# Derive Galilean transformation. (The meaning of the relativity)

In the book The meaning of the relativity Einstein says that in classic mechanics two postulates are previously supposed:

1.- The time is absolute.

2.- The longitude is absolute.

And this implies that two inertial frames are related by a transformation: $$\left\{\begin{array}{lll} x'=x-a+vt\\ t'=t+\lambda \end{array}\right.$$ called Galilean transformation. How I can prove this implication? Is trivial?

• You can not prove it, since it's not true. What you can do is to take a crude clock and yard stick around your lab or the entire planet and make measurements, which will tell you that the above relationship does, indeed, hold within the precision of your crude equipment. If you repeat the same measurements with better equipment, then you will find that the relationship does not hold. Apr 7, 2016 at 19:06
• @CuriousOne: "prove" means "given a set of axioms, show that a conclusion is true". It does not mean "show that conclusion corresponds to the physical world". The OP gives the postulates and is asking to show logical equivalence of two statements. Apr 7, 2016 at 19:20
• @barrycarter: That's a common misconception that is usually voiced by people who have done neither science nor mathematics. The two are about as related as literature and calligraphy. Apr 7, 2016 at 20:38
• And what I said is that all physics are machines that are self-explanatory because their form and function are identical. In contrast all mathematics is just symbols on paper that have no meaning whatsoever without a reader. When the last person is dead who understood them, they will be meaningless. You don't have to trust me about that. Just ask the mathematicians who are dealing with mathematical logic. :-) Apr 8, 2016 at 3:44
• "aaabbabbbaababbaaa" is a valid mathematical theorem. Please explain its meaning to me. OTOH, if I put a simple machine made from a pivoted lever with two unequal but balanced masses at the ends on the town square, everybody with a minimum of physics knowledge understands that not only was Kilroy here, but that he also knew about the concepts of mass, weight, force, work and energy conservation. A future alien excavating the remains of the LHC will be able to tell its approximate particle energy, the particle types it could accelerate and they could derive that we were hunting for the Higgs. Apr 8, 2016 at 17:50

The question is: how does one derive the Galilean transform from the requirement that "time" and "length" (not "longtitude") be invariant. The comments by Einstein first need to be made more precise.

To do so, we'll consider coordinate differentials $$d𝐫 = (dx,dy,dz)$$ and $$dt$$ and the velocity $$𝐯$$ and ask what transforms they may have; and we'll focus primarily on the infinitesimal version of the transforms and then, afterwards, convert them to finite form.

For compatibility, we require the following for the transform of the velocity: $$Δ(d𝐫 - 𝐯 dt) = Λ(d𝐫 - 𝐯 dt),$$ with a Lagrange multiplier $$Λ$$ as a $$3×3$$ matrix, to ensure that the kinematic relation $$d𝐫 = 𝐯 dt$$ remains invariant.

The statements are then:

• Time is absolute: $$Δ(dt) = 0$$.
• Length is absolute: $$Δ\left(|d𝐫 - 𝐯 dt|^2\right) = 0$$.
The context makes it clear that he is referring to lengths in motion. There really is no length invariant in the space-time geometry of non-relativistic theory, because the lengths can move. The actual invariants that go with the Galilean transform are: $$dt, \quad dt \frac{∂}{∂t} + dx \frac{∂}{∂x} + dy \frac{∂}{∂t} + dz \frac{∂}{∂z}, \quad ∇² ≡ \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2.$$ To be able to talk about a length invariant, you need to include the motion in the length so you can track the length. This is why the velocity $$𝐯$$ is also included with the coordinate differentials $$d𝐫$$ and $$dt$$.

From the second of his statements, we get: $$(d𝐫 - 𝐯 dt)^TΛ(d𝐫 - 𝐯 dt) = 0,$$ which means $$Λ$$ is an anti-symmetric $$3×3$$-matrix. Therefore, the vector-matrix product, without loss of generality, can be written as: $$Λ(d𝐫 - 𝐯 dt) = 𝛚×(d𝐫 - 𝐯 dt).$$ This reduces the compatibility condition to: $$𝛚×(d𝐫 - 𝐯 dt) = Δ(d𝐫) - (Δ𝐯) dt - 𝐯 Δ(dt).$$ From the first of his statements, this may be reduced further to: $$𝛚×d𝐫 - 𝛚×𝐯 dt = Δ(d𝐫) - (Δ𝐯) dt,$$ or $$(Δ𝐯 - 𝛚×𝐯) dt = Δ(d𝐫) - 𝛚×d𝐫.$$

Define $$𝛖 ≡ Δ𝐯 - 𝛚×𝐯.$$ Then, our infinitesimal transforms become: $$Δ𝐯 = 𝛚×𝐯 + 𝛖, \quad Δ(d𝐫) = 𝛚×d𝐫 + 𝛖 dt, \quad Δ(dt) = 0.$$ Finally, integrate the coordinate differentials: $$Δ𝐫 = 𝛚×𝐫 + 𝛖 t + 𝛆, \quad Δt = τ.$$

The constants of integration $$𝛆$$ and $$τ$$ may be recognized as the infinitesimal forms, respectively, of spatial translation and time translation. The other parameters $$𝛚$$ and $$𝛖$$ are the infinitesimal forms, respectively, of spatial rotation and the Galilean boost; which is in reverse, since it's more usual to use $$-𝛖$$ in place of $$𝛖$$.

How does one convert this to finite form? Write it out as a Taylor expansion. This process is known as "exponentiation". For an infinitesimal transform $$ΔQ$$ of a quantity $$Q$$, the finite form is the transform: $$Q → \exp(λΔ)Q ≡ Q + λΔQ + \frac{λ^2}{2!}Δ^2Q + \frac{λ^3}{3!}Δ^3Q + ⋯.$$ Ignore the spatial rotations, for the moment. Then the series will be finite, since $$Δ(𝐫,𝐯,t) = (𝛖 t + 𝛆, 𝛖, τ), \quad Δ^2(𝐫,𝐯,t) = (𝛖 Δt, 𝟬, 0) = (𝛖 τ, 𝟬, 0), \quad Δ^3(𝐫,𝐯,t) = (𝟬, 𝟬, 0).$$ Therefore, $$(𝐫,𝐯,t) → (𝐫,𝐯,t) + (𝛖 t + 𝛆, 𝛖, τ)λ + (𝛖 τ, 𝟬, 0)\frac{λ^2}{2} = (𝐫 + 𝐮 t + 𝐚, 𝐯 + 𝐮, t + b),$$ where the finite forms of the transform parameters are given here by $$(𝐮, 𝐚, b) ≡ \left(λ𝛖, λ𝛆 + \frac{λ^2}{2}𝛖τ, λτ\right).$$

That's the Galilean transform you were asking for: the boosts and the translations together.

You could do this with the rotations woven into the mix, but it's a little easier to do it separately, to just look at the transform on $$𝐫$$ (since the case for $$𝐯$$ is similar) and on $$t$$, and (without loss of generality) assume the infinitesimal rotation is along the $$z$$ axis so that you can write it as: $$Δ((x,y,z),t) = Δ(𝐫,t) = (𝛚×𝐫,0) = ((0,0,ω)×(x,y,z),0) = ((-ωy,ωx,0),0).$$ Therefore, $$Δ(x,y) = (-ωy,ωx), \quad Δ^2(x,y) = (-ω^2x,-ω^2y), \quad Δ(z,t) = (0,0).$$ The Taylor series for the transforms on $$(z,t)$$ is trivial: $$(z,t) → (z,t),$$ while the Taylor series for the transforms on $$(x,y)$$ wraps around by the second order, leading to: $$Δ^{2n}(x,y) = (-1)^nω^{2n}(x,y), \quad Δ^{2n+1}(x,y) = (-1)^nω^{2n+1}(-y,x) \quad (n = 0, 1, 2, ⋯).$$ Thus \begin{align} (x,y) &→ (x,y)\left(1 - \frac{(ωλ)^2}{2!} + \frac{(ωλ)^4}{4!} - \frac{(ωλ)^6}{6!} + ⋯\right) + (-y,x)\left(ωλ - \frac{(ωλ)^3}{3!} + \frac{(ωλ)^5}{5!} - ⋯\right)\\ &= (x,y)\cos{ωλ} + (-y,x)\sin{ωλ}\\ &= (x\cos{θ} - y\sin{θ}, y\cos{θ} + x\sin{θ}), \end{align} where the finite form of the transform parameter is given by $$θ = ωλ.$$ Thus, $$(x, y, z, t) → (x\cos{θ} - y\sin{θ}, y\cos{θ} + x\sin{θ}, z, t).$$ That's also part of the Galilean transform. He excluded it, explicitly, by saying "If the axes ... are parallel".

Finally, you can do the whole transform together, in a similar way, rather than piecemeal; but I won't do the analysis here.

The relativistic form of these two invariants comes out of the transforms, starting with: $$Δdt = -\frac{𝞄·d𝐫}{c^2}, \quad Δd𝐫 = 𝞈×d𝐫 - 𝞄dt.$$ Again, the transform $$Δ𝐯$$ is found by imposing the same condition as before; but this time it leads to the following: $$Δ𝐯 = 𝞈×𝐯 - 𝞄 + \frac{𝞄·𝐯𝐯}{c^2}.$$ The invariants, it ultimately turns out, are: $$\frac{|d𝐫 - 𝐯dt|^2 - |𝐯×d𝐫|^2/c^2}{1 - |𝐯|^2/c^2} = |d𝐫 - 𝐯dt|^2 - \frac{(𝐯·(d𝐫 - 𝐯dt))^2}{c^2 - |𝐯|^2},\\ \frac{dt - 𝐯·d𝐫/c^2}{\sqrt{1 - |𝐯|^2/c^2}} = \sqrt{1 - \frac{|𝐯|^2}{c^2}} dt - \frac{1}{c}\frac{𝐯·(d𝐫 - 𝐯dt)}{\sqrt{c^2 - |𝐯|^2}}.$$