﻿ The Relativistic Formulas for Linear and Angular Momenta Based Upon Lagrangian Dynamics
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 The Relativistic Formulas for Linear and Angular Momenta Based Upon Lagrangian Dynamics

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 apparent mass of a body in motion relative to an observer is given by

#### m = m0/(1−β²)½

where m0 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 usually 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. Instead there is an additional correction which must be taken into account.

## Lagrangian Dynamics

If K is the kinetic energy of a system and V is the potential energy then the Lagrangian of the system is defined as

#### L = K − V

If the Lagrangian of a system is a function of a set of variables {qi; i=1,2,…,n} and their time derivatives {dqi/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

#### d(∂L/∂vi)/dt − (∂L/∂qi) = 0

where vi=dqi/dt.

The expression (∂L/∂vi) is the momentum with respect to the variable qi.

## The Newtonian Case

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

#### L = ½mv² − V and therefore ∂L/∂v = mv and since ∂L/∂x = 0 d(mv)/dt = 0 and thus linear momentum is constant over time

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.

## The Relativistic Case

Now consider the relativistic case. The kinetic energy is then (mc²−m0c²) where m=m0/(1−β²)½ and β=v/c.

The expression ½mv² is not the kinetic energy; it is just the first order approximation of the kinetic energy. For more on this point see Relativistic Mechanics.

The Lagrangian is

#### L = (mc²−m0c²) − V and the partial derivative of L with respect to v is −½(m0c²/(1−β²)3/2)(−2β(1/c)) which reduces to (m0c²/(1−β²)3/2)(v/c)(1/c) or, equivalently (m0/(1−β²)3/2)v which can be expressed as mv/(1−β²)

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,

#### d(mv/(1−β²))/dt = 0

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

#### −½(m0c²/(1−β²)3/2)(−2β(r/c)) which reduces to (m0c²/(1−β²)3/2)(ωr/c)(r/c) and further to (m0/(1−β²)3/2)(ωr²) which can be expressed as mvr/(1−β²)

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.

## Precedents

Herbert Goldstein, in his book Classical Mechanics, refers to the longitudinal and the transverse masses for a body; i.e.,

#### ml = m0/(1−β²)3/2 mt = m0/(1−β²)1/2

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

#### d[m0v/(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. 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.

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 masses. J.J. Thompson pointed this out in 1881 and Oliver Heavyside worked out the mathematics in 1897. The formulas for longitudinal and tranverse mass were the same as those given by Goldstein.

## An Illustration and Its Implications

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 1.8343×103 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.

## Conclusion

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.

#### P = m0v/(1−β²)3/2 = mv//(1−β²)

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.)

## Appendix

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 r1>r2. The values of v1 and v2 are determined from

#### m0v1r1/(1−β1²)3/2 = L and m0v2r2/(1−β2²)3/2 = L

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.

## Computation of Relative Velocity from Momentum

If P represents linear momentum and

#### P = m0v/(1−β²)3/2or, equivalently P = m0cβ/(1−β²)3/2

Then the solution for β requires solving

#### β/(1−β²)3/2 = P/(m0c)

For algebraic convenience let P/(m0c) be denoted as 1/ν and (1−β²) as ζ. Squaring the previous equation then gives

#### (1−ζ)/ζ³ = 1/ν² or, equivalently, the cubic equation ζ³ − ν²ζ + ν²sup2; = 0.

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/(m0c) gives a value that is small compared to unity then the relativistic corrections will be negligible.

## Computation of Relative Velocity from Angular Momentum

The same procedure may be followed as with linear momentum and the result is:

#### β/(1−β²)3/2 = L/(rm0c)

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=rm0c/L (and hence L/(rm0c)=1/ν) and simplification results in the same cubic equation

#### ζ³ − ν²ζ + ν² = 0or, equivalently ζ³ = (ζ−1)ν²

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). 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. 