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Relativistic Linear Momentum |
In Newtonian (non-relativistic) mechanics the linear momentum of a body is defined as the product of its mass m and its velocity v. It was found that this quantity is appaently conserved over time. Albert Einstein in his Special Theory of Relativity found that the apparent mass of a body in motion relative to an observer is given by
where m_{0} is the rest mass of the body and β is the velocity of the body relative to the speed of light c; i.e., β=v/c.
It is almost universally assumed that under Special Relativity linear momentum is still given by the formula p=mv and the only relativistic correction is that the velocity-dependence of mass must be taken into account. It will be shown below that according to Lagrangian analysis this is not the case. Instead there is an additional correction which must be taken into account.
Lagrangian dynamics provides a way to derive the formula for relativistic linear momentum rather than just assuming it. If K is the kinetic energy of a system and V is the potential energy then the Lagrangian of the system is defined as
If the Lagrangian of a system is a function of a set of variables {q_{i}; i=1,2,…,n} and their time derivatives {dq_{i}/dt; i=1,2,…,n} and the system is not subject to external forces then the dynamics of the system is given by the set of equations
where v_{i}=dq_{i}/dt.
The expression (∂L/∂v_{i}) is the momentum with respect to the variable q_{i}.
In Newtonian mechanics the kinetic energy of a body is ½mv² and v=dx/dt for some x. If the potential energy function does not depend upon x or v then
Now consider the relativistic case. The kinetic energy is then (mc²−m_{0}c²) where m=m_{0}/(1−β²)^{½} and β=v/c.
The expression ½mv² is not the full kinetic energy; it is just the first order approximation of the kinetic energy.
The Lagrangian for a particle moving in a potential field V is
In other words, relativistic linear momentum requires not only the relativistic adjustment of mass but also division by a factor of (1−β²).
Thus if the kinetic and potential energies are independent of the spatial variable that defines velocity then it is mv/(1−β²) which is constant over time.
That is to say,
The expression m_{0}v/(1−β²)^{3/2} asymptotically approaches m_{0}v as β→0 just as m_{0}v/(1−β²)^{½} does.
Herbert Goldstein's works on mechanics have long been considered the definitive source on the topic of classical mechanics. In the two editions of his Classical Mechanics, he refers to the longitudinal and the transverse masses for a body; i.e.,
where longitudinal means in the direction of the motion and tranverse is perpendicular to the direction of motion. Transverse mass is just the relativistic mass and longitudinal mass is just the relativistic mass divided by (1−β²). This is the same results that came out of the analysis above. Goldstein says however that the use of these concepts of mass has been decreasing because they obscure the physics. This may be true that there is no reason for defining mass in this way, but that is no reason not to take into account the (1−β²) term in the computation of momentum. The other interpretation of the result of Lagrangian analysis is that the correct formula for relativistic momentum is
Goldstein adopts a different approach to the issue. He argues that the true equation of motion is just d[m_{0}v/(1−β²)^{½}]/dt = F
and to get that equation of motion he changes the nature of the Lagrangian so that it is not the difference of kinetic and potential energies. This appears to be an unjustified distortion of the universal principle of Lagrangian Dynamics. In the 1950's edition of his book Golstein takes the definition, of the Lagrangian to be whatever formula gives relativistic momentum to be mv under Lagrangian analysis. Other authors of books on classical mechanics, such as Jerry Marion and Stephen Thornton, adopt the same approach as Goldstein. For a particle in a potential field V that is
Intellectually this is invalid. There is no support for the mv formula for relativistic momentum provided by showing that there is a pseudo-Lagrangian from which it can be derived by Lagrangian analysis. In the 1981 edition Goldstein repeats this invalid exercise (page 321, equation 7-136) but gives the formulas for longitudinal and transverse only in a suggested exercise.
Here is a graph of the supposed kinetic energy from Goldstein's pseudo-Lagrangian as a function of relative velocity β. expressed in terms of its ratio to m_{0}c².
