Covariance of Newton's 2nd Law under Galilean boosts

[Shirish]

I'm reading a section in a textbook on the explanation of covariance of Newton's 2nd law under Galilean boosts. It's explained that $\mathbf{a}=\mathbf{a'}$ (where we're considering two frames $S$ and $S'$ moving inertially w.r.t. each other). Mass is assumed to not vary across the frames, i.e. $m=m'$: if an observer in $S$ claims that the mass of some particle is $m$, then an observer in $S'$ also claims the same mass.

So essentially the RHS of Newton's 2nd law is invariant between $S$ and $S'$: $m\mathbf{a}=m'\mathbf{a'}$.

Do we implicitly assume here that a given force acting on a given body does not change depending on the frame of reference? Without this assumption, i.e. $\mathbf{F}=\mathbf{F'}$, I don't see how we can claim that Newton's 2nd law is covariant.

 

[PeroK]

I'm reading a section in a textbook on the explanation of covariance of Newton's 2nd law under Galilean boosts. It's explained that $\mathbf{a}=\mathbf{a'}$ (where we're considering two frames $S$ and $S'$ moving inertially w.r.t. each other). Mass is assumed to not vary across the frames, i.e. $m=m'$: if an observer in $S$ claims that the mass of some particle is $m$, then an observer in $S'$ also claims the same mass.

So essentially the RHS of Newton's 2nd law is invariant between $S$ and $S'$: $m\mathbf{a}=m'\mathbf{a'}$.

Do we implicitly assume here that a given force acting on a given body does not change depending on the frame of reference? Without this assumption, i.e. $\mathbf{F}=\mathbf{F'}$, I don't see how we can claim that Newton's 2nd law is covariant.

If we assume we know what we mean by the mass of the particle and we can measure $\mathbf a$, then $\mathbf F$ is defined by Newton's second law.

The point of this exercise, I guess, is to show that if $\mathbf F$ is defined as $\mathbf F = m \mathbf a$, then $\mathbf F$ is invariant and the second law is covariant.

 

[Shirish]

If we assume we know what we mean by the mass of the particle and we can measure $\mathbf a$, then $\mathbf F$ is defined by Newton's second law.

The point of this exercise, I guess, is to show that if $\mathbf F$ is defined as $\mathbf F = m \mathbf a$, then $\mathbf F$ is invariant and the second law is covariant.

Sorry if I sound rude, but are we sure about that? If we define the force as $m\mathbf{a}$, then let's say we shift to any other non-inertial reference frame $S''$, under whatever transformation, such that $\mathbf{a}\neq\mathbf{a''}$.

Even in that case, an observer in $S''$ can say that $\mathbf{F''}$ is defined as $\mathbf{F''}=m\mathbf{a''}$, and so Newton's second law has the same form in $S''$ => it should be covariant across all frames and should be a universal physical law (which, from what I've studied, it isn't outside of inertial reference frames).

 

[PeroK]

Sorry if I sound rude, but are we sure about that? If we define the force as $m\mathbf{a}$, then let's say we shift to any other non-inertial reference frame $S''$, under whatever transformation, such that $\mathbf{a}\neq\mathbf{a''}$.

Even in that case, an observer in $S''$ can say that $\mathbf{F''}$ is defined as $\mathbf{F''}=m\mathbf{a''}$, and so Newton's second law has the same form in $S''$ => it should be covariant across all frames and should be a universal physical law (which, from what I've studied, it isn't outside of inertial reference frames).

I don't have access to your textbook, so I can't say exactly what the author is assuming as postulates or definitions. That's why I used the phrase "I guess".

One way out of this is to define inertial references frames (IRF) as special. Then the force on a particle, for example, is defined by what you measure in an IRF. And, you show that the force you measure in one IRF is the same in them all.

I have no way of knowing how your book intends to proceed.

 

[Shirish]

I don't have access to your textbook, so I can't say exactly what the author is assuming as postulates or definitions. That's why I used the phrase "I guess".

One way out of this is to define inertial references frames (IRF) as special. Then the force on a particle, for example, is defined by what you measure in an IRF. And, you show that the force you measure in one IRF is the same in them all.

I have no way of knowing how your book intends to proceed.

The section I'm reading is "Coordinate transformations and the principle of covariance". Then it explains two specific types of Galilean transformations: boosts and rotation about the $z$-axis. Till this point, force isn't mentioned anywhere, just the transformation matrices are derived which is standard stuff.

