Finding Christoffel Symbols using via variational method.
$begingroup$
I'm trying to find the Christoffel Symbols for the Lorentz metric $${rm d}s^2 = cos(2pi x)({rm d}x^2-{rm d}y^2) - 2sin(2pi x),{rm d}x,{rm d}y$$by looking at the Euler-Lagrange equations for $$L(x,dot{x},y,dot{y}) = cos(2pi x)(dot{x}^2-dot{y}^2) - 2sin(2pi x),dot{x},dot{y}.$$I have already done my fair share of computations like this, but I must be making some algebraic mistake that I cannot find for the life of me. If we write $$begin{align}frac{partial L}{partial x} - frac{{rm d}}{{rm d}t}&left(frac{partial L}{partial dot{x}}right) = -2pisin(2pi x)(dot{x}^2-dot{y}^2)-4picos(2pi x)dot{x}dot{y} \ &qquad - frac{{rm d}}{{rm d}t}left(2dot{x}cos(2pi x) - 2dot{y}sin(2pi x)right),end{align} $$and we will have a term with $ddot{y}$. This is a problem, since as far as I understand the geodesic equation corresponding to the $x$-coordinate should have the form $$ddot{x} + Gamma(dot{x},dot{y})=0,$$maybe after dividing by something. What is going on?
multivariable-calculus differential-geometry riemannian-geometry calculus-of-variations
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
$endgroup$
add a comment |
$begingroup$
I'm trying to find the Christoffel Symbols for the Lorentz metric $${rm d}s^2 = cos(2pi x)({rm d}x^2-{rm d}y^2) - 2sin(2pi x),{rm d}x,{rm d}y$$by looking at the Euler-Lagrange equations for $$L(x,dot{x},y,dot{y}) = cos(2pi x)(dot{x}^2-dot{y}^2) - 2sin(2pi x),dot{x},dot{y}.$$I have already done my fair share of computations like this, but I must be making some algebraic mistake that I cannot find for the life of me. If we write $$begin{align}frac{partial L}{partial x} - frac{{rm d}}{{rm d}t}&left(frac{partial L}{partial dot{x}}right) = -2pisin(2pi x)(dot{x}^2-dot{y}^2)-4picos(2pi x)dot{x}dot{y} \ &qquad - frac{{rm d}}{{rm d}t}left(2dot{x}cos(2pi x) - 2dot{y}sin(2pi x)right),end{align} $$and we will have a term with $ddot{y}$. This is a problem, since as far as I understand the geodesic equation corresponding to the $x$-coordinate should have the form $$ddot{x} + Gamma(dot{x},dot{y})=0,$$maybe after dividing by something. What is going on?
multivariable-calculus differential-geometry riemannian-geometry calculus-of-variations
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
$endgroup$
$begingroup$
I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example
$endgroup$
– Rafa Budría
Nov 29 '18 at 18:57
$begingroup$
I understand and I don't, at the same time. I understand that the coefficient of $ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $dot{x}dot{y}$ will always generate a $ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{ell k}ddot{x}^k + g_{ell k}Gamma_{ij}^k dot{x}^idot{x}^j = 0$$ we could still have $ddot{x}_m$ terms for $m neq ell$ since the metric coefficient matrix is not diagonal?
$endgroup$
– Ivo Terek
Nov 29 '18 at 19:23
1
$begingroup$
I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $cos(2pi x)>0$.
$endgroup$
– Ted Shifrin
Nov 29 '18 at 19:55
add a comment |
$begingroup$
I'm trying to find the Christoffel Symbols for the Lorentz metric $${rm d}s^2 = cos(2pi x)({rm d}x^2-{rm d}y^2) - 2sin(2pi x),{rm d}x,{rm d}y$$by looking at the Euler-Lagrange equations for $$L(x,dot{x},y,dot{y}) = cos(2pi x)(dot{x}^2-dot{y}^2) - 2sin(2pi x),dot{x},dot{y}.$$I have already done my fair share of computations like this, but I must be making some algebraic mistake that I cannot find for the life of me. If we write $$begin{align}frac{partial L}{partial x} - frac{{rm d}}{{rm d}t}&left(frac{partial L}{partial dot{x}}right) = -2pisin(2pi x)(dot{x}^2-dot{y}^2)-4picos(2pi x)dot{x}dot{y} \ &qquad - frac{{rm d}}{{rm d}t}left(2dot{x}cos(2pi x) - 2dot{y}sin(2pi x)right),end{align} $$and we will have a term with $ddot{y}$. This is a problem, since as far as I understand the geodesic equation corresponding to the $x$-coordinate should have the form $$ddot{x} + Gamma(dot{x},dot{y})=0,$$maybe after dividing by something. What is going on?
