Mixing
Problem of coordinate transformation $x^alpha$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
This is intuitively very simple problem but I am unable to complete it with Mathematical rigor. Here is the deal:
A coordinate system $(u,v,w,p)$ in which the metric tensor has the following non-zero components, $g_uv= g_ww=g_pp=1$. Find the coordinate transformation between $(u,v,w,p)$ to the regular coordinates $(t,x,y,z)$.
So in the first part of the same question you are supposed to prove that coordinate space $(u,v,w,p)$ is flat which is easy enough to see because the metric components are constant, all Riemann tensor components are zero. Hence the space is Minkowski. Which, intuitively, implies that the transformation between the coordinates will be linear because standard $(t,x,y,z)$ form a Minkowski space. But I am unable to prove it Mathematically.
Here is what I did. The question had a hint to calculate $e_u.e_u$ and $e_v.e_v$ (e represents the basis in either coordinate system and subscript implies which coordinates they belong to. I am removing the vector heads or the circumflex). From the metric tensor in this coordinate they are going to be zero.
So, I did write the conversion between the basis of the two systems,
$$
e_alpha ' = Lambda^alpha_alpha ' = fracpartial x^alphapartial x^alpha 'e_alpha
$$
Here the primed indices represent the $(u,v,w,p)$ coordinates and unprimed represents $(t,x,y,z)$. This gives the following dot product,
$$
e_alpha '.e_beta '= Lambda^alpha_alpha 'Lambda^beta_beta 'e_alpha.e_beta = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
And I can use the orthonormality of the $(t,x,y,z)$ coordinates to simplify $e_alpha.e_beta$ but even then using the metric all,
$$
e_alpha '.e_beta ' = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
gives me is a bunch of (actually 16) partial differential equations but I do not know where to go from there.
How do I prove that the partial differential equations have a solution,
$x^alpha = A^alpha_alpha 'x^alpha '$ where $A^alpha_alpha '$ are constants? Surely it can be seem that it is ONE of the solution of the system of PDEs but is that the only solution(?). I arrived at the answer intuitively but want to be able to SEE it mathematically being a solution. Also, how do I end up calculating the values of $A^alpha_alpha '$?
linear-transformations coordinate-systems transformation
asked Jul 15 at 7:56
Shaz
629
add a comment |Â
up vote
1
down vote
favorite
This is intuitively very simple problem but I am unable to complete it with Mathematical rigor. Here is the deal:
A coordinate system $(u,v,w,p)$ in which the metric tensor has the following non-zero components, $g_uv= g_ww=g_pp=1$. Find the coordinate transformation between $(u,v,w,p)$ to the regular coordinates $(t,x,y,z)$.
So in the first part of the same question you are supposed to prove that coordinate space $(u,v,w,p)$ is flat which is easy enough to see because the metric components are constant, all Riemann tensor components are zero. Hence the space is Minkowski. Which, intuitively, implies that the transformation between the coordinates will be linear because standard $(t,x,y,z)$ form a Minkowski space. But I am unable to prove it Mathematically.
Here is what I did. The question had a hint to calculate $e_u.e_u$ and $e_v.e_v$ (e represents the basis in either coordinate system and subscript implies which coordinates they belong to. I am removing the vector heads or the circumflex). From the metric tensor in this coordinate they are going to be zero.
So, I did write the conversion between the basis of the two systems,
$$
e_alpha ' = Lambda^alpha_alpha ' = fracpartial x^alphapartial x^alpha 'e_alpha
$$
Here the primed indices represent the $(u,v,w,p)$ coordinates and unprimed represents $(t,x,y,z)$. This gives the following dot product,
$$
e_alpha '.e_beta '= Lambda^alpha_alpha 'Lambda^beta_beta 'e_alpha.e_beta = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
And I can use the orthonormality of the $(t,x,y,z)$ coordinates to simplify $e_alpha.e_beta$ but even then using the metric all,
$$
e_alpha '.e_beta ' = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
gives me is a bunch of (actually 16) partial differential equations but I do not know where to go from there.
