a lightning bolt. You can't ignore EM. We know that single relativistic particles doesn't make it that far through the atmosphere.

For the most part, cosmic rays are charged particles. Ie, already disassociated.

Uncharged cosmic rays are, as far as I know, gamma rays: but they are also a poor model.

The chances of hitting nuclei are too low to matter, thus EM must work to stop even higly relativistic particles.

That isn't true of Gamma rays -- they have too much energy to be bothered much by electrons.

The problem I see is that we don't know how much EM interaction there would be with relativistic neutral atomic matter, and extrapolating from the many examples of relativistic ionized atomic matter is not a very valid thing to do. Neutron radiation makes a poor model, because relativistic atomic matter is much bulkier than single neutrons.

Hence me being uncertain.

[gmalivuk]

It will spread out fast, because not all the collisions will be head-on, and thus not all the particles will stay close to the initial path of the ball.

[rmsgrey]

It will spread out fast, because not all the collisions will be head-on, and thus not all the particles will stay close to the initial path of the ball.

The glancing collisions make it spread out faster, but they're also less efficient at stealing momentum.

In order for the ball to stop within a mere 1000km (a seventh of the way to the edge of space at 100km altitude, setting out at a tangent to the surface) the cloud would have to have spread to a cross-section with diameter about the height of a six-storey building and have shared momentum with everything in its path. To stop within 10km, it would have to have spread to a ~200m diameter, still mopping up everything in its path

[gmalivuk]

A spread of 1 degree is still pretty narrow, though.

[Himself]

Okay, I don't really know much about relativity or the physics of subatomic particles. But from reading this thread I've seen it mentioned that charged particles don't travel very far in the atmosphere and I believe it's been mentioned that the electrons in the ball would be stripped away. Would that leave the ball with enough positive charge to sap an significant amount of energy from it as it travels through matter? I would also expect the ball would have the walls of the stadium, a few nearby buildings and perhaps a few hills to travel through. I realize that state of matter doesn't really play in here, but these obstacles would mean a lot more matter to interact with than just the atmosphere.

[rmsgrey]

Okay, I don't really know much about relativity or the physics of subatomic particles. But from reading this thread I've seen it mentioned that charged particles don't travel very far in the atmosphere and I believe it's been mentioned that the electrons in the ball would be stripped away. Would that leave the ball with enough positive charge to sap an significant amount of energy from it as it travels through matter? I would also expect the ball would have the walls of the stadium, a few nearby buildings and perhaps a few hills to travel through. I realize that state of matter doesn't really play in here, but these obstacles would mean a lot more matter to interact with than just the atmosphere.

Above, I mentioned that, to get down to the speed of sound, the initial momentum of the ball would have to be shared amongst 3 kilotonnes of matter - that's 3 million kg. Hills could be significant, but buildings are mostly walls separated by a lot of air - until the ball drops from ludicrous speed down to something around the speed of sound in the medium, it's going to punch through solid objects without dumping significant quantities of energy into the surrounding structure - there just isn't time for the force to propagate through the structure before it gets torn apart...

[Tass]

Okay if we assume that it does mop up everything in its path (I agree that this is not a completely given yet). Then how is this heavily simplified scenario?: The ball is about a thousand times denser than air, so it will cover its first ball mass of air in the first thousand ball diameters, or about 80 meters. I am going to ignore relativity here, it is not the worst approximation here. The mass of plasma is now twice as heavy, but has the same momentum, it is therefore moving forward at half the speed, or .45c. It still has all the energy so the average speed of the component particles are 0.9c/sqrt(2), it is thus also spreading sideways at ~0.45c. So now after 80 meters, even though we have so far magically kept it together as a ball, now it is spreading at a 45degree cone. Again ignoring further spread, in the next hundred meters it sweeps 1/3*pi*(50m)²*100m=25,000m³ weighing about a quarter of a million balls. This enourmous plasmaball is now moving at ~700m/s, still faster than a riffle bullet, but there is energy enough that the individual particles are moving at a speed of 0.9c/sqrt(250000)=0.0045c, about 500 times faster than the ball as a whole. The plasma/fire-ball is now spreading in all directions, and soon the initial momentum of the ball is completely irrelevant, and only the energy is left. When the diameter of the ball is a kilometer, its mass (still assuming it is an air blast), is about a million tons, the average speed of particles are down to 300m/s - just a regular fireball, while the momentum only gives an ordered movement of ~0.03m/s, completely negligible. From now on it is indistinguishable from a nuclear explosion or any other intense localised discharge of energy.

