The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
I'm looking for a totally mathematical description of Minkowski space, and I'm unable to find one. The wikipedia page says that its a 4-dimensional vector space equipped with a bilinear norm. However, I want a mathematical definition of what that vector space is. Is it R^4? Taabibtaza (talk) 08:11, 24 February 2024 (UTC)[reply]
No. Minkowksi space is a vector space, period. It is defined by the vectors that live in it; to start with, these are abstract objects. They can be related to through an isomorphism, i.e. a mapping from those abstract vectors and operations to, I'll say, matrices ( or ) in . In fact there are (infinitely) many such isomorphisms distinguished by the choice of a vector basis; physically that corresponds to the choice of reference system. This distinction is in my view essential for a proper understanding of relativity: there is an absolute space-time, mathematically represented by Minkowski space, in which physical quanities, e.g. four-momentum, are given as vectors, geometrical objects, that are what they are independent of any observer. When we measure these things we measure vector components in a specific basis, our reference system, given by the experimental setup. These components (e.g. energy or one of the three spatial momentum components) are scalar products of the vector with one the basis vectors and they are relative in the sense that they depend on the choice of the basis vector, i.e. the reference system. --Wrongfilter (talk) 08:24, 24 February 2024 (UTC)[reply]
I'm just starting to learn Minkowski space, so I'm assuming we have already chosen a coordinate system for Minkowski space such that it's origin corresponds to an inertial observer who has three axes attached to him such that the 3 spatial coordinates of any event correspond to the coordinates he will measure for that event using the axes that are attched to him, and such that the time at any event in spacetime corresponds to the time on our inertial observer's clock (whose clock was set to 0 at an arbitrary time) when that event happens. This is how the wikipedia introduces minkowski space (I'll learn how to shift to a different perspective later). Also, changing the basis while not changing the vector space won't won't allow us to change to an arbitrary frame of reference since our origin is still the same. Each vector space has a UNIQUE origin by the very definition of a vector space. The origin is the vector you get by multiplying any vector by the scalar "0". Anyway, now that we are done with the physical interpretation, think of Minkowski space mathematically.
Is that vector space R^4? If not, you have to give a definition of that vector space and its elements, you can't just say that vectors in that vector space are abstract objects. That's not how math works, even abstract objects have definitions. For example the elements of R^4 are defined as ordered quadruples of real numbers. Taabibtaza (talk) 11:01, 24 February 2024 (UTC)[reply]
First up, I'm a physicist with an incomplete and probably not too solid a background in the relevant mathematics, so I may be shaky in my use of terminology and incomplete in my perspective on these things. With that out of the way, back to pretending that I understand something. A vector is simply defined as a member of a vector space. It is defined by how it behaves, not by what it is. For instance you can show that the ordered quadruples (I called them matrices above, but ordered quadruples is better) with the appropriate definition of addition and multiplication by real numbers form a vector space. Therefore is a vector space and the ordered quadruples are vectors. Add the appropriate definition for the metric and you can show that with that metric is a Minkowski space. Next comes the physical assumption that space-time with events and separations between events can be described as a Minkowski space (if you insist in thinking about the origin, maybe as a class of Minkowski spaces because the origin is physically irrelevant by virtue of the relativity principle). Thus we have two spaces that are both Minkowski, therefore these two spaces are isomorphic, therefore we can map space-time to , and it will turn out that we can do that in infinitely many ways (through Lorentz transformations). Maybe more directly to your question: (1) Mathematically, a Minkowski space is an abstract space with certain properties. (2) Physically, in special relativity, Minkowski space is space-time, the vectors are separations between events. I guess that on this abstract level, four-momenta form yet another Minkowski space. (3) is another, extremely useful, Minkowski space. This is of course the main mathematical structure to apply the theory.
I don't think, actually, that you need to worry too much about these things (subtleties?) while learning, but it's good be aware that they exist. Note incidentally that in quantum mechanics it is more common to say things like "Spin states form a Hilbert space", implying that Hilbert space is an abstract mathematical structure that can be filled with physical content in various ways. In relativity "Minkowski space" more uniquely associated with a particular physical context, viz. space-time. --Wrongfilter (talk) 12:05, 24 February 2024 (UTC)[reply]
Helium would make it a bit more opaque by adding absorption lines. Also HeH+ would add opacity. More energy could be derived from pure hydrogen. A pure hydrogen star could form bigger than when mixed with helium. Graeme Bartlett (talk) 21:21, 24 February 2024 (UTC)[reply]
And note that some Helium (and possibly a trace of Lithium) was formed by the 'Big Bang' (the last I heard), so a completely pure Hydrogen star is never going to have happened. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 22:03, 24 February 2024 (UTC)[reply]
Yes I was aware. This does bring up the question of 1st generation stars vs. non-existent ones that are exactly the same except they ignited with exactly 100% isotopes of H and He in 1st gen star ratios. Could current tech tell them apart and from what distance? Sagittarian Milky Way (talk) 22:26, 24 February 2024 (UTC)[reply]
All stars have spectral lines and specs: Mass, luminosity and magnitude, radius, variability, isotope and element ratios, color indexes like blue minus V and U minus blue... Different starting compositions would have different specs or spectra at the same age and starting masses, the question is how easy could they be told apart with 6 orders of magnitude less non-1st row element content than the Sun vs infinite orders of magnitude less. Sagittarian Milky Way (talk) 06:46, 25 February 2024 (UTC)[reply]
BB is correct. An arthropod is invertebrate animal of the phylum Arthropoda, characterized by a chitinous exoskeleton and multiple jointed appendages. Humans do not qualify. Philvoids (talk) 12:37, 26 February 2024 (UTC)[reply]
Is the doppler effect effected by any acceleration of the reciever and/or the source, according to special relativity. In other words, let's say you have two point objects S and R (for source and reciever) which will only move along a straight line. Let's say that at t=0 (from the perspective of R) S emits a photon of wavelength l. So, when the photon will reach R, will the photon be of wavelength l(1+ v/c)γ where γ is the lorentz factor, and v is the velocity between the inertial frame of refrence in which R is stationary when it recieves the photon and the inertial frame of reference in which S was stationary when it released the photon. Taabibtaza (talk) 19:43, 26 February 2024 (UTC)[reply]
When it receives the photon. So, when there's a stream of photons being emitted from S, their wavelengths as observed by R will change as the velocity of R changes. Also: "Is the Doppler effect affected...". --Wrongfilter (talk) 20:27, 26 February 2024 (UTC)[reply]
I'm only talking about one photon, can we use the same formula for it, as what we use for the relativistic doppler effect if there is 0 acceleration. Taabibtaza (talk) 02:13, 27 February 2024 (UTC)[reply]
I literally learned the effect and the formula I presented from that article, I'm just wanting to know if we can use the same formula if there is acceleration (and what would v mean in the formula in that case). Taabibtaza (talk) 02:17, 27 February 2024 (UTC)[reply]
But S will accelerate after the the release of the photon, so shouldn't "v" be the velocity of R with respect to the inertial frame of reference in which S was stationary when it released the photon? Taabibtaza (talk) 20:06, 2 March 2024 (UTC)[reply]
February 27
Bulletin of the Seismological Society of America in online libraries?
I'm trying to make the cited sources in White Wolf Fault as accessible to readers as possible but I haven't been able to find this journal in the Internet Archive, JSTOR or Wiley. HathiTrust only has the issues from 1911 to 1926. The Seismological Society of America's website has has the issues cited, but it requires library or institutional access. If there are other journal repositories in our Library that have it, I've missed them. I'm stumped. If someone can help out, future readers will be better off...
I do not at all mind having to hunt down the article in the proper repository (keeps me from getting lazy, lol). Just point me in the right direction and off I'll go! Oona Wikiwalker (talk) 07:19, 27 February 2024 (UTC)[reply]
The cited articles are also available online for a (hefty) fee at GeoScienceWorld; see here and here. The fee imposed suggests that the publisher enforces its copyright. --Lambiam11:38, 27 February 2024 (UTC)[reply]
But Library access is within copyright restrictions because they pay for access for their patrons. That's what I'm inquiring after; did I miss a collection within our library that contains this journal?