Immediately after there's a subsection on "Form invariance". I'll quote:

The idea of form invariance can be illustrated using the Galilean transformation, because acceleration is invariant under that transformation: $\mathbf{a'}\equiv(d^2/dt'^2)\mathbf{r'}=(d^2/dt^2)(\mathbf{r}-\mathbf{v}t)=(d^2/dt^2)\mathbf{r}=\mathbf{a}$. Observers in $S$ and $S'$ agree on the form of Newton's second law: $\mathbf{F'}=m\mathbf{a'}=m\mathbf{a}=\mathbf{F}$, where mass is the same in all IRFs. What about electromagnetism? (then starts discussing about form invariance of Maxwell's equations).

Hope this gives the exhaustive context. The reason I didn't find the above paragraph satisfactory is that one can't claim the form invariance of the equation $\mathbf{F}=m\mathbf{a}$ without asserting (for some reason that I'm trying to understand) that $\mathbf{F}=\mathbf{F'}$. Without that assertion, one can only claim that the quantity $m\mathbf{a}$ is invariant across IRFs under a Galilean boost and nothing more (it's true for rotations too but the above paragraph is concerning boosts only).

 

[PeroK]

The section I'm reading is "Coordinate transformations and the principle of covariance". Then it explains two specific types of Galilean transformations: boosts and rotation about the $z$-axis. Till this point, force isn't mentioned anywhere, just the transformation matrices are derived which is standard stuff.

Immediately after there's a subsection on "Form invariance". I'll quote:

The idea of form invariance can be illustrated using the Galilean transformation, because acceleration is invariant under that transformation: $\mathbf{a'}\equiv(d^2/dt'^2)\mathbf{r'}=(d^2/dt^2)(\mathbf{r}-\mathbf{v}t)=(d^2/dt^2)\mathbf{r}=\mathbf{a}$. Observers in $S$ and $S'$ agree on the form of Newton's second law: $\mathbf{F'}=m\mathbf{a'}=m\mathbf{a}=\mathbf{F}$, where mass is the same in all IRFs. What about electromagnetism? (then starts discussing about form invariance of Maxwell's equations).

Hope this gives the exhaustive context. The reason I didn't find the above paragraph satisfactory is that one can't claim the form invariance of the equation $\mathbf{F}=m\mathbf{a}$ without asserting (for some reason that I'm trying to understand) that $\mathbf{F}=\mathbf{F'}$. Without that assertion, one can only claim that the quantity $m\mathbf{a}$ is invariant across IRFs under a Galilean boost and nothing more (it's true for rotations too but the above paragraph is concerning boosts only).

It seems clear to me that one can say nothing about Newton's laws, their covariance or otherwise until force has been defined in some way. Fundamentally, an equation only makes sense if you can define everything in it.

That said, I assume the author must cover this somewhere. He must say something about what he means by force by this stage in the book.

 

[Shirish]

It seems clear to me that one can say nothing about Newton's laws, their covariance or otherwise until force has been defined in some way. Fundamentally, an equation only makes sense if you can define everything in it.

That said, I assume the author must cover this somewhere. He must say something about what he means by force by this stage in the book.

Totally agree with that statement. As for mentions of force before that point:
1. Force is mentioned as the reason a massive object changes its state of inertial motion.
2. In IRFs, acceleration is caused solely by forces. Forces arise from physical interactions.

Apart from that, no concrete definition of force - it's only mentioned as the reason for acceleration / change in state of motion up until that point.

 

[PeroK]

Totally agree with that statement. As for mentions of force before that point:
1. Force is mentioned as the reason a massive object changes its state of inertial motion.
2. In IRFs, acceleration is caused solely by forces. Forces arise from physical interactions.

Apart from that, no concrete definition of force - it's only mentioned as the reason for acceleration / change in state of motion up until that point.

I guess that counts as being defined by $m \mathbf a$. Note that any function would do. If $m$ and $\mathbf a$ are invariant, then anything that depends solely on $m$ and $\mathbb a$ must be invariant.

 

[Shirish]

I guess that counts as being defined by $m \mathbf a$. Note that any function would do. If $m$ and $\mathbf a$ are invariant, then anything that depends solely on $m$ and $\mathbb a$ mus be invariant.

And I guess we can define $\mathbf{F}$ that way for IRFs because the only source of acceleration in IRFs is force? I mean, even if I go to the non-inertial frame $S''$, I'm not allowed to define $\mathbf{F''}=m\mathbf{a''}$ (in that case Newton's second law would be covariant across non-IRFs too) since forces aren't the only sources of acceleration (more specifically coordinate acceleration)?

