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In the Large Hadron Collider (LHC) the particles move close to the speed of light.

The LHC accelerates beams of particles, usually protons, around and around a 17-mile ring until they reach 99.9999991 percent the speed of light.

(Source)

Mass of proton: $1.6726219 \times 10^{-27}$.
At $99.9999991 \%$ speed of light, the total energy (kinetic energy plus rest energy) for a proton is:

$$ E=\frac{m c^{2}}{\sqrt[2]{1-\frac{v^{2}}{c^{2}}}}=\frac{1.6726219 \times 10^{-27} \times\left(300000 \times 10^{3}\right)^{2}}{\sqrt[2]{1-\frac{\left(\frac{99.9999991}{100} \times 300000 \times 10^{3}\right)^{2}}{\left(300000 \times 10^{3}\right)^{2}}}}=1.122 \times 10^{-6} J $$

As a proton is provided with more and more energy, it's energy will tend toward infinite energy. I understand that I've used a proton which is made up of more fundamental particles, i.e. quarks. Anyway, how are we so sure that as the proton gains more and more energy, it wouldn't break into more fundamental particles on its own without colliding into another particle? In other words, is it possible that as energy becomes quite large (or, as its speed tends closer to speed of light) the proton disintegrates into other stuff on its own. Though, electron is considered to be a fundamental particle, as science history is witness, one can never be too sure about it. Have they ever accelerated the proton to such speeds where its energy becomes, say, 2000 kJ?

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    $\begingroup$ The speed does not change the way it could decay, there is always an inertial observer that sees the proton at rest. $\endgroup$
    – user65081
    Aug 13 at 4:26
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    $\begingroup$ Possible duplicate: If a mass moves close to the speed of light, does it turn into a black hole? $\endgroup$
    – Qmechanic
    Aug 13 at 5:12
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    $\begingroup$ Hello! There seem to be some typesetting errors in the denominator of your large fraction (note for example the two brackets in the square root). You might want to edit that. $\endgroup$
    – Jonas
    Aug 14 at 13:17
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    $\begingroup$ @Jonas Thanks for pointing this out. I've fixed it. $\endgroup$
    – PG1995
    Aug 15 at 7:49
  • $\begingroup$ @Wolphramjonny: But it decays in that reference frame like a stationary proton would in ours. The dilation differences mean we would observe it to decay differently than a proton that's stationary in our frame. More slowly, as at least one test has shown for other particle types. I can imagine the question presuming more energy instead means faster decay, and therefore having a point where it's practically impossible for a proton to complete the entire loop without decaying. $\endgroup$
    – MichaelS
    Aug 16 at 2:35
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Since the energy of a particle is frame dependent, it is not particularly meaningful to discuss the energy of a single particle, by itself. The reason for this is that one can find a reference frame where that particle has any arbitrary kinetic energy, and since the laws of physics are the same in all inertial reference frames, we cannot have a proton disintegrate in one frame and continue existing in another. This is why the protons must collide with something else (another proton moving the opposite direction, in the case of the LHC) in order to produce an interesting reaction. As such, the more interesting quantity to look at is the energy of the collision, which is the same in all reference frames, given by the sum of the energies of the protons in the centre of mass frame.

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    $\begingroup$ One must add however, that a free neutron decays regardless of the reference frame $\endgroup$
    – wander95
    Aug 13 at 19:46
  • $\begingroup$ " As such, the more interesting quantity to look at is the energy of the collision, which is the same in all reference frames, given by the sum of the energies of the protons in the centre of mass frame." Isn't the energy of a collision in the reference frame of its center of mass smaller than its energy in all other reference frames? $\endgroup$ Aug 13 at 23:48
  • $\begingroup$ Thanks for the answer but I was thinking more in terms of disintegration of internal structure of proton. $\endgroup$
    – PG1995
    Aug 14 at 0:13
  • $\begingroup$ @Acccumulation Perhaps I might not be using exactly the right terminology, but by the "energy of the collision," I am referring to the energy released in the collision; i.e. the Lorentz invariant quantity $\sqrt{p_\mu p^\mu}$, where $p$ is the sum of the 4-momenta of the particles. $\endgroup$
    – Sandejo
    Aug 14 at 2:36
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    $\begingroup$ @PG1995 Your question here seems to focus specifically on how the speed of the protons affects them, which I think I've addressed in my answer (if not, ask for clarification). If you want to ask about proton decay more generally, then I would recommend asking a separate question. $\endgroup$
    – Sandejo
    Aug 14 at 2:43
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how are we so sure that as the proton gains more and more energy, it wouldn't break into more fundamental particles on its own without colliding into another particle?

No, it would not. No experiment has ever shown that a free proton decays$^1$.

