Minkowski space
$begingroup$
In Minkowski space, coordinates which satisfy
$x^2 = t^2 - X^2 > 0$
are in the region of spacetime that is time-like.
If it's
$x^2 = t^2 - X^2 < 0$
the region is space-like.
But if
$x^2 = t^2 - X^2 > 0$
then its "trajectory of light-like particles".
I have understood the first two points about time- and space-like regions but I could not get the third one about "light-like particles".
My confusion is - why just light-like particles? There are many other particles at quantum level.
special-relativity mass metric-tensor causality
edited Apr 7 at 10:02
Gallifreyan
1056
asked Apr 6 at 20:07
sk9298sk9298
846
$endgroup$
add a comment |
$begingroup$
In Minkowski space, coordinates which satisfy
$x^2 = t^2 - X^2 > 0$
are in the region of spacetime that is time-like.
If it's
$x^2 = t^2 - X^2 < 0$
the region is space-like.
But if
$x^2 = t^2 - X^2 > 0$
then its "trajectory of light-like particles".
I have understood the first two points about time- and space-like regions but I could not get the third one about "light-like particles".
My confusion is - why just light-like particles? There are many other particles at quantum level.
special-relativity mass metric-tensor causality
edited Apr 7 at 10:02
Gallifreyan
1056
asked Apr 6 at 20:07
sk9298sk9298
846
$endgroup$
$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
Apr 6 at 22:38
add a comment |
$begingroup$
In Minkowski space, coordinates which satisfy
$x^2 = t^2 - X^2 > 0$
are in the region of spacetime that is time-like.
If it's
$x^2 = t^2 - X^2 < 0$
the region is space-like.
But if
$x^2 = t^2 - X^2 > 0$
then its "trajectory of light-like particles".
I have understood the first two points about time- and space-like regions but I could not get the third one about "light-like particles".
My confusion is - why just light-like particles? There are many other particles at quantum level.
special-relativity mass metric-tensor causality
edited Apr 7 at 10:02
Gallifreyan
1056
asked Apr 6 at 20:07
sk9298sk9298
846
$endgroup$
In Minkowski space, coordinates which satisfy
$x^2 = t^2 - X^2 > 0$
are in the region of spacetime that is time-like.
If it's
$x^2 = t^2 - X^2 < 0$
the region is space-like.
But if
$x^2 = t^2 - X^2 > 0$
then its "trajectory of light-like particles".
I have understood the first two points about time- and space-like regions but I could not get the third one about "light-like particles".
My confusion is - why just light-like particles? There are many other particles at quantum level.
special-relativity mass metric-tensor causality
special-relativity mass metric-tensor causality
edited Apr 7 at 10:02
Gallifreyan
1056
asked Apr 6 at 20:07
sk9298sk9298
846
edited Apr 7 at 10:02
Gallifreyan
1056
asked Apr 6 at 20:07
sk9298sk9298
846
edited Apr 7 at 10:02
Gallifreyan
1056
edited Apr 7 at 10:02
Gallifreyan
1056
edited Apr 7 at 10:02
Gallifreyan
1056
1056
asked Apr 6 at 20:07
sk9298sk9298
846
asked Apr 6 at 20:07
sk9298sk9298
846
asked Apr 6 at 20:07
sk9298sk9298
846
846
$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
Apr 6 at 22:38
add a comment |
$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
Apr 6 at 22:38
$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
Apr 6 at 22:38
$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
Apr 6 at 22:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 6 at 20:17
DaleDale
6,6971829
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
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$begingroup$
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.
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
$endgroup$
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.
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
answered Apr 6 at 20:13
LucashWindowWasherLucashWindowWasher
38812
38812
add a comment |
add a comment |
$begingroup$
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.
answered Apr 6 at 20:17
DaleDale
6,6971829
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 6 at 20:17
DaleDale
6,6971829
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 6 at 20:17
DaleDale
6,6971829
$endgroup$
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.
answered Apr 6 at 20:17
DaleDale
6,6971829
answered Apr 6 at 20:17
DaleDale
6,6971829
answered Apr 6 at 20:17
DaleDale
6,6971829
answered Apr 6 at 20:17
DaleDale
6,6971829
6,6971829
add a comment |
add a comment |
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$begingroup$
In the future, please use MathJax, not HTML markup, to display math. Thanks.
$endgroup$
– G. Smith
Apr 6 at 22:38