Symmetry under boosts gives what conserved quantity?

John Baez

March 4, 2006

In article <8fc591$eu4$>, wrote:

>The invariance of the Lagrangian under boosts (either Lorentzian or
>Galilean) gives conservation of what?
Heh. People keep asking this question, so by now I just pull out my handy little file of one of the previous discussions we've had about this... see below.
>At least I searched in many books and they didn't even mention that.
Right. That's why everybody keeps asking this question....
>Is Noether's theorem not valid in such cases or what?
It's valid, all right.
>and if Noether's theorem still
>works in these cases, why the heck do books not talk about the conserved
>quantities corresponding to such symmetries?
Either 1) the textbook writers are too stupid to have thought about this issue, or 2) they have decided it's better to let everybody figure this out for themselves, or 3) they feel the answer is not sufficiently important to waste a precious paragraph on it. I don't know. When I become king of the universe, I will make all books on mechanics mention this issue.

But for now, read these old articles and emails:

Newsgroups: sci.physics,sci.math
Subject: Re: Noether's Theorem and the Poincare Group
From: John Baez
In article <4o1019$r59@darkstar.UCSC.EDU> (Jacob Alexnder Mannix) writes:
>what do those
>funny lorentz "rotations" in the x-t, y-t, z-t planes (the boosts)
>give rise to as conserved quantities? 
This question comes up repeatedly on sci.physics because for some stupid reason most textbooks are lazy to discuss these conserved quantities.

Hopefully Matt McIrvin and others will repost the enlightening replies they gave to this question the last time it rolled around. And maybe I can get up the energy to work this into a FAQ for the next time it rolls around.

Let me just say this:

It is amusing to note that analogous observables also show up in nonrelativistic mechanics: not from the Lorentz boosts, of course, but from the "Galilei boosts". The Galilei group is the limit of the Poincare group as c → ∞, and it's generated by spacetime translations, spatial rotations, and "boosts" of the form

t |→ t

x |→ x + vt

y |→ y

z |→ z

and similarly for y and z.

If we consider a single nonrelativistic free particle - in one-dimensional space, to keep life simple - and describe its state by its position q and momentum p at t = 0, we see that the Galilei boost

t |→ t

x |→ x + vt

has the following effect on its state:

p |→ p + mv

q |→ q

In other words, a Galilei boost is just a translation in momentum space.

In nonrelativistic quantum mechanics this should be familiar, though somewhat disguised. Here it is a commonplace that the momentum observable p generates translations in position space; in the Schroedinger representation it's just -i hbar d/dx. But by the same token the position observable q generates translations in momentum space. As we've seen, a translation in momentum space is just a Galilei boost. So basically, the generator of Galilei boosts is the observable q.

Ugh, but there is a constant "m" up there in the formula for a Galilei boost. So I guess the observable that generates Galilei boosts is really mq. If we generalize this to a many-particle system we'd get the sum over all particles of their mass times their position, or in other words: the total mass times the center of mass.

Now this seems weird at first because it's not a conserved quantity! Wasn't Noether's theorem supposed to give us a conserved quantity? Well, it does, but since our symmetry (the boost) was explicitly time-dependent - it involved "t" - our conserved quantity will also be explicitly time-dependent. What I was just doing now was working out its value at t = 0.

If we work out its value for arbitrary t, I guess we get: the total mass times the center of mass minus t times the total momentum.

Using the fact that total mass is conserved we can turn this conserved quantity into something perhaps a bit simpler: the center of mass minus t times the velocity of the center of mass.

Something similar applies to the relativistic case.

From: Toby Bartels
Newsgroups: sci.physics
Subject: Re: Noether's Theorem and the Poincare Group
Date: 24 May 1996 17:49:49 GMT
Organization: California Institute of Technology, Pasadena, CA USA
Jacob Alexnder Mannix wrote in part:
>I have the infinitesimal rotations in the x-y, y-z, and z-x planes,
>giving me conservation of each component of angular momentum, and by
>now, you probably see what I am leaving for last, but what do those
>funny lorentz "rotations" in the x-t, y-t, z-t planes (the boosts)
>give rise to as conserved quantities?  Does it coorespond to some
>sort of relationship between the momentum and energy, like 
>E^2 = (pc)^2 - (mc^2)^2 give or take a minus sign or power of 2, or
>something completely strange?
Symmetry under rotation in the xy plane corresponds to conservation of

Lz = rx py - ry px

(where r and p are linear position and momentum).

In the same way, symmetry under hyperbolic rotation in the tx plane corresponds to conservation of

rt px + rx pt = t px - x E.

In other words, the system's speed in the x direction is px / E.


