8 ms·

Physics Has Demoted Mass

157 points by AliCollins 9y ago

dude01 9y ago
I was just explaining to my teenage son what quarks are -- he hadn't learned about them yet. So I have to upvote this. tl;dr If you decompose matter all the way down, mass doesn't just "add up the parts to the whole". Not by several orders of magnitude!
SAI_Peregrinus 9y ago
It's very helpful to use natural units (or QCD units) when working with particle physics. Set the speed of light, the reduced Planck's constant, and Boltzmann's constant equal to 1. For QCD units also set the mass of the proton equal to 1. Then compute all the other units based on this. (There are several other systems of "natural" units also in use. All of them set the speed of light to 1, but differ in what other units are included.) A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m. Energy and mass are completely equivalent. Strong force binding energy in a proton is mass. The only bit that's not straight from binding energy is from the Higgs mechanism, which is where the mass of the electron (and the quarks themselves) comes from. Matt Strassler has an excellent explanation of that: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/how-the-higgs-field-works-with-math/ https://profmattstrassler.com/articles-and-posts/particle-ph...
- JPLeRouzic 9y ago
  This comment is much better than the article, because it tells that some of the complex relations between conceptual entities in physic, could be presented in a simplified but equivalent way. So it hints that the "problem" presented in the article is only an artifact. Henry Poincaré wrote exactly this in 1902 in "the science and hypothesis" Nous sommes acculés à la définition suivante, qui n'est qu'un aveu d'impuissance : les masses sont des coefficients qu'il est commode d'introduire dans les calculs.[0] I am also happy for the reference to the excellent blog of Matt Strassler [0] https://en.wikipedia.org/wiki/Science_and_Hypothesis https://en.wikipedia.org/wiki/Science_and_Hypothesis
- emmelaich 9y ago
  E=m ... and I=m. Or so I was taught. Inertia is not merely something that depends on mass. It is mass. (though from a brief web search, it is not typically presented as such)
  - pducks32 9y ago
    Yea physicists are in a weird spot when it comes to inertia. Everyone’s heard of it but it’s not real. To say something has mass is to say it requires a force to move it which is what we call inertia. And also the E=m equivalence isn’t as simple as may be believed. The famous equivalence is cutting out a piece that is important when talking about photons.
    - SAI_Peregrinus 9y ago
      There are certainly many subtleties. For example, I said E=M. But that's for objects at rest. For an object in motion it's E^2=m^2c^4+p^2c^2. In natural units that turns into E=m+p. The energy of a photon is hf, where h is Planck's constant and f is the frequency. That energy is entirely the photon's momentum. And that's without getting into the issues of invariant mass or relativistic mass increases...
      - evanb 9y ago
        You dropped a bunch of important squares... E^2 = m^2 + p^2 does not simplify to E=m+p.
        SAI_Peregrinus 9y ago
        You're right, of course. I really shouldn't do math in my head after midnight.
      - jdmichal 9y ago
        Unicode's got a full supplement of sub- and superscript numbers now. E² = m²c⁴ + p²c² https://en.wikipedia.org/wiki/Unicode_subscripts_and_superscripts https://en.wikipedia.org/wiki/Unicode_subscripts_and_supersc...
- bambax 9y ago
  > A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m. Energy and mass are completely equivalent. Yes but that is still true (E=m) if c<>1, since c is a constant. It's just a question of scale, isn't it?
  - throwaway287391 9y ago
    In the sense that GP states, yes, but I disagree with GP: "E=m" is wrong even if c=1 in the units you're working with since c is not unitless; it has speed units. You can't have a kilogram of energy. You need to multiply a mass by something with squared-speed units to get something with energy units.
    - chriswarbo 9y ago
      > "E=m" is wrong even if c=1 in the units you're working with since c is not unitless In natural units, c is "unitless" (AKA dimensionless), by definition.
      - throwaway287391 9y ago
        I won't say you're wrong since I stopped taking physics in high school, but this makes absolutely no sense to me. How does defining the Planck speed unit to be the speed of light suddenly make c a unitless constant? It's not "c=1" (except in informal shorthand), it's "c = 1 Planck speed unit". It seems like all sorts of things would "break" very quickly when you start dropping units. Maybe someone has a link that goes into more depth on why all of this is okay?
      - raattgift 9y ago
