Are Proportional Variables ALWAYS Multiplied/Divided from One Another?

  • #1
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To me, it seems like variables that are proportional to the same result are almost always (if not always) multiplied or divided by each other. I've noticed that this is the case in all the physics equations that I've seen so far.

A good example of this is Newton's Second Law:

[tex]{F}\propto{m}[/tex]

[tex]{F}\propto{a}[/tex]

[tex]{F}\propto{ma}[/tex]

What's the reasoning behind multiplying the two? Is it just how a proportional relationship is defined? Is it because of the Multiplicative Property of Zero? Could this be re-written any other way? (And I don't mean using calculus)

I can see how multiplicative properties can be a lot more useful than properties of addition but it just seems kind of crazy to me that multiplication/division are the only ways to quantify some physical quantities :/


Thanks for taking the time to read!
 

Answers and Replies

  • #2
symbolipoint
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Physical number qualities are not all either connected by addition-subtraction or all connected by multiplication-division. Some related qualities or numeric properties are connected by multiplication, some are connected by division, some are connected by addition.

In the proportionality situation which you gave example for, the language is "directly proportional", and this would be F = k*a, where k is a constant (and the asterisk I use as multiplication). F is directly proportional to a. Also, F is directly proportional to m, so maybe say, F=j*m (I'm using j as a constant but different from k).
 
  • #3
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But still, [tex]\textbf{F=ma}[/tex] seems the only possible way (again, not including its calc form/notation) that it can be written. And the only reasons for this (that I can think of at least) are because of the special cases where either [tex]\textbf{m}[/tex] or [tex]\textbf{a}[/tex] = 0, [tex]\textbf{m}[/tex] or [tex]\textbf{a}[/tex] is a fraction/decimal, or if one of the variables is negative (negative acceleration/deceleration).


And of course, this is all disregarding dimensions and units :P

Is this truly the only way it can be written? (basic notation, not calculus notation)
 
  • #4
Mute
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[itex]F \propto m[/itex] means that [itex]F = c_1 m[/itex], where c_1 may depend on as many other parameters as you like, but it cannot depend on m. Similarly, [itex]F \propto a[/itex] means that [itex]F = c_2 a[/itex], where c_2 does not depend on a. If you divide these two equations, since F is the same in both, you get

[tex]1= \frac{c_1(a) m}{c_2(m) a},[/tex]
where I have noted that c1 depends on a somehow and c_2 depends on m somehow (as well as both possibly depending on other parameters). Rearranging:

[tex]\frac{c_1(a)}{a} = \frac{c_2(m)}{m}.[/tex]
This equation states that the left hand side, which depends only on a (or at least not on m) is equal to the right hand side, which depends only on m (or at least not a). The only way a function f(x) can equal something independent of x (for all values of x), is for the function to be a constant. Hence, for this equality to hold, both sides must be equal to the same constant, say, [itex]c_3[/itex]. Hence,[itex]c_1(a) = c_3 a[/itex] and [itex]c_2(m) = c_3 m[/itex]. This tells us

[tex]F = c_3 ma[/tex]

or [itex]F \propto ma[/itex]. Hence it is proven in this two variable case. The proof could be extended pairwise to cases of higher numbers of variables.
 
  • #5
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[itex]F \propto m[/itex] means that [itex]F = c_1 m[/itex], where c_1 may depend on as many other parameters as you like, but it cannot depend on m. Similarly, [itex]F \propto a[/itex] means that [itex]F = c_2 a[/itex], where c_2 does not depend on a. If you divide these two equations, since F is the same in both, you get

[tex]1= \frac{c_1(a) m}{c_2(m) a},[/tex]
where I have noted that c1 depends on a somehow and c_2 depends on m somehow (as well as both possibly depending on other parameters). Rearranging:

[tex]\frac{c_1(a)}{a} = \frac{c_2(m)}{m}.[/tex]
Sorry I'm a little confused. When you noted that c1 depends on a and c2 depends on m, was that as if you were taking a guess or is there a reason behind that?

Also, (not trying to be clever) why is c1 and c2 multiplied to the variables? Could they be added or subtracted by the variables instead and still result in the same formula?
 