From: Charles Torre
Newsgroups: sci.physics
Subject: Re: Noether's Theorem and the Poincare Group
Date: 23 May 96 12:01:28 MDT
Organization: Utah State University

In article <4o1019$r59@darkstar.UCSC.EDU>, (Jacob Alexnder Mannix) writes:
> I have a question concerning a physical interpretation of some of
> the conserved quantities associated with the Poincare group (lorentz
> boosts, rotations in 3-space, and translations).  


> but what do those
> funny lorentz "rotations" in the x-t, y-t, z-t planes (the boosts)
> give rise to as conserved quantities?  
This is a good question which comes up with some regularity and is rarely given much treatment in the texts I am familiar with. I will make a brief attempt to explain it.

From the point of view of relativity there is no invariant way to separate the boosts from the rotations because, in effect, different inertial observers will not agree on the definitions of the various x-t, x-y, etc. planes. So, ultimately, one should simply view the 6 conserved quantities associated with the boosts and rotations as the relativistic form of angular momentum. This is completely analogous to the fact that there is no invariant way to separate energy from momentum - different inertial observers will see different mixtures-and so one speaks only of the "4-momentum". Still, this does not quite answer the question. In a given inertial reference frame I can identify 3 of the six conservation laws, the ones coming from spatial rotational symmetry in the given frame, as corresponding to familiar notions of angular momentum. What of the other 3 coming from boosts? Actually this question can be asked already in Newtonian physics which employs "Galilean relativity". One can show that a Galilean invariant Lagrangian will be symmetric (up to a total time derivative) with respect to boosts (the non-relativistic boosts, not Einstein's). The corresponding conserved quantity is an example of an "explicitly time dependent constant of the motion" and has the physical interpretation of the initial position of the center of mass of the system. So, one can view the extra 3 conservation laws associated with boosts in a Poincare invariant theory as being a relativistic generalization of this. The Einstein-relativistic generalization of Galilean-relativistic angular momentum and initial center of mass is a bit complicated since the Galilean version of these quantities involves, in effect, a preferred notion of simultaneity, which is of course absent in Einstein relativity. You can usually find discussion of bits and pieces of these issues in a mechanics book which treats special relativity, such as Goldstein and/or which bothers to discuss Galilean relativity such as Landau and Lifshitz.

Charles Torre
Department of Physics
Utah State University
Logan, UT 84322-4415 USA

From: John Baez
Subject: Re: Lorentzian or Galilean boosts & Noether theorem
Date: 23 Jun 2000 00:00:00 GMT
Newsgroups: sci.physics.research

squark wrote:

>> (John Baez) wrote:

>>><> wrote:

>>>I'm doing research on some of the properties of [the generator of
>>>Lorentz or Galilei boosts] for my Master's degree, and for the sake 
>>>of convenience, have given it a working name:  I call it the boost 

>> How about "the center of mass at time zero?"  Not very elegant, but
>> that's basically what it is.    
Well, at least after we divide it by the total energy of the system!!!
>This is literally true only in the non-relativistic case, right? 
Yes - but only because in the relativistic case we need to make up our minds what we mean by "center of mass".
>Or do you suggest the quantity as a definition of relativistic 

For a nonrelativistic system of point particles interacting by central forces, the generator of Galilei boosts is their center of mass at time zero, times the sum of their masses. So let's divide this generator by the sum of the masses and call this quantity "the center of mass at time zero".

For a single free relativistic point particle, the components of the generator of Lorentz boosts are:

x E - t px

y E - t py

z E - t pz

If we divide these by the energy of the particle we get the components of the particle's position at time zero.

So it seems sensible to generalize and call the generator of Lorentz boosts, divided by the total energy of the system, the "center of mass at time zero".

We can define a similar quantity at any other time (in our chosen frame), and if we use this to think of the center of mass as a function of time, it moves along at constant velocity. That sounds like what a center of mass should do.

But perhaps "center of energy" would be a better word for it, in the relativistic case. (Apparently particle physicists use the term "center of momentum".)

From: Zach Taylor
To: John Baez
Subject: galilean conservation laws...faq/ reference
Date: Sat, 25 Feb 2006 17:40:07 -0600

Dear Professor Baez,

In the past while searching for an explanation of conservation laws in Newtonian mechanics, I encountered commentary related to this topic on your site. Anyway, if anyone would need a reference to a book that does contain an additional explanation, here is one:

I imagine that all is well taken care of without this comment, but this does provide an explanation of:

Center of Mass = [ t ( Momentum)/ Mass ] + c.

W.Z.P. Taylor

© 2006 John Baez