        The value of c is arbitrary and depends on one's choice of coordinates. However the presence of c in the line element of a spacetime like ours is mandatory. There is, however, some conceptual value in setting c to some value other than 1, which I'll return to in the last paragraph below. When using a pseudo-Riemannian manifold one uses a metric signature where (keeping it simple, cf. "metric signature" on wikipedia) coordinates on orthogonal dimensions of can take one sign, or the opposite sign and a constant multiplier. In the Lorentzian case there will be one dimension taking one sign, and one or more taking the other; the choice of whether the solitary dimension takes a + or - sign is a matter of convention or preference. Conventionally the solitary dimension also takes the constant multiplier and is called the timelike dimension. A Lorentzian signature (minuses, 1) or (1, plusses) guarantees that one can describe paths through the manifold as null, timelike, or spacelike, and this gives one a causal structure. Our universe can be well represented by a Lorentzian manifold (3, 1) or (1, 3), and this has been tested to exquisite precision. It does not tell us the value of the constant c, but we can determine that from tests of causal relations, the boundaries of which will be null. Alternatively, a massless pointlike object will always travel on null geodesics. One runs into c being set to unity in systems geometrized units in relativity often; it's very handy to have mark off coordinates as e.g. "-seconds" vs "c seconds" (-,+,+,+ aka (1,3)) as long as one doesn't mess up one's dimensional analyses (which is unfortunately easy). As a concrete example, the line element for (1,3) flat spacetime using Cartesian coordinates is dS^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2. Compare the formula for Euclidean distance in the (Euclidean flat) plane between points p = (p1,p2) and q = (q1,q2) for (x,y) coordinates: d(p,q) = sqrt((q1-p1)^2 + (q2-p2)^2), which we could rewrite as dx = q1-p1, dy = q2-p2, ds = sqrt(dx^2 + dy^2) or ds^2 = dx^2 + dy^2. In Euclidean 3-space, we add another axis: ds^2 = dx^2 + dy^2 + dz^2. In Lorentzian 4-spacetime, we have to change the sign, so ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2. Note that we are not restricted to use any particular unit of distance or system of units; dx could be in metres, miles, astronomical units, light-years, gigaparsecs or practically anything else, while dt could be in seconds or fortnights or any other handy unit of time. When setting c to unity, we do need to choose appropriate units for dx (and dy and dz ...) vs dt. In SI units, that's seconds and light-seconds, but we could use another system if we wanted. Notably we aren't restricted to Cartesian coordinates, however if we were to change to some other system of coordinates (e.g. polar ones) the line element would need to reflect that. For example, in the Lorentzian 4-spacetime case, we would write ds^2 = dr^2 + r^2 * dtheta^2 + r^2sin^2(theta)dphi^2 -c^2dt^2. Finally, when using SI units (for example), one can see very clearly that a path taken by an object moving much slower than the speed of light is totally dominated by the amount of time between starting point and finishing point, because the value of c is large. Using the (+,-,-,-) metric signature [ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2], a large c helps make it clear that lightlike paths through spacetime are shorter, and purely timelike paths (where dx=0, dy=0, dz=0, dt != 0) have extremized length, since we don't subtract anything from cdt. This is the root of the explanation of the twin paradox in flat spacetime: the twin moving quickly compared to the speed of light takes a shorter path between together1 = (x1,y1,z1,t1) and together2 = (x2,y2,z2,t1) than the twin moving slowly compared to the speed of light. In the extreme, twin A holds x=const,y=const,z=const, whereas twin B's x coordinate is only equal to twin A's at the start and end of the journey. This holds up under any system of coordinates; we could consider r=const vs changing r in spherical coordinates, for instance.
- davidmanescu 9y ago
  Equalities that involve units aren't straightforward. Not a physicist of any sort but don't all the 1's still fundamentally change the equality since they change the units? I would guess they therefore implicitly change the intuition behind these equations?
  - beojan 9y ago
    That's what makes natural units natural. Certain quantities (like mass and energy, or distance and time) are physically related such that the distinction is, in a sense, artificial. The conversion factors between the units used for these pairs of quantities are the fundamental physical constants that get set to unity in a natural system of units.
    - jessriedel 9y ago
      The distinction between distance and time is not artificial at all. It is encoded in the metric, at the most fundamental level. The existence of symmetries is not the same as equivalence.
      - coldtea 9y ago
        Parent is not talking about the "distinction between distance and time" but about the multipliers that come into play into their equations (which are artificial and based on the base units selected).
        jessriedel 9y ago
        The distinction between feet and seconds is just as real as the distinction between distance and time, and is much more real than the distinction between feet and meters. Look at the parent's literal words and pretend you did not already know what's going on. I'm well aware of the practical value and conceptual clarity of natural units. They are just not being explained well in this thread. "What the teacher really meant was..." is not a good defense when the student doesn't understand.