  • #6
Mute
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Sorry I'm a little confused. When you noted that c1 depends on a and c2 depends on m, was that as if you were taking a guess or is there a reason behind that?
It is logically impossible that c_1 is independent of a and c_2 is independent of m, provided the variables a and m are independent of each other. If I do two experiments to measure the quantity F, one in which I vary m and hold all other variables constant, and it gives me the relation [itex]F = c_1 m[/itex], and other experiment in which I vary a and hold all other variables constant, giving me the relation [itex]F = c_2 a[/itex], then these two formulas actually represent the same formula; it's just that in each experiment I didn't get the full information on how F varies with its variables because I only tuned one variable at a time. I thus know that F depends on both m and a. Thus, F is really F(m,a). One relation says [itex]F(m,a) = c_1 m[/itex]. m does not depend on a by assumption of the independence of m and a, so the only way F can depend on a according to this relation is if c_1 depends on a. A similar arguments works for c_2 depending on m.

Also, (not trying to be clever) why is c1 and c2 multiplied to the variables? Could they be added or subtracted by the variables instead and still result in the same formula?
The notation "[itex]A \propto B[/itex]" means, by definition, "[itex]A = cB[/itex], for some "constant" c (which is only constant in that it doesn't depend on B).
 
  • #7
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Oh okay I think I gotcha. So you're saying the only way the two constants (c_1 & c_2), multiplied by their respective variables, can stay consistent with the same F is if c_1 = the variable on the other side of the equation and vice versa?
 
  • #8
Mute
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They can only be consistent if c_1 is proportional to the other variable, and vice versa. The mathematical argument in my first post in the thread shows the result precisely.
 
  • #9
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I've wondered about the same thing.

For example, in physics, "a is proportional to b" is defined as "if I multiply a with a number, b gets multiplied with that same number". An obvious example is b = 2a.

But anyway, if that's what proportionality means, isn't it also valid to say "force is proportional to speed"? Okay I can hear you shout "no!", but what about the fact that [tex]F = m \frac{\mathrm d v}{\mathrm d t}[/tex]? This implies that if we multiply v with a number, F gets multiplied with that same number! ...

I'm not arguing for the fact F is actually proportional to v (it doesn't sound physically sound), but the working definition seems to imply this! For example, in my Introduction to Electrodynamics book (Griffiths), he somewhere says "seeing as this quantity zilch is = something times [tex]\int I [/tex] something, zilch must be proportional to I", but that is the same reasoning that leads me to say F is proportional to v (but with the differentation replaced with an integration)!
 
  • #10
Mute
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I've wondered about the same thing.

For example, in physics, "a is proportional to b" is defined as "if I multiply a with a number, b gets multiplied with that same number". An obvious example is b = 2a.

But anyway, if that's what proportionality means, isn't it also valid to say "force is proportional to speed"? Okay I can hear you shout "no!", but what about the fact that [tex]F = m \frac{\mathrm d v}{\mathrm d t}[/tex]? This implies that if we multiply v with a number, F gets multiplied with that same number! ...
Your starting definition is imprecise, though - it needs a mathematical formulation. I would say the definition of proportionality is "[itex]A \propto B \Rightarrow A = cB[/itex], where c is a non-zero number, that may in principle depend on other quantities, but not B. The requirement that c be a number rules out saying F is proportional to velocity, as that would require the proportionality factor to actually be an operator. So, you could amend your definition of "proportional" to allow c to be an operator, and then F would be proportional to v, but I don't think anyone else's definition would agree with yours. :)

I'm not arguing for the fact F is actually proportional to v (it doesn't sound physically sound), but the working definition seems to imply this! For example, in my Introduction to Electrodynamics book (Griffiths), he somewhere says "seeing as this quantity zilch is = something times [tex]\int I [/tex] something, zilch must be proportional to I", but that is the same reasoning that leads me to say F is proportional to v (but with the differentation replaced with an integration)!
There's a different line of reasoning that goes into that, though. Saying [itex]\int I = 0 \Rightarrow I = 0[/itex] (proportionality sign is unnecessary) typically follows from the integral being zero for all integration regions, hence the only way for [itex]\int I[/itex] to be zero is if I is zero. A similar situation is integrals of the form [itex]\int_a^b Ig = 0 \Rightarrow I = 0[/itex] for fixed limits but any arbitrary function g. If you do something like fix the range and the function g, then you have some clear counterexamples to [itex]\int I = 0 \Rightarrow I = 0[/itex]! e.g.,

[tex]\int_0^{2\pi} dx I(x) = 0[/tex]
by no means implies I = 0. We could have [itex]I(x) = \sin(nx + \phi)[/itex], for instance. (n an integer, of course). :)
 
  • #11
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Mute, where did you get the zero from?

