[Eternal Student (OP)]

Hi.
    Has anyone looked at this sort of situation and would be willing to provide the benefit of their own experience?

General outline of the question:
      For a multi-particle system in Newtonian mechanics a COM (Centre of Momentum) frame can be constructed.  This is just a frame where the net momentum of the particles will be 0.  Out of many COM frames that exist (all rotations and translations of each other) there is one that is often called the centre of mass frame, it has an origin that is always located at the centre of mass and the x,y,z axis just run parallel to the axis of the original lab frame.   Can we do the same in special relativity?

Formal statement of the Question:
      In a multi-particle system with some particles possibly having relativistic speeds in the lab frame,  can a centre of momentum frame be constructed where the origin is just the centre of mass in the lab frame?    Would the new axis x', y' and z' axis remain parallel to the original lab frame x,y, z axis?
     Alternatively, can you use relativistic mass as determined in the lab frame and construct a COM frame where the origin is the centre of relativistic mass?   (Also how would the new x'y', z' axis now compare to the lab frame?)

      Just for clarification, the particles are moving in ordinary flat space, it's only special relativity that needs to be considered.   The particles may undergo collisions where new particles might be formed but conservation of energy and of momentum would apply as usual (note that realtivistic momentum is what is being conserved not the Newtonian momentum etc.).   I'm not sure it matters at all but there is no significant potential field contributing to the energy:   The particles don't exert gravity or electrostatic force on each other and they don't move in a some externally sourced potential either.   So the Energy of a particle is just γ(v).mc².

My thoughts:    I don't think the centre of mass found in the lab frame will be suitable to generate a COM frame, it will be of incredibly little use.   The centre of mass (as determined in the lab frame) does not seem to move with the right velocity (relative to the lab frame) to make the net momentum of the system 0, so it's not going to be any fixed point (like the origin) of any COM frame.   I don't think that using relativistic mass (determined in the lab frame) is going to change the situation either.
     I suspect you'd have to identify the velocity of a COM frame first (which can be done).  Only then do you stand some chance of picking one COM frame out of many by fixing the origin at the centre of mass (OR relativistic mass) that could be determined in a COM frame (any COM frame).    But I really don't know.

Best Wishes to everyone.

[Halc]

For a multi-particle system in Newtonian mechanics a COM (Centre of Momentum) frame can be constructed.  This is just a frame where the net momentum of the particles will be 0.  Out of many COM frames that exist (all rotations and translations of each other) there is one that is often called the centre of mass frame, it has an origin that is always located at the centre of mass and the x,y,z axis just run parallel to the axis of the original lab frame.

First, some terminology quips. As I understand the words, a frame is just a velocity reference. Once can say 'the frame in which X is stationary, or in which X has a velocity V. Thus your comment "many COM frames that exist (all rotations and translations of each other)" seems confusing. These are all the exact same frame. If on the other hand, one assigns an origin and orientation of the three spatial axes, then one has a coordinate system (CS), and each of these coordinate systems is indeed unique. Rotations are different assignments of axes, and translations are different assignments of origin.  So I'd say that many CSs exist, and while one can assign the origin to the COM of a given system, there is no necessary orientation of the axes of this 'lab', and thus no objective orientation for the axes for this one 'center of mass frame'. The choice of axis orientation is completely arbitrary, but the choice of the 4th time axis is not, since it is always aligned parallel to the worldline of the inertial COM of your system.

In actuality, the lab frame cannot be used for axis orientation. If the lab is moving relative to the COM of the system, then at least some of the axes will be oriented differently than those in any inertial frame in which the system is stationary. You can't rotate the time axis without also rotating the other axes, at least not if you intend to keep them orthogonal.

Yes, we can do this in SR. I don't see anything that needs to change. I assume we're talking about linear momentum here.

In a multi-particle system with some particles possibly having relativistic speeds in the lab frame

What the lab is doing relative to the system seems entirely irrelevant. The existence of a high speed lab somewhere has zero effect on the COM frame of say my car.

can a centre of momentum frame be constructed where the origin is just the centre of mass in the lab frame?

Not as you defined it. You said the system (and not the lab) needed to be stationary in any COMCS (center of momentum CS, as distinct from COMF since the latter does not define an origin by common usage of the term), and your lab is moving relative to the system.

Would the new axis x', y' and z' axis remain parallel to the original lab frame x,y, z axis?

One might. You're right, Newtonian mechanics (Galilean transforms in particular) would allow preservation of all 3 spatial axes. Relativity does not.

