# I Prove equation of motion is unchanged under Galilean transformation

1. Jul 15, 2017

### Happiness

Is the attached solution complete? In particular, do we need to prove that $V'(r_{12}')=V(r_{12})$, where $V'(r_{12}')$ is the potential energy function in the reference frame $S'$, moving at a uniform velocity with respect to the reference frame $S$, and $r_{12}'$ is the distance between particles $1$ and $2$ in reference frame $S'$.

Another question is would a central force remain central under a change of reference frame under Galilean transformation? Why?

The solution to example 4:

Last edited: Jul 15, 2017
2. Jul 15, 2017

### hilbert2

The second time derivative of a trajectory $\mathbf{r}(t)$ doesn't change when a term $\mathbf{v}t + \mathbf{x}_0$ is added to it, neither does the distance $|\mathbf{r}_a (t) - \mathbf{r}_b (t) |$ change when it's added to both $\mathbf{r}_a$ and $\mathbf{r}_b$ (all forces are functions of this kind of distances unless there's friction forces that define a preferred inertial frame).

3. Jul 16, 2017

### Happiness

For TL;DR version, skip to the bottom of the post. But I implore you to read this post in its entirety as I believe it contains many subtle concepts that we may have overlooked most of the time.

What the author wants to prove
I can't seem to understand the solution. It is not obvious to me how the answer (1.7b) in post #1 achieves what the author set out to do, which I have reproduced in the following image:

The statement that is required to be proven
One interpretation of "the equations of motion" is the equation $F=ma$. That means that we want to show that this equation is true even in another reference frame S', i.e., we want to show $F'=m'a'$. The book is clear in showing why $m=m'$ and $a=a'$. But I have a hard time understanding why $F=F'$. We need to show $F=F'$ so that we can change the LHS of $F=ma$ into $F'=ma$ and then independently (independent from $F=F'$) change the RHS using $m=m'$ and $a=a'$ to get $F'=m'a'$. This way, we would then have shown that the law of mechanics $F=ma$ is also true in S'. For the purpose of this question, I am treating $F=ma$ as a law or theorem rather than as a definition.A (See point A in the footnotes below for my reason.)

Why the solution is lacking and what can be added to complete it
As given, $F_x=-\frac{\partial U}{\partial x}$.B Then the equation of motion, in meaningful terms (free of the use of the concept of "force", which is a vague, ill-defined concept on its own), is $-\frac{\partial U}{\partial x}=ma_x$. It is clear that $m=m'$ and $a_x=a'_x$. So it remains to show, by using a method independent from $m=m'$ and $a_x=a'_x$, that $-\frac{\partial U}{\partial x}=-\frac{\partial U'}{\partial x'}$. But this is NOT shown by the book at all! The book only shows $-\frac{\partial U}{\partial x}=-\frac{\partial U}{\partial x'}$ (the difference from what is required is that the $U$ on the RHS is unprimed). Nevertheless, let's continue to figure out how to prove what is required. If $U(r_{12})=U'(r'_{12})$ is true, then $\frac{\partial U(r_{12})}{\partial x_1}=\frac{\partial U'(r'_{12})}{\partial x'_1}$is also true.C Thus if we prove $U(r_{12})=U'(r'_{12})$, then we would have proven $-\frac{\partial U}{\partial x}=-\frac{\partial U'}{\partial x'}$ and thus $-\frac{\partial U'}{\partial x'}=m'a'_x$, i.e., $F'=m'a'$, and we are done! Since we know that $r_{12}=r'_{12}$, it follows that $U(r_{12})=U(r'_{12})$. What remains to be shown is $U(r'_{12})=U'(r'_{12})$: the potential-energy function is invariant under a Galilean reference-frame transformation, if $U'(r'_{12})$ even exists in the new reference frame S' in the first place. Please take a moment to ponder over the meanings of $U(r_{12})$ and $U'(r'_{12})$ and how they are different.D But after much thought, it is still unclear to me why $U(r_{12})=U'(r'_{12})$ is true:E I cannot see how this can be derived from the only two premises of the questions: (1) $F_x=-\frac{\partial U}{\partial x}$ and (2) the Galilean transformation equations (1.3).