 

[PeroK]

And I guess we can define $\mathbf{F}$ that way for IRFs because the only source of acceleration in IRFs is force? I mean, even if I go to the non-inertial frame $S''$, where $\mathbf{a''}\neq\mathbf{a}$, I'm not allowed to define $\mathbf{F''}=m\mathbf{a''}$ (in that case Newton's second law would be covariant across non-IRFs too) since forces aren't the only sources of acceleration (more specifically coordinate acceleration)?

I think the point is that $\mathbf a$ is definitely not invariant across all reference frames. Whatever else you do $F = ma$ must vary from frame to frame in terms of the quantities involved. Also, ultimately, all that has really been shown here is that $\mathbf a$ is invariant under across IRFs. Nothing really follows from that alone. But, if you make assumptions about mass and force and the relationship between them, then you have a covariant second law.

 

[Shirish]

I think the point is that $\mathbf a$ is definitely not invariant across all reference frames. Whatever else you do $F = ma$ must vary from frame to frame in terms of the quantities involved. Also, ultimately, all that has really been shown here is that $\mathbf a$ is invariant under across IRFs. Nothing really follows from that alone. But, if you make assumptions about mass and force and the relationship between them, then you have a covariant second law.

Makes sense. One last thing: Newton's 2nd law seems to deal with coordinate acceleration. If I defined some other quantity as $\mathbf{K}\equiv m\mathbf{a_p}$, where $\mathbf{a_p}$ is the proper acceleration, then this "law" $\mathbf{K}=m\mathbf{a_p}$ would hold even in non-inertial frames in Newtonian mechanics, right?

 

[PeroK]

Makes sense. One last thing: Newton's 2nd law seems to deal with coordinate acceleration. If I defined some other quantity as $\mathbf{K}\equiv m\mathbf{a_p}$, where $\mathbf{a_p}$ is the proper acceleration, then this "law" $\mathbf{K}=m\mathbf{a_p}$ would hold even in non-inertial frames in Newtonian mechanics, right?

I honestly don't know the role of proper acceleration in classical mechanics.

 

[wrobel]

For the sole particle

1)For the sole particle in the 2 Newton law $m\boldsymbol a=\boldsymbol F$ the force can depend on $\boldsymbol r,\boldsymbol{\dot r},t$ only
2)The Galilean invariance implies $\boldsymbol F =0$.
This is exactly the 1 Newton law: in an inertial frame the acceleration of the sole particle is equal to zero

For the closed system of $N$ particles $$m_1,\boldsymbol r_1,\ldots, m_N,\boldsymbol r_N\quad m_i\boldsymbol a_i=\boldsymbol F_i;$$
the Galilean invariance implies that the forces $\boldsymbol F_i$ do not depend on $t$ and depend on all kinds of differences $\boldsymbol r_j-\boldsymbol r_k,\quad \boldsymbol {\dot r}_j-\boldsymbol {\dot r}_k$ only

 

[Shirish]

I'm sorry to be stubborn, but I don't think I'm satisfied on an answer to my original question.

I revisited this question recently and just to recap, under a Galilean transformation, we notice that $a=a'$. We assume that $m=m'$ (Newtonian mechanics).

Let's say Newton's second law holds in one of the frames: $F=ma$ and we want to prove the covariance of this law under the Galilean transformation, i.e., that even in the other frame, $F'=m'a'(=ma')$ holds.

It's easy to show that $ma=m'a'$, so now we must show that $F=F'$ to actually show that $F'=m'a'$, since if $F'=F$, then $F'=F=ma=m'a'$.

I'm very sure we cannot define $F=ma$ or $F=m\frac{dv}{dt}$, because then I can trivially claim Newton's second law for all frames - inertial or non-inertial - and for all sorts of transformations.

I have tried my level best to see if $F\neq F'$ produces a contradiction (without assuming any variation of $F=ma$), but haven't been successful. Would appreciate any help, thanks in advance!

 

[etotheipi]

I'm very sure we cannot define $F=ma$ or $F=m\frac{dv}{dt}$, because then I can trivially claim Newton's second law for all frames - inertial or non-inertial - and for all sorts of transformations.

I might have misunderstood, but I will try my best! Newton's first law asserts the existence of IRFs, and Newton's second law can count as a definition of force via. $\vec{F} = m\vec{a}$ in an IRF. Here $\vec{F}$ are real forces (between interacting material bodies) and they fully account for $\vec{a}$ in the IRF. Under a Galilean boost, this $\vec{a}$ is, as you say, an invariant.

