# Proving Galilean Transformation for Inertial Frames

• I
• lriuui0x0

#### lriuui0x0

I know we can prove that a Galilean transformation sends one inertial frame to another inertial frame, by proving ##\frac{d^2 f(\vec{r})}{d(f(t))^2} = \frac{d^2 \vec{r}}{dt^2}##, but can we prove the reverse? Can we prove that if the acceleration seen in two frames are the same, then the transformation function ##f## must be a Galilean transformation?

Not all the Frame of References(FRs) that satisfy Galilean transformation are Inertial Frame of References (IFRs).
For an example say a toy train is running on the floor of an accelerating rocket with constant speed v. Train FR and Rocket FR satisfy Galilean transformation with speed v, but both the FRs are not IFRs.

Last edited:
Not all the Frame of References(FRs) that satisfy Galilean transformation are Inertial Frame of References (IFRs).
For an example say a toy train is running on the floor with constant speed v of an accelerating rocket. Train FR and Rocket FR satisfy Galilean transformation with speed v, but both the FRs are not IFRs.
That's not what I'm asking. I'm asking that we know a Galilean transformation brings one inertial frame to another inertial frame. However is converse true that given two inertial frames, they can always be related with a Galilean transformation? How can we prove this?

However is converse true that given two inertial frames, they can always be related with a Galilean transformation? How can we prove this?
Given two IFRs, one is moving with constant speed to the other where Galilean transformation applies.

However Newton's first laws of motion, also called the "law of inertia", and Galilean transformation are two things of different nature, IFRs are closed subset in Galilean transformation as a result.

Last edited:
That's not what I'm asking. I'm asking that we know a Galilean transformation brings one inertial frame to another inertial frame. However is converse true that given two inertial frames, they can always be related with a Galilean transformation? How can we prove this?
No. See e.g.

https://arxiv.org/abs/math-ph/0102011

In this paper the authors perform a finite general coordinate transformation (i.e. not an infinitesimal one to derive the algebra, but a finite to derive the group at once) on the action of a point particle, and solve the resulting differential equation stating that the Lagrangian should be kept invariant up to a total time derivative. It's been a while for me (so correct me if I'm wrong), but I'd say this group is the Schrodinger group.

To give a simple counter example: take the action of the point particle, and rescale the time as

$$t \rightarrow t'= z^{\alpha} t, \ \ \ \ x^i \ \rightarrow x^{'i} = z^{\beta} x^i$$.

and then solve for the coefficients ##\alpha## and ##\beta## by the demand that the action stays invariant. This transformation, called a dilation, is not in the Galilei-group. Another transformation would be the one given by a rescaling of the mass. This transformation is not described by a spacetime transformation, but by the action of the so-called central extension of the Galilei-algebra called the Bargmann algebra.

• • • dextercioby, andresB, vanhees71 and 1 other person

Also in electrodynamics, if you ask for the symmetry group of the free Maxwell equations (##\rho=0##, ##\vec{j}=0##) you also get a larger group than the Poincare group.

• dextercioby
No. See e.g.

https://arxiv.org/abs/math-ph/0102011

In this paper the authors perform a finite general coordinate transformation (i.e. not an infinitesimal one to derive the algebra, but a finite to derive the group at once) on the action of a point particle, and solve the resulting differential equation stating that the Lagrangian should be kept invariant up to a total time derivative. It's been a while for me (so correct me if I'm wrong), but I'd say this group is the Schrodinger group.

To give a simple counter example: take the action of the point particle, and rescale the time as

$$t \rightarrow t'= z^{\alpha} t, \ \ \ \ x^i \ \rightarrow x^{'i} = z^{\beta} x^i$$.

and then solve for the coefficients ##\alpha## and ##\beta## by the demand that the action stays invariant. This transformation, called a dilation, is not in the Galilei-group. Another transformation would be the one given by a rescaling of the mass. This transformation is not described by a spacetime transformation, but by the action of the so-called central extension of the Galilei-algebra called the Bargmann algebra.
Does that mean what's here on wikipedia is incorrect? https://en.wikipedia.org/wiki/Galilean_invariance#Newton's_theory_versus_special_relativity

If we additionally require the basis vector in the frame of reference as orthonormal, does that rule out the dilation transforms? Also I don't think relabeling mass counts here, we only talk about transformation on the spacetime.

Also in electrodynamics, if you ask for the symmetry group of the free Maxwell equations (##\rho=0##, ##\vec{j}=0##) you also get a larger group than the Poincare group.
What do you mean by that's for free particles only? I think inertial frames are defined to be the frames in which the free particles have no acceleration?

• vanhees71
No. See e.g.

https://arxiv.org/abs/math-ph/0102011

In this paper the authors perform a finite general coordinate transformation (i.e. not an infinitesimal one to derive the algebra, but a finite to derive the group at once) on the action of a point particle, and solve the resulting differential equation stating that the Lagrangian should be kept invariant up to a total time derivative. It's been a while for me (so correct me if I'm wrong), but I'd say this group is the Schrodinger group.