Alternatively, can you use relativistic mass as determined in the lab frame and construct a COM frame where the origin is the centre of relativistic mass?

No. Can't use the lab frame for that since the momentum of the system isn't zero in that frame. It violated your definition. Sure, the system has a frame dependent moving center of mass, but it isn't the same location in the system in all frames like it is under Newton's laws.

The particles don't exert gravity or electrostatic force on each other and they don't move in a some externally sourced potential either.

How are collisions or other interactions supposed to take place without electrostatic force?

I don't think the centre of mass found in the lab frame will be suitable to generate a COM frame, it will be of incredibly little use.

Agree. Also, somewhere you switched from center of momentum to center of mass. The center of momentum, as you defined it above, is more frame independent. Hard to say with relativity of simultaneity involved and particles are accelerating.

The centre of mass (as determined in the lab frame) does not seem to move with the right velocity (relative to the lab frame) to make the net momentum of the system 0

In a frame where the system momentum is zero, the center of mass is stationary. Relativity doesn't change that.

[Eternal Student (OP)]

Hi and thanks for your time Halc.

Frames and Co-ordinate systems:
      There are a lot of different uses of these terms.
Here's what Wikipedia says:  In physics and astronomy, a frame of reference (or reference frame) is an abstract coordinate system with an origin, orientation, and scale specified by a set of reference points―geometric points whose position is identified both mathematically (with numerical coordinate values) and physically (signaled by conventional markers)

   Ignore most of the description the important bit that I have tended to focus on is just the first part:  A frame  is a co-ordinate system.     While in the usage of the terms that I employ, a co-ordinate system isn't always a frame.   In the Wiki definition, the orientation and origin is also important and must be specified.   We can then see that this location and set of directions have both a mathematical representation and can also be identified with something of physical meaning.   For example, a unit vector along the x -direction may be written (1,0,0) but it would also mean a physical direction such as true North using a compass.   What is not clear in the Wiki definition is that the frame is a Cartesian co-ordinate system but I do tend to assume this as a requirement.

   I'm a simple person and my idea of a frame is that you make it by taking 3 rigid rods and weld them together so that the rods are mutually orthogonal in space.  I'd always assumed the word "frame" has the same etymology as the word "framework".  To put that into more coventional mathematical terms:  A frame will be a Cartesian co-ordinate system,   while something like a spherical polar co-ordinate system is a co-ordinate system but it's not a frame.  So for my frames, I need to know where the origin is located.  I can choose to use mathematical representation where the axis are a right-handed axis system.  So the only other piece of information I need to know is about orientation, i.e. what the three directions , and are supposed to be.

   Obviously people sometimes say a "Cartesian frame" which throws my idea out of the window, as if there could be some other sort of frame.   But that's people and that's life.  There is such a lack of consistency in definitions that places like NASA publish their own internal definitions so that everyone ends up navigating correctly.
    I've attached a picture of the relevant snippet of some suitable navigational information,  it came from https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/Tutorials/pdf/individual_docs/17_frames_and_coordinate_systems.pdf

   Anyway, none of this matters too much.   To establish the COM frame all that is usually important is to establish the velocity of that frame relative to the lab frame.   So many references will describe this as "the COM frame" instead of "a COM frame".   
    Here's Wikipedias take on defining the COM frame:
In physics, the center-of-momentum frame (also zero-momentum frame or COM frame) of a system is the unique (up to velocity but not origin) inertial frame in which the total momentum of the system vanishes.

Personally, I think their definintion should also have had "orientation" mentioned but the sentence was already getting too long.    I think they should have said   "....is the unique (up to velocity but not origin or orientation) inertial frame in which......"

- - - - - - - - -
    I'm generally very happy with what you've said and can only apologise if my use of the term  "frame"  differs.   I'm going to comment on some other sections in another post,  this one's already too long and is nothing more than a side-line note on the various different uses of the term "frame".

Best Wishes.

[Halc]

Frames and Co-ordinate systems:
      There are a lot of different uses of these terms.
Here's what Wikipedia says:  In physics and astronomy, a frame of reference (or reference frame) is an abstract coordinate system with an origin, orientation, and scale specified by a set of reference points―geometric points whose position is identified both mathematically (with numerical coordinate values) and physically (signaled by conventional markers)

Not the way I most often see the terms used, but I'm fine with this. OK, so you have a bunch of different frames, differing only in origin and/or spatial orientation, but not differing in temporal orientation. This COMF has probably got its origin at the unmoving center of mass of the system, but its orientation seems still a completely arbitrary choice since not even the moving lab (not part of the system) seems to define one.

