Why is a particle non-relativistic when its kinetic energy is small compared to its rest energy?












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$begingroup$


For example, nucleons in nucleus are in motion with kinetic energies of 10 MeV. Their rest energies are about 1000 MeV. Kinetic energy of nucleons is small compared to rest energy. They are hence considered non-relativistic.






special-relativity energy speed-of-light speed approximations







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$endgroup$

















    4












    $begingroup$


    For example, nucleons in nucleus are in motion with kinetic energies of 10 MeV. Their rest energies are about 1000 MeV. Kinetic energy of nucleons is small compared to rest energy. They are hence considered non-relativistic.






    special-relativity energy speed-of-light speed approximations







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      2



      $begingroup$


      For example, nucleons in nucleus are in motion with kinetic energies of 10 MeV. Their rest energies are about 1000 MeV. Kinetic energy of nucleons is small compared to rest energy. They are hence considered non-relativistic.






      special-relativity energy speed-of-light speed approximations







      share|cite|improve this question











      $endgroup$




      For example, nucleons in nucleus are in motion with kinetic energies of 10 MeV. Their rest energies are about 1000 MeV. Kinetic energy of nucleons is small compared to rest energy. They are hence considered non-relativistic.






      special-relativity energy speed-of-light speed approximations




      special-relativity energy speed-of-light speed approximations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 8 hours ago









      Qmechanic♦

      104k121881193




      104k121881193










      asked 12 hours ago









      TaeNyFanTaeNyFan

      3417




      3417






















          3 Answers
          3






          active

          oldest

          votes


















          13












          $begingroup$

          When we say a particle is non-relativistic we mean the Lorentz factor $gamma$ is close to one, where $gamma$ is given by:



          $$ gamma = frac{1}{sqrt{1 - v^2/c^2}} $$



          So saying $gamma$ is close to one means that the velocity $v$ must be much less than $c$.



          With a bit of algebra we can show that the kinetic energy of a particle is given by:



          $$ T =(gamma - 1)mc^2 $$



          And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:



          $$ frac{T}{E} = frac{(gamma - 1)mc^2}{mc^2} = gamma - 1 $$



          And if this ratio is small that means $gamma approx 1$, which was our original criterion for non-relativistic behviour.






          share|cite|improve this answer











          $endgroup$





















            13












            $begingroup$

            I would like to add something to the already great answers posted.



            Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".



            In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated
            by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-7}$, and would otherwise give errors of kilometers.



            The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
              $endgroup$
              – Cort Ammon
              6 hours ago










            • $begingroup$
              It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
              $endgroup$
              – Salvador Villarreal
              2 hours ago










            • $begingroup$
              Are you sure of that 10^-7?? I thought it was more like 10^-12.
              $endgroup$
              – Loren Pechtel
              1 hour ago










            • $begingroup$
              I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
              $endgroup$
              – Salvador Villarreal
              1 hour ago





















            10












            $begingroup$

            'Non-relativistic' means $vll c$.



            That is effectively the same as $gamma approx 1$ as $gamma={1 over sqrt{1-v^2/c^2}}$.



            But also $gamma={E_{tot}over E_{rest}} equiv 1+{E_{kin} over E_{rest}}$



            So if the kinetic energy is small compared to the rest mass, $gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.






            share|cite|improve this answer











            $endgroup$













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              3 Answers
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              active

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              3 Answers
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              13












              $begingroup$

              When we say a particle is non-relativistic we mean the Lorentz factor $gamma$ is close to one, where $gamma$ is given by:



              $$ gamma = frac{1}{sqrt{1 - v^2/c^2}} $$



              So saying $gamma$ is close to one means that the velocity $v$ must be much less than $c$.



              With a bit of algebra we can show that the kinetic energy of a particle is given by:



              $$ T =(gamma - 1)mc^2 $$



              And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:



              $$ frac{T}{E} = frac{(gamma - 1)mc^2}{mc^2} = gamma - 1 $$



              And if this ratio is small that means $gamma approx 1$, which was our original criterion for non-relativistic behviour.






              share|cite|improve this answer











              $endgroup$


















                13












                $begingroup$

                When we say a particle is non-relativistic we mean the Lorentz factor $gamma$ is close to one, where $gamma$ is given by:



                $$ gamma = frac{1}{sqrt{1 - v^2/c^2}} $$



                So saying $gamma$ is close to one means that the velocity $v$ must be much less than $c$.



