0
$\begingroup$

We define an inertial frame, as a frame of reference where:

Newton's 1st law holds.

It is then stated that a reference frame moving with constant velocity with respect to an inertial frame is also inertial, but there is no further justification.

My justification is the following:

We can perform a Galilean transformation and show that:

$$\mathbf{a} = \mathbf{a}'$$

where prime refers to the frame of reference under consideration. Since we showed that accelerations are the same (both zero), we only need to show that:

$$ \mathbf{F} = \mathbf{F}'=0$$

But this means we have to accept that forces (net force more precisely), don't depend on the frame of reference. Is that postulated in Newtonian mechanics?

Comment

One could argue that the forces are the same if we knew that the primed frame of reference is inertial (since Newton's second law holds in all inertial frames), which entails:

$$\mathbf{F} = m\mathbf{a} = m\mathbf{a}' = \mathbf{F}'$$

But this doesn't make sense since we already assumed that the primed frame is also inertial (this must be shown).

$\endgroup$

4 Answers 4

2
$\begingroup$

Unfortunately, your question is a typical question impossible to answer without a clear preliminary statement about the exact form of definitions and principles one is assuming as the basis of the dynamics.

Just saying that we use Newton's laws is not enough due to the long history of criticisms and re-formulations they have undergone in more than three centuries.

As of today, I would say that at least two different points of view are possible. Each of them is acceptable, but it is not a good idea to mix them.

From a modern "Newtonian" point of view, forces are not defined by ${\bf F}=m {\bf a}$, but are primitive quantities the Dynamics depends on, characterized by some properties (the relevant ones in the present context is that they are vectors and do not depend on the reference frame). The first principle can be seen as the statement that at least one reference frame exists such that all the free particles move uniformly along straight lines.

Then, in this approach, the proof that a reference frame moving uniformly with respect to an inertial reference frame is inertial too is straightforward: we have just to show that every free particle moving uniformly in the first frame is free and moves uniformly in the second frame too. Indeed, if force is zero in the inertial frame, it will remain zero in any other reference frame. Now, if a free particle moves uniformly with respect to the inertial frame, and the new system moves uniformly with respect to the same inertial frame, also the motion of the particle with respect to the new frame will be uniform (Galilean transformations).

Schematically, in such an approach, the chain of implications is (symbols with and without a prime sign indicate the same quantity in the two reference frames $S$ and $S'$: $$ F=0 \Rightarrow F'=0 \Rightarrow a = a' =0 \Rightarrow {\mathrm {uniform~ motion~ of~ free~ particles~in~ both~ references~ frames}} $$

In a point of view influenced by Mach, forces are defined by ${\bf F}=m {\bf a}$, therefore a particle moving uniformly is a free particle by definition. But accelerations in two frames with a uniform relative motion are equal. Therefore every free particle in the inertial system is a free particle in the system with a uniform relative motion and moves uniformly. Then, also the other system is an inertial reference frame.

Here, the chain of implications is: $$ {\mathrm {uniform~ motion~ in~ S}}\Rightarrow {\mathrm {uniform~ motion~ in~ S'}} \Rightarrow F=0; F'=0 \Rightarrow{\mathrm {~~a~free~ particle~ in~S~ is ~also ~a~ free ~particle~in~S'}} $$

To summarize, in the first approach, the equality $$ {\bf F}={\bf F'} $$ is a property of the force (by assumption).

In the second approach, it is a consequence of the equality of accelerations, starting with positions related through a Galilean transformation.

$\endgroup$
2
  • $\begingroup$ This is what I was looking for. I should add that I never had in mind the point of view of Mach (I am considering $\mathbf{F} = m \mathbf{a}$ a postulate). So if I understand it correctly, forces are assumed to be frame independent. Since accelerations are invariant under a Galilean transformation then every free particle in the first frame is also free on the second. Is it though necessary to check every free particle? If we have just one free particle that doesn't accelerate, doesn't this guarantees that the frame is inertial? $\endgroup$ Commented Aug 31, 2022 at 23:28
  • $\begingroup$ @AntoniosSarikas Yes, in what I call the modern Newtonian point of view, (real) forces are frame-independent. In this approach, we have not to check every free particle. But, as soon as we know $F=0$ (remember that in such an approach forces are given) we can be sure that the uniform motion will hold for every free particle. $\endgroup$ Commented Sep 1, 2022 at 4:31
1
$\begingroup$

We can perform a Galilean transformation and show that:

$$\mathbf{a} = \mathbf{a}'$$

we only need to show that:

$$ \mathbf{F} = \mathbf{F}'=0$$

But this means we have to accept that forces (net force more precisely), don't depend on the frame of reference.

There are a lot of caveats one could make in answering this question. Perhaps most importantly, it is not always true that $\vec F=\vec F'$. For example, if there is a fixed external potential that creates a force then the force is not invariant (unless you also translate the external potential).