At the LHC, protons have been accelerated to speeds up to about $3 \frac{m}{s}$ less than the speed of light. This corresponds to a Lorentz factor of $\gamma \approx 7000$ corresponding to a speed $v=0.99999999c$. But even at this speed, the proton still does not decay, and it should not, since even at this speed you can still find a frame of reference where the proton's kinetic energy is arbitrarily small (since kinetic energy is a frame dependent quantity).

In other words, is it possible that as energy becomes quite large (or, as its speed tends closer to speed of light) the proton disintegrates into other stuff on its own.

Again, kinetic energy is frame dependent. You can always find a frame where its kinetic energy is arbitrary. This is why the proton will only decay into other particles if it is collided with another proton, moving in the opposite direction.

$^1$Proton decay was first hypothesized Sakharov in 1967. Even after many experiments, proton decay has never been observed. Some experimentalists have concluded that the lifetime of a proton has to be greater than $10^{34}$ years. That is, $10,000,000,000,000,000,000,000,000,000,000,000$ years.

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    $\begingroup$ A corollary of this is that protons in the LHC moving at $\gamma \approx 7000$ are less likely to decay than protons at rest with respect to the Earth, due to relativistic time dilation. (If they decay at all, of course.) $\endgroup$ Aug 13 at 13:37
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    $\begingroup$ You can also find a frame where its energy is arbitrarily large. There are frames where every proton in my body has the energy of a freight train. Why don't I disintegrate? $\endgroup$
    – user253751
    Aug 13 at 14:11
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    $\begingroup$ yes @llama, it was rhetorical $\endgroup$
    – user253751
    Aug 13 at 14:28
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    $\begingroup$ $\gamma = 7000$ is peanuts compared to the $\gamma$ of the solar neutrinos streaming right through us. We're not sure of the rest mass energy of the neutrino, but it's probably around 0.1 to 1 eV. Our best neutrino detectors can't see neutrinos with less energy than 233 keV (so $\gamma > 233,000$) and many neutrino detection experiments operate at much higher energies. $\endgroup$
    – PM 2Ring
    Aug 14 at 0:23
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    $\begingroup$ @joseph h $$ \begin{array}{l} \text { Sorry, I earlier said that } \gamma=70711 \text { . That was a mistake and you were correct. }\\ \gamma=\frac{1}{\sqrt[2]{1-\frac{\left(\frac{99.9999991}{100} \times 300000 \times 10^{3}\right)^{2}}{\left(300000 \times 10^{3}\right)^{2}}}}=7453.6 \approx 7000 \end{array} $$ $\endgroup$
    – PG1995
    Aug 15 at 7:48
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For entities as particles , atoms, molecules, when talking of energy special relativity and its four vectors have to be used.

The "length" of the energy momentum four vector is the invariant mass describing the entities $$\sqrt{p_\mu p^\mu}=\sqrt{E^2-(pc)^2}=m_0c^2$$

The energy of the proton you are concerned about is $E$. When there is no momentum, the particle is at rest, one gets the invariant mass, $m_0$ which characterizes a particle for any momentum. Thus , as there are only limits about proton decay, there is no way within mainstream physics to suppose that a proton will disintegrate.

The momentum it may be moving with is irrelevant, the mass in $E=mc^2$ is the relativistic mass which is no longer used as it gives rise to such confusions.

Within mainstream physics the momentum of the proton has nothing to do with its internal structure. The proton is the same in all inertial frames .

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    $\begingroup$ Thank you! You said, Within mainstream physics the momentum of the proton has nothing to do with its internal structure. The proton is the same in all inertial frames . I like your wording. I was thinking of disintegration of proton's internal structure. $\endgroup$
    – PG1995
    Aug 14 at 0:06
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The decay rates of particles have to be independent of the reference frame in which they are viewed (taking time dilation into account). It follows that simply accelerating a particle to a very high velocity cannot induce it to decay more rapidly. If that were not the case, then you immediately encounter impossible contradictions. Consider a pool of a trillion protons, say, stationary here on Earth. Now imagine they are passed by ten separate observers, each travelling at a different speed very close to the speed of light. Since each of those observers will consider the particles to have a different energy, each observer would expect to see a different number of the protons had decayed, which is clearly impossible.

If you factor in time dilation, the position becomes even more paradoxical. As the observers move at increasing speed past our pool of a trillion protons, they see the time experience by the protons to be dilated, so if anything the lifetimes of the protons would be prolonged, not shortened by their increased speed.