I've checked the Wikipedia Library and it doesn't provide access. I have added the doi and bibcode to the cite to make it a bit easier for readers to click to the journal. Mike Turnbull (talk) 16:48, 28 February 2024 (UTC)[reply]
The version with c2 is formally correct. Another matter is whether that should be in the lede of the article (and who is "some authors"?). I don't think I've ever seen that convention — whenever I've seen it was due to setting , in which case . --Wrongfilter (talk) 18:13, 27 February 2024 (UTC)[reply]
I admit I'm quite surprised. The four momentum is (p0, p1, p2, p3), traditionally written also as (E/c, px, py, pz), hence: p0 = E/c. On the other hand, both sides of the current expression (in the lede): p0 = E/c2 don't have the same units, so how can it be correct? HOTmag (talk) 18:29, 27 February 2024 (UTC)[reply]
You are allowed to avoid answering my last question, which has nothing to do with the article. Actually, regardless of the article, I was quite curious to know whether you also considered the third statement p0 = E/c (which had been my correction before it was reverted): as "weird", while I assumed you accepted the two statements it followed, and that's why I asked you my last question, regardless of the article. But again, nobody forces you to answer my last question. HOTmag (talk) 19:44, 27 February 2024 (UTC)[reply]
So (back to article), my correction p0 = E/c in the article was right, and it didn't have to be reverted to p0 = E/c2. That's what I wanted to know from the very beginning: whether the user who reverted my correction was wrong (as I thought). HOTmag (talk) 22:26, 27 February 2024 (UTC)[reply]
No you were wrong in that context. That bit has been removed now from the article because what was there isn't common and as shown here will just cause confusion. It is best to have all the elements using the same units. NadVolum (talk) 23:25, 27 February 2024 (UTC)[reply]
I haven't asked about whether the context justified mentioning this formula there, nor about why the whole formula was eventually removed. I've only asked about whether, my replacing the original formulation p0 = E/c2 by the correct formula p0 = E/c, had been a better version than the original version p0 = E/c2. In other words, I've wanted to know if the user who reverted my correction p0 = E/c back to the original version p0 = E/c2 made the article worse. I think they did, and I've wanted to get your opinion about what I think. All of that is quite clear in my first post, that had actually been posted before the whole formula was eventually removed. HOTmag (talk) 08:07, 28 February 2024 (UTC)[reply]
The original statement was a conditional, with an "if... then..." structure. The formula that you changed was in the "then..." part, and was correct under the condition stated in the "if..." part. By changing the "then..." part without taking into account the "if..." part, you introduced an error, and the other user was right in reverting your edit. --Wrongfilter (talk) 08:38, 28 February 2024 (UTC)[reply]
I got it now. Thank you. I understand now that the original "then" part (before I changed it) would be correct assuming that the "if" part were correct. However, the "if" part really introduces a definition I've never come across, and should be omitted. HOTmag (talk) 09:52, 28 February 2024 (UTC)[reply]
Parachutes and terminal velocity
Hi, I'm currently doing a lab using a small model parachute. As I currently understand it, the terminal velocity can be calculated by equating the weight of the object to the drag force, given by the drag equation. My question is, how can I use this terminal velocity to estimate the time it would take an object to fall a set distance? Do I assume the object starts out at terminal velocity? Do I use a SUVAT equation? Thanks! ARandomName123 (talk)Ping me!23:05, 27 February 2024 (UTC)[reply]
Take the height of the fall, divide by the terminal velocity. That gives you a rough answer. So if the falling object gets close to terminal velocity after just a short fraction of the fall (for a human with a parachute, around 50 metres) and the fall is short compared to the scale height of the atmosphere (around 7 kilometres on Earth; else the changing density changes the terminal velocity) and you don't care too much about the second digit of your answer, this may be good enough. Otherwise, you have to integrate the equations of motion. PiusImpavidus (talk) 11:18, 28 February 2024 (UTC)[reply]
In the following, we only consider the motion of an idealized point particle along a vertical line. We use the variables for time, for the particle's position above , for its velocity and for its acceleration. Furthermore, stands for its mass and for the acceleration of gravity.
The force exerted on the particle is the sum of the forces due to drag and to gravity:
in which whose form given by the drag equation, contains a yet unknown constant Its value may be determined if the terminal velocity is known, as follows. Terminal velocity is the limit velocity, approximated arbitrarily closely as the acceleration tends to which means (since then tends to 0) that the upward drag force cancels the downward gravitational force, or It follows that so now we have
This differential equation, combined with the condition that at time the moment of release, is solved by:
Integrating again, with the condition that at time the particle is released from a given height results in:
Solving for gives the time it takes the particle to reach ground level:
in which stands for the inverse hyperbolic cosine. Asymptotically, as tends to infinity, this is equal to
Two runways in series instead of in parallel (i.e. 30 Back for landing 30 Front for takeoff 12 Back for landing 12F takeoff if wind azimuth near 120) and the paths never have to cross even in the air. And another two in series at right angles that they switch to if crosswind's too strong. In extreme both engines exploded on fire emergencies they clear all runways much easier landing.
Space at existing airports within metropolitan areas is obviously one limiting factor. With completely new builds in rural areas, it could work. Airports are also always looking at ways of minimising noise on the ground. Your proposal would stretch the area affected by aircraft noise to twice the distance. HiLo48 (talk) 02:48, 28 February 2024 (UTC)[reply]
I assume you mean noise due to the flight paths of the airplanes. But besides noise, different flight paths from different runway directions may also limit construction height near the airport or simply not be possible due to current constructions. See also [1][2] and consider whether this works so well under the proposal. Nil Einne (talk) 09:43, 28 February 2024 (UTC)[reply]
Well Denver had forced land, forced altitude and enormous property in the middle of nowhere but they made this (one of the most efficient layouts possible). DenverSagittarian Milky Way (talk) 04:21, 1 March 2024 (UTC)[reply]
Even away from metropolitan areas flat land suitable for an airport may be at a premium and the shape of the suitable area limits the possibilities. The angles of runway exits were not much of an issue until the introduction of heavywide-body aircraft, which take longer to reduce speed and don't turn as readily as lighter aircraft. If the space is available, an L-shaped two-runway pattern offers the same crosswind-avoidance advantage as a cross while having a decreased collision risk and giving an easier task in designing the whole airport plan. --Lambiam10:03, 28 February 2024 (UTC)[reply]
Also consider what happens on two inline runways when there's a departure on one and a go-around on the other: the aircraft going around will end up in the wake vortex of the departing aircraft. So the capacity of two in-line runways isn't really much better than that of a single runway. It's better to put them in a staggered configuration.
High-speed runway exits appeared in the 1950s, right at the start of the jet age with its high landing speeds and long runways. And at that time, air traffic wasn't so dense that they needed to land 25 planes per hour per runway. PiusImpavidus (talk) 11:38, 28 February 2024 (UTC)[reply]
Corporate execs want bragging rights: "My bizjet can do Mach 0.92," not "I paid the airport $600k for a high-speed taxiway at exactly the right spot, so now I save 30 seconds each time I visit HQ." PiusImpavidus (talk) 23:02, 28 February 2024 (UTC)[reply]
Depending on the quality of your surveillance system, and ignoring wakes, you need a horizontal separation of 2.5 to 3NM, or a vertical separation of 300m/1000ft. A modern jet climbs about 1000m/30000ft per minute, i.e. it has the necessary vertical separation in 20 seconds. Takeoff speed is 150knots or higher, so at least 2.5NM per minute, and rapidly increasing. So you would have the required horizontal separation in less than a minute. IIRC, the very best airports manage about one takeoff per minute, so separation is not a problem. --Stephan Schulz (talk) 21:14, 28 February 2024 (UTC)[reply]
Parallel runways usually have about 500m separation (like 7C/25C and 7R/25L at EDDF). How much you need depends on the operational flexibility you want; parallel departures require less separation than parallel arrivals. 500m additional taxi distance is less than a minute travel time. Or you could put the runways on opposite sides of the terminal (like at EGLL), so there's no additional taxi distance at all. If you have them slightly staggered (like at EDDB) instead of exactly side by side (like at EDDL), you save some taxi distance and you gain some safety during parallel approaches. BTW, LEMD gets close to your solution. It has two parallel departure runways and two parallel arrival runways, pretty much end to end, but with an angle between them. The new terminal 4 is where the runways almost join. PiusImpavidus (talk) 23:40, 28 February 2024 (UTC)[reply]
So when I see 9,000 feet on Google that won't help too often vs 500 m? We have 300 yards apart in KEWR slightly staggered and 250 yards in KSFO but those are compromises with limited land. I think KSFO might be the record for big planes in my country. LEMD looks pretty cool though I wonder why they didn't join them, Space Shuttle accidents and Gimli Gliders would have an easier land and maybe they could avoid a tire explosion or gear breakdown from trying to stop so hard. Sagittarian Milky Way (talk) 00:55, 29 February 2024 (UTC)[reply]
I suppose KEWR normally uses one runway for arrivals, one for departures, with no parallel arrivals. KSFO does parallel arrivals, but has some restrictions. The aircraft must fly side by side, so that neither can get into the wake of the other (wakes are angled in a crosswind) and the pilots must see the other plane before aligning with the runway, so they can only do parallel arrivals of planes with the same landing speed in good weather. My local airport (EHAM) has 3 parallel runways with 2.1–2.8km separation and 1.5–3km stagger, allowing uncoordinated parallel arrivals on centre and right (left is departures only, right arrivals only) even in poor weather. They didn't need all that separation for that, but the terminal had to fit in the east gap and the west gap was needed for noise reasons.
