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- How stable is the photon?
Julian Heeck
Phys. Rev. Lett. 111, 021801 (2013)
arXiv:1304.2821 [hep-ph]
If the photon is unstable and decays into other particles, then the number density of photons in the cosmic microwave background (CMB) should decrease while the photons are propagating. But then, the energy density of the spectrum would no longer fit the almost perfectly thermal Planck curve that we observe. One can thus use the CMB measurements to constrain the photon lifetime.
If one uses the largest photon mass presently consistent with experiment, the photon lifetime is 3 years in the restframe of the photon. If one calculates the γ-factor (ie, the time dilatation) to obtain the lifetime of light in the visible spectrum it turns out to be at least 1018 years.
It’s rare to find such a straight-forward and readable paper addressing an interesting question in particle physics. “Cute” was really the only word that came to my mind.
Hm, so such a photon would travel only at 0.9999999999999999999999999999999999955 of light speed? interesting...
ReplyDeleteWhat about pair production of an electron and anti electron? That process is allowed for a massless photon. Or am i wrong?
ReplyDeleteHow stable is the graviton? What are the corresponding limits?
ReplyDeleteIt’s rare to find such a straight-forward and readable paper addressing an interesting question in particle physics.
ReplyDeleteIndeed. Anybody who can produce a particle physics paper that a biophotonics person like me can easily read and understand is a person with a bright future in teaching, if he desires that. (I mean that as a very high compliment.)
Nidnus,
ReplyDeleteA massless photon can't make an electron positron pair, it's kinematically not possible. That process only exists if the photon is off-shell, ie comes from another process, couples to something else etc. If the photon is massive it can decay into an electron positron pair, but only if the mass would be twice as large as the electron mass, which we know it isn't. If it was, it would decay... Best,
B.
Good explanation 🤔...
DeleteAlexis,
ReplyDeleteThat photon would travel at less than the speed of massless particles that is a constant entering Lorentz-transformations, which is commonly but erroneously referred to as the speed of light. If photons are indeed massless, these speeds are both the same. If they are not, light still travels at the speed of light. Best,
B.
Phillip,
ReplyDeleteNobody's ever detected a graviton, so the question is pointless. You can ask what are the constraints on a mass-term in the linear approximation of GR. Best,
B.
Nidnus:
ReplyDeletePS Above a certain threshold it's also possible if Lorentz-invariance is violated. Best,
B.
I was thinking when you said cute you may have meant beauty?:)
ReplyDeleteSo before the Higg's boson....mass less entities... representative in the WMAP, as photon exchange?
ummmm......
ReplyDeleteIf the sun and its temperature were spread out all over the sky, it would be completely useless to us. It’s because you have not just the sun, but the cold background of the sky that you get this entropy difference between the smaller number of solar high-energy photons being turned into a greater number of low-energy photons going away. All the entropy, all the randomness, is transferred into those photons that the earth sends back into spaceRoger Penrose
A photon is its own anti-particle. Off the shelf... Omit the pig-in-a-blanket holoraum at the National Ignition Facility: 192 beams, 500 terawatts of 351 nm photons; engineer a (lambda/2)^3 focus. See what comes out of the vacuum that is not a photon, extrapolate backwards, parameterizing to a single photon decaying. A single 10^44 Hz photon would also do it. (Either way, researchers should wear SPF N_A sun block plus Dolce & Gabanna DG2027B sunglasses.)
ReplyDeleteDoes time really exist for a photon? I always thought that what we see as before and after does not mean the same thing for a photon, and the same photon is measured over and over again in our experiments.
ReplyDeleteI know it does not make much sense, because we "See" lots of photons at the same time, but we don't live at the speed of light eikther.
Hi Plato,
ReplyDeleteNo, I didn't mean beautiful. While it's an interesting analysis and useful, it isn't much of an insight in the sense that it doesn't really explain anything because nobody suspects the photon to be unstable to begin with. Best,
B.
Joey_Blau,
ReplyDeletePlease read what I wrote: The photon has to be massive. Best,
B.
"Nobody's ever detected a graviton, so the question is pointless. You can ask what are the constraints on a mass-term in the linear approximation of GR."
ReplyDeleteYes, no-one has detected a graviton, but I don't see that this makes the question pointless. One can still derive limits on the mass. For example, if there is a finite mass, then the inverse-square law will break down.
