Is the acceleration of two objects with different temperature same with the same force?











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$F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?










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    $F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?










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      $F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?










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      $F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?







      general-relativity acceleration






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      asked 2 hours ago









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          At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
          $m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]



          The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
          $$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
          You'll recognise the left hand side as the Lorentz force.



          A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.



          You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.






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            down vote













            You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.



            The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.



            In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by



            $$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$



            where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.






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              At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
              $m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]



              The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
              $$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
              You'll recognise the left hand side as the Lorentz force.



              A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.



              You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.






              share|cite|improve this answer



























                up vote
                3
                down vote













                At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
                $m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]



                The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
                $$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
                You'll recognise the left hand side as the Lorentz force.



                A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.



                You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.






                share|cite|improve this answer

























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
                  $m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]



                  The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
                  $$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
                  You'll recognise the left hand side as the Lorentz force.



                  A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.



                  You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.






                  share|cite|improve this answer














                  At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
                  $m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]



                  The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
                  $$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
                  You'll recognise the left hand side as the Lorentz force.



                  A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.



                  You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 41 mins ago

























                  answered 1 hour ago









                  Philip Wood

                  7,4223616




                  7,4223616






















                      up vote
                      0
                      down vote













                      You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.



                      The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.



                      In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by



                      $$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$



                      where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.






                      share|cite

























                        up vote
                        0
                        down vote













                        You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.



                        The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.



                        In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by



                        $$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$



                        where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.






                        share|cite























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.



                          The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.



                          In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by



                          $$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$



                          where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.






                          share|cite












                          You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.



                          The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.



                          In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by



                          $$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$



                          where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.







                          share|cite












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