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In special relativity is mass just a measure of all other energy than kinetic?
Is it part of special relativity that mass possessing energy is more dense?Potential energy in Special RelativityPropulsion Energy in Special RelativityTranslational and Rotational Energy in Special RelativityEnergy definition in special relativityDifference between kinetic energy “relativistic” mass and heat massHypothetical special relativity with mass conservationConsequence of _adding a constant to potential energy_ in the context of mass in special relativityWhy is Mass-Energy Equivalence more explicit for high speed particles than low-speed ones? (since it comes from Relativity)Reconciling classical mechanics with special relativity - energy lost as heat in an inelastic collision
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The energy momentum equation in special relativity is:
$$E^2=(pc)^2+(mc^2)^2.$$
and it holds for a moving but not accelerating object.
One special case is the massless photon:
$$E=pc.$$
And another one is a resting object:
$$E=mc^2.$$
The first term in the energy momentum equation seems to be kinetic energy of the object as a whole?
That should mean that the second term is all other forms of energy?
This could include kinetic energy of the constituents of the object.
So a full battery for instance has more energy than an empty battery and therefore has a larger mass?
Likewise a hot kilogram prototype has more energy than a cold one and therefore has larger mass? This thermic energy is kinetic energy at constituent level but not for the object as a whole.
In classical mechanics mass is a property of an object that has to do with its inertia. We could define $m=fracFa$.
In classical mechanics an empty battery has the same mass as a full one and a cold kilogram prototype has the same mass as a hot one.
Does this not mean that the concept of mass has changed in special relativity and now is a measure of all energy except kinetic energy of an object?
If so $E=mc^2$ does not necessarily predict the atomic bomb?
Instead of it saying that all objects have some intrinsic energy due to their mass it could say that an absolutely cold still object has no mass since it has no energy. Which sounds absurd.
So I am guessing that the old concept of mass from classical mechanics is an approximation to the new concept of mass in special relativity?
special-relativity energy mass momentum mass-energy
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
asked Jul 15 at 13:49
AndyAndy
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show 1 more comment
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The energy momentum equation in special relativity is:
$$E^2=(pc)^2+(mc^2)^2.$$
and it holds for a moving but not accelerating object.
One special case is the massless photon:
$$E=pc.$$
And another one is a resting object:
$$E=mc^2.$$
The first term in the energy momentum equation seems to be kinetic energy of the object as a whole?
That should mean that the second term is all other forms of energy?
This could include kinetic energy of the constituents of the object.
So a full battery for instance has more energy than an empty battery and therefore has a larger mass?
Likewise a hot kilogram prototype has more energy than a cold one and therefore has larger mass? This thermic energy is kinetic energy at constituent level but not for the object as a whole.
In classical mechanics mass is a property of an object that has to do with its inertia. We could define $m=fracFa$.
In classical mechanics an empty battery has the same mass as a full one and a cold kilogram prototype has the same mass as a hot one.
Does this not mean that the concept of mass has changed in special relativity and now is a measure of all energy except kinetic energy of an object?
If so $E=mc^2$ does not necessarily predict the atomic bomb?
Instead of it saying that all objects have some intrinsic energy due to their mass it could say that an absolutely cold still object has no mass since it has no energy. Which sounds absurd.
So I am guessing that the old concept of mass from classical mechanics is an approximation to the new concept of mass in special relativity?
special-relativity energy mass momentum mass-energy
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
asked Jul 15 at 13:49
AndyAndy
1888 bronze badges
$endgroup$
1
$begingroup$
Have a look hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . The third equation is not used in particle physics, because it is misleading, it is the inertial mass acquired when particles reach close to the velocity of light. What is used is the "invariant mass" , the "length" of the four vector of a particle , or a system of particles. Vector algebra is used.
$endgroup$
– anna v
Jul 15 at 14:12
2
$begingroup$
So far you have only considered an isolated particle. If you extend the consideration to system you’ll find that the mass of a system can include some of the kinetic energy of its constituents.
$endgroup$
– dmckee♦
Jul 15 at 14:26
1
$begingroup$
In relation to @dmckee's point, when you say that a hot object has more mass than an otherwise identical cold object, you are actually pointing to the fact that some fraction of the mass of a system comes from some of the kinetic energy of its constituents.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:29
$begingroup$
Yes you are right. I have updated my question. The kinetic energy due to the momentum of the system as a whole is the firsts term.
$endgroup$
– Andy
Jul 15 at 14:33
1
$begingroup$
@ZachMcDargh We do not empirically see that the mass of an object does not depend on temperature. It very much does.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:40
|
show 1 more comment
$begingroup$
The energy momentum equation in special relativity is:
$$E^2=(pc)^2+(mc^2)^2.$$
and it holds for a moving but not accelerating object.
One special case is the massless photon:
$$E=pc.$$
And another one is a resting object:
$$E=mc^2.$$
The first term in the energy momentum equation seems to be kinetic energy of the object as a whole?
That should mean that the second term is all other forms of energy?
This could include kinetic energy of the constituents of the object.
So a full battery for instance has more energy than an empty battery and therefore has a larger mass?
Likewise a hot kilogram prototype has more energy than a cold one and therefore has larger mass? This thermic energy is kinetic energy at constituent level but not for the object as a whole.
In classical mechanics mass is a property of an object that has to do with its inertia. We could define $m=fracFa$.
In classical mechanics an empty battery has the same mass as a full one and a cold kilogram prototype has the same mass as a hot one.