Other issues arising from this
So it seems to me that we have to assume that $U(r_{12})=U'(r'_{12})$, i.e., the invariance of the potential-energy function,F in order to prove that $F'=m'a'$, i.e., that the laws of mechanics is invariant under a Galilean reference-frame transformation, since it cannot be derived from the premises. And if $U(r_{12})=U'(r'_{12})$ cannot be derived from the premises, then a corollary is that we also cannot derived from the premises that a force is invariant, in both magnitude and direction, under a Galilean reference-frame transformation. Consequently, it wouldn't seem far fetched to contemplate the possibility that certain central forces may not remain central under a Galilean reference-frame transformation, or even if they do remain central, their magnitude may change, e.g., the gravitational constant $G$ may increase with the $v$ in equation (1.3), i.e., the velocity frame S' moves with respect to frame S.

Why the solution and the question are actually absurd
Let's now consider what happens if we do in fact require the assumption of $U(r_{12})=U'(r'_{12})$ in solving this question. Firstly, the assumption of $U(r_{12})=U'(r'_{12})$ is stronger than the assumption of $F=F'$, i.e., a force is invariant, in both magnitude and direction, under a Galilean reference-frame transformation (since the latter assumption does not require the assumption that there exists a scalar function $U'(r'_{12})$ from which $F'$ can be derived, i.e., $F'_x=-\frac{\partial U'(r'_{12})}{\partial x'}$, but the former assumption does). By differentiating $U(r_{12})=U'(r'_{12})$ on both sides of the equation, we get $F=F'$. That means $U(r_{12})=U'(r'_{12})$ implies $F=F'$. But after a considerable amount of thought on this implication, I find the given solution rather silly! This is because the solution requires the assumption of $U(r_{12})=U'(r'_{12})$ in its proof and then spends a considerable amount of effort in differentiating $U(r_{12})$ and $U(r'_{12})$. But if we already need to assume $U(r_{12})=U'(r'_{12})$, which automatically implies $F=F'$, then why not just use $F=F'$ straight away? Then the proof would be made laughably exceedingly simpler: since $F=F'$, $m=m'$ and $a=a'$, then clearly $F=ma$ implies $F'=m'a'$.

Footnotes
A. Is F=ma a law or a definition?
If I treat $F=ma$ as a law or theorem, then what I am saying is that I am taking $F$ to have a definition or meaning independent from $m\times a$. As given by the question, if the force is central, then the force $F_x=-\frac{\partial U}{\partial x}$. So I am taking this to be the "definition" of $F_x$.B Then the equation of motion, in meaningful terms, is $-\frac{\partial U}{\partial x}=ma_x$. If I treat $F=ma$ as a definition, what I mean is that $F$ has no independent definition or meaning of its own; $F$ basically just means $m\times a$, the product of $m$ and $a$, nothing more, nothing less. In other words, $F$ is just a shorthand of writing $m\times a$. If so, then it cannot be a law of mechanics that gives us any information about the motion of objects. For instances, I can choose to shorthand New York as N. Y. or something "nonsensical" like 3c1. The equations "N. Y. = New York" or "3c1 = New York" are just definitions; they contain no extra information, let alone information about the motion of objects. Then it would be meaningless to prove $F'=m'a'$ because it is just another definition, and definitions require no proof at all: it is always true. This interpretation of $F=ma$ as a definition would make the question redundant and silly because $F'=m'a'$ would always be true even if the force is not central or the transformation is not Galilean. (In this case then, the equation of motion would have morphed into $-\frac{\partial U}{\partial x}=F_x$. But I would rather not take this approach.)