However, as long as the force is dependent only on the magnitude of the difference between interacting particles, then it is straightforward to show that $\vec F=\vec F'$.

To convince yourself, have a look at the example of a gravitational interaction force between a swarm of $N$ particles. The force on the $i$th particle in the swarm is: $$ \vec F^{(i)} = \sum_{j\neq i}^N\frac{-Gm_i m_j(\vec x^{(i)} - \vec x^{(j)})}{|\vec x^{(i)} - \vec x^{(j)}|^3}\;. $$

Newton's 2nd law for the $i$th particle reads: $$ m_i \frac{d^2 \vec x^{(i)}}{dt^2} = \sum_{j\neq i}^N\frac{-Gm_i m_j(\vec x^{(i)} - \vec x^{(j)})}{|\vec x^{(i)} - \vec x^{(j)}|^3}\;, $$ where the $\vec x^{(i)}$ represents the coordinates of the $i$th particle in one particular frame.

Now transition to a different frame, related to the first by: $$ \vec {\tilde x} = \vec x + \vec v_0 t\;. $$

We have: $$ m_i\frac{d^2 \vec{\tilde x}^{(i)}}{dt^2} = m_i\frac{d^2 \vec{x}^{(i)}}{dt^2} = \sum_{j\neq i}^N\frac{-Gm_i m_j(\vec x^{(i)} - \vec x^{(j)})}{|\vec x^{(i)} - \vec x^{(j)}|^3} = \sum_{j\neq i}^N\frac{-Gm_i m_j(\vec{\tilde x}^{(i)} -\vec v_0t - \vec{\tilde x}^{(j)} +\vec v_0t)}{|\vec{\tilde x}^{(i)} -\vec v_0t - \vec{\tilde x}^{(j)} + \vec v_0 t|^3} = \sum_{j\neq i}^N\frac{-Gm_i m_j(\vec{\tilde x}^{(i)} - \vec{\tilde x}^{(j)} )}{|\vec{\tilde x}^{(i)} - \vec{\tilde x}^{(j)}|^3}\;. $$

Or, comparing the far LHS with the far RHS of the above equation, we see: $$ m_i\frac{d^2 \vec{\tilde x}^{(i)}}{dt^2} = \sum_{j\neq i}^N\frac{-Gm_i m_j(\vec{\tilde x}^{(i)} - \vec{\tilde x}^{(j)})}{|\vec{\tilde x}^{(i)} - \vec{\tilde x}^{(j)}|^3}\;, $$ which shows us that the 2nd law has the same form in both frames.


Update:

The example given above of Galilean invariance of the form of the equation of motion obviously also applied to any force that can be written as the gradient of a potential that depends only on the distance between particles.

But, in the comments, a question arose about the existence of velocity dependent forces, such as the Lorentz force. Indeed, considerations of this type eventually lead to the abandonment of the principle of Galilean invariance!

Over 100 years ago, Einstein was curious about a similar issue. He eventually concluded that there is a relativity principle that is different from Galilean relativity that applies to both classical mechanics and classical electrodynamics. This was not immediately obvious, and for a long time many physicists continued to support Galilean relativity along with the postulate of an "ether" that provided a preferred frame for electromagnetism. However, as we all know, eventually Einstein's theory of relativity prevailed.

In Einstein's theory, we no longer have exact Galilean invariance, but rather, the equations of physics are invariant under Lorentz transformations rather than the Galilean transformation.


Update 2

The concept of "inertial frames" is still good and still remains whether we deal with relativistic speeds of not. The thing that Einstein/Lorentz discovered is that we need to use a different transformation to change coordinates from one frame to another.

In non-relativistic mechanics, for relative velocity $v$ much less than the speed of light, we used the Galilean transformation: $$ t' = t $$ $$ x' = x - vt $$

The correct transformation (correct for all relative velocities) is the Lorentz transformation: $$ t' = \gamma t - v\gamma x/c^2 $$ $$ x' = \gamma x - v \gamma t\;, $$ where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$.

You can show that for $v<<c$ the two transformations are very close to the same, since in this limit $\gamma \to 1$.

