Dividing by m to make conclusions for m=0

[A.T.]

In the physics section there is doubt, that a mathematician would object to deriving something by dividing an equation by a variable $m$, and then applying that derived result to all values of $m$ including $m = 0$:

Even if some mathematician would object (which I doubt), it is clear that in physics canceling variables on both sides of an equation is a common and well-established technique. It is part of nearly every non-trivial proof that I have ever seen. So even if a mathematician would object to $mx=my$ for all m implying $x=y$ I don't think that any physicist would object to $ma=mg$ implying $a=g$.

To context is this proof that Newtonian gravity affects particles with $m = 0$:

It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.

Newtonian gravity is a force that exists between masses $(F=G m_1 m_2 / r^2)$. In his equations there would be zero gravitational force acting on zero mass. Since $F=ma$, $F/m=a$, but in this case that is $0/0=a$ which is undefined.

$F=ma$

$GMm/r^2=ma$

$gm=ma$

$g=a$The acceleration is independent of mass, and is well defined.

My question is not about physics, which can postulate whatever it wants. I would like to hear a mathematicians opinion, on whether you can apply something, that you derived by doing this:

$gm=ma$

$g=a$

to a situation where $m = 0$. In my opinion no, because the above step implicitly assumes that $m ≠ 0$, otherwise it would not be a valid operation. So whatever you derive by doing this step cannot be applied when $m = 0$.

 

[pbuk]

My question is not about physics, which can postulate whatever it wants. I would like to hear a mathematicians opinion, on whether you can apply something, that you derived by doing this:

The problem does not arise - physics tells us that $a = F/m$ and this clearly does not hold where $m = 0$ so when you write $gm = ma$ this already only holds when $m \ne 0$.

 

[Fredrik]

If I understand the question correctly, the answer is simply this: If mg=ma for all m, then g=1g=1a=a.

One can argue that "for all m" is ambiguous if we haven't specified what set we're talking about. So the statement should be something like "for all m in S, we have mg=ma". But what set S should we use? It doesn't really matter. All we need for the conclusion g=a to be valid is that S contains the real number 1.

Note that Newtonian mechanics doesn't even attempt to describe zero-mass particles, so the most natural choice for S is the set of positive real numbers, $S=\{x\in\mathbb R|x>0\}$.

 

[A.T.]

The problem does not arise - physics tells us...

As I said, I'm asking about the purely mathematical correctness of the above derivation, not about physics.

...that $a = F/m$ and this clearly does not hold where $m = 0$

But that is not used in the above derivation. What is used is $F=ma$, which is a valid statement for $m=0$, in which case it merely implies that $F=0$ regardless what $a$ is.

so when you write $gm = ma$ this already only holds when $m \ne 0$.

I disagree. Mathematically $gm = ma$ is a valid (and trivially true) statement for $m = 0$. The restriction to $m \ne 0$ is introduced by dividing both sides by $m$ to obtain $g=a$. That's why we cannot use $g=a$ for cases where $m = 0$ based on this derivation.

 

[pbuk]

I have read the thread I think you are referring to: DaleSpam's derivation is clearly flawed, however note Vanadium 50's contribution.

Of course Newtonian mechanics is based only on observations of massive objects and algebraic manipulation of its formulae cannot tell us anything about massless particles (which is just as well otherwise we would have trouble keeping their velocity at $c$ while accelerating at $g$): the only way to see if, for instance $a = g$ holds for massless particles is to observe them. Which of course we have done, and we see that it is true, and that both this phenomenon and Newtonian mechanics are explained by general relativity where we see that $a = F / m$ is not even true in general when $m \ne 0$

 

[A.T.]

If mg=ma for all m, then g=1g=1a=m.

Is that last 'm' supposed to be an 'a'? How do you go from mg=ma to g=a without dividing by m, which is not valid for all m.

 

[Fredrik]

I don't think I understood the question correctly. I'm still not sure I understand it. I tried to edit my post, but my computer bluescreened. Since several new posts have appeared, I will write a new post instead.

If we naively apply Newton's theory of gravity to a mass M=0, then it says that the acceleration of an object with mass m>0 is given by $ma=GMm/r^2=0$. Clearly this implies that a=0. But what about the acceleration of the object with mass M? We can't cancel M from $Ma=GMm/r^2=0$. But there's no physics in that equation anyway. It's only telling us that 0=0, and that doesn't tell us anything about either of the objects.