- lkrubner 9y ago
  A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m. That is clever, but you've already adjusted the mass to take into account the fact that you are now setting c=1. This reminds me of computer programming where, for the sake of an easy-to-read clarity, I decide to take one line of code and make it two. Basically, instead of doing this: e=mc^2 You have decided to make it two lines of code: m=mc^2 e=m I often do that, especially if I'm working with junior devs and I want the code to be as easy as possible for them to read (90% of the time). It's clever, but everyone should remember the assumption that allows this: the mass has already been adjusted to reflect the reality that of c=1.
  - johncolanduoni 9y ago
    Except you would otherwise have at least one of c, hbar, boltzman’s constant etc. in virtually every equation and now you don’t. You get a lot more mileage out of it than just simplifying one line.
- matt_wulfeck 9y ago
  > A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m Since mass also determines the speed of time to the observer, isn't a time machine basically just a thing that converts energy into mass?
  - deleted 9y ago
    [deleted]
  - lisper 9y ago
    No. Mass doesn't "determine" the speed of time (a lovely phrase, BTW). It is the observer's motion through space that does that. However, for an observer to move through time at a speed faster than 0 it has to move through space at a speed slower than c, and to do that it needs to have mass. A massless particle can only transfer energy if it moves at the speed of light. Massless particles moving slower than the speed of light cannot transfer energy, and so cannot be detected experimentally, and so do not exist.
- alphaalpha101 9y ago
  >Since c is 1, and 1^2 is 1, e=m. Energy and mass are completely equivalent. E=m doesn't mean 'energy and mass are completely equivalent' at all. Where do you get that idea? Energy and mass have a certain equivalence regardless of the units you use.
averagewall 9y ago
The conclusion that energy is more fundamental or easier to have an intuition for has been the case since, what 100 years ago? It certainly was when I studied physics 20 years ago. But he doesn't need to go into quarks for that. Simply comparing the masses of protons and neutrons (+ electrons) to the mass of an atom shows they're different. There's a little bit less mass due to the lower energy of the bound nucleons. Quarks certainly make the effect more dramatic though. A more practical objection to using mass is that nobody can agree on what the word means. Does a photon have mass? Yes or no, depending on if you're thinking of relativistic or rest mass. If you call it energy, there's no ambiguity. If I could recommend he change anything though, it would be the horrible mixture of units - MeV/c^2, atomic mass units, and grams. The author surely has the time to convert them all to the same unit so the reader can easily compare them. That would eliminate the need to explain Avogadro's law. It's a completely redundant complication of chemistry and has nothing to do with the fundamental concepts he's focusing on.
- davrosthedalek 9y ago
  There is a difference though: A hydrogen atom is slightly lighter than it's constituents. That's why it is stable. However, the nucleons are much, much heavier than the sum of the quark masses. Which is strange, because it means that naively, the unbound state should be preferred. But it isn't, instead, you have confinement!
ars 9y ago
For some reason people are still caught up in finding some difference between mass and energy. There is no difference. They are two words that mean the same thing. There are different kinds of mass-energy, some types are easy to convert into others, some types move at the speed of light, some don't. But there is nothing distinguishing mass from energy. It was a revelation to see photons attracted by gravity - but once you realize it's energy that has a gravitational field [not just the particles we call mass], it would be surprising if photons were not attracted by gravity. (Although photons moving only at the speed of light have different equations governing their motion under acceleration.) Now, all that said, There is a distinction between things that only move at the speed of light, and things that never do. But the words energy and mass [as commonly used] do not properly fit those two categories.
- dukwon 9y ago
  Meanwhile, in real physics, there's a well-defined difference between the norm of a vector and one of its components. Mass is Lorentz-invariant, energy is not.
  - raattgift 9y ago
    Certainly, but both are encoded in the energy-momentum tensor and for better or worse that's usually considered to be what generates the metric (or perturbations thereof, if you want to do things that way). In Newtonian mechanics, kinetic energy is rotationally but not Galilean invariant, sure. But in GR in a local inertial frame the pressure T^{ij}, i=j, i!=0 sure looks like kinetic energy \gamma m(v^i)^2. [1] Should we really strongly distinguish between the pressure and T^00 just because in a local inertial frame the latter looks like \gamma mc^2 ? Sharpen the question by considering vastly different frames of reference. - -- [1] https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressure_and_kinetic_energy https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressu...