Maybe my question is the one you suggest I ask. Do you know what those limits are?
/*most importantly a lower direct bound of 3 yr on the photon lifetime*/
ReplyDeleteThe article is nonsensical as it considers only photon-photon interactions in vacuum. In dense aether model the photons are massive and they decay just after few nanoseconds/microseconds due the quantum decoherence and they recombine after while. What we are observing from distant stars are another photons than those emitted.
Hi Phillip,
ReplyDeleteThe short answer is no I don't know. The long answer is: it's complicated. It's complicated because it's difficult to make sense of massive gravitons without ruining GR altogether. Roughly speaking, the limit to zero isn't continuous because you have different degrees of freedom. To fix that you have to use the Stückelberg fields also mentioned in Julian's paper. There is however some controversy about whether that works or doesn't work (without introducing ghosts) or exactly how it does work. I'm not an expert on that, but if you check the arxiv for massive gravity, you'll find loads of literature... This is a field which has been very active in the recent years: After it has long been believed to be just not possible to make massive gravity consistent, now many people think it is possible after all. Best,
B.
/*..If the photon is unstable and decays into other particles, then the number density of photons in the cosmic microwave background (CMB) should decrease while the photons are propagating..*/
ReplyDeleteI presume, everyone can see the hole in this line of reasoning. For example, inside of hot mixture of oxygen and hydrogen the average life of water molecules is just few seconds, despite the equillibrium remain steady state. This is just an example, how the math models can be as exact and robust, as is the logics and assumptions used in their derivations.
Wouldn't a massive photon have three polarizations? Why hasn't this been detected?
ReplyDeleteI presume, everyone can see the hole in this line of reasoning. For example, inside of hot mixture of oxygen and hydrogen the average life of water molecules is just few seconds, despite the equillibrium remain steady state. This is just an example, how the math models can be as exact and robust, as is the logics and assumptions used in their derivations.
ReplyDeleteOver a long enough time scale, photons and their daughter particles should equilibrate. However, if the cross-sections are small enough, and the number density of photons to bump into each other is low enough, the time scale for that could be many, many multiples of the age of the universe.
B.. wait what? Where do you say it has to be massive.. you use the term "mere possibilty".
ReplyDeleteAre you saying we now think photons have mass? Cause yeah I missed that.
Phillip, Sabine,
ReplyDeleteyes, massive gravity seems to be quite involved. At least within a certain radius (e.g. Vainshtein radius) it may be possible to recover GR.
Just recently there was a very nice talk at Perimeter Institute about massive gravity, including an interesting summary of its very rich history:
https://perimeterinstitute.ca/videos/massive-gravity-and-cosmology
Best
"... it is interesting to note that nowhere in twentieth century physics has it been proved that the photon indeed has no mass."
- B. G. Sidharth
/*Are you saying we now think photons have mass? Cause yeah I missed that.*/
ReplyDeleteWe should realize, that the relativity doesn't support concept of energy quantization, so it actually excludes the existence of photon. In relativity the photons should be always unstable. It deals just with harmonic light wave, the whole existence of photon violates it. So it has no deeper meaning to ask for mass of photon from relativistic perspective, for rest mass of photon the less, as the photons never stay at rest.
In my opinion the photons should have dynamic mass, because during explosion of star their matter gets evaporated just in the form of photons. Such a photons will therefore interact gravitationally each other during their flight - on this assumption for example my theory of distant gamma ray burst stability is based.
Bee, I think the author of this paper left out the principle of you-no-what. (See your fourth article back.) If you are trying to consolidate changes in time flow from early universe to now then there is a much, much simpler explanation than photons having mass and then decaying.
ReplyDeleteWhen particles are accelerated they take on energy. You could think of it as absorbing photons. There is nothing to stop you from applying this directly to the early universe.
In this case photons would be taken up as mass as particles accelerated outward in the big bang. The higher the energy of the photons the higher the acceleration. But since time is measured by observation of events, usually via photons, time flow will be affected by the density of photons, not the energy of the photons. The acceleration will correspond exactly to the energy of the photons absorbed by matter.
The clincher is that when you reverse the cycle and collide matter together at high accelerations then photons are emitted. It would be interesting to see if the energy of the photons emitted correspond to the decceleration at the center of mass.
BTW As Arun already informed you, your tribesman Alexander Unzicker has published book, which reflects the dull situation of contemporary physics quite well - at least by publisher review.
