Can a center of momentum frame always be found for just the mechanical momentum of an electromagnetic system?

[black head]

My question is:

Can a center of momentum frame always be found for just the mechanical
momentum of an electromagnetic system?

My answer would be not always, even in the case of a closed
electromagnetic system. An obvious exception is at the instant an
electrostatic system is released by the external forces holding it
together. But even here, this is only true at the instant the forces
are released because the increasing velocity of the charges means an
increase in electromagnetic momentum, and it's the electromagnetic +
mechanical momentum which is conserved.

On the other hand, a center of momentum frame can always be found for
the mechanical + electromagnetic momentum of a closed electromagnetic
system.

Thanks in advance for any corrections and comments to the above,

Larry.

[xxein]

xxein: Otoh, what is closed? The universe acts (at the present time)
as an open system. How do you relate to that?

[black head]

Closed, as in the total energy inside a volume enclosing the system
remains constant.

Regards, Larry.

[xxein]

xxein: You don't even know if a volume is changing or what its volume
restrictions are.

A simple example of this is the Boston marathon bombing. It is
believed that the pressure containment vessel exceeded the pressure
that could be contained within that volume. So boom and lives hurt or
lost.

Go play with the idea that all energy can be contained within a volume
and I'll read about it in your obituary.

[Alfred Einstead]

On Apr 17, 1:12 pm, black head <larryhar...@softhome.net> wrote:
> My question is:
>
> Can a center of momentum frame always be found for just the mechanical
> momentum of an electromagnetic system?

You means a rest frame? Take the 4x4 stress/energy/momentum tensor and
do a Lorentz transform that renders the {10,20,30,01,02,03} components
0. It should work if those components are small enough relative to the
{00} component.

If the field dynamics is described by a Lagrangian function L(B, E) of
the B and E fields, then the D and H fields can be defined as the
partial derivatives D = dL/dE, H = -dL/dB. The usual expression used
for L is L = 1/2 epsilon_0 (E^2 - B^2 c^2).

For a Lorentz-invariant and gauge-invariant Lagrangian, the most
general form is a function
L = L(I, J)
where
I = (E^2 - B^2 c^2), J = B.E.
Then, defining derivatives epsilon = dL/dI, theta = dL/dJ (the
coefficients, themselves, being functions of I and J), you get the
relations D = epsilon E + theta B, H = B/mu - theta E, where mu = 1/
(epsilon c^2).

Regardless of what L is, the stress/energy/momentum tensor have
components:
T^0_0 = D.E - L
(T^1_0, T^2_0, T^3_0) = (E x H)^i
(T^0_1, T^0_2, T^0_3) = (B x D)_j
(T^i_j: i,j = 1, 2, 3) = (D^i E_j + B^i H_j - delta^i_j (B.H + L))
This has a decomposition into the form
L = epsilon I + theta J - Delta/4, for a suitably defined function
"Delta"
T^0_0 = n U + Delta/4
T^0_j = -n P K
T^i_j = n m c^2 (XX - YY - U/(mc^2) ZZ)^i_j + delta^i_j Delta/4
where (X, Y, Z) for an orthonormal triad, with dyad notation used
(e.g. the ^i_j component of XX is X^i X_j); and with functions n, U
and m given by solving
n epsilon_0 = epsilon
B x D = -n/c root(U^2 - (mc^2)^2) Z
E x H = nc root(U^2 - (mc^2)^2) Z
D.E + B.H = 2n U.
E + i c B = (X + i Y)/root(epsilon_0) * e^{i phi}
I + i c J = mc^2/epsilon_0 e^{2i phi}
for a suitably defined phase phi.

For the frame (X,Y), one can define it by the conditions:
E = (x X cos phi - y Y sin phi)/root(epsilon_0)
B = root(mu_0) (y Y cos phi - x X sin phi)
provided x and y are chosen appropriately (x^2 = U + mc^2 and y^2 = U
- mc^2 should work).

I think that covers the basis. Check the calculations.

If you take components relative to the frame (X,Y,Z) you should be
able to readily define the Lorentz transformation that yields 0 for B
x D and E x H.

I think the relevant velocity u is the "Lorentz 1/2" of the velocity v
that emerges from
D x B = E x H/c^2 = m v/root(1 - (v/c)^2).
So it would be a solution to
v = 2 u/root(1 - (u/c)^2).

[Dono.]

