If infinitesimal transformations commute why dont the generators of the Lorentz group commute?
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If infinitesimal transformations commute as proved eg here, why are the commutators for the generators of the Lorentz group nonzero?
general-relativity special-relativity lorentz-symmetry
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
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$begingroup$
If infinitesimal transformations commute as proved eg here, why are the commutators for the generators of the Lorentz group nonzero?
general-relativity special-relativity lorentz-symmetry
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
$endgroup$
add a comment |
$begingroup$
If infinitesimal transformations commute as proved eg here, why are the commutators for the generators of the Lorentz group nonzero?
general-relativity special-relativity lorentz-symmetry
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
$endgroup$
If infinitesimal transformations commute as proved eg here, why are the commutators for the generators of the Lorentz group nonzero?
general-relativity special-relativity lorentz-symmetry
general-relativity special-relativity lorentz-symmetry
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
asked 9 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1187
1187
add a comment |
add a comment |
1 Answer
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I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.
Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.
When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
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$begingroup$
I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.
Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.
When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
add a comment |
$begingroup$
I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.
Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.
When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
add a comment |
$begingroup$
I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.
Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.
When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.
Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.
When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
12.4k21540
add a comment |
add a comment |
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