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Minkowski Space in Group Theory


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#1 geordief

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Posted 29 August 2016 - 12:01 AM

 I understand that Minkowski (Poincare?) Space  is a Group in Group Theory.(am I right so far?)

 

Well I have (re) learned  that for  a Group to be a Group  there are one or two (4 ?) basic preconditions and that these are (1)  that the set must have a operator  and (2) must also include an identity element  , (3) be commutative  in the operations and  that (4) each element must  have an inverse.

 

Oh and (5) it must exhibit "closure"  

 

How do these conditions apply to Minkowski Space?

 

Are the elements of the set spacetime  vectors? Are all the vectors  unit lengths or can they be any length?  What is the operation ?

 

Is the  set infinite?

 

Have I got the wrong end of the stick somewhere?


Edited by geordief, 29 August 2016 - 12:14 AM.

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#2 MigL

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Posted 29 August 2016 - 01:53 AM

Minkowsky space-time ( flat, ignoring gravity effects ) is not a group.

 

It is certain transformational operations ( isometries IIRC ) that are part of a non-Abelian Lie group.

The Lorentz transforms  of SR are a sub-group of the Poincaire group.


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#3 studiot

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Posted 29 August 2016 - 09:05 AM

Take another look at group axioms.

 

http://mathworld.wol....com/Group.html

 

It is not necessary for a group to be commutative, only associative see axiom 2 in the link.

 

The defined operation between the group members is called the group operation.

This is part of the definition of the particular group in question, not an axiom.

So we say multiplication over the real numbers forms a group.

Such a statement specifies the elements and the operation.

This operation must be between two elements of the group only and produce another member of the group.

It is not necessary for the operation to be commutative; if it is the group is called abelian, but there are non commutative groups.

Because it is between two elements it is called a binary operation (often it is called multiplication, but this depends upon what the elements are)

Yes it has also been found most useful to require that the operation has an inverse and that there is an identity element.

Note the the inverse and identity are both required, one will not do. 

 

Note that this is different from the definition of a vectors which requires two sets of elements, the set of vectors and the set of scalars.


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#4 geordief

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Posted 29 August 2016 - 09:16 AM

Thanks. I will reset the satnav :embarass:

 

"La culture connaissance c'est comme le beurre, moins on en a plus on l'étale"
 
It is Poincaré  not Poincaire  ,  MigL ;)
 
Crossposted with Studiot. Thanks ,as well. "commutative" was a missrecall. I knew it was "associative"  but had a lapse.

Edited by geordief, 29 August 2016 - 09:29 AM.

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#5 blue89

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Posted 29 August 2016 - 10:23 AM

 I understand that Minkowski (Poincare?) Space  is a Group in Group Theory.(am I right so far?)   (???)

basic preconditions and that these are

(1)  that the set must have a operator  ✓  and

(2) must also include an identity element ✓ ,

(3) be commutative (!)

(4) each element must  have an inverse. 

Oh and (5) it must exhibit "closure"  ✓

 

5)How do these conditions apply to Minkowski Space?

Is the  set infinite?

 

 

 

Hi ;

 

there is no compulsory requirement such a being commutative for any set so as to be a group.

other requirements which are marked with " ✓ " are right for every group. so,

5) simply, to be able to say that " Minkowski space/set" was a "group"   , Minkowski Space MUST Ensure the properties which are marked with  ""

we are unable to say for any undefined group whether it was infinite or finite. for instance ;

Z is a group for "+" operator , and infinite ,however

G= {-1,1} is a group acorrding to multiplicaton operator.

as you see while Z is infinite G is finite .


Edited by blue89, 29 August 2016 - 03:25 PM.

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#6 geordief

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Posted 29 August 2016 - 11:15 AM

 

 

Hi ;

 

there is no compulsory requirement such a being commutative for any set so as to be a group.

other requirement which are marked with " ✓ " are right for every group. so,

5) if Minkowski space/set is a group , then it must ensure the properties which are marked with  ""

we are unable to say for any undefined group whether it was infinite or finite. for instance ;

Z is a group for "+" operator , and infinite ,however

G= {-1,1} is a group acorrding to multiplicaton operator.

as you see while Z is infinite G is finite .

You should look at MigL's post#2 where he says   "Minkowsky space-time ( flat, ignoring gravity effects ) is not a group" and goes on to elaborate.

 

My initial assumption  was apparently wrong.


Edited by geordief, 29 August 2016 - 12:27 PM.

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#7 blue89

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Posted 29 August 2016 - 12:41 PM

You should look at MigL's post#2 where he says   "Minkowsky space-time ( flat, ignoring gravity effects ) is not a group" and goes on to elaborate.

 

My initial assumption  was apparently wrong.

 

I request you to check my sentence more carefully ,I have not said that it as a group ,I said if it is a group,then it should ensure all that requirements.

I am too busy and cannot control other threads now. sorry.


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#8 studiot

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Posted 29 August 2016 - 01:37 PM

Usually the best way to test for conformity with the axioms is to concentrate on the operation.

