# Galilean invariance: Newton's 2nd Law of motion

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## Main Question or Discussion Point

Consider a frame S' moving with speed u along +ve x direction with respect to another frame S. Consider a body moving with speed v along +ve x direction with respect to frame S . Both frame are inertials.

here,force acting in S frame on the body is $$F\hat x=\frac {dp} {dt}\hat x,$$
$$F\hat x=v\frac {dm} {dt}\hat x+m\frac {dv} {dt}\hat x,$$
According to Galilean transformation,
force acting in S' frame on the body is $$F'\hat x'=\frac {dp'} {dt'}\hat x',$$
where $$\hat x'=\hat x,$$
and $$t'=t,$$
$$F'\hat x={(v-u)}\frac {dm} {dt}\hat x+m\frac {dv} {dt}\hat x,$$

So, under Galilean transformation, Newton's 2nd law of motion is invariant only when mass of the system is constant with respect to time; otherwise it is not co-variant,too. Is this right?

Is it only Maxwell wave equation, which is not co-variant under Galilean transformation or all wave equations derived using laws of classical mechanics are not co-variant under Galilean transformation?

Even if Maxwell wave equation is not not co-variant under Galilean transformation,why should this lead to special relativity?

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## Answers and Replies

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Ibix
What does "mass changing with time" actually mean? In Newtonian physics there's only one option - you are dumping matter somewhere. That's fine (it's how a rocket works, for example) as long as you keep accounting for the mass that's being dumped. But you forgot about the mass and any momentum it might carry away.

Implicitly, your first form assumes that it carries no momentum away. So that lost mass must have zero velocity in S. But that means that it has velocity -u in S'. And that's where your extra term is coming from - the maths did not forget about the missing momentum. :grin:

Maxwell's equations lead to relativity because the way to make them covariant is to replace the Galilean transforms with the Lorentz transforms.

PeroK, vanhees71 and Pushoam
Maxwell's equations lead to relativity because the way to make them covariant is to replace the Galilean transforms with the Lorentz transforms.
Why is there a need of making Maxwell's equations covariant?
By using Lorentz transformation, do we make only Maxwell wave equation covariant or all the four equations of electrodynamics?

Is it only Maxwell wave equation, which is not co-variant under Galilean transformation or all wave equations derived using laws of classical mechanics are not co-variant under Galilean transformation?

Ibix
Why is there a need of making Maxwell's equations covariant?
By using Lorentz transformation, do we make only Maxwell wave equation covariant or all the four equations of electrodynamics?
All of Maxwell's equations are Lorentz covariant and not Galileo covariant. Historically, the problem was that you could describe a charge moving past a magnet exactly in accord with experiment. And you could describe a magnet moving past a charge exactly in accord with experiment. But you couldn't take one description and transform it into the other using the Galilean transforms. You got nonsense - magnetic monopoles all over the place, which we simply don't see.

Everyone assumed that Maxwell had found a special case - perhaps for observers that were at rest with respect to some medium. They set about hunting for extra terms. Ultimately the problem turned out to be that Maxwell's equations were exactly right and Newton and Galileo were wrong - although in a way they had no hope of detecting with the equipment of their day.

You asked about other wave equations. The answer is that it depends how deep you go. Most areas of physics are based on Newtonian assumptions. So the equations governing water waves, for example, are covariant under Galilean transforms. But this means they are subtly wrong: if you ever find water under conditions where the speed of particle motion is comparable to the speed of light (not sure chemistry will permit that, but never mind) the equations will be wrong. The point about Maxwell's equations was that we went direct to a full relativistic treatment of electromagnetism without having a non-relativistic treatment or the theory of special relativity. That was a bit confusing, but Einstein sorted it out in the end.

Pushoam
vanhees71
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Why is there a need of making Maxwell's equations covariant?
By using Lorentz transformation, do we make only Maxwell wave equation covariant or all the four equations of electrodynamics?
The spacetime symmetry is the most fundamental property underlying all of physis. If physical laws like the Maxwell Equations (there are four of them consisting of Gauss's Laws for the electric and the magnetic field components, the Faraday Law of Induction, and the Ampere-Maxwell Law) are not invariant under the assumed spacetime symmetry there are thus only two possibilities: Either the Maxwell Equations are wrong and have to be adapted such that they are invariant under the spacetime-symmetry group (that's what almost all physicists believed in the 19th century) or the assumed spacetime symmetry is wrong. As it turned out, the latter is right, because there are no known exceptions to Maxwell's theory (or its quantum version, QED) and (local) Poincare symmetry of relativistic spacetime is among the best empirically confirmed assumptions in contemporary theoretical physics.