multivariable-calculus differential-geometry riemannian-geometry calculus-of-variations
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
$endgroup$
I'm trying to find the Christoffel Symbols for the Lorentz metric $${rm d}s^2 = cos(2pi x)({rm d}x^2-{rm d}y^2) - 2sin(2pi x),{rm d}x,{rm d}y$$by looking at the Euler-Lagrange equations for $$L(x,dot{x},y,dot{y}) = cos(2pi x)(dot{x}^2-dot{y}^2) - 2sin(2pi x),dot{x},dot{y}.$$I have already done my fair share of computations like this, but I must be making some algebraic mistake that I cannot find for the life of me. If we write $$begin{align}frac{partial L}{partial x} - frac{{rm d}}{{rm d}t}&left(frac{partial L}{partial dot{x}}right) = -2pisin(2pi x)(dot{x}^2-dot{y}^2)-4picos(2pi x)dot{x}dot{y} \ &qquad - frac{{rm d}}{{rm d}t}left(2dot{x}cos(2pi x) - 2dot{y}sin(2pi x)right),end{align} $$and we will have a term with $ddot{y}$. This is a problem, since as far as I understand the geodesic equation corresponding to the $x$-coordinate should have the form $$ddot{x} + Gamma(dot{x},dot{y})=0,$$maybe after dividing by something. What is going on?
multivariable-calculus differential-geometry riemannian-geometry calculus-of-variations
multivariable-calculus differential-geometry riemannian-geometry calculus-of-variations
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
asked Nov 29 '18 at 18:30
Ivo TerekIvo Terek
46.1k953142
46.1k953142
$begingroup$
I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example
$endgroup$
– Rafa Budría
Nov 29 '18 at 18:57
$begingroup$
I understand and I don't, at the same time. I understand that the coefficient of $ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $dot{x}dot{y}$ will always generate a $ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{ell k}ddot{x}^k + g_{ell k}Gamma_{ij}^k dot{x}^idot{x}^j = 0$$ we could still have $ddot{x}_m$ terms for $m neq ell$ since the metric coefficient matrix is not diagonal?
$endgroup$
– Ivo Terek
Nov 29 '18 at 19:23
1
$begingroup$
I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $cos(2pi x)>0$.
$endgroup$
– Ted Shifrin
Nov 29 '18 at 19:55
add a comment |
$begingroup$
I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example
$endgroup$
– Rafa Budría
Nov 29 '18 at 18:57
$begingroup$
I understand and I don't, at the same time. I understand that the coefficient of $ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $dot{x}dot{y}$ will always generate a $ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{ell k}ddot{x}^k + g_{ell k}Gamma_{ij}^k dot{x}^idot{x}^j = 0$$ we could still have $ddot{x}_m$ terms for $m neq ell$ since the metric coefficient matrix is not diagonal?
$endgroup$
– Ivo Terek
Nov 29 '18 at 19:23
1
$begingroup$
I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $cos(2pi x)>0$.
$endgroup$
– Ted Shifrin
Nov 29 '18 at 19:55
$begingroup$
I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example
$endgroup$
– Rafa Budría
Nov 29 '18 at 18:57
$begingroup$
I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example
$endgroup$
– Rafa Budría
Nov 29 '18 at 18:57
$begingroup$
I understand and I don't, at the same time. I understand that the coefficient of $ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $dot{x}dot{y}$ will always generate a $ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{ell k}ddot{x}^k + g_{ell k}Gamma_{ij}^k dot{x}^idot{x}^j = 0$$ we could still have $ddot{x}_m$ terms for $m neq ell$ since the metric coefficient matrix is not diagonal?