Your 'SPICE' picture says a coordinate system 'specifies a mechanism for locating points within a reference frame' which I find weird. The frame, if it already defines an origin and axes, has already defined unique coordinates for every spacetime event. Not sure what's left for this 'mechanism' to do.

In the Wiki definition, the orientation and origin is also important and must be specified.   We can then see that this location and set of directions have both a mathematical representation and can also be identified with something of physical meaning.   For example, a unit vector along the x -direction may be written (1,0,0) but it would also mean a physical direction such as true North using a compass.

I prefer x=forward, y=right, z=up. North has connotations of a polar frame such that your North, East, and up are completely different directions than mine, even at a fixed moment in time.

What is not clear in the Wiki definition is that the frame is a Cartesian co-ordinate system but I do tend to assume this as a requirement.

I don't. Polar coordinates provide unique coordinates for every event, as do rotating coordinates. Accelerating frames are iffy since there are definitely events without defined coordinate values.

Yes, orthogonal Cartesian seems best for the discussion you're trying to have here. Not sure where you're going with it all.

Anyway, none of this matters too much.   To establish the COM frame all that is usually important is to establish the velocity of that frame relative to the lab frame.

Why? The speed of the lab has no effect on the definition of the COM frame.

So many references will describe this as "the COM frame" instead of "a COM frame".

Except you said there were many of them (different orientations), so 'the' seems misleading.

Here's Wikipedias take on defining the COM frame:
In physics, the center-of-momentum frame (also zero-momentum frame or COM frame) of a system is the unique (up to velocity but not origin) inertial frame in which the total momentum of the system vanishes.

Personally, I think their definintion should also have had "orientation" mentioned but the sentence was already getting too long.    I think they should have said   "....is the unique (up to velocity but not origin or orientation) inertial frame in which......"

I thought we put the origin at the system's center of mass.

[Eternal Student (OP)]

Hi again.

Sorry it's taken so long to reply.   Stuff and things happen, you know how it is.    Halc, you've spent a lot of time here and I'm gratefull but you need to know that this isn't urgent, it's just for my own interest.

Not sure where you're going with it all.

    I wasn't planning on presenting anything new for people to read or developing an idea.   I was just hoping to check something I'd thought about before I do anything else with it.   So I was simply asking a question as this forum was originally intended for instead of making a lengthy discussion.   (I know that's rare for me but that's how it was).   Obviously I'm always willing to discuss most things, so I might as well tell you what prompted the question in the first place.

What prompted the question?
      Imagine two particles, a proton and an anti-proton, accelerated to high speed v and made to have a collision.  The original question I was looking at was to try and determine the most massive particle that could be created.
Here's a diagram:

    speed v                v
*  ----------->       <----------- *
mass m                             m

Now when they collide a whole set of things could happen:  They might create some new particles;  one might remain intact while the other breaks into smaller parts ...  a whole load of different things.    However, if you're trying to determine the most massive particle that could be produced it takes a moments thought to realise that this will happen if there was only one particle at the end of the process.   In practice there often isn't a particle with the right mass, so there's no way you actually get that, you'd have to get something else  BUT  we're just evaluating a theoretical upper limit for the mass of the most massive particle that could be produced.
     Anyway, with the constraint that the initial particles had the same speed and same mass, you're going to get the most massive particle if they have a collision as shown in the diagram above -  you want a head on collision where the velocity of each particle is precisely aligned etc.   It would be nice if the final particle produced was just stationary because then all of the energy is converted to rest mass in the final particle (none is wasted as kinetic energy for the final products).   You can get this if the particles have a head on collision - there is 0 net momentum initially and so you can have 0 momentum at the end.
     So, you can do those calculations and it turns out the greatest Mass M that you could create is given by this formula:
      M  =   2. γ . m              Where γ = the usual gamma factor = 

    I then needed to consider a slightly different situation.   The two particles were the same but this time one was just a stationary target and only the other partcile could be accelerated in a LINAC and fired at it.

     speed v                  stationary
 *   ---------->                  *
mass m                        m

     You can re-work the calculation to determine the most massive particle produced.   It turns out to be as follows:
 M =

   Examining that formula shows that quite a lot of the initial energy is not available for rest mass production.   You'd have to get the proton moving at much, much higher speeds to try and produce a particle of the same mass as in the first situation.   It's an extremely inefficient way of trying to produce a massive particle.
    This is easily explainable:  The final particle produced can't be stationary.  There was non-zero momentum initially and so the final product must have momentum.   This is wasted energy that can't be made available for the rest mass.