So a "theory" that tells us that that's the equation from which the acceleration is to be obtained, fails to make predictions and can't be considered a theory. This means that Newton's theory of gravity, in its usual form, is not a theory of zero mass particles.

However, there's nothing that prevents us from changing the "force law" to an "acceleration law" that says that the accelerations of the two masses m and M are given respectively by $a_m=GM/r^2$ and $a_M=Gm/r^2$. If all we're interested in is gravity, then there's no need to even introduce the concept of "force". What is "force" anyway? The way I see it, classical mechanics is built up around the theorem that says that a differential equation of the form $x''(t)=f(x'(t),x(t),t)$, where f is a nice enough function, has a unique solution for each initial condition. The "force" on an object with mass m>0 is then defined as $f(x'(t),x(t),t)/m$. Clearly, it wouldn't make sense to use this formula when m=0.

So I would say that the issue here isn't whether Newtonian gravity makes sense when one of the masses is 0. It's whether the concept of "force" does. And I would say that it doesn't.

 

[A.T.]

DaleSpam's derivation is clearly flawed, however note Vanadium 50's contribution.

Vanadium 50 misunderstood what the controversy was about. Nobody in the other thread had a problem with postulating $g=a$ as a fundamental physical law, which also applies when $m=0$. The controversy was about the math in DaleSpam's derivation based on $gm=am$ and it's applicability to a case where $m=0$.

 

[Fredrik]

Is that last 'm' supposed to be an 'a'? How do you go from mg=ma to g=a without dividing by m, which is not valid for all m.

Sorry about that typo. I have edited it in my post. I don't understand the question. I didn't divide by m to get g=a. I just used the defining property of the number 1, and the fact that the assumption was a "for all" statement.

 

[pbuk]

As I said, I'm asking about the purely mathematical correctness of the above derivation, not about physics.

No, you are asking about the correctness of the interpretation, which has nothing to do with mathematics.

But that is not used in the above derivation. What is used is $F=ma$

We have observed that $a = F/m$ for massive objects traveling at low speed. This enables us to conclude that $F = ma$ for massive objects traveling at low speed. It does not enable us to make any statements about massless objects, or objects which are traveling at high speed. Attempting to use it to predict the motion of a massless, light-speed photon is therefore doubly wrong.

I disagree. Mathematically $gm = ma$ is a valid (and trivially true) statement for $m = 0$. The restriction to $m \ne 0$ is introduced by dividing both sides by $m$ to obtain $g=a$. That's why we cannot use $g=a$ for cases where $m = 0$ based on this derivation.

I didn't state myself clearly - I agree with what you are saying.

 

[A.T.]

This means that Newton's theory of gravity, in its usual form, is not a theory of zero mass particles.

That was the point of those objected to DaleSpam's derivation. That you cannot algebraically derive the acceleration of zero mass particles from the Newton's force law of gravity.

However, there's nothing that prevents us from changing the "force law" to an "acceleration law"...

That was never contended in the other thread, and is not the topic of this thread. The issue was purely with the mathematical correctness of the derivation for m=0 from the "force law".

 

[Fredrik]

That was never contended in the other thread, and is not the topic of this thread. The issue was purely with the mathematical correctness of the derivation for m=0 from the "force law".

I still don't understand what this thread is about. Did someone say that the mg=ma law predicts that zero mass objects would have acceleration g?

mg=ma is based on the empirical observation that every time we drop something, the acceleration is approximately g. So it makes more sense to think of "mg=ma" as "g=a multiplied by m" than to think of "g=a" as "mg=ma divided by m".

 

[pbuk]

Better format for vectors

I think we are all saying the same thing - you cannot derive $\mathbf a = \mathbf g$ for massless objects using $\mathbf F = m \mathbf a$.

Frederik and I are going one stage further, by saying that $\mathbf F$ (and therefore $\mathbf F = m \mathbf a$ doesn't have any meaning for massless objects.

 

[A.T.]

We have observed that ...

Please note that I'm not asking about consistency with observation here. That's why I posted that under math not physics.

The restriction to $m \ne 0$ is introduced by dividing both sides by $m$ to obtain $g=a$. That's why we cannot use $g=a$ for cases where $m = 0$ based on this derivation.

I agree with what you are saying.

Thanks, that was the core of my question.

 

[A.T.]

I think we are all saying the same thing - you cannot derive $\mathbf a = \mathbf g$ for massless objects using $\mathbf F = m \mathbf a$.

