6 ms·

I can make a classical analogy. The Higgs explains why fundamental particles have mass in the "F=ma" Newton's-Laws-Of-Motion sense. But you can formulate clas

by evanb 9y ago

I can make a classical analogy. The Higgs explains why fundamental particles have mass in the "F=ma" Newton's-Laws-Of-Motion sense. But you can formulate classical mechanics without gravity at all. The m in this equation is sometimes referred to as the inertial mass. F_gravity = G M m / r^2 on its face looks like it is talking about the same inertial mass (or, by adjusting G, something proportional to the inertial mass). But, this is an assumption---a body could equally well have a gravitational mass independent of its inertial mass. If inertial and gravitational masses are proportional to one another then when you calculate the acceleration due to gravity you can cancel the m from both sides, and you find that all bodies fall the same way (even light, which has m=0 and might cause concern that the cancellation isn't valid). This is a repercussion of inertial = gravitational mass. Einstein promoted this observation to the Equivalence Principle (and people do experimental searches for violations of this assumption). The Higgs gives particles inertial mass. But the Higgs doesn't cause things to fall "down" (assuming you can define "down" without a gravitational field). You need the gravitons to communicate the gravitational attraction, which is independent of the existence of mass.

daxfohl 9y ago
Oh, duh, because while "physics" (general relativity) shows that mass is gravity, the whole point of this exploration is see if we can derive that equivalence (or something extremely close to it) from core QM principles rather than assuming it a-priori. Because the a-priori assumption (general relativity) contradicts QM. So some deeper principle needs to unfold to present both Higgs bosons (inertial mass) and gravitons (gravitational mass) and their relationship, from which something very very very close to general relativity can be derived. Is that right? If so, has any actual string-theoretical mathematical relationship between gravitons and Higgs bosons been developed that would explain general relativity? Or is it stuck at "in string theory, something like 'gravitons' could exist"?
- wayn3 9y ago
  the "problem" here is that the graviton "has to" exist. why? Forces have to be conveyed by something. If you want to push someone, you have to physically touch them. you have to REACH them. you cant just push the air, across the room, and expect them to feel the push. this is true on the microscopic level as well. when particles exert gravity on each other, they do so at a distance. but something has to cross this distance. it is not ordinarily obvious how the sun traps our planet in its gravitational field - at a distance. there has to be something that exchanges the gravitational force between sun and earth. we call this thing the graviton. it is important to notice that something that does the "job" of the graviton HAS TO exist. maybe its not a particle. but something causes the exchange of gravitation and thats what we're looking for. you know that "in space, no one can hear you scream". thats because there is no air that could make the sound waves travel. in a similar way, without the graviton, there would be no "medium" that conveys the gravitational force. the problem with detecting the graviton is that it is very weak. we would have to build really expensive machines to "observe" it. explaining this theoretically is not difficult. we have plenty of "theories". the problem is confirming them with experiments.
  - AstralStorm 9y ago
    Nothing in physics "has" to be something or other. It is well known that QM is incomplete, so forcing forces to be quantum might not be the right way forward at all. Strings are not quanta to begin with. They behave somewhat similarly at lower energies at most. Forces are essentially a model too, and are compatible with both wave, quantum and string physics. Each substantially different.
    - wayn3 9y ago
      hence the quotation marks. heres your "well, actually" cookie.
  - justin_vanw 9y ago
    I'm not an expert, but I simply don't understand your reasoning here. For example, lets suppose some mass is moving at a constant inertial velocity through empty space. To me it seems that your reasoning would require gravitons to communicate to the mass that there are not other masses nearby and so to 'tell it' that 'straight line' for it means to go in a euclidian straight line. Otherwise, how does it know? I think baked into your logic is that there is something special about geodesics in the presence of masses and so you need to tell the moving mass to 'curve', but to the mass it is just going straight, even if to an external observer it appears that it is curving or even orbiting.
    - wayn3 9y ago
      - its not my reasoning, i'm just trying to explain quantum field theory. - whether you call it "gravitational force" or "curvature of spacetime along whose geodesics massive objects slide" - the effect has to be mediated by a "particle". for popular media, "particle" is too big a word, because people tend to think of protons or atoms. subatomic particles are just excitations of quantum fields. little blips of localized energy, of which we are only able to see the top layer. ^ this has nothing to do with general relativity. general relativity describes the macroscopic world pretty well. it generally breaks down on very small scales. how planets move is described very well by general relativity. how they mediate the involved forces is not described at all. edit: i just thought about that straight line statement. there seems to be a misconception that a geodesic is a "generalized straight line". That is not remotely true. Geodesics, in mathematics, are "shortest paths". While that happens to coincide with what a straight line does in a plane, generalizing that meaning in the other direction doesn't work. In general relativity, we talk about geodesics when we mean "out of all the possible paths we can take, we are choosing the one that minimizes energy loss". That is, then, a geodesic. But a geodesic is far from a straight line in terms of movement. Its the path of least resistance in the energy picture. If you ask "whats the difference?" - the difference is that a straight line in energy space is not a straight line in regular space. Earth, for example, is travelling along a geodesic. But it is clearly accelerated towards the sun. There is nothing "straight line" about it. When you fall into a black hole, you travel along a geodesic. But it wont feel like a straight line to you at all. That you happen to be travelling along a straight line in the absence of forces is just a tautological truth. Applying differential geometry to that statement just makes it way more complicated to state the obvious.

