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












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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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    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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      3








      3


      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






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      edited 3 hours ago









      Qmechanic♦

      104k121881193




      104k121881193










      asked 7 hours ago









      TaeNyFanTaeNyFan

      3367




      3367






















          3 Answers
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          active

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          11












          $begingroup$

          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











          $endgroup$





















            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











            $endgroup$





















              9












              $begingroup$

              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
                50 mins ago











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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              11












              $begingroup$

              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











              $endgroup$


















                11












                $begingroup$

                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











                $endgroup$
















                  11












                  11








                  11





                  $begingroup$

                  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 6 hours ago

























                  answered 6 hours ago









                  John RennieJohn Rennie

                  274k43543790




                  274k43543790























                      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 5 hours ago









                          Chris

                          9,29872942




                          9,29872942










                          answered 7 hours ago









                          RogerJBarlowRogerJBarlow

                          2,507416




                          2,507416























                              9












                              $begingroup$

                              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
                                50 mins ago
















                              9












                              $begingroup$

                              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
                                50 mins ago














                              9












                              9








                              9





                              $begingroup$

                              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 6 hours ago

























                              answered 6 hours ago









                              Salvador VillarrealSalvador Villarreal

                              1417




                              1417












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


















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