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Is this relativistic mass?
Relativistic Black Hole?Would an object lose physical mass if it accelerated to a relativistic speed (would an object burn it's own mass)?Is the energy of momentum stored physically?If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light?Will objects heat up and become hidden at relativistic speed?Dark Matter vs. Mass from Kinetic EnergyCan relativistic mass be treated as rest mass?Questions on MassProper mass and space-time wrap questionAre relativistic momentum and relativistic mass conserved in special relativity?
$begingroup$
I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc. But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object. This confuses me. Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases? Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?
special-relativity
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
$endgroup$
add a comment |
$begingroup$
I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc. But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object. This confuses me. Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases? Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?
special-relativity
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
$endgroup$
add a comment |
$begingroup$
I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc. But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object. This confuses me. Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases? Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?
special-relativity
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
$endgroup$
I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc. But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object. This confuses me. Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases? Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?
special-relativity
special-relativity
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
asked 5 hours ago
Achilles' AdvisorAchilles' Advisor
538
538
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object.
Yes, and this is not in contradiction with the convention of invariant mass. Mass is defined by the identity $m^2=E^2-p^2$ (in units where $c=1$), which implies that it isn't additive. So say I have two electrons, each with mass $m$. If one is moving to the right at $0.9c$, and the other is moving to the left at $-0.9c$, then the mass of the whole system is greater than $2m$. However, each electron individually still has mass $m$.
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
$endgroup$
add a comment |
$begingroup$
Yes, you can obtain alternatives to the ordinary Einstein equivalence relation, for instance, Max Planck suggested a correction of the form
$E = mc^2 + PV$
Which would take into account internal thermal contributions to the rest mass. The constituent particles which a system is also subject to kinetic energy (they are in motion) and as predicted from the theory of systems being heated, the particles gain energy and so contribute to rest mass. It's sort of similar to when a photon enters a box, the box's mass will increase according to the energy gained. In the same way, kinetic theory of heat involves the excitation of many particles and so contribute to larger mass. But it certainly is not a relativistic mass for the system contribution.
answered 4 hours ago
Gareth MeredithGareth Meredith
1
$endgroup$
2
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object.
Yes, and this is not in contradiction with the convention of invariant mass. Mass is defined by the identity $m^2=E^2-p^2$ (in units where $c=1$), which implies that it isn't additive. So say I have two electrons, each with mass $m$. If one is moving to the right at $0.9c$, and the other is moving to the left at $-0.9c$, then the mass of the whole system is greater than $2m$. However, each electron individually still has mass $m$.
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
$endgroup$
add a comment |
$begingroup$
But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object.
Yes, and this is not in contradiction with the convention of invariant mass. Mass is defined by the identity $m^2=E^2-p^2$ (in units where $c=1$), which implies that it isn't additive. So say I have two electrons, each with mass $m$. If one is moving to the right at $0.9c$, and the other is moving to the left at $-0.9c$, then the mass of the whole system is greater than $2m$. However, each electron individually still has mass $m$.
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
$endgroup$
add a comment |
$begingroup$
But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object.
Yes, and this is not in contradiction with the convention of invariant mass. Mass is defined by the identity $m^2=E^2-p^2$ (in units where $c=1$), which implies that it isn't additive. So say I have two electrons, each with mass $m$. If one is moving to the right at $0.9c$, and the other is moving to the left at $-0.9c$, then the mass of the whole system is greater than $2m$. However, each electron individually still has mass $m$.
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
$endgroup$
But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object.
Yes, and this is not in contradiction with the convention of invariant mass. Mass is defined by the identity $m^2=E^2-p^2$ (in units where $c=1$), which implies that it isn't additive. So say I have two electrons, each with mass $m$. If one is moving to the right at $0.9c$, and the other is moving to the left at $-0.9c$, then the mass of the whole system is greater than $2m$. However, each electron individually still has mass $m$.
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
answered 4 hours ago
Ben CrowellBen Crowell
53.9k6165313
53.9k6165313
add a comment |
add a comment |
$begingroup$
Yes, you can obtain alternatives to the ordinary Einstein equivalence relation, for instance, Max Planck suggested a correction of the form
$E = mc^2 + PV$
Which would take into account internal thermal contributions to the rest mass. The constituent particles which a system is also subject to kinetic energy (they are in motion) and as predicted from the theory of systems being heated, the particles gain energy and so contribute to rest mass. It's sort of similar to when a photon enters a box, the box's mass will increase according to the energy gained. In the same way, kinetic theory of heat involves the excitation of many particles and so contribute to larger mass. But it certainly is not a relativistic mass for the system contribution.
answered 4 hours ago
Gareth MeredithGareth Meredith
1
$endgroup$
2
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
add a comment |
$begingroup$
Yes, you can obtain alternatives to the ordinary Einstein equivalence relation, for instance, Max Planck suggested a correction of the form
$E = mc^2 + PV$
Which would take into account internal thermal contributions to the rest mass. The constituent particles which a system is also subject to kinetic energy (they are in motion) and as predicted from the theory of systems being heated, the particles gain energy and so contribute to rest mass. It's sort of similar to when a photon enters a box, the box's mass will increase according to the energy gained. In the same way, kinetic theory of heat involves the excitation of many particles and so contribute to larger mass. But it certainly is not a relativistic mass for the system contribution.
answered 4 hours ago
Gareth MeredithGareth Meredith
1
$endgroup$
2
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
add a comment |
$begingroup$
Yes, you can obtain alternatives to the ordinary Einstein equivalence relation, for instance, Max Planck suggested a correction of the form
$E = mc^2 + PV$
Which would take into account internal thermal contributions to the rest mass. The constituent particles which a system is also subject to kinetic energy (they are in motion) and as predicted from the theory of systems being heated, the particles gain energy and so contribute to rest mass. It's sort of similar to when a photon enters a box, the box's mass will increase according to the energy gained. In the same way, kinetic theory of heat involves the excitation of many particles and so contribute to larger mass. But it certainly is not a relativistic mass for the system contribution.
answered 4 hours ago
Gareth MeredithGareth Meredith
1
$endgroup$
Yes, you can obtain alternatives to the ordinary Einstein equivalence relation, for instance, Max Planck suggested a correction of the form
$E = mc^2 + PV$
Which would take into account internal thermal contributions to the rest mass. The constituent particles which a system is also subject to kinetic energy (they are in motion) and as predicted from the theory of systems being heated, the particles gain energy and so contribute to rest mass. It's sort of similar to when a photon enters a box, the box's mass will increase according to the energy gained. In the same way, kinetic theory of heat involves the excitation of many particles and so contribute to larger mass. But it certainly is not a relativistic mass for the system contribution.
answered 4 hours ago
Gareth MeredithGareth Meredith
1
edited 4 hours ago
answered 4 hours ago
Gareth MeredithGareth Meredith
1
answered 4 hours ago
Gareth MeredithGareth Meredith
1
answered 4 hours ago
Gareth MeredithGareth Meredith
1
1
2
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
add a comment |
2
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
2
2
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This seems obscure, historical, or speculative. The OP isn't asking anything obscure. They're just asking a question about how mass behaves in standard SR.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
This isn't obscure, or speculative. Give some reasons why?
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
$begingroup$
And no they are not, they are asking how thermal contributions from the constituent particles of a system, may contribute to the rest mass of a system, including also if this is a case of relativistic mass, which I explained it wasn't. This is good old classical physics and equipartition.
$endgroup$
– Gareth Meredith
4 hours ago
add a comment |
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