That entails knowing precisely what the elements are.

 

You seem unclear as to what the elements of your proposed group are

 

"I understand that Minkowski space is a group..."

 

What elements are you proposing?

 

Vectors?

 

 

So what is the operation?

 

(vector) multiplication ?

 

Does the dot product produce another vector?

 

Is there an inverse vector for this operation for every vector?

 

Does the cross product produce another vector?

 

Is there an inverse vector for this operation for every vector?


Edited by studiot, 29 August 2016 - 01:52 PM.

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#9 geordief

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Posted 29 August 2016 - 02:28 PM

Usually the best way to test for conformity with the axioms is to concentrate on the operation.

That entails knowing precisely what the elements are.

 

You seem unclear as to what the elements of your proposed group are

 

"I understand that Minkowski space is a group..."

 

What elements are you proposing?

 

Vectors?

 

 

So what is the operation?

 

(vector) multiplication ?

 

Does the dot product produce another vector?

 

Is there an inverse vector for this operation for every vector?

 

Does the cross product produce another vector?

 

Is there an inverse vector for this operation for every vector?

 I thought I had accepted MigL's correction in post#2 where he said: "Minkowsky space-time ( flat, ignoring gravity effects ) is not a group."

 

You are not asking me to lay out and expand on  my previous misconception ,surely?

 

There was a cross-post between us when I replied to MigL in post #4 . Perhaps that threw you?

 

I understand now that the elements in the Minkowski space are (acc MigL) "certain transformational operations ( isometries IIRC ) "

 

This is now my point of departure for  learning about the subject and I am  at the beginning of that process.(I don't imagine vectors can any longer be (in my mind) a candidate for elements in the set in question)


 

I request you to check my sentence more carefully ,I have not said that it as a group ,I said if it is a group,then it should ensure all that requirements.

I am too busy and cannot control other threads now. sorry.

You are right and  you did preface  your statement with an "if".

 

Even so  ,it is a bit confusing  to set up a condition that everyone in the thread agrees is wrong. In my opinion you should  have made it clear that the statement following the "if" was incorrect to avoid potential misunderstandings. (I am open to correction,of course).


Edited by geordief, 29 August 2016 - 02:08 PM.

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#10 blue89

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Posted 29 August 2016 - 03:43 PM

actually ,I do not know what the minkowski group is .

 

but I know at functional analysis there exists an inequality of minkowski. and it is

   

 

( ∑ (ai + bi ) p )1/p  ≤  ( ∑ ai )1/p  +  ( ∑ bi p )1/p

 

 there, i= 1,2,..,n   p >1

     

but to show this we need Cauchy - Schwars inequality.

 

furthermore ,I reformed my that comment ,is everything ok about it?   :-)


Edited by blue89, 29 August 2016 - 03:53 PM.

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#11 geordief

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Posted 29 August 2016 - 03:49 PM

 

 

furthermore ,I reformed my that comment ,is everything ok about it?   :-)

Yes thanks. I am fine with that.


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#12 blue89

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Posted 29 August 2016 - 03:51 PM

Yes thanks. I am fine with that.

 

:)


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#13 wtf

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Posted 30 August 2016 - 10:55 PM

I thought I had accepted MigL's correction in post#2 where he said: "Minkowsky space-time ( flat, ignoring gravity effects ) is not a group."
...
I understand now that the elements in the Minkowski space are (acc MigL) "certain transformational operations ( isometries IIRC ) "


Let me see if I can provide some context for MigL's remark.

We have a mathematical object that we think of as some kind of geometrical space. For the moment forget Minkowski space and just think about the familiar Euclidean plane or perhaps Euclidean 3-space.

Imagine rotating the standard Euclidean plane around the origin. It's clear that if you do one rotation then another, it's the same as if you'd combined them from the start. In other words the collection of rotations is closed under composition of rotations. Composition is associative (needs proof). The identity is the rotation through 0 degrees, which leaves the plane unchanged. If you rotate the plane you can just rotate it back to where you started, so we have inverses. Therefore the set of rotations of the plane forms a group under composition of rotations.

In the case of the plane, the group is Abelian, after Niels Henrik Abel.

In general, geometrical operations are not commutative. In Euclidean 3-space we can rotate around one of the standard axes or around some arbitrary line. There are more ways for commutativity to fail. If you have good 3D visualization (which I never did) you can see that 3D rotations do not in general commute.

In the late 19th and early 20th centuries, mathematicians figured out that to study a geometric space, it was useful to study the groups of geometrical transformations that operated on a space. This is the general pattern. We have a space and we have various groups associated with geometrical transformations of that space. We study the groups to better understand the space.

Minkowski space is (as I understand it) is the mathematical model of relativity theory. The Wiki page would take some time to work through, after which you'd know a lot of differential geometry and relativity. But basically it's just 4-dimensional spacetime with the funny metric that combines time with space to model modern relativity theory. (Apologies to the physicists for anything I've mangled here).

