Centre of Mass-Energy: Griffiths Electrodynamics 4th Ed.

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In summary: The equation does not hold if electromagnetic energy and momentum are included, or if there are external forces.
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This is actually from Griffiths Electrodynamics 4th edition, page 546.

He defines the centre of mass-energy of a system of particles as:
$$\vec{R} = \frac{1}{E} \sum E_i \vec{r_i}$$
And gives that the total momentum of the system is:
$$\vec{P} = \frac{E}{c^2} \frac{d \vec{R}}{dt}$$
In a footnote he says that the proof of this is non-trivial and refers to a couple of papers.
I took a look at this for two particles and got:
$$\frac{d \vec{R}}{dt} = \frac{1}{(E_1 + E_2)^2}[E_1^2 \vec{u_1} + E_2^2 \vec{u_2} + E_1E_2(\vec{u_1} + \vec {u_2}) + (E_1\frac{dE_2}{dt} - E_2\frac{dE_1}{dt})(\vec{r_2} - \vec{r_1})]$$
$$= \frac{1}{(E_1 + E_2)^2}[(E_1 + E_2)(E_1 \vec{u_1} + E_2 \vec{u_2}) + (E_1\frac{dE_2}{dt} - E_2\frac{dE_1}{dt})(\vec{r_2} - \vec{r_1})]$$
$$= \frac{c^2}{E} \vec{P} + \frac{1}{E^2}[(E_1\frac{dE_2}{dt} - E_2\frac{dE_1}{dt})(\vec{r_2} - \vec{r_1})]$$
The total momentum came out but I can't see how the cross terms in the position vectors disappear. I couldn't find any specific references to this online. Any ideas about what's wrong?
 
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  • #2
I might be able to answer my own question. If the particles are accelerating owing to an EM field that they themselves create, then energy and momentum are stored in the fields. This relation is only true when you include the energy-momentum from the fields. Unlike the classical case, using centre of mass, where it's generally true regardless of how the particles are accelerated.

It's clear now. Especially after reading the next page.
 
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  • #3
However the ancient references are both wrong.
Example 12.13 and Eq. (12.72) are also wrong, as is the paper in Ref. 21.
The only correct statement is "this is not a very realistic model."
 
  • #4
Of course, in the case of pure neutral particle dynamics, with energy differentials zero except for collisions, the result is trivially true ...
 
  • #5
The equation does not hold if electromagnetic energy and momentum are included, or if there are external forces.
 
  • #6
Meir Achuz said:
The equation does not hold if electromagnetic energy and momentum are included, or if there are external forces.
Consistent with what I said ...
 

1. What is the concept of "centre of mass-energy" in Griffiths Electrodynamics?

The centre of mass-energy is a concept in Griffiths Electrodynamics that refers to the point in space where the total energy and momentum of a system is concentrated. It is the point at which an external force can be applied to the system without causing it to rotate or accelerate.

2. How is the centre of mass-energy calculated in Griffiths Electrodynamics?

The centre of mass-energy is calculated using the formula Xcm = (1/M)∫xρdV, where Xcm is the location of the centre of mass, M is the total mass of the system, x is the position vector, ρ is the mass density, and dV is the volume element. This formula can be applied to both classical and relativistic systems.

3. What is the significance of the centre of mass-energy in Griffiths Electrodynamics?

The centre of mass-energy is significant because it allows us to simplify the analysis of complex systems by treating them as a single point. It also helps us understand the behavior of a system under external forces and how energy and momentum are conserved in the system.

4. Can the centre of mass-energy be located outside of the physical boundaries of a system?

Yes, the centre of mass-energy can be located outside of the physical boundaries of a system. This can occur if the system has an asymmetrical distribution of mass or if there are external forces acting on the system.

5. How does the centre of mass-energy relate to the concept of equilibrium in Griffiths Electrodynamics?

The centre of mass-energy is closely related to the concept of equilibrium in Griffiths Electrodynamics. In a system at equilibrium, the centre of mass-energy remains stationary and any external forces acting on the system must pass through this point. This allows us to analyze the system using the principles of statics, making the calculations simpler and more intuitive.

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