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

1. Jan 27, 2013

### Hakkinen

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

1. The problem statement, all variables and given/known data
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

2. Relevant equations
The Galilean transformation.
Newtons 2nd law

3. The attempt at a solution
This is a non-required problem in the first hw 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!

2. Jan 29, 2013

### ehild

It is not true: The velocity transforms as

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

ehild

3. Jan 29, 2013

### Hakkinen

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

4. Jan 29, 2013

### ehild

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

ehild

5. Jan 29, 2013

### 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?

6. Jan 29, 2013

### 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