   Now, hopefully, since my original post was about a COM frame, you might recognise that situation 1 where both particles were moving and situation 2 where only one particle is moving aren't actually all that different.   Stop viewing situation 2 from the lab frame and instead view it from the COM frame.   Then both particles are moving and come together in a head on collision with equal but opposite velocities.   Now since rest mass is an invariant in any inertial frame, we should be able to use the easier formula  M = 2 γ(u) m  instead of the more complicated formula.   I've marked the gamma factor as γ(u)   since it would use a new velocity u  which is the velocity both particles have in the COM frame,   while v is just the velocity of one particle in the lab frame.
   Another advantage of switching to the COM frame is that the velocity of the end product in the lab frame is just given to you on a plate, you don't have to calculate it by some other means.   The COM frame is such that net momentum is always 0 in that frame.   So the final particle must be at rest in the COM frame and therefore clearly has velocity u in the lab frame.

   Anyway,  that's what inspired the interest in COM frames.   General application to multi-particle systems undergoing collisions, although it's useful enough just for 2 particle systems.   It's just a new set of ideas I was thinking about and when I say "new" I just mean new to me,   it's not new to the world of particle physicists.

     So, to make the method really fast and useful I need to be able to find  u,   the velocity of the COM frame as measured in the lab frame,     given only information about the velocity of particles in the lab frame and their mass.    One practical method would have been to just identify the centre of mass in the lab frame and watch it move as time progresses.    In Newtonian mechanics, the centre of mass would have been a fixed point in a COM frame (actually it's often taken as the origin of the most natural COM frame, the centre of mass frame),  so it would have moved with velocity u that would have been precisely the velocity of the COM frame as measured in the lab frame.    However, in special relativity I don't think that holds.

    Hence the original question in the first post:

In a multi-particle system with some particles possibly having relativistic speeds in the lab frame,  can a centre of momentum frame be constructed where the origin is just the centre of mass in the lab frame?

     The comment about relativistic speeds wasn't meant to be arbitrary and suggest the lab frame could be moving fast, although I did appreciate the humour - thank you Halc.   It was just intended to advise that you must be prepared to work with special relativity and not Newtonian mechanics  because the low speed Newtonian approximations are not going to be sufficient.     I didn't want to prejudice anyone who might look at the problem but one of the first consequences is that momentum is not determined as  m.v  = rest mass x  velocity of a particle.     In  SR  the momentum should be determined as   γ(v) . m. v    because only that quantity is guaranteed to be conserved when it's summed for all the particles at any given time.    Newtonian momentum is not conserved in SR (except in some degenerate cases such as when all the particles were stationary anyway).    Hence the COM frame is not identified by setting   = 0    which leads quickly to the idea that a centre of mass could be used to fix an origin of a COM frame.   
    Instead you have to use    = 0     which at a glance could have lead to using the relativistic mass, γm, instead of the mass and finding the centre of relativistic mass to act as the origin of a COM frame.

    Also you can hopefully see why I'm interested in picking out just one suitable COM frame from many.   I do only need to determine u, the velocity of the COM frame as measured in the lab frame (which is usually all you need to characterise "the" COM frame).   However, it's much easier to calculate quantities like momentum and energy if I can use just one COM frame where the orientation and translation of that frame is such that the COM frame is just a simple Lorentz boost of the lab frame.  Specifically, since the mass and velocity of the particles is specified in the lab frame, the energy and momentum of every particle is easily determined in the lab frame, so the 4-momentum vector of every particle is known.   Applying a Lorentz transformation with parameter u,  to the 4-momentum vector of each particle gives the 4-momentum vector in the COM frame.... from that we can sum the momentum and set it to 0 and this gives a simple set of equations from which the parameter u can be determined.   Anyway, this can be done and it does seem to work in the two particle case.
      Sadly the equations get longer and messier with more particles.   That's why I'm seeking a faster method to determine u,  the velocity of the COM frame in the lab frame.   Finding the centre of relativistic mass in the lab frame and just watching it move might be a way forward.

Best Wishes.

[Halc]

Imagine two particles, a proton and an anti-proton, accelerated to high speed v and made to have a collision.  The original question I was looking at was to try and determine the most massive particle that could be created.