                With a bit of algebra we can show that the kinetic energy of a particle is given by:



                $$ T =(gamma - 1)mc^2 $$



                And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:



                $$ frac{T}{E} = frac{(gamma - 1)mc^2}{mc^2} = gamma - 1 $$



                And if this ratio is small that means $gamma approx 1$, which was our original criterion for non-relativistic behviour.






                share|cite|improve this answer











                $endgroup$
















                  13












                  13








                  13





                  $begingroup$

                  When we say a particle is non-relativistic we mean the Lorentz factor $gamma$ is close to one, where $gamma$ is given by:



                  $$ gamma = frac{1}{sqrt{1 - v^2/c^2}} $$



                  So saying $gamma$ is close to one means that the velocity $v$ must be much less than $c$.



                  With a bit of algebra we can show that the kinetic energy of a particle is given by:



                  $$ T =(gamma - 1)mc^2 $$



                  And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:



                  $$ frac{T}{E} = frac{(gamma - 1)mc^2}{mc^2} = gamma - 1 $$



                  And if this ratio is small that means $gamma approx 1$, which was our original criterion for non-relativistic behviour.






                  share|cite|improve this answer











                  $endgroup$



                  When we say a particle is non-relativistic we mean the Lorentz factor $gamma$ is close to one, where $gamma$ is given by:



                  $$ gamma = frac{1}{sqrt{1 - v^2/c^2}} $$



                  So saying $gamma$ is close to one means that the velocity $v$ must be much less than $c$.



                  With a bit of algebra we can show that the kinetic energy of a particle is given by:



                  $$ T =(gamma - 1)mc^2 $$



                  And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:



                  $$ frac{T}{E} = frac{(gamma - 1)mc^2}{mc^2} = gamma - 1 $$



                  And if this ratio is small that means $gamma approx 1$, which was our original criterion for non-relativistic behviour.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 11 hours ago

























                  answered 12 hours ago









                  John RennieJohn Rennie

                  274k43543790




                  274k43543790























                      13












                      $begingroup$

                      I would like to add something to the already great answers posted.



                      Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".



                      In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated
                      by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-7}$, and would otherwise give errors of kilometers.



                      The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
                        $endgroup$
                        – Cort Ammon
                        6 hours ago










                      • $begingroup$
                        It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
                        $endgroup$
                        – Salvador Villarreal
                        2 hours ago










                      • $begingroup$
                        Are you sure of that 10^-7?? I thought it was more like 10^-12.
                        $endgroup$
                        – Loren Pechtel
                        1 hour ago










                      • $begingroup$
                        I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
                        $endgroup$
                        – Salvador Villarreal
                        1 hour ago


















                      13












                      $begingroup$

                      I would like to add something to the already great answers posted.



                      Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".



                      In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated
                      by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-7}$, and would otherwise give errors of kilometers.



                      The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
                        $endgroup$
                        – Cort Ammon
                        6 hours ago










                      • $begingroup$
                        It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
                        $endgroup$
                        – Salvador Villarreal
                        2 hours ago










                      • $begingroup$
                        Are you sure of that 10^-7?? I thought it was more like 10^-12.
                        $endgroup$
                        – Loren Pechtel
                        1 hour ago










                      • $begingroup$
                        I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
                        $endgroup$
                        – Salvador Villarreal
                        1 hour ago
















                      13












                      13








                      13





                      $begingroup$

                      I would like to add something to the already great answers posted.



                      Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".



                      In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated
                      by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-7}$, and would otherwise give errors of kilometers.



                      The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.






                      share|cite|improve this answer











                      $endgroup$



                      I would like to add something to the already great answers posted.



                      Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".



                      In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated
                      by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-7}$, and would otherwise give errors of kilometers.