Does this not mean that the concept of mass has changed in special relativity and now is a measure of all energy except kinetic energy of an object?
If so $E=mc^2$ does not necessarily predict the atomic bomb?
Instead of it saying that all objects have some intrinsic energy due to their mass it could say that an absolutely cold still object has no mass since it has no energy. Which sounds absurd.
So I am guessing that the old concept of mass from classical mechanics is an approximation to the new concept of mass in special relativity?
special-relativity energy mass momentum mass-energy
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
asked Jul 15 at 13:49
AndyAndy
1888 bronze badges
$endgroup$
The energy momentum equation in special relativity is:
$$E^2=(pc)^2+(mc^2)^2.$$
and it holds for a moving but not accelerating object.
One special case is the massless photon:
$$E=pc.$$
And another one is a resting object:
$$E=mc^2.$$
The first term in the energy momentum equation seems to be kinetic energy of the object as a whole?
That should mean that the second term is all other forms of energy?
This could include kinetic energy of the constituents of the object.
So a full battery for instance has more energy than an empty battery and therefore has a larger mass?
Likewise a hot kilogram prototype has more energy than a cold one and therefore has larger mass? This thermic energy is kinetic energy at constituent level but not for the object as a whole.
In classical mechanics mass is a property of an object that has to do with its inertia. We could define $m=fracFa$.
In classical mechanics an empty battery has the same mass as a full one and a cold kilogram prototype has the same mass as a hot one.
Does this not mean that the concept of mass has changed in special relativity and now is a measure of all energy except kinetic energy of an object?
If so $E=mc^2$ does not necessarily predict the atomic bomb?
Instead of it saying that all objects have some intrinsic energy due to their mass it could say that an absolutely cold still object has no mass since it has no energy. Which sounds absurd.
So I am guessing that the old concept of mass from classical mechanics is an approximation to the new concept of mass in special relativity?
special-relativity energy mass momentum mass-energy
special-relativity energy mass momentum mass-energy
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
asked Jul 15 at 13:49
AndyAndy
1888 bronze badges
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
asked Jul 15 at 13:49
AndyAndy
1888 bronze badges
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
edited Jul 15 at 18:42
Qmechanic♦
111k12 gold badges215 silver badges1315 bronze badges
111k12 gold badges215 silver badges1315 bronze badges
asked Jul 15 at 13:49
AndyAndy
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asked Jul 15 at 13:49
AndyAndy
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asked Jul 15 at 13:49
AndyAndy
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1
$begingroup$
Have a look hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . The third equation is not used in particle physics, because it is misleading, it is the inertial mass acquired when particles reach close to the velocity of light. What is used is the "invariant mass" , the "length" of the four vector of a particle , or a system of particles. Vector algebra is used.
$endgroup$
– anna v
Jul 15 at 14:12
2
$begingroup$
So far you have only considered an isolated particle. If you extend the consideration to system you’ll find that the mass of a system can include some of the kinetic energy of its constituents.
$endgroup$
– dmckee♦
Jul 15 at 14:26
1
$begingroup$
In relation to @dmckee's point, when you say that a hot object has more mass than an otherwise identical cold object, you are actually pointing to the fact that some fraction of the mass of a system comes from some of the kinetic energy of its constituents.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:29
$begingroup$
Yes you are right. I have updated my question. The kinetic energy due to the momentum of the system as a whole is the firsts term.
$endgroup$
– Andy
Jul 15 at 14:33
1
$begingroup$
@ZachMcDargh We do not empirically see that the mass of an object does not depend on temperature. It very much does.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:40
|
show 1 more comment
1
$begingroup$
Have a look hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . The third equation is not used in particle physics, because it is misleading, it is the inertial mass acquired when particles reach close to the velocity of light. What is used is the "invariant mass" , the "length" of the four vector of a particle , or a system of particles. Vector algebra is used.
$endgroup$
– anna v
Jul 15 at 14:12
2
$begingroup$
So far you have only considered an isolated particle. If you extend the consideration to system you’ll find that the mass of a system can include some of the kinetic energy of its constituents.
$endgroup$
– dmckee♦
Jul 15 at 14:26
1
$begingroup$
In relation to @dmckee's point, when you say that a hot object has more mass than an otherwise identical cold object, you are actually pointing to the fact that some fraction of the mass of a system comes from some of the kinetic energy of its constituents.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:29
$begingroup$
Yes you are right. I have updated my question. The kinetic energy due to the momentum of the system as a whole is the firsts term.
$endgroup$
– Andy
Jul 15 at 14:33
1
$begingroup$
@ZachMcDargh We do not empirically see that the mass of an object does not depend on temperature. It very much does.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:40
1
1
$begingroup$
Have a look hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . The third equation is not used in particle physics, because it is misleading, it is the inertial mass acquired when particles reach close to the velocity of light. What is used is the "invariant mass" , the "length" of the four vector of a particle , or a system of particles. Vector algebra is used.
$endgroup$
– anna v
Jul 15 at 14:12
$begingroup$
Have a look hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . The third equation is not used in particle physics, because it is misleading, it is the inertial mass acquired when particles reach close to the velocity of light. What is used is the "invariant mass" , the "length" of the four vector of a particle , or a system of particles. Vector algebra is used.
$endgroup$
– anna v
Jul 15 at 14:12
2
2
$begingroup$
So far you have only considered an isolated particle. If you extend the consideration to system you’ll find that the mass of a system can include some of the kinetic energy of its constituents.