Note that at β=0 this supposed kinetic energy is −m_{0}c² and at at β=1 this supposed kinetic energy is zero. Negative kinetic energy is of course complete nonsense. Apparentl conventional physicists are so set on the formula mv for relativistic momentum that they are willing to accept a derivation of it from nonsense. The negativity and the nonzero value for the supposed kinetic energy at β=0 can be remedied by added the term m_{0}c² to it, as does S.W. McCuskey in his An Introduction to Advaand anced Dynamics. But that does not take care of the finiteness of the value at β=1 and that finiteness is also nonsense.
The standard formula for kinetic energy is
Here is a graph of K/(m_{0}c²).
The issue can only be resolved empirically; i.e., which momentum, mv or mv/(1−β²) is conserved? In the limit as v→0 there is no difference between the two, so the settlement of the issue requires results for the case of β near unity.
Surprisingly the concepts and terms longitudinal mass and transverve mass preceded Special Relativity. In the late 19th century various theorists noticed that charged bodies resisted acceleration more than is acounted for by their rest masses; i.e.,
where F is force and a is acceleration
J.J. Thompson pointed this out in 1881 and Oliver Heavyside worked out the mathematics in 1897. These formulas for longitudinal and tranverse mass are the same as those given by Goldstein. Note that they were based upon empirical evidence.
Mass is a coefficient connecting two physical quantities in dynamics: force and acceleration, kinetic energy and a function of velocity and momentum and velocity. In Newtonian mechanics these three coefficients are the same. In general this is not true. Let kinetic energy be given by
where γ(v) is an unspecified function of velocity.
Then by Lagrangian analysis momentum p is given by
The mass given by the momentum/velocity relationship is then
and this generally cannot be the same as the mass involved in the kinetic-energy/velocity relationship. It is significant that the mass concept derived from on relationship does not have to be the same as that derived from the other relationships. Thus the acceptance of longitudinal mass as the mass in the momomentum/velocity relationship does mean that it should apply to mass in the other relationships.
Suppose a nucleus of Ba^{137} emits an electron (beta decay). Suppose the velocity of the electron is one half the speed of light, β=0.5. The rest mass of the electron is 9.10938215×10^{-31} kg. Its mass at β=0.5 is then 10.5186085×10^{-31} kg. Its linear momentum based upon the formula mv is then 3.1556×10^{-22} kg m/sec. On the basis of the formula mv/(1−β²) it is 4.2074×10^{-22} kg m/sec, 1/3 larger. The momentum of the Ba^{137} nucleus would be this same value but in the opposite direction.
The mass of the barium 137 nucleus would be about 252,000 times that of an electron, or about 2.29×10^{-25} kg. Its recoil momentum would be 4.2074×10^{-22} kg m/sec and hence its recoil velocity would then be about 1.8343×10^{3} which is only 6.114×10^{−6} times the velocity of light, so the relativistic correction is insignificant. Thus the barium nucleus has a momentum of 4.2074×10^{-22} kg m/sec, whereas the electron's momentum computed from the formula mv is only 3.1556×10^{-22} kg m/sec. There thus appears to be a discrepancy of 1.0518×10^{-22} kg m/sec when in fact there is no discrepancy at all, just the use of an incorrect formula for computing the linear momenta. This computation does not take into account the matter of the neutrino.
Thus when an electron is ejected from a nucleus at a speed close to that of light its momentum computed from the product of its relativistic mass and velocity would be seriously less than its true value. The recoil of the nucleus would have a linear momentum equal to the true momentum of the electron. The recoil of the nucleus would involve a much smaller velocity and thus its momentum computed from the incorrect formula would be closer to the true momentum. Thus the net momentum of the nucleus/particle system after the ejection of the electron would appear to be different from the zero momentum before the ejection.
There appears to be no rigorous derivation of relativistic momententum as relativistic mass times velocity, mv. Based upon Lagrangian analysis relativistic linear momentum apparently requires not only the relativistic adjustment of mass but also division by a factor of (1−β²). The correct formula for relativistic momentum is then
This is equivalent to relativistic momentum being equal to the product of longitudinal relativistic mass and velocity.
If the kinetic and potential energies are independent of the spatial variable that defines velocity then it is mv/(1−β²)=m_{l}v which is constant over time.
Not taking into account the division by (1−β²) means that a momentum of a fast particle is underestimated. The extent of the underestimation depends upon velocity.
(To be continued.)
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