# Why does Newton's third law exist even in non-inertial reference frames?

While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?

• I know this isn't what you're looking for, but it is reasonable: "Because if it were otherwise the laws of physics would be non-discoverable." – Joshua Jan 25 '15 at 23:42

If we define a rest-frame such that $$\mathbf{r} = \mathbf{R}_0 + \mathbf{r}' \\ \mathbf{v} = \mathbf{V}_0 + \mathbf{v}' \\ \mathbf{a} = \mathbf{A}_0 + \mathbf{a}'$$ where $\mathbf{R}_0$ represents the distance from the rest-frame origin to the moving-frame origin (and similarly for $\mathbf{V}_0$ and $\mathbf{A}_0$). If $\mathbf{A}_0=0$, then $\mathbf{F}=m\mathbf{a}=m\mathbf{a}'$. However, if $\mathbf{A}_0\neq0$, then in this case, the force becomes $$\mathbf{F} = m\mathbf{A}_0+m\mathbf{a}'$$ which we can re-write as $$\mathbf{F}-m\mathbf{A}_0 = m\mathbf{a}'$$ Which we can then define $\mathbf{F}'=\mathbf{F}-m\mathbf{A}_0$ to get $$\mathbf{F}'=m\mathbf{a}'$$ which is similar to Newton's second law. If object A acts on object B, then the forces are $\mathbf{F}'_{AB}=-\mathbf{F}'_{BA}$. Since both have the same $-m\mathbf{A}_0$ term, then this reduces to $\mathbf{F}_{AB}=-\mathbf{F}_{BA}$ which is Newton's 3rd law.

The force is given by $$F=ma=m\frac{d^2x}{dt^2}$$ If we move to a inertial frame (and assuming non-relativistic speeds), we are really letting $x\to x+Vt$ where $V$ denotes the moving velocity. The time derivatives then become $$\frac{dx}{dt} \to \frac{dx}{dt}+V$$ $$\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{dx^2}{dt^2} \to \frac{d^2x}{dt^2}$$ Thus the force is not changed under this change of frame.

• But, non-inertial reference frames are accelerating frames and not those going with a uniform velocity.In frames moving with uniform velocity all the three laws are valid. – Rajath Krishna R Oct 18 '13 at 14:39
• Oops, I misread the question. – Kyle Kanos Oct 18 '13 at 14:41
• @RajathKrishnaR: I edited my answer to actually answer your question this time. – Kyle Kanos Oct 18 '13 at 14:49
• the v3 answer is wrong. The transformation $x \to x +Vt$ takes you to another inertial frame (like riding a bus does). Your explanation does not go through if you allow arbitrary (non-inertial) changes of frame. – Brian Moths Oct 18 '13 at 16:06
• @NowIGetToLearnWhatAHeadIs: Huh? The first part covers the arbitrary non-inertial changes. The second part covers my incorrect reading of the question for an arbitrary inertial change. – Kyle Kanos Oct 18 '13 at 16:09

While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?

Interesting interpretation. I would put it exactly the other way around: in a noninertial frame, the first and second laws hold, but the third law doesn't.

Let's say we're in a rotating frame, and in that frame, a baseball experiences a centrifugal force. There is no third-law partner for this force: the baseball doesn't create a force back on any other object. This is because the centrifugal force is not an interaction between two objects, so we can't have the third-law pattern of A on B, B on A.

On the other hand, the first and second laws certainly apply to the baseball, provided that we include the centrifugal and Coriolis forces as forces. These fictitious forces also obey the law of vector addition, which is a fundamental law of Newtonian mechanics, although not traditionally considered one of Newton's laws.

I suppose the opposite interpretation, as given in the question, occurs if you refuse to consider fictitious forces as forces. Then they don't violate Newton's third law, because they're not forces. (Dogs can't violate the law against murder, because the law only applies to people, and dog's are not considered people.) The first and second laws are then violated, because we refuse to put in the inertial forces that would have been needed in order to make them work.

The cutest way to see this is to restate Newton's third law as "no interaction can change the total momentum of the universe." Then, note that since an accelerating reference frame is accelerating with respect to whatever "base" inertial reference frame you're using, everything else seems to be accelerating away. Therefore, the net momentum of the universe is changing. Therefore, Newton's Third Law does not hold in this reference frame.

$\newcommand{fp}[0]{\vec{F}_\textrm{phys}}$ $\newcommand{fn}[0]{\vec{F}_\textrm{non-inertial}}$ $\newcommand{fab}[0]{\vec{F}_{AB}}$ $\newcommand{fba}[0]{\vec{F}_{BA}}$In a non-inertial frame, every object feels the physical force $\fp$, that it felt in the inertial frame, plus a force $\fn$. The non intertial force felt by an object may depend on its mass, position, time, and possibly other things. An objects acceleration is then given by $m \vec{a} = \fn + \fp$. Thus newton's second law, $m \vec{a} = \fp$, breaks down, and you need a correction for the non-inertial forces.

Let's look at newton's third law. It says $\fab= -\fba$. We know this holds true in the inertial frame. If we transform these forces to a non-inertial frame, the transformed coordinates will be different, but because of the way coordinate transformations work, it will still be true that $\fab = -\fba$ in the transformed coordinate system. Thus newton's third law still holds.