How do I prove that the partial differential equations have a solution,
$x^alpha = A^alpha_alpha 'x^alpha '$ where $A^alpha_alpha '$ are constants? Surely it can be seem that it is ONE of the solution of the system of PDEs but is that the only solution(?). I arrived at the answer intuitively but want to be able to SEE it mathematically being a solution. Also, how do I end up calculating the values of $A^alpha_alpha '$?
linear-transformations coordinate-systems transformation
asked Jul 15 at 7:56
Shaz
629
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is intuitively very simple problem but I am unable to complete it with Mathematical rigor. Here is the deal:
A coordinate system $(u,v,w,p)$ in which the metric tensor has the following non-zero components, $g_uv= g_ww=g_pp=1$. Find the coordinate transformation between $(u,v,w,p)$ to the regular coordinates $(t,x,y,z)$.
So in the first part of the same question you are supposed to prove that coordinate space $(u,v,w,p)$ is flat which is easy enough to see because the metric components are constant, all Riemann tensor components are zero. Hence the space is Minkowski. Which, intuitively, implies that the transformation between the coordinates will be linear because standard $(t,x,y,z)$ form a Minkowski space. But I am unable to prove it Mathematically.
Here is what I did. The question had a hint to calculate $e_u.e_u$ and $e_v.e_v$ (e represents the basis in either coordinate system and subscript implies which coordinates they belong to. I am removing the vector heads or the circumflex). From the metric tensor in this coordinate they are going to be zero.
So, I did write the conversion between the basis of the two systems,
$$
e_alpha ' = Lambda^alpha_alpha ' = fracpartial x^alphapartial x^alpha 'e_alpha
$$
Here the primed indices represent the $(u,v,w,p)$ coordinates and unprimed represents $(t,x,y,z)$. This gives the following dot product,
$$
e_alpha '.e_beta '= Lambda^alpha_alpha 'Lambda^beta_beta 'e_alpha.e_beta = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
And I can use the orthonormality of the $(t,x,y,z)$ coordinates to simplify $e_alpha.e_beta$ but even then using the metric all,
$$
e_alpha '.e_beta ' = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
gives me is a bunch of (actually 16) partial differential equations but I do not know where to go from there.
How do I prove that the partial differential equations have a solution,
$x^alpha = A^alpha_alpha 'x^alpha '$ where $A^alpha_alpha '$ are constants? Surely it can be seem that it is ONE of the solution of the system of PDEs but is that the only solution(?). I arrived at the answer intuitively but want to be able to SEE it mathematically being a solution. Also, how do I end up calculating the values of $A^alpha_alpha '$?
linear-transformations coordinate-systems transformation
asked Jul 15 at 7:56
Shaz
629
This is intuitively very simple problem but I am unable to complete it with Mathematical rigor. Here is the deal:
A coordinate system $(u,v,w,p)$ in which the metric tensor has the following non-zero components, $g_uv= g_ww=g_pp=1$. Find the coordinate transformation between $(u,v,w,p)$ to the regular coordinates $(t,x,y,z)$.
So in the first part of the same question you are supposed to prove that coordinate space $(u,v,w,p)$ is flat which is easy enough to see because the metric components are constant, all Riemann tensor components are zero. Hence the space is Minkowski. Which, intuitively, implies that the transformation between the coordinates will be linear because standard $(t,x,y,z)$ form a Minkowski space. But I am unable to prove it Mathematically.
Here is what I did. The question had a hint to calculate $e_u.e_u$ and $e_v.e_v$ (e represents the basis in either coordinate system and subscript implies which coordinates they belong to. I am removing the vector heads or the circumflex). From the metric tensor in this coordinate they are going to be zero.
So, I did write the conversion between the basis of the two systems,
$$
e_alpha ' = Lambda^alpha_alpha ' = fracpartial x^alphapartial x^alpha 'e_alpha
$$
Here the primed indices represent the $(u,v,w,p)$ coordinates and unprimed represents $(t,x,y,z)$. This gives the following dot product,
$$
e_alpha '.e_beta '= Lambda^alpha_alpha 'Lambda^beta_beta 'e_alpha.e_beta = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
And I can use the orthonormality of the $(t,x,y,z)$ coordinates to simplify $e_alpha.e_beta$ but even then using the metric all,
$$
e_alpha '.e_beta ' = fracpartial x^alphapartial x^alpha 'fracpartial x^betapartial x^beta 'e_alpha.e_beta
$$
gives me is a bunch of (actually 16) partial differential equations but I do not know where to go from there.
