Validity of Newton's 2nd Law in accelerating ref. frame and constant v.

[Hakkinen]

First of all I'm not sure if this is the right forum for this problem.

Homework Statement

Given that \overline{F}=m\overline{a} 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 \overline{v} in part (a) and with an acceleration \overline{κ} 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
\overline{F}=m(d²x/dt²)=m(d\overline{u}/dt) (Where \overline{u} is the velocity measured in S)

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

\overline{u}=\overline{u'}+\overline{v}

In S'
\overline{F}=m(d(\overline{u'}+\overline{v}/dt)

if \overline{u}=\overline{u'}+\overline{v} then \overline{a'}= (d(\overline{u'}+\overline{v})/dt)) should equal the same acceleration \overline{a} 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 \overline{v}.



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

Similarly in S
\overline{F}=m(d²x/dt²)=m(d\overline{u}/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!

 

[ehild]

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

Similarly in S
\overline{F}=m(d²x/dt²)=m(d\overline{u}/dt)

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

It is not true: The velocity transforms as

\overline{u}=\overline{u'}+\overline{v}

ehild

 

[Hakkinen]

If that is the case then how is the result different from part (a)?

 

[ehild]

The velocity of the frame of reference changes with time: its time derivative is equal to κ

ehild

 

[Hakkinen]

Ah I Think I understand now.

$\overline{u'}=\overline{u}-\overline{v}$

$\frac{d\overline{u'}}{dt} = \frac{d\overline{u}}{dt}-\overline{κ}$

$\overline{F}=m\frac{d\overline{u'}}{dt}+ m \overline{κ}$

Is this correct?

 

[ehild]

Yes, there is a new force in the accelerating frame, which does not belong to any interaction between bodies. That force pushes you forward in a braking car.

ehild