LEMD was extensively rebuilt about 20 years ago. It used to have 4 crossing runways, 3 of which were closed. The original 18/36 was almost in-line with the new 18L/36R, but was closed. I suppose interference of go-arounds with departures and available space (both land and noise) was the reason. A runway in-line with 14R/32L (the old runway left) to the NW would have led up to La Moraleja, a hill with some expensive housing. PiusImpavidus (talk) 11:41, 29 February 2024 (UTC)[reply]
Close but each direction's only one runway in a row not two. If Denver doubled like this the longest strip would be at least 32,000 feet long (16,000 like the current runway for takeoff and a downwind 16,000 for landing). Sagittarian Milky Way (talk) 21:41, 29 February 2024 (UTC)[reply]
Does a given body's total rest energy, have any other meaning, besides the body's ability to totally vanish - becoming energetic radiation?
A "kilobar" gold bar, with a rest energy of some 90 PJ, has a market value of about €60,000 or $65,000, twice that of a goldbar with half that rest energy. Transformed into "usable energy", the market value of this energy would be much higher, but we do not have the technology for the necessary transformation. The main reason for physicists to use the concept is to avoid conversions in calculating the energy balance of particle interactions, being able to use eV units throughout. --Lambiam10:31, 28 February 2024 (UTC)[reply]
The main reason to introduce the concept of rest energy (corresponding to the mass of the body) is that it is necessary to include it in the energy balance to satisfy conservation of energy. This is what enables the conversion into radiation, for example (in practice, other conservation laws may stand in the way of that process). In addition, it determines the inertia and the gravitation of the body (as mass, rather than rest energy). In addition, the internal energy (which is part of the rest energy in the case of composite systems) plays a role in the thermodynamics of the thing (where this is relevant). --Wrongfilter (talk) 10:43, 28 February 2024 (UTC)[reply]
Note I asked about total rest energy. I can understand how it's helpful in determining the value of energy of the radiation - the body has turned into, whereas the value of the body's mass is already known. However, does the body's total rest energy - also have any meaning - in determining the inertia or the gravitation (or anything else) of the body whose mass is already known?
To sum up: besides a Gedanken experiment in which a given body having a given value of mass - totally vanishes - becoming energetic radiation, can you think about any other Gedanken experiment, which involves a body having a given value of mass, and which shows how the body's total rest energy can be helpful? HOTmag (talk) 11:28, 28 February 2024 (UTC)[reply]
After the formula E=m^2 was discovered, all of us know there's no difference between a total rest energy and a total mass (times c^2). However, before discovering the formula E=m^2, we couldn't predict the value of energy - of the radiation a given body turned into, even when we did know the value of total mass - the body had had - before it became energetic radiation. This fact proves, that the meaning of a given body's total rest energy, lies (for example) in the body's ability to totally vanish - becoming energetic radiation - whereas the value of the body's total mass is already known. That said, I still wonder if this is the only meaning of the total rest energy, whereas the value of the body's total mass is already known. HOTmag (talk) 12:24, 28 February 2024 (UTC)[reply]
The terms "rest energy" and "rest mass" are practically interchangeable, except that some physicists feel compelled to express the latter in units of MeV/c2. Others have no such compunction and are happy to write, for example, "a K0 particle of a mass of about 650 Mev".[3] --Lambiam16:05, 28 February 2024 (UTC)[reply]
Yes, the terms "rest energy" and "rest mass" are practically interchangeable, due to the discovery of the formula E=m^2. However, they are not conceptually interchangeable, as I explained in my previous response you've just responded to. HOTmag (talk) 18:01, 28 February 2024 (UTC)[reply]
You keep writing . The formula is . The units don't make any sense the other way.
As Lambiam has pointed out, in some contexts, physicists will adopt a system of units where c = 1 (and sometimes h = 1, too), but you still don't square the value of the mass in that case.
I would argue that the terms "rest energy" and "rest mass" are wholly synonymous. The former indicates nothing other than, "what is the rest mass of this thing, expressed in units of energy?"
Note I'd written to Wrongfilter: "After the formula E=m^2 was discovered, all of us know there's no difference between a total rest energy and a total mass (times c^2)". So, you can figure out that E=m^2 was of course a repeated typo (in which I only kept omitting the "c" before the "2") in my reciting the best known scientific formula, which I was reciting (correctly) by heart - before I knew to count (and before I knew what it meant)...
As for your second remark: Well, it seems you didn't read all of my response (mentioned above) to Wrongfilter. If I'm wrong and you did read all of it, so let me put it this way: If Einstein thought like you that those terms were synonymous, he wouldn't feel he had to prove the equivalence between those terms (up to c^2). HOTmag (talk) 23:21, 28 February 2024 (UTC)[reply]
But he DID prove it, and it's been experimentally verified continuously since then. We don't have to behave as though that hasn't already been thoroughly established. PianoDan (talk) 23:26, 28 February 2024 (UTC)[reply]
Yes, he proved it. But this fact doesn't mean those terms are synonynous. Just as the terms "101+10" and "110+1" are not synonymous, even though it's provable that 101+10 = 110+1. Unless we disagree about what "synonymous" means. It seems that you're coming from a practical standpoint, whereas I'm coming from a conceptual standpoint. HOTmag (talk) 23:34, 28 February 2024 (UTC)[reply]
If you combine the definition "having the same meaning" of synonymous with the definition "the denotation, referent, or idea connected with a word, expression, or symbol" given for meaning, we can observe that the mathematical expressions "101+10" and "110+1" are two names for the same referent (the integer 111) and so are synonymous in at least some sense. Viewed as strings, formulae, or numerical recipes, they can obviously be distinguished, in which case they have – in another domain than the integers – different referents. --Lambiam01:07, 1 March 2024 (UTC)[reply]
Yes, the same argument could be made for the apparent synonymy between tg and sin/cos. Anyway the OP (i.e. me) posted this thread while assuming that tg and sin/cos are not synonymous. And if you wonder now whether, then...no, synonymy is not the shortest word containing at least three Ys. syzygy is shorter :) HOTmag (talk) 09:43, 1 March 2024 (UTC)[reply]
For two pairwise interacting point particles, the gravitational potential energy is given by
without making any distinction between Classical mechanics and Relativistic mechanics.
I wonder if this definition leads to a vicious circle in Relativistic mechanics, since: According to the formula the whole mass of a given body already contains also the body's gravitational potential energy defined above by the whole mass whose content contains also so isn't it a vicious circle?
Apparently, it's like when someone mentioning A - who is asked "what do you mean by A" - defines A by B, but when asked "what do you mean by B" - they define B by A. In our case, the whole mass depends (according to the formula also on the energy depending (according to the formula ) also on right? Unless I miss something here.
However, if our article (in the first chapter) did have to make a distinction between Classical mechanics and Relativistic mechanics, then how does the latter avoid the vicious circle mentioned above? HOTmag (talk) 08:42, 29 February 2024 (UTC)[reply]
I think the answers are yes and yes. This is what makes General Relativity nonlinear and very hard. This is related to the question whether gravitational waves could exist in GR — Einstein showed that linear waves could exist in the weak-gravity limit (that's easy, and in this regime your questions are quantitatively irrelevant), but had doubts himself whether waves could actually be generated in the strong-gravity regime. This question was only settled in the 1960s. --Wrongfilter (talk) 12:17, 29 February 2024 (UTC)[reply]
If (as you say) the answers are yes and yes, i.e. a given body's gravitational potential energy, both contributes to the body's total mass, and also depends on the body's total mass, so it follows logically that the gravitational potential energy, both contributes to itself (it being a part of the total mass), and also depends on itself (for the same reason). Isn't it like Baron Münchhausen who saved himself from drowning by pulling up on his own hair? In other words, isn't it a vicious circle? All of that reminds me of the algebraic equation: x=x+1...
Indeed, sometimes defining an object by itself does not lead to any vicious circle, e.g. in the algebraic equation: 2x+1=x+2, but that's only because we can prove that this equation has a solution (actually a unique one). However, how can we be sure that the interdependence between the total mass and the gravitational potential energy leads to no contradiction, as opposed to the case of the interdependence of the sides of the equation x=x+1? In both cases, an object depends on itself !