$endgroup$
– dmckee♦
Jul 15 at 14:26
$begingroup$
So far you have only considered an isolated particle. If you extend the consideration to system you’ll find that the mass of a system can include some of the kinetic energy of its constituents.
$endgroup$
– dmckee♦
Jul 15 at 14:26
1
1
$begingroup$
In relation to @dmckee's point, when you say that a hot object has more mass than an otherwise identical cold object, you are actually pointing to the fact that some fraction of the mass of a system comes from some of the kinetic energy of its constituents.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:29
$begingroup$
In relation to @dmckee's point, when you say that a hot object has more mass than an otherwise identical cold object, you are actually pointing to the fact that some fraction of the mass of a system comes from some of the kinetic energy of its constituents.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:29
$begingroup$
Yes you are right. I have updated my question. The kinetic energy due to the momentum of the system as a whole is the firsts term.
$endgroup$
– Andy
Jul 15 at 14:33
$begingroup$
Yes you are right. I have updated my question. The kinetic energy due to the momentum of the system as a whole is the firsts term.
$endgroup$
– Andy
Jul 15 at 14:33
1
1
$begingroup$
@ZachMcDargh We do not empirically see that the mass of an object does not depend on temperature. It very much does.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:40
$begingroup$
@ZachMcDargh We do not empirically see that the mass of an object does not depend on temperature. It very much does.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:40
|
show 1 more comment
2 Answers
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Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a measure of all energy except the kinetic energy is correct with a couple of caveats:
- As you already notice in the updated version of your question, a many-particle system can have motion of its constituents which do not contribute to the overall momentum of the system but do contribute to the overall energy. And thus, they contribute to the mass of the system as a whole. Thus, the mass of the system as a whole does include contributions from kinetic energy but only from the kinetic energy that doesn't contribute to the overall momentum of the system.
- Due to the quadratic nature of the relation $E^2=p^2+m^2$, it is a bit problematic to directly identify $p^2$ with the kinetic energy of the system. Rather, the kinetic energy would be $sqrtp^2+m^2-m$ which can be approximated to be $fracp^22m$ as usual for values of $p$ that are very small as compared to $m$. If you naively identify the $p^2$ as the kinetic energy, you wouldn't recover the correct non-relativistic limit.
Now, all your claims such as a hot cup of coffee having more mass than an otherwise identical but cold cup of coffee are true. However, this doesn't mean that the notion of mass doesn't anymore correspond to the property of inertia. Relativity doesn't change the notion of mass completely--it rather corrects it while unifying it with the notion of energy. In particular, it would be more difficult to accelerate a hot cup of coffee than a cold one if you can measure all the minuscule effects. So, the notion of mass in relativity is yet very much representative of the quality of inertia. The way to see this is to write down the expressions for momentum and energy in relativity. As you probably know, in relativity, $$p=fracmvsqrt1-v^2$$$$E=fracmsqrt1-v^2$$As you can see, it is the same $m$ that enters the formula for the energy also enters the formula for momentum. Thus, the same $m$ that represents both the measure of the energy in the rest frame (via entering the formula for energy) and the property of inertia (via entering the formula for momentum). This is the most basic conceptual unification represented in relativity and the genius of Einstein--the (rest) energy of a system is not independent of its inertia but two are the very same thing.
Now, finally, all of this doesn't mean that a cold object at rest shouldn't have any energy at all. It can very well have all sorts of reasons to have rest energy (and thus, mass). For example, even if all the constituents of a system are at rest and there is no interaction potential energy among them, the system as a whole would still have mass but it would simply be the sum of the masses of all its constituents. So, an object whose constituents are at rest simply means that its mass would not have contributions from the kinetic energy of its constituents. More importantly, there cannot be a massless system with no momentum (i.e., a system known to be rest must have mass). If something has no mass then having no momentum implies that its energy is also zero and this simply means that there is nothing.
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
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Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
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– Andy
Jul 15 at 15:45
1
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
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– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
2
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[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
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– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
add a comment |
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As per our currently accepted model, the SM, there are elementary point particles, with no internal structure, no spatial extension.
There are two types:
massless, like the photon, gluon
particles with rest mass, like the electron and quarks
Now the rest mass of the electron is an intrinsic property, theoretical, calculated, nobody actually measured an electron at rest.
If you put a photon into a box with massless walls, the box gains rest mass. Why? because the photons in it are exerting pressure on the walls.
Now matter, is built up by elementary particles. Quarks, and massless gluons build up neutrons and protons. Only 1% of the mass of the neutrons and protons comes from the rest mass of the quarks. 99% is the gluons (massless) bonding energy, because they are confined into the neutrons and protons (like photons in a box).
Neutrons and protons build up nucleuses. Some of the mass of the nucleus comes from the bonding energy between the neutrons and protons (massless pions bonding energy, like photons in a box). The rest comes from the rest mass of the neutrons and protons. As per the correct comments, as you go to higher levels (nucleus, atoms, molecules), the binding energy actually becomes negative in the mass of the total QM system.
But the main point is, that these nuclei's rest mass is mostly from the binding energy of the gluons inside the nucleons.
Atoms are built up by nuclei and electrons. The rest mass of these adds some part of the atom, but the bonding energy is again there (actually it is negative).
But here too, most of the rest mass of the atoms are made up of the binding energy of the gluons inside the nucleons.
As you build atoms into molecules, the rest mass of the atoms is there, but the covalent bond adds to the rest mass of the molucule too (negative).
Here too, most of the rest mass of the molecule is made up of the binding energy of the gluons inside the nucleons.