What do you consider to be a particle? A planet maybe? You've got arbitrarily large energy to work with, but also you seem to be ignoring conservation of baryon number. All you're going to get is light, and at best some lighter stuff. Your scenario doesn't seem designed to make something bigger than a proton, despite the energy involved.

I've marked the gamma factor as γ(u)  since it would use a new velocity u  which is the velocity both particles have in the COM frame,   while v is just the velocity of one particle in the lab frame.

OK. Newton would say u is half v. Relativity requires one to solve the velocity addition formula for u:
This gets you u, but it doesn't say where the COM is at a given moment in the frame of the lab. Sure, the point is stationary in the COM frame, but the particles are not yet present there, and their locations are different when you switch between COM and lab frames.
It gets more complicated with systems of more than two particles, or systems where the masses aren't all the same.

In Newtonian mechanics, the centre of mass would have been a fixed point in a COM frame (actually it's often taken as the origin of the most natural COM frame, the centre of mass frame),  so it would have moved with velocity u that would have been precisely the velocity of the COM frame as measured in the lab frame.    However, in special relativity I don't think that holds.

It does hold. Conservation of momentum is not lost under relativity, so the COM frame still moves at speed u, even if it's not the same u that Newton would have computed.

Hence the original question in the first post:

In a multi-particle system with some particles possibly having relativistic speeds in the lab frame,  can a centre of momentum frame be constructed where the origin is just the centre of mass in the lab frame?

Yes, there is a COM frame of the system, and it is expressible as a moving but inertial worldline in the lab frame.

Hence the COM frame is not identified by setting   = 0  which leads quickly to the idea that a centre of mass could be used to fix an origin of a COM frame.   
Instead you have to use   = 0

Yes. Works either way if there's just two equal mass particles.

Also you can hopefully see why I'm interested in picking out just one suitable COM frame from many.

I don't actually. I don't see what makes one more useful than any other. I suppose it would be convenient to align the motion of the two particles on one of the axes, but the orientation of the other two axes are completely arbitrary, especially if your lab isn't rectangular, or your accelerator isn't aligned with the walls.

[Eternal Student (OP)]

Hi again.

What do you consider to be a particle? A planet maybe? You've got arbitrarily large energy to work with, but also you seem to be ignoring conservation of baryon number. All you're going to get is light, and at best some lighter stuff. Your scenario doesn't seem designed to make something bigger than a proton, despite the energy involved.

    I agree with all of that.  I was only calculating theoretical upper limits for the mass of a particle that could be formed.  Using protons and anti-protons was just an example and they're easy to accelerate since they are charged particles.   I think we've had discussions before about proton and anti-proton collisions in the thread "Even more questions that anti-matter to me" originally posted by Aeris some months ago.
    I will refer again to an article from Fermilab:
The most common outcome of a proton-antiproton collision is that the two hadrons simply break apart — the two bags of marbles break — weakly scattering the internal quarks and gluons. This is called a “soft” interaction. The collisions that we are usually interested in are those in which a parton from the proton interacts directly with a parton from the antiproton in a “hard” scattering process that can produce new particles such as Higgs, W and Z bosons and other quarks.

[ https://news.fnal.gov/2014/02/seeing-double-in-proton-antiproton-collisions/ ]

   Note that      Z boson mass ~ 90 GeV/c²   >   proton mass ~ 0.9 GeV/c²     so it's possible that particles with more mass than protons can be created from such collisions, although rare.

   Also, just so that we're clear, I don't actually do these experiments, I just read some books and pass some time considering more extreme situations while I'm supposed to be doing housework.

    The remainder of this post is getting too long.  I'm going to end this post here, it stands on it's own adequately.

Best Wishes.

[Eternal Student (OP)]

Hi again.

   This is the next bit you (Halc) said:

OK. Newton would say u is half v. Relativity requires one to solve the velocity addition formula for u

   I agree.  This is excellent, well done Halc.  It's what I I did first but it turns out to be the hard way (that wasn't apparent until after I tried it).  Anyway, I cannot begin to tell you how glad I am that someone else thought of solving the problem this way.

Why isn't this a good method?
   Using the velocity addition formula you generate a quadratic expression in the unknown  u and the algebra is messy.   It will be just as fast to do it rather than talk about it, so here it is:

   Halc gave this expression which is produced by the velcoity addition formula in special relativity:

       where  v = velocity given for the moving proton in the lab frame;  u = velocity of the C.Of.Mom. frame in the lab frame.

(I'm saying C.of.Mom frame for the Centre of Momentum frame because I think the acronym COM is sometimes being used by Halc for the centre of mass).

