Galilean transformation of non-inertial frame

In summary: As pointed out, it’s not actually clear what you mean by a Galilean transformation between two non-inertial frames.
  • #1
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It's frequently discussed Galilean transformation brings one inertial frame to another inertial frame, and such a transformation leaves Newton's second law invariant (of the same form). I wonder what happens for non-inertial frame? If we start with a non-inertial frame, and Galilean transform it, does it leave Newton's second law invariant?
 
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  • #2
I am not sure that “Galilean transform” even makes sense to apply to a non inertial frame. The Galilean transform is the transform between inertial frames.

However, if you simply mean ##x \rightarrow x+vt## for some spatial coordinate ##x##, then no, this will not in general preserve the same fictitious forces.

Consider a rotating frame in Cartesian coordinates. If ##x## is one of the in-plane coordinates then the fictitious forces get weird.
 
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  • #3
Dale said:
However, if you simply mean ##x \rightarrow x+vt## for some spatial coordinate ##x##, then no, this will not in general preserve the same fictitious forces.
The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?
 
  • #4
A.T. said:
The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?
Is not preserved either! If the net real force is ##\mathbf{F}##,
and in non-inertial frame ##S_1## the net inertial force is ##\mathbf{f}_1## and acceleration is ##\mathbf{a} |_{S_1}##
and in a different non-inertial frame the net inertial force is ##\mathbf{f}_2## and the acceleration is ##\mathbf{a}|_{S_2}##
then

##m \mathbf{a}|_{S_1} = \mathbf{F} + \mathbf{f}_1##
and
##m \mathbf{a}|_{S_2} = \mathbf{F} + \mathbf{f}_2##

so the difference in the net fictitious force due to the transformation is ##\Delta \mathbf{f} = m(\mathbf{a}|_{S_2} - \mathbf{a}|_{S_1}) \neq 0## in the general case.

(note that the notation ##\mathbf{a}|_{S} \equiv \ddot{x}_i \mathbf{e}_i## where ##x_i## are the coordinates of the frame ##S##).
 
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  • #5
Dale said:
I am not sure that “Galilean transform” even makes sense to apply to a non inertial frame. The Galilean transform is the transform between inertial frames.
If we take the definition from Arnold's

The galilean group is the group of all transformations of a galilean space which preserve its structure. The elements of this group are called galilean transformations. Thus, galilean transformations are affine transformations of A^4 which preserve intervals oftime and the distance between simultaneous events.​

This is a just a spacetime transformation that can transform one observer to another observer, no matter if the observer is inertial or not?
 
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  • #6
lriuui0x0 said:
It's frequently discussed Galilean transformation brings one inertial frame to another inertial frame, and such a transformation leaves Newton's second law invariant (of the same form). I wonder what happens for non-inertial frame? If we start with a non-inertial frame, and Galilean transform it, does it leave Newton's second law invariant?
I'm a bit unsure what you mean by "Galilean transform" of a non-inertial frame. I can make sense of the question, what happens when you change the inertial reference frame to describe the non-inertial frame. Then of course, nothing changes, because all inertial reference frames are equivalent due to Galilei invariance of Newtonian mechanics. What of course changes are the expressions of the inertial forces in terms of the non-inertial coordinates.
 
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  • #7
ergospherical said:
Is not preserved either! If the net real force is ##\mathbf{F}##,
and in non-inertial frame ##S_1## the net inertial force is ##\mathbf{f}_1## and acceleration is ##\mathbf{a} |_{S_1}##
and in a different non-inertial frame the net inertial force is ##\mathbf{f}_2## and the acceleration is ##\mathbf{a}|_{S_2}##
then

##m \mathbf{a}|_{S_1} = \mathbf{F} + \mathbf{f}_1##
and
##m \mathbf{a}|_{S_2} = \mathbf{F} + \mathbf{f}_2##

so the difference in the net fictitious force due to the transformation is ##\Delta \mathbf{f} = m(\mathbf{a}|_{S_2} - \mathbf{a}|_{S_1}) \neq 0## in the general case.

(note that the notation ##\mathbf{a}|_{S} \equiv \ddot{x}_i \mathbf{e}_i## where ##x_i## are the coordinates of the frame ##S##).
Actually given the transformation is Galilean, shouldn't the ##\mathbf{a}_1## and ##\mathbf{a}_2## be the same, up to an orthogonal transformation?
 
