# What exactly is an inertial reference frame?

1. Jan 20, 2013

### Mangoes

Due to my job and other classes, I've been studying ahead of my class by myself to not fall behind and I'm not sure if I'm oversimplifying this in my head and not really grasping the idea.

Newton's First Law of Motion states that an object with a net force of zero stays in constant motion (or rest). So, if an object appears to be accelerating, there must be a net force acting on it.

The way I understand it, if I were to get in my car and accelerate to some object A, that object A would appear to be accelerating towards me, even though there isn't a force acting on it, and this is why Newton's First Law doesn't apply to accelerating reference frames. Am I oversimplifying this?

While I can understand how objects would seem to accelerate due to an accelerating reference frame and you wouldn't want that, what particularly confuses me is that we use the earth as a reference frame yet the earth accelerates around the sun.

Are we just using the earth as a good enough approximation of an inertial reference frame? Would it be correct in thinking that there isn't an absolute inertial reference frame?

2. Jan 20, 2013

### Philip Wood

Your understanding of inertial frames is, imo, excellent. And you're right about the Earth as an approximation to an inertial frame.

3. Jan 20, 2013

### vanhees71

First of all, one should be clear about the fact that the existence of a (globel) inertial frame (and then arbitrarily many inertial frames which all move uniformly against each other) is an assumption. It's part of Newton's postulate on space, according to which space is independently from all physical processes a three-dimensional Euclidean space, and time, which is a one-dimensional oriented (Euclidean) space, independent from all physical processes.

Then by definition, an inertial frame is one, in which the First Law, i.e., Galilei's principle of inertia holds: If there is no force acting on a body, it moves with constant velocity (note that velocity has to be understood as a vector quantity here).

Thinking a bit about the mathematics of how to calculate the space-time coordinates of a particle wrt. an inertial frame S' given its space-time coordinates wrt. another inertial frame S, you come to the Galileo transformations, which form a continuous Lie group of ten paramaters.

First of all, as a Euclidean affine space space is translation invariant, i.e., there is nothing in space that distinguishes one point from any other point, i.e., you can choose your origin of the spatial coordinate system, determining an inertial frame as you wish. The laws of nature must not change when doing such a transformation, i.e., if $\vec{x}$ is the position vector wrt. S, then you can introduce another system S', which is at rest wrt. to S and simply has it's origin shifted to $\vec{a}$ (wrt. S). Time stays as it is. Then the coordinates wrt. S' are
$$\vec{x}'=\vec{x}-\vec{a}, \quad t'=t.$$
Further, the same argument holds for time, i.e., there is no point in time distinguished from any other, so that you can choose your origin of the time axis as you wish:
$$t'=t-b, \quad b=\text{const}.$$
This is the translation group of space-time, which must be a symmetry of the natural laws within Newton's postulates.

In space you further are free to orient your coordinate system as you wish. Of course, the Cartesian coordinate systems are especially simple to use, so that we'll do this from now on. All distances and angles between vectors are fixed quantities and independent from any spatial rotation. A spatial rotation is described by an orthogonal Matrix $\hat{R}$ and is parametrized by three parameters (e.g., two angles for the orientation of the rotation axis and one angle of rotation around this axis, or the three Euler angles). If S' is simply given by a rigid rotation wrt. S, and if the new basis vectors are given by the rotation matrix $\hat{R}$, i.e., if
$$\vec{e}_j=\sum_{k=1}^3 R_{jk} \vec{e}_k'$$
then the coordinates of the position vector $\vec{x}$ wrt. the new Cartesian reference frame are given by
$$x_i'=\vec{e}_i' \cdot \vec{x}=\vec{e}_i' \cdot \sum_{j=1}^3 \vec{e}_j x_j = \sum_{j=1}^3 R_{ij} x_j.$$
For the coordinates we thus get
$$\vec{x}'=\hat{R} \vec{x}, \quad t'=t.$$
Since $\hat{R}$ must be an orthogonal matrix, we have the constraint
$$\hat{R}^{-1}=\hat{R}^T.$$
This leads to the rotations (and spatial reflections, which we however exclude from the symmetry group of space time because we are only interested in such transformations that can be reached by continuous operations).

Last but not least we can also let the frame S' move with constant velocity $\vec{v}$ wrt. S. Then the transformation, a "Galileo boost", reads
$$\vec{x}'=\vec{x}-\vec{v} t, \quad t'=t.$$
So the Galileo group is generated by all compositions of
-translations in space and time (4 parameters),
-rotations in space (3 parameters),
-Galileo boosts (3 parameters for the boost velocity $\vec{v}$).

All together we have indeed a 10-dimensional group, under which the Laws of motion are invariant. One can analyze this constraint with help of analytical mechanics, using variational principles (particularly Hamilton's principle of least action), which admits an elegant derivation the possible dynamical laws of motion given the constraints of the space-time symmetry. The most simple realization of such a dynamical law leads to Newton's 2nd and 3rd law.

4. Jan 20, 2013

### DiracPool

Although Einstein liked to talk about moving trains on Earth, he liked to think about elevators in outer space. And, in fact, that is where most of the thinking about inertial reference frames should take place when you are learning about it. To try to figure this stuff out in a cosmic sense initially using your rotating bedroom on earth as your frame of reference is working harder, not smarter. Start thinking about your 4-vectors in the intergalactic space between the milky way and the andromeda galaxy.