You can tell if you are in an accelerated frame of reference since if you hold an accelerometer it will read a non-zero value. Then $\vec{F} = m\vec{a}$ no longer applies, and you would have a means of knowing this via. the accelerometer. Of course you can make Newton II work by introducing inertial forces.

 

[Shirish]

I might have misunderstood, but I will try my best! Newton's first law asserts the existence of IRFs, and Newton's second law can count as a definition of force via. $\vec{F} = m\vec{a}$ in an IRF. Here $\vec{F}$ are real forces (between interacting material bodies) and they fully account for $\vec{a}$ in the IRF. Under a Galilean boost, this $\vec{a}$ is, as you say, an invariant.

You can tell if you are in an accelerated frame of reference since if you hold an accelerometer it will read a non-zero value. Then $\vec{F} = m\vec{a}$ no longer applies, and you would have a means of knowing this via. the accelerometer. Of course you can make Newton II work by introducing inertial forces.

Fair enough, for force defined via the accelerometer reading (to ensure that we're not measuring fictitious forces on top of the physical force), in non-inertial frames Newton's 2nd Law doesn't apply.

But still the main point of the question remains: if we define $\vec{F} = m\vec{a}$ even for IRFs, I can still trivially claim the covariance of Newton's 2nd Law across inertial frames, irrespective of the transformation we use to switch between them, which I don't think is correct.

There's certainly a reason we go through the trouble of proving that $a=a'$ under a Galilean transformation. But that has to be supplemented with a justification of why $F=F'$. So the point of the question is: what makes us say that $F=F'$ across inertial reference frames?

 

[etotheipi]

So the point of the question is: what makes us say that $F=F'$ across inertial reference frames?

Forces are frame invariant; i.e. both observers must agree on the extension of a spring, or the distance between two point charges. The physical laws describing these real forces are invariant in all frames (even in accelerating frames!).

Of course, things like magnetic forces throw a bit of a spanner in the works (e.g. $\vec{F} = q\vec{v}\times\vec{B}$), but then you're really into the realm of relativistic field theory! This isn't really related to your question.

But still the main point of the question remains: if we define $\vec{F} = m\vec{a}$ even for IRFs, I can still trivially claim the covariance of Newton's 2nd Law across inertial frames, irrespective of the transformation we use to switch between them, which I don't think is correct.

I don't quite understand what you mean by this. If you transform into another IRF (be it via a rotation, or a translation, or a boost, or all three!), then Newton II will hold in that frame too. $\vec{F}$ is the same in all inertial frames, and so is $\vec{a}$!

 

[Shirish]

Forces are frame invariant; i.e. both observers must agree on the extension of a spring, or the distance between two point charges. The physical laws describing these real forces are invariant in all frames (even in accelerating frames!).

Fair enough. So then, "The physical laws describing these real forces are invariant in all frames" - is another assumption/postulate, correct? So then ultimately it all boils down to empirically justifying that forces are frame invariant, right?

I don't quite understand what you mean by this. If you transform into another IRF (be it via a rotation, or a translation, or a boost, or all three!), then yes, Newton II will hold in that frame too. $\vec{F}$ is the same in all inertial frames, and so is $\vec{a}$!

So what I meant was in an argument where we want to show that Newton's 2nd Law is covariant under a Galilean transformation, we should not define $F=ma$. If, of course, it's true - by assumption or empirical justification - that forces are frame invariant as you said, then my question is resolved.

Obviously I'm aware that forces are frame invariant, but my purpose was to know for sure why we say that. Is it a postulate in Newtonian mechanics, or is it experimentally observed, etc.

 

[etotheipi]

For one thing, there is no way of choosing a "preferred" IRF, so if $\vec{F} = m\vec{a}$ holds in one IRF it better hold in all of the other ones (you have proven that if $x' = x + vt$ then it follows $\dot{x'} = \dot{x} + v$ and $\ddot{x'} = \ddot{x}$, so the force must evidently be the same under a Galilean boost!).

Generally forces in classical mechanics (i.e. gravitational, electric, ...with the exception of the weird magnetic force!) depend only on the relative separation of two particles i.e. $\vec{F}_{12} = f(\vec{r}_1 - \vec{r}_2)$. The vector $\vec{r}_1 - \vec{r}_2$ is of course invariant in all frames of reference (inertial and non-inertial), so it follows that the force on anyone particle due to a given interaction is also frame invariant.

