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Linear and Angular Momenta |
In Newtonian (non-relativistic) mechanics the linear momentum of a body is defined as the product of its mass m and its velocity v. Angular momentum is similarly defined as mvr, where r is the radial distance to the center of revolution. Einstein's Special Theory of Relativity found that the mass of a body 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 presumed that linear and angular momenta are still given by the formulas mv and mvr and the only relativistic correction is that the velocity-dependence of mass must be taken into account. It will be shown below that this is not the case.
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
If a body is revolving around a center at an angular rotation rate ω then its velocity is rω. Its kinetic energy is ½mr²ω². If θ is the angular position and the potential energy does not depend upon θ then, since ω=dθ/dt.
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 just the first order approximation of the kinetic energy. For more on this point see Relativistic Mechanics.
The Lagrangian is
In other words, the relativistic linear momentum requires not only the relativistic adjustment of mass but also division by a factor of (1−β²). 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,
For a body revolving around a center at a distance r at a rate of rotation of ω then v=ωr and β=ωr/c. The partial derivative of the Langrangian with respect to ω is
This is the momenum with respect to θ; i.e., the angular momentum. So again the relativistic momentum requires not only the mass to be dependent upon velocity but also division by the factor (1−β²). Also if the potential energy is not dependent upon θ then this angular momentum, mvr/(1−β²), is conserved.
Herbert Goldstein, in his book Classical Mechanics, 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 is correct because 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.
Goldstein adopts a different approach to the issue. He argues that the true equation of motion is just
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. 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, so the settlement of the issue requires results for the case of β near unity.
If relativistic momentum P is conserved then
This says that relativistic mass decreases with velocity despite the definition that m=m_{0}/(1−β²)^{½}. This occurs because relativistic momentum has to be conserved. ****
In the case of linear momentum there would be no occasion for the change in the velocity of a particle without some interaction such as the elastic collision of two particles. In the case of angular momentum there is the opportunity to compare two situations in which the velocities are different but angular momentum is conserved. This would occur if the radius of the orbit of the particle changes.
The same relationship prevails conserning relativistic angular momentum and relativistic mass as in the linear case. Let L be the relativistic angular momentum and let the velocities and radii for two situations be identified by the subscripts 1 and 2. Let r_{1}>r_{2}. The values of v_{1} and v_{2} are determined from
Before proceding further it is necessary to consider how the velocities would be determined from these equations. It is convenient to first consider the case of linear momentum.
If P represents linear momentum and
Then the solution for β requires solving
For algebraic convenience let P/(m_{0}c) be denoted as 1/ν and (1−β²) as ζ. Squaring the previous equation then gives
For purposes of exploring the nature of the cubic function ζ³ − ν²ζ + ν² let it be denoted as g(ζ).
Here is what this function looks like over the range [-5, +5] for ν²=12.
There is a standard formula for the solution of a cubic polynomial equation that will give three values of ζ but possibly only one of them may be physically relevant. For that value of ζ to be relevant it must be positive and less than unity. Then β=±(1−ζ)^{½}.
However, if a trial estimate of β as P/(m_{0}c) gives a value that is small compared to unity then the relativistic corrections will be negligible.
The same procedure may be followed as with linear momentum and the result is:
The function on the left is a monotonically increasing function of β. If r decreases the term on the right increases and β must also increase to conserve relativistic angular momentum. Because β increases the relativistic mass must also increase.
An explicit solution for β can be obtained by squaring both sides of the above equation and letting (1−β²)=ζ and ν=rm_{0}c/L (and hence L/(rm_{0}c)=1/ν) and simplification results in the same cubic equation
Since only values of ζ that are between 0 and 1 are relevant, a convenient way to find solutions is to plot the two sides of the above equation for ζ in the interval [0, 1]. The function on the right is equal to 0 at ζ=1 for all values of ν² and its slope is equal to ν². Thus the line for the function (ζ−1)ν² pivots about the point (1, 0).
The value of ν increases with decreasing r. When ν increases the solution value of ζ increases which means that the value of β increases. If β increases then the relativistic mass increases. Thus a mass increment emerges with decreasing radius.
This is in contradiction to one of the greatest enigmas of nuclear physics; i.e., the mass deficits of nuclei. When a neutron and proton come together to form a deuteron the mass of the deuteron is less than the combined mass of the neutron and the proton. According to the above the neutron and the proton widely separated have a certain amount of relativistic angular momentum. As the particles come together they revolve about their center of mass faster and faster, but the relatistic angular momentum is maintained. For it to be maintained the relativistic masses of the particles must decrease. There thus develops a mass deficit.****
There is a critical value of ν where the solution is a tangency solution; i.e., g(ζ)=0 and g'(ζ)=0. This occurs for ζ=3/2 and ν²=3(3/2)²=27/4=6.75. For values of ν² below this level there are no solutions for ζ in the vicinity of 1.
In order for g(ζ)=0 to have a solution for 0<ζ<1 it is necessary that g'(0)<0 and g'(1)>0, which means ν² must be less than 3.
The minimum of g(ζ) occurs for ζ=ν/√3 and its value is
Suppose a nucleus of Cs 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 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 appears to have a momentum of 4.2074×10^{-22} kg m/sec, whereas the electrons 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. ****************** One of the surprising implications is that if a force acts on a particle in the direction of its motion the relativistic momentum increases but the increases in velocity become less and less. However the relativisitic mass decreases. If the velocity were to reach the speed of light the relativistic mass would be zero. Its relativistic momentum would be infinite but its relativistic mass would be zero even though its rest mass is not zero. Particles such as photons have zero rest mass and a zero relativistic momentum but a nonzero momentum. ****
Relativistic linear momentum requires not only the relativistic adjustment of mass but also division by a factor of (1−β²). 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.
Not taking into account the division by (1−β²) means that a momentum is underestimated. The extent of the underestimation depends upon velocity. For example, 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 after the ejection of the electron would appear to be different from the zero momentum before the ejection.
(To be continued.)
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