Combining two Lorentz boostsGeneral matrix Lorentz transformationComposition of Lorentz transformations using generators and the Wigner rotationLorentz boost matrix in terms of four-velocityComposition of Lorentz TransformationsOrder of index in Lorentz transformLorentz transformation of Gamma matrices $gamma^mu$Lorentz boosts in some combined directionUnderstanding the Dirac spinor representation $(1/2,0) bigoplus (0,1/2)$ of the Lorentz group?Derivation of Lorentz boostsTrouble calculating Wigner rotation for photonRindler Coordinates DerivationError with generators of Lorentz group (basis of Lorentz Lie algebra)
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Combining two Lorentz boosts
General matrix Lorentz transformationComposition of Lorentz transformations using generators and the Wigner rotationLorentz boost matrix in terms of four-velocityComposition of Lorentz TransformationsOrder of index in Lorentz transformLorentz transformation of Gamma matrices $gamma^mu$Lorentz boosts in some combined directionUnderstanding the Dirac spinor representation $(1/2,0) bigoplus (0,1/2)$ of the Lorentz group?Derivation of Lorentz boostsTrouble calculating Wigner rotation for photonRindler Coordinates DerivationError with generators of Lorentz group (basis of Lorentz Lie algebra)
$begingroup$
Is it possible to express two Lorentz boosts $A_x(beta)$ and $A_y(beta)$ along the x/y-axis as one boost described by $A(overrightarrow delta)$?
To answer this, I start by defining $theta equiv arctanleft(betaright)$ and $beta equiv v/c ,$ then:
$$
beginalign
A_x(beta) &= beginbmatrix cosh(theta) & sinh(theta) & 0 \ sinh(theta) & cosh(theta) & 0 \ 0 & 0 & 1 endbmatrix\[10px]
A_y(beta) &= beginbmatrix cosh(theta) & 0 & sinh(theta) \ 0 & 1 & 0 \ sinh(theta) & 0 & cosh(theta) endbmatrix \[10px]
A(overrightarrow delta) &= beginbmatrixgamma & gamma delta^1 & gamma delta^2 \ gamma delta^1 & 1+B_11 & B_12 \ gamma delta^2 & B_21 & 1+B_22 endbmatrix \[10px]
B_ij &= fracgamma -1delta^2beta_idelta_j
endalign
$$
I first calculated:
$$
A_y(beta)*A_x(beta)
=
beginbmatrix cosh^2(beta) & cosh(beta)*sinh(beta) & sinh(beta) \ sinh(beta) & cosh(beta) & 0 \ cosh(beta)* sinh(beta) & sinh^2(beta) & cosh(beta) endbmatrix
$$
then since $B_ij=B_ji ,$ I get
$$
sinh^2left(thetaright) = 0
quadRightarrowquad
theta=0
quadRightarrowquad
v=0
,.
$$
That would mean, that it is not possible to express two subsequent boosts in $xtext-$ and $ytext-$ direction as a combined boost. This does seem a bit odd to me and I wonder if I made a mistake in my computation.
special-relativity inertial-frames lorentz-symmetry
edited May 16 at 22:43
Nat
3,77242033
asked May 16 at 16:32
KekksKekks
504
$endgroup$
add a comment |
$begingroup$
Is it possible to express two Lorentz boosts $A_x(beta)$ and $A_y(beta)$ along the x/y-axis as one boost described by $A(overrightarrow delta)$?
To answer this, I start by defining $theta equiv arctanleft(betaright)$ and $beta equiv v/c ,$ then:
$$
beginalign
A_x(beta) &= beginbmatrix cosh(theta) & sinh(theta) & 0 \ sinh(theta) & cosh(theta) & 0 \ 0 & 0 & 1 endbmatrix\[10px]
A_y(beta) &= beginbmatrix cosh(theta) & 0 & sinh(theta) \ 0 & 1 & 0 \ sinh(theta) & 0 & cosh(theta) endbmatrix \[10px]
A(overrightarrow delta) &= beginbmatrixgamma & gamma delta^1 & gamma delta^2 \ gamma delta^1 & 1+B_11 & B_12 \ gamma delta^2 & B_21 & 1+B_22 endbmatrix \[10px]
B_ij &= fracgamma -1delta^2beta_idelta_j
endalign
$$
I first calculated:
$$
A_y(beta)*A_x(beta)
=
beginbmatrix cosh^2(beta) & cosh(beta)*sinh(beta) & sinh(beta) \ sinh(beta) & cosh(beta) & 0 \ cosh(beta)* sinh(beta) & sinh^2(beta) & cosh(beta) endbmatrix
$$
then since $B_ij=B_ji ,$ I get
$$
sinh^2left(thetaright) = 0
quadRightarrowquad
theta=0
quadRightarrowquad
v=0
,.