  - ars 9y ago
    > Mass is Lorentz-invariant, energy is not. That's not completely accurate. Chemical energy and binding energy are also Lorentz-invariant. The only type of energy is not Lorentz-invariant is velocity energy, so you are putting your distinction in the wrong place. On top of that, there are other violated invariants. The weight of a lump of iron near a magnetar is greater than the weight of the same lump of iron near an identically massing neutron start. This is because the potential energy of the iron is greater near the magnetic field, so its mass (as seem by the magnetar) is greater, and so is the gravitational attraction between them. This means you can't just say "No velocity, the mass is identical", it's not - the extra potential energy means extra mass. Or in other words there is no such thing as mass as distinguishable from energy.
ComputerGuru 9y ago
I love reading about quantum mechanics at all different levels, and have read and appreciated articles well beyond what I could properly comprehend in fullness; but man was this article all over the place. I've never seen an author use metaphors or similies so poorly before. To draw parallels between something hard to grasp and something made up and vaguely defined in an effort to explain the former is.. ill advised at best. Then there's the fact that the author spends forever to explain basic chemistry then jumps into color charge in such a way that anyone not already intimately familiar with at least the terminology of quantum physics would never be able to understand, then the author jumps from topic to topic seemingly in a race to drop references to as many different concepts as possible without actually explaining any of them, almost like a student writing an essay then going back and swapping words with a thesaurus to seem better informed. Even the science aside, the writing itself is rather atrocious. The author "answers" mysteries he never even asked or previously posed, and expects readers to already know what he's trying to say so he can refer to that in his explanations of why he said it. Then the author has a tendency to jump from field to field, converting apples into oranges with the help of a long-dead scientist only so he can add them together in the most basic way and then convert them back to apples again. But he got to prove that he knows of Avagdro, so obviously there's that. Usually Nautilus articles are written much better than this. If you value your sanity or actually care to understand the topic discussed, do yourself a favor and look elsewhere.
- em3rgent0rdr 9y ago
  Exactly...and even though he spends so much time discussing basic chemistry as if he is talking to an average joe, he jumps into use the term "mole" without explaining it.
- hamilyon2 9y ago
  I respectfully disagree. He is imprecise, metaphorical, and offers entertainment, rather than knowlege. And jet, he knows his reader very well, and he delivers important bits of knowlege in very digestable form. Sure, this is not a substitution for reading a book. But his target audience will never read that book, nor perform any research on their own. Now i firmly remeber what QCD exist and firmly believe I don't know what it is about. Isn't it good, by itself?
mirimir 9y ago
I mean, sure, mass is energy. We all know that. And still, this is a great article. It ties together many concepts that I've read about.
rurban 9y ago
I would rather describe as a very weak, long-ranging, attracting force. A force of a very special kind, because all other forces are strong, short-ranging (they need to interact, the only not-mechanical force) and repelling.
- davrosthedalek 9y ago
  Sorry, but that's quite wrong: We have gravitational, weak, e/m and strong interaction. Electromagnetic has the same range, only most objects seem to be neutral. "Mechanical" force is normally electromagnetic. The weak force is weak, but not as weak as gravitation (assuming normal charges) The strong force has indeed short range but is attractive.
  - evanb 9y ago
    Indeed, the reason most objects seem to be neutral is because of how strong the electromagnetic force is! Were it weaker it would be much easier to separate charges on longer length / time scales.
cyberfart 9y ago
Veritasium had a video explaining this a few years back, it's one of the videos responsible for sparking my interest in physics and cosmology. https://www.youtube.com/watch?v=Ztc6QPNUqls&feature=youtu.be&t=4m47s https://www.youtube.com/watch?v=Ztc6QPNUqls&feature=youtu.be...
angry_octet 9y ago
"But what is matter, exactly? Imagine a cube of ice, measuring a little over one inch (or 2.7 centimeters) in length." Gasping. For. Air. I love that an article on units gets a unit conversion massively wrong.
- angry_octet 9y ago
  Touched a raw nerve obviously. Lost due to use of imperial units: 1998 Solar Heliospheric Observatory 1999 Mars Climate Orbiter