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    $\begingroup$ What about the muon decay rate? Isn't the longer decay time for muons entering Earth's atmosphere considered to be one of the proofs of relativistic time dilation? $\endgroup$
    – D. Halsey
    Aug 14 at 0:32
  • $\begingroup$ Thanks for the answer! I believe the energy conservation law is true for every inertial reference frame but the quantity of measured energy can differ. If an observer measures 10 J in his/her reference frame then another observer in different inertial reference frame could measure the same energy to be 5 J. $\endgroup$
    – PG1995
    Aug 14 at 1:22
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    $\begingroup$ @PG1995 But the conservation law does hold.in each of the inertial frame. Comparing kinetic energies of the same object in different inertial frames makes no sense, though. Almost like apples and oranges. $\endgroup$
    – Vladimir F
    Aug 14 at 8:20
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    $\begingroup$ @D.Halsey Good point! Time dilation is another argument. I will update my answer to add it. $\endgroup$ Aug 14 at 8:46
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Several misconceptions in this question.

  1. Kinetic energy is a property of particle in given coordinate system. Perform the following experiment: imagine a coordinate system moving with 99.9999991% speed of light with respect to you. In that coordinate system you are moving with 99.9999991% speed of light and your kinetic energy is enormous. However, while performing the experiment you will notice no change in your body or your surroundings, and performing this imagination is entirely safe.

  2. Kinetic energy is not internal energy of an object. Consider the following examples:

    a) You throw a piece of ice. You increase its kinetic energy, but nothing bad happens to it while its flying.

    b) A flying piece ice hits the ground. It may break, because its kinetic energy transforms into other forms of energy, like mechanical waves or heat.

    c) You put a piece of ice on a stove and turn on heating. Soon the ice will melt. The piece doesn't move, you don't increase its kinetic energy. You rather increase its thermal energy. Thermal energy is related to kinetic energy of individual water molecules, but total kinetic energy of the whole piece of ice remains 0.

    The fact that a proton is moving fast doesn't change anything about it. The quarks and gluons forming a quickly moving proton behave the same as those in a stationary one. Only if our proton collides with something, e.g. with another proton moving in the opposite direction, their kinetic energies can be transformed into something else, e.g. breaking them and forming new particles.

    A fast moving proton is something quite different than what's called excited proton or more often excited nucleon which has elevated internal energy, in form of angular momentum and spin configuration. Excited nucleons can be created in high energy collisions, such as those in LHC, and are very unstable; after very short time they release their extra energy (in form of new particles) and decay back to "regular" proton or neutron state.

  3. Several people mentioned decays. I think this is not what you asked about. You speak about disintegration of a proton due to its large internal energy, e.g. similarly to breaking or melting piece of ice. The only way we have to increase internal energy is by colliding the proton with something else. That's something completely different from spontaneous decay.

    Some particles are unstable and decay into other particles after some time. Decay of proton is theorized, however it hasn't been discovered experimentally. Therefore proton is either stable, or its lifetime is extremely long.

    Particle decays are actually affected by their movement, due to time dilation. Time passes slower for a moving object. E.g. muons produced by interactions of cosmic rays high in the atmosphere reach the ground, despite their mean lifetime of 2.2µs multiplied by the maximum speed they may have, speed of light is only 660m. So if proton could actually decay, it would even less likely do so if it moved with relativistic speed in an accelerator.

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  • $\begingroup$ Thanks for the detailed answer. You've touched upon subtle but important points. You said, "*Several people mentioned decays. I think this is not what you asked about. You speak about disintegration of a proton due to its large internal energy, e.g. similarly to breaking or melting piece of ice. *" You got it correct. The space around is not just empty; it has all those quantum fields, space-time, quantum fluctuations, etc. What they used to call in old days ''aether'. So, I was thinking once matter is moving really, really fast, its Cont'd $\endgroup$
    – PG1995
    Aug 15 at 7:59
  • $\begingroup$ interaction with all-pervasive space would affect matter. Possibly, I had 'aether wind' somewhere in my mind! :) $\endgroup$
    – PG1995
    Aug 15 at 8:00
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    $\begingroup$ @PG1995 1. There is no aether. See: Michelson-Morley experiment. There is Cosmic Background Radiation but LHC protons are way to slow to interact with it inelastically. 2. Proton does interact with external fields, for instance with the electric and magnetic fields which accelerate it and keep in the ring. These forces are way smaller than those bonding the proton together. 3. Protons in LHC interact sometimes with rare molecules of gas that vacuum pumps failed to remove (see: beam-gas interactions). 4. If you have any specific question about any of that, ask it as a separate question. $\endgroup$ Aug 15 at 16:04
  • $\begingroup$ Thank you. I agree that there is no aether but after aether it's been space-time, quantum fields, quantum fluctuations, etc.! Anyway, I get your point and once again thanks for providing the question with more context. $\endgroup$
    – PG1995
    Aug 16 at 5:04
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No, it can't.