ReplyDeleteZephir,
ReplyDeleteI assure you it has not escaped my attention. As you can guess, I heard of that when it was published in German. I just chose not to waste my time on it since the reviews unambiguously said there's no insights to be gained from this book. Best,
B.
Joey:
ReplyDeleteMaybe we're having a grammar problem here. I wrote: the possibility that the photons may have a mass brings up another question, which is what is their lifetime. This is the question that has been addressed in the paper. Best,
B.
/*I just chose not to waste my time on it since the reviews unambiguously said there's no insights to be gained from this book.*/
ReplyDeleteYes, this is an example of informational singularity: the close sectarian community of physicists inside of human society represents the analogy of boson condensate inside of black hole. They're separated causally from the information outside. It has therefore no meaning to critize them, because they even don't admit such a critique. The only feedback which works is not to pay their useless "re-search".
First,
ReplyDeleteYes, you need to get rid of the third polarization somehow, same as with massive gravity (see comment above). That's what the Stückelberg mechanism is good for, see introduction of paper. Best,
B.
(Sorry, of course for massive gravity it's the fifth polarization that you need to get rid of.)
ReplyDeleteSabine, this isn't cute, this is junk. It's as if E=mc² never happened. The mass of a body is a measure of its energy-content. A radiating body loses mass. The amount of E-mc² energy it loses is the same as the E=hf energy of the emitted photon. When you trap that massless photon in a mirror-box body it increases the mass of the body. Because the mass of a body is a measure of its energy-content. By the way gravitons are virtual. Any concentration of energy causes gravity, including a photon. A photon is not surrounded by gravitons like flies round a cow.
ReplyDeleteZephir: I bought the Unzicker book.
Johnduffieldblog,
ReplyDeleteThe amount of garble in your comment is amazing. You're a strong competitor for Zephir. Let me just pick out one thing: the mass of a body is in general not a measure of its energy. Let me be very clear here that when I say mass I mean of course the relativistically invariant restmass. Best,
B.
Sabine, do you even know any physics? See this:
ReplyDeletehttp://www.fourmilab.ch/etexts/einstein/E_mc2/www/
That's Einstein's E=mc² paper. Note this line:
"The mass of a body is a measure of its energy-content".
Now, are you trying to tell everybody that E=mc² is wrong?
And whilst I've got you on the ropes, let me advise you that in general relativity, invariant mass varies. See this:
ReplyDeletehttps://en.wikipedia.org/wiki/Mass_in_general_relativity#Questions.2C_answers.2C_and_simple_examples_of_mass_in_general_relativity
When you lift a brick you do work on it. You expend energy, and the brick gets it. Its mass increases as a result. When you let it fall "gravitational potential energy" within the brick is converted into kinetic energy. You know this is true, because you know about the mass deficit. You know that if you drop a 1kg brick into a black hole, the mass of the black hole increases by 1kg. Not 10000000kg. At all times during its descent, the "relativistic mass" of the brick is 1kg, which means the relativistically invariant rest mass is reducing by virtue of conservation of energy.
Gravitational potential energy is only negative by convention, because the zero level is deemed to be at an infinite distance. But no bricks are comprised of negative energy. Gravitational field energy is positive, hence "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy".
Confusion here in the comments, some of it honest, some not.
ReplyDeleteIn the past, "mass" often meant total mass. For at least a few decades now, "mass" means "rest mass" or "relativistically invariant mass".
There's no dishonesty on my part Phillip. I put "relativistic mass" in quotes because it's a measure of energy, and Einstein didn't favour it. In a special relativity context it's the rest mass (aka mass) of a moving body plus its kinetic energy. Rest mass is said to be invariant, but when you move to the general relativity context, it isn't. It's the same for the -13.6eV binding energy of the ground-state hydrogen atom. There's a mass deficit.
ReplyDelete@johnduffieldblog
ReplyDeletePerhaps you could tone it down a wee bit. As they say, you can attract more flies with honey than vinegar. Being correct isn't always the end all and be all. In the end we still need to try to work together on common problems.
I hear you Eric.
ReplyDeleteSabine: sorry for coming down heavy on you. We're all in this together.
Sabine,
ReplyDeleteThis is something that's been bothering me since I started reading your (awesome) blog: there is a huge disparity between the quality of your posts (great)and the quality of readers' comments (some verging on insane). Your patience really amazes me.