Frederik and I are going one stage further, by saying that $\mathbf F$ (and therefore $\mathbf F = m \mathbf a$ doesn't have any meaning for massless objects.

Yes, thanks for the clarification.

 

[A.T.]

Did someone say that the mg=ma law predicts that zero mass objects would have acceleration g?

That is how I interpret DaleSpam's post. He responds to CKH, who states that the force law leads to undefined a for m=0, by deriving a=g from mg=ma, and concluding that a is well defined.

It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.

Newtonian gravity is a force that exists between masses $(F=G m_1 m_2 / r^2)$. In his equations there would be zero gravitational force acting on zero mass. Since $F=ma$, $F/m=a$, but in this case that is $0/0=a$ which is undefined.

$F=ma$

$GMm/r^2=ma$

$gm=ma$

$g=a$The acceleration is independent of mass, and is well defined.

 

[Nugatory]

Family disputes should happen indoors, not in the yard in front of the neighbors. This thread is evolving into something that does not belong in the math sub forum.

Moving, and closing pending moderation.

 

[Dale]

In the physics section there is doubt, that a mathematician would object to deriving something by dividing an equation by a variable $m$, and then applying that derived result to all values of $m$ including $m = 0$:

A.T., you are again misstating my position. Setting aside the physics and concentrating on the math only, the mathematical question is the following: Given the proposition “$mx=my$ for all m”* what can you conclude about the relationship between x and y?

Note that the proposition contains two parts, “$mx=my$” and “for all m”. It is important to remember both parts of the proposition, since both parts must be satisfied. A.T., you consistently move the second part "for all m" to the conclusions instead of to the givens which results in an incorrect inference and which I am NOT claiming (it is a strawman argument). That is partly my fault since I did not consistently write it in the previous thread, I wrote it often, but not consistently every time.

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$, however that procedure is under dispute because of the concern about division by zero. So we seek another way to determine the relationship without using division at all. One other way to determine the relationship is to simply plug in all possible values for x, y, and m, and see which values satisfy the proposition.

Try, x=1 and y=1. That satisfies $mx=my$ for m=0, m=1, m=2, and in fact for all m, therefore it satisfies the proposition “$mx=my$ for all m”. Similarly, x=3 and y=3, satisfies $mx=my$ for m=0, m=1, m=2, and for all m, and therefore it also satisfies the proposition “$mx=my$ for all m”.

In contrast, x=1 and y=3 satisfies $mx=my$ for m=0, but not for m=1, nor m=2, nor any other m, therefore x=1 and y=3 does not satisfy the proposition “$mx=my$ for all m”.

A little thought shows that the only values for x and y which satisfy the proposition “$mx=my$ for all m” are the values $x=y$. Therefore, given the proposition “$mx=my$ for all m” we can indeed conclude that the relationship between x and y is unambiguously and without further qualification $x=y$. This is shown by examination of possible solutions and testing them directly against the proposition, without any use of division.

* we should also specify the set of admissible values for m, i.e. "for all m" should technically be "for all m in S" where S is some set appropriate for the problem. In the question that spawned this thread S would be the non-negative real numbers, but similar proofs would still be valid for other S.

 

[Dale]

OK, the thread is reopened for now. A.T. do you understand the two proofs given, how both make use of the "for all m" part of the proposition, and how neither is subject to criticism from division by 0?

 

[pbuk]

A.T., you are again misstating my position. Setting aside the physics and concentrating on the math only, the mathematical question is the following: Given the proposition “$mx=my$ for all m in the real numbers” what can you conclude about the relationship between x and y?

We have observed that $F = ma$ for massive objects traveling slowly. We can therefore make the proposition $mg = ma \ \mathrm{for} \ m > 0$. We CANNOT make the proposition that $mg = ma \mathrm{\ for\ } m \in \mathbb R$ any more than we can make the proposition that momentum of a massless particle is given by $P = mv = 0c = 0$.

 

[Dale]

I disagree, but the OPs question is only about the validity of the reasoning, not about the validity of the proposition.

 

[pbuk]

A little thought shows that the only values for x and y which satisfy the proposition “$mx=my$ for all m” are the values $x=y$.

That is true if x and y are values independent of m, but what if x (or in this case a) is a function of m: $a(m) = g \ (m > 0), 0 \ (m = 0)$? Your proof has to exclude that possibility - so in order to prove that a = g holds independent of mass, you have to assume that a is independent of mass.

 

[pbuk]

I disagree, but the OPs question is only about the validity of the reasoning, not about the validity of the proposition.

