Yhyjyk

Is it improper etiquette to ask your opponent what his/her rating is before the game?

What was this official D&D 3.5e Lovecraft-flavored rulebook?

Closed-form expression for certain product

How to indicate a cut out for a product window

Did arcade monitors have same pixel aspect ratio as TV sets?

Aragorn's "guise" in the Orthanc Stone

Are paving bricks differently sized for sand bedding vs mortar bedding?

Find the Primitive Roots Mod 31

Why a symmetric relation is defined: ∀x∀y( xRy⟹yRx) and not ∀x∀y (xRy⟺yRx)?

Why should universal income be universal?

Multiplicative persistence

Is this toilet slogan correct usage of the English language?

Count the occurrence of each unique word in the file

The screen of my macbook suddenly broken down how can I do to recover

Is Witten's Proof of the Positive Mass Theorem Rigorous?

Why does the Sun have different day lengths, but not the gas giants?

How to create ADT in Haskell?

What does routing an IP address mean?

Fear of getting stuck on one programming language / technology that is not used in my country

The probability of Bus A arriving before Bus B

Loading commands from file

Biological Blimps: Propulsion

What if a revenant (monster) gains fire resistance?

If a character has darkvision, can they see through an area of nonmagical darkness filled with lightly obscuring gas?

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

2

1

$begingroup$

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

share|cite|improve this question

asked 4 hours ago

KALLE THE BAWSMANKALLE THE BAWSMAN

1207

$endgroup$

add a comment | 

2

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked 4 hours ago

KALLE THE BAWSMANKALLE THE BAWSMAN

1207

$endgroup$

share|cite|improve this question

asked 4 hours ago

KALLE THE BAWSMANKALLE THE BAWSMAN

1207

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

5

$begingroup$

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

answered 4 hours ago

Chiral AnomalyChiral Anomaly

12.3k21540

$endgroup$

add a comment | 

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

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

Name

Email

Required, but never shown

5

$begingroup$

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

answered 4 hours ago

Chiral AnomalyChiral Anomaly

12.3k21540

$endgroup$

add a comment | 

5

$begingroup$

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

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 $epsilon$.

share|cite|improve this answer

answered 4 hours ago

Chiral AnomalyChiral Anomaly

12.3k21540

$endgroup$

share|cite|improve this answer

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