Conclusion: Mopping up and spreading is a superexponential proces. For the ball to stand just a slight chance of making it out of the atmospere, that fact that the high speed air atoms brings electrons along has to make their penetrationpower incredibly much larger. A little bit will not be enough.

Charged particles lose much energy as cherenkov radiation. Do you really think that an electron cloud with size measured in Ångstrøms are going to matter to the radiation from a nucleus of size in femtometers? The nuclei does not need to hit, if they pass close (measured in picometers) the EM interaction between them does not care about electrons a thousand times further out.

To get the average velocity of the particles sharing that momentum down to Mach-1 (roughly 300 m s-1) you're looking at about 3 kilotonnes of matter - roughly a trillionth of the whole atmosphere - or the amount contained in a vertical cylinder about 40 metres in diameter, or a tangential cylinder about 5 metres in diameter (assuming my quick calculations are correct).

If you assume the ball's momentum is averaged out over all the particles it interacts with, getting it to the point where it's interacting with the air as a gas rather than as independent particles requires it to have spread into a cloud about the size of a house before it leaves the atmosphere. Getting it to "stop" closer to the launch point requires it to spread out faster...

We just need to get it to a point where it is no longer neutral particles, because we know ions do not have this penetration power, even at hyperrelativistic speeds. And as soon as it stops quite a bit, energy's v² scaling quickly overtakes momentums v scaling, making the spread impressive.

[sparehead3]

I'm too new to post URLs, but it looks like someone took this idea heart;

Go to the video site WIMP ("period" cee oh emm) and scroll down the list to: "Sep 28 - The best baseball commercial ever." (this is not a link).

If someone with more posts can put the link here (or embed the video if possible) I'd appreciate it.

[Fieari]

This would be the link you're looking for. And it is a pretty neat commercial, yet still not as explosive as the relativistic baseball would be. That ball is merely causing explosions sufficient to tear up the ground below its path, but not extending more than a meter or so to the left or right of the path.

[sparehead3]

That ball is merely causing explosions sufficient to tear up the ground below its path, but not extending more than a meter or so to the left or right of the path.

True, true... By the same token, the vast destruction described in Randall's analysis (describing the total annihilation of the players on both teams, the fans in the stands, as well as the stadium and surrounding city) probably wouldn't make for a very good advertisement (it also makes you wonder if that 5-story tall sheep of plexiglass behind the batter would really be all that effective).

Nevertheless I do find the release of the video so soon after Randall's first "What if..." posting to be interestingly coincidental.

[ivnja]

(it also makes you wonder if that 5-story tall sheep of plexiglass behind the batter would really be all that effective).

[pitareio]

Seen in this month's edition of "Science & Vie Junior", a french scientific magazine aimed at teenagers, an article titled "Et si... mon vélo allait à la vitesse de la lumière ?", literally : "What if... My bicycle went at the speed of light?".

The article begins with the bicycle being magically propelled at 0.9c, and the consequences (as well as what you see and how you are seen at these speeds).

Inspiration or coincidence? I wouldn't go as far as calling it plagiarism, as the ensuing kaboom was only a part of the article, and well, they had to change the baseball part anyways, as few teenagers here have more than a remote idea of what it is. But I think it's reasonable to assume the author, at least, got the idea from xkcd.

I didn't read the article thoroughly, nor bought the magazine, but you can see the cover here : http://pvsamplersla5.immanens.com/fr/pv ... =1&nu=last .

[sehkzychic]

I'll be the first to admit that I know little and less about baseball, but regarding the ruling of the batter being "hit by pitch," I have my doubts. Considering that the article explicitly states that the ball crosses home plate before the batter is affected by the pitch, wouldn't the batter being hit become irrelevant? Isn't that what crowding the plate means? If part of your body is in the strike zone, it's fair game for the pitcher? And at any rate, since the ball actually hits the bat (goodness knows how the batter managed to either swing that fast or that early so as to have the bat in the right place at the right time), and then the ball continues towards the backstop, wouldn't it be a foul ball?