Pushoam
So, under Galilean transformation, Newton's 2nd law of motion is invariant only when mass of the system is constant with respect to time; otherwise it is not co-variant,too. Is this right?
No. Newton's 2nd law of motion is Galilean invariant. Force is equal to the change of motion in all inertial frames. Your calculation is based on this fact:

force acting in S' frame on the body is $$F'\hat x'=\frac {dp'} {dt'}\hat x',$$
Not the 2nd law but the force is frame dependent if mass changes. That's something different.

Not the 2nd law but the force is frame dependent if mass changes. That's something different.
Will you please illustrate it as I didn't get what you meant by the above statement?

Will you please illustrate it as I didn't get what you meant by the above statement?
The second law says that the force is proportional to the change of motion. That applies to all inertial frames. No matter how you change the frame of reference - the second law remains unchanged. But that doesn't mean that the force, which is definied by the laws of motion, must be frame-independent. In classical mechanics it can be frame-dependent in case of open systems (as shown in your calculation above).

TL;DR: Always F=dp/dt and F'=dp'/dt for all inertial frames but not necessarely F=F'.

What I have got so far is:
It is not Newton's 2nd law of motion which is invariant ; it is Force which remains invariant under Galilean transformation i.e. F= F' when mass is not changing with respect to time.
Newton's 2nd law of motion remains always covariant i.e. F=dp/dt and F'=dp'/dt for all inertial frames .
A law of physics cannot be invariant, it can be covariant, while a physical quantity can be invariant but it cannot be covariant.
We apply the concept of invariance to physical quantity, while the concept of covariance to laws of physics.
Is this right?

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It is not Newton's 2nd law of motion which is invariant ; it is Force which remains invariant under Galilean transformation i.e. F= F' when mass is not changing with respect to time.
Newton's 2nd law of motion remains always covariant i.e. F=dp/dt and F'=dp'/dt for all inertial frames .
Is this right?
Yes, that's what I mean.

Thank you.
Thanks to all.

vanhees71
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No. Newton's 2nd law of motion is Galilean invariant. Force is equal to the change of motion in all inertial frames. Your calculation is based on this fact:

Not the 2nd law but the force is frame dependent if mass changes. That's something different.
The problem with your attempt to create a problem with Galilean invariance is that you don't use the correct 2nd Law, which is
$$\dot{\vec{p}}=\vec{F}.$$
To analyze symmetry principles the most elegant way is to use the Hamiltonian formalism which lives in phase space, i.e., to work with position and conjugate-momentum variables. Usually (if no magnetic force is present) the momentum is given by
$$\vec{p}=m \vec{v}.$$
A Galilei boost is given by
$$t \mapsto t'=t, \quad \vec{x} \mapsto \vec{x}'=\vec{x}+\vec{v} t, \quad \vec{p} \mapsto \vec{p}'=\vec{p}+m \vec{v}, \quad m \mapsto m'=m, \quad \vec{v}=\text{const}.$$
Now obviously
$$\dot{\vec{p}}'=\dot{\vec{p}},$$
and thus the force must also transform trivially. This restricts the possible forms of forces, i.e., the dynamical laws of Nature to the class of Galilei-invariant laws.

$$\dot{\vec{p}}'=\dot{\vec{p}},$$

This is correct only when m is constant with respect to time.

vanhees71
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If ##m## is not constant then you have an open system and have to include the parts of matter taken in and out of the system. Galilean invariance, analyzed in context of quantum theory, implies the conservation of total mass (super selection rule since mass is a non-trivial central charge of the quantum Galileo Lie algebra).

weirdoguy
If ##m## is not constant then you have an open system and have to include the parts of matter taken in and out of the system.
Yes, that's what we are talking about.

Galilean invariance, analyzed in context of quantum theory, implies the conservation of total mass
Not only in context of quantum theory.

vanhees71
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In classical Newtonian physics it's an assumption, as far as I know.

In classical Newtonian physics it's an assumption, as far as I know.
Changing the total mass would violate the laws of motion.

vanhees71
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Can you prove that? I'm not aware of any symmetry principle or other argument in classical Newtonian mechanics that leads to mass conservations.

Can you prove that?
According to the second law the force acting on a single system in a given inertial frame K is

$F_i = m_i \cdot \dot v_i + v_i \cdot \dot m_i$

and

$F'_i = m_i \cdot \dot v_i + v'_i \cdot \dot m_i = m_i \cdot \dot v_i + v_i \cdot \dot m_i - u \cdot \dot m_i = F_i - u \cdot \dot m_i$

in another inertial frame K' moving with u relative to K. According to the third law all forces in an isolated global system cancel each other out. That means in K

$\sum\limits_i {F_i } = \sum\limits_i {m_i \cdot \dot v_i } + \sum\limits_i {v_i \cdot \dot m_i } = 0$

and in K'

$\sum\limits_i {F'_i } = \sum\limits_i {F_i } - u \cdot \sum\limits_i {\dot m_i } = 0$

This is only possible with

$\sum\limits_i {\dot m_i } = 0$

vanhees71, Buffu and Pushoam