                      The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 11 hours ago

























                      answered 11 hours ago









                      Salvador VillarrealSalvador Villarreal

                      1819




                      1819












                      • $begingroup$
                        The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
                        $endgroup$
                        – Cort Ammon
                        6 hours ago










                      • $begingroup$
                        It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
                        $endgroup$
                        – Salvador Villarreal
                        2 hours ago










                      • $begingroup$
                        Are you sure of that 10^-7?? I thought it was more like 10^-12.
                        $endgroup$
                        – Loren Pechtel
                        1 hour ago










                      • $begingroup$
                        I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
                        $endgroup$
                        – Salvador Villarreal
                        1 hour ago




















                      • $begingroup$
                        The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
                        $endgroup$
                        – Cort Ammon
                        6 hours ago










                      • $begingroup$
                        It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
                        $endgroup$
                        – Salvador Villarreal
                        2 hours ago










                      • $begingroup$
                        Are you sure of that 10^-7?? I thought it was more like 10^-12.
                        $endgroup$
                        – Loren Pechtel
                        1 hour ago










                      • $begingroup$
                        I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
                        $endgroup$
                        – Salvador Villarreal
                        1 hour ago


















                      $begingroup$
                      The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
                      $endgroup$
                      – Cort Ammon
                      6 hours ago




                      $begingroup$
                      The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.
                      $endgroup$
                      – Cort Ammon
                      6 hours ago












                      $begingroup$
                      It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
                      $endgroup$
                      – Salvador Villarreal
                      2 hours ago




                      $begingroup$
                      It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites.
                      $endgroup$
                      – Salvador Villarreal
                      2 hours ago












                      $begingroup$
                      Are you sure of that 10^-7?? I thought it was more like 10^-12.
                      $endgroup$
                      – Loren Pechtel
                      1 hour ago




                      $begingroup$
                      Are you sure of that 10^-7?? I thought it was more like 10^-12.
                      $endgroup$
                      – Loren Pechtel
                      1 hour ago












                      $begingroup$
                      I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
                      $endgroup$
                      – Salvador Villarreal
                      1 hour ago






                      $begingroup$
                      I need to check my notes from back then, but a quick calculation of $gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $sqrt{gamma}$, but I honestly don't remeber. I will check and change the answer if necessary.
                      $endgroup$
                      – Salvador Villarreal
                      1 hour ago













                      10












                      $begingroup$

                      'Non-relativistic' means $vll c$.



                      That is effectively the same as $gamma approx 1$ as $gamma={1 over sqrt{1-v^2/c^2}}$.



                      But also $gamma={E_{tot}over E_{rest}} equiv 1+{E_{kin} over E_{rest}}$



                      So if the kinetic energy is small compared to the rest mass, $gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.






                      share|cite|improve this answer











                      $endgroup$


















                        10












                        $begingroup$

                        'Non-relativistic' means $vll c$.



                        That is effectively the same as $gamma approx 1$ as $gamma={1 over sqrt{1-v^2/c^2}}$.



                        But also $gamma={E_{tot}over E_{rest}} equiv 1+{E_{kin} over E_{rest}}$



                        So if the kinetic energy is small compared to the rest mass, $gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.






                        share|cite|improve this answer











                        $endgroup$
















                          10












                          10








                          10





                          $begingroup$

                          'Non-relativistic' means $vll c$.



                          That is effectively the same as $gamma approx 1$ as $gamma={1 over sqrt{1-v^2/c^2}}$.



                          But also $gamma={E_{tot}over E_{rest}} equiv 1+{E_{kin} over E_{rest}}$



                          So if the kinetic energy is small compared to the rest mass, $gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.






                          share|cite|improve this answer











                          $endgroup$



                          'Non-relativistic' means $vll c$.



                          That is effectively the same as $gamma approx 1$ as $gamma={1 over sqrt{1-v^2/c^2}}$.



                          But also $gamma={E_{tot}over E_{rest}} equiv 1+{E_{kin} over E_{rest}}$



                          So if the kinetic energy is small compared to the rest mass, $gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 10 hours ago









                          Chris

                          9,29872942




                          9,29872942










                          answered 12 hours ago









                          RogerJBarlowRogerJBarlow

                          2,507416




                          2,507416






























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