This is my wonder from a logical viewpoint. But, let's put logic aside, and get back to physics. My practical question is: Can General relativity describe the gravitational energy, in such a way, that this gravitational energy will only depend on the mass's components other than its gravitational potential energy component? Just as we can do the same in the algebraic equation: 2x+1=x+2, by replacing it by a direct equation x=1, i.e. so that x will only depend on a pure number. For simplicity, let's focus on the two object case, assuming that the whole universe only contains two electrons alone (or any two uncharged objects alone). HOTmag (talk) 13:15, 29 February 2024 (UTC)[reply]
Gravitational potential energy is not associated with just some of the several bodies involved; it is a quantity associated with a system of bodies in gravitational interaction. Inasmuch as one may want to consider an increase in gravitational energy as an increase in mass, it is the mass of the whole multi-body system. --Lambiam11:09, 1 March 2024 (UTC)[reply]
Of course, but I don't see how all of that has anything to do with how General relativity avoids the vicious circle I asked about. 15:09, 1 March 2024 (UTC) HOTmag (talk) 15:09, 1 March 2024 (UTC)[reply]
I'm still not sure whether I fully understand what question B actually means. I'll start with what I think I know for sure (this is essentially weak fields), and then indicate what I think is the case in the general, strong-field case. Taking the equation for the gravitational energy as it is, U is the energy that we need to supply to separate two bodies of mass M and m, initially at distance R from each other, i.e. move both to "infinity". This is a binding energy, it is negative, and causes a mass defect, so that the total mass of the system is less than the sum M + m. If the system (say a binary black hole) is gravitationally bound, the mass defect is — only half of the gravitational binding energy because there is kinetic energy that keeps the system from collapsing, and the balance between gravitational and kinetic energy is given by the virial theorem. Therefore the total mass is . This is not a fully general relativistic treatment of the problem; I don't know what the terms "gravitational energy" and "total mass" would mean in GR; they are integrated quantities and integration in GR is a bit of a mystery to me. Having said that, the gravitational field (the metric) carries energy (gravitational waves certainly do transfer energy) so it enters the energy-momentum tensor, and the field/metric consequently appears on both sides of Einstein's equation. There are self-consistent solutions for this, which are hard to find, but they exist and the question is not a logical conundrum or "vicious circle". If you're not happy with this sort of answer (I expect your aren't), you'll have to find an actual expert or dive deep into the GR literature. --Wrongfilter (talk) 16:06, 1 March 2024 (UTC)[reply]
I don't get why the mass equivalent of the potential energy is not added to the rest masses of the components, like ? --Lambiam11:11, 2 March 2024 (UTC)[reply]
The binding energy is negative, as energy needs to be lost for the system to form a bound state, or conversely you need to supply energy to break a bound system apart. Take a comet that falls in from the Oort cloud ("from infinity"). If unperturbed, it will follow a hyperbolic orbit past the sun and vanish back to infinity. If it passes close to Jupiter, say, it may transfer some of its energy to that planet and be perturbed onto a tighter elliptical orbit. This bound orbit has a lower energy than the previous unbound orbit, hence the binding energy is negative. Similarly for a collapsing gas cloud that forms a star or a galaxy: The collapse leads to a tighter configuration only if the gas cloud loses energy, mostly through radiation (radiative cooling). This loss of gravitational energy is also what powers active galactic nuclei, quasars and such. Gravitational systems that cannot cool radiatively (e.g. dark matter) can cool through other processes like violent relaxation or in a simpler way by accelerating and losing some particles while the remainder are decelerated and become more tightly bound. In black hole mergers, this is quantitatively relevant: The models for the observed gravitational-wave events always have and end state (single black hole) of lower mass than the sum of the original black holes. Long story short, bound systems have lower total energy compared to unbound systems, hence binding energy is negative, hence the total mass is less than the sum of the masses of the constituents. --Wrongfilter (talk) 12:14, 2 March 2024 (UTC)[reply]
Using the formula we have
which means that the energy increases and decreases with distance. The highest energy is at infinity. Isn't that consistent with a tighter orbit having a lower energy? --Lambiam21:43, 2 March 2024 (UTC)[reply]
Energy of one period for alternating current
In alternating current, as at the output of an amplifier for a dipole antenna, I cannot find a direct formula giving the energy over a period. Is it simply accepted that this value is given by dividing the power by the frequency? Or is there a direct formula and if so which one? Malypaet (talk) 12:10, 29 February 2024 (UTC)[reply]
Energy is power per unit time. So "energy over a period" is just power.
The formula for that isn't so much "accepted" as it is simply the definition of the terms. Energy defined to be power per unit time. (Power / unit time) * (time period) = power. Just basic dimensional analysis. PianoDan (talk) 15:32, 29 February 2024 (UTC)[reply]
I assume they mean "energy emitted over a period", i.e. power * (time period). For a monochromatic sinusoidal oscillation this would indeed be power / frequency. --Wrongfilter (talk) 15:39, 29 February 2024 (UTC)[reply]
If you use "P" for power in watt or joule/s, "E" for energy in joule, for the unit of time of 1s and T for a period in second (), "Energy is power per unit time" is like and as what I know is , you write ...People have a great problem in distinguishing Energy (kinetic energy of an object or energy accounted for in an interval of time in an energy flow), and Power that is an energy rate (flow,etc...).
This is why I am looking for an equation directly giving the energy accounted for over a period () of a sinusoidal electric current. Energy in joule equal to W-s (watt⋅second). Malypaet (talk) 21:30, 29 February 2024 (UTC)[reply]
OK, mea culpa, I wrote that wrong. Power is energy per unit time, not the other way around. That said, it's STILL just basic dimensional analysis. (Admittedly, using the correct units does help.)
Could Neanderthals and other premodern humans been handicapped by being tongue tied? How could we find out or rule it out if it isn't preserved by fossils? Maybe by DNA?Rich (talk) 06:57, 1 March 2024 (UTC)[reply]
Neanderthal genomic DNA has successfully been sequenced and analyzed. To use this for testing hypotheses regarding ankyloglossia impacting Neanderthal speech capability, researchers need to identify the genes involved in the development of a human untethered tongue. I do not know whether this has been achieved, but I doubt it has even been attempted. Mutations (possibly epigenetic) connected to ankyloglossia are reported for the MTHFR gene and the TBX22 gene, but these have much more general roles than the development of a normal tongue. (According to sources, most cases of the anomaly (in humans) are sporadic, so I do not understand why the infobox in our article has "SpecialtyMedical genetics".) Finally, the handicap impedes the production of coronal consonants, but not of vowels and the labial, dorsal and laryngeal consonants, leaving enough phonemic space for a rich vocabulary. --Lambiam10:53, 1 March 2024 (UTC)[reply]
Manually igniting a star
In theory, if you were to go to the center of a cloud of hydrogen/helium gas in the middle of space, an area of gas that hasn't undergone nuclear fission, would it be possible to light the gas and start burning it? Assuming you have the oxygen required to actually light the initial flame, could you, in theory, start the process of fission through a simple handheld lighter? Or would the gas not ignite and simply remain how it is until it's compressed and undergoes fission that way? 71.38.197.194 (talk) 23:08, 1 March 2024 (UTC)[reply]
You mean fusion, which (like fission, its opposite) is a nuclear reaction. A chemical reaction, such as oxidation, has too little energy density to start fusion; else life would be very different. —Tamfang (talk) 00:31, 2 March 2024 (UTC)[reply]
Stars run on fusion which is multiple times more powerful per gram than fission. The scales of either are absolutely insane compared to fire (https://xkcd.com/1162/). By the time fusion starts the star would average about 4 times the density of osmium (the densest material), have a center of 5,400,000 Fahrenheit, 1,000,000 times sea level pressure and thousands and thousands of times the pressure a human could survive & 1,000 times denser than water (522 pounds per 8 fluid ounces), 44 times denser than osmium, 52 times denser than gold, 11,600 times denser than unboiled hydrogen, 816,000 times denser than air and 12,000,000 times denser than the boiled hydrogen in a party balloon. With tens of thousands of Earth masses of hydrogen crushing down with a human-fatal surface gravity 165 times Earth. Only then does the least massive protostar ignite (it's very hard to crush hydrogen to 0.01% the size of mundane maximally compressed hydrogen cause matter takes up space/void-free liquids+solids are hard to compress/electron degeneracy pressure). It's amazing science is even able to fuse thin gas density hydrogen in a medium-size building though that's cause fusion increases exponentially with temperature they just heat the easiest fuel to hundreds of millions of Fahrenheits. There's fusion reactions where if the temperature doubles the burn rate rises 68,719,476,736 times. Whatever the x is watts rise x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times (x times itself 35 times). So the lowest mass 0.06-0.07 Sun mass stars won't "use up their fuel" for 380,000,000,000,000 years while the biggest stars are thousands of times more massive but die 190,000,000 times sooner with core temperatures only about a hundred times hotter. When the core of an 8 to 30 Sun mass star "runs out of fuel" the starstuff falls almost as fast as freefall, fusion accelerates exponentially till 11 to 18 billion Fahrenheit which is when large fractions of the energy it'll ever fuse in its life (few million years) fuse in like 2 seconds and even the 100,000,000,000 to 1,000,000,000,000 Earth surface gravity 1.4 to 3 Sun mass center that remains is powerless to stop the core bounce and runway fusion explosion from escaping the gravity forever. With the light of billions of Suns despite 99% of the explosion energy "wasted" not on light but ghost particles that'd average 6,000,000,000,000 miles of solid lead before the slightest interaction. Sagittarian Milky Way (talk) 00:42, 2 March 2024 (UTC)[reply]