$endgroup$
– Ivo Terek
Nov 29 '18 at 19:23
$begingroup$
I understand and I don't, at the same time. I understand that the coefficient of $ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $dot{x}dot{y}$ will always generate a $ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{ell k}ddot{x}^k + g_{ell k}Gamma_{ij}^k dot{x}^idot{x}^j = 0$$ we could still have $ddot{x}_m$ terms for $m neq ell$ since the metric coefficient matrix is not diagonal?
$endgroup$
– Ivo Terek
Nov 29 '18 at 19:23
1
1
$begingroup$
I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $cos(2pi x)>0$.
$endgroup$
– Ted Shifrin
Nov 29 '18 at 19:55
$begingroup$
I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $cos(2pi x)>0$.
$endgroup$
– Ted Shifrin
Nov 29 '18 at 19:55
add a comment |
1 Answer
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The infinitesimal variation of the Lagrangian
$$ L(x,dot{x})~=~ g_{ij}(x)~ dot{x}^idot{x}^j tag{1} $$
is
$$ frac{1}{2}delta L~=~ -left{ color{blue}{ g_{kell}ddot{x}^{ell}} +color{red}{Gamma_{k,ij} dot{x}^idot{x}^j}right}delta x^k
+frac{mathrm{d}}{mathrm{d}t}left{ color{green}{ g_{kell}dot{x}^{ell} delta x^k}right},tag{2} $$
where we have introduced the lowered Levi-Civita Christoffel symbols
$$Gamma_{k,ij}~:=~g_{kell}Gamma^{ell}_{ij}. tag{3} $$
Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $t$-derivatives are distributed.
In particular we see that the geodesic equations are multiplied with the metric, cf. OP's last question.
Example. OP's Lagrangian reads
$$L~=~c(dot{x}^2-dot{y}^2) - 2sdot{x}dot{y}, qquad c~:=~cos(2pi x), qquad s~:=~sin(2pi x) ,tag{4}$$
corresponding to the metric
$$ begin{pmatrix} g_{xx} & g_{xy} cr g_{yx} & g_{yy} end{pmatrix} ~=~begin{pmatrix} c & -s cr -s & -c end{pmatrix}. tag{5}$$
We calculate the infinitesimal variation:
$$begin{align}frac{1}{2} delta L
~=~&left{ color{blue}{sddot{y}-cddot{x}}
+ color{red}{pi s (dot{x}^2+dot{y}^2)} right} delta x
+ left{ color{blue}{sddot{x}+cddot{y}}
+ color{red}{2pi (cdot{x}^2-sdot{x}dot{y})} right} delta y cr
& + frac{mathrm{d}}{mathrm{d}t}left{ color{green}{(cdot{x}-sdot{y})delta x -(sdot{x}+cdot{y})delta y}right},
end{align}tag{6}$$
which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols
$$ Gamma_{x,xx}~=~-pi s ~=~Gamma_{x,yy}~=~-Gamma_{y,xy}, qquad Gamma_{y,xx}~=~-2pi c,tag{7} $$
cf. OP's title.
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
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add a comment |
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$begingroup$
The infinitesimal variation of the Lagrangian
$$ L(x,dot{x})~=~ g_{ij}(x)~ dot{x}^idot{x}^j tag{1} $$
is
$$ frac{1}{2}delta L~=~ -left{ color{blue}{ g_{kell}ddot{x}^{ell}} +color{red}{Gamma_{k,ij} dot{x}^idot{x}^j}right}delta x^k
+frac{mathrm{d}}{mathrm{d}t}left{ color{green}{ g_{kell}dot{x}^{ell} delta x^k}right},tag{2} $$
where we have introduced the lowered Levi-Civita Christoffel symbols
$$Gamma_{k,ij}~:=~g_{kell}Gamma^{ell}_{ij}. tag{3} $$
Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $t$-derivatives are distributed.
In particular we see that the geodesic equations are multiplied with the metric, cf. OP's last question.
