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In the Talk page of Wikipedia Coefficient I read this comment:

As far as I can tell, the mathematical definition should imply that coefficients are unitless, however, the physical sciences have been using "coefficient" for factors that include dimensions for a long time, where "constant" or "factor" would be a more informative term.

In the same page, examples of "physical coefficients" have both dimensionless and dimensionful coefficients. I am trying to understand the difference of how "coefficient" is used in mathematics and physics.

Thanks.

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Assigning units to mathematical constants can be a useful way to find errors. Taking the quadratic $ax^2+bx+c=0$, if you imagine $x$ having units of length then $a$ must be length$^{-2}$, $b$ must be inverse length, and $c$ must be unitless. This would catch certain errors in the derivation (or memory) of the quadratic formula. It won't catch all errors, but neither does casting out $9$'s. –  Ross Millikan Jan 14 '12 at 5:29
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2 Answers

Like Dustan has stated, there isn't a significant difference between them. It's more of a subtlety, if anything, that separates them.

In physics, coefficients refer to actual physical things. In mathematics, it's not like that. Mathematics tends to characterize general abstractions. For example, in the quadratic equation $ax^2+bx+c=0$, there is nothing physical about $a$,$b$, and $c$. They don't represent any physical thing. However, in $F_{\text{friction}}=\mu N$, $\mu$ represents a very real thing: The coefficient of friction for a particular substance.

Perhaps an equivalent way of looking at this illustrates things better: Physical coefficients are applied mathematical coefficients. For example, $ax^2+bx+c=0$ is a quadratic equation that can become the equation for distance: $d=d_0+v_0t+\frac{1}{2}at^2$