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total energy of a system of particles (Definition)

Let us multiply the equation of motion of the kth particle scalarly with ddrt, and sum over all the particles. Then

∑      d2r dr     d 1 ∑      ( dr )2    ∑     dr     ∑  ∑         dr
   mk  --2k--k-=  ----    mk   --k-  =      Fk---k+         𝜖jkFjk--k-
 k     dt   dt    dt2  k       dt        k     dt     k  j         dt
(1)

Integrating between the times t0 and t:

1∑      ( drk )2   1 ∑      ( drk)2    ∫  rk(t)∑           ∫ rk(t)∑   ∑
--   mk   ----   − --    mk   ----   =           Fkdrk  +              Fjkdrk
2 k        dt  t   2  k       dt   t0    rk(t0)  k           rk(t0) k   j
(2)

The left member represents the total change in kinetic energy of the system, the right member gives the work done by the internal and external forces. But it is by no means the case that the work done by the internal forces cancels out in calculating the energy, as one might expect it to do. The kinetic energy may be divided into two parts, each of which has a physical meaning. If we introduce a second coordinate system, whose origin O is at the center of gravity of the system, and if we denote all radius vectors referred to this system by primes, we have

          ′
rk = ¯r + rk

Then, identically,

∑        (    )2     (    )2 ∑           ∑        ′     ∑      (   ′ )2
    1mk   drk-   =  1- d¯r-      mk  + d¯r    mk drk-+  1-   mk   drk-
 k  2      dt       2  dt     k       dt  k      dt    2 k        dt
(3)

The second sum on the right vanishes, however, since mkrk∕M is, by equation (3), the radius vector of the center of gravity, and this, by hypothesis, is zero in the primed coordinates. The first term on the right represents the kinetic energy of the system, considering the entire mass to be concentrated at the center of gravity. The last term gives the kinetic energy of motion of the system referred to the center of gravity, when considered at rest. Thus, we may say:

The total kinetic energy is equal to the translational kinetic energy of the entire mass, considered concentrated at the center of gravity, plus the energy of motion of the parts of the system relative to the center of gravity.

We further assume that the internal forces are such that they are derivable from a potential. The potential of the force operating between the points j and k is a function of the distance between the two points, and therefore of their coordinates:

                     ( ∘ ----------------------------------)
Ujk =  Ujk(rjk) = Ujk    (xj − xk)2 + (yj − yk)2 + (zj − zk)2
(4)

The force acting on k is obtained by taking j to be fixed, and considering k to move in the potential field given by the point function Ujk; i.e. we consider the coordinates of j to be fixed, those of k to be variable. Then

F   =  −ˆi∂Ujk- − ˆj∂Ujk- − ˆk∂Ujk- = − ∇  U
  jk      ∂xk      ∂yk      ∂zk        k jk
(5)

in like manner,

         ∂Ujk     ∂Ujk     ∂Ujk
Fkj =  −ˆi-∂x-- − ˆj-∂y-- − ˆk-∂z-- = − ∇jUjk =  − Fjk
             j       j        j
(6)

The work done in causing small displacements of j and k is

                    (                                                               )
Fjkdrk+Fkjdrj  =  −   ∂Ujkdxk +  ∂Ujk-dyk + ∂Ujk-dzk + ∂Ujkdxj +  ∂Ujk-dyj + ∂Ujkdzj   = − dUjk
                      ∂xk        ∂yk        ∂zk        ∂xj        ∂yj        ∂zj
(7)

The negative of the sum of Fjkdrk and Fkjdrj is therefore obtained by forming the total differential of Ujk, defined as a funtion of the six coordinates of the two points, in (11). If, then, we wish to introduce the internal potential into the right member of equation (9), we must write

∑  ∑                 1-   ∑  ∑
       𝜖jkFjkdrk = − 2 𝜖jk        dUjk
 k   j                     k   j
(8)

It is readily seen that the factor 12 enters: If we start with point 1, and calculate the mutual energy Ujk between this and all the other points, k runs from 2 to N; but when we take point 2, we must start counting with 3, since the mutual effect of points 1 and 2 was already taken into account in dealing with point 1, and so on. Thus, in extending the summation over all combinations j and k, we must divide by two.

If the external forces have also a potential, the energy equation (9) becomes

    ∑         1∑   ∑             (0)  ∑     (0)   1∑   ∑       (0)
T +     Uk +  2        𝜖jkUjk =  T   +     Uk  +  2        𝜖jkU jk = const.
      k          k  j                  k            k  j
(9)

where T denotes the kinetic energy. The sum of the kinetic energy and of the external and internal potential energy of a system is constant, if the external as well as the internal forces are conservative.

0.1 References

[1] Joos, Georg. ”Theoretical physics” 3rd Edition, Hafner Publishing Company; New York, 1954.

This entry is a derivative of the Public domain work [1].


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Cross-references: domain, theoretical physics, field, function, force, translational kinetic energy, mass, radius vectors, center of gravity, energy, internal forces, external forces, work, system, kinetic energy, motion

This is version 2 of total energy of a system of particles, born on 2009-03-29, modified 2009-03-29.
Object id is 613, canonical name is TotalEnergyOfASystemOfParticles.
Accessed 1902 times total.

Classification:
Physics Classification45.40.Cc (Rigid body and gyroscope motion)
 45.50.Dd (General motion)
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