$\endgroup$
6
  • $\begingroup$ In this specific case the force depends on the relative position between the particles, which is the same in all frames of reference. But what if the force depends on velocity for example? My question is how starting from an inertial frame, making a Galilean transformation (jumping to a frame which moves with constant velocity with respect to the inertial frame), we end up with another inertial frame? We must show that Newton's first law holds also in this frame. This is what I am trying to show. $\endgroup$ Commented Aug 31, 2022 at 21:47
  • $\begingroup$ "But what if the force depends on velocity for example?" Do you have a specific example in mind? There are four known force types: electromagnetic, gravitational, strong force, and weak force. Please pick an example from one of these buckets. Or are you asking a question outside of mainstream physics? $\endgroup$
    – hft
    Commented Aug 31, 2022 at 22:21
  • $\begingroup$ What comes to my mind is Lorentz force. $\endgroup$ Commented Aug 31, 2022 at 22:40
  • $\begingroup$ Yes, a very good example indeed! Einstein had the same question in mind about 100 years ago. I will update my answer to explain. $\endgroup$
    – hft
    Commented Aug 31, 2022 at 23:33
  • $\begingroup$ Enlightening update. Is Galilean invariance an additional assumption in Newtonian mechanics or it can be derived? I am asking this because, for example Newton's second law is postulated to hold in all inertial frames which seems to make this principle irrelevant. Can you please elaborate on "This Galilean invariance of non-relativistic classical mechanics is already implicit in Newton's First Law"? $\endgroup$ Commented Sep 1, 2022 at 8:09
1
$\begingroup$

The mathematical answer is that if we replace $\vec{x}$ with $R\vec{x}-\vec{x}_0-t\vec{v}_0$, where $R^T=R^{-1}$, then $\vec{F}=m\ddot{\vec{x}}$ is invariant.

The much more satisfying physics answer is that if two reference frames differ by a constant relative velocity (and rotation due to $R$), you can't tell which one should be inertial. In the (translated) words of Galileo, explaining the case for $R=I_3$:

Shut yourself up with some friend in the main cabin below decks on some large ship, and have with you there some flies, butterflies, and other small flying animals. Have a large bowl of water with some fish in it; hang up a bottle that empties drop by drop into a wide vessel beneath it. With the ship standing still, observe carefully how the little animals fly with equal speed to all sides of the cabin. The fish swim indifferently in all directions; the drops fall into the vessel beneath; and, in throwing something to your friend, you need throw it no more strongly in one direction than another, the distances being equal; jumping with your feet together, you pass equal spaces in every direction. When you have observed all these things carefully (though doubtless when the ship is standing still everything must happen in this way), have the ship proceed with any speed you like, so long as the motion is uniform and not fluctuating this way and that. You will discover not the least change in all the effects named, nor could you tell from any of them whether the ship was moving or standing still. In jumping, you will pass on the floor the same spaces as before, nor will you make larger jumps toward the stern than toward the prow even though the ship is moving quite rapidly, despite the fact that during the time that you are in the air the floor under you will be going in a direction opposite to your jump. In throwing something to your companion, you will need no more force to get it to him whether he is in the direction of the bow or the stern, with yourself situated opposite. The droplets will fall as before into the vessel beneath without dropping toward the stern, although while the drops are in the air the ship runs many spans. The fish in their water will swim toward the front of their bowl with no more effort than toward the back, and will go with equal ease to bait placed anywhere around the edges of the bowl. Finally the butterflies and flies will continue their flights indifferently toward every side, nor will it ever happen that they are concentrated toward the stern, as if tired out from keeping up with the course of the ship, from which they will have been separated during long intervals by keeping themselves in the air. And if smoke is made by burning some incense, it will be seen going up in the form of a little cloud, remaining still and moving no more toward one side than the other. The cause of all these correspondences of effects is the fact that the ship's motion is common to all the things contained in it, and to the air also. That is why I said you should be below decks; for if this took place above in the open air, which would not follow the course of the ship, more or less noticeable differences would be seen in some of the effects noted.

(I've added emphasis to two excerpts.)

$\endgroup$
0
$\begingroup$

You can rethink it as following: Inertial frame of reference can be viewed as a frame of reference where 2nd Newton law takes it form as we know it ${\bf F}=m{\bf a}$ (and vise-versa). In non-inertial frames of reference this equation differs by inertia forces.

That said, if you have two frames of reference, in which we have ${\bf F}=m{\bf a}$ and ${\bf F'}=m{\bf a'}$ respectively, these are both inertial. If you try to find transition between these inertial frames, you will find that only Galilean transforms (3 'boosts' + 3 'rotations'), spatial and time translations respect this transition.

Relating specifically to what bothers you: Forses are not supposed to be invariant under Galilean transforms (i.e. no such requirement as $F=F'=0$), but rather equation of motion is supposed to be invariant under them.

Let me know if that is still confusing.

$\endgroup$
2
  • $\begingroup$ How could the equations of motion be invariant under Galilean transforms if forces were not invariant? $\endgroup$ Commented Aug 31, 2022 at 17:41
  • $\begingroup$ Requiring $F=F'$ is different from requiring $F=ma -> F'=ma'$. In the case of e.g. rotational transform, your forces change direction; Therefore force have changed. At the same time, however, acceleration changed direction as well. Force is not preserved, equation of motion is preserved. P.S. Speaking of pure 'constant velocity boost' forces indeed get preserved. But that's more of a consequence of equation of motion invariance rather than physically independent fact. $\endgroup$ Commented Aug 31, 2022 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.