 

[Shirish]

For one thing, there is no way of choosing a "preferred" IRF, so if $\vec{F} = m\vec{a}$ holds in one IRF it better hold in all of the other ones (you have proven that if $x' = x + vt$ then it follows $\dot{x'} = \dot{x} + v$ and $\ddot{x'} = \ddot{x}$, so the force must evidently be the same under a Galilean boost!).

Generally forces in classical mechanics (i.e. gravitational, electric, ...with the exception of the weird magnetic force!) depend only on the relative separation of two particles i.e. $\vec{F}_{12} = f(\vec{r}_1 - \vec{r}_2)$. The vector $\vec{r}_1 - \vec{r}_2$ is of course invariant in all frames of reference (inertial and non-inertial), so it follows that the force on anyone particle due to a given interaction is also frame invariant.

The second paragraph sounds sensible to me for sure, in the context of Newtonian mechanics (which is what we're focusing on anyway).

I'm not entirely convinced by the first paragraph. All we can say, if $F=ma$ didn't hold in all frames, is that this equation wouldn't represent a physical law. As a parallel, briefly stepping into SR, simultaneity doesn't hold across all IRs and is hence discarded as a physical law. Violation of simultaneity doesn't imply any preferred frame and by the same token, a potential violation of the 2nd law in Newtonian mechanics wouldn't imply a preferred frame either - just that it would have to be thrown out as a physical law.

But again, what you wrote in the second paragraph sounds like a convincing argument so far in favor of the invariance of force magnitude in IRs.

 

[etotheipi]

As a parallel, briefly stepping into SR, simultaneity doesn't hold across all IRs and is hence discarded as a physical law. Violation of simultaneity doesn't imply any preferred frame and by the same token, a potential violation of the 2nd law in Newtonian mechanics wouldn't imply a preferred frame either.

In special relativity the 4-force is a frame invariant quantity, whilst the spatial component (that which we usually call force) is frame variant. Likewise, coordinate accelerations in special relativity are also frame variant between IRFs.

However, the physical law $\vec{F} = \frac{d\vec{P}}{dt}$ is of course valid across all inertial frames.

This is key; a physical law is invariant in all inertial frames of reference. (And classical mechanics, we can also add in 'non-inertial forces' in accelerated frames so that physical laws regain their original form).

 

[Shirish]

In special relativity the 4-force is a frame invariant quantity, whilst the spatial component (that which we usually call force) is frame variant. Likewise, accelerations in special relativity are also frame variant.

However, the physical law $\vec{F} = \frac{d\vec{P}}{dt}$ is of course valid across all inertial frames.

Fair enough, but I didn't claim that the physical law $\vec{F} = \frac{d\vec{P}}{dt}$ isn't valid across IRs in SR. In any case, SR is a completely different ballgame so I'll refrain from getting deeper into that for this thread at least (another reason is that I've yet to properly study relativistic dynamics and want to clear up my concepts in the Newtonian domain first ). Still don't think the principle of relativity is sufficient to claim invariance of force across IRs in Newtonian mechanics.

 

[etotheipi]

Fair enough, but I didn't claim that the physical law $\vec{F} = \frac{d\vec{P}}{dt}$ isn't valid across IRs in SR. In any case, SR is a completely different ballgame so I'll refrain from getting deeper into that for this thread at least (another reason is that I've yet to properly study relativistic dynamics and want to clear up my concepts in the Newtonian domain first ).

Ak okay, yes apologies if I misinterpreted your post!

Still don't think the principle of relativity is sufficient to claim invariance of force across IRs in Newtonian mechanics.

No, perhaps not. What I was trying to say is that if you accept that accelerations are invariant under a Galilean boost, and that if you also accept $\vec{F} = m\vec{a}$ as a definition for IRFs, then it must follow that $\vec{F}$ be invariant under a Galilean boost too! In other words, if it works in my frame, then it must still work for someone running at 5ms^-1 alongside me, since they are both equally valid as choices of IRF.

Maybe someone else can provide further insight?

 

[Shirish]

No, perhaps not. What I was trying to say is that if you accept that accelerations are invariant under a Galilean boost, and that if you also accept $\vec{F} = m\vec{a}$ as a definition for IRFs, then it must follow that $\vec{F}$ be invariant under a Galilean boost too! In other words, if it works in my frame, then it must still work for someone running at 5ms^-1 alongside me, since they are both equally valid as choices of IRF.

Maybe someone else can provide further insight?