$$
That would mean, that it is not possible to express two subsequent boosts in $xtext-$ and $ytext-$ direction as a combined boost. This does seem a bit odd to me and I wonder if I made a mistake in my computation.
special-relativity inertial-frames lorentz-symmetry
edited May 16 at 22:43
Nat
3,77242033
asked May 16 at 16:32
KekksKekks
504
$endgroup$
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
May 16 at 16:58
$begingroup$
Also related.
$endgroup$
– J.G.
May 16 at 18:45
$begingroup$
Related : General matrix Lorentz transformation
$endgroup$
– Frobenius
May 16 at 21:15
$begingroup$
Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation.
$endgroup$
– Nat
May 16 at 22:46
add a comment |
$begingroup$
Is it possible to express two Lorentz boosts $A_x(beta)$ and $A_y(beta)$ along the x/y-axis as one boost described by $A(overrightarrow delta)$?
To answer this, I start by defining $theta equiv arctanleft(betaright)$ and $beta equiv v/c ,$ then:
$$
beginalign
A_x(beta) &= beginbmatrix cosh(theta) & sinh(theta) & 0 \ sinh(theta) & cosh(theta) & 0 \ 0 & 0 & 1 endbmatrix\[10px]
A_y(beta) &= beginbmatrix cosh(theta) & 0 & sinh(theta) \ 0 & 1 & 0 \ sinh(theta) & 0 & cosh(theta) endbmatrix \[10px]
A(overrightarrow delta) &= beginbmatrixgamma & gamma delta^1 & gamma delta^2 \ gamma delta^1 & 1+B_11 & B_12 \ gamma delta^2 & B_21 & 1+B_22 endbmatrix \[10px]
B_ij &= fracgamma -1delta^2beta_idelta_j
endalign
$$
I first calculated:
$$
A_y(beta)*A_x(beta)
=
beginbmatrix cosh^2(beta) & cosh(beta)*sinh(beta) & sinh(beta) \ sinh(beta) & cosh(beta) & 0 \ cosh(beta)* sinh(beta) & sinh^2(beta) & cosh(beta) endbmatrix
$$
then since $B_ij=B_ji ,$ I get
$$
sinh^2left(thetaright) = 0
quadRightarrowquad
theta=0
quadRightarrowquad
v=0
,.
$$
That would mean, that it is not possible to express two subsequent boosts in $xtext-$ and $ytext-$ direction as a combined boost. This does seem a bit odd to me and I wonder if I made a mistake in my computation.
special-relativity inertial-frames lorentz-symmetry
edited May 16 at 22:43
Nat
3,77242033
asked May 16 at 16:32
KekksKekks
504
$endgroup$
Is it possible to express two Lorentz boosts $A_x(beta)$ and $A_y(beta)$ along the x/y-axis as one boost described by $A(overrightarrow delta)$?
To answer this, I start by defining $theta equiv arctanleft(betaright)$ and $beta equiv v/c ,$ then:
$$
beginalign
A_x(beta) &= beginbmatrix cosh(theta) & sinh(theta) & 0 \ sinh(theta) & cosh(theta) & 0 \ 0 & 0 & 1 endbmatrix\[10px]
A_y(beta) &= beginbmatrix cosh(theta) & 0 & sinh(theta) \ 0 & 1 & 0 \ sinh(theta) & 0 & cosh(theta) endbmatrix \[10px]
A(overrightarrow delta) &= beginbmatrixgamma & gamma delta^1 & gamma delta^2 \ gamma delta^1 & 1+B_11 & B_12 \ gamma delta^2 & B_21 & 1+B_22 endbmatrix \[10px]
B_ij &= fracgamma -1delta^2beta_idelta_j
endalign
$$
I first calculated:
$$
A_y(beta)*A_x(beta)
=
beginbmatrix cosh^2(beta) & cosh(beta)*sinh(beta) & sinh(beta) \ sinh(beta) & cosh(beta) & 0 \ cosh(beta)* sinh(beta) & sinh^2(beta) & cosh(beta) endbmatrix
$$
then since $B_ij=B_ji ,$ I get
$$
sinh^2left(thetaright) = 0
quadRightarrowquad
theta=0
quadRightarrowquad
v=0
,.