I suppose I have to apologize now for having made an off-topic comment myself.
Pamplemousse,
ReplyDeleteThanks for your comment... You see, I myself have given up reading comments on blogs or news items, and I very rarely add one myself. The reason is that comment sections tend to be populated by people who have an abundance of time to complain or to spread their own believes, and I don't have that time. The same is true, I am guessing, for most scientists during working hours. (And I'm not often online after hours.)
When it comes to my own blog, I have a choice. I can either entirely block all of the low quality comments, or I can treat it as an opportunity for education. It's not that I thought about this as a deliberate decision in advance, but it just seems I've taken the latter option. It also helps me to not lose perspective. Sometimes when I write a post I think everybody knows that anyway, why bother explaining it? Then the comments come... Best,
B.
johnduffieldblog:
ReplyDelete"Now, are you trying to tell everybody that E=mc^2 is wrong?"
The complete equation is E^2 = m^2 c^4 + p^2 c^2. The rest mass of a body is an invariant, also in general relativity. It's a scalar; it's *constructed* to be invariant. As I said above, and as Phillip also explained, the word "mass" is today almost exclusively used to mean "rest mass". Gravitational potential energy is an ill-defined concept in general. All these are facts that you can easily find in any standard textbook. If you have trouble understanding basic physics, I suggest you consult one of the forums that address such problems.
Unless you have something to comment on the content of this post, I'll delete all further off-topic comments of yours. Best,
B.
Eric: Thanks.
ReplyDelete/*When you trap that massless photon in a mirror-box body it increases the mass of the body.*/
ReplyDeleteAfter all, it was proven experimentally many times: when we absorb the gamma ray photon in atom nuclei, then the atom nuclei become heavier, which can be measured with mass spectrometer.
There is absolutely no discussion about it.
Nobody doubts that energy can be converted into mass or the other way round. What I'm saying is that the energy of a body is not the same as its rest mass. I've now said it twice and this is the last time I'll repeat it: The mass of the photon that the paper is referring to is of course the relativistically invariant restmass. It is not the same as the "energy content" of the photon. Energy isn't an invariant, the invariant mass is an invariant.
ReplyDeleteSabine: with respect, my understanding of relativity is very good. For example there's a flip-flop between the momentum and mass terms of E^2 = m^2 c^4 + p^2 c^2 in gamma-gamma pair production and annihilation. The initial photons have no mass, then in the rest frame of the electron and positron, there is no momentum. When they annihilate, it's back to momentum only. If you'd prefer to talk about such matters offline to avoid unseemly on-blog dissent, please feel free to email me on my name (all one word) at btconnect dot com.
ReplyDeleteYour comment noted Zephir.
John Duffield
Hm. I have an observation, though. Historically, the relativistic derivation of the velocity-dependence of mass was obtained in the context of electrodynamics, that is, for the electron mass (it actually arises when you define the relativistic 4-force, based on the Lorentz force). The derivation was later generalized to arbitrary forces in relativistic mechanics, by Lewis and Tolman in 1909. But then you should *postulate* that a particle must have a momentum vector parallel to its velocity and a scalar kinetic energy in *such a way* as to obtain conservation laws, and for that to happen, no momentum, energy or heat should arise during interaction between particles. I am not sure how to interpret/reconcile the case for a massive photon, specially on the issue of self-energy, because then the force per unit volume will seem to imply a non-invariance of the rest mass... I could be confused here, though. I have not read the paper in question yet.
ReplyDeleteBest,
Christine
"no momentum, energy or heat should arise" -> "no transfer of momentum, energy or heat should arise"
ReplyDeleteJohn,
ReplyDelete"with respect, my understanding of relativity is very good. "
After you proclaim that you doubt that I "know any physics," you want "respect" from me? Sorry, but you've lost my respect in your first two comments, and you'll have a hard time getting it back. If your understanding of relativity is "very good" then I think it should be very clear to you that there's nothing wrong with the treatment in the paper and that it's not "junk" as you called it. It would probably help if you'd read the paper. Respectfully,
B.
Hi Christine,
ReplyDeleteSorry, I don't know what you mean. The kinetic energy isn't a scalar, or possibly you mean something else by scalar than I do. When I say scalar I mean a scalar in the tensorial sense and that is invariant by definition. Best,
B.
Hi Sabine,
ReplyDeleteSure, kinetic energy is not a scalar, it changes by coordinate transformations.
