In Kronecker delta notation, what if any is the significance of using subscript or superscript indices?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Studying special relativity for an exam on an Italian book (RelativitÃÂ - Barone for those interested), before introducing a geometrical analysis for Lorentz transformations and Minkowski spacetime, the author gives some reminders on the classical laws of physics and their geometry.
He explains all the properties of the Euclidian space $mathbb E^3$, and what interested me for this question was the definition of distance that was given, mainly
$$mathbb dmathbf x^2 = delta_ijmathbb dx^imathbb dx^j$$
which is very clear.
Then the author proceeds to describe the property of the Minkowski spacetime $mathbb M$, and defines the distance by the metric tensor $g_munu$ as follows
$$mathbb d s^2=g_munumathbb dx^mumathbb dx^nu$$
for completeness sake the metric tensor is chosen with signature $(+,-,-,-)$, but it won't be necessary.
He then goes by describing the properties of the metric tensor $g_munu$ and here's where my problem lays: he states $$g^munug_nurho = delta^mu_rho $$
Question: Why is now the Kronecker delta shown with one upper index and one lower index? Is there any difference?
I know that clearly the first and the second Kronecker delta have to be the same, but if they are, why showing them differently? Is there any nuances that I should know? Is is just because the first Kronecker delta is in $3$-dimensions and the second in $4$?
notation tensors special-relativity
add a comment |Â
up vote
1
down vote
favorite
Studying special relativity for an exam on an Italian book (RelativitÃÂ - Barone for those interested), before introducing a geometrical analysis for Lorentz transformations and Minkowski spacetime, the author gives some reminders on the classical laws of physics and their geometry.
He explains all the properties of the Euclidian space $mathbb E^3$, and what interested me for this question was the definition of distance that was given, mainly
$$mathbb dmathbf x^2 = delta_ijmathbb dx^imathbb dx^j$$
which is very clear.
Then the author proceeds to describe the property of the Minkowski spacetime $mathbb M$, and defines the distance by the metric tensor $g_munu$ as follows
$$mathbb d s^2=g_munumathbb dx^mumathbb dx^nu$$
for completeness sake the metric tensor is chosen with signature $(+,-,-,-)$, but it won't be necessary.
He then goes by describing the properties of the metric tensor $g_munu$ and here's where my problem lays: he states $$g^munug_nurho = delta^mu_rho $$
Question: Why is now the Kronecker delta shown with one upper index and one lower index? Is there any difference?
I know that clearly the first and the second Kronecker delta have to be the same, but if they are, why showing them differently? Is there any nuances that I should know? Is is just because the first Kronecker delta is in $3$-dimensions and the second in $4$?
notation tensors special-relativity
2
It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way
â quidâ¦
Aug 27 at 16:13
1
Thank you for the help @quid, really appreciate it!
â Davide Morgante
Aug 27 at 16:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Studying special relativity for an exam on an Italian book (RelativitÃÂ - Barone for those interested), before introducing a geometrical analysis for Lorentz transformations and Minkowski spacetime, the author gives some reminders on the classical laws of physics and their geometry.
He explains all the properties of the Euclidian space $mathbb E^3$, and what interested me for this question was the definition of distance that was given, mainly
$$mathbb dmathbf x^2 = delta_ijmathbb dx^imathbb dx^j$$
which is very clear.
Then the author proceeds to describe the property of the Minkowski spacetime $mathbb M$, and defines the distance by the metric tensor $g_munu$ as follows
$$mathbb d s^2=g_munumathbb dx^mumathbb dx^nu$$
for completeness sake the metric tensor is chosen with signature $(+,-,-,-)$, but it won't be necessary.
He then goes by describing the properties of the metric tensor $g_munu$ and here's where my problem lays: he states $$g^munug_nurho = delta^mu_rho $$
Question: Why is now the Kronecker delta shown with one upper index and one lower index? Is there any difference?
I know that clearly the first and the second Kronecker delta have to be the same, but if they are, why showing them differently? Is there any nuances that I should know? Is is just because the first Kronecker delta is in $3$-dimensions and the second in $4$?
notation tensors special-relativity
Studying special relativity for an exam on an Italian book (RelativitÃÂ - Barone for those interested), before introducing a geometrical analysis for Lorentz transformations and Minkowski spacetime, the author gives some reminders on the classical laws of physics and their geometry.
He explains all the properties of the Euclidian space $mathbb E^3$, and what interested me for this question was the definition of distance that was given, mainly
$$mathbb dmathbf x^2 = delta_ijmathbb dx^imathbb dx^j$$
which is very clear.
Then the author proceeds to describe the property of the Minkowski spacetime $mathbb M$, and defines the distance by the metric tensor $g_munu$ as follows
$$mathbb d s^2=g_munumathbb dx^mumathbb dx^nu$$
for completeness sake the metric tensor is chosen with signature $(+,-,-,-)$, but it won't be necessary.
He then goes by describing the properties of the metric tensor $g_munu$ and here's where my problem lays: he states $$g^munug_nurho = delta^mu_rho $$
Question: Why is now the Kronecker delta shown with one upper index and one lower index? Is there any difference?
I know that clearly the first and the second Kronecker delta have to be the same, but if they are, why showing them differently? Is there any nuances that I should know? Is is just because the first Kronecker delta is in $3$-dimensions and the second in $4$?
notation tensors special-relativity
edited Aug 27 at 16:23
quidâ¦
36.4k85091
36.4k85091
asked Aug 27 at 16:07
Davide Morgante
2,538623
2,538623
2
It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way
â quidâ¦
Aug 27 at 16:13
1
Thank you for the help @quid, really appreciate it!
