If infinitesimal transformations commute why don't the generators of the Lorentz group commute? The 2019 Stack Overflow Developer Survey Results Are InWhy do we use the complexification of the Lorentz group?Difference Between Algebra of Infinitesimal Conformal Transformations & Conformal AlgebraSubgroup of Lorentz Group Generated by BoostsLorentz Group Generators: Two MethodsMeaning of Lorentz GeneratorsFinite 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

Protecting Dualbooting Windows from dangerous code (like rm -rf)

Does coating your armor in silver add any effects?

How to type this arrow in math mode?

Is this app Icon Browser Safe/Legit?

Looking for Correct Greek Translation for Heraclitus

What do the Banks children have against barley water?

Return to UK after having been refused entry years ago

Is there a symbol for a right arrow with a square in the middle?

Am I thawing this London Broil safely?

What do hard-Brexiteers want with respect to the Irish border?

Can we generate random numbers using irrational numbers like π and e?

What tool would a Roman-age civilization have for the breaking of silver and other metals into dust?

Landlord wants to switch my lease to a "Land contract" to "get back at the city"

How come people say “Would of”?

What does ひと匙 mean in this manga and has it been used colloquially?

FPGA - DIY Programming

Identify boardgame from Big movie

Why do we hear so much about the Trump administration deciding to impose and then remove tariffs?

Who coined the term "madman theory"?

Why is the Constellation's nose gear so long?

Multiply Two Integer Polynomials

How to support a colleague who finds meetings extremely tiring?

Aging parents with no investments

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?



If infinitesimal transformations commute why don't the generators of the Lorentz group commute?



The 2019 Stack Overflow Developer Survey Results Are InWhy do we use the complexification of the Lorentz group?Difference Between Algebra of Infinitesimal Conformal Transformations & Conformal AlgebraSubgroup of Lorentz Group Generated by BoostsLorentz Group Generators: Two MethodsMeaning of Lorentz GeneratorsFinite 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










2












$begingroup$


If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?










      share|cite|improve this question











      $endgroup$




      If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?







      special-relativity group-theory lorentz-symmetry commutator lie-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 24 at 17:44









      Qmechanic

      107k122001241




      107k122001241










      asked Mar 23 at 15:23









      KALLE THE BAWSMANKALLE THE BAWSMAN

      1207




      1207




















          3 Answers
          3






          active

          oldest

          votes


















          8












          $begingroup$

          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 $epsilon$.






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            1. Illustrative example: It is straightforward to prove that the Lie group
              $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
              of 3D rotations is generated by the corresponding Lie algebra
              $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
              of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


            2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!






            share|cite|improve this answer









            $endgroup$




















              1












              $begingroup$

              While
              beginalign
              e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
              &= 1+ epsilon (A+B)+ frac12
              epsilon^2 (A^2+AB +B^2)+ldots
              endalign

              we have
              beginalign
              e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
              &= 1+ epsilon (B+A)+ frac12
              epsilon^2 (A^2+BA +B^2)+ldots
              endalign

              so:



              1. To order $epsilon$, the transformations are the same,

              2. To order $epsilon^2$ they are different.

              Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






              share|cite|improve this answer









              $endgroup$













                Your Answer





                StackExchange.ifUsing("editor", function ()
                return StackExchange.using("mathjaxEditing", function ()
                StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
                StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                );
                );
                , "mathjax-editing");

                StackExchange.ready(function()
                var channelOptions =
                tags: "".split(" "),
                id: "151"
                ;
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function()
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled)
                StackExchange.using("snippets", function()
                createEditor();
                );

                else
                createEditor();

                );

                function createEditor()
                StackExchange.prepareEditor(
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: false,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: null,
                bindNavPrevention: true,
                postfix: "",
                imageUploader:
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                ,
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                );



                );













                draft saved

                draft discarded


















                StackExchange.ready(
                function ()
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468216%2fif-infinitesimal-transformations-commute-why-dont-the-generators-of-the-lorentz%23new-answer', 'question_page');

                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                8












                $begingroup$

                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 $epsilon$.






