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Minkowski space


Minkowski MetricMinkowski Metric SignatureMinkowski metric — why does it follow from the constancy of the speed of light?Minkowski spacetime: Is there a signature (+,+,+,+)?Minkowski spacetime vs Euclidian spaceMinkowski metric: Why does it look like it does?Understanding space-like hyperplanes in Minkowski spaceCausality in general relativityCan an a distance in Minkowski space, based on a Euclidean plane, be time-like?How to Construct Proper Spherical Coordinates in Minkowski Spacetime?













1












$begingroup$


1) In Minkowski space, coordinates which satisfy



x2=t2-X2>0



Region of spacetime is time-like.



2) If its



x2=t2-X2<0



Region is space-like.



3) But if



x2=t2-X2=0



then its "trajectory of light-like particles".



I have understood the first two points about time and space like region but I could not get the third one about "light like particles".
My confusion is about why just light like particles? there are many other particles at quantum level.










share|cite|improve this question









New contributor




sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    In the future, please use MathJax, not HTML markup, to display math. Thanks.
    $endgroup$
    – G. Smith
    1 hour ago















1












$begingroup$


1) In Minkowski space, coordinates which satisfy



x2=t2-X2>0



Region of spacetime is time-like.



2) If its



x2=t2-X2<0



Region is space-like.



3) But if



x2=t2-X2=0



then its "trajectory of light-like particles".



I have understood the first two points about time and space like region but I could not get the third one about "light like particles".
My confusion is about why just light like particles? there are many other particles at quantum level.










share|cite|improve this question









New contributor




sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    In the future, please use MathJax, not HTML markup, to display math. Thanks.
    $endgroup$
    – G. Smith
    1 hour ago













1












1








1





$begingroup$


1) In Minkowski space, coordinates which satisfy



x2=t2-X2>0



Region of spacetime is time-like.



2) If its



x2=t2-X2<0



Region is space-like.



3) But if



x2=t2-X2=0



then its "trajectory of light-like particles".



I have understood the first two points about time and space like region but I could not get the third one about "light like particles".
My confusion is about why just light like particles? there are many other particles at quantum level.










share|cite|improve this question









New contributor




sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




1) In Minkowski space, coordinates which satisfy



x2=t2-X2>0



Region of spacetime is time-like.



2) If its



x2=t2-X2<0



Region is space-like.



3) But if



x2=t2-X2=0



then its "trajectory of light-like particles".



I have understood the first two points about time and space like region but I could not get the third one about "light like particles".
My confusion is about why just light like particles? there are many other particles at quantum level.







special-relativity mass metric-tensor causality






share|cite|improve this question









New contributor




sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Qmechanic

107k121991239




107k121991239






New contributor




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asked 3 hours ago









sk9298sk9298

595




595




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New contributor





sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






sk9298 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    In the future, please use MathJax, not HTML markup, to display math. Thanks.
    $endgroup$
    – G. Smith
    1 hour ago
















  • $begingroup$
    In the future, please use MathJax, not HTML markup, to display math. Thanks.
    $endgroup$
    – G. Smith
    1 hour ago















$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
1 hour ago




$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
1 hour ago










2 Answers
2






active

oldest

votes


















3












$begingroup$


My confusion is about why just light like particles? there are many other particles at quantum level.




You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.



An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.



    The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$


      My confusion is about why just light like particles? there are many other particles at quantum level.




      You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.



      An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.






      share|cite|improve this answer









      $endgroup$

















        3












        $begingroup$


        My confusion is about why just light like particles? there are many other particles at quantum level.




        You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.



        An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.






        share|cite|improve this answer









        $endgroup$















          3












          3








          3





          $begingroup$


          My confusion is about why just light like particles? there are many other particles at quantum level.




          You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.



          An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.






          share|cite|improve this answer









          $endgroup$




          My confusion is about why just light like particles? there are many other particles at quantum level.




          You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.



          An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          DaleDale

          6,5971829




          6,5971829





















              1












              $begingroup$

              Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.



              The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.



                The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.



                  The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.






                  share|cite|improve this answer









                  $endgroup$



                  Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.



                  The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  LucashWindowWasherLucashWindowWasher

                  30312




                  30312




















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