The chain of reasoning I'm looking for is in a different direction, i.e. if we accept that accelerations are invariant under a Galilean boost, and if we have a justification for the claim that $\vec{F}$ is invariant under a Galilean boost too, then it must follow that $\vec{F} = m\vec{a}$ holds in IRFs (which is essentially the covariance of Newton's second law).

That justification is what I'm looking for. From our discussions (and thanks so much for the help btw) so far, the justification is empirical / the fact that physical laws giving rise to the forces depend only on invariant quantities like the length separation.

 

[etotheipi]

The justification I wanted was in the other direction, i.e. if we accept that accelerations are invariant under a Galilean boost, and if we have a justification for the claim that $\vec{F}$ is invariant under a Galilean boost too, then it must follow that $\vec{F} = m\vec{a}$ holds in IRFs.

I see; however I think this is backward. Even if you could justify that forces and accelerations are invariant under Galilean boosts, it still doesn't tell you that the constant of proportionality is $m$.

IMO the most consistent way to view it is to take Newton's three laws as axioms for classical mechanics in inertial reference frames. You can prove that accelerations are invariant under the Galilean boost, and it pretty much then follows that forces are frame invariant too.

 

[PeroK]

The chain of reasoning I'm looking for is in a different direction, i.e. if we accept that accelerations are invariant under a Galilean boost, and if we have a justification for the claim that $\vec{F}$ is invariant under a Galilean boost too, then it must follow that $\vec{F} = m\vec{a}$ holds in IRFs (which is essentially the covariance of Newton's second law).

That justification is what I'm looking for. From our discussions (and thanks so much for the help btw) so far, the justification is empirical / the fact that physical laws giving rise to the forces depend only on invariant quantities like the length separation.

It's up to you how you choose to try to learn physics, but with this approach you seem to be spending a lot of time going round in circles.

Newton's laws are supposed to be conceptually simple. I really can't understand the problem here.

 

[Shirish]

It's up to you how you choose to try to learn physics, but with this approach you seem to be spending a lot of time going round in circles.

Newton's laws are supposed to be conceptually simple. I really can't understand the problem here.

You're probably right and it probably doesn't matter in the grand scheme of Physics understanding, but the fact still remains I didn't like the unsatisfactory reasoning for covariance of Newton's 2nd Law from wherever I looked it up from (for example, https://www.quora.com/How-do-you-pr...nertial-frames-under-Galilean-transformations).

It's very important to me to have a very clear understanding of what's considered an axiom/postulate based on empirical observations, and what all are mathematically/logically derived consequences.

Specifically for this type of question, I felt that distinction wasn't there.

 

[Ibix]

If $F=ma$ holds in an inertial frame and we can show that $a$ is invariant under a Galilean boost then either $F$ is also invariant or Galilean relativity does not hold. By the latter, I mean that we could measure the force required to accelerate an object at $a$ in two different frames and get a different value, violating the principle of relativity and allowing us to deduce an absolute speed for the frame.

We have never been able to detect an absolute speed in reality. That eliminates the second possibility, leading to the claim that $F$ is also invariant under Galilean boost.

 

[Shirish]

If $F=ma$ holds in an inertial frame and we can show that $a$ is invariant under a Galilean boost then either $F$ is also invariant or Galilean relativity does not hold. By the latter, I mean that we could measure the force required to accelerate an object at $a$ in two different frames and get a different value, violating the principle of relativity and allowing us to deduce an absolute speed for the frame.

We have never been able to detect an absolute speed in reality. That eliminates the second possibility, leading to the claim that $F$ is also invariant under Galilean boost.

Could you elaborate a bit more on this? How do we measure the force required to accelerate an object at $a$ without assuming $F=ma$ in $S'$?

And how can we deduce an absolute speed for the frame in this scenario?

 

[Ibix]

Could you elaborate a bit more on this? How do we measure the force required to accelerate an object at $a$ without assuming $F=ma$ in $S'$?

You use a force meter (e.g., a spring) and an accelerometer (e.g. a clock and a ruler). $F=ma$ is a statement about the relationship between their readings, and assuming Galilean relativity it tells you that the relationship will always be the same independent of speed. If relativity does not hold then the same devices will give different readings when in different states of motion.

And how can we deduce an absolute speed for the frame in this scenario?

Absolute speed was possibly not the right word, but if $F'$ is not the same value as $F$ then it must be larger or smaller. You could then keep track of $F'$ to measure your speed. With the additional assumption that there's a frame where the force takes a macimum or minimum value, it would be reasonable to conclude that this corresponds to a universal state of "at rest".