B. Definitions of force
My actual definition of force is the invisible "thing" or physical quantity that causes objects to accelerate. But this definition is rather vague. So if you like, you may take $F_x=-\frac{\partial U}{\partial x}$ to be the definition of force, for now.

C. A proof
If $U(r_{12})=U'(r'_{12})$, then $\frac{\partial U'(r'_{12})}{\partial x'_1}=\frac{\partial U(r_{12})}{\partial x'_1}=\frac{\partial U(r'_{12})}{\partial x'_1}=\frac{\partial U(r_{12})}{\partial x_1}$ The first equality is due to our assumption $U(r_{12})=U'(r'_{12})$; the second equality is due to $r_{12}=r'_{12}$; the third equality is shown by the given solution in the equations immediately preceding (1.7b).

D. Why we cannot assume the potential-energy function must be invariant
Suppose $U(r_{12})=-\frac{GMm}{r_{12}}$. Then what makes us believe that the potential-energy function $U'(r'_{12})$ in frame S' must necessarily be $U'(r'_{12})=-\frac{GMm}{r'_{12}}$? Have you ever considered the possibility that when frame S' moves at a velocity $v$ with respect to frame S, it could be that $U'(r'_{12})=-\frac{GMm+kv}{r'_{12}}$ instead? In this case then, $U\neq U'$. In other words, objects may obey Newton's law of gravitation in frame S but not in frame S'. If this seems nonsensical to you because, I suppose, you argue that objects must obey Newton's law of gravitation in all inertial frames, then please be reminded that in this question we are proving that the equations of motion are invariant under a Galilean reference-frame transformation, and then from this invariance we conclude that the laws of mechanics are the same in all inertial frames. So using the argument that objects must obey Newton's law of gravitation in all inertial frames, in our solution, would mean that you are already assuming to be true the very conclusion we want to prove in the first place! This would make the argument a circular one. In other words, if you are already assuming the conclusion to be true, then you are not actually proving it.

E. What we really need to prove and alternative reasons why the potential-energy function is invariant
The point of this question, as I see it, is to figure out what can be deduced from a given set of premises. So for the purpose of this question, we are discussing how the two premises, (1) that the force is central and (2) that the reference-frame transformation is Galilean, implies that the potential-energy function is invariant under such a Galilean transformation, and not discussing whether or not the statement "the potential-energy function is invariant under a Galilean transformation" per se is true. Nonetheless, I have explored alternative reasons as to why the statement may be true. One axiom that we may subscribe to is that the velocity at which an observer is moving should have no effect on the potential energy or the force experienced by an object because these quantities are "intrinsic" properties of the object. So it would make no sense how the velocity of something extrinsic to the object could in any way influence the object. So with this axiom, we can explain why the potential-energy function is invariant under a Galilean transformation, but this still doesn't change the fact that the truth of the statement does not follow from the question's two premises. On the other hand, the axiom may not even be true. This is because we know from special relativity that the velocity of the observer changes the relativistic mass of the object even though mass is an "intrinsic" property. So after all, it doesn't seem far fetched to contemplate that the potential-energy function may NOT be invariant under a Galilean transformation.

F. The meaning of the invariance of the potential-energy function
Strictly speaking, the invariance of the potential-energy function only means $U(r'_{12})=U'(r'_{12})$ and not $U(r_{12})=U'(r'_{12})$, but since our reference-frame transformation is Galilean, it is distance preserving, i.e., $r_{12}=r'_{12}$, and in this case, $U(r_{12})=U(r'_{12})=U'(r'_{12})$.

TL;DR version
For the equation of motion in reference frame $S'$ to have identical form as that in $S$, it must be true that $m_1a_{x1}=-\frac{\partial U}{\partial x_1}$ implies $m_1'a_{x1}'=-\frac{\partial U'}{\partial x_1'}$. But this is not shown by the book's answer in (1.7b), shown in post #1.

Last edited: Jul 16, 2017