How do I prove that the partial differential equations have a solution,
$x^alpha = A^alpha_alpha 'x^alpha '$ where $A^alpha_alpha '$ are constants? Surely it can be seem that it is ONE of the solution of the system of PDEs but is that the only solution(?). I arrived at the answer intuitively but want to be able to SEE it mathematically being a solution. Also, how do I end up calculating the values of $A^alpha_alpha '$?
linear-transformations coordinate-systems transformation
asked Jul 15 at 7:56
Shaz
629
edited Jul 15 at 11:48
asked Jul 15 at 7:56
Shaz
629
asked Jul 15 at 7:56
Shaz
629
asked Jul 15 at 7:56
Shaz
629
629
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
It is not required to prove that the solution is unique. It is possible that the solution is not unique as a rotation or boost would still be another solution for the same criteria. So it is just required, as far as the scope of the problem is concerned, that $x^alpha ' = A^alpha '_alpha x^alpha$ is A solution not THE solution.
Secondly, the trick used is to not change the coordinates which are orthonormal, which are $w$ and $p$. So just let them $y=w$ and $z=p$. That reduces the number of unknowns and the number of equations.
answered Jul 22 at 12:24
Shaz
629
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It is not required to prove that the solution is unique. It is possible that the solution is not unique as a rotation or boost would still be another solution for the same criteria. So it is just required, as far as the scope of the problem is concerned, that $x^alpha ' = A^alpha '_alpha x^alpha$ is A solution not THE solution.
Secondly, the trick used is to not change the coordinates which are orthonormal, which are $w$ and $p$. So just let them $y=w$ and $z=p$. That reduces the number of unknowns and the number of equations.
answered Jul 22 at 12:24
Shaz
629
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
add a comment |Â
up vote
1
down vote
accepted
It is not required to prove that the solution is unique. It is possible that the solution is not unique as a rotation or boost would still be another solution for the same criteria. So it is just required, as far as the scope of the problem is concerned, that $x^alpha ' = A^alpha '_alpha x^alpha$ is A solution not THE solution.
Secondly, the trick used is to not change the coordinates which are orthonormal, which are $w$ and $p$. So just let them $y=w$ and $z=p$. That reduces the number of unknowns and the number of equations.
answered Jul 22 at 12:24
Shaz
629
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It is not required to prove that the solution is unique. It is possible that the solution is not unique as a rotation or boost would still be another solution for the same criteria. So it is just required, as far as the scope of the problem is concerned, that $x^alpha ' = A^alpha '_alpha x^alpha$ is A solution not THE solution.
Secondly, the trick used is to not change the coordinates which are orthonormal, which are $w$ and $p$. So just let them $y=w$ and $z=p$. That reduces the number of unknowns and the number of equations.
answered Jul 22 at 12:24
Shaz
629
It is not required to prove that the solution is unique. It is possible that the solution is not unique as a rotation or boost would still be another solution for the same criteria. So it is just required, as far as the scope of the problem is concerned, that $x^alpha ' = A^alpha '_alpha x^alpha$ is A solution not THE solution.
Secondly, the trick used is to not change the coordinates which are orthonormal, which are $w$ and $p$. So just let them $y=w$ and $z=p$. That reduces the number of unknowns and the number of equations.
answered Jul 22 at 12:24
Shaz
629
edited Jul 22 at 12:36
answered Jul 22 at 12:24
Shaz
629
answered Jul 22 at 12:24
Shaz
629
answered Jul 22 at 12:24
Shaz
629
629
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
add a comment |Â
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
Thanks for updating with your own solution. You can accept it too so the question doesn't remain on an open queue.
– Ethan Bolker
Jul 22 at 12:27
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%2f2852284%2fproblem-of-coordinate-transformation-x-alpha%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