- #1

lriuui0x0

- 101

- 25

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter lriuui0x0
- Start date

- #1

lriuui0x0

- 101

- 25

- #2

Dale

Mentor

- 33,856

- 11,549

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.

- #3

A.T.

Science Advisor

- 11,753

- 3,033

The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?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.

- #4

ergospherical

- 888

- 1,222

Is not preserved either! If the net real force is ##\mathbf{F}##,The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?

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##).

- #5

lriuui0x0

- 101

- 25

If we take the definition from Arnold'sI 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.

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?

- #6

- 22,463

- 13,378

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.

- #7

lriuui0x0

- 101

- 25

Actually given the transformation is Galilean, shouldn't the ##\mathbf{a}_1## and ##\mathbf{a}_2## be the same, up to an orthogonal transformation?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##).

- #8

ergospherical

- 888

- 1,222

As pointed out, it’s not actually clear what you mean by a Galilean transformation between two non-inertial frames.Actually given the transformation is Galilean, shouldn't the ##\mathbf{a}_1## and ##\mathbf{a}_2## be the same, up to an orthogonal transformation?

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.

- #9

Dale

Mentor

- 33,856

- 11,549

It certainly isn’t clear to me how to apply that to a non-inertial frameIf 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?

- #10

A.T.

Science Advisor

- 11,753

- 3,033

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)?

My question wasn't about the general case, but specifically about the transformation mentioned by @Dale in the part I quoted.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.

- #11

ergospherical

- 888

- 1,222

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}##.My question wasn't about the general case, but specifically about the transformation mentioned by @Dale in the part I quoted.

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

haushofer

Science Advisor

- 2,731

- 1,200

- #13

A.T.

Science Advisor

- 11,753

- 3,033

The individual fictitious forces will not be preserved, but what about their sum (net fictitious force)?

The question was about their sum (net fictitious force).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.

- #14

ergospherical

- 888

- 1,222

Yeah, what I mean is that you have for this sum:\begin{align*}The question was about their sum (net fictitious force).

\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.

- #15

lriuui0x0

- 101

- 25

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

A.T.

Science Advisor

- 11,753

- 3,033

I also don't see why not. I understood your question as described here:Hmmmm...there must be something I understood wrongly on this (why Galilean transformation is not applicable to non-inertial frame).

However, if you simply mean ##x \rightarrow x+vt## for some spatial coordinate ##x##,

- #17

- 22,463

- 13,378

$$m \ddot{\vec{r}}=\vec{F}(\vec{r})-2 m \vec{\omega} \times \dot{\vec{r}} -m \vec{\omega}\times (\vec{\omega} \times \vec{r}) - \dot{\vec{\omega}}\times \vec{r},$$

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##.

- #18

ergospherical

- 888

- 1,222

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.I also don't see why not. I understood your question as described here:

- #19

A.T.

Science Advisor

- 11,753

- 3,033

Maybe we are interpreting the boost in velocity differently. To me theYeah, 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.

So for your initial description:##x \rightarrow x+vt## for some spatial coordinate ##x##,

we have: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.

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

And thus:

##\Delta \mathbf{f} = 0##

- #20

ergospherical

- 888

- 1,222

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... :)Maybe we are interpreting the boost in velocity differently. To me thevin the transformation below is constant in the initial non-inertial frameS(not in some inertial frame)._{1}

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

##\Delta \mathbf{f} = 0##

- #21

- 22,463

- 13,378

- #22

ergospherical

- 888

- 1,222

- #23

A.T.

Science Advisor

- 11,753

- 3,033

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.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... :)

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

@lriuui0x0 Is this what you had in mind?

- #24

lriuui0x0

- 101

- 25

@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:

(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:

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?

Last edited:

- #25

A.T.

Science Advisor

- 11,753

- 3,033

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).But there's also the question if the real force changes its form or not.

- #26

valenumr

- 467

- 191

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.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.

- #27

lriuui0x0

- 101

- 25

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...

- #28

A.T.

Science Advisor

- 11,753

- 3,033

Neither the total momentum nor total energy are frame invariant under Galilean transformations between inertial frames.if you try to take a system of relative inertial observers with Galilei invariance, everyone will agree on net momentum and energy

- #29

valenumr

- 467

- 191

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.Neither the total momentum nor total energy are frame invariant under Galilean transformations between inertial frames.

- #30

valenumr

- 467

- 191

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...Neither the total momentum nor total energy are frame invariant under Galilean transformations between inertial frames.

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.

- #31

A.T.

Science Advisor

- 11,753

- 3,033

So A,B,C are not just frames. but massive objects at rest in the frames A,B,C respectively?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.

No, the rest frames of A,B,C will not agree on the combined total momentum and energy of the objects A,B,C.But they will agree on the total energy and momentum of the system.

Last edited:

- #32

valenumr

- 467

- 191

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.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.

- #33

valenumr

- 467

- 191

To be more clear, they would agree that momentum was conserved.

- #34

A.T.

Science Advisor

- 11,753

- 3,033

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.To be more clear, they would agree that momentum was conserved.

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.

- #35

valenumr

- 467

- 191

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?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.

Share:

- Replies
- 14

- Views
- 689

- Last Post

- Replies
- 3

- Views
- 998

- Last Post
- Mechanics

- Replies
- 3

- Views
- 497

- Replies
- 94

- Views
- 4K

- Replies
- 2

- Views
- 408

- Replies
- 14

- Views
- 1K

- Replies
- 12

- Views
- 2K

- Replies
- 8

- Views
- 420

- Last Post

- Replies
- 8

- Views
- 567

- Replies
- 7

- Views
- 463