EDIT: Oh I see, you presumed "zilch" meant zero, I just meant it to be an unspecified variable
 
  • #12
Mute
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Mute, where did you get the zero from?

EDIT: Oh I see, you presumed "zilch" meant zero, I just meant it to be an unspecified variable
Ah. You'll have to forgive me for the misunderstanding, then, as I have only ever heard of zilch being used to mean zero. If Griffith's says [itex] A \propto \int I \Rightarrow A \propto I[/itex], then I would be inclined to disagree, and say that Griffith's is using a non-standard definition of proportional (or perhaps I is a constant, in which case the statement is true. ;) )
 
  • #13
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Well thank you for returning my confidence in my notion of proportionality (in the sense that I'm right to ignore any other definition except "a and b are proportional <=> a/b = constant"); maybe I was too quick on interpreting Griffiths, I must watch out for said statement next time around, maybe I was indeed constant or maybe I misread something, thank you.
 
  • #14
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This is a little off topic but since Euler is credited as being the one that coined the actual formula "F = ma", did he use the same procedure in creating the formula? If he never figured mass as the constant between acceleration and force, would he still have been able to create this relationship?

Or am I thinking it in reverse? Was Euler able to make this formula only BECAUSE Newton laid down the experimental data & proportionality factors beforehand?
 
  • #15
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I just realized something; if Euler did indeed use Newton's experimental data, wouldn't he need established units of force? Obviously there's something wrong in my thinking because how can there be units of force without an already-made formula of its components? (mass and distance/square time)

Also, I'm not the greatest at math so I'm having trouble understanding how the result of one ratio (say [tex]\frac{F}{m}[/tex]) shows anything about the other ([tex]\frac{F}{a}[/tex]).

And couldn't this support my above statement as well? Wouldn't you need units of force in order to find equivalencies between force, mass and acceleration? ([tex]\frac{F}{m}={a}[/tex] and [tex]\frac{F}{a}={m}[/tex])
 
  • #16
Mute
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As my earlier argument showed, we found [itex]F = c_3 ma[/itex], for some constant c_3. Force is not known to depend on anything other than mass and acceleration, so c_3 is truly a constant. The units of force are simply chosen such that c_3 = 1.

As for Euler, I had not previously heard of his contributions to F=ma laws. I looked it up on wikipedia, and it looks like Newton originally wrote down F = ma as a law for point particles, and Euler proved it also holds for rigid bodies, where a is taken to be the centre of mass acceleration of the object.
 
  • #17
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Sorry for all the trouble Mute and thanks for being so helpful and responsive. But I'm kinda stuck on what you meant on your original post.

Are you saying that c_1 must be equal to a (and vice versa) OR are you saying that c_1 must be PROPORTIONAL to a (and vice versa)?

I've looked at your math from your original post and I've been trying to figure it out in math that is a little easier for me to understand (the [tex]\textfb{c_1(a)}[/tex] and [tex]\textfb{c_2(m)}[/tex] sorta confuse me :/ )

Following your explanation though, what I've got so far is that if you have [tex]{c_1 m}\propto{c_2 a}[/tex], then that could mean [tex]\textfb{c_1 = a}[/tex] and [tex]\textfb{c_2 = m}[/tex]. Obviously, this is not always the case so this can't be true. The only other way I can think of writing this down is [tex]\frac{c_1}{c_2}=\frac{a}{m}[/tex], but that's it. I don't know how to get where you did in saying that [tex]\textfb{F = c_3 ma}[/tex].

Here's an example of why this is confusing me:

[tex]{3}\times{5} = {5}\times{3}[/tex]

... But it could also be...

[tex]{3}\times{5} = {7.5}\times{2}[/tex]

I don't understand how you can get [tex]{F = ma}[/tex] without explicitly stating that [tex]{c_1 = a}[/tex] and [tex]{c_2 = m}[/tex]

:cry:

Again, sorry for all the trouble and thanks so much for helping me to understand this!
 
  • #18
Mute
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My mathematical argument shows that [itex]c_1(a) = c_3 a[/itex] and [itex]c_2(m) = c_3m[/itex]. This follows from the fact I showed [itex]c_1(a)/a = c_2(m)/m[/itex]. The only way f(x) = g(y), where x and y are independent, is if f(x) = constant and g(y) = that same constant.

Then, we know, [itex]F = c_1(a)m = c_3 ma[/itex]. We then choose units such that c_3 = 1 to get F = ma.
 

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