so my books says

The transformation that allows us to go from one inertial frame $O$ with coordinates $x_i$ to another inertial frame $O'$ with coordinates $x_i'$ is the Galilean transformation. If the relative velocity of the two frames is given to be $\vec{v}=const$ and their relative orientations are specified by three angles $\alpha, \beta$ and $\gamma$, the new coordinates are related to the old ones by $x_i \to > x_i' = \mathbf{R}_{ij} x_j - v_it$, where $\mathbf{R}=\mathbf{R}(\alpha, \beta, \gamma)$ is the rotation matrix. In newtonian physics, the time coordinate is assumed to be absoute, i.e. it is the same in every coordinate framge.

For the Galilean transformation, we are mainly interested in coordinate transformations among inertial frames with the same orientation, $\mathbf{R}(0,0,0)_{ij}=\delta_{ij}$. Such a transformation is called a (Galilean) boost:

\begin{align} x_i \to x_i' &= x_i - v_it \tag{1.1}\\ t \to t' &= t \tag{1.2} \end{align}

and it also states the velocity addition rule

$u_i \to u_i' = u_i - v_i \tag{2}$

Now I wanted to solve the following problem:

Newtonian relativity: Consider a few definite examples of Newtonian mechanics that are unchanged by the Galilean transformation of (1.1, 1.1) and (2):

Show that force as given by the product of acceleration and mass $\vec{F}=m\vec{a}$, as well as a force law such as Newton's law of gravity $\vec{F}=G_n\frac{m_1m_2}{r^2}\hat{r}$ remain the same in every inertial frame.

What I did is

  1. $F=ma=m\frac{d^2}{dt^2}x=m\frac{d^2}{dt'^2}(x'+vt)=m\frac{d^2}{dt'^2}x'=ma'$
  2. Now assume $v$ only acts in x direction. I want to show that the $\vec{F}=G_n m_1m_2 \frac{1}{r^2}\hat{r}$ is the same in every inertial frame. So basically I want to show $\frac{1}{r^2}\hat{r} = \frac{1}{r'^2}\hat{r'}$ resp. $\frac{\vec{r}}{r^3}=\frac{\vec{r'}}{r'^3}$

$\frac{\vec{r}}{r^3}=\frac{1}{\sqrt{x^2+y^2+z^2}^3}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\frac{1}{\sqrt{(x'+vt)^2+y'^2+z'^2}^3}\begin{pmatrix} x'+vt \\ y' \\ z' \end{pmatrix}$

Now 1. works great. Everything seems fine but 2. doesn't work and I don't understand why it should work to begin with.

Now 1. basically tells me: If I measure the force of a spring at home or in my car that moves with a constant velocity: I get the same result.

Now what does 2. tell me? I interpret it like that: Assume we have a mini-earth that we can take with us. We take it to our room. We measure the gravitational force acting on some point mass in a specific location $\vec{x_0}$. We get some value. We then drive around in our car, again with a constant velocity and measure again: Same result.

Now I think what I do in my attempt to proof 2. is fixing the earth. I leave it at home, sit in my car, drive around and try to measure stuff.

I'm very very unsure though so I opted for a simpler example: Hook's Law.

We have $F_h=-kx$. It's easy to imagine: If I squeeze my spring for $x=1m$ I get some kind of force that pushes back. It doesn't matter if I do that at home or in the car, so let's try to show that:

$F_h=-kx=-k(x'+vt)\neq F_h' = -kx' \tag{3}$

So again, in 3 we basically leave the spring at home, no?

How do I show that a force law doesn't change under Galilean transformation?


1 Answer 1


I see two problems in your question. The first is that the force laws in both of your examples make certain assumptions: the quantities ($\mathbf{r}$ in the first example and $x$ in the second example) are not absolutes, they are distances measured from some origin which represents something physical. In the case of the gravitational force, the centre usually represents a "larger" mass (like the Sun in the solar system), while in the case of Hooke's law the centre represents the point at which the spring is attached.

So the actual force laws that you should be considering (in the general case) are the following:

  1. Gravitational force: consider two masses $m_1$ and $m_2$ at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ respectively. Then the force on $1$ due to $2$ is: $$\mathbf{F}_{12} = -\frac{Gm_1 m_2}{(\mathbf{r}_1 - \mathbf{r}_2)^2}\hat{\mathbf{r}}_{12}$$

  2. The force on the mass attached to the end of the spring is $F = - k(x-x_0)$ where $x_0$ is the position of the point at which the spring is attached.

Secondly, your interpretation of Galilean Invariance is not strictly correct. When you do a Galilean boost, you should imagine looking at the entire system as a whole from a moving car (or rocket, if you prefer). Therefore, you must boost all the relevant quantities, otherwise you're just giving one of them a velocity, and not the other, and that's certainly not the same thing as observing the system from a different inertial frame. That you are doing this is reflected in the fact that your answers aren't showing the equations to be invariant!

Let's look at the second example to see this properly: if you observe the spring from a different frame of reference, the point at which it is fixed will also appear to be moving, and therefore you would have: \begin{align}x' &= x - vt\\x_0' &= x_0 - v t,\end{align} so that $x'-x_0' = x - x_0$, and you can easily show that the force law is invariant.

See also this very similar question: Galilean invariance of Newton's universal gravitation law.

  • $\begingroup$ After thinking about it for ages I came up with I think what you pointed out. 1. I didn't consider intervals! But I should consider intervals! Leading to what you said above. 2. The Galilean boost is "I take the spring with in my car" and not "I look at my experiment in my room from my car", that's because of 1. - we boost the origin as well! So thanks a lot for clarification, I'll compute everything later today. :) $\endgroup$
    – handy
    Commented Apr 11, 2021 at 12:48

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