To the 3 questions in short: No, No, agreeing No. Less respectfully, I ask the OP to review the wrong assumptions in their question. It's reasonable to suppose that given a handheld lighter, the OP can ignite the gas from a gas cooker. The gas goes on burning with a flame that emits light and heat, and you can assume that the atmosphere supplies sufficient oxygen under pressure to maintain the required stoichiometric ratio of fuel/oxygen. Each of the underlined terms that are relevant to a cooker in your kitchen are irrelevant and as assumptions misleading to understanding the fusion process in stars. That is a subject of intense research, both because it may become a practical power source if researchers solve difficult challenges of initiating fusion (needs high pressure and energy to start) and containing the subsequent reaction, and because science predicts that we can find the origins of chemical elements in star fusion. For that we are indebted to a landmark 1957 paper called B2FH after the initials of its authors. The best use for your hand lighter at this point would be to heat cups of tea for the hard-working fusion researchers. Philvoids (talk) 11:14, 2 March 2024 (UTC)[reply]
The momenta that light carries are observed to be largely independent of its speed. For example, when light, which are particle-waves, passes from one medium to another and then back again into the original medium such as through a window pane, unless it gets scattered (see Compton scattering), it emerges with the same speed, frequencies, energies and momenta it started with. This is because the photons' momenta and energies are conserved throughout although their wave velocities differ within the different media. Conversely, the momenta of masses are known to depend on their speed. Modocc (talk) 19:55, 2 March 2024 (UTC)[reply]
Yes, I know that the momenta that light carries are observed to be largely independent of its speed. But I was asking about the invalidity of the formula of relativistic momentum: Please notice this formula does not warn: "Don't use me for calculating the positive momentum of slow light not moving in a vacuum", does it? As long as it does not, what makes it void/invalid, for such a calculation, from a formal viewpoint? HOTmag (talk) 20:16, 2 March 2024 (UTC)[reply]
On one hand, individual photons don't have rest mass. So there is nothing really to calculate. On the other hand, two or more counterposing photons do have an invariant mass to which the formula can be applied. Modocc (talk) 20:53, 2 March 2024 (UTC)[reply]
Isn't is common to regard the photon's rest mass as zero? Note that the formula of relativistic momemtum does not warn: "Don't use me when the rest mass is zero", does it? HOTmag (talk) 21:02, 2 March 2024 (UTC)[reply]
The formula + zero mass yields, simply, zero momentum. When we measure lights' momenta, its not zero, unless the system as a whole has a mass. The formula is only used for calculating the momenta of mass. Modocc (talk) 21:18, 2 March 2024 (UTC)[reply]
An isolated silvered box containing coherent photons will oscillate with respect to transference of the photons' constant momenta even if these are considerably slowed down. Clearly, the formula is not valid for modeling their momenta. Modocc (talk) 21:47, 2 March 2024 (UTC)[reply]
I agree with you that the formula is not valid for modeling those momenta. However, my question is about the formula per se: Why doesn't it warn: "Don't use me for any zero mass"? HOTmag (talk) 21:53, 2 March 2024 (UTC)[reply]
A warning is needed for the formula to be valid. Just as the well known algebraic formula "If ab=ac then b=c" must warn: "Don't use me if a=o", for this algebraic formula to be valid. Without such a warning, this algebraic formula may be wrong (e.g. in case a=0), whereas a warning makes this algebraic formula valid because it will never be used in cases mentioned in the warning. The same is true for the relativistic formula of momentum: Without the warning "Don't use me if the rest mass is zero", this physical formula may be wrong (e.g. in case the rest mass is zero), whereas a warning makes this physical formula valid because it will never be used in cases mentioned in the warning. HOTmag (talk) 00:47, 3 March 2024 (UTC)[reply]
The fact it's shown to be invalid and said to be invalid, repeatedly by me here, and by others in the literature: suffices. Modocc (talk) 01:25, 3 March 2024 (UTC)[reply]
What? Is it really "said to be invalid...by others in the literature"? I will be glad to see any source ascribing "invalidity" to this formula. HOTmag (talk) 01:46, 3 March 2024 (UTC)[reply]
"not valid for individual photons"? I've never seen this formula warn: "Don't use me for individual photons" (What about gluons?), but I've only seen this formula warn: "Don't use me for velocities not slower than C (because arithmetically one is not allowed to devide by zero nor to extract square roots of negative numbers)".
I didn't understand your second remark about the invariance: Note this formula does not demand, of the frequency or of the energy or of the momentum, to be invariant. It only demands of the rest mass (needed for calculating the relativistic momentum) to be invariant. Apparently, every photon's rest mass is invariant, so apparently this formula doesn't warn against using it for slow photons, because no division by zero (nor any extraction of square root of any negative number) is made for them. HOTmag (talk) 09:52, 3 March 2024 (UTC)[reply]
and light's momentum is thus given by E/c in accordance with its energy which is related to its frequency per the Planck relation, Furthermore, don't use for light because its momenta are not zero. From what I've read and recall at this point, slow light interaction models typically invoke either group waves, their photons' coupling to quasi-particles and/or lattice calculations involving Quantum electrodynamics. Stuff that is far too complex to expound on. Modocc (talk) 14:17, 3 March 2024 (UTC)[reply]
I already know I shouldn't use the formula of relativistic momentum, for any slow photon, because (as you explain well): "its momenta are not zero". That's obvious, and I've known all of that. I remind of you, that my question was only about why this formula doesn't warn: "Don't use me for slow photons", while all other formulas I know - do warn againt using them in forbidden cases. HOTmag (talk) 14:41, 3 March 2024 (UTC)[reply]
Thank you for the article, which actually makes a distinction between the kinetic momentum and the canonical momentum.
On the other hand, a given slow photon's momentum could be easily determined (hence defined) - as zero, if we were allowed to use the formula of relativistic momemta, which does not warn: "Don't use me when the rest mass is zero". The question is, why aren't we allowed to use this formula, when it comes to a given slow photon? HOTmag (talk) 21:02, 2 March 2024 (UTC)[reply]
(ec) What makes you think that a photon in a medium has mass zero? Please don't forget that a photon is not a simple classical particle but an excitation of a quantum field which interacts with the medium in an a priori rather complicated way. I assume that we can effectively continue to treat it as a particle, certainly with non-zero momentum, possibly with an effective mass due to the interaction with the medium, but I'm out of my depth here. --Wrongfilter (talk) 21:35, 2 March 2024 (UTC)[reply]
Don't forget, that by "mass" we actually mean "invariant mass", so if it's zero in a vacuum, it's apparently zero in any medium. Just as my own invariant mass does not change when I pass from a vacuum to any medium. HOTmag (talk) 21:50, 2 March 2024 (UTC)[reply]
Besides photons, about which we can argue whether their invariant mass may change when passing from a given medium to another, do you have in mind any other example of a particle whose invariant mass may change when passing from a given medium to another? HOTmag (talk) 22:28, 2 March 2024 (UTC)[reply]
To be clear, both Minkowski and Abraham arguments seem to assume some sort of complicated transference of the photons' momenta to and from the propagation media. Personally, FWIW, I've come to reject both arguments per Occam's razor in favor of any models in which the photons retain their momenta... But that's on me. Modocc (talk) 21:30, 2 March 2024 (UTC)[reply]
The photon, with it's rest mass of zero, always travels at the speed of light. Note that even inside a medium most of the space is empty. If "light" travels more slowly in a medium, that is not because of a slower speed of individual photons, but either because the interaction with the medium causes photons to travel a more complicated path ("bouncing between atoms"), or because the photon is absorbed and re-emitted. --Stephan Schulz (talk) 14:30, 3 March 2024 (UTC)[reply]
Perhaps individual photons do slow down, or if not at least couple with polaritons, and gain effective mass and attract each otherdue to QED interactions. Upon reading their article they used an effective field theory and I quote: "...where vg is the group velocity inside the medium,..." and a function of vg "...is the effective photon mass...". It is somewhat interesting though. Modocc (talk) 16:17, 3 March 2024 (UTC)[reply]
The Gravitational constant is usually written in terms of mass, but it seems to me it would be more naturally written in terms of energy, because after all it's energy that gravitates, not [just] mass.