Example. OP's Lagrangian reads
$$L~=~c(dot{x}^2-dot{y}^2) - 2sdot{x}dot{y}, qquad c~:=~cos(2pi x), qquad s~:=~sin(2pi x) ,tag{4}$$
corresponding to the metric
$$ begin{pmatrix} g_{xx} & g_{xy} cr g_{yx} & g_{yy} end{pmatrix} ~=~begin{pmatrix} c & -s cr -s & -c end{pmatrix}. tag{5}$$
We calculate the infinitesimal variation:
$$begin{align}frac{1}{2} delta L
~=~&left{ color{blue}{sddot{y}-cddot{x}}
+ color{red}{pi s (dot{x}^2+dot{y}^2)} right} delta x
+ left{ color{blue}{sddot{x}+cddot{y}}
+ color{red}{2pi (cdot{x}^2-sdot{x}dot{y})} right} delta y cr
& + frac{mathrm{d}}{mathrm{d}t}left{ color{green}{(cdot{x}-sdot{y})delta x -(sdot{x}+cdot{y})delta y}right},
end{align}tag{6}$$
which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols
$$ Gamma_{x,xx}~=~-pi s ~=~Gamma_{x,yy}~=~-Gamma_{y,xy}, qquad Gamma_{y,xx}~=~-2pi c,tag{7} $$
cf. OP's title.
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
$endgroup$
add a comment |
$begingroup$
The infinitesimal variation of the Lagrangian
$$ L(x,dot{x})~=~ g_{ij}(x)~ dot{x}^idot{x}^j tag{1} $$
is
$$ frac{1}{2}delta L~=~ -left{ color{blue}{ g_{kell}ddot{x}^{ell}} +color{red}{Gamma_{k,ij} dot{x}^idot{x}^j}right}delta x^k
+frac{mathrm{d}}{mathrm{d}t}left{ color{green}{ g_{kell}dot{x}^{ell} delta x^k}right},tag{2} $$
where we have introduced the lowered Levi-Civita Christoffel symbols
$$Gamma_{k,ij}~:=~g_{kell}Gamma^{ell}_{ij}. tag{3} $$
Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $t$-derivatives are distributed.
In particular we see that the geodesic equations are multiplied with the metric, cf. OP's last question.
Example. OP's Lagrangian reads
$$L~=~c(dot{x}^2-dot{y}^2) - 2sdot{x}dot{y}, qquad c~:=~cos(2pi x), qquad s~:=~sin(2pi x) ,tag{4}$$
corresponding to the metric
$$ begin{pmatrix} g_{xx} & g_{xy} cr g_{yx} & g_{yy} end{pmatrix} ~=~begin{pmatrix} c & -s cr -s & -c end{pmatrix}. tag{5}$$
We calculate the infinitesimal variation:
$$begin{align}frac{1}{2} delta L
~=~&left{ color{blue}{sddot{y}-cddot{x}}
+ color{red}{pi s (dot{x}^2+dot{y}^2)} right} delta x
+ left{ color{blue}{sddot{x}+cddot{y}}
+ color{red}{2pi (cdot{x}^2-sdot{x}dot{y})} right} delta y cr
& + frac{mathrm{d}}{mathrm{d}t}left{ color{green}{(cdot{x}-sdot{y})delta x -(sdot{x}+cdot{y})delta y}right},
end{align}tag{6}$$
which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols
$$ Gamma_{x,xx}~=~-pi s ~=~Gamma_{x,yy}~=~-Gamma_{y,xy}, qquad Gamma_{y,xx}~=~-2pi c,tag{7} $$
cf. OP's title.
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
$endgroup$
add a comment |
$begingroup$
The infinitesimal variation of the Lagrangian
$$ L(x,dot{x})~=~ g_{ij}(x)~ dot{x}^idot{x}^j tag{1} $$
is
$$ frac{1}{2}delta L~=~ -left{ color{blue}{ g_{kell}ddot{x}^{ell}} +color{red}{Gamma_{k,ij} dot{x}^idot{x}^j}right}delta x^k
+frac{mathrm{d}}{mathrm{d}t}left{ color{green}{ g_{kell}dot{x}^{ell} delta x^k}right},tag{2} $$
where we have introduced the lowered Levi-Civita Christoffel symbols
$$Gamma_{k,ij}~:=~g_{kell}Gamma^{ell}_{ij}. tag{3} $$
Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $t$-derivatives are distributed.
In particular we see that the geodesic equations are multiplied with the metric, cf. OP's last question.
