Dividing by m to make conclusions for m=0

[pbuk]

I don't think it's unreasonable to describe the theorem

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\subseteq S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

as "the acceleration is independent of mass". It's obviously not as clear a statement as the theorem, but it sounds like an attempt to summarize the content of the theorem.

Are you saying that that theorem holds even where a = a(m) may depend on m?

 

[A.T.]

The whole point of the math derivation is only to show that given $mx=my,\forall m \in S$ you do indeed get $x=y$.

You get it only under the assumption of independence of m, as you explained in post #57.

Since we were not trying to separate them in the other thread, the independence I mentioned there is obtained not only from this mathematical step...

Why "not only"?. Shouldn't that be a simple "not"? A mathematical step that assumes independence from m, cannot be used at all to show the independence from m.

...but also from other knowledge of the physics (specifically that g is independent of mass).

So "acceleration is independent of mass" was a premise of your derivation, not its result?

 

[Dale]

Are you saying that that theorem holds even where a = a(m) may depend on m?

It obviously holds as long as a is a vector and m is an element of its field. Did you not understand post 42?

That said, I cannot think how to make a function a vector over its own domain. It seems like the equivalent operation of vector multiplication would convert a function to a real, an inconsistency in your approach that I have mentioned several times.

 

[Dale]

So "acceleration is independent of mass" was a premise of your derivation, not its result?

"g is independent of mass" was an unstated premise of my derivation. "a is independent of mass" was a conclusion, obtained through the fact that a=g.

A.T., in 42 I posted a solid proof of the mathematical topic under discussion for THIS thread. 22 posts later, you have never even acknowledged it even though it is central to the topic of this thread. Instead of commenting on that proof you have begun to reach back to a different discussion in a different thread that is (unsurprisingly) different from the one here. You point out the obvious differences in my comments in what seems like an attempt to discredit the proof in 42 by discrediting me (for the heinous crime of having different comments in different threads on different topics).

Instead of avoiding the issue, please directly examine the proof in 42 and directly respond to it. If the math in 42 is solid then it is solid regardless of my own inconsistencies. Do you agree with it or disagree with it? If you disagree, on what grounds do you do so? Address the proof on its own merits, regardless of what my personal inconsistencies and failings may be.

 

[Alien8]

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 $m = 0$:

F= G \frac {m1m2}{r^2} resulting in a number greater than zero when mass m1 or m2 is zero seems about as likely as F= k_e \frac {q1q2}{r^2} resulting in a number greater than zero when charge q1 or q2 is zero.

 

[Fredrik]

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\in S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.

Are you saying that that theorem holds even where a = a(m) may depend on m?

I find it difficult to process that question. If a depends on m, then we're not talking about that theorem. Apparently you want to change something in the assumptions, but it's not entirely clear what. So all I can say is that the following two statements are both theorems with trivial proofs.

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\in S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

For all $g,a,S$ such that $S\subseteq \mathbb R$ and $g,a:S\to\mathbb R$, if $mg(m)=ma(m)$ for all $m\in S$ then $g(m)=a(m)$ for all $m\in S-\{0\}$.​

Feel free to quote, copy, paste and edit if you want to ask about a different statement.

Edit: Corrected $1\subseteq S$ to $1\in S$.

 

[pbuk]

I find it difficult to process that question. If a depends on m, then we're not talking about that theorem. Apparently you want to change something in the assumptions, but it's not entirely clear what. So all I can say is that the following two statements are both theorems with trivial proofs.

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\subseteq S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

For all $g,a,S$ such that $S\subseteq \mathbb R$ and $g,a:S\to\mathbb R$, if $mg(m)=ma(m)$ for all $m\in S$ then $g(m)=a(m)$ for all $m\in S-\{0\}$.​

Feel free to quote, copy, paste and edit if you want to ask about a different statement.

Thank you, that's exactly what I wanted to clarify.

 

[Fredrik]

I see that I've been writing $1\subseteq S$ in a few places where I meant $1\in S$. I hope that hasn't caused any confusion. I have edited my recent posts above to correct this mistake.

 

[A.T.]

Instead of avoiding the issue, please directly examine the proof in 42 and directly respond to it.

I have no problem with the proof in #42, under the assumption you state in post #57: That the relationship between x and y doesn't depend on m.

"a is independent of mass" was a conclusion, obtained through the fact that a=g.

But you have derived that a=g by doing a step which assumes that the relationship between a and g doesn't depend on m. Aren't you just concluding your previous assumption?

 

[Dale]

I have no problem with the proof in #42, under the assumption you state in post #57: That the relationship between x and y doesn't depend on m.

If x and y could depend on m and still be elements of a vector space then the proof in 42 would still hold. It is the vector nature that is important to the proof, not the independence.

That said, if you agree with the proof in 42, then I think that the math discussion is complete, and the rest is necessarily a physics discussion.

 

[A.T.]

It is the vector nature that is important to the proof, not the independence.

Okay, it depends on the type of relationship. So your derivation doesn't show that "a is independent of m", but it rules out certain types of relationships.

That said, if you agree with the proof in 42, then I think that the math discussion is complete, and the rest is necessarily a physics discussion.

I have no problem with the physics part. A physicist can always state that the potential relationships not covered by the proof are "too weird to occur in nature", and discard them on that basis.

 

[Dale]

Okay, it depends on the type of relationship. So your derivation doesn't show that "a is independent of m", but it rules out certain types of relationships.

Yes. Or rather, it "rules in" certain types of relationships.

It says that if x and y are vectors*, then then you can factor out the m without any problem including for m=0. It does not say that you definitely cannot factor out m if they are not vectors*, it just says that you definitely can factor it out if they are.

*with all of the detailed requirements on x, y, and m explicitly spelled out in 42

I have no problem with the physics part. A physicist can always state that the potential relationships not covered by the proof are "too weird to occur in nature", and discard them on that basis.

OK, then I think it is done.

 

[A.T.]

OK, then I think it is done.

Yes, thanks for the patient explanations to everyone.