Ultimately, this fact is decided by observation - but it fits within a general property of the laws of physics that is known even from circumstances closer to everyday human scale - that motion is relative.

What this statement means is the following. If you consider a particular object and its surroundings, the laws of physics will generate the same future history(*) in the situation where that object is moving and the surroundings are stationary, as where the object is stationary and the surroundings are moving in the complementary way (i.e. with the opposite direction and speed), provided of course we also likewise adjust the future-generated history from the first scenario to bias motion accordingly to translate it into the second one.

Suppose that there existed a particle with the property that it could spontaneously burst, like you imagine, at a suitably high speed. Say that when it gets suitably close to the speed of light (about 300 megameters a second [Mm/s]), it becomes unstable and bursts after a short flight time, say 4 seconds. (Note: for anyone who knows about real particle decays - which I'll touch on later - this is obviously "too deterministic" but I'm trying to keep things simple.)

Now imagine what the history might look like: at second 1 it travels 300 Mm, at second 2 it has traveled 600 Mm, at second 3 it has traveled 900 Mm, then at second 4, just as its odometer finally hits ~1200 Mm (about 3 times further than the Moon!) it suddenly bursts into a shower of quarks.

Now take that this same behavior was not observed were it "staying home". 10 seconds, 100, 1000 later it would still be there, whereas the moving particles would reliably burst at 4 seconds and ~1200 Mm displacement once at or above, say, 99.99% of the speed of light.

Now consider two such particles of the same type, starting out from the same point. One is "resting", the other is moving "near the speed of light". The second particle bursts at 4 s in, while the first remains integrate forever. Remember what I said above about transformation. Suppose I transform this history so that the "resting" particle is now moving backwards at 300 Mm/s, and the "mover" particle is resting. If the motion-is-relative property is to be upheld, then we must have one of two things:

  1. if the "particle decays in 4 seconds" property is to be upheld, then in this transformed history we must have that the now-resting "mover" particle bursts at some point. But resting particles cannot burst,

  2. if the "particle is stable" property is to be upheld, then the now-moving "at home" particle, 100, 1000 s, or even many ages of the universe, later will still be intact, in contradiction to the rule that "near light speed" particles must burst after 4 s.

Or to think of it another way, simply shifting the motions of things mathematically cannot create or remove a complex phenomenon like the rupture of an object into a shower of smaller objects. Thus, if such a thing did indeed occur at one speed but not another, we can use that to infer our speed: we just have a particle of the given type at rest with us, and if we see it burst in front of our eyes, then we know we must be travelling at 99.99% of the speed of light - absolutely, and with enough measurements of this phenomenon we can extrapolate a universal reference frame with regard to which we can say everything is .

Ultimately though, as mentioned, the undetectability of any putative universal reference frame in this form is still only an empirical truth. Maybe there are such particles out there - but we haven't seen one, and it gets less and less likely the more evidence we get, just as with anything in science. And likewise with scientific reasoning, the strong track record of the relative-motion principle lets us say that

All the above said, there is an important subtlety here that deserves mention. We know of many particles that do spontaneously "burst" as just described: they are called unstable particles and, in fact, most of the particles on the Standard Model are unstable as are most composites of quarks except the proton, and the neutron when bound up in a nucleus (again, as far as we can observe!). But there is also something we can observe with them that may make you question this: if we take, say, a muon - essentially a heavy version of an electron that the Standard Model provides - which "at rest" decays in an average of 2.2 μs, then we boost it to near the speed of light, we will see it decay slower than that - not faster, but slower: e.g. suitably close it will now take ~2200 μs, say. Does this mean speed "stabilizes" particles?

The answer, again, is no: first, keep in mind that in one case we have no decay at all versus the presence of a decay, whereas here we only have a shift in the amount of time. But second, and more comprehensively, we can see that this can be fully accounted for by only slightly more complicating the transformation that we use when relating histories at rest versus motion - i.e. we need to do a few more adjustments to convert one history to another as in what I described prior before the laws of physics will "accept" it as truthful, but they are still relatively simple and, more importantly, universally applicable mathematical transformations, meaning that even in completely different situations the same transformation will still preserve the same physics. This a basis of special relativity, but the point is the laws of physics still have a motion symmetry, just "shaped" a little different.


(*) If you prefer, another way to think of laws of physics is that they "verify" particular candidate histories as being "physically plausible" or not, and we are saying that if we change each history suitably into the other by a regular mathematical transformation, the laws of physics will once again verify it as valid.

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  • $\begingroup$ Thanks for the answer and clarifying many subtle points and give it more clarity. $\endgroup$
    – PG1995
    Aug 15 at 8:18

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