Example. OP's Lagrangian reads
$$L~=~c(dot{x}^2-dot{y}^2) - 2sdot{x}dot{y}, qquad c~:=~cos(2pi x), qquad s~:=~sin(2pi x) ,tag{4}$$
corresponding to the metric
$$ begin{pmatrix} g_{xx} & g_{xy} cr g_{yx} & g_{yy} end{pmatrix} ~=~begin{pmatrix} c & -s cr -s & -c end{pmatrix}. tag{5}$$
We calculate the infinitesimal variation:
$$begin{align}frac{1}{2} delta L
~=~&left{ color{blue}{sddot{y}-cddot{x}}
+ color{red}{pi s (dot{x}^2+dot{y}^2)} right} delta x
+ left{ color{blue}{sddot{x}+cddot{y}}
+ color{red}{2pi (cdot{x}^2-sdot{x}dot{y})} right} delta y cr
& + frac{mathrm{d}}{mathrm{d}t}left{ color{green}{(cdot{x}-sdot{y})delta x -(sdot{x}+cdot{y})delta y}right},
end{align}tag{6}$$
which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols
$$ Gamma_{x,xx}~=~-pi s ~=~Gamma_{x,yy}~=~-Gamma_{y,xy}, qquad Gamma_{y,xx}~=~-2pi c,tag{7} $$
cf. OP's title.
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
$endgroup$
The infinitesimal variation of the Lagrangian
$$ L(x,dot{x})~=~ g_{ij}(x)~ dot{x}^idot{x}^j tag{1} $$
is
$$ frac{1}{2}delta L~=~ -left{ color{blue}{ g_{kell}ddot{x}^{ell}} +color{red}{Gamma_{k,ij} dot{x}^idot{x}^j}right}delta x^k
+frac{mathrm{d}}{mathrm{d}t}left{ color{green}{ g_{kell}dot{x}^{ell} delta x^k}right},tag{2} $$
where we have introduced the lowered Levi-Civita Christoffel symbols
$$Gamma_{k,ij}~:=~g_{kell}Gamma^{ell}_{ij}. tag{3} $$
Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $t$-derivatives are distributed.
In particular we see that the geodesic equations are multiplied with the metric, cf. OP's last question.
Example. OP's Lagrangian reads
$$L~=~c(dot{x}^2-dot{y}^2) - 2sdot{x}dot{y}, qquad c~:=~cos(2pi x), qquad s~:=~sin(2pi x) ,tag{4}$$
corresponding to the metric
$$ begin{pmatrix} g_{xx} & g_{xy} cr g_{yx} & g_{yy} end{pmatrix} ~=~begin{pmatrix} c & -s cr -s & -c end{pmatrix}. tag{5}$$
We calculate the infinitesimal variation:
$$begin{align}frac{1}{2} delta L
~=~&left{ color{blue}{sddot{y}-cddot{x}}
+ color{red}{pi s (dot{x}^2+dot{y}^2)} right} delta x
+ left{ color{blue}{sddot{x}+cddot{y}}
+ color{red}{2pi (cdot{x}^2-sdot{x}dot{y})} right} delta y cr
& + frac{mathrm{d}}{mathrm{d}t}left{ color{green}{(cdot{x}-sdot{y})delta x -(sdot{x}+cdot{y})delta y}right},
end{align}tag{6}$$
which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols
$$ Gamma_{x,xx}~=~-pi s ~=~Gamma_{x,yy}~=~-Gamma_{y,xy}, qquad Gamma_{y,xx}~=~-2pi c,tag{7} $$
cf. OP's title.
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
edited Dec 2 '18 at 11:23
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
answered Nov 30 '18 at 21:35
QmechanicQmechanic
5,03211856
5,03211856
add a comment |
add a comment |
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$begingroup$
I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example
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– Rafa Budría
Nov 29 '18 at 18:57
$begingroup$
I understand and I don't, at the same time. I understand that the coefficient of $ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $dot{x}dot{y}$ will always generate a $ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{ell k}ddot{x}^k + g_{ell k}Gamma_{ij}^k dot{x}^idot{x}^j = 0$$ we could still have $ddot{x}_m$ terms for $m neq ell$ since the metric coefficient matrix is not diagonal?
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– Ivo Terek
Nov 29 '18 at 19:23
1
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I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $cos(2pi x)>0$.
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– Ted Shifrin
Nov 29 '18 at 19:55