Christoffel symbol transformation
$begingroup$
The Christoffel symbols transform like
$$Gamma^{prime a}_{bc} = frac{partial x^{prime a}}{partial x^d} frac{partial x^e}{partial x^{prime b}} frac{partial x^f}{partial x^{prime c}} Gamma^{d}_{ef} - frac{partial x^d}{partial x^{prime b}} frac{partial x^e}{partial x^{prime c}} frac{partial^2 x^{prime a}}{partial x^d partial x^e}$$
Now the second term can be written as
$$ frac{partial x^e}{partial x^{prime c}} frac{partial x^d}{partial x^{prime b}} frac{partial}{partial x^d} left(frac{partial x^{prime a}}{ partial x^e}right) = frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{prime b}} left(frac{partial x^{prime a}}{ partial x^e}right) color{red}{=} frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{e}} left(frac{partial x^{prime a}}{ partial x^{prime b}}right) = frac{partial}{partial x^{prime c}} delta^a_b = 0$$
I don't think this is correct. I guess that the problem is that derivatives of different coordinates don't commute, and so the mistake is in the step $2 to 3$, where the equal sign is red, but that the previous are correct. Is this right?
differential-geometry change-of-basis general-relativity
asked Dec 7 '18 at 15:33
AlexAlex
63
$endgroup$
add a comment |
$begingroup$
The Christoffel symbols transform like
$$Gamma^{prime a}_{bc} = frac{partial x^{prime a}}{partial x^d} frac{partial x^e}{partial x^{prime b}} frac{partial x^f}{partial x^{prime c}} Gamma^{d}_{ef} - frac{partial x^d}{partial x^{prime b}} frac{partial x^e}{partial x^{prime c}} frac{partial^2 x^{prime a}}{partial x^d partial x^e}$$
Now the second term can be written as
$$ frac{partial x^e}{partial x^{prime c}} frac{partial x^d}{partial x^{prime b}} frac{partial}{partial x^d} left(frac{partial x^{prime a}}{ partial x^e}right) = frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{prime b}} left(frac{partial x^{prime a}}{ partial x^e}right) color{red}{=} frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{e}} left(frac{partial x^{prime a}}{ partial x^{prime b}}right) = frac{partial}{partial x^{prime c}} delta^a_b = 0$$
I don't think this is correct. I guess that the problem is that derivatives of different coordinates don't commute, and so the mistake is in the step $2 to 3$, where the equal sign is red, but that the previous are correct. Is this right?
differential-geometry change-of-basis general-relativity
asked Dec 7 '18 at 15:33
AlexAlex
63
$endgroup$
$begingroup$
I don't even know what the stuff in the second expression means. You're differentiating a function of $x$ with respect to $x'{}^b$.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 19:48
$begingroup$
@TedShifrin in the second expression I differentiate one set of coordinates with respect to the others, for example if these were cartesian and polar, we will have the derivative of $r$ and $theta$ with respect to $x$ and $y$, where we can interpret $r = r(x,y)$, $theta = theta(x,y)$ and also $x = x(r,theta)$, $y = y(r,theta)$
$endgroup$
– Alex
Dec 8 '18 at 12:13
add a comment |
$begingroup$
The Christoffel symbols transform like
$$Gamma^{prime a}_{bc} = frac{partial x^{prime a}}{partial x^d} frac{partial x^e}{partial x^{prime b}} frac{partial x^f}{partial x^{prime c}} Gamma^{d}_{ef} - frac{partial x^d}{partial x^{prime b}} frac{partial x^e}{partial x^{prime c}} frac{partial^2 x^{prime a}}{partial x^d partial x^e}$$
Now the second term can be written as
$$ frac{partial x^e}{partial x^{prime c}} frac{partial x^d}{partial x^{prime b}} frac{partial}{partial x^d} left(frac{partial x^{prime a}}{ partial x^e}right) = frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{prime b}} left(frac{partial x^{prime a}}{ partial x^e}right) color{red}{=} frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{e}} left(frac{partial x^{prime a}}{ partial x^{prime b}}right) = frac{partial}{partial x^{prime c}} delta^a_b = 0$$
I don't think this is correct. I guess that the problem is that derivatives of different coordinates don't commute, and so the mistake is in the step $2 to 3$, where the equal sign is red, but that the previous are correct. Is this right?
