How do you check if a given force is Galilean invariant?

Question

If I have a given force like $$F_A = \dot{\vec{x}}_A- \dot{\vec{x}}_B = -F_B$$ or $$F_A = ((\vec x_A - \vec x_B) \times (\dot{\vec{x}}_A- \dot{\vec{x}}_B))^2 (\vec x_A - \vec x_B) = -F_B,$$ how do I check wether the force is Galilei invariant?

Answer 1

The inhomogeneous Galilei group contains space-time translations, space rotations and boosts (see this PSE post). In the following, I will not include space and/or time reflections, see this answer.

• Position differences (e.g., $x_A-x_B$, as in the original question) are invariant under space-time translations (because of the difference) and boosts (because positions are unaffected by them).

• Velocity differences (e.g., $\dot{x}_A-\dot{x}_B$) are invariant under space-time translations (velocities are unaffected by them) and boosts (because of the difference).

• For "exotic" force models (the force is a prescribed model to close the equations of motion) like the Abraham–Lorentz force you can also consider time derivatives of the velocity: they are space-time translation and boost invariants (the derivative kills the constant boost parameter).

Now you just have to deal with rotations. To this end, you can consider the two invariant symbols of the rotation group (also known as isotropic tensors: the Kronecker delta symbol and the Levi-Civita symbol) and use them to build Galilei-invariant models for the force.

Finally, it depends if you want to implement the action-reaction principle (you should, as it is equivalent to the conservation of total momentum that, in turn, derives from space translation invariance). Similarly for the conservation of energy, see this. Understanding this "construction" provides a fast "visual" way of checking the invariance of the force model at hand.

Example: $F_i \propto \epsilon_{ijk} {(x^j_A-x^j_B)}(v^k_A-v^k_B) \, G\left( (x^l_A-x^l_B)\delta_{lm}(x^m_A-x^m_B) \right)$ where $G$ is a scalar real function. For the reader who is not acquainted with the index notation, this is just a way of writing $F \propto (x_A-x_B)\times (v_A-v_B) \, G\left( |x_A-x_B|^2 \right)$. However, the action-reaction principle is broken: exchanging the labels $A$ and $B$ results in the same $F$ rather than $-F$.

Answer 2

In general, to check if any object $f$ is invariant under a given transformation $T$, the object should maintain the same value for any such transformation. In the case of Galilean transformations, which does a non-relativistic boost in any direction, mapping coordinate systems to each other like

$$(t,x,y,z)\to(t,\,x-v_xt,\,y-v_yt,\,z-v_zt),$$

you just need to check if this transformation changes $f$ at all for arbitrary $v_x$, $v_y$, $v_z$ at any $t$, $x$, $y$, $z$. In general, it will take being constant in space (which makes the spatial derivatives vanish, $\partial_xf=\partial_yf=\partial_zf=0$) to be Galilei invariant.

Intuitively you see this if you imagine the mass density of a planet - if you’re in one Galilei frame, you will see the planet’s mass density function as having certain shapes to it, which are guaranteed to change if you shift to another frame.

Answer 3

Define force in the new reference frame $F'$, making substitution $x'=x+vt$, then check for the resulting mapping- you should get the same force if it is Galilean invariant. For example in the first case calculate: $$ \begin{align} F_A' &= dx_A'-dx_B'\\ &=d(x_A+vt)-d(x_B+vt)\\ &=dx_A+v-dx_B-v\\ &=dx_A-dx_B\\ &=F_A \end{align} $$ which will show that force transfers between frames with no modifications, hence is Galilean invariant.

Bonus, check that second Newton law is also Galilean invariant, $$ \begin{align} F'&=m~\text {dd} x'\\ &=m~\text{dd}(x+vt)\\ &=m~\text{dd}x+ m~\text{dd}(vt)\\ &=m~\text{dd}x+ m~\text{d}v\\ &=m~\text{dd}x~~~~~~~~\scriptsize\text {(Since for inertial reference frame dv=0)}\\ &=F \end{align}$$

BTW, in Galilean invariancy test it is assumed that $t'=t$, since all inertial reference frames shares same universal time according to Newtonian relativity. I.e. time is assumed to be absolute.