157 points by AliCollins 9y ago

I was just explaining to my teenage son what quarks are -- he hadn't learned about them yet. So I have to upvote this. tl;dr If you decompose matter all the way down, mass doesn't just "add up the parts to the whole". Not by several orders of magnitude!

SAI_Peregrinus 9y ago

It's very helpful to use natural units (or QCD units) when working with particle physics. Set the speed of light, the reduced Planck's constant, and Boltzmann's constant equal to 1. For QCD units also set the mass of the proton equal to 1. Then compute all the other units based on this. (There are several other systems of "natural" units also in use. All of them set the speed of light to 1, but differ in what other units are included.) A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m. Energy and mass are completely equivalent. Strong force binding energy in a proton is mass. The only bit that's not straight from binding energy is from the Higgs mechanism, which is where the mass of the electron (and the quarks themselves) comes from. Matt Strassler has an excellent explanation of that: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/how-the-higgs-field-works-with-math/ https://profmattstrassler.com/articles-and-posts/particle-ph...

JPLeRouzic 9y ago
This comment is much better than the article, because it tells that some of the complex relations between conceptual entities in physic, could be presented in a simplified but equivalent way. So it hints that the "problem" presented in the article is only an artifact. Henry Poincaré wrote exactly this in 1902 in "the science and hypothesis" Nous sommes acculés à la définition suivante, qui n'est qu'un aveu d'impuissance : les masses sont des coefficients qu'il est commode d'introduire dans les calculs.[0] I am also happy for the reference to the excellent blog of Matt Strassler [0] https://en.wikipedia.org/wiki/Science_and_Hypothesis https://en.wikipedia.org/wiki/Science_and_Hypothesis
emmelaich 9y ago
E=m ... and I=m. Or so I was taught. Inertia is not merely something that depends on mass. It is mass. (though from a brief web search, it is not typically presented as such)
- pducks32 9y ago
  Yea physicists are in a weird spot when it comes to inertia. Everyone’s heard of it but it’s not real. To say something has mass is to say it requires a force to move it which is what we call inertia. And also the E=m equivalence isn’t as simple as may be believed. The famous equivalence is cutting out a piece that is important when talking about photons.
  - SAI_Peregrinus 9y ago
    There are certainly many subtleties. For example, I said E=M. But that's for objects at rest. For an object in motion it's E^2=m^2c^4+p^2c^2. In natural units that turns into E=m+p. The energy of a photon is hf, where h is Planck's constant and f is the frequency. That energy is entirely the photon's momentum. And that's without getting into the issues of invariant mass or relativistic mass increases...
    - evanb 9y ago
      You dropped a bunch of important squares... E^2 = m^2 + p^2 does not simplify to E=m+p.
      - SAI_Peregrinus 9y ago
        You're right, of course. I really shouldn't do math in my head after midnight.
    - jdmichal 9y ago
      Unicode's got a full supplement of sub- and superscript numbers now. E² = m²c⁴ + p²c² https://en.wikipedia.org/wiki/Unicode_subscripts_and_superscripts https://en.wikipedia.org/wiki/Unicode_subscripts_and_supersc...
bambax 9y ago
> A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m. Energy and mass are completely equivalent. Yes but that is still true (E=m) if c<>1, since c is a constant. It's just a question of scale, isn't it?
- throwaway287391 9y ago
  In the sense that GP states, yes, but I disagree with GP: "E=m" is wrong even if c=1 in the units you're working with since c is not unitless; it has speed units. You can't have a kilogram of energy. You need to multiply a mass by something with squared-speed units to get something with energy units.
  - chriswarbo 9y ago
    > "E=m" is wrong even if c=1 in the units you're working with since c is not unitless In natural units, c is "unitless" (AKA dimensionless), by definition.
    - throwaway287391 9y ago
      I won't say you're wrong since I stopped taking physics in high school, but this makes absolutely no sense to me. How does defining the Planck speed unit to be the speed of light suddenly make c a unitless constant? It's not "c=1" (except in informal shorthand), it's "c = 1 Planck speed unit". It seems like all sorts of things would "break" very quickly when you start dropping units. Maybe someone has a link that goes into more depth on why all of this is okay?
    - raattgift 9y ago
      The value of c is arbitrary and depends on one's choice of coordinates. However the presence of c in the line element of a spacetime like ours is mandatory. There is, however, some conceptual value in setting c to some value other than 1, which I'll return to in the last paragraph below. When using a pseudo-Riemannian manifold one uses a metric signature where (keeping it simple, cf. "metric signature" on wikipedia) coordinates on orthogonal dimensions of can take one sign, or the opposite sign and a constant multiplier. In the Lorentzian case there will be one dimension taking one sign, and one or more taking the other; the choice of whether the solitary dimension takes a + or - sign is a matter of convention or preference. Conventionally the solitary dimension also takes the constant multiplier and is called the timelike dimension. A Lorentzian signature (minuses, 1) or (1, plusses) guarantees that one can describe paths through the manifold as null, timelike, or spacelike, and this gives one a causal structure. Our universe can be well represented by a Lorentzian manifold (3, 1) or (1, 3), and this has been tested to exquisite precision. It does not tell us the value of the constant c, but we can determine that from tests of causal relations, the boundaries of which will be null. Alternatively, a massless pointlike object will always travel on null geodesics. One runs into c being set to unity in systems geometrized units in relativity often; it's very handy to have mark off coordinates as e.g. "-seconds" vs "c seconds" (-,+,+,+ aka (1,3)) as long as one doesn't mess up one's dimensional analyses (which is unfortunately easy). As a concrete example, the line element for (1,3) flat spacetime using Cartesian coordinates is dS^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2. Compare the formula for Euclidean distance in the (Euclidean flat) plane between points p = (p1,p2) and q = (q1,q2) for (x,y) coordinates: d(p,q) = sqrt((q1-p1)^2 + (q2-p2)^2), which we could rewrite as dx = q1-p1, dy = q2-p2, ds = sqrt(dx^2 + dy^2) or ds^2 = dx^2 + dy^2. In Euclidean 3-space, we add another axis: ds^2 = dx^2 + dy^2 + dz^2. In Lorentzian 4-spacetime, we have to change the sign, so ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2. Note that we are not restricted to use any particular unit of distance or system of units; dx could be in metres, miles, astronomical units, light-years, gigaparsecs or practically anything else, while dt could be in seconds or fortnights or any other handy unit of time. When setting c to unity, we do need to choose appropriate units for dx (and dy and dz ...) vs dt. In SI units, that's seconds and light-seconds, but we could use another system if we wanted. Notably we aren't restricted to Cartesian coordinates, however if we were to change to some other system of coordinates (e.g. polar ones) the line element would need to reflect that. For example, in the Lorentzian 4-spacetime case, we would write ds^2 = dr^2 + r^2 * dtheta^2 + r^2sin^2(theta)dphi^2 -c^2dt^2. Finally, when using SI units (for example), one can see very clearly that a path taken by an object moving much slower than the speed of light is totally dominated by the amount of time between starting point and finishing point, because the value of c is large. Using the (+,-,-,-) metric signature [ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2], a large c helps make it clear that lightlike paths through spacetime are shorter, and purely timelike paths (where dx=0, dy=0, dz=0, dt != 0) have extremized length, since we don't subtract anything from cdt. This is the root of the explanation of the twin paradox in flat spacetime: the twin moving quickly compared to the speed of light takes a shorter path between together1 = (x1,y1,z1,t1) and together2 = (x2,y2,z2,t1) than the twin moving slowly compared to the speed of light. In the extreme, twin A holds x=const,y=const,z=const, whereas twin B's x coordinate is only equal to twin A's at the start and end of the journey. This holds up under any system of coordinates; we could consider r=const vs changing r in spherical coordinates, for instance.
davidmanescu 9y ago
Equalities that involve units aren't straightforward. Not a physicist of any sort but don't all the 1's still fundamentally change the equality since they change the units? I would guess they therefore implicitly change the intuition behind these equations?
- beojan 9y ago
  That's what makes natural units natural. Certain quantities (like mass and energy, or distance and time) are physically related such that the distinction is, in a sense, artificial. The conversion factors between the units used for these pairs of quantities are the fundamental physical constants that get set to unity in a natural system of units.
  - jessriedel 9y ago
    The distinction between distance and time is not artificial at all. It is encoded in the metric, at the most fundamental level. The existence of symmetries is not the same as equivalence.
    - coldtea 9y ago
      Parent is not talking about the "distinction between distance and time" but about the multipliers that come into play into their equations (which are artificial and based on the base units selected).
      - jessriedel 9y ago
        The distinction between feet and seconds is just as real as the distinction between distance and time, and is much more real than the distinction between feet and meters. Look at the parent's literal words and pretend you did not already know what's going on. I'm well aware of the practical value and conceptual clarity of natural units. They are just not being explained well in this thread. "What the teacher really meant was..." is not a good defense when the student doesn't understand.
lkrubner 9y ago
A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m. That is clever, but you've already adjusted the mass to take into account the fact that you are now setting c=1. This reminds me of computer programming where, for the sake of an easy-to-read clarity, I decide to take one line of code and make it two. Basically, instead of doing this: e=mc^2 You have decided to make it two lines of code: m=mc^2 e=m I often do that, especially if I'm working with junior devs and I want the code to be as easy as possible for them to read (90% of the time). It's clever, but everyone should remember the assumption that allows this: the mass has already been adjusted to reflect the reality that of c=1.
- johncolanduoni 9y ago
  Except you would otherwise have at least one of c, hbar, boltzman’s constant etc. in virtually every equation and now you don’t. You get a lot more mileage out of it than just simplifying one line.
matt_wulfeck 9y ago
> A particular advantage to intuition comes with the famous e=mc^2. Since c is 1, and 1^2 is 1, e=m Since mass also determines the speed of time to the observer, isn't a time machine basically just a thing that converts energy into mass?
- deleted 9y ago
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- lisper 9y ago
  No. Mass doesn't "determine" the speed of time (a lovely phrase, BTW). It is the observer's motion through space that does that. However, for an observer to move through time at a speed faster than 0 it has to move through space at a speed slower than c, and to do that it needs to have mass. A massless particle can only transfer energy if it moves at the speed of light. Massless particles moving slower than the speed of light cannot transfer energy, and so cannot be detected experimentally, and so do not exist.
alphaalpha101 9y ago
>Since c is 1, and 1^2 is 1, e=m. Energy and mass are completely equivalent. E=m doesn't mean 'energy and mass are completely equivalent' at all. Where do you get that idea? Energy and mass have a certain equivalence regardless of the units you use.

