The Direction of Force and Acceleration

by Richard Chace Tolman

Non-Newtonian Mechanics :— The Direction of Force and Acceleration.

By Richard C. Tolman, Ph.D., Instructor in Physical Chemistry at the University of Michigan^[1].

If force is defined as the rate of increase of momentum, the equation

${\mathsf {F}}={\frac {d}{dt}}(m{\mathsf {u}})=m{\frac {d{\mathsf {u}}}{dt}}+{\frac {dm}{dt}}{\mathsf {u}}$

(1)

allows for a change in mass as well as a change in velocity. This is the fundamental equation of non-Newtonian mechanics^[2].

It has been shown from the principle of relativity^[3] that the mass of a moving body is given by the equation

$m={\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}},$

where $m_{0}$ is the mass of the body at rest and $c$ is the velocity of light. Substituting in equation (1) we obtain

${\mathsf {F}}={\frac {d}{dt}}\left({\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}{\mathsf {u}}\right)={\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}{\frac {d{\mathsf {u}}}{dt}}+{\frac {d}{dt}}\left({\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}\right){\mathsf {u}}$

(2)

From an inspection of equations (1) and (2) it is evident that the force acting on a body is equal to the sum of two vectors, one of which is in the direction of the acceleration $d{\mathsf {u}}/dt$ and the other in the direction of the existing velocity u, so that in general the force and the acceleration it produces are not in the same direction. If the force which does produce acceleration in a given direction he resolved perpendicular and parallel to the acceleration, it may be shown that the two components are connected by a definite relation.

Relation between the Components of Force Parallel and Perpendicular to the Acceleration.

Consider a body (fig. 1) moving with the velocity

${\mathsf {u}}=u_{x}{\mathsf {i}}+u_{y}{\mathsf {j}}.$

Let it be accelerated in the Y direction by the action of the component forces ${\mathsf {F}}_{y}$ and ${\mathsf {F}}_{x}$.

From equation (2) we have

${\mathsf {F}}_{x}={\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}{\frac {du_{x}}{dt}}+{\frac {d}{dt}}\left({\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}\right)u_{x}$

(3)

${\mathsf {F}}_{y}={\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}{\frac {du_{y}}{dt}}+{\frac {d}{dt}}\left({\frac {m_{0}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}\right)u_{y}$

(4)

Introducing the condition that there is no acceleration in the X direction, which makes $du_{x}/dt=0$, further noting that $u^{2}=u_{x}^{2}+u_{y}^{2}$, by the division of equation (3) by (4) we obtain

${\frac {{\mathsf {F}}_{x}}{{\mathsf {F}}_{y}}}={\frac {u_{x}u_{y}}{c^{2}-u_{x}^{2}}},$

${\mathsf {F}}_{x}={\frac {u_{x}u_{y}}{c^{2}-u_{x}^{2}}}{\mathsf {F}}_{y}$

(5)

Hence in order to accelerate a body in a given direction, we may apply any force ${\mathsf {F}}_{y}$ in the desired direction, but must at the same time apply at right angles another force ${\mathsf {F}}_{x}$ whose magnitude is given by equation (5).

From a qualitative consideration, it is also possible to see the necessity of a component of force, perpendicular to the desired acceleration. Referring again to fig. 1, since the body is being accelerated in the Y direction, its total velocity and hence its mass are increasing. This increasing mass is accompanied by increasing momentum in the X direction even when the velocity in that direction remains constant. The component force ${\mathsf {F}}_{x}$ is necessary for the production of this increase in X-momentum.

In predicting the path of moving electrons with the help of the fifth equation of electromagnetic theory, ${\mathsf {F}}={\mathsf {E}}+{\frac {1}{c}}{\mathsf {v}}\times {\mathsf {H}},$, we find an interesting application of equation (5).

Application in Electromagnetic Theory.

Consider a charge $\epsilon$ constrained to move in the X direction with the velocity $v$ and let it be the origin of a system of moving coordinates Y$\epsilon$X (fig. 2). Suppose now a test electron $t$, of unit charge, situated at the point $x=0$, $y=y$,

moving in the X direction with the same velocity $v$ as the charge $\epsilon$, and also having a component velocity in the Y direction $u_{y}$. Let us predict the nature of its motion under the influence of the charge $\epsilon$.

The moving charge $\epsilon$ will be surrounded by electric and magnetic fields whose intensities at any point are given by the following expressions^[4], obtained by integrating Maxwell’s four field equations, for the case of a moving point charger,-

${\mathsf {E}}=\left(1-{\frac {v^{2}}{c^{2}}}\right){\frac {\epsilon {\mathsf {R}}}{{\mathsf {R}}^{3}\left(1-{\frac {v^{2}}{c^{2}}}\sin ^{2}\psi \right)^{\frac {3}{2}}}}$

(6)

${\mathsf {H}}={\frac {1}{c}}{\mathsf {v}}\times {\mathsf {E}},$

(7)

where R is the radius vector connecting the moving charge with the point in question and $\psi$ is the angle between R and v.

For the field acting on the test electron $t$, situated at the point $x=0$, $y=y$, we may substitute ${\mathsf {R}}=y{\mathsf {j}}$ and $\sin \psi =1$, giving us,

${\mathsf {E}}={\frac {\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}{\mathsf {j}}$

(8)

and

${\mathsf {H}}={\frac {v}{c}}{\frac {\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}{\mathsf {k}},$

(9)

substituting into the fifth fundamental equation of electromagnetic theory,

${\mathsf {F}}={\mathsf {E}}+{\frac {1}{c}}{\mathsf {v}}\times {\mathsf {H}},$

(10)

we obtain the force acting on the unit test electron $t$.

[Note in the above equation that v, the velocity of the electron, is for our case $v{\mathsf {i}}+u_{y}{\mathsf {j}}$.]

${\mathsf {F}}={\frac {\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}{\mathsf {j}}-{\frac {1}{c^{2}}}{\frac {v^{2}\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}{\mathsf {j}}+{\frac {1}{c^{2}}}{\frac {v^{2}u_{y}\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}i,$

(11)

or

${\mathsf {F}}_{x}={\frac {\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}{\frac {vu_{y}}{c^{2}}},$

(12)

${\mathsf {F}}_{y}={\frac {\epsilon }{y^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}}}\left(1-{\frac {v^{2}}{c^{2}}}\right)$

(13)

Under the action of the component force ${\mathsf {F}}_{x}$ we might at first sight expect the electron $t$ to aquire an acceleration in the X direction: Such condition, however, would not be in agreement with the principle of relativity, since from the point of view of an observer who is moving along with the charge $\epsilon$, the phenomenon is merely one of ordinary electrostatic repulsion and the test electron should experience no change in velocity in the X direction but should be accelerated merely in the Y direction.

If, however, we divide equation (12) by (13) we obtain

${\mathsf {F}}_{x}={\frac {vu_{y}}{c^{2}-v^{2}}}{\mathsf {F}}_{y},$

(14)

which agrees with equation (5), the necessary relation for zero acceleration in the X direction. The application of equation (5) thus removes a discrepancy which could not be accounted for in any system of mechanics in which force and acceleration are in the same direction.

Summary.

For non-Newtonian mechanics, it has been pointed out that force and the acceleration it produces are not in general in the same direction. A definite relation (equation 5) has been derived connecting the components of force parallel and perpendicular to the acceleration. For a special problem, the application of this relation has removed an apparent discrepancy between the predictions based on the electromagnetic theory and on the principle of relativity.

Ann Arbor, Mich.

March 25th, 1911.

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1. Communicated by the Author.
2. This definition of force was first used by Lewis (Phil. Mag. xvi. p. 705 (1908)). In Einstein‘s later treatment of the principle of relativity, Jahrbuch der Radioktivität, iv. p. 411 (1907), he defines force by the equations

   ${\mathsf {F}}_{x}={\frac {d}{dt}}\left({\frac {m_{0}u_{x}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}\right),\ {\mathsf {F}}_{y}={\frac {d}{dt}}\left({\frac {m_{0}u_{y}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}\right),\ {\mathsf {F}}_{z}={\frac {d}{dt}}\left({\frac {m_{0}u_{z}}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}\right).$

   He there states that this definition has in general no physical meaning. We see, however, that these are merely the scalar equations corresponding to equation (2) above and hence derivable from equation (1), which is an obvious definition of force and has a physical meaning. In further support of this definition of force, it has recently been pointed out by the writer, Phil. Mag. xxi. p. 296 (1911), that, combined with the principle of relativity, it leads to a derivation of the fifth fundamental equation of electromagnetic theory in its exact form

   ${\mathsf {F}}={\mathsf {E}}+{\frac {1}{c}}{\mathsf {v}}\times {\mathsf {H}},$

   there being no necessity for distinguishing between longitudinal and transverse mass.

3. Lewis & Tolman, Proc. Amer. Acad. xliv. p. 711 (1909); Phil. Mag. xviii. p. 510 (1909).
4. Abraham, Theorie der Elektrizität, vol. ii. p. 86 et seq. (B. G. Teubner, Leipzig and Berlin, 1908).