I'm posing a different question, which is rarely mentioned in textbooks. As I wrote, the relativistic derivation of the velocity-dependence of mass was originally obtained in the context of electrodynamics. So it is a question of how to generalize it for the whole mechanics, not necessarily charged particles like the electron under the Lorentz force.
So there is an argument by Lewis and Tolman, which you can find in the book by Pauli, section 38. It does refer to a scalar kinetic energy as an initial postulate in order to conciliate with the electrodynamics results, namely they show how Lorentz invariance is necessary at the end.
For me the "thought experiment" by Lewis & Tolman is not trivial and a bit confusing. But re-phrasing my question as an exercise: show that there is no conflict on the general derivation of Lewis & Tolman in the case of a massive photon.
That is either a highly trivial exercise, or a rather deep result.
Best,
Christine
A clarification can be seen here (see item 4).
ReplyDelete(Which is a translation of a paper by Epstein, 1911)...
Hm. I'll have to take a look opportunely.
Christine
John, Philip, Sabine,
ReplyDeletethere is still a lot of confusion of what is meant by mass.
Maybe this paper helps to clarify things a bit.
"The complete equation is E^2 = m^2 c^4 + p^2 c^2. The rest mass of a body is an invariant, also in general relativity. It's a scalar; it's *constructed* to be invariant. As I said above, and as Phillip also explained, the word "mass" is today almost exclusively used to mean "rest mass"."
ReplyDeleteExactly !
Christine: perhaps the Lewis and Tolman derivation doesn’t lend itself well to photon mass because it features length contraction, which doesn’t work the way most people think. And because photons are waves wherein E=hf and p=hf/c applies: the λ=h/p photon wavelength is not length-contracted to zero because that photon is moving at c relative to you.
ReplyDeleteHowever you can perhaps see a hint of something “deep” in § 4. Appendix: Momentum and mass transformation. Have a read of Light is Heavy by van der Mark and ‘t Hooft (not the Nobel laureate) for something that might make it clearer. Also google on “photon effective mass”.
When you slow down a massless photon to less than c in say glass, it exhibits a small “effective mass”. When you trap a photon in a mirror-box, it’s still moving at c, but in aggregate, its overall speed with respect to you is zero. Now all of its energy-momentum is indeed effective as mass. The box is harder to move. Then when you open the box, it’s a radiating body that loses mass. Hence in Einstein’s E=mc² paper you can read ”if a body gives off the energy L in the form of radiation, its mass diminishes by L/c²” and ”if the theory corresponds to the facts, radiation conveys inertia between the emitting and absorbing bodies”. Hence light is “heavy”. Try to envisage a symmetry between momentum and mass wherein momentum is resistance to change-in-motion for a wave propagating linearly at c, and mass is resistance to change-in-motion for a “standing” wave going round and round at c, but initially going nowhere with respect to you. Don’t forget the wave nature of matter, or that in atomic orbitals electrons “exist as standing waves”. And keep an eye out for Frank Close referring to the Higgs field as a relativistic aether , the idea being that the Higgs field is responsible for 100% of proton mass rather than 1%.
Sabine: apologies for my unkind remark.
Markus: that’s a good paper. I’ve seen it before, but I confess I tend to just say E=mc² because people are generally more comfortable with it.
This experiment gives somewhat more realistic take on life-time of photon.
ReplyDeleteJohn:
ReplyDelete"that’s a good paper. I’ve seen it before"
I mention it because it helped to resolve my own confusion on the topic.
"but I confess I tend to just say E=mc² because people are generally more comfortable with it."
I see.
I have decided to follow Einstein's 1948 advice because things are much clearer to me this way.
In particular I do not have to worry about the interpretation of the mass contraction formula any more, because contrary to length and time contractions it's hard to understand it in geometric terms.
Best.
Markus: Me too. Hence my use of quotes round "relativistic mass". This article by Philip Gibbs gives a fair account of mass in special relativity. It mentions Einstein's 1948 advice:
ReplyDelete"It is not good to introduce the concept of the mass M = m/(1-v2/c2)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than 'the rest mass' m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion."
You do have to be cautious about invariant mass with respect to the mass deficit though. All interesting stuff.
John,
ReplyDeletethanks for the interesting link.