In those terms the value would be: 8.2627195×10^-45 m/J. (i.e. G / c^4) Which is a much simpler unit. With this formulation of G you can just plug in the energy of a photon, for example, and get results, no need to resort to complicated curvature of space explanations.
The article on the Gravitational constant does not have it written this way - before I add it, I just wanted to see what others thought. Ariel. (talk) 22:50, 3 March 2024 (UTC)[reply]
You should only add this if you can cite reliable sources that present the constant in units of this dimension. See WP:OR. Is your unit correct? Shouldn't it be m·kg−1? Given the uncertainty in the value of G, the number of digits in the numerical value you give is unwarranted. I think it should be something more like 8.26271(19) × 10−45. --Lambiam00:18, 4 March 2024 (UTC)[reply]
is the correct unit for , although it's not clear to me why G is being divided by . It seems this is introducing a new constant G', which complicates other things, like Newton's law of gravitation would change from to As to the OP question, it should definitely not be added to the article unless there is a reliable source that presents it in this way; otherwise it would be original research. CodeTalker (talk) 01:08, 4 March 2024 (UTC)[reply]
Agreed about the significant digits, I was going to do that if I added it.
It's not a new constant, rather instead of the mass you would put in the energy of the object, and then use this form of G instead of the original. It's more correct that way because it's energy that gravitates, not just mass.
Is it really possible that this is OR, and that no one else thought of defining G in terms of energy instead of mass? That would surprise me. Ariel. (talk) 02:06, 4 March 2024 (UTC)[reply]
You can express the gravitational constant in the unit , but that doesn't change its numerical value. As long as you use SI units, the numerical value remains the same. When you divide the gravitational constant by , you get a new constant G', and if you want to that instead of G, you have to compensate every formula with another . For Newton's law of gravity, that makes things more complex. Newton's theory, which is is older and more widely used than General Relativity, suddenly gets a ridiculously small constant and a weird lightspeed to the fourth, despite having nothing to do with the speed of light. In General Relativity, these factors may compensate for others, but the theory is still full of powers of , most of which get cleverly absorbed somewhere.
That mass is equivalent to energy doesn't mean that expressing things in energy is more correct. You can just as well express energy as relativistic mass, which is the more common method in GR (not in particle physics, but that doesn't deal with gravity). Physicists tend to use what's most practical and I've never seen anyone use this constant G'.
BTW, have you noticed how the accuracy of G is improved when expressed in parsec per solar mass times (km/s) squared? That's because the uncertainty in the solar mass compensates for the uncertainty in G, the product of which is known much more accurately than either of them separately. PiusImpavidus (talk) 10:48, 4 March 2024 (UTC)[reply]
I see now that there is something called the Einstein gravitational constant, which has the same dimension as your energetic constant but differs by a scalar factor:
Accelerating (massless) gluons by the strong force
Here are two facts, AFAIK:
Fact #1. When it comes to photons, the impact of gravity on them - which actually must change their momentum (according to the very definition of "force" in mechanics), is reflected, either by accelerating them - i.e. by changing their direction (if their motion has just been perpendicular to the direction of the gravitational field), or by their redshifting or blueshifting (if their motion is parallel to the direction of the gravitational field).
Fact #2. When it comes to gluons, they have a color charge; while this color charge - with respect to the strong force, plays the same role, as the electric charge does - with respect to the electric force. Hence, the strong force has an impact on gluons.
My question is: What's the impact of the strong force on gluons, bearing in mind that gluons have no mass, so apparently the strong force can't accelerate them by changing - their speed - i.e. the absulote value of their velocity.
In other words: Is the way gluons are influenced by the strong force, identical to the way photons are influneced by gravity, even though the explanation of the impact of gravity is usually ascribed to the curved spacetime - which is not the case when it comes to the strong force?
But maybe I'm wrong, so although the gluon is massless, its speed (i.e the absolute value of the gluon's velocity) is not as "constant" as the photon's speed is, so also the gluon's speed can be changed by the strong force. Can it? HOTmag (talk) 05:07, 4 March 2024 (UTC)[reply]
Basically you are asking: Since Gluons are massless they always travel at the speed of light, so how can they accelerate/change energy? But so far as I know Gluons are virtual only, there are no free Gluons, and virtual particals do not need to conserve momentum or energy in isolation. Ariel. (talk) 06:06, 4 March 2024 (UTC)[reply]
Gluons were first detected in 1978, and I guess they were detected as free. Quark–gluon plasma, in which gluons are free, was first detected in the year 2000. Additionally, glueballs are assumed to have their gluons free.
The observations that could best be explained as involving gluons, made starting in 1978, did not "detect" gluons. One can say they were "discovered". The researchers studied the hadronicdecay of the Y bosonϒ meson and identified that this decay was probably mediated by gluons. Only in 1979 had enough evidence been collected to assign a sufficiently high statistical significance to the gluon explanation to confirm the physical existence of this thus far theoretical particle. --Lambiam12:49, 4 March 2024 (UTC)[reply]
Fact 1 doesn't work. A photon has no mass, so the gravitational force acting on it is zero. The resulting acceleration is zero divided by zero, which is undefined. In other words, classical mechanics doesn't apply to photons. Newton's laws can be bent in such a way that they predict deflection of photons in a gravitational field, but the predicted deflection is only half that predicted by General Relativity. Observations are consistent with GR. PiusImpavidus (talk) 11:18, 4 March 2024 (UTC)[reply]
Due to your comment, I realized I had to strike out the content of parentheses containing the word "force" in my original post, and that's what I've just done. However, I also realized that your comment had no impact on my question, being about gluons, rather than about photons. HOTmag (talk) 12:11, 4 March 2024 (UTC)[reply]
Before asking my question, let me begin with three facts regarding it:
1. By the classical definitions in Mechnaics, a mass of a given particle, reflects the particle's resistance to accelaration: The bigger mass the particle carries, the bigger its resistance to acceleration is.
2. By the classical definitions in Mechnaics, the angle of deflecting the light by gravitational lensing, can be regarded as a kind of acceleration, actually a normal/radial one: Indeed, it does not change the speed (the absolute value of the veloicity), yet it does change the velocity's direction.
3. By General relativity, the angle of deflecting the light by gravitational lensing, only depends on properties of the gravitational lens (actually its mass), and on the distance between the lens and the light, yet not on any propery of the light itself (e.g. its momentum/color).
Question: Am I allowed to infer - from these three facts alone (regardless of special relativity), that light has a constant mass (e.g. zero mass), independent of the light's momentum/color? For, if the light's mass varied - e.g. depended on the light's momentum/color - so that blue photons were "heavier" than red ones, then by #1 - a red photon's resistance to this photon's normal/radial accelaration - would be smaller than a blue photon's resistance to this photon's normal/radial accelaration, and so by #2 - the angle of deflecting red photons by a given gravitational lens - would be wider than the angle of deflecting blue photons by that very gravitational lens, as opposed to #3. Hence (apparently), light has a constant mass, independent of the light's momentum/color. Is my deduction correct, from a logical point of view (regardless of special relativity)? HOTmag (talk) 10:19, 4 March 2024 (UTC)[reply]
Light has no mass. But any masses going in the same direction and speed as each other would be deflected the same. The orbit of a satellite around the earth does not depend on its mass - well not unless we had one that had a mass comparable to the earth! NadVolum (talk) 10:54, 4 March 2024 (UTC)[reply]
Oops, I forgot that a given body's gravitational acceleration does not depend on the body's mass. Sorry for this huge (contemporary) mistake... HOTmag (talk) 11:05, 4 March 2024 (UTC)[reply]
Do you have in mind examples, of an exceedingly weak gravitational field (if not the weakest one), ever measured?
Regardless of the following four formulas, is there any other way (whether an empirical one or a theoretical one) to prove that light has no mass?
Denoting: the speed of light by the velocity by the gamma factor by the mass by the momentum by and the energy by the four formulas I'm currently ignoring are:
1.
2.
3. (bearing in mind that
4. (while denotes the so-called "relativistic mass").
I'm asking the question in the title, because the formulas mentioned above were originally developed/deduced for bodies slower than light, so these formulas won't be reliable (in my eyes) for determining whether light has any mass - e.g. a mass for creating a gravitational field.
