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The following question occurred to me while reading a proof of the following statement:

If K is an inertial frame of reference, then a K’ frame of reference, which is moving with a constant velocity compared to K, is also an inertial frame of reference.

In the proof $\sum F=\sum F'$ is used and I'd like to know why is this equation valid.


Here goes the proof:

Let $\underline{r}(t)$ be the position vector of the point $P$ in the $K$ frame. Let $\underline{r'}(t)$ be the position vector of the point $P$ in the $K'$ frame. Let $\underline{r}_{K'}(t)$ be the position vector of the point $K'$ in the $K$ frame.

$K’$ is moving with a constant velocity compared to $K$, therefore: $$\begin{equation} \underline{r}_{K'}(t)=\underline{w}\cdot t+\underline{r_0} \end{equation}$$ where $\underline{w}$ is the velocity vector of $K'$ and $\underline{r_0}$ is the position vector of the origo of $K'$ (in the $K$ reference frame) in the $t=0$ moment. The connection between the two position vectors: $$ \underline{r}(t)=\underline{r'}(t)+\underline{r}_{K'}(t)=\underline{r'}(t)+\underline{w}\cdot t+\underline{r_0}$$ This is the Galileian-transformation. After derivating with respect to time twice we get: $$ \begin{aligned} \underline{v}(t) &=\underline{v'}(t)+\underline{w} \\ Eq. 1: \underline{a}(t) &=\underline{a'}(t) \end{aligned} $$ Newton II. in the inertial frame of reference $K$ $$ m \underline{a}=\sum\underline{F} $$ where $\sum\underline{F}$ is the net force in the $K$ frame. Combining this formulae with $Eq. 1$ and using that $\sum\underline{F}=\sum\underline{F}'$ ( forces are independent from the frame of reference) we get: $$ \sum \underline{F}'=m\underline{a'}$$ which implies that $K'$ is an inertial reference frame. $ \blacksquare $

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  • $\begingroup$ Equation 1 says it all. Accelerations/Forces in both reference frames are the same. This is because $\underline{w}$ is constant. Can you explain what exactly is not clear to you? $\endgroup$ – Aziraphale Jan 19 '15 at 6:56
  • $\begingroup$ @Aziraphale My question is how does ΣF'=ma' follow from Eq. 1. With multiplying both sides by m we get ma=ma'. Since K is inertial we know that ΣF=ma=ma'. So we showed that ΣF=ma' but in order to prove the statement we would have to show that ΣF'=ma' (note the apostrophe). Using ΣF=ΣF' this would be pretty simple. My question is why does ΣF equal ΣF' (Note that we don't know if ma'=ΣF' is true or not, since we don't know if K' is inertial or not.) $\endgroup$ – naroslife Jan 19 '15 at 7:29
  • $\begingroup$ No. It is the other way round: We derive the last equation and from that we conclude that $K'$ is an inertial reference frame. ma' is the sum of all forces in K'. There are no other forces (Eq. 1). This is expressed in the equation SUM(F')=ma'. Not more and not less. $\endgroup$ – Aziraphale Jan 19 '15 at 8:21
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If an object has no acceleration in one inertial frame of reference that means no real forces acting on it. now suppose you observe the same object from a different inertial frame,its not possible that just because you are observing the same object from a different inertial frame somehow a real force will start acting on the object. But if you observe now the same object from an noninertial frame of reference then fictitious forces will act on the object and here the object can have acceleration without some real forces acting on it.

That means you first calculate the net force acting on a particle in one inertial frame of reference and then calculate net force from a different frame of reference(Inertial or noninertial) . If the net force is not same(for both frames) means some fictitious forces are acting on the particle. If there are fictitious forces then you are observing now from a non inertial frame of reference.

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  • $\begingroup$ But it does not mean that the magnitude of the forces acting on the body are the same if we consider your argument. I mean all the forces might be stronger in the other ref. frame, but still add up to zero as the net force. $\endgroup$ – onurcanbektas Oct 5 '18 at 3:54
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If the forces are determined, say, by the relative position of two objects, then the relative position of the two objects is the same in the two frames. If the force is determined, say, by the relative velocity of two objects, then the relative velocity of the two objects is the same in the two frames.

Many forces are like this; a spring, friction due to air resistance, Newton's gravitational force, etc.

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Suppose an object is experiencing a certain total force $\mathbf F$ applied to it. If you change to a different frame of reference which is inertial w.r.t. to the previous one (i.e. they are related by a Galilei transformation, that is to say one is moving with constant velocity w.r.t. the other), then there is no apparent acceleration added to the object, since $\mathbf a = \ddot{\mathbf x}$, and the relation between coordinates in the two frame of reference are at most linear in $t$, whence the extra acceleration is $\mathbf a=0$. Therefore forces, whether real or apparent, are conserved when passing to a different frame of reference through a Galilei transformation.

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  • $\begingroup$ What do you mean by both real and apparent? $\endgroup$ – Paul Jan 19 '15 at 11:56
  • $\begingroup$ apparent forces are those due to the fact that the original frame of reference might fail to be inertial, like any observer on Earth for instance. Real forces are those that are really applied to an object. $\endgroup$ – Phoenix87 Jan 19 '15 at 12:14
  • $\begingroup$ an object can have an acceleration in a non inertial frame even though no real forces acting on it ,on the other hand if we observe it from inertial frame it has no acceleration. then what you mean by,''Therefore forces, whether real or apparent, are conserved when passing to a different frame of reference through a Galilei transformation.'' $\endgroup$ – Paul Jan 19 '15 at 12:18
  • $\begingroup$ that a Galilei transformation cannot change accelerations. $\endgroup$ – Phoenix87 Jan 19 '15 at 12:20
  • $\begingroup$ You're saying "If you change to a different frame of reference which is inertial w.r.t. to the previous one..." But the proof itself is about if the second reference frame is inertial or not. You can't suppose it's inertial to prove the statement itself. $\endgroup$ – naroslife Jan 19 '15 at 12:38

protected by Qmechanic May 13 '16 at 14:24

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