averagewall 9y ago

The conclusion that energy is more fundamental or easier to have an intuition for has been the case since, what 100 years ago? It certainly was when I studied physics 20 years ago. But he doesn't need to go into quarks for that. Simply comparing the masses of protons and neutrons (+ electrons) to the mass of an atom shows they're different. There's a little bit less mass due to the lower energy of the bound nucleons. Quarks certainly make the effect more dramatic though. A more practical objection to using mass is that nobody can agree on what the word means. Does a photon have mass? Yes or no, depending on if you're thinking of relativistic or rest mass. If you call it energy, there's no ambiguity. If I could recommend he change anything though, it would be the horrible mixture of units - MeV/c^2, atomic mass units, and grams. The author surely has the time to convert them all to the same unit so the reader can easily compare them. That would eliminate the need to explain Avogadro's law. It's a completely redundant complication of chemistry and has nothing to do with the fundamental concepts he's focusing on.

davrosthedalek 9y ago
There is a difference though: A hydrogen atom is slightly lighter than it's constituents. That's why it is stable. However, the nucleons are much, much heavier than the sum of the quark masses. Which is strange, because it means that naively, the unbound state should be preferred. But it isn't, instead, you have confinement!

ars 9y ago

For some reason people are still caught up in finding some difference between mass and energy. There is no difference. They are two words that mean the same thing. There are different kinds of mass-energy, some types are easy to convert into others, some types move at the speed of light, some don't. But there is nothing distinguishing mass from energy. It was a revelation to see photons attracted by gravity - but once you realize it's energy that has a gravitational field [not just the particles we call mass], it would be surprising if photons were not attracted by gravity. (Although photons moving only at the speed of light have different equations governing their motion under acceleration.) Now, all that said, There is a distinction between things that only move at the speed of light, and things that never do. But the words energy and mass [as commonly used] do not properly fit those two categories.