differential-geometry change-of-basis general-relativity
asked Dec 7 '18 at 15:33
AlexAlex
63
$endgroup$
The Christoffel symbols transform like
$$Gamma^{prime a}_{bc} = frac{partial x^{prime a}}{partial x^d} frac{partial x^e}{partial x^{prime b}} frac{partial x^f}{partial x^{prime c}} Gamma^{d}_{ef} - frac{partial x^d}{partial x^{prime b}} frac{partial x^e}{partial x^{prime c}} frac{partial^2 x^{prime a}}{partial x^d partial x^e}$$
Now the second term can be written as
$$ frac{partial x^e}{partial x^{prime c}} frac{partial x^d}{partial x^{prime b}} frac{partial}{partial x^d} left(frac{partial x^{prime a}}{ partial x^e}right) = frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{prime b}} left(frac{partial x^{prime a}}{ partial x^e}right) color{red}{=} frac{partial x^e}{partial x^{prime c}} frac{partial}{partial x^{e}} left(frac{partial x^{prime a}}{ partial x^{prime b}}right) = frac{partial}{partial x^{prime c}} delta^a_b = 0$$
I don't think this is correct. I guess that the problem is that derivatives of different coordinates don't commute, and so the mistake is in the step $2 to 3$, where the equal sign is red, but that the previous are correct. Is this right?
differential-geometry change-of-basis general-relativity
differential-geometry change-of-basis general-relativity
asked Dec 7 '18 at 15:33
AlexAlex
63
asked Dec 7 '18 at 15:33
AlexAlex
63
asked Dec 7 '18 at 15:33
AlexAlex
63
asked Dec 7 '18 at 15:33
AlexAlex
63
asked Dec 7 '18 at 15:33
AlexAlex
63
63
$begingroup$
I don't even know what the stuff in the second expression means. You're differentiating a function of $x$ with respect to $x'{}^b$.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 19:48
$begingroup$
@TedShifrin in the second expression I differentiate one set of coordinates with respect to the others, for example if these were cartesian and polar, we will have the derivative of $r$ and $theta$ with respect to $x$ and $y$, where we can interpret $r = r(x,y)$, $theta = theta(x,y)$ and also $x = x(r,theta)$, $y = y(r,theta)$
$endgroup$
– Alex
Dec 8 '18 at 12:13
add a comment |
$begingroup$
I don't even know what the stuff in the second expression means. You're differentiating a function of $x$ with respect to $x'{}^b$.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 19:48
$begingroup$
@TedShifrin in the second expression I differentiate one set of coordinates with respect to the others, for example if these were cartesian and polar, we will have the derivative of $r$ and $theta$ with respect to $x$ and $y$, where we can interpret $r = r(x,y)$, $theta = theta(x,y)$ and also $x = x(r,theta)$, $y = y(r,theta)$
$endgroup$
– Alex
Dec 8 '18 at 12:13
$begingroup$
I don't even know what the stuff in the second expression means. You're differentiating a function of $x$ with respect to $x'{}^b$.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 19:48
$begingroup$
I don't even know what the stuff in the second expression means. You're differentiating a function of $x$ with respect to $x'{}^b$.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 19:48
$begingroup$
@TedShifrin in the second expression I differentiate one set of coordinates with respect to the others, for example if these were cartesian and polar, we will have the derivative of $r$ and $theta$ with respect to $x$ and $y$, where we can interpret $r = r(x,y)$, $theta = theta(x,y)$ and also $x = x(r,theta)$, $y = y(r,theta)$
$endgroup$
– Alex
Dec 8 '18 at 12:13
$begingroup$
@TedShifrin in the second expression I differentiate one set of coordinates with respect to the others, for example if these were cartesian and polar, we will have the derivative of $r$ and $theta$ with respect to $x$ and $y$, where we can interpret $r = r(x,y)$, $theta = theta(x,y)$ and also $x = x(r,theta)$, $y = y(r,theta)$
$endgroup$
– Alex
Dec 8 '18 at 12:13
add a comment |
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$begingroup$
I don't even know what the stuff in the second expression means. You're differentiating a function of $x$ with respect to $x'{}^b$.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 19:48
$begingroup$
@TedShifrin in the second expression I differentiate one set of coordinates with respect to the others, for example if these were cartesian and polar, we will have the derivative of $r$ and $theta$ with respect to $x$ and $y$, where we can interpret $r = r(x,y)$, $theta = theta(x,y)$ and also $x = x(r,theta)$, $y = y(r,theta)$
$endgroup$
– Alex
Dec 8 '18 at 12:13