Additionally, for defining force, I only use the formula (which actually defines the force as the change in momentum with respect to time), rather than the formula (which actually defines the force as the product of mass and acceleration).
I would also like to ignore the theory of elementary particles (including gauge theory), according to which photons have no mass. HOTmag (talk) 15:33, 4 March 2024 (UTC)[reply]
It cannot be proved on theoretical grounds, but in general: Rest mass is invariant (it's measured to be the same quantity in all inertial reference frames) and massless energy is not. A cloud of photons has both gravity and rest mass. Taken individually a photon's frequency is not invariant, it's arbitrary; its blueness/redness depends on the observer. There are empirical constraints on its mass, see Photon#Experimental_checks_on_photon_mass. Modocc (talk) 17:52, 4 March 2024 (UTC)[reply]
Has it ever been empirically proved that a cloud of photons has gravity/mass? I emphasize I'm asking from an empirical point of view, because I guess it would be impossible to theoretically prove this - without relying on any of the formulas mentioned above which I currently ignore. HOTmag (talk) 18:22, 4 March 2024 (UTC) HOTmag (talk) 18:22, 4 March 2024 (UTC)[reply]
Are you referring now to my new question about the cloud of photons? If you are, then - using the conservation law you've mentioned - along with the annihilation you've mentioned, how do you prove that a cloud of photons has gravity - without relying on any of the formulas I've mentioned in my first post? Note that Relativity theory allows mass to disappear and to re-appear, as long as the total energy is conserved. HOTmag (talk) 18:38, 4 March 2024 (UTC)[reply]
I didn't notice your question(s) earlier. Certainly, the photons' gravity can and should be empirically demonstrated: for example, by measuring the deflection of ultra-thin beams of energetic light which cross paths in extremely close proximity to each other. The gravitational deflection(s) will be small and these can be measured by redirecting the beams into an interferometer. In the confines of a small Earth-bound setting a ring of mirrors can be added such there are a multitude of repeated crossings of the beams (in a vacuum) that would accumulate their deflections if necessary. One would need to calibrate the beams independently and then turn them on together to see if there is an observable effect. Ideally one would repeat the experiment and vary its parameters to get a more complete picture of their interactions if any. Modocc (talk) 22:31, 4 March 2024 (UTC)[reply]
That showed that the sun gravitates light. The unanswered question is whether or not photons reciprocate and gravitate. The difficulty with that is photons are not known to interact with each other via gravity, and if they do their angular deflections would be very very small anyway, perhaps too small to detect. Modocc (talk) 02:45, 5 March 2024 (UTC)[reply]
The two words "reciprocate and" in your recent response, are needless, in my opinion. Even without them, the "unswered question" (as you name it) basically remains the same question. HOTmag (talk) 08:55, 5 March 2024 (UTC)[reply]
You cannot use relativity to derive the mass of the photon or anything else; relativity per se doesn't know what a friggin photon is. You tell relativity: "Here's a particle with mass and energy " (info comes from other theories, e.g. de Broglie for the energy), and then you ask "How does it behave?", and relativity will tell you that. For instance, it tells you that the momentum of the photon is , using equation (3), which is the only relevant equation here. --Wrongfilter (talk) 12:40, 5 March 2024 (UTC)[reply]
1. Do you think that a given frigging photon has no mass?
If you do, then my original question was:
2. How do you know that? Do you only know that by (some of) the equations I've indicated in my first post?
I can only repeat myself. All these equations are part of the theory of relativity. Relativity doesn't know what a photon is, therefore these equations cannot tell you what the mass of a photon is. You need to look elsewhere for that. --Wrongfilter (talk) 13:25, 5 March 2024 (UTC)[reply]
Let's leave personal opinions. Can science determine, that a given body - whose speed does not depend on any observer - has no mass? HOTmag (talk) 13:40, 5 March 2024 (UTC)[reply]
I've read this chapter many times over the recent years. Yet, I don't see how it answers my question, about "no mass", rather than about a mass smaller than a given tiny limit. HOTmag (talk) 15:58, 5 March 2024 (UTC)[reply]
1. Systems whose four-momentum is a null vector (for example, a single photon or many photons moving in exactly the same direction) have zero invariant mass and are referred to as massless. But unfortunately, the article doesn't explain why "a single photon" mentioned in the parentheses is an "example" of that. In other words, the article doesn't explain why a photon's four momentum is a nul vector. It wouldn't have been a null vector, had we assumed that a photon does have a (positive) mass.
2. Thus, the mass of a system of several photons moving in different directions is positive, which means that an invariant mass exists for this system even though it does not exist for each photon...For example, rest mass and invariant mass are zero for individual photons. But unfortunately, the article doesn't explain why this is true.
Consider two red photons a and b from stationary sources approaching you with velocities +|v| and -|v| from opposite directions. Now switch to another reference frame such that the source of a is moving towards you and the source of b is receding. a is bluer and b is redder. Doppler shift changes what energy is measured for each photon, but it doesn't change their total energy. That way the invariant mass of the proton and my newspapers didn't change if they were on my bike, I was walking with them, or being read by the numerous customers I delivered them to. Physicists are simply tallying (empirically) an invariant mass-energy for each particle. Modocc (talk) 19:18, 5 March 2024 (UTC)[reply]
I understand your thought experiment (a well known one), but I don't understand why you think it proves that the mass is zero. Why zero? HOTmag (talk) 20:50, 5 March 2024 (UTC)[reply]
I don't think that at all. Your 1st quote about zero mass is simply empirical, and I explained the 2nd quote to give it some context. Modocc (talk) 21:53, 5 March 2024 (UTC)[reply]
When an electron and a positron collide - annihilating each other, does the gravitational field - having been created by them - disappear?
The photons carry away the mass-energy from the collision and redistributes it to other parts of the system. So the field itself has to change (in accordance with its mass-energy distribution). Modocc (talk) 18:53, 4 March 2024 (UTC)[reply]
Their combined energy and momentum do not change but are now carried by photons. In general relativity, the gravitational field is not ascribed to individual bodies but a universal field entirely determined by the stress–energy tensor, which only depends on the distribution of energy and momentum. --Lambiam18:56, 4 March 2024 (UTC)[reply]
Is it a response to me or to Modocc?
Anyway, bearing in mind that the two original massive particles no longer exist after the light was emitted, is it carrying now the original gravitational field? If this light is not, then which entity is, in this case? Is this gravitational field carried now by no entity? In other words: does spacetime get curved, yet by no entity? I'm quite surprised... HOTmag (talk) 19:01, 4 March 2024 (UTC)[reply]
It was meant to be a response to the original question; I had not noticed Modocc's response. Your question above ignores the nature of the gravitational field iin general relativity, which is not ascribed to individual bodies. So the photons do not carry a gravitational field, but they influence the gravitational field. --Lambiam19:44, 4 March 2024 (UTC)[reply]
I thought your first response had been meant to be a response to Modocc, because of its indent. Anyway, back to your main response: Assuming theoretically - that the only gravitational field that existed in the whole universe before the collision of the two massive particles - was the gravitational field created by them, you actually claim it still exists after the collision - although the whole universe (in our theoretical case) currently contains light only - actually being the light emitted after the collision, so you actually claim we've got a curved spacetime - without any mass - we assuming that light has no mass. That's why I was so surprised, having read your first response. I'd always thought, that spacetime could only be curved by a presence of mass. Had I been wrong? HOTmag (talk) 19:56, 4 March 2024 (UTC)[reply]
I'm surprised again. If (as you actually claim) light curves spacetime, i.e. creates a gravitational field, i.e. has (by defintion) an active gravitational mass, so what's the physical meaning of a given photon having "no mass" - as commonly accepted to claim? On the second hand, if - by saying that light has "no mass" - one actually means light has no passive gravitational mass, then one actually says nothing, because the change in a given body's momentum due to the gravitatoinal force (while defining any force as rather than as ) - as well as the body's gravitational normal/radial acceleration, are not influenced by the body's passive gravitational mass - nor even by whether it's a positive one or a zero one. On the third hand, if - by saying that light has "no mass" - one actually means light has no inertial mass, then what does a given photon's inertial mass mean, bearing in mind that light is not influenced by any force other than the gravitational one (e.g. when a given photon's path is deflected when it approaches the sun)...