The first point follows from a failure of the principle of relativity. The latter does require an additional assumption about the mathematical form of that failure, yes.

 

[Shirish]

You use a force meter (e.g., a spring) and an accelerometer (e.g. a clock and a ruler). $F=ma$ is a statement about the relationship between their readings, and assuming Galilean relativity it tells you that the relationship will always be the same independent of speed. If relativity does not hold then the same devices will give different readings when in different states of motion.

Absolute speed was possibly not the right word, but if $F'$ is not the same value as $F$ then it must be larger or smaller. You could then keep track of $F'$ to measure your speed. With the additional assumption that there's a frame where the force takes a macimum or minimum value, it would be reasonable to conclude that this corresponds to a universal state of "at rest".

The first point follows from a failure of the principle of relativity. The latter does require an additional assumption about the mathematical form of that failure, yes.

I'm not sure about the first bold line to be honest (I'm sorry I know it must be a torture explaining things to me). I don't think just assuming Galilean relativity necessarily implies that the relationship should always be the same.

The principle of relativity combined with the assumption that the relationship between $F$ and $a$ is a physical law, on the other hand, should definitely imply that the relationship remains constant.

As for the second paragraph, that's really helpful! So the fact that $F\neq F'$ means that the force would have to be functionally dependent on the velocity. Even assuming no special properties of this functional dependence, two observers would still be able to distinguish between their inertial states of motion by measuring that force.

I guess an apology to @etotheipi is in order. He/she was probably trying to convey the same thing yesterday but I didn't quite get it at the time.

 

[etotheipi]

I guess an apology to @etotheipi is in order. He/she was probably trying to convey the same thing yesterday but I didn't quite get it at the time.

No need! Best to clear it out with the experts.

In any case, I think you're getting to the stage where you're overanalysing things. Why not step back for a little while, and come back to it with a fresh outlook in a few days?

 

[Ibix]

I don't think just assuming Galilean relativity necessarily implies that the relationship should always be the same.

You just agreed that if $F'\neq ma$ then I can use my spring, mass, clock and ruler to measure my speed from inside a sealed box.

Mathematically, what we're saying is that the only difference between frames is their velocity (assuming Galilean relativity). So if $F=ma$ holds in one frame then the only possible option for other frames (given the invariance of $m$ and $a$) is that there are some velocity-dependent terms that just happened to be zero or one (depending if they are additive or multiplicative) in the first frame. If those terms don't have those same values in another frame then I can measure them by playing around with different accelerations and masses, and hence deduce my velocity.

I'm sorry I know it must be a torture explaining things to me

This stuff is subtle. You aren't asking stupid questions, so don't worry.

 

[wrobel]

Strictly speaking, the 2Newton law is not covariant. Vectors are transformed by the contravariant law. The covariant law is for
$$\frac{d}{dt}\frac{\partial L}{\partial \dot x^i}-\frac{\partial L}{\partial x^i}$$

 

[vanhees71]

It's up to you how you choose to try to learn physics, but with this approach you seem to be spending a lot of time going round in circles.

Newton's laws are supposed to be conceptually simple. I really can't understand the problem here.

I think, it's an important issue to learn, how symmetry arguments work in physics, and Newtonian mechanics is a very good example to be thoroughly studied.

Newtonian mechanics starts (as all of physics) with an assumption about how to describe time and space, and as we know from the 19th century developments in math and particularly geometry, one can most efficiently describe the structure of space (and as we know since the early 20th century with the discovery of relativistic space-time models, rather the structure of spacetime) by investigating its symmetries.

For Newtonian mechanics one has by assumption a fixed 3D Euclidean affine space with its symmetries being homogeneity (translation invariance), isotropy (invariance under rotations). Time and space are completely independent of each other by assumption (Newton's concept of absolute time and absolute space). Time thus is simply a one-dimensional directed continuum with homogeneity (time-translation invariance) as the assumed symmetry. Last but not least the physics is invariant under Galilei boosts, i.e., there is a class of preferred reference frames, where Newton's 1st Law holds, which is the special principle of relativity.

Now you have a mathematical framework you can use to make assumptions about the physical laws, which all must obey these symmetry principles. The most appropriate framework is the Hamilton principle of least action (equivalently in Lagrange and Hamiltonian formulation) to write down an action of, say, a system of interacting point particles which obey all these symmetries. This leads to the usual laws with instantaneous conservative interaction forces between point particles (in the most simplest cases covering almost all practical applications of particle-pair forces).