You cannot separate the reasoning from the proposition; in particular if the proposition is the equivalence of two functions of $m$ whose domains include $m = 0$ then division by $m$ is not a valid step of reasoning.

 

[Fredrik]

Is there really anything to discuss here other than this?

Theorem: For all $x,y\in\mathbb R$ such that $mx=my$ for all $m>0$, we have $x=y$.

Proof: Let $x,y$ be arbitrary real numbers such that $mx=my$ for all $m>0$. Since $1>0$ and $mx=my$ for all $m>0$, we have $x=1x=1y=y$.

This is the theorem that A.T. asked about, and its complete proof. Sure we can consider different theorems, but then we're going off topic again.

 

[Dale]

if the proposition is the equivalence of two functions of $m$ whose domains include $m = 0$ then division by $m$ is not a valid step of reasoning.

Neither of the proofs used division by m in their reasoning. They are not subject to that criticism.

 

[Dale]

Is there really anything to discuss here other than this?

The only other thing that I can see to discuss is the generalization of the proof to handle $\forall m \in S$ where S is different from "positive reals". Your same proof would also hold for S is "non-negative reals" (which is of particular interest to the OP), "all reals", "all complex numbers", "all integers", ... I think that the only requirement is that $1 \in S$.

 

[A.T.]

We have observed that...

DISCLAIMER: Consistency with observation is irrelevant here. This thread is not about physics. It was moved from math to physics sub forum against the wish of the topic starter.

 

[A.T.]

Theorem: For all $x,y\in\mathbb R$ such that $mx=my$ for all m>0, we have $x=y$.

No, that is not the claim by DaleSpam. His proposition contains "for all m" not "for all m > 0":

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$

Therefore, given the proposition “$mx=my$ for all m” we can indeed conclude that the relationship between x and y is unambiguously and without further qualification $x=y$.

 

[A.T.]

About Misunderstandings:

Note that the proposition contains two parts, “$mx=my$” and “for all m”.

I always understood it like that, even when “for all $m$” wasn't stated explicitly. Unless stated otherwise, I assume no constraints on the value of $m$.
About dividing by m:

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$,...

Unless stated otherwise, I assume no constraints on the value of $m$. So I assume that the last $x=y$ means "$x=y$ for all $m$". Is that correct?

...however that procedure is under dispute because of the concern about division by zero.

Do you dispute the following?:

Dividing an equation by m is only valid under the assumption that m≠0. So whatever is derived by dividing by m is also only valid under the assumption that m≠0.

Explicitly written that means:

$mx=my$ for all $m$ | divide both sides by $m$

$x=y$ for all $m≠0$
​

About the workaround that avoids dividing by m:

Therefore, given the proposition “$mx=my$ for all m” we can indeed conclude that the relationship between x and y is unambiguously and without further qualification $x=y$.

I disagree that it's unambiguous, because aside of the relationship

$$R_1: \hspace{10 mm} x=y$$

there is an infinite number of relationships between $x$ and $y$, that satisfy the proposition. For example:

$$R_2: \hspace{10 mm} x =\begin{cases} y & m \neq 0 \\ 3y & m = 0 \end{cases}$$

Your objection in the previous thread was that $x = 3y$ doesn't satisfy the proposition for all $m$, but that is irrelevant, because $x = 3y$ only comes into play for $m=0$. The relationship $R_2$ as a whole does sastify $mx=my$ for all $m$.

 

[Fredrik]

No, that is not the claim by DaleSpam. His proposition contains "for all m" not "for all m > 0":

OK, but what does "for all" mean? In mathematics based on set theory, it should mean "for all sets" unless we say otherwise, but it can't mean "for all sets" here, because multiplication isn't defined for arbitrary sets. So the person who writes the statement must have some specific set in mind. Let's denote that set by S. When he writes "for all m", it means "for all m in S".

It makes no difference if S is the set of positive real numbers or the set of non-negative real numbers, because if S is any subset of $\mathbb R$ that contains the number 1, we get the result x=y (without using division):

Theorem: $\forall x,y\in\mathbb R\ ((\forall m\in S\ mx=my)\ \Rightarrow\ x=y))$

Proof: Let x,y be arbitrary real numbers. We want to prove the implication $(\forall m\in S\ mx=my)\ \Rightarrow\ x=y$, so suppose that mx=my for all m in S. Since $1\in S$, this assumption implies that $x=1x=1y=y$. So the implication is true. Since x,y are arbitrary real numbers, this proves the theorem.