"You do have to be cautious about invariant mass with respect to the mass deficit though." - I had to think that through, but my guess is that you can consider it in a rest frame (only involving rest masses) where the two world views on mass coincide anyway ($E = mc^2 = E_0 = m_0 c^2$), so in a way this is separate problem. But yes, a constant rest mass isn't "holy", e.g. also when you allow for particle creation and annihilation.
John,
ReplyDeleteThanks. Apology accepted.
Heeck gives the lower bound of the photon lifetime as 3 years in the restframe of the photon. This corresponds to a mass of 10^(-18) eV.
ReplyDeleteShouldn't 3 years be the upper bound?
Doesn't a massless photon decay immediately in the restframe of a photon? This is based on the idea that time does not exist for a massless photon.
Correction: I understand that mass is required for a particle to decay. It might have been better to say in reference to Heeck's letter that 10^18 years is zero time in the reference frame of a massless photon. Therefore, isn't 3 years the upper bound rather than the lower bound?
ReplyDeleteShouldn't 3 years be the upper bound?
ReplyDeleteCan anyone answer this and provide an explanation?
The lifetime is 3 years or longer. The longer the lifetime, the less relevant the difference to infinite lifetime, thus the bound gets weaker. The conversion from the restframe of the massive photon to our restframe depends on the mass.
ReplyDeletePhillip,
ReplyDeleteRe the graviton mass, you might find this useful. Best,
B.
Bee:While it's an interesting analysis and useful, it isn't much of an insight in the sense that it doesn't really explain anything because nobody suspects the photon to be unstable to begin with
ReplyDeleteIt's extremely hard for me as layman to see only the sun in it's glory and what we see in the daylight is somehow not answerable while we see photosynthesis work from the plants at night. So photosynthesis as an example of the photon process, as a converse side of entropic valuation in the WMAP.
So while reductionism is seen as energetic valuations down to such limits, an examination as a decay product, seeks to see expressions in photon exchange of some energetic valuation, from which an emergent point as an entangled process is exemplified?
Such frequency establishes an signature of elemental consideration then, as cosmological configuration is in expression from that emergent point?
Of course here, the combination of light and gravity is being sought here as an established process toward recognition of such signatures, as a "gravitational emergence" in photon expression. Consciousness even.
Best,
So what emerges?
ReplyDeleteThoughts and statements that occur to me while looking at a reductionist world, need a comparative analogy that exists in nature? So statements people make are sometimes revelatory to such a young mind examining, for why nature can confuse us when anomalies emerge through or observations. Can that even happen, then, that by consensus such examinations are not considered. So one said that is not science.
So I have to look a statements that cover qualitative correlations and the message that it would imply.
Any chemist will tell you that with no chemistry, there would be no biology, and without biology there would be no psychology.....The periodic table would have only one entry, chemistry wouldn't exist, animals made of biochemicals wouldn't exist Why there would be no chemistry without relativity:)Borrowed words.
Best,
The elementary signature, is it's rest mass?:)
ReplyDeleteSabine,
ReplyDeleteThanks for your reply.
So, if the upper bound of a photon's mass was decreased to 10^(-27) eV from 10^(-18) eV, its lifetime would increase to over 3 yrs? I guess I am not following the math.
What if the upper bound decreased further to 10^(-60) eV or to zero mass, would the lifetime keep increasing in the photon's restframe? What is the lifetime of a massless photon in it's own restframe? I understand it to be infinite in Earth's restframe, but I thought it was zero in it's own restframe. Are you saying it's infinite? That's screwing up my understanding of special relativity (massless twin leaves Earth at the speed of light...).
I'm a lay person, so I completely understand if you would rather not waste your time on me. Thanks again for your reply.
James,
ReplyDeleteNo, it's the other way round. The bound is 3 years in the restframe of the photon. To convert the 3 years into a frame where the photon is not in rest but has a certain energy, you need to know the mass. If you want to know eg the lifetime of photons in the visible range, you know the approximate energy. The smaller the mass, the larger the boost you need to get to that energy, and the larger the lifetime at that energy. Best,
B.
Could it be possible for a massless particle to "decay" to some other massless particle (or for it to oscillate between these two particles)? I know that massless particles lack rest frames (and don't experience the passing of time), and lifetimes are measured in rest frames, but still, is that completely ruled out?
ReplyDeleteThough I guess everyone would have to measure the same lifetime in their frame of reference (unless there is a preferred frame), which isn't compatible with special relativity.
ReplyDelete