As you go closer to the macro world, less comes from bonding energy (confined massless particles like the photon in the box) and more from the rest mass of the constituents. The closer to QM, the more comes from the confinement of massless particles.
In relativistic QM, gluons are traveling at speed c. Their bonding energies give a lot to the mass of neutrons and protons. You are right, SR is compatible with QM.
Mass at the lowest scale is mostly just these massless particles (moving at relativistic speeds) confinement (bonding). Quarks rest mass adds only very little. As per our current model, the SM, these quarks and electrons do have an intrinsic rest mass (that we explain by Higgs). Maybe one day when we will have figured out what these are made up of, we will know that the rest mass of these are just confined massless particles too (strings).
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
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A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
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– Andy
Jul 15 at 15:19
3
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This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
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– dmckee♦
Jul 15 at 18:10
1
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@Andy yes you are correct, most of the mass of the matter comes from binding energy.
$endgroup$
– Árpád Szendrei
Jul 15 at 18:14
1
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
2
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
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$begingroup$
Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a measure of all energy except the kinetic energy is correct with a couple of caveats:
- As you already notice in the updated version of your question, a many-particle system can have motion of its constituents which do not contribute to the overall momentum of the system but do contribute to the overall energy. And thus, they contribute to the mass of the system as a whole. Thus, the mass of the system as a whole does include contributions from kinetic energy but only from the kinetic energy that doesn't contribute to the overall momentum of the system.
- Due to the quadratic nature of the relation $E^2=p^2+m^2$, it is a bit problematic to directly identify $p^2$ with the kinetic energy of the system. Rather, the kinetic energy would be $sqrtp^2+m^2-m$ which can be approximated to be $fracp^22m$ as usual for values of $p$ that are very small as compared to $m$. If you naively identify the $p^2$ as the kinetic energy, you wouldn't recover the correct non-relativistic limit.
Now, all your claims such as a hot cup of coffee having more mass than an otherwise identical but cold cup of coffee are true. However, this doesn't mean that the notion of mass doesn't anymore correspond to the property of inertia. Relativity doesn't change the notion of mass completely--it rather corrects it while unifying it with the notion of energy. In particular, it would be more difficult to accelerate a hot cup of coffee than a cold one if you can measure all the minuscule effects. So, the notion of mass in relativity is yet very much representative of the quality of inertia. The way to see this is to write down the expressions for momentum and energy in relativity. As you probably know, in relativity, $$p=fracmvsqrt1-v^2$$$$E=fracmsqrt1-v^2$$As you can see, it is the same $m$ that enters the formula for the energy also enters the formula for momentum. Thus, the same $m$ that represents both the measure of the energy in the rest frame (via entering the formula for energy) and the property of inertia (via entering the formula for momentum). This is the most basic conceptual unification represented in relativity and the genius of Einstein--the (rest) energy of a system is not independent of its inertia but two are the very same thing.
Now, finally, all of this doesn't mean that a cold object at rest shouldn't have any energy at all. It can very well have all sorts of reasons to have rest energy (and thus, mass). For example, even if all the constituents of a system are at rest and there is no interaction potential energy among them, the system as a whole would still have mass but it would simply be the sum of the masses of all its constituents. So, an object whose constituents are at rest simply means that its mass would not have contributions from the kinetic energy of its constituents. More importantly, there cannot be a massless system with no momentum (i.e., a system known to be rest must have mass). If something has no mass then having no momentum implies that its energy is also zero and this simply means that there is nothing.
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
$endgroup$
$begingroup$
Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
$endgroup$
– Andy
Jul 15 at 15:45
1
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
2
$begingroup$
[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
add a comment |
$begingroup$
Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a measure of all energy except the kinetic energy is correct with a couple of caveats:
- As you already notice in the updated version of your question, a many-particle system can have motion of its constituents which do not contribute to the overall momentum of the system but do contribute to the overall energy. And thus, they contribute to the mass of the system as a whole. Thus, the mass of the system as a whole does include contributions from kinetic energy but only from the kinetic energy that doesn't contribute to the overall momentum of the system.
- Due to the quadratic nature of the relation $E^2=p^2+m^2$, it is a bit problematic to directly identify $p^2$ with the kinetic energy of the system. Rather, the kinetic energy would be $sqrtp^2+m^2-m$ which can be approximated to be $fracp^22m$ as usual for values of $p$ that are very small as compared to $m$. If you naively identify the $p^2$ as the kinetic energy, you wouldn't recover the correct non-relativistic limit.
Now, all your claims such as a hot cup of coffee having more mass than an otherwise identical but cold cup of coffee are true. However, this doesn't mean that the notion of mass doesn't anymore correspond to the property of inertia. Relativity doesn't change the notion of mass completely--it rather corrects it while unifying it with the notion of energy. In particular, it would be more difficult to accelerate a hot cup of coffee than a cold one if you can measure all the minuscule effects. So, the notion of mass in relativity is yet very much representative of the quality of inertia. The way to see this is to write down the expressions for momentum and energy in relativity. As you probably know, in relativity, $$p=fracmvsqrt1-v^2$$$$E=fracmsqrt1-v^2$$As you can see, it is the same $m$ that enters the formula for the energy also enters the formula for momentum. Thus, the same $m$ that represents both the measure of the energy in the rest frame (via entering the formula for energy) and the property of inertia (via entering the formula for momentum). This is the most basic conceptual unification represented in relativity and the genius of Einstein--the (rest) energy of a system is not independent of its inertia but two are the very same thing.