â Davide Morgante
Aug 27 at 16:26
add a comment |Â
2
It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way
â quidâ¦
Aug 27 at 16:13
1
Thank you for the help @quid, really appreciate it!
â Davide Morgante
Aug 27 at 16:26
2
2
It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way
â quidâ¦
Aug 27 at 16:13
It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way
â quidâ¦
Aug 27 at 16:13
1
1
Thank you for the help @quid, really appreciate it!
â Davide Morgante
Aug 27 at 16:26
Thank you for the help @quid, really appreciate it!
â Davide Morgante
Aug 27 at 16:26
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I believe the point is that in Einstein's convention, whenever a symbol shows up as both a lower and upper index, there is an implicit summation.
With this convention it is common to use $x^1,dots,x^n$ as your coordinates so that, for instance,
$$ delta_i,j rm dx^i rm dx^j text means sum_i = 1^n sum_j = 1^n delta_i,j rm dx^i rm dx^j.$$
If this is the case, that means that your first equation is written incorrectly: $x_i$ instead of $x^i$.
In this convention, the symbols with an upper subscript act on symbols with a lower subscript and vice versa (i.e. they are elements of dual vector spaces). That is, the standard dot product would be written as
$$ x^1y_1 + dots + x^ny_n = x^i y_i. $$
In particular, upper indices are for "contravariant" vectors and lower indices are for "covariant" vectors (aka covectors) (see e.g. Wikipedia).
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
1
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
1
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I believe the point is that in Einstein's convention, whenever a symbol shows up as both a lower and upper index, there is an implicit summation.
With this convention it is common to use $x^1,dots,x^n$ as your coordinates so that, for instance,
$$ delta_i,j rm dx^i rm dx^j text means sum_i = 1^n sum_j = 1^n delta_i,j rm dx^i rm dx^j.$$
If this is the case, that means that your first equation is written incorrectly: $x_i$ instead of $x^i$.
In this convention, the symbols with an upper subscript act on symbols with a lower subscript and vice versa (i.e. they are elements of dual vector spaces). That is, the standard dot product would be written as
$$ x^1y_1 + dots + x^ny_n = x^i y_i. $$
In particular, upper indices are for "contravariant" vectors and lower indices are for "covariant" vectors (aka covectors) (see e.g. Wikipedia).
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
1
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
1
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
add a comment |Â
up vote
2
down vote
accepted
I believe the point is that in Einstein's convention, whenever a symbol shows up as both a lower and upper index, there is an implicit summation.
With this convention it is common to use $x^1,dots,x^n$ as your coordinates so that, for instance,
$$ delta_i,j rm dx^i rm dx^j text means sum_i = 1^n sum_j = 1^n delta_i,j rm dx^i rm dx^j.$$
If this is the case, that means that your first equation is written incorrectly: $x_i$ instead of $x^i$.
In this convention, the symbols with an upper subscript act on symbols with a lower subscript and vice versa (i.e. they are elements of dual vector spaces). That is, the standard dot product would be written as
$$ x^1y_1 + dots + x^ny_n = x^i y_i. $$
In particular, upper indices are for "contravariant" vectors and lower indices are for "covariant" vectors (aka covectors) (see e.g. Wikipedia).
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
1
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
1
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I believe the point is that in Einstein's convention, whenever a symbol shows up as both a lower and upper index, there is an implicit summation.
With this convention it is common to use $x^1,dots,x^n$ as your coordinates so that, for instance,
$$ delta_i,j rm dx^i rm dx^j text means sum_i = 1^n sum_j = 1^n delta_i,j rm dx^i rm dx^j.$$
If this is the case, that means that your first equation is written incorrectly: $x_i$ instead of $x^i$.
In this convention, the symbols with an upper subscript act on symbols with a lower subscript and vice versa (i.e. they are elements of dual vector spaces). That is, the standard dot product would be written as
$$ x^1y_1 + dots + x^ny_n = x^i y_i. $$
In particular, upper indices are for "contravariant" vectors and lower indices are for "covariant" vectors (aka covectors) (see e.g. Wikipedia).
I believe the point is that in Einstein's convention, whenever a symbol shows up as both a lower and upper index, there is an implicit summation.
With this convention it is common to use $x^1,dots,x^n$ as your coordinates so that, for instance,
$$ delta_i,j rm dx^i rm dx^j text means sum_i = 1^n sum_j = 1^n delta_i,j rm dx^i rm dx^j.$$
If this is the case, that means that your first equation is written incorrectly: $x_i$ instead of $x^i$.
In this convention, the symbols with an upper subscript act on symbols with a lower subscript and vice versa (i.e. they are elements of dual vector spaces). That is, the standard dot product would be written as
$$ x^1y_1 + dots + x^ny_n = x^i y_i. $$
In particular, upper indices are for "contravariant" vectors and lower indices are for "covariant" vectors (aka covectors) (see e.g. Wikipedia).
edited Aug 27 at 16:20
answered Aug 27 at 16:13
Trevor Gunn
12.9k32045
12.9k32045
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
1
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
1
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
add a comment |Â
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
1
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
1
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ?
â Davide Morgante
Aug 27 at 16:15
1
1
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
@DavideMorgante It depends on what kind of tensor it is. See quid's link for instance.
â Trevor Gunn
Aug 27 at 16:21
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help!
â Davide Morgante
Aug 27 at 16:25
1
1
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
@DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf).
â Trevor Gunn
Aug 27 at 19:16
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2896348%2fin-kronecker-delta-notation-what-if-any-is-the-significance-of-using-subscript%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way
â quidâ¦
Aug 27 at 16:13
1
Thank you for the help @quid, really appreciate it!
â Davide Morgante
Aug 27 at 16:26