by evanb 9y ago

daxfohl 9y ago

Oh, duh, because while "physics" (general relativity) shows that mass is gravity, the whole point of this exploration is see if we can derive that equivalence (or something extremely close to it) from core QM principles rather than assuming it a-priori. Because the a-priori assumption (general relativity) contradicts QM. So some deeper principle needs to unfold to present both Higgs bosons (inertial mass) and gravitons (gravitational mass) and their relationship, from which something very very very close to general relativity can be derived. Is that right? If so, has any actual string-theoretical mathematical relationship between gravitons and Higgs bosons been developed that would explain general relativity? Or is it stuck at "in string theory, something like 'gravitons' could exist"?

wayn3 9y ago
the "problem" here is that the graviton "has to" exist. why? Forces have to be conveyed by something. If you want to push someone, you have to physically touch them. you have to REACH them. you cant just push the air, across the room, and expect them to feel the push. this is true on the microscopic level as well. when particles exert gravity on each other, they do so at a distance. but something has to cross this distance. it is not ordinarily obvious how the sun traps our planet in its gravitational field - at a distance. there has to be something that exchanges the gravitational force between sun and earth. we call this thing the graviton. it is important to notice that something that does the "job" of the graviton HAS TO exist. maybe its not a particle. but something causes the exchange of gravitation and thats what we're looking for. you know that "in space, no one can hear you scream". thats because there is no air that could make the sound waves travel. in a similar way, without the graviton, there would be no "medium" that conveys the gravitational force. the problem with detecting the graviton is that it is very weak. we would have to build really expensive machines to "observe" it. explaining this theoretically is not difficult. we have plenty of "theories". the problem is confirming them with experiments.
- AstralStorm 9y ago
  Nothing in physics "has" to be something or other. It is well known that QM is incomplete, so forcing forces to be quantum might not be the right way forward at all. Strings are not quanta to begin with. They behave somewhat similarly at lower energies at most. Forces are essentially a model too, and are compatible with both wave, quantum and string physics. Each substantially different.
  - wayn3 9y ago
    hence the quotation marks. heres your "well, actually" cookie.
- justin_vanw 9y ago
  I'm not an expert, but I simply don't understand your reasoning here. For example, lets suppose some mass is moving at a constant inertial velocity through empty space. To me it seems that your reasoning would require gravitons to communicate to the mass that there are not other masses nearby and so to 'tell it' that 'straight line' for it means to go in a euclidian straight line. Otherwise, how does it know? I think baked into your logic is that there is something special about geodesics in the presence of masses and so you need to tell the moving mass to 'curve', but to the mass it is just going straight, even if to an external observer it appears that it is curving or even orbiting.
  - wayn3 9y ago
    - its not my reasoning, i'm just trying to explain quantum field theory. - whether you call it "gravitational force" or "curvature of spacetime along whose geodesics massive objects slide" - the effect has to be mediated by a "particle". for popular media, "particle" is too big a word, because people tend to think of protons or atoms. subatomic particles are just excitations of quantum fields. little blips of localized energy, of which we are only able to see the top layer. ^ this has nothing to do with general relativity. general relativity describes the macroscopic world pretty well. it generally breaks down on very small scales. how planets move is described very well by general relativity. how they mediate the involved forces is not described at all. edit: i just thought about that straight line statement. there seems to be a misconception that a geodesic is a "generalized straight line". That is not remotely true. Geodesics, in mathematics, are "shortest paths". While that happens to coincide with what a straight line does in a plane, generalizing that meaning in the other direction doesn't work. In general relativity, we talk about geodesics when we mean "out of all the possible paths we can take, we are choosing the one that minimizes energy loss". That is, then, a geodesic. But a geodesic is far from a straight line in terms of movement. Its the path of least resistance in the energy picture. If you ask "whats the difference?" - the difference is that a straight line in energy space is not a straight line in regular space. Earth, for example, is travelling along a geodesic. But it is clearly accelerated towards the sun. There is nothing "straight line" about it. When you fall into a black hole, you travel along a geodesic. But it wont feel like a straight line to you at all. That you happen to be travelling along a straight line in the absence of forces is just a tautological truth. Applying differential geometry to that statement just makes it way more complicated to state the obvious.