[mathmannix]

I'll be the first to admit that I know little and less about baseball, but regarding the ruling of the batter being "hit by pitch," I have my doubts. Considering that the article explicitly states that the ball crosses home plate before the batter is affected by the pitch, wouldn't the batter being hit become irrelevant? Isn't that what crowding the plate means? If part of your body is in the strike zone, it's fair game for the pitcher? And at any rate, since the ball actually hits the bat (goodness knows how the batter managed to either swing that fast or that early so as to have the bat in the right place at the right time), and then the ball continues towards the backstop, wouldn't it be a foul ball?

Good point. According to Wikipedia, hit by pitch is only called when (A) the ball hits outside the strike zone, (B) the player attempts to avoid the ball (or had no opportunity to avoid it), and (C) he did not swing the bat.

For reference, the strike zone is a three-dimensionally bounded area, directly over home plate, extending from (simplifying) the batter's knees to his shoulders; the ball could be directly over home plate and outside the strike zone if it is either too low or too high. However, the bat would not be directly over (or under!) the strike zone (or behind and over the strike zone); it would be off to the side - unless the player had swung the bat. So, a relativistic baseball pitched directly over home plate could not meet both conditions A and C above (although it would presumably meet condition B, as the player would have no opportunity to avoid the ball after it left the pitcher!)

[Max85]

My understanding is that when a nucleus is bombarded and split, the sum of the parts is less than the total of protons and neutrons. each element in turn, has a unique "packing factor".

[ehque]

Okay if we assume that it does mop up everything in its path (I agree that this is not a completely given yet). Then how is this heavily simplified scenario?: The ball is about a thousand times denser than air, so it will cover its first ball mass of air in the first thousand ball diameters, or about 80 meters. I am going to ignore relativity here, it is not the worst approximation here. The mass of plasma is now twice as heavy, but has the same momentum, it is therefore moving forward at half the speed, or .45c. It still has all the energy so the average speed of the component particles are 0.9c/sqrt(2), it is thus also spreading sideways at ~0.45c. So now after 80 meters, even though we have so far magically kept it together as a ball, now it is spreading at a 45degree cone.

I have thought a bit about this (over the past 2 years. (wow has this been really long.)). The approximation is probably very wrong.

Lets say the ball sweeps up everything. In 80m, each ball-atom strikes, lets say, one air-atom.

Modelling these atoms as billiard balls will help shed light. In a dead on hit, the energy is transferred from ball-atom to air-atom, and the net result is still a coherent beam - there has been no gain in sideways momentum. In a glancing hit, the air-atom will be travelling 90 degs out sideways from the beam, but because almost no energy is transferred from ball-atom to air-atom, the majority of the momentum is still conserved in the ball-atom, still going straight downrange. Since the collisions will vary from glancing to dead-on to glancing again, in actual fact, only a small fraction of atoms will be coming out of the side of the beam at relativistic velocities (the actual percentage calculation has to do with integrals of sine and I don't have enough data to write these equations out).

Another point to note is that the proton beam dump at LHC is 7m long and 2000 times denser than air, but only 0.7m across. You would imagine if it was going to spread out in a 45 deg angle after hitting solid matter, the beam dump would need to be shaped like a half-sphere, if not a massive cone pointed towards the beam.

I just don't think that the r-ball is going to spread out that much.

Not to say it won't kill you, though. The atoms blasting off the side at significant fractions the speed of light will crash into other atoms off the main beam path, releasing interesting (sub)atomic particles and copious amounts of radiation. I'm going to guess that if you're close enough to see it, you'll probably die.

[Keybounce]

It seems that Dragon Ball Super has gotten into this action. We had a super sayan level pitch and batter. And yes, it's as silly as it sounds.

(A challenge from Champa! This Time, a Baseball Game).

EDIT: Ok, just re-watched this episode. The first pitch was the relativistic one, complete with a comment about how that was a nuclear pitch, and needing gods to restore the planet/baseball field. The Vegeta pitch/Goku at-bat was tame in comparison.

[SuicideJunkie]

Say; what are the rules in the situation where the umpires are unable to make a call, and the instant replay video equipment is not available/sufficient?