Re-arranging that expression we obtain the following quadratic in u:

     

Solving that quadratic is a bit of mess but we obtain the following:
    u =    =          =     

[Equation 1]

  Where  y = γ(v)  =    = usual gamma factor with v as velocity parameter.

We need to consider which of the + and - roots we can keep.  From [Equation 1]  we can see that the +ve solution from the root of the quadratic can be discarded:   Since  v < c  would otherwise produce  u > c which is a non-physical solution, no velocity can be greater than the speed of light.
   So we obtain:

u =     

[Equation 2]

Expressing u as a multiple of c was useful to eliminate the +ve root of the quadratic but in general it's more useful to express u directly as a multiple of v,     u = k.v.   We can proceed from [Equation 2]  by multiplying by and using the difference of two squares to simplify:

u =     =   

Then note the following identity (which took a half hour to find but looks so simple now):
  - 1     ≡         -  1      ≡          ≡        and replace this in the numerator of the above expression.

Thus, u =    

[Equation 3]

To check, at low speeds γ ≈ 1   and  [Equation 3] reduces to   u  =  v/2    which is the Newtonian solution.

Anyway, having done all of that, it should now be apparent that this was a flippin' great shed load of algebra.   In particular we started with a quadratic to solve.   If we were going to generalise this to a system of N particles then we'll have multiple non-linear equations to solve.  None of us have the days for that.

Anyway, that's why it's the hard method.  A faster and easier method is to use the 4-momentum vector of each particle directly and just see what happens to it under a Lorentz boost.   This was suggested in earlier posts but it's probably worth actually doing it to see the difference,  that will be the next post.

Best Wishes.

[Eternal Student (OP)]

Hi again.

   Here we go with the easier method.  Recall we still just have two particles,  one at rest in the lab frame and one moving with constant speed v.

The 4-momentum method:
   We'll have the particle that was moving made to move along the x-axis in the lab frame (W.L.O.G. - if it wasn't, use a different lab frame so that it is).
   Then we set up for relativity and Lorentz boosts as usual, with two frames F and F' where F' moves at a constant velocity u along the x-axis as measured in Frame F.    Take F = the lab frame and we want the primed frame F' to act as our  C.of.Momentum Frame.   So we require that the sum of the (spatial component of) momentums of the particles = 0  in F'.   

[Requirement for F' = C.of Momentum frame ]

    There are only two particles.  The one which was stationary in the lab frame will have velocity -u along the x' axis in the primed frame, so it has the x'-component of 4-momentum  =  -γ(u).m.u  in  F'.       

[Momentum of particle 1 in F' ]

    The other particle had speed v in the lab frame, so it had momentum γ(v). m.v.  in F (that's the x component of the 4- momentum vector in the unprimed frame = the lab frame)   and  it turns out we'll need to know the Energy of that particle in the lab frame =  γ(v). m. c²  (because that's the time component of the 4-momentum vector).  The y- and z- components of momentum are just 0.  That will give us a complete 4-momentum vector for particle 2 in Frame F.
    We then find the x' component of the 4-momentum vector for the particle in the primed frame F' as usual by applying a Lorentz boost.  (Either actually write out the matrix multiplication for yourself or else use 9.1.3 of this reference where it's all been done for you, noting that their parameter v is our parameter u.   https://phys.libretexts.org/Bookshelves/Nuclear_and_Particle_Physics/Book%3A_Nuclear_and_Particle_Physics_(Walet)/09%3A_Relativistic_Kinematics/9.01%3A_Lorentz_Transformations_of_Energy_and_Momentum)

   We obtain,  p'_x' =     =    

[Momentum of particle 2 in F' ]

    The other spatial components of momentum (in the y and z directions or  y' and z' directions of the frames F and F' ) were always 0   -   there was never any motion in those directions in either frame.

   Anyway,  applying our [requirement for F' to be the C.of.Mom frame]  to the sum of  [Momentum for particle 1 in F']  and [Momentum of particle 2 in F']  and cancelling common factors of m and γ(u) we obtain:

      The main feature of that expression is that this little fellow is LINEAR in u NOT QUADRATIC in u.

   Re-arranging that to solve for u, we obtain:     
u =  
Which is precisely  [Equation 3]  we saw in the previous post,  Hooray!    But it took a lot less work to get here.  (I know the words took a lot of space but that was just clearly explaining what we were doing, the mathematics was a few lines).