And there are a number of interesting groups associated with various classes of geometrical transformations on it. For example in this page on Lie groups (Lie pronounced "Lee") we find MigL's example:
 

The Lorentz group is a 6-dimensional Lie group of linear isometries of the Minkowski space.


An isometry is just a rigid motion, like a rotation or reflection or translation. Any transformation that preserves distances.

So even though we may not know every detail of the Lorenz group; we can understand it as some group of rigid transformations of 4D spacetime. That's what we mean when we talk about groups in conjunction with geometry. We're considering collections of transformations that preserve some geometric property we care about. As long as the individual transformations are reversible, we'll have a group.

Edited by wtf, 30 August 2016 - 11:01 PM.

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#14 geordief

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Posted 30 August 2016 - 11:17 PM

Thanks very much for that wtf.

 

Actually I had made a little inroad into that territory (had opened those Wiki pages more than a few times)   but it is great to  have my initial  understandings  broadly confirmed .

 

It is amazing  ,actually what these earlier mathematicians were able to put together. They were the  Olympic athletes of their day  and even now it is hard to keep up with them despite all the scientific progress we have gone through since.


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#15 Mordred

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Posted 31 August 2016 - 02:46 AM

Pretty decent coverage wtf. +1. Here is a sidenote tidbit. In Minkowskii all observers will see his frame of reference as Euclidean. Define each observer as an event then the transformation is from one event (Euclidean "at rest") to the other event which the previous observer will not see in the Euclidean frame. Even though observer B sees herself in that frame. The ds^2 line element shows the wordline between those two events (separation distance).

Edited by Mordred, 31 August 2016 - 03:17 AM.

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#16 ydoaPs

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Posted 31 August 2016 - 04:00 AM

But basically it's just 4-dimensional spacetime with the funny metric that combines time with space to model modern relativity theory.


It's not "funny", it's amazing and elegant. It gives a simply stunning expression for the invariant spacetime distance (well, its square, anyway).

So, the diagonal of the Minkowski metric is 1, -1, -1, -1. This gives our distance measure as s2 = (cdt)2 - (dx)2 - (dy)2 - (dz)2. If you factor out a negative 1, you get s2 = (cdt)2 - [(dx)2 + (dy)2 + (dz)2]. The part in the square brackets is simply the Euclidean distance in the frame being analyzed. So, our metric gives us an amazing equation:

(invariant spacetime distance)2 = (distance through time in a given frame)2 - (distance through space in the same frame)2.

To me, that's pretty freakin rad.
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#17 studiot

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Posted 31 August 2016 - 08:08 AM

 

​wtf

 

Therefore the set of rotations of a plane forms a group under composition of rotations.

 

Sorry if I miscopied this but wonderful windows10 will again not let me copy and paste or use the quote function.

 

Although his profile does not indicate much wtf posts as though he is or was a professional mathematician so it was good to see my words in post 3 echoed and much extended, in a particularly understandable and helpful way.

 

+1

 

geordief,

 

A really good place to look for this is the Cambridge University Text for SMP (school mathematics project) by E A Maxwell

 

Geometry by Transformations

 

This connects old and new style geometry and stirs in a little group and isometry theory rather well.


Edited by studiot, 31 August 2016 - 08:09 AM.

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#18 wtf

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Posted 31 August 2016 - 08:36 AM

Thanks very much for that wtf.


You're very welcome.
 

Although his profile does not indicate much wtf posts as though he is or was a professional mathematician so it was good to see my words in post 3 echoed and much extended, in a particularly understandable and helpful way.


Just a couple of years of grad school a long time ago and a lot of Internet surfing. The physicists in this thread certainly know far more differential geometry than I do.

Edited by wtf, 31 August 2016 - 08:44 AM.

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#19 studiot

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Posted 31 August 2016 - 09:10 AM

It's not "funny", it's amazing and elegant. It gives a simply stunning expression for the invariant spacetime distance (well, its square, anyway).

So, the diagonal of the Minkowski metric is 1, -1, -1, -1. This gives our distance measure as s2 = (cdt)2 - (dx)2 - (dy)2 - (dz)2. If you factor out a negative 1, you get s2 = (cdt)2 - [(dx)2 + (dy)2 + (dz)2]. The part in the square brackets is simply the Euclidean distance in the frame being analyzed. So, our metric gives us an amazing equation:

(invariant spacetime distance)2 = (distance through time in a given frame)2 - (distance through space in the same frame)2.

To me, that's pretty freakin rad.

 

 

A niggle here.

 

It is not dimensionally true to say you can subtract distance (or its square) from time (or its square).

 

Time has dimension T

The constant, c, has dimensions LT-1

 

When multiplied together the result has dimension, L


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#20 geordief

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Posted 31 August 2016 - 09:56 AM

So if the Poincaré group is the set of transformations,what is the operator ** ? There are not  different assigned operations for each class of transformation ,are there?

 

If that is the case , how many types of transformation are there in Minkowski space and how many in Euclidean (that is the same as "Galilean" ,isn't it ?) space?

 

Just a few?

 

**so we could actually talk about sets of operators if it served a purpose......? 


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