- #1

Hakkinen

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## Homework Statement

Given that [itex]\overline{F}[/itex]=m[itex]\overline{a}[/itex] is valid in the lab frame S, show that:

(a) it is also valid in a moving frame S' with a constant velocity relative to S,

but (b) invalid in a moving frame with a changing velocity (ie, a frame S' accelerating relative to S).

Assume the force and mass are invariant

## Homework Equations

The Galilean transformation.

Newtons 2nd law

## The Attempt at a Solution

This is a non-required problem in the first homework set for an applied modern physics (sophomore level) course I'm taking. We have only had one lecture and I don't have the textbook so I'm not sure if the conclusions I've drawn are correct

So we have two reference frames S and S', with respective coordinates (x,y,z,t) and (x',y',z',t'). But we can use just (x,t) and (x',t') and draw the same conclusions. S' is moving relative to S with a constant velocity [itex]\overline{v}[/itex] in part (a) and with an acceleration [itex]\overline{κ}[/itex] in part (b). I believe t=t' must also be assumed as the concept of absolute time still exists in Newtonian mechanics.

For part (a)

In S

[itex]\overline{F}[/itex]=m(d

^{2}x/dt

^{2})=m(d[itex]\overline{u}[/itex]/dt) (Where [itex]\overline{u}[/itex] is the velocity measured in S)

The Galilean transformation gives this relationship between [itex]\overline{u}[/itex] and [itex]\overline{u'}[/itex] (velocity as measured in S')

[itex]\overline{u}=\overline{u'}+\overline{v}[/itex]

In S'

[itex]\overline{F}[/itex]=m(d([itex]\overline{u'}+\overline{v}/dt[/itex])

if [itex]\overline{u}=\overline{u'}+\overline{v}[/itex] then [itex]\overline{a'}[/itex]= ([itex]d(\overline{u'}+\overline{v})[/itex]/dt)) should equal the same acceleration [itex]\overline{a}[/itex] measured in S. Thus if the force and mass are invariant as well, the 2nd law holds for a moving reference frame with some constant velocity [itex]\overline{v}[/itex].

For part (b) when S' is accelerating relative to S

Similarly in S

[itex]\overline{F}[/itex]=m(d

^{2}x/dt

^{2})=m(d[itex]\overline{u}[/itex]/dt)

##\overline{u}=\overline{u'}+\overline{κ}##

##\overline{F}=m(d(\overline{u'}+\overline{κ})/dt)##

Thus ##\overline{F}=m\overline{a'}+m(d\overline{κ}/dt)## where ##(d\overline{κ}/dt)## is the jerk of ##\overline{κ}##

I believe this shows that the 2nd Law is not valid for an accelerating reference frame.

I would greatly appreciate any corrections and comments you can provide!