3
$\begingroup$

A naive calculation suggests that a proton moving in a vacuum at $(1-10^{-38})c$ would have relativistic momentum equal to $\gamma mv \approxeq 2\mathrm{kgms^{-1}}$ if we say $m = 10^{-27}\mathrm{kg}$ (verified using https://keisan.casio.com/calculator). This is obviously a ridiculous amount of momentum for something measuring femtometres across.

However, at this speed, what would happen if, say, a $1$m x $1$m x $0.2$m piece of wood was placed in the direct path of the proton?

Since atomic nuclei are so small compared to the distance between them in the wood, and the protons immense inertia, I would expected it to pass straight through most of the time - but what if it struck a nucleus head-on? Would we see the piece of wood break akin to some sort of bullet impact, or would we see something more like a fission chain reaction as the target nucleus instantly disintegrates into more highly energetic particles?

Further, how would the miniscule de Broglie wavelength of such a particle affect this scenario, if at all?

I realise this is a highly speculative question, but I would like to know if my conjectures seem plausible.

$\endgroup$
2
  • 1
    $\begingroup$ Why would this be a "speculative question"? Do you think in all the experiments with proton beams we've never actually shot them at any materials? Even a human has been hit by proton beams. $\endgroup$
    – ACuriousMind
    Aug 29, 2021 at 13:01
  • $\begingroup$ @ACuriousMind I included that phrase to emphasise that I don't really have any solid idea what I'm talking about, but I realise now it is sort of untrue. Thanks. $\endgroup$
    – Poo2uhaha
    Aug 29, 2021 at 14:53

2 Answers 2

2
$\begingroup$

First, I want to advise against direct use of any calculator for finding the Lorentz factor $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$, because in your case this would involve calculating the difference of two nearly equal numbers ($1$ and $v^2/c^2$). This is prone to loss of precision. You can circumvent this issue by doing some algebra first:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{(1+\frac{v}{c})(1-\frac{v}{c})}} = \frac{1}{\sqrt{(2+10^{-38})\cdot 10^{-38}}} \approx \frac{1}{\sqrt{2\cdot 10^{-38}}} = 7.1\cdot 10^{18}$$

Then the momentum is $$p = \gamma mv \approx \gamma mc = 7.1\cdot 10^{18} \cdot 1.7\cdot 10^{-27}\text{ kg}\cdot 3\cdot 10^8\text{ m/s} = 3.6\text{ kg}\cdot\text{m/s}$$ which differs significantly from your result ($p=2\text{ kg}\cdot\text{m/s}$).

Let's also calculate the energy: $$E=\gamma mc^2 = 7.1\cdot 10^{18} \cdot 1.7\cdot 10^{-27}\text{ kg} \cdot (3\cdot 10^8\text{ m/s})^2 = 1.1\cdot 10^9\text{ J}$$

With this high energy you are way beyond the realm of ultra-high-energy cosmic rays ($E > 8\text{ J}$) and even the Oh-My-God particle ($E=51\text{ J}$). When such a high-energy particle hits your block of wood, it will most probably just pass through without causing much ionization of the atoms. The particle would interact not much with the electrons and nuclei, just because it flies by very fast and hence there is not enough time to transfer much momentum and energy to any of these.

In the rare event of a head-on collision with a nucleus, it would smash the nucleus into pieces and also produce some particle/antiparticle pairs (see pair production).

enter image description here
(image from Wikipedia - Cosmic rays)

However, the primary high-energy particle still would only loose a small percentage of its energy in this collison. So it can do many more such collisions on its way after leaving the block of wood. And also the fragments produced by the collision will later collide with other nuclei on their way (most probably also after leaving the block of wood). So you get a cascade of secondary rays. But the block of wood as a whole would not take any visible damage.

$\endgroup$
2
  • $\begingroup$ Note: the reason for the result differing so much from mine would be the fact that I used $m = 10^{-27}\mathrm{kg}$ as a rough estimate, instead of the more accurate value which you have used (this detail was included in the question). So our results were almost identical in the end, although your difference-of-two-squares trick is a neat one indeed, thank you. $\endgroup$
    – Poo2uhaha
    Aug 29, 2021 at 14:55
  • 1
    $\begingroup$ Note that, even though such collisions are rare, the mostly-empty cosmic microwave background is opaque to to protons with this energy on galactic distance scales. It takes some finesse to have a good intuition about the interaction between enormous energies and tiny cross-sections and tiny densities and enormous volumes. $\endgroup$
    – rob
    Aug 29, 2021 at 16:58
1
$\begingroup$

First of all it is a little strange to start a naive calculation with the ad hoc speed of $(1-10^{-38})c$. This is obviously very close to speed of light, and the closer you get, the higher the energy of the particle is. Particle physicists usually give the energy of a particle in electronvolts, and if I did not make a mistake, the momentum you gave corresponds to an energy (via $E=pc$, neglecting the mass at such high momentum) of around $10^{27}\mathrm{eV}$. This is even ten million times more energy than the highest energy cosmic rays (also mostly protons) that have been found, while these already have much higher energies than the particles in colliders. There is no known mechanism that could give protons the energy you consider, actually there is not even a known mechanism that explains the highest energy cosmic rays. Cosmic rays hit us all the time. (Although most of them are of 'medium' energy.)

Nevertheless, the process that happens if these particles hit matter, e.g. living materials like wood or tissue is the similar for different energy ranges: There is a certain probability that it will interact with another particle inside the wood/metal/tissue, via the rules of relativistic quantum mechanics, i.e. quantum field theory, and by that the resulting particles produce a lot of secondary and tertiary (and so on) particles, since all the kinetic energy can be converted to mass of other particles. This is similar to the chain reaction you mentioned, but especially in wood I would not know how this could start a 'fission' i.e. nuclear chain reaction. As you mentioned, the high energy leads to a small deBroglie wavelength, which I would think implies that the interaction will happen with individual nucleons, like another proton or a neutron, and not with a whole nucleus. In living tissue these highly kinetic particles can disintegrate molecules and therefore actually destroy cells, which is of course not healthy if it happens in too many cells at once, but as I mentioned, it basically happens all the time. I would not know how a single particle could cause a macroscopic visible 'bullet hole', no matter how high it's energy is, since then it would need to interact with many atoms that are close kind of in sync, and I do not know how that would be possible.

EDIT: I just looked up that the usual beam in the LHC contains all together (it consists of a LOT of particles) around 350 MJ. This is actually roughly the amount of energy you considered for your ONE particle in your question. When the beam is dumped into many tons of metal, the room they shoot it in heats to $800^{\circ}C$. See here for a reference in German. So I think such a ridiculous high energetic particle could lead to so much kinetic energy in the resulting particles that just the heat, which in the end of course comes from the particle interactions, destroys the whole wood block, but this only happens if enough of the resulting particles actually interact within the woodblock.

$\endgroup$
1
  • $\begingroup$ In regard to your first paragraph, I realised that this is proton velocity is somewhat nonsensical to attain, which is why I added the note at the bottom of the question: "I realise this is a highly speculative question... [etc]" $\endgroup$
    – Poo2uhaha
    Aug 29, 2021 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.