Now, finally, all of this doesn't mean that a cold object at rest shouldn't have any energy at all. It can very well have all sorts of reasons to have rest energy (and thus, mass). For example, even if all the constituents of a system are at rest and there is no interaction potential energy among them, the system as a whole would still have mass but it would simply be the sum of the masses of all its constituents. So, an object whose constituents are at rest simply means that its mass would not have contributions from the kinetic energy of its constituents. More importantly, there cannot be a massless system with no momentum (i.e., a system known to be rest must have mass). If something has no mass then having no momentum implies that its energy is also zero and this simply means that there is nothing.
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
$endgroup$
$begingroup$
Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
$endgroup$
– Andy
Jul 15 at 15:45
1
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
2
$begingroup$
[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
add a comment |
$begingroup$
Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a measure of all energy except the kinetic energy is correct with a couple of caveats:
- As you already notice in the updated version of your question, a many-particle system can have motion of its constituents which do not contribute to the overall momentum of the system but do contribute to the overall energy. And thus, they contribute to the mass of the system as a whole. Thus, the mass of the system as a whole does include contributions from kinetic energy but only from the kinetic energy that doesn't contribute to the overall momentum of the system.
- Due to the quadratic nature of the relation $E^2=p^2+m^2$, it is a bit problematic to directly identify $p^2$ with the kinetic energy of the system. Rather, the kinetic energy would be $sqrtp^2+m^2-m$ which can be approximated to be $fracp^22m$ as usual for values of $p$ that are very small as compared to $m$. If you naively identify the $p^2$ as the kinetic energy, you wouldn't recover the correct non-relativistic limit.
Now, all your claims such as a hot cup of coffee having more mass than an otherwise identical but cold cup of coffee are true. However, this doesn't mean that the notion of mass doesn't anymore correspond to the property of inertia. Relativity doesn't change the notion of mass completely--it rather corrects it while unifying it with the notion of energy. In particular, it would be more difficult to accelerate a hot cup of coffee than a cold one if you can measure all the minuscule effects. So, the notion of mass in relativity is yet very much representative of the quality of inertia. The way to see this is to write down the expressions for momentum and energy in relativity. As you probably know, in relativity, $$p=fracmvsqrt1-v^2$$$$E=fracmsqrt1-v^2$$As you can see, it is the same $m$ that enters the formula for the energy also enters the formula for momentum. Thus, the same $m$ that represents both the measure of the energy in the rest frame (via entering the formula for energy) and the property of inertia (via entering the formula for momentum). This is the most basic conceptual unification represented in relativity and the genius of Einstein--the (rest) energy of a system is not independent of its inertia but two are the very same thing.
Now, finally, all of this doesn't mean that a cold object at rest shouldn't have any energy at all. It can very well have all sorts of reasons to have rest energy (and thus, mass). For example, even if all the constituents of a system are at rest and there is no interaction potential energy among them, the system as a whole would still have mass but it would simply be the sum of the masses of all its constituents. So, an object whose constituents are at rest simply means that its mass would not have contributions from the kinetic energy of its constituents. More importantly, there cannot be a massless system with no momentum (i.e., a system known to be rest must have mass). If something has no mass then having no momentum implies that its energy is also zero and this simply means that there is nothing.
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
$endgroup$
Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a measure of all energy except the kinetic energy is correct with a couple of caveats:
- As you already notice in the updated version of your question, a many-particle system can have motion of its constituents which do not contribute to the overall momentum of the system but do contribute to the overall energy. And thus, they contribute to the mass of the system as a whole. Thus, the mass of the system as a whole does include contributions from kinetic energy but only from the kinetic energy that doesn't contribute to the overall momentum of the system.
- Due to the quadratic nature of the relation $E^2=p^2+m^2$, it is a bit problematic to directly identify $p^2$ with the kinetic energy of the system. Rather, the kinetic energy would be $sqrtp^2+m^2-m$ which can be approximated to be $fracp^22m$ as usual for values of $p$ that are very small as compared to $m$. If you naively identify the $p^2$ as the kinetic energy, you wouldn't recover the correct non-relativistic limit.
Now, all your claims such as a hot cup of coffee having more mass than an otherwise identical but cold cup of coffee are true. However, this doesn't mean that the notion of mass doesn't anymore correspond to the property of inertia. Relativity doesn't change the notion of mass completely--it rather corrects it while unifying it with the notion of energy. In particular, it would be more difficult to accelerate a hot cup of coffee than a cold one if you can measure all the minuscule effects. So, the notion of mass in relativity is yet very much representative of the quality of inertia. The way to see this is to write down the expressions for momentum and energy in relativity. As you probably know, in relativity, $$p=fracmvsqrt1-v^2$$$$E=fracmsqrt1-v^2$$As you can see, it is the same $m$ that enters the formula for the energy also enters the formula for momentum. Thus, the same $m$ that represents both the measure of the energy in the rest frame (via entering the formula for energy) and the property of inertia (via entering the formula for momentum). This is the most basic conceptual unification represented in relativity and the genius of Einstein--the (rest) energy of a system is not independent of its inertia but two are the very same thing.
Now, finally, all of this doesn't mean that a cold object at rest shouldn't have any energy at all. It can very well have all sorts of reasons to have rest energy (and thus, mass). For example, even if all the constituents of a system are at rest and there is no interaction potential energy among them, the system as a whole would still have mass but it would simply be the sum of the masses of all its constituents. So, an object whose constituents are at rest simply means that its mass would not have contributions from the kinetic energy of its constituents. More importantly, there cannot be a massless system with no momentum (i.e., a system known to be rest must have mass). If something has no mass then having no momentum implies that its energy is also zero and this simply means that there is nothing.
