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Is this relativistic mass?


Does relativistic mass exhibit gravitiational effects?Would an object lose physical mass if it accelerated to a relativistic speed (would an object burn it's own mass)?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?Can relativistic mass be treated as rest mass?Questions on MassProper mass and space-time wrap questionGravitational Field of a Photon compared to that of Massive MatterDoes the mass of object really increase?Are relativistic momentum and relativistic mass conserved in special relativity?













3












$begingroup$


I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc.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?










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc.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?










    share|cite|improve this question











    $endgroup$














      3












      3








      3


      1



      $begingroup$


      I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc.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?










      share|cite|improve this question











      $endgroup$




      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 mass inertial-frames mass-energy






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Qmechanic

      107k121991239




      107k121991239










      asked 7 hours ago









      Achilles' AdvisorAchilles' Advisor

      538




      538




















          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$


          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 $m$.






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$


            Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




            Outdated does not mean wrong. It means confusing, since we have much better tools to study the microcosm of atoms, since it was realized that special relativity in the motion of particles is completely and cleanly described by defining the relativistic four vectors, which obey vector equations.



            Here is the energy momentum four vector :



            E,pinvarmass



            and to the right the definition of the invariant mass:




            The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.




            As with the everyday length of vectors, lengths are not addivive, one has to use vector addition, in the special relativity case as defined on the right.




            Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?




            Note what you said:




            Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




            bold mine.



            When you hold a solid, is the solid moving with respect to your observation? The statement holds mathematically for each individual electron and atom with respect to the other, but statistically there is no motion that an external observer can measure.
            The four vector formalism simplifies this. The addition of all the four vectors in a solid will give the total four vector whose length is the mass you can measure in the laboratory. Hotter items have higher momenta and the total addition of four vectors will give higher invariant mass for a hot object than it has when cold.






            share|cite|improve this answer









            $endgroup$




















              -4












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$








              • 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
                7 hours ago










              • $begingroup$
                This isn't obscure, or speculative. Give some reasons why?
                $endgroup$
                – Gareth Meredith
                7 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
                7 hours ago











              Your Answer





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              7












              $begingroup$


              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 $m$.






              share|cite|improve this answer









              $endgroup$

















                7












                $begingroup$


                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 $m$.






                share|cite|improve this answer









                $endgroup$















                  7












                  7








                  7





                  $begingroup$


                  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 $m$.






                  share|cite|improve this answer









                  $endgroup$




                  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 $m$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  Ben CrowellBen Crowell

                  53.9k6165313




                  53.9k6165313





















                      0












                      $begingroup$


                      Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                      Outdated does not mean wrong. It means confusing, since we have much better tools to study the microcosm of atoms, since it was realized that special relativity in the motion of particles is completely and cleanly described by defining the relativistic four vectors, which obey vector equations.



                      Here is the energy momentum four vector :



                      E,pinvarmass



                      and to the right the definition of the invariant mass:




                      The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.




                      As with the everyday length of vectors, lengths are not addivive, one has to use vector addition, in the special relativity case as defined on the right.




                      Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?




                      Note what you said:




                      Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                      bold mine.



                      When you hold a solid, is the solid moving with respect to your observation? The statement holds mathematically for each individual electron and atom with respect to the other, but statistically there is no motion that an external observer can measure.
                      The four vector formalism simplifies this. The addition of all the four vectors in a solid will give the total four vector whose length is the mass you can measure in the laboratory. Hotter items have higher momenta and the total addition of four vectors will give higher invariant mass for a hot object than it has when cold.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$


                        Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                        Outdated does not mean wrong. It means confusing, since we have much better tools to study the microcosm of atoms, since it was realized that special relativity in the motion of particles is completely and cleanly described by defining the relativistic four vectors, which obey vector equations.



                        Here is the energy momentum four vector :



                        E,pinvarmass



                        and to the right the definition of the invariant mass:




                        The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.




                        As with the everyday length of vectors, lengths are not addivive, one has to use vector addition, in the special relativity case as defined on the right.




                        Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?




                        Note what you said:




                        Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                        bold mine.



