# Dividing by m to make conclusions for m=0

1. Sep 14, 2014

### 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 $m = 0$:

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

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:

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 $m = 0$.

Last edited: Sep 14, 2014
2. Sep 14, 2014

### MrAnchovy

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 $m \ne 0$.

3. Sep 14, 2014

### Fredrik

Staff Emeritus
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, $S=\{x\in\mathbb R|x>0\}$.

Last edited: Sep 14, 2014
4. Sep 14, 2014

### A.T.

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

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.

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.

5. Sep 14, 2014

### MrAnchovy

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 $m \ne 0$

6. Sep 14, 2014

### A.T.

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.

7. Sep 14, 2014

### Fredrik

Staff Emeritus
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.

8. Sep 14, 2014

### A.T.

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 $m=0$.

9. Sep 14, 2014

### Fredrik

Staff Emeritus
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.

10. Sep 14, 2014

### MrAnchovy

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

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

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

11. Sep 14, 2014

### A.T.

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.

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".

12. Sep 14, 2014

### Fredrik

Staff Emeritus
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".

13. Sep 14, 2014

### MrAnchovy

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.

14. Sep 14, 2014

### A.T.

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

Thanks, that was the core of my question.

15. Sep 14, 2014

### A.T.

Yes, thanks for the clarification.

16. Sep 14, 2014

### A.T.

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.

17. Sep 14, 2014

### Staff: Mentor

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.

18. Sep 14, 2014

### Staff: Mentor

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.

Last edited: Sep 14, 2014
19. Sep 14, 2014

### Staff: Mentor

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?

20. Sep 14, 2014

### MrAnchovy

We have observed that $F = ma$ for massive objects travelling 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 $P = mv = 0c = 0$.