If infinitesimal transformations commute why dont the generators of the Lorentz group commute?Why do we use the complexification of the Lorentz group?Subgroup of Lorentz Group Generated by BoostsLorentz Group Generators: Two MethodsMeaning of Lorentz GeneratorsLorentz Transformations Vs Coordinate TransformationsFinite lorentz transform for 4-vectors in terms of the generatorsOn the generators of the Lorentz groupLie group compactness from generatorsRelation between the Dirac Algebra and the Lorentz groupCommutation relations of the generators of the Lorentz group
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If infinitesimal transformations commute why dont the generators of the Lorentz group commute?
Why do we use the complexification of the Lorentz group?Subgroup of Lorentz Group Generated by BoostsLorentz Group Generators: Two MethodsMeaning of Lorentz GeneratorsLorentz Transformations Vs Coordinate TransformationsFinite lorentz transform for 4-vectors in terms of the generatorsOn the generators of the Lorentz groupLie group compactness from generatorsRelation between the Dirac Algebra and the Lorentz groupCommutation relations of the generators of the Lorentz group
$begingroup$
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 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
$endgroup$
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 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
$endgroup$
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 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
$endgroup$
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 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
asked 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
asked 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
asked 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
asked 4 hours ago
KALLE THE BAWSMANKALLE THE BAWSMAN
1207
1207
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 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
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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 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
$endgroup$
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 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
$endgroup$
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 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
$endgroup$
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 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
answered 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
answered 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
answered 4 hours ago
Chiral AnomalyChiral Anomaly
12.3k21540
12.3k21540
add a comment |
add a comment |
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