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Forgive me if I'm duplicating the question - I was searching through the Phys.StackEx. but I could not find an answer to a question, which seems relatively simple.

When the form-invariance of the Newton's $2^{nd}$ law of motion under Galilean transformations is proved, it is usually shown that the acceleration is invariant under these transformations - this is clear to me. I have a difficulty with a system with time-varying mass, which - for a 1D problem, for simplicity - is governed by the following equation:

$\dot{m}\dot{x} + m \ddot{x} = F \; \; \; \; (*)$

Form-invariance of the above equation with respect to time-shift, translation and rotation in space is clear for me. I have a problem with the "Galilean boost". If I consider the following transformation:

$x \rightarrow X + vt$

$\dot{x} \rightarrow \dot{X} + v$

$\ddot{x} \rightarrow \ddot{X}$

then the equation of motion takes the following form:

$\dot{m}(\dot{X}+v)+m\ddot{X} = F$

which is obviously different from (*). Where do I miss the point? Is it because I should consider the WHOLE physical system in which the total mass is constant? Or should I somehow "add" the $\dot{m}v$ term to the force term? I would be very grateful for explanation.

Qmechanic
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P. Szeptynski
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3 Answers3

1

The additional $\dot mv$ term arises only because of the transformation. i.e., it is an inertial term.

What you are doing is applying a Galilean boost $$\dot x\rightarrow \dot X+v$$ to Newton's second law, and in "roundabout" way getting $$\tag1 F=\frac{dp}{dt}=\frac{d(m\dot X +mv)}{dt}=\dot m\dot X + m\ddot X + \dot mv + m\dot v$$ where the transformed momentum $$p=m\dot x=m\dot X+mv$$ and the final term in eq. (1), $m\dot v$ vanishes because $v$ is constant (boost velocity). That is $$F=\dot m\dot X + m\ddot X + \dot mv$$ which is your result. That is, form invariance is restored if the additional term $\dot m\vec v =0$.

Your first equation considers a variable mass system, but in your analysis, no consideration is given to the fact that this lost mass carries away momentum. This is like saying we assume that the lost mass moves away with velocity $v=0$ in the initial frame. So of course if you boost the system with a velocity $v$ you will pick up the extra term $ m v$ in the other. Again, $\dot mv$ is an inertial term that arises only because of the boost.

You could try various methods to retain the form of Newton's second law, but it will require additional assumptions and mathematics. One such attempt is made here, and the claim is that form invariance is retained, though I have not read the paper and cannot vouch for its validity and rigor.

joseph h
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1

Newton's second law is the same in all inertial frames, but this law is dp/dt=F. If this equation is true in all inertial frames, then Newton's second law is form-invariant under Galelian transformations. It's also called "Galelian covariant". However, this doesn't mean that everything you write on paper on one frame will be a carbon copy of the things you white in all other inertial framee. The details will differ, but the law, i.e. F=dp/dt, stays the same.

Ahmed Samir
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-1

lets write it like this

$$F=\frac{dp}{dt}=\frac{d}{dt}\left(m(t)\,v(t)\right)\tag 1$$

if

$~x(t)\mapsto x(t)+V\,t~\Rightarrow ~v(t)\mapsto v(t)+V~$

thus equation (1)

$$F=\frac{d}{dt}\left[m(t)\,(v(t)+V)\right]\quad\Rightarrow$$ $$F-\dot m\,V=\frac{d}{dt}\left[m(t)\,v(t)\right]\tag 2$$

compare equation (1) with (2), we got a fictitious force $~F_f=\dot m\,V~$

This additional term in the neu reference frame, is purely a consequence of the time depending mass $~m=m(t)~$. It is "analog" to the equations of motion in rotating system.

Eli
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