                share|cite|improve this answer









                $endgroup$

















                  8












                  $begingroup$

                  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 $epsilon$.






                  share|cite|improve this answer









                  $endgroup$















                    8












                    8








                    8





                    $begingroup$

                    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 $epsilon$.






                    share|cite|improve this answer









                    $endgroup$



                    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 $epsilon$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 23 at 15:57









                    Chiral AnomalyChiral Anomaly

                    13.5k21845




                    13.5k21845





















                        1












                        $begingroup$

                        1. Illustrative example: It is straightforward to prove that the Lie group
                          $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                          of 3D rotations is generated by the corresponding Lie algebra
                          $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                          of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                        2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          1. Illustrative example: It is straightforward to prove that the Lie group
                            $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                            of 3D rotations is generated by the corresponding Lie algebra
                            $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                            of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                          2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            1. Illustrative example: It is straightforward to prove that the Lie group
                              $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                              of 3D rotations is generated by the corresponding Lie algebra
                              $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                              of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                            2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!






                            share|cite|improve this answer









                            $endgroup$



                            1. Illustrative example: It is straightforward to prove that the Lie group
                              $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                              of 3D rotations is generated by the corresponding Lie algebra
                              $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                              of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                            2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 24 at 17:44









                            QmechanicQmechanic

                            107k122001241




                            107k122001241





















                                1












                                $begingroup$

                                While
                                beginalign
                                e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                &= 1+ epsilon (A+B)+ frac12
                                epsilon^2 (A^2+AB +B^2)+ldots
                                endalign

                                we have
                                beginalign
                                e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                &= 1+ epsilon (B+A)+ frac12
                                epsilon^2 (A^2+BA +B^2)+ldots
                                endalign

                                so:



                                1. To order $epsilon$, the transformations are the same,

                                2. To order $epsilon^2$ they are different.

                                Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






                                share|cite|improve this answer









                                $endgroup$

















                                  1












                                  $begingroup$

                                  While
                                  beginalign
                                  e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                  &= 1+ epsilon (A+B)+ frac12
                                  epsilon^2 (A^2+AB +B^2)+ldots
                                  endalign

                                  we have
                                  beginalign
                                  e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                  &= 1+ epsilon (B+A)+ frac12
                                  epsilon^2 (A^2+BA +B^2)+ldots
                                  endalign

                                  so:



                                  1. To order $epsilon$, the transformations are the same,

                                  2. To order $epsilon^2$ they are different.

                                  Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






                                  share|cite|improve this answer









                                  $endgroup$















                                    1












                                    1








                                    1





                                    $begingroup$

                                    While
                                    beginalign
                                    e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                    &= 1+ epsilon (A+B)+ frac12
                                    epsilon^2 (A^2+AB +B^2)+ldots
                                    endalign

                                    we have
                                    beginalign
                                    e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                    &= 1+ epsilon (B+A)+ frac12
                                    epsilon^2 (A^2+BA +B^2)+ldots
                                    endalign

                                    so:



                                    1. To order $epsilon$, the transformations are the same,

                                    2. To order $epsilon^2$ they are different.

                                    Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






                                    share|cite|improve this answer









                                    $endgroup$



                                    While
                                    beginalign
                                    e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                    &= 1+ epsilon (A+B)+ frac12
                                    epsilon^2 (A^2+AB +B^2)+ldots
                                    endalign

                                    we have
                                    beginalign
                                    e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                    &= 1+ epsilon (B+A)+ frac12
                                    epsilon^2 (A^2+BA +B^2)+ldots
                                    endalign

                                    so:



                                    1. To order $epsilon$, the transformations are the same,

                                    2. To order $epsilon^2$ they are different.

                                    Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Mar 24 at 18:02









                                    ZeroTheHeroZeroTheHero

                                    21.2k53364




                                    21.2k53364



























                                        draft saved

                                        draft discarded
















































                                        Thanks for contributing an answer to Physics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid


                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.

                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function ()
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468216%2fif-infinitesimal-transformations-commute-why-dont-the-generators-of-the-lorentz%23new-answer', 'question_page');

                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        Popular posts from this blog

                                        How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

                                        random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

                                        Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye