0

I'm a math student who's studying classical mechanics for the first time, so forgive me if this question sounds pedantic, but I find it hard to think about such problems without stating assumptions clearly.

If a frame K' is accelerating away from an inertial frame K, we wish to transform from the inertial to the non-inertial coordinates. Now assuming we know the displacement between the origins $\vec{R}$, we can write:

$$\vec{r} = \vec{r}{}' + \vec{R} \qquad \Longrightarrow \qquad \vec{r}{}' = \vec{r} - \vec{R}$$

This is usually accompanied by a drawing illustrating the vector algebra. However, don't we assume implicitly here that both frames will observe the same displacement? In other words, that displacement is invariant in all frames of reference (not just inertial ones)?

All references I am able to find giving an axiomatic approach to classical mechanics take great pains to emphasize that, in the Galilean relativity of classical mechanics, one can only assume quantities to be preserved under transformations of the Galilean group, which is then shown to exclude accelerating frames (translation velocity must be constant), e.g. in Arnold:

'The galilean group is the group of all transformations of a galilean space which preserve its structure (...) every Galilean transformation of the space R×R$^3$ can be written in a unique way as the composition of a rotation, a translation, and a uniform motion.'

So, does a non-inertial observer measure the same displacement? And if not, does it make sense to transform into that frame from our non-inertial displacement measurement?

Elio Fabri
  • 5,780
Othman El Hammouchi
  • 103

2 Answers2

1

Yes that is being implicitly assumed. In any non relativistic physics, length and time intervals are invariant under coordinate transformation.

Brain Stroke Patient
  • 1,931
1

The primary assumption that you are talking about: I like to call that the 'inertial dead reckoning' assumption.

So the metaphor is 'dead reckoning'.
Quick recapitulation of 'dead reckoning'.
Two ships are on a motionless body of water. The two ships depart from each other, both keep an exact log of their velocity and direction of velocity. The two ships can plan to rejoin at some future point in time. As long as a ship keeps an accurate log it can plot a course to the point of departure.

In theory of motion we have the following three tiers:
-position
-velocity
-acceleration

'Inertia dead reckoning' is analogous to dead reckoning, but the measurement is one tier higher. (The usual name for such navigation is: 'inertial guidance system')

Two spaceships in space, far enough from sources of gravity that gravitational effect is negligable. The two spaceships can depart from each other and plan to rejoin at some point in the future. By maintaining an accurate log of acceleration and direction of acceleration the two ships can at al times plot a course back to the rendez-vous point.


So there is this assumption that all of the space of the solar system is a uniform space. More specifically, we assume that the space presents itself as a perfectly uniform space because inertia is uniform throughout space.


We have the following corroborating evidence for the uniformity of inertia:

If inertia would be in any way non-uniform then planets of the Solar system will not move according to a simple motion law.

Wat we observe is that the planets of the solar system do move along highly regular orbits. The orbits of the planets of the Solar system are described to a high level of accuracy by the inverse square law of gravity.

Once planet-planet interactions are accounted for only an anomalous precession of Mercury's orbit remains (which is in the order of seconds of arc per century).


So:
For the time being we can count the Solar system as 'our system' (Because it is for the extent of the Solar system that we have observation of motion that can serve as corroborating evidence.)

We have strong corroborating evidence that inertia is uniform throughout the Solar system.

That evidence for the uniformity of inertia is the justification for relying on the Principle of Galilean relativity.

Cleonis
  • 20,795
  • What a great answer! As a follow-up: would this then imply that transforming into an accelerating frame is a galilean transform? Because all references I can find require the translation velocity to be constant. – Othman El Hammouchi Aug 10 '21 at 12:13
  • Or perhaps we should consider transition maps between galilean charts (leaving distance and time invariant, which includes transformations to non-inertial frames) as seperate from the more special galilean transformations of the galilean group, which additionally require the equations of motions to remain the same (thus excluding accelerating frames)? (all of this of course in the classical context) – Othman El Hammouchi Aug 10 '21 at 12:16