dukwon 9y ago
Meanwhile, in real physics, there's a well-defined difference between the norm of a vector and one of its components. Mass is Lorentz-invariant, energy is not.
- raattgift 9y ago
  Certainly, but both are encoded in the energy-momentum tensor and for better or worse that's usually considered to be what generates the metric (or perturbations thereof, if you want to do things that way). In Newtonian mechanics, kinetic energy is rotationally but not Galilean invariant, sure. But in GR in a local inertial frame the pressure T^{ij}, i=j, i!=0 sure looks like kinetic energy \gamma m(v^i)^2. [1] Should we really strongly distinguish between the pressure and T^00 just because in a local inertial frame the latter looks like \gamma mc^2 ? Sharpen the question by considering vastly different frames of reference. - -- [1] https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressure_and_kinetic_energy https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressu...
- ars 9y ago
  > Mass is Lorentz-invariant, energy is not. That's not completely accurate. Chemical energy and binding energy are also Lorentz-invariant. The only type of energy is not Lorentz-invariant is velocity energy, so you are putting your distinction in the wrong place. On top of that, there are other violated invariants. The weight of a lump of iron near a magnetar is greater than the weight of the same lump of iron near an identically massing neutron start. This is because the potential energy of the iron is greater near the magnetic field, so its mass (as seem by the magnetar) is greater, and so is the gravitational attraction between them. This means you can't just say "No velocity, the mass is identical", it's not - the extra potential energy means extra mass. Or in other words there is no such thing as mass as distinguishable from energy.

ComputerGuru 9y ago

I love reading about quantum mechanics at all different levels, and have read and appreciated articles well beyond what I could properly comprehend in fullness; but man was this article all over the place. I've never seen an author use metaphors or similies so poorly before. To draw parallels between something hard to grasp and something made up and vaguely defined in an effort to explain the former is.. ill advised at best. Then there's the fact that the author spends forever to explain basic chemistry then jumps into color charge in such a way that anyone not already intimately familiar with at least the terminology of quantum physics would never be able to understand, then the author jumps from topic to topic seemingly in a race to drop references to as many different concepts as possible without actually explaining any of them, almost like a student writing an essay then going back and swapping words with a thesaurus to seem better informed. Even the science aside, the writing itself is rather atrocious. The author "answers" mysteries he never even asked or previously posed, and expects readers to already know what he's trying to say so he can refer to that in his explanations of why he said it. Then the author has a tendency to jump from field to field, converting apples into oranges with the help of a long-dead scientist only so he can add them together in the most basic way and then convert them back to apples again. But he got to prove that he knows of Avagdro, so obviously there's that. Usually Nautilus articles are written much better than this. If you value your sanity or actually care to understand the topic discussed, do yourself a favor and look elsewhere.

em3rgent0rdr 9y ago
Exactly...and even though he spends so much time discussing basic chemistry as if he is talking to an average joe, he jumps into use the term "mole" without explaining it.
hamilyon2 9y ago
I respectfully disagree. He is imprecise, metaphorical, and offers entertainment, rather than knowlege. And jet, he knows his reader very well, and he delivers important bits of knowlege in very digestable form. Sure, this is not a substitution for reading a book. But his target audience will never read that book, nor perform any research on their own. Now i firmly remeber what QCD exist and firmly believe I don't know what it is about. Isn't it good, by itself?

mirimir 9y ago

I mean, sure, mass is energy. We all know that. And still, this is a great article. It ties together many concepts that I've read about.

rurban 9y ago

I would rather describe as a very weak, long-ranging, attracting force. A force of a very special kind, because all other forces are strong, short-ranging (they need to interact, the only not-mechanical force) and repelling.

davrosthedalek 9y ago
Sorry, but that's quite wrong: We have gravitational, weak, e/m and strong interaction. Electromagnetic has the same range, only most objects seem to be neutral. "Mechanical" force is normally electromagnetic. The weak force is weak, but not as weak as gravitation (assuming normal charges) The strong force has indeed short range but is attractive.
- evanb 9y ago
  Indeed, the reason most objects seem to be neutral is because of how strong the electromagnetic force is! Were it weaker it would be much easier to separate charges on longer length / time scales.

cyberfart 9y ago

Veritasium had a video explaining this a few years back, it's one of the videos responsible for sparking my interest in physics and cosmology. https://www.youtube.com/watch?v=Ztc6QPNUqls&feature=youtu.be&t=4m47s https://www.youtube.com/watch?v=Ztc6QPNUqls&feature=youtu.be...

angry_octet 9y ago

"But what is matter, exactly? Imagine a cube of ice, measuring a little over one inch (or 2.7 centimeters) in length." Gasping. For. Air. I love that an article on units gets a unit conversion massively wrong.

angry_octet 9y ago
Touched a raw nerve obviously. Lost due to use of imperial units: 1998 Solar Heliospheric Observatory 1999 Mars Climate Orbiter