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
edited Jul 15 at 15:18
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
answered Jul 15 at 15:03
Feynmans Out for Grumpy CatFeynmans Out for Grumpy Cat
6,4294 gold badges28 silver badges78 bronze badges
6,4294 gold badges28 silver badges78 bronze badges
$begingroup$
Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
$endgroup$
– Andy
Jul 15 at 15:45
1
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
2
$begingroup$
[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
add a comment |
$begingroup$
Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
$endgroup$
– Andy
Jul 15 at 15:45
1
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
2
$begingroup$
[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
$begingroup$
Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
$endgroup$
– Andy
Jul 15 at 15:45
$begingroup$
Thank you very much for explaining this! I take it what Einstein did could be summarized somewhat simplified this way: throw out mass conservation but fix Newtons equations by introducing a gamma correction factor + modifying the definition of mass slightly? That way he managed to unify classical mechanics and electromagnetics?
$endgroup$
– Andy
Jul 15 at 15:45
1
1
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
$begingroup$
@Andy Yes, you could say that. However, in relativity, mass conservation still applies--just not with the naive understanding of mass. For example, naively, when an electron and a proton combine into an atom, the mass of the atom is lower than the sum of the masses of the proton and the electron before they combined. But, the mass of the electron+proton system would be lower than the sum of their masses even before getting combined to form an atom due to the potential energy between them. [...]
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
2
2
$begingroup$
[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
$begingroup$
[...] Naively, we ignore this correction and take the mass of the whole system to be the sum of the masses of the free proton and the free electron. But, when they combine to form an atom, we can directly measure the mass of an atom as a single particle and this correction (which we should have considered all along) becomes manifest. In mathematical language, you can see that mass is obviously conserved in relativity in this way: the momentum four-vector $p^mu=(E,vecp)$ is conserved. Thus, its norm, $E^2-p^2$ is conserved. But, that is precisely the mass as $E^2=p^2+m^2$.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 15:59
add a comment |
$begingroup$
As per our currently accepted model, the SM, there are elementary point particles, with no internal structure, no spatial extension.
There are two types:
massless, like the photon, gluon
particles with rest mass, like the electron and quarks
Now the rest mass of the electron is an intrinsic property, theoretical, calculated, nobody actually measured an electron at rest.
If you put a photon into a box with massless walls, the box gains rest mass. Why? because the photons in it are exerting pressure on the walls.
Now matter, is built up by elementary particles. Quarks, and massless gluons build up neutrons and protons. Only 1% of the mass of the neutrons and protons comes from the rest mass of the quarks. 99% is the gluons (massless) bonding energy, because they are confined into the neutrons and protons (like photons in a box).
Neutrons and protons build up nucleuses. Some of the mass of the nucleus comes from the bonding energy between the neutrons and protons (massless pions bonding energy, like photons in a box). The rest comes from the rest mass of the neutrons and protons. As per the correct comments, as you go to higher levels (nucleus, atoms, molecules), the binding energy actually becomes negative in the mass of the total QM system.
But the main point is, that these nuclei's rest mass is mostly from the binding energy of the gluons inside the nucleons.
Atoms are built up by nuclei and electrons. The rest mass of these adds some part of the atom, but the bonding energy is again there (actually it is negative).
But here too, most of the rest mass of the atoms are made up of the binding energy of the gluons inside the nucleons.
As you build atoms into molecules, the rest mass of the atoms is there, but the covalent bond adds to the rest mass of the molucule too (negative).
Here too, most of the rest mass of the molecule is made up of the binding energy of the gluons inside the nucleons.
As you go closer to the macro world, less comes from bonding energy (confined massless particles like the photon in the box) and more from the rest mass of the constituents. The closer to QM, the more comes from the confinement of massless particles.
In relativistic QM, gluons are traveling at speed c. Their bonding energies give a lot to the mass of neutrons and protons. You are right, SR is compatible with QM.
Mass at the lowest scale is mostly just these massless particles (moving at relativistic speeds) confinement (bonding). Quarks rest mass adds only very little. As per our current model, the SM, these quarks and electrons do have an intrinsic rest mass (that we explain by Higgs). Maybe one day when we will have figured out what these are made up of, we will know that the rest mass of these are just confined massless particles too (strings).
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
$endgroup$
2
$begingroup$
A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
$endgroup$
– Andy
Jul 15 at 15:19
3
$begingroup$
This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
$endgroup$
– dmckee♦
Jul 15 at 18:10
1
$begingroup$
@Andy yes you are correct, most of the mass of the matter comes from binding energy.
$endgroup$
– Árpád Szendrei
Jul 15 at 18:14
1
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
2
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
|
show 3 more comments
$begingroup$
As per our currently accepted model, the SM, there are elementary point particles, with no internal structure, no spatial extension.
There are two types:
massless, like the photon, gluon
particles with rest mass, like the electron and quarks
Now the rest mass of the electron is an intrinsic property, theoretical, calculated, nobody actually measured an electron at rest.
If you put a photon into a box with massless walls, the box gains rest mass. Why? because the photons in it are exerting pressure on the walls.
Now matter, is built up by elementary particles. Quarks, and massless gluons build up neutrons and protons. Only 1% of the mass of the neutrons and protons comes from the rest mass of the quarks. 99% is the gluons (massless) bonding energy, because they are confined into the neutrons and protons (like photons in a box).