The symmetries imply the conservation laws (thanks to Emmy Noether), and the symmetries of Newtonian spacetime lead to conservation of energy (homogeneity of time), conservation of momentum (homogeneity of space), angular momentum (isotropy of space), center-of-mass motion (Galilei-boost invariance).

As you rightly say, in Newtonian physics one assumes mass to be a scalar under the full Galilei group. For the acceleration you get that it is Galilei invariant too. From Newton's 2nd Law this indeed implies that the force must be a Galilei invariant either. This constrains the proper force laws to the ones you usually use, e.g., the Newton model of gravitational interactions, which implies that a point particle system can be described by central conservative pair forces, i.e., they are derivable from a potential
$$V(\vec{r}_1,\ldots,\vec{r}_2)=\frac{1}{2} \sum_{j \neq k} V_2(|\vec{r}_j-\vec{r}_k|.$$
Then the force on particle $j$ is given by
$$\vec{F}_j=-\vec{\nabla}_j V,$$
and it's pretty straight forward to see that this obeys all the symmetries of Newton spacetime. For Galilei boosts it's pretty simple. You have
$$\vec{r}_j'=\vec{r}_j+\vec{v} t, \quad \vec{v}=\text{const}.$$
Then
$$\vec{r}_j-\vec{r}_k=\vec{r}_j'-\vec{r}_k',$$
and thus
$$V'(\vec{r}_1',\ldots,\vec{r}_2')=V(\vec{r}_1,\ldots,\vec{r}_2),$$
i.e., the interaction potential is a scalar and thus $\vec{F}$ a vector under the full Galilei symmetry transformation group.

 

[Shirish]

You just agreed that if $F'\neq ma$ then I can use my spring, mass, clock and ruler to measure my speed from inside a sealed box.

Mathematically, what we're saying is that the only difference between frames is their velocity (assuming Galilean relativity). So if $F=ma$ holds in one frame then the only possible option for other frames (given the invariance of $m$ and $a$) is that there are some velocity-dependent terms that just happened to be zero or one (depending if they are additive or multiplicative) in the first frame. If those terms don't have those same values in another frame then I can measure them by playing around with different accelerations and masses, and hence deduce my velocity.

This stuff is subtle. You aren't asking stupid questions, so don't worry.

Thanks a ton for the help so far. One more question pops up from this - how do we carry over or modify the argument in bold for special relativity. I think it was mentioned in this thread that the standard spatial force is no longer invariant, but the 4-force is.

I won't pursue that question too much right now because I haven't studied relativistic dynamics yet. My guess is that even in SR, the force still can't be a function of velocity (since that'll allow us to again play around and deduce the velocity), but the key difference might be that spatial intervals change across IRs, hence the usual 3-force also changes.

 

[PeroK]

Thanks a ton for the help so far. One more question pops up from this - how do we carry over or modify the argument in bold for special relativity. I think it was mentioned in this thread that the standard spatial force is no longer invariant, but the 4-force is.

I won't pursue that question too much right now because I haven't studied relativistic dynamics yet. My guess is that even in SR, the force still can't be a function of velocity (since that'll allow us to again play around and deduce the velocity), but the key difference might be that spatial intervals change across IRs, hence the usual 3-force also changes.

In SR we do not have $t' = t$.

 

[Ibix]

how do we carry over or modify the argument in bold for special relativity.

$F\neq ma$ in relativity - force isn't even typically parallel to acceleration. With relativity, you've reached the point where "force" isn't a particularly useful concept, really. You can define a thing called the four-momentum (or the energy-momentum four vector) and take the derivative with respect to proper time, which gets you a thing called the four-force, and you can relate that to three-force, which is the thing your force meter measures. But it's a lot messier than in Newtonian physics, and people tend to use Lagrangian methods instead.

 

[vanhees71]

In SR we have no more "forces" in Newton's sense, i.e., no more "actions at a distance" but the interactions are mediated by fields, such that all the "forces" get local concepts through the interaction between fields (which themselves become dynamical entities in their own right) and "particles". One should however note that a fully consistent classical dynamics between fields and point particles seems not to work. The classical picture roughly breaks down at spatial resolutions smaller than the de Broglie wavelength for a given particle, and then you have to use quantum theory. Also here the local field concept leads to the most successful description yet, i.e., local relativistic QFTs. In that sense today all there is in the most fundamental description are quantum fields, no more point particles.