Generalising the method to N particles
   For N particles all with different masses....  if it's the method based on the velocity addition formula this generates non-linear equations and everyone will go home angry.    If it's the 4-momentum method generating a system of linear equations then life's good.

Best Wishes.

[Eternal Student (OP)]

Hi again,

ES said:   In Newtonian mechanics, the centre of mass would have been a fixed point in a COM frame (actually it's often taken as the origin of the most natural COM frame, the centre of mass frame),  so it would have moved with velocity u that would have been precisely the velocity of the COM frame as measured in the lab frame.    However, in special relativity I don't think that holds.
Halc replied:
It does hold. Conservation of momentum is not lost under relativity, so the COM frame still moves at speed u, even if it's not the same u that Newton would have computed.

  I think we might have been talking at cross-purposes here.   I agree that the centre of mass as determined in the C.of.Mom frame will be a fixed point in the C.of.Mom. frame at all times.      I was actually talking about the Centre of mass as determined in the lab frame - but I can see I didn't make that at all clear.   

   For the two particle situation we've been discussing, where we have  one stationary and one moving particle in the lab frame,  we both agree that the Centre of Mass as determined in the lab frame  will move with velocity  v/2    in the lab frame.   Meanwhile you and I both agree that the C.of.Mom frame moves with velocity u ≠ v/2 as viewed from the lab frame.    So there is no way the centre of mass as determined in the lab frame represents one fixed place in the C.of.Mom frame,  they move at different speeds and will therefore drift apart as time progresses.

    I'm not all that interested in finding the centre of mass in the  C.of.Mom. frame.   If I already had the C.of.Mom. frame then I'd probably know what it's offset velocity, u, was from the lab frame.   Instead, I was starting from just the information we have initially.  We have the lab frame and the masses and all the velocities specified in the lab frame.  I was interested to see if you can identify the centre of mass in the lab frame and use that as a guide for how fast the C.of.Mom. frame would move when viewed from the lab frame.   You can't (see above, something with velocity v/2 is not going to be equal to the velocity u ).

   I think there's been a few instances of us talking at cross-purposes but I'm not going to raise all of those.  A lot of it is my fault for not being clear and there simply is too much here to expect anyone to read it carefully anyway.

   I'm very grateful for your time, @Halc and I think I've sorted out most of my concerns.  Some of that was just due to writing down the ideas as carefully as possible but I don't need to bore everyone else and do more of that on the forum.

Best Wishes.




   

[Halc]

It does hold. Conservation of momentum is not lost under relativity, so the COM frame still moves at speed u, even if it's not the same u that Newton would have computed.

  I think we might have been talking at cross-purposes here.   I agree that the centre of mass as determined in the C.of.Mom frame will be a fixed point in the C.of.Mom. frame at all times.      I was actually talking about the Centre of mass as determined in the lab frame - but I can see I didn't make that at all clear.

   For the two particle situation we've been discussing, where we have  one stationary and one moving particle in the lab frame,  we both agree that the Centre of Mass as determined in the lab frame  will move with velocity  v/2    in the lab frame.

I agree with no such thing. It moves at velocity u in the lab frame. v/2 is just a Newtonian thing that is a good approximation at low velocities. This COMass that moves at u will be inertial: that is, its velocity in the lab frame will not change so long as the system is closed. This is not true of any hypothetical thing computed as v/2, and thus the laws of motion are contradicted.

So there is no way the centre of mass as determined in the lab frame represents one fixed place in the C.of.Mom frame,  they move at different speeds and will therefore drift apart as time progresses.

Disagree. Maybe they're not the same point (they might differ from one frame to the next say), but not by the argument you're presenting. The two (CoMass, CoMom) do have to have the same velocity.

[Eternal Student (OP)]

Hi again.

    Working in the lab frame throughout all of this:

1.  The position vector of the centre of mass is, by definition, the vector  r   such that:



     = 

   where  r ₁  = position vector of particle 1   ;     r ₂ = position vector of particle 2.


2.   Hence,       =         This is differentiation w.r.t.   time t in the lab frame.

     However, particle 2 is stationary in the lab frame,   so    =  0,    while  particle 1 has the velocity v  in the lab frame,   so    =  v.

3.    Hence,   the velocity of the  centre of mass position =      =   ½ .v


   How are you (Halc) defining the velocity of the Centre of mass in the lab frame?   