                        When you hold a solid, is the solid moving with respect to your observation? The statement holds mathematically for each individual electron and atom with respect to the other, but statistically there is no motion that an external observer can measure.
                        The four vector formalism simplifies this. The addition of all the four vectors in a solid will give the total four vector whose length is the mass you can measure in the laboratory. Hotter items have higher momenta and the total addition of four vectors will give higher invariant mass for a hot object than it has when cold.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$


                          Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                          Outdated does not mean wrong. It means confusing, since we have much better tools to study the microcosm of atoms, since it was realized that special relativity in the motion of particles is completely and cleanly described by defining the relativistic four vectors, which obey vector equations.



                          Here is the energy momentum four vector :



                          E,pinvarmass



                          and to the right the definition of the invariant mass:




                          The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.




                          As with the everyday length of vectors, lengths are not addivive, one has to use vector addition, in the special relativity case as defined on the right.




                          Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?




                          Note what you said:




                          Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                          bold mine.



                          When you hold a solid, is the solid moving with respect to your observation? The statement holds mathematically for each individual electron and atom with respect to the other, but statistically there is no motion that an external observer can measure.
                          The four vector formalism simplifies this. The addition of all the four vectors in a solid will give the total four vector whose length is the mass you can measure in the laboratory. Hotter items have higher momenta and the total addition of four vectors will give higher invariant mass for a hot object than it has when cold.






                          share|cite|improve this answer









                          $endgroup$




                          Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                          Outdated does not mean wrong. It means confusing, since we have much better tools to study the microcosm of atoms, since it was realized that special relativity in the motion of particles is completely and cleanly described by defining the relativistic four vectors, which obey vector equations.



                          Here is the energy momentum four vector :



                          E,pinvarmass



                          and to the right the definition of the invariant mass:




                          The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.




                          As with the everyday length of vectors, lengths are not addivive, one has to use vector addition, in the special relativity case as defined on the right.




                          Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?




                          Note what you said:




                          Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?




                          bold mine.



                          When you hold a solid, is the solid moving with respect to your observation? The statement holds mathematically for each individual electron and atom with respect to the other, but statistically there is no motion that an external observer can measure.
                          The four vector formalism simplifies this. The addition of all the four vectors in a solid will give the total four vector whose length is the mass you can measure in the laboratory. Hotter items have higher momenta and the total addition of four vectors will give higher invariant mass for a hot object than it has when cold.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 22 mins ago









                          anna vanna v

                          161k8153453




                          161k8153453





















                              -4












                              $begingroup$

                              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.






                              share|cite|improve this answer











                              $endgroup$








                              • 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
                                7 hours ago










                              • $begingroup$
                                This isn't obscure, or speculative. Give some reasons why?
                                $endgroup$
                                – Gareth Meredith
                                7 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
                                7 hours ago















                              -4












                              $begingroup$

                              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.






                              share|cite|improve this answer











                              $endgroup$








                              • 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
                                7 hours ago










                              • $begingroup$
                                This isn't obscure, or speculative. Give some reasons why?
                                $endgroup$
                                – Gareth Meredith
                                7 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
                                7 hours ago













                              -4












                              -4








                              -4





                              $begingroup$

                              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.






                              share|cite|improve this answer











                              $endgroup$



                              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.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 7 hours ago

























                              answered 7 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
                                7 hours ago










                              • $begingroup$
                                This isn't obscure, or speculative. Give some reasons why?
                                $endgroup$
                                – Gareth Meredith
                                7 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
                                7 hours ago












                              • 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
                                7 hours ago










                              • $begingroup$
                                This isn't obscure, or speculative. Give some reasons why?
                                $endgroup$
                                – Gareth Meredith
                                7 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
                                7 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
                              7 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
                              7 hours ago












                              $begingroup$
                              This isn't obscure, or speculative. Give some reasons why?
                              $endgroup$
                              – Gareth Meredith
                              7 hours ago




                              $begingroup$
                              This isn't obscure, or speculative. Give some reasons why?
                              $endgroup$
                              – Gareth Meredith
                              7 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
                              7 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
                              7 hours ago

















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