Your question seems to be ill-formed, as such it doesn't have an
answer.
You can always find the center of momentum frame:

Given a system characterized by the four vectors (E_i, \vec{p_i}) ,
i=1,n the center of momentum frame is the frame that moves at the
velocity

\vec{V}=c^2* \frac{\Sigma \vec{p_i}}{\Sigma E_i}

[black head]

> \vec{V}=c^2* \frac{\Sigma \vec{p_i}}{\Sigma E_i}- Hide quoted text -
>

When you transform to the COM frame using that V, the energy and
momentum are no longer being summed at the same instant of time,
because of the relativity of simultaneity.

This doesn't matter in the case of a number of particles interacting
at the same space-time point. Neither does it matter in a closed
system where you have a number of particles interacting locally at
different space-time points, because of the conservation of the four-
momentum. But in a system where they interact non-locally via a field,
I don't see how the formula you quoted can generally work.

Regards, Larry.

[black head]

On May 16, 1:31 am, Alfred Einstead <federation2...@netzero.com>
wrote:

Thanks, the above is a little above what I understand about
electrodynamics, but useful as something for me to study over the next
few weeks.

I can see there's always a rest frame when the momentum of the field
is included, but the question was asking if there's always a rest
frame just for the mechanical momentum of the electromagentic system.

Regards, larry.

[Dono.]

The (E,p) four-vectors have no time component, so, your bringing in
relativity of simultaneity makes absolutely no sense.

[black head]

We use the LT on the (ct, x) of the particles to get their new
position four-vectors. But since the particles are separated in space,
the transformed ct' for one particle isn't equal to the other
particle's ct', meaning their transformed (E', p') are no longer
simultaneous either. We're therefore no longer summing (E',p')
simultaneously for both particles and therefore \Sigma (E',p') done
simultaneously may not equal zero .

Regards, Larry.

[Dono.]

I don't think you understand what is going on, let me try one more
time:

Given a system characterized by the four vectors (E_i, \vec{p_i})

i=1,n, where the components E_i, \vec{p_i} are taken with respect ONE
GIVEN FRAME,

the center of momentum frame is the frame that moves at the
velocity

\vec{V}=c^2* \frac{\Sigma \vec{p_i}}{\Sigma E_i} with respect to the
ONE GIVEN FRAME.

Does this help?

[black head]

You were very clear to start with, but I don't think you understand
the point I'm trying to make. I'll try to make things clearer for you:

The formula you've given comes from summing the (E, p) four-vectors of
all the particles at THE SAME TIME in one given frame, and then
solving for V in the LT for \Sigma\vec p'_i = 0. Do you agree?

But after boosting to the COM frame using this V, \Sigma\vec p'_i that
was used to define this frame, isn't the same as \Sigma\vec p'_i done
simultaneously here, if the particles are separated in space. Do you
agree?

Therefore, \Sigma\vec p'_i does not necessarily equal zero after
boosting here. Do you agree?

Regards, Larry.

[Dono.]

Not exactly, I now understand your confusion.
1. In frame F, \Sigma \vec{p_i} is non-zero
2. The \vec{V }value does indeed correspond to a boosted frame F'
where \Sigma \vec{p'_i}=0
3. Energy-momentum (E, \vec{p}) is NOT frame invariant, so the above
should come is no surprise.

4. " \Sigma \vec{p'_i} done simultaneously here" (I assume in F') has
no meaning.

> Therefore, \Sigma\vec p'_i does not necessarily equal zero after
> boosting here. Do you agree?
>

No, see above,

5. \Sigma \vec{p'_i}=Lorentz_Transform (\Sigma \vec{p_i}) = 0

The above is true by virtue of the way \vec{V} was calculated.

[black head]

Yeah

> 2. The \vec{V }value does indeed correspond to a boosted frame F'
> where \Sigma \vec{p'_i}=0

Yeah

> 3. Energy-momentum (E, \vec{p}) is NOT frame invariant, so the above
> should come is no surprise.

Yeah

> 4. " \Sigma \vec{p'_i} done simultaneously here" (I assume in F') has
> no meaning.
> > Therefore, \Sigma\vec p'_i does not necessarily equal zero after
> > boosting here. Do you agree?
>
> No, see above,
>
> 5. \Sigma \vec{p'_i}=Lorentz_Transform (\Sigma \vec{p_i}) = 0
>
> The above is true by virtue of the way \vec{V} was calculated.