    As it happens there are other definitions for C.of. Mass you could be using.   In special relativity the Centre of Mass is a difficult thing to define and there's no way to do it so that it has all the properties of the Newtonian situation with just Galilean transformations.    See:  https://en.wikipedia.org/wiki/Center_of_mass_(relativistic) ,   although it's a poor discussion without enough references backing it up.

Best Wishes.

[Halc]

1.  The position vector of the centre of mass is, by definition, the vector  r   such that:


How are you (Halc) defining the velocity of the Centre of mass in the lab frame?

Pretty much like that, but at least using relativistic mass and relativistic velocity addition of the components, and not proper mass and Newtonian addition as you're doing.

Alternatively, I could use the proper center of mass, which is worldline of the CoM in the frame where said CoM is stationary. Both of these have the intuitive property that the speed of the lab in the frame of the system is the same as the speed of the system in the frame of the lab. Yours does not.
My former definition identifies a frame dependent (pseudo-)worldline that is parallel but potentially spatially displaced from the latter proper definition.

Both definitions seem to have some seemingly desirable properties: The CoM speed cannot change for a closed system. That doesn't work with your definition which allows the CoM to accelerate without reaction, in defiance of Newton's 3rd law. That seems distasteful to me, hence my disagreement in the prior post.

As it happens there are other definitions for C.of. Mass you could be using.   In special relativity the Centre of Mass is a difficult thing to define and there's no way to do it so that it has all the properties of the Newtonian situation with just Galilean transformations.    See:  https://en.wikipedia.org/wiki/Center_of_mass_(relativistic) ,   although it's a poor discussion without enough references backing it up.

I looked through that and I'm sufficiently over my head with the terminologies to determine which method either of us is using. My CoM pseudo-worldline is definitely frame dependent, as is yours, so I suspect I'm using perhaps Newton–Wigner–Pryce or Møller, but I can't tell which, or if neither.
The second (proper) definition I mention above is an objective worldline, not frame dependent. So maybe that qualifies it as Fokker–Pryce. Again, not sure since I cannot follow the language in the article.

[Eternal Student (OP)]

Hi again,

Pretty much like that, but at least using relativistic mass and relativistic velocity addition of the components, and not proper mass and Newtonian addition as you're doing.

   OK.   Well, I wasn't most of the time, which you can probably tell.   I'm looking at stuff about COM frames for the first time.   In the days at the beginning of this post I was very much using the Newtownian method to determine and define the Centre of mass (but very quickly realised that had all sorts of strange behaviours in SR - see below).

    In the first post I mentioned the positibility of using relativistic mass instead of mass and determining what I described as the "centre of relativistic mass".     With  the benefit of several extra days, I would now classify that as being the Moller definition which Wikipedia calls the centre of energy.   As far as I can see the distinction between using the energy and the relativistic mass is....  political.... it just avoids using the phrase "relativistic mass" which isn't popular these days.    Using Energy you have terms like  γmc²  in the numerator and the denominator,    cancelling out the c² you do just have terms like γm  appearing.

The postion 3-vector of the centre of energy by the M∅ller definition:

       =   

   From which, it is apparent that the c²  can be cancelled  and you have effectively taken the Newtonian definition and replaced the mass with the relativistic mass.

My former definition identifies a frame dependent (pseudo-)worldline that is parallel but potentially spatially displaced from the latter proper definition.

   Agreed.  A fixed (time independant) displacement wouldn't be a problem.   (The displacement may be frame dependant but it's not time dependant in any frame).    A fixed displacement doesn't affect speed determination so that  the C.of. Mass determined by that method  and also   the C.of Mom. frame  would both have speed u  in the lab frame.

That doesn't work with your definition which allows the CoM to accelerate without reaction, in defiance of Newton's 3rd law.

   I agree.   That's what would happen if the Newtonian method for determining the Centre of Mass was used.   As soon as the particles collide and coalesce the  Centre of Mass would appear to have undergone an abrupt change in velocity without any apparent external force acting on the system.   (This is the best example of the strange behaviour that the centre of mass can show under S.R. that does not appear in Newtonian Mechanics).

Conclusion
   I realised a day ago (or thereabouts)  that you can calculate the position of the centre of energy (I would have called it the centre of relativistic mass) in the lab frame and it does what I wanted.   That position will move in the lab frame as time progresses and you can use that velocity as the velocity, u, of the C.of.Mom frame in the lab frame.   Well, it worked nicely for two particles anyway.   Before I bothered checking it in the more general situation I had already stumbled upon the various definitions of centre of mass that exist  (Moller,  Pryce and friends)  and the work that other people have already done.

Best Wishes.