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Physics LibreTexts

15.29: Force

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Force is defined as rate of change of momentum, and we wish to find the transformation between forces referred to frames in uniform relative motion such that this relation holds on all such frames.

Suppose that, in Σ, a mass has instantaneous mass m and velocity whose instantaneous components are ux and uy. If a force acts on it, then the velocity and hence also the mass are functions of time. The x-component of the force is given by

Fx=ddt(mux).

We want to express everything on the right hand side in terms of unprimed quantities. Thus from Equation 15.21.8 and the inverse of Equation 15.16.2, we obtain

mux=mγ(uxv).

Also

ddt=dtdtddt

Let us first evaluate ddt(mγuxmγv). In this expression, v and γ are independent of time (the frame Σ is moving at constant velocity relative to Σ), and ddt of mux is the x-component of the force in Σ, that is Fx. Thus

ddt(mγuxmγv)=γ(Fxvdmdt).

Now we need to evaluate ddt in terms of unprimed quantities. If we start with

dt=(tx)tdx+(tt)xdt

and we’ll evaluate dtdt which, being a total derivative, is the reciprocal of dtdt. The partial derivatives are given by Equations 15.15.3j,k and l, while dxdt=ux. Hence we obtain

dtdt=1γ(1uxvc2).

Thus we arrive at

Fx=Fxv(dmdt)1uxvc2

The mass is not constant (i.e. dmdt is not zero) because there is a force acting on the body, and we have to relate the term dmdt to the force. At some instant when the force and velocity (in Σ) are F and u, the rate at which F is doing work on the body is Fu =Fxux+Fyuy+Fzuz and this is equal to the rate of increase of energy of the body, which is ˙mc2. (In Section 15.24, in deriving the expression for kinetic energy, I wrote that the rate of doing work was equal to the rate of increase of kinetic energy. Now I have just written that it is equal to the rate of increase of (total) energy. Which is right?)

dmdt=1c2(Fxux+Fyuy+Fzuz).

Substitute this into Equation ??? and, after a very little more algebra, we finally obtain the transformation for Fx:

Fx=Fxvc2uxv(uyFy+uzFz).

The y and z components are a little easier, and I leave it as an exercise to show that

Fy=1v2c21uxvcFy

Fz=1v2c21uxvcFz.

As usual, the inverse transformations are found by interchanging the primed and unprimed quantities and changing the sign of v.

The force on a particle and its resultant acceleration are not in general in the same direction, because the mass is not constant. (Newton’s second law is not F=ma; it is F=˙p) Thus

F=ddt(mu)=ma+˙mu.

Here

m=m0(1u2c2)12

and so

˙m=m0uac2(1u2c2)32.

Thus

F=m0(1u2c2)12(a+uac2u2u).


This page titled 15.29: Force is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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