OK, just one question:

In the expression you gave for the COM frame, do you agree that the
p_i of the particles in frame F are simultaneous with one another,
whereas the corresponding p'_i of the particles in frame F' aren't?

Regards, Larry.

[Dono.]

Once again, the above question doesn't make sense. The only thing we
know is the relationship between (E'p') and (E,p) as given by the
Lorentz transforms:

p'=\gamma(V)(p+EV/c^2)
E'=\gamma(V)(E+Vp)

time doesn't enter into the transforms

[black head]

Yes, time doesn't enter into these transforms.

When using the COM formula, you're also right when saying that time
doesn't play a part. Yes, we can choose any set of p_i however we
like, and find a frame where p'_i sum to zero. The p_i don't have to
be simultaneous for the formula to work.

The problem for me is when you claim that the formula enables us to
always find the center of momentum frame. This frame has a precise
definition:

The COM frame is that frame where the sum of the 3-momentum, each
evaluated
AT THE SAME TIME, sum to zero.

So while I may stick in a set of simultaneous p_i, and find that V
where the total p_i' sums to zero, I can't assume it's a COM frame
because the p-i' aren't simultaneous with one another in F'. I've
ended up with a frame where the p_i' sum to zero -- yes, but each p_i'
evaluated at a different time to one another.

Regards, Larry.

[Dono.]

On May 18, 2:23 pm, black head <larryhar...@softhome.net> wrote:
>
> The problem for me is when you claim that the formula enables us to
> always find the center of momentum frame. This frame has a precise
> definition:
>
>    The COM frame is that frame where the sum of the 3-momentum, each
> evaluated
>    AT THE SAME TIME, sum to zero.
>

What gives you this idea?

http://en.wikipedia.org/wiki/Center-of-momentum_frame

[black head]

At the top is says:

"The center of momentum of a system is not a location, but usually
refers to the coordinate reference frame in which the momenta of a
system's components add to zero"

In the properties section:

"In the centre of momentum frame, the total linear momentum of the
system is zero - by definition"

It doesn't explicitly say the momentum are evaluated at the same time
before being summed, I guess because they're relying on the common
sense of the reader to assume this.

Regards, Larry.

[Dono.]

On May 18, 5:16 pm, black head <larryhar...@softhome.net> wrote:
>
> It doesn't explicitly say the momentum are evaluated at the same time
> before being summed, I guess because they're relying on the common
> sense of the reader to assume this.
>
> Regards, Larry.

Like I said, I do not know what gives you this bright idea.

[black head]

This is how it's always done in standard text books on basic classical
mechanics because it's a conserved quantity for a closed system. I'm
surprised you've never come across this, unless you can provide an
example where summing momentum at different times is physically
useful.

Regards, Larry.

[Dono.]

Why don't you provide a citation from one of the textbooks? Look, I
tried to help you, you seem to have some fixed ideas, I don't think I
can waste any more time on you.

[black head]

OK, thanks for your input anyway.

Regards, Larry.

[Dono.]

I finally understood the nature of your confusion:

(E_i, \vec{p_i}) are NOT time dependent , so, it does NOT matter
whether they are evaluated simultaneously or NOT. Can you figure out
why (E_i, \vec{p_i}) are time INdependent?

[shuba]

Larry wrote:

> Can a center of momentum frame always be found for just the
> mechanical momentum of an electromagnetic system?

I don't see the usefulness of defining such as thing as the center
of mechanical momentum, given that mechanical momentum is not
generally conserved.

> On the other hand, a center of momentum frame can always be found for
> the mechanical + electromagnetic momentum of a closed electromagnetic
> system.

Yes, as the total 3-momentum (canonical momentum) is conserved.


---Tim Shuba---

[black head]

On May 20, 6:47 pm, shuba <t...@sh.uba> wrote:
> Larry wrote:
> > Can a center of momentum frame always be found for just the
> > mechanical momentum of an electromagnetic system?
>
> I don't see the usefulness of defining such as thing as the center
> of mechanical momentum, given that mechanical momentum is not
> generally conserved.

That's what I thought, but some people think that it's always possible
so I just wanted to make sure I wasn't being careless.

> > On the other hand, a center of momentum frame can always be found for
> > the mechanical + electromagnetic momentum of a closed electromagnetic
> > system.
>
> Yes, as the total 3-momentum (canonical momentum) is conserved.

Good, I'm not losing my marbles after all ;)

Regards, Larry.