By the way: regarding your first ten words: "Yes, if you only consider only matter to have mass", please see the first paragraph in our article Einstein field equations: "In the general theory of relativity, the Einstein field equations (EFE; also known as Einstein's equations) relate the geometry of spacetime to the distribution of matter within it". This first claim in the lede, is sourced by a reliable source. So, would you suggest that the lede in this article be fixed somehow, e.g. by replacing "matter" by "mass" (or by "mass-energy"), along with replacing the source (ibid.) by another one? HOTmag (talk) 08:49, 5 March 2024 (UTC)[reply]
In your post above my last post, you asked about the physical meaning of "a photon has no mass", followed by some discussion on active or passive gravitational mass and inertial mass. I answered that it's the rest mass that's zero. In other words, I gave a direct answer to the question you asked just above. If that's irrelevant, you asked the wrong question.
Yes, I read your previous thread. It sounds like you don't care too much about actually understanding physics, so I assume the answer to my question is no. PiusImpavidus (talk) 18:43, 5 March 2024 (UTC)[reply]
Yes, I know that the photon's rest mass must be zero (because the photon can't be at rest). That's said, my question you've quoted, which should have been quoted in its full version: "what's the physical meaning of a given photon having 'no mass' - as commonly accepted to claim?", was actually meant to be addressed to all those physicists (not me but for example Okun) who claim that there's only one kind of mass, which I and you call "rest mass", and which they call "the mass" - i.e. "the only one mass possible". They don't make any distinction between, what I and you call "rest mass", and any other kind of mass. If you think (like me) that there is more than one kind of mass, then we can become good friends. However, as far as they are concerned, they actually claim that "a photon has no mass", without making any distinction between a photon's rest mass and a photon's relativistic mass. In their opinion, if a photon's mass is zero (e.g. when considering the "rest mass"), then any mass of the photon must be zero, simply because there's only one mass possible.
As for your second remark ("It sounds like you don't care too much about actually understanding physics"). Unfortunately, you didn't explain why you thought so (about a person who thought she could become your friend), so I can't respond to this remark with the same elaboration appearing in my previous thread. HOTmag (talk) 19:08, 5 March 2024 (UTC)[reply]
What do you mean by "the system"? The only system has been the original pair of electron-positron, which turned out to be light. HOTmag (talk) 19:00, 4 March 2024 (UTC)[reply]
Thank you for this link. I was not aware of this theoretical effect. I still wonder, though, if it has ever been empirically detected. I also wonder if it's mentioned in Wikipedia. HOTmag (talk) 19:43, 4 March 2024 (UTC)[reply]
Gravitational waves can only be created, if the curvature in spacetime is already given - the waves characterizing the manner of how this given curvature propogates in spacetime.
That said, I'm asking if the very curvature really still exists after the collision, assuming that the whole universe had only contained the electron and the positron before they collided and annihilated each other while emitting light only, about which I'm actually asking whether it can curve spacetime (along with creating gravitational waves with respect to this curvature), assuming that the whole universe no longer contains matter after the collision. HOTmag (talk) 09:50, 5 March 2024 (UTC)[reply]
In theory, all massless particles gravitate. Consider protons: Their gluons are massless but these have enough energy to account for the proton's mass which gravitates. Modocc (talk) 11:45, 5 March 2024 (UTC)[reply]
A given proton is composed of quarks and gluons. Every quark "inside" the proton (i.e. contrary to a "free" quark) carries, not only what would be its mass if the quark were free, but also another component of mass which reflects the potential energy associated with the strong force carried by the gluons. So why do you think the proton gravitates due to the gluons, rather than due to those two components of mass of the quarks contained in the proton? HOTmag (talk) 12:40, 5 March 2024 (UTC)[reply]
Oops. It was in my thoughts but I didn't type "most of" when I said they have "...enough energy to account for the proton's mass...". Note, I didn't specify why or how. Modocc (talk) 15:24, 5 March 2024 (UTC)[reply]
See Proton: "Using lattice QCD calculations, the contributions to the mass of the proton are the quark condensate (~9%, comprising the up and down quarks and a sea of virtual strange quarks), the quark kinetic energy (~32%), the gluon kinetic energy (~37%), and the anomalous gluonic contribution (~23%, comprising contributions from condensates of all quark flavors)". Modocc (talk) 16:35, 5 March 2024 (UTC)[reply]
Note, that the quarks this experiment refers to, have become "free" quarks, after the stationary proton split into several moving particles carrying "kinetic energy", like: "up and down quarks", a sea of "virtual strange quarks", and moving gluons. On the other hand, my question to you was about the quarks "inside" the stationary proton, rather than about "free" quarks outside it. The question also explains why there's a difference between, the mass of a quark "inside" the proton, and the mass of a "free" quark. HOTmag (talk) 17:53, 5 March 2024 (UTC)[reply]
Yes, the mass of the proton is the mass of the stationary proton, but the masses of the quarks and gluons are masses of those components after they became free, i.e. after the proton split into them while they (including "sea of virtual strange quarks") became particles carrying "kinetic energy". HOTmag (talk) 18:49, 5 March 2024 (UTC)[reply]
My sandwich contains numerous bound and free particles. That fact doesn't change their contributions to its mass. Modocc (talk) 19:38, 5 March 2024 (UTC)[reply]
The difference between a quark inside a proton (i.e. a bound quark) and a quark outside a proton (i.e. a free quark) is as follows: All agree that a given proton is composed of quarks and gluons. However, contrary to a "free" quark, every quark "inside" the proton carries, not only what would be its mass if the quark were free, but also another component of mass which reflects the potential energy associated with the strong force carried by the gluons.
Note that the article equates the mass of the stationary proton (i.e. before it split into plenty of particles) with the sum of the masses/energies of those many particles (including a "see of strange" ones) into which the proton split. My question to you was: Why do you rule out the following option: If the calculation had only referred to the quarks "inside" the proton before it split, the mass of the proton would have turned out to be exactly the sum of all the bound quarks inside it alone, so that its gravity would have only depended on this sum alone, without relying on any gluon. HOTmag (talk) 20:00, 5 March 2024 (UTC)[reply]
Yes, but after a proton splits into moving quarks (including "a sea of strange" ones), they become free - even for a millionth of second - until they create new hadrons. Without their being free - even for a millionth of second, they couldn't have been detected, but they have! Further, the calculation of the mass of a proton - as the sum of the masses/energies of those many particles (including a "see of strange" ones) into which the proton split, actually relies on this millionth of second - during which those particles were free (untill they created new hadrons because of the color confinement). HOTmag (talk) 20:37, 5 March 2024 (UTC)[reply]
Pumped with enough energy we have observed Quark matter that decays. That doesn't change their contributions to the proton mass. BTW, I've other things to tend to so I will be disappearing soon. Modocc (talk) 21:16, 5 March 2024 (UTC)[reply]
Let a given photon, move perpendicular to a given gravitational field created by a given star. According to General relativity, the photon's trajectory will be deflected by the angle toward the star, whereas: denotes the photon's velocity (i.e. ), denotes the universal constant of gravitation, denotes the star's mass, and denotes the distance between the star and the photon.
Question: What will the angle be, if we replace the photon by a massive particle, its properties being the same as before (except its velocity which will be slower than of course). Will the angle be a half of the angle mentioned above? HOTmag (talk) 15:11, 5 March 2024 (UTC)[reply]
There's no distinction - energy and mass are equivalent. The deflection of the light is just a much greater version of the same effect as the prcession of Mercury. NadVolum (talk) 18:54, 5 March 2024 (UTC)[reply]
Thx.
Here you've written "there's no distinction", but you've also written "No big difference" in the edit summary (see the history page), so I wonder what's more exact.
Now I'm a bit confused, because of the contradiction between your reply and the previous reply above yours. Of course, if anyone of you could supply a source (or any argument analogous to a source), I would be much less confused (if at all). HOTmag (talk) 20:23, 5 March 2024 (UTC)[reply]
The general formula, for massless and massive particles, is:[5]
where is the velocity at a large distance, before any acceleration due to the gravitational attraction. When the factor in parenthesis equals When it approaches --Lambiam10:47, 6 March 2024 (UTC)[reply]
March 6
Pathogens that increase crop yields
Is there any plant virus/bacterium/whatever that has been associated with increased crop yields? I remember seeing a paper a long time ago that a few plant pathogens (I think it was viruses) increase the fruit/seed yield of their hosts, along with the usual disease symptoms, and discussed reasons why this might be so. Note that I am asking about "pathogens"; symbiotic or commensal relationships or endophytes and mycorrhiza are not what I am asking about. Jo-Jo Eumerus (talk) 08:14, 6 March 2024 (UTC)[reply]
Not symbiosis, precisely not symbiosis. Pathogens, viruses and diseases. I saw years ago a paper discussing that sometimes, plants produce more/bigger fruits/seeds after an infection with viruses or phytoplasms, but I can't find it again. Jo-Jo Eumerus (talk) 10:00, 9 March 2024 (UTC)[reply]
Has the cubic correction to redshift versus luminosity been tested, or is it likely to be?