To give a simple counter example: take the action of the point particle, and rescale the time as

$$t \rightarrow t'= z^{\alpha} t, \ \ \ \ x^i \ \rightarrow x^{'i} = z^{\beta} x^i$$.

and then solve for the coefficients ##\alpha## and ##\beta## by the demand that the action stays invariant. This transformation, called a dilation, is not in the Galilei-group. Another transformation would be the one given by a rescaling of the mass. This transformation is not described by a spacetime transformation, but by the action of the so-called central extension of the Galilei-algebra called the Bargmann algebra.
I think with some further restrictions, the transformation can only be a Galilean transformation.

1) We require transformation preserves inertial frame, this means the transformation has to be affine at least. Say the spatial transformation is ##f(\mathbf{x}, t) = \mathbf{A}\mathbf{x} + \mathbf{B}t + C## and temporal transformation is ##g(\mathbf{x}, t) = \mathbf{a}\mathbf{x} + bt + c##.

2) If we then require time intervals of events to be the same in every frame, then it must be ##\mathbf{a} = 0, b = 1##.

3) If we require that two events at the same time (##\mathbf{B}t## cancels) have the same Euclidean distance, then it must be that ##\mathbf{A}## is orthogonal.

Thus we arrive at the general Galilean transformation, with ##\mathbf{b}, C, c## being the degree of freedom. If we change 2) and 3) to be the constancy of light speed, then we get Poincare group.

Does that mean what's here on wikipedia is incorrect?
Wikipedia should not be taken as an authoritative source, so one should not be surprised that something on Wikipedia might be incorrect.

Does that mean what's here on wikipedia is incorrect? https://en.wikipedia.org/wiki/Galilean_invariance#Newton's_theory_versus_special_relativity

If we additionally require the basis vector in the frame of reference as orthonormal, does that rule out the dilation transforms? Also I don't think relabeling mass counts here, we only talk about transformation on the spacetime.
Well, that depends what they mean exactly by "an inertial frame". Is switching coordinates also switching inertial frame? Is performing a dilation "switching coordinates frames"? As Peter says, I wouldn't spend too much time on the precise meanings of Wikipedia statements ;)

What do you mean by that's for free particles only? I think inertial frames are defined to be the frames in which the free particles have no acceleration?
That's of course true, and indeed the Galilei group in classical and the central extension of its covering group in quantum mechanics, is a symmetry of the free particle, but if you consider the full symmetry group of the resulting Hamiltonian, you find a larger group with the Galilei group as its subgroup. Something similar also happens for the Kepler problem, which has a larger symmetry and more conserved quantitities than the space-time symmetry and the related Noether conservation laws of this group. There's an additional dynamical symmetry, leading to the additional conservation law for the Runge-Lenz vector.

Also in electrodynamics, if you ask for the symmetry group of the free Maxwell equations (##\rho=0##, ##\vec{j}=0##) you also get a larger group than the Poincare group.
Yes, that's the conformal group. Its non-relativistic counterpart is the Schrodinger group. However, there's a subtlety here. You can perform an Inönü-Wigner contraction on the Poincaré algebra, giving you the Galilei algebra (if you want to obtain bargmann, you have to add a u(1) part to the algebra). But you can't obtain the Schrodinger algebra likewise from the conformal algebra. The physical reason is obvious: massless representations are, nonrelativistically speaking, problematic. There is a so-called Galilean conformal algebra, which hower does not contain the central extension of the bargmann algebra. It does play a role in nonrelativistic versions of ads/cft; see e.g. papers by Bagchi and Gopakumar.

In the past I tried to obtain a nonrelativistic "superconformal tensor calculus" by gauging the galilean conformal algebra to obtain a non-relativistic theory of supergravity. That doesn't work, and leads you to the (supersymmetric) Schrodinger algebra. But now I'm starting to reminisce about my good old PhD period, so let me stop there.

• vanhees71
Well, that depends what they mean exactly by "an inertial frame". Is switching coordinates also switching inertial frame? Is performing a dilation "switching coordinates frames"? As Peter says, I wouldn't spend too much time on the precise meanings of Wikipedia statements ;)
Good comment on the wikipedia statement :)

The definition of inertial frame I have in mind is "the frame in which particle with no force experiences no acceleration". In under this definition, a dilated frame is still an inertial frame.

Do you think the reasoning I had with restrictions on inertial frames is correct?

• vanhees71
Good comment on the wikipedia statement :)

The definition of inertial frame I have in mind is "the frame in which particle with no force experiences no acceleration". In under this definition, a dilated frame is still an inertial frame.

Do you think the reasoning I had with restrictions on inertial frames is correct?
Yes. It would be a good exercise to solve the constraints yourself when you applied the most general transformation to Newton's 2nd law and see what you get.

• vanhees71