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  • #8
lriuui0x0 said:
Actually given the transformation is Galilean, shouldn't the ##\mathbf{a}_1## and ##\mathbf{a}_2## be the same, up to an orthogonal transformation?
As pointed out, it’s not actually clear what you mean by a Galilean transformation between two non-inertial frames.

Those two accelerations in the expression have velocity-dependent, position-dependent and angular-velocity-dependent terms, so for the vast majority of transformations they will not be the same.
 
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  • #9
lriuui0x0 said:
If we take the definition from Arnold's

The galilean group is the group of all transformations of a galilean space which preserve its structure. The elements of this group are called galilean transformations. Thus, galilean transformations are affine transformations of A^4 which preserve intervals oftime and the distance between simultaneous events.​

This is a just a spacetime transformation that can transform one observer to another observer, no matter if the observer is inertial or not?
It certainly isn’t clear to me how to apply that to a non-inertial frame
 
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  • #10
Dale said:
However, if you simply mean ##x \rightarrow x+vt## for some spatial coordinate ##x##, then no, this will not in general preserve the same fictitious forces.
A.T. said:
The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?
ergospherical said:
Is not preserved either! If the net real force is ##\mathbf{F}##,
and in non-inertial frame ##S_1## the net inertial force is ##\mathbf{f}_1## and acceleration is ##\mathbf{a} |_{S_1}##
and in a different non-inertial frame the net inertial force is ##\mathbf{f}_2## and the acceleration is ##\mathbf{a}|_{S_2}##
then

##m \mathbf{a}|_{S_1} = \mathbf{F} + \mathbf{f}_1##
and
##m \mathbf{a}|_{S_2} = \mathbf{F} + \mathbf{f}_2##

so the difference in the net fictitious force due to the transformation is ##\Delta \mathbf{f} = m(\mathbf{a}|_{S_2} - \mathbf{a}|_{S_1}) \neq 0## in the general case.
My question wasn't about the general case, but specifically about the transformation mentioned by @Dale in the part I quoted.
 
  • #11
A.T. said:
My question wasn't about the general case, but specifically about the transformation mentioned by @Dale in the part I quoted.
I still don't think it's that simple. Say two non-inertial frames have for example a common non-zero angular velocity ##\boldsymbol{\omega}##, but otherwise are related by a boost ##\mathbf{v}##.

The velocity-dependent and position-dependent terms in the acceleration ##\mathbf{a} |_{S}## of the particle will be completely different relative to the two frames.
 
  • #12
Well, e.g. the Coriolis-force is velocity dependent. So Galilei-boosting such an expression (i.e. Newton's 2nd law in a rotating frame) changes it.
 
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  • #13
A.T. said:
The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?
ergospherical said:
I still don't think it's that simple. Say two non-inertial frames have for example a common non-zero angular velocity ##\boldsymbol{\omega}##, but otherwise are related by a boost ##\mathbf{v}##.

The velocity-dependent and position-dependent terms in the acceleration ##\mathbf{a} |_{S}## of the particle will be completely different relative to the two frames.
The question was about their sum (net fictitious force).
 
  • #14
A.T. said:
The question was about their sum (net fictitious force).
Yeah, what I mean is that you have for this sum:\begin{align*}
\Delta \mathbf{f} = m(\mathbf{a}|_{S_2} - \mathbf{a}|_{S_1})
\end{align*}If ##\mathbf{a}^*## is the acceleration of the particle in an inertial frame, then the accelerations in the non-inertial frame take the form\begin{align*}
\mathbf{a}_S = \mathbf{a}^* - \mathbf{A} - 2\boldsymbol{\omega} \times \mathbf{v}|_{S} - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}|_{S})
\end{align*}so that, if ##S_1## and ##S_2## have identical angular velocity but are related purely by a boost in velocity (i.e. they have also identical linear acceleration ##\mathbf{A}##), then \begin{align*}\dfrac{1}{m} \Delta \mathbf{f} = -2\boldsymbol{\omega} \times (\mathbf{v}|_{S_2}-\mathbf{v}|_{S_1}) - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}|_{S_2}) + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}|_{S_1})
\end{align*}and it's clear that this difference in net fictitious force will not be zero even if the frames have identical angular velocity ##\boldsymbol{\omega}## and are related only by a boost in velocity.
 