Neutrons and protons build up nucleuses. Some of the mass of the nucleus comes from the bonding energy between the neutrons and protons (massless pions bonding energy, like photons in a box). The rest comes from the rest mass of the neutrons and protons. As per the correct comments, as you go to higher levels (nucleus, atoms, molecules), the binding energy actually becomes negative in the mass of the total QM system.
But the main point is, that these nuclei's rest mass is mostly from the binding energy of the gluons inside the nucleons.
Atoms are built up by nuclei and electrons. The rest mass of these adds some part of the atom, but the bonding energy is again there (actually it is negative).
But here too, most of the rest mass of the atoms are made up of the binding energy of the gluons inside the nucleons.
As you build atoms into molecules, the rest mass of the atoms is there, but the covalent bond adds to the rest mass of the molucule too (negative).
Here too, most of the rest mass of the molecule is made up of the binding energy of the gluons inside the nucleons.
As you go closer to the macro world, less comes from bonding energy (confined massless particles like the photon in the box) and more from the rest mass of the constituents. The closer to QM, the more comes from the confinement of massless particles.
In relativistic QM, gluons are traveling at speed c. Their bonding energies give a lot to the mass of neutrons and protons. You are right, SR is compatible with QM.
Mass at the lowest scale is mostly just these massless particles (moving at relativistic speeds) confinement (bonding). Quarks rest mass adds only very little. As per our current model, the SM, these quarks and electrons do have an intrinsic rest mass (that we explain by Higgs). Maybe one day when we will have figured out what these are made up of, we will know that the rest mass of these are just confined massless particles too (strings).
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
$endgroup$
2
$begingroup$
A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
$endgroup$
– Andy
Jul 15 at 15:19
3
$begingroup$
This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
$endgroup$
– dmckee♦
Jul 15 at 18:10
1
$begingroup$
@Andy yes you are correct, most of the mass of the matter comes from binding energy.
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– Árpád Szendrei
Jul 15 at 18:14
1
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
2
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
|
show 3 more comments
$begingroup$
As per our currently accepted model, the SM, there are elementary point particles, with no internal structure, no spatial extension.
There are two types:
massless, like the photon, gluon
particles with rest mass, like the electron and quarks
Now the rest mass of the electron is an intrinsic property, theoretical, calculated, nobody actually measured an electron at rest.
If you put a photon into a box with massless walls, the box gains rest mass. Why? because the photons in it are exerting pressure on the walls.
Now matter, is built up by elementary particles. Quarks, and massless gluons build up neutrons and protons. Only 1% of the mass of the neutrons and protons comes from the rest mass of the quarks. 99% is the gluons (massless) bonding energy, because they are confined into the neutrons and protons (like photons in a box).
Neutrons and protons build up nucleuses. Some of the mass of the nucleus comes from the bonding energy between the neutrons and protons (massless pions bonding energy, like photons in a box). The rest comes from the rest mass of the neutrons and protons. As per the correct comments, as you go to higher levels (nucleus, atoms, molecules), the binding energy actually becomes negative in the mass of the total QM system.
But the main point is, that these nuclei's rest mass is mostly from the binding energy of the gluons inside the nucleons.
Atoms are built up by nuclei and electrons. The rest mass of these adds some part of the atom, but the bonding energy is again there (actually it is negative).
But here too, most of the rest mass of the atoms are made up of the binding energy of the gluons inside the nucleons.
As you build atoms into molecules, the rest mass of the atoms is there, but the covalent bond adds to the rest mass of the molucule too (negative).
Here too, most of the rest mass of the molecule is made up of the binding energy of the gluons inside the nucleons.
As you go closer to the macro world, less comes from bonding energy (confined massless particles like the photon in the box) and more from the rest mass of the constituents. The closer to QM, the more comes from the confinement of massless particles.
In relativistic QM, gluons are traveling at speed c. Their bonding energies give a lot to the mass of neutrons and protons. You are right, SR is compatible with QM.
Mass at the lowest scale is mostly just these massless particles (moving at relativistic speeds) confinement (bonding). Quarks rest mass adds only very little. As per our current model, the SM, these quarks and electrons do have an intrinsic rest mass (that we explain by Higgs). Maybe one day when we will have figured out what these are made up of, we will know that the rest mass of these are just confined massless particles too (strings).
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
$endgroup$
As per our currently accepted model, the SM, there are elementary point particles, with no internal structure, no spatial extension.
There are two types:
massless, like the photon, gluon
particles with rest mass, like the electron and quarks
Now the rest mass of the electron is an intrinsic property, theoretical, calculated, nobody actually measured an electron at rest.
If you put a photon into a box with massless walls, the box gains rest mass. Why? because the photons in it are exerting pressure on the walls.
Now matter, is built up by elementary particles. Quarks, and massless gluons build up neutrons and protons. Only 1% of the mass of the neutrons and protons comes from the rest mass of the quarks. 99% is the gluons (massless) bonding energy, because they are confined into the neutrons and protons (like photons in a box).
Neutrons and protons build up nucleuses. Some of the mass of the nucleus comes from the bonding energy between the neutrons and protons (massless pions bonding energy, like photons in a box). The rest comes from the rest mass of the neutrons and protons. As per the correct comments, as you go to higher levels (nucleus, atoms, molecules), the binding energy actually becomes negative in the mass of the total QM system.
But the main point is, that these nuclei's rest mass is mostly from the binding energy of the gluons inside the nucleons.
Atoms are built up by nuclei and electrons. The rest mass of these adds some part of the atom, but the bonding energy is again there (actually it is negative).
But here too, most of the rest mass of the atoms are made up of the binding energy of the gluons inside the nucleons.
As you build atoms into molecules, the rest mass of the atoms is there, but the covalent bond adds to the rest mass of the molucule too (negative).
Here too, most of the rest mass of the molecule is made up of the binding energy of the gluons inside the nucleons.
As you go closer to the macro world, less comes from bonding energy (confined massless particles like the photon in the box) and more from the rest mass of the constituents. The closer to QM, the more comes from the confinement of massless particles.
In relativistic QM, gluons are traveling at speed c. Their bonding energies give a lot to the mass of neutrons and protons. You are right, SR is compatible with QM.
Mass at the lowest scale is mostly just these massless particles (moving at relativistic speeds) confinement (bonding). Quarks rest mass adds only very little. As per our current model, the SM, these quarks and electrons do have an intrinsic rest mass (that we explain by Higgs). Maybe one day when we will have figured out what these are made up of, we will know that the rest mass of these are just confined massless particles too (strings).
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
edited Jul 15 at 18:29
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
answered Jul 15 at 14:56
Árpád SzendreiÁrpád Szendrei
6,9941 gold badge11 silver badges33 bronze badges
6,9941 gold badge11 silver badges33 bronze badges
2
$begingroup$
A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
$endgroup$
– Andy
Jul 15 at 15:19
3
$begingroup$
This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
$endgroup$
– dmckee♦
Jul 15 at 18:10
1
$begingroup$
@Andy yes you are correct, most of the mass of the matter comes from binding energy.
$endgroup$
– Árpád Szendrei
Jul 15 at 18:14
1
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
2
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
|
show 3 more comments
2
$begingroup$
A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
$endgroup$
– Andy
Jul 15 at 15:19
3
$begingroup$
This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
$endgroup$
– dmckee♦
Jul 15 at 18:10
1
$begingroup$
@Andy yes you are correct, most of the mass of the matter comes from binding energy.
$endgroup$
– Árpád Szendrei
Jul 15 at 18:14
1
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
2
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
2
2
$begingroup$
A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
$endgroup$
– Andy
Jul 15 at 15:19
$begingroup$
A hydrogen atom has slightly less mass than a free electron and a proton. That means that there is energy to "harvest" by combining free electron and protons into hydrogen: fusion? If I understand you correctly there is a lot more energy to be "harvested" by combining quarks into electrons and protons? Alternatively it takes a crazy amount of energy to split into quarks: CERN collider?
$endgroup$
– Andy
Jul 15 at 15:19
3
3
$begingroup$
This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
$endgroup$
– dmckee♦
Jul 15 at 18:10
$begingroup$
This answer feels like a bunch of disconnected facts thrown against the wall in hopes that some will stick, and it has some wrong (or at least misleading) statements in it. Pointedly, for nuclear, atomic, and molecular systems the binding energy is negative (reduces the system mass to less than the sum of the masses of the components). This makes lines like " Most of the mass of the nucleus comes from the bonding energy between the neutrons and protons" simply wrong.
$endgroup$
– dmckee♦
Jul 15 at 18:10
1
1
$begingroup$
@Andy yes you are correct, most of the mass of the matter comes from binding energy.
$endgroup$
– Árpád Szendrei
Jul 15 at 18:14
$begingroup$
@Andy yes you are correct, most of the mass of the matter comes from binding energy.
$endgroup$
– Árpád Szendrei
Jul 15 at 18:14
1
1
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
$begingroup$
@Andy There's a lot of confusion in your statements :) Combining a free electron and a proton into hydrogen isn't fusion, it's just chemistry. Both quarks and electrons are different elementary particles. This still releases energy, mind you. But the same principle does apply to fusion as well - combine a free proton and a free neutron, and you get a deuteron and a lot of energy. Combine two deuterons, and you get helium-4, and huge loads of energy. It works for fission in reverse - split a Uranium atom, and you get two daughter nuclei, some free neutrons and lots of energy.
$endgroup$
– Luaan
Jul 16 at 8:13
2
2
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
$begingroup$
This doesn't even seem to try to answer the question. OP wants to know about mass in SR, but you're giving a response on, as others noted, a hodgepodge of SM factoids.
$endgroup$
– Kyle Kanos
Jul 16 at 12:48
|
show 3 more comments
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Have a look hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . The third equation is not used in particle physics, because it is misleading, it is the inertial mass acquired when particles reach close to the velocity of light. What is used is the "invariant mass" , the "length" of the four vector of a particle , or a system of particles. Vector algebra is used.
$endgroup$
– anna v
Jul 15 at 14:12
2
$begingroup$
So far you have only considered an isolated particle. If you extend the consideration to system you’ll find that the mass of a system can include some of the kinetic energy of its constituents.
$endgroup$
– dmckee♦
Jul 15 at 14:26
1
$begingroup$
In relation to @dmckee's point, when you say that a hot object has more mass than an otherwise identical cold object, you are actually pointing to the fact that some fraction of the mass of a system comes from some of the kinetic energy of its constituents.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:29
$begingroup$
Yes you are right. I have updated my question. The kinetic energy due to the momentum of the system as a whole is the firsts term.
$endgroup$
– Andy
Jul 15 at 14:33
1
$begingroup$
@ZachMcDargh We do not empirically see that the mass of an object does not depend on temperature. It very much does.
$endgroup$
– Feynmans Out for Grumpy Cat
Jul 15 at 14:40