$$
That would mean, that it is not possible to express two subsequent boosts in $xtext-$ and $ytext-$ direction as a combined boost. This does seem a bit odd to me and I wonder if I made a mistake in my computation.
special-relativity inertial-frames lorentz-symmetry
special-relativity inertial-frames lorentz-symmetry
edited May 16 at 22:43
Nat
3,77242033
asked May 16 at 16:32
KekksKekks
504
edited May 16 at 22:43
Nat
3,77242033
asked May 16 at 16:32
KekksKekks
504
edited May 16 at 22:43
Nat
3,77242033
edited May 16 at 22:43
Nat
3,77242033
edited May 16 at 22:43
Nat
3,77242033
3,77242033
asked May 16 at 16:32
KekksKekks
504
asked May 16 at 16:32
KekksKekks
504
asked May 16 at 16:32
KekksKekks
504
504
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
May 16 at 16:58
$begingroup$
Also related.
$endgroup$
– J.G.
May 16 at 18:45
$begingroup$
Related : General matrix Lorentz transformation
$endgroup$
– Frobenius
May 16 at 21:15
$begingroup$
Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation.
$endgroup$
– Nat
May 16 at 22:46
add a comment |
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
May 16 at 16:58
$begingroup$
Also related.
$endgroup$
– J.G.
May 16 at 18:45
$begingroup$
Related : General matrix Lorentz transformation
$endgroup$
– Frobenius
May 16 at 21:15
$begingroup$
Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation.
$endgroup$
– Nat
May 16 at 22:46
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
May 16 at 16:58
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
May 16 at 16:58
$begingroup$
Also related.
$endgroup$
– J.G.
May 16 at 18:45
$begingroup$
Also related.
$endgroup$
– J.G.
May 16 at 18:45
$begingroup$
Related : General matrix Lorentz transformation
$endgroup$
– Frobenius
May 16 at 21:15
$begingroup$
Related : General matrix Lorentz transformation
$endgroup$
– Frobenius
May 16 at 21:15
$begingroup$
Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation.
$endgroup$
– Nat
May 16 at 22:46
$begingroup$
Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation.
$endgroup$
– Nat
May 16 at 22:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not possible. Two boosts are, however, equivalent to one boost combined with one rotation! This surprising rotation is known as a Thomas-Wigner rotation.
This rotation arises because two infinitesimal boosts do not commute; their commutator is an infinitesimal rotation. In terms of rotation generators $J_i$ and boost generators $K_i$, the Lorentz algebra is
$$[J_i,J_j]=epsilon_ijkJ_k$$
$$[K_i,K_j]=-epsilon_ijkJ_k ,text(!)$$
$$[J_i,K_j]=epsilon_ijkK_k$$
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
$endgroup$
$begingroup$
Thank you for the explanation and the link.
$endgroup$
– Kekks
May 16 at 16:42
1
$begingroup$
This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 17:12
$begingroup$
@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
$endgroup$
– G. Smith
May 16 at 17:21
$begingroup$
I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 18:02
$begingroup$
@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
$endgroup$
– G. Smith
May 16 at 19:21
add a comment |
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$begingroup$
It is not possible. Two boosts are, however, equivalent to one boost combined with one rotation! This surprising rotation is known as a Thomas-Wigner rotation.
This rotation arises because two infinitesimal boosts do not commute; their commutator is an infinitesimal rotation. In terms of rotation generators $J_i$ and boost generators $K_i$, the Lorentz algebra is
$$[J_i,J_j]=epsilon_ijkJ_k$$
$$[K_i,K_j]=-epsilon_ijkJ_k ,text(!)$$
$$[J_i,K_j]=epsilon_ijkK_k$$
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
$endgroup$
$begingroup$
Thank you for the explanation and the link.
$endgroup$
– Kekks
May 16 at 16:42
1
$begingroup$
This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 17:12
$begingroup$
@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
$endgroup$
– G. Smith
May 16 at 17:21
$begingroup$
I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 18:02
$begingroup$
@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
$endgroup$
– G. Smith
May 16 at 19:21
add a comment |
$begingroup$
It is not possible. Two boosts are, however, equivalent to one boost combined with one rotation! This surprising rotation is known as a Thomas-Wigner rotation.
This rotation arises because two infinitesimal boosts do not commute; their commutator is an infinitesimal rotation. In terms of rotation generators $J_i$ and boost generators $K_i$, the Lorentz algebra is
$$[J_i,J_j]=epsilon_ijkJ_k$$
$$[K_i,K_j]=-epsilon_ijkJ_k ,text(!)$$
$$[J_i,K_j]=epsilon_ijkK_k$$
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
$endgroup$
$begingroup$
Thank you for the explanation and the link.
$endgroup$
– Kekks
May 16 at 16:42
1
$begingroup$
This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 17:12
$begingroup$
@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
$endgroup$
– G. Smith
May 16 at 17:21
$begingroup$
I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 18:02
$begingroup$
@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
$endgroup$
– G. Smith
May 16 at 19:21
add a comment |
$begingroup$
It is not possible. Two boosts are, however, equivalent to one boost combined with one rotation! This surprising rotation is known as a Thomas-Wigner rotation.
This rotation arises because two infinitesimal boosts do not commute; their commutator is an infinitesimal rotation. In terms of rotation generators $J_i$ and boost generators $K_i$, the Lorentz algebra is
$$[J_i,J_j]=epsilon_ijkJ_k$$
$$[K_i,K_j]=-epsilon_ijkJ_k ,text(!)$$
$$[J_i,K_j]=epsilon_ijkK_k$$
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
$endgroup$
It is not possible. Two boosts are, however, equivalent to one boost combined with one rotation! This surprising rotation is known as a Thomas-Wigner rotation.
This rotation arises because two infinitesimal boosts do not commute; their commutator is an infinitesimal rotation. In terms of rotation generators $J_i$ and boost generators $K_i$, the Lorentz algebra is
$$[J_i,J_j]=epsilon_ijkJ_k$$
$$[K_i,K_j]=-epsilon_ijkJ_k ,text(!)$$
$$[J_i,K_j]=epsilon_ijkK_k$$
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
edited May 16 at 16:56
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
answered May 16 at 16:36
G. SmithG. Smith
13.4k12145
13.4k12145
$begingroup$
Thank you for the explanation and the link.
$endgroup$
– Kekks
May 16 at 16:42
1
$begingroup$
This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 17:12
$begingroup$
@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
$endgroup$
– G. Smith
May 16 at 17:21
$begingroup$
I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 18:02
$begingroup$
@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
$endgroup$
– G. Smith
May 16 at 19:21
add a comment |
$begingroup$
Thank you for the explanation and the link.
$endgroup$
– Kekks
May 16 at 16:42
1
$begingroup$
This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 17:12
$begingroup$
@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
$endgroup$
– G. Smith
May 16 at 17:21
$begingroup$
I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
$endgroup$
– Feynmans Out for Grumpy Cat
May 16 at 18:02
$begingroup$
@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
$endgroup$
– G. Smith
May 16 at 19:21
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Thank you for the explanation and the link.
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– Kekks
May 16 at 16:42
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Thank you for the explanation and the link.
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– Kekks
May 16 at 16:42
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This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
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– Feynmans Out for Grumpy Cat
May 16 at 17:12
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This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $mathbbso(1,3)$ as $mathbbsu(2)timesmathbbsu(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"?
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– Feynmans Out for Grumpy Cat
May 16 at 17:12
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@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
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– G. Smith
May 16 at 17:21
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@DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it.
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– G. Smith
May 16 at 17:21
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I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
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– Feynmans Out for Grumpy Cat
May 16 at 18:02
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I just meant that when we go to $mathbbsu(2)timesmathbbsu(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=fracJ_i+iK_i2$ and $B_i=fracJ_i-iK_i2$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality?
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– Feynmans Out for Grumpy Cat
May 16 at 18:02
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@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
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– G. Smith
May 16 at 19:21
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@DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question.
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– G. Smith
May 16 at 19:21
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Related.
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– Cosmas Zachos
May 16 at 16:58
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Also related.
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– J.G.
May 16 at 18:45
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Related : General matrix Lorentz transformation
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– Frobenius
May 16 at 21:15
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Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation.
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– Nat
May 16 at 22:46