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  • #15
Hmmmm...there must be something I understood wrongly on this (why Galilean transformation is not applicable to non-inertial frame). I will try to be a bit more explicit:

Let's say the world is the 4 dimensional space with Galilean structure defined in the Arnold's. A Galilean transformation is just an affine transformation. We say an observer is a particular world line, together with a continuous 3d orthonormal basis frame at each time. This definition of observer can be both inertial and non-inertial. And we can apply that Galilean transform function that transforms the worldline to different worldline, as a different observer.

My original question is whether Newton's laws have the same form (invariant) across these two potentially non-inertial observers.
 
  • #16
lriuui0x0 said:
Hmmmm...there must be something I understood wrongly on this (why Galilean transformation is not applicable to non-inertial frame).
I also don't see why not. I understood your question as described here:
Dale said:
However, if you simply mean ##x \rightarrow x+vt## for some spatial coordinate ##x##,
 
  • #17
The EoM in a rotating frame reads
$$m \ddot{\vec{r}}=\vec{F}(\vec{r})-2 m \vec{\omega} \times \dot{\vec{r}} -m \vec{\omega}\times (\vec{\omega} \times \vec{r}) - \dot{\vec{\omega}}\times \vec{r},$$
where all vectors are column vectors with the components referring to the rotating frame. Here, for simplicity I assume that the non-inertial frame (NIF) is only rotating but is not in addition in accelerated translation wrt. the inertial frame (IF), and the origins of both frames coincide. The more general case of also translationally accelerates NIFs is straight-forward.

If you now insert a formal Galilei transformations of these non-inertial components, ##\vec{r}'=\vec{r}-\vec{v} t## with ##\vec{v}=\text{const}## you immediately see that the EoM is not forminvariant. You get additional terms involving ##\vec{v}## from all the inertial forces even when you consider a free particle with ##\vec{F}=0##.
 
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  • #18
A.T. said:
I also don't see why not. I understood your question as described here:
you're ignoring that if the frames have some (common, say) angular velocity, then even if they're related by a formal Galilean transformation then the accelerations w.r.t. those two frames are no longer the same, c.f. the expression in #14.
 
  • #19
ergospherical said:
Yeah, what I mean is that you have for this sum:\begin{align*}
\Delta \mathbf{f} = m(\mathbf{a}|_{S_2} - \mathbf{a}|_{S_1})
\end{align*}If ##\mathbf{a}^*## is the acceleration of the particle in an inertial frame, then the accelerations in the non-inertial frame take the form\begin{align*}
\mathbf{a}_S = \mathbf{a}^* - \mathbf{A} - 2\boldsymbol{\omega} \times \mathbf{v}|_{S} - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}|_{S})
\end{align*}so that, if ##S_1## and ##S_2## have identical angular velocity but are related purely by a boost in velocity (i.e. they have also identical linear acceleration ##\mathbf{A}##), then \begin{align*}\dfrac{1}{m} \Delta \mathbf{f} = -2\boldsymbol{\omega} \times (\mathbf{v}|_{S_2}-\mathbf{v}|_{S_1}) - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}|_{S_2}) + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}|_{S_1})
\end{align*}and it's clear that this difference in net fictitious force will not be zero even if the frames have identical angular velocity ##\boldsymbol{\omega}## and are related only by a boost in velocity.
Maybe we are interpreting the boost in velocity differently. To me the v in the transformation below is constant in the initial non-inertial frame S1 (not in some inertial frame).
Dale said:
##x \rightarrow x+vt## for some spatial coordinate ##x##,
So for your initial description:
ergospherical said:
If the net real force is ##\mathbf{F}##,
and in non-inertial frame ##S_1## the net inertial force is ##\mathbf{f}_1## and acceleration is ##\mathbf{a} |_{S_1}##
and in a different non-inertial frame the net inertial force is ##\mathbf{f}_2## and the acceleration is ##\mathbf{a}|_{S_2}##
then

##m \mathbf{a}|_{S_1} = \mathbf{F} + \mathbf{f}_1##
and
##m \mathbf{a}|_{S_2} = \mathbf{F} + \mathbf{f}_2##

so the difference in the net fictitious force due to the transformation is ##\Delta \mathbf{f} = m(\mathbf{a}|_{S_2} - \mathbf{a}|_{S_1}) \neq 0## in the general case.
we have:

##\mathbf{a}|_{S_2} = \mathbf{a}|_{S_1}##

And thus:

##\Delta \mathbf{f} = 0##
 
  • #20
A.T. said:
Maybe we are interpreting the boost in velocity differently. To me the v in the transformation below is constant in the initial non-inertial frame S1 (not in some inertial frame).
##\mathbf{a}|_{S_2} = \mathbf{a}|_{S_1}##
##\Delta \mathbf{f} = 0##
I think this would indeed hold true, although it definitely wasn't what I had in mind. Then again, the OP didn't really specify... :)
 
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  • #21
But that's what I considered in #17, and if you substitute ##\vec{r}=\vec{r}'+\vec{v} t## everywhere in the given formula, you'll see that the fictitious forces contain ##\vec{v}## (the external real force ##\vec{F}## does so anyway, but that has nothing to do with using a NIF). This explicitly shows that even for free particles, where ##\vec{F}=0## the EoM for NIF coordinates is not forminvariant under Galilei transformations.
 
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  • #22
I think what A.T. might be thinking of is a situation where the frames have relative velocity ##\mathbf{v} + \boldsymbol{\omega} \times \mathbf{v}## as viewed from a non-rotating frame.
 
  • #23
ergospherical said:
I think this would indeed hold true, although it definitely wasn't what I had in mind. Then again, the OP didn't really specify... :)
To me it seems like the obvious interpretation: Take the description in some non-inertial frame, and apply ##x \rightarrow x+vt## to it. Since this preserves the coordinate accelerations, the sum of all inertial forces must also be preserved.

So the "correction" to Newton's 2nd Law is preserved under this type of transformation.

@lriuui0x0 Is this what you had in mind?
 
  • #24
A.T. said:
@lriuui0x0 Is this what you had in mind?

What I originally had in mind is below:

So we have an affine spacetime ##N^4##, the associated vector space ##V^4##, with the simultaneity subspace ##V^3##. An oberserver in my mind is the combination of a smooth worldline ##\gamma: \mathbb{R} \to N^4## and a smooth orthonormal basis assignment in ##\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3: \mathbb{R} \to V^3##. The Galilean transformation transforms this worldline and the basis assignment. It will be an affine map ##f: N^4 \to N^4##, with associated linear map ##g: V^4 \to V^4##, and ##g## can be applied to the orthonormal basis vector to let the observer "look at different direction".

Please help check if the above idea of transforming an observer makese sense!

For transforming among inertial observers, it will conceptually look like:

1637944849802.png

(The three big coordinate arrows was just there to make the drawing look better, they're not part of the entities under discussion).

I think this transformation has nothing specific to the observer being inertial, right? It can be similarly be applied to non-inertial observers, like:

1637945167814.png


The Newton's equation will have a particular form on the left, involving ficticious forces. My original question is will it be of the same form on the right? Based on the calculation above, ##\mathbf{a}_2 = \mathbf{a}_1##. Or at least ##\mathbf{a}_2 = R(\mathbf{a}_1)## with ##R## being a rotation, if we take the general transformation ##(x, t) \mapsto (Rx + vt + d, t + s)##. So I think the fictious force doesn't change apart from a rotation?

But there's also the question if the real force changes its form or not. I was thinking for gravity and Columb's force, since they depend on only the distance of particles, they will not change its form under Galilean transformation, at least if you start with an inertial observer. But I also want to confirm this and also what about non-inertial observer? If they don't change form, then can we conclude that the entire Newton's equation doesn't change form either?
 
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  • #25
lriuui0x0 said:
But there's also the question if the real force changes its form or not.
The net real force cannot change, as it's related to directly observable frame invariant proper acceleration (if we exclude gravity or treat it as inertial).
 
  • #26
vanhees71 said:
I'm a bit unsure what you mean by "Galilean transform" of a non-inertial frame. I can make sense of the question, what happens when you change the inertial reference frame to describe the non-inertial frame. Then of course, nothing changes, because all inertial reference frames are equivalent due to Galilei invariance of Newtonian mechanics. What of course changes are the expressions of the inertial forces in terms of the non-inertial coordinates.
Im not sure I understand the question either, but if you try to take a system of relative inertial observers with Galilei invariance, everyone will agree on net momentum and energy (I think that's right). But if you take a non inertial observer who makes measurements before and after the acceleration, they will not be able to come up with conservation of momentum or energy. Where as the inertial observers should, if they account for translation of potential to kinetic energy. For an accelerated observer, at least in the context in question, I think they would see a net change in energy and momenta of their "universe" that would be very different from what they measured wrt their frame of reference.
 
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  • #27
I think I started to understand why it's an unnatural thing to think about Galilean transformation on non-inertial observers, because the Galilean group naturally applies to the set of all inertial observers and is in one-to-one relationship to it (not sure what's the right math word here?).

I initially was thinking that since Galilean transformation is defined so fundamentally as the transformation that preserves the spacetime structure, it may be sensible for physical laws to be invariant under it. But I will think a bit more on this, and try to clarify what exactly my question is in my head...
 
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  • #28
valenumr said:
if you try to take a system of relative inertial observers with Galilei invariance, everyone will agree on net momentum and energy
Neither the total momentum nor total energy are frame invariant under Galilean transformations between inertial frames.
 
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  • #29
A.T. said:
Neither the total momentum nor total energy are frame invariant under Galilean transformations between inertial frames.
Um, that's not what I said said. You can start small, and realize two observers can compute the *net* energy and momentum of a system, by choosing one or the other to be at rest. It kind of extrapolates from there.
 
  • #30
A.T. said:
Neither the total momentum nor total energy are frame invariant under Galilean transformations between inertial frames.
You might consider a system of three observers in a single dimension. A is our chosen reference frame , with B moving left (negative velocity) and C moving right (positive velocity). Firstly, no one will agree on energy or momentum of A, B, C. But they will agree on the total energy and momentum of the system. Pick one reference frame as stationary and do the math...

So now consider that A is a spring loaded cannon, and shoots a cannon ball that is a small fraction of the mass. Now the cannon and the cannon ball have been accelerated. B and C, as well as the center of momentum frame A, will see that energy and momentum have been conserved. They will even all agree in the amount of potential energy released by the spring.

But trying to add up the energy or momentum of the whole universe, from the bullet or cannon frame after acceleration will not work. It will show that the net change in energy from the potential released by the cannon is not the same as the observed change in energy of every other massive observer.
 
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  • #31
valenumr said:
You might consider a system of three observers in a single dimension. A is our chosen reference frame , with B moving left (negative velocity) and C moving right (positive velocity). Firstly, no one will agree on energy or momentum of A, B, C.
So A,B,C are not just frames. but massive objects at rest in the frames A,B,C respectively?

valenumr said:
But they will agree on the total energy and momentum of the system.
No, the rest frames of A,B,C will not agree on the combined total momentum and energy of the objects A,B,C.
 
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  • #32
A.T. said:
So A,B,C are not just frames. but massive objects at rest in the frames A,B,C respectively?No, the rest frames of A,B,C will not agree on the combined total momentum and energy of the objects A,B,C.
That's correct, I worded that poorly. I meant they will agree on a net change in energy and momentum before and after. They would usually have different values for the totals.
 
  • #33
To be more clear, they would agree that momentum was conserved.
 
  • #34
valenumr said:
To be more clear, they would agree that momentum was conserved.
Yes, inertial frames have momentum conserved. And non-inertial frames don't have momentum conserved, because the inertial forces there don't obey Newtons 3rd Law.

We already established that applying a Galilean transformation to a non-inertial frame will preserve the total inertial force, so the amount of momentum conservation violation is also preserved.
 
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  • #35
A.T. said:
Yes, inertial frames have momentum conserved. And non-inertial frames don't have momentum conserved, because the inertial forces there don't obey Newtons 3rd Law.

We already established that applying a Galilean transformation to a non-inertial frame will preserve the total inertial force, so the amount of momentum conservation violation is also preserved.

A.T. said:
Yes, inertial frames have momentum conserved. And non-inertial frames don't have momentum conserved, because the inertial forces there don't obey Newtons 3rd Law.

We already established that applying a Galilean transformation to a non-inertial frame will preserve the total inertial force, so the amount of momentum conservation violation is also preserved.
After rereading all of this, is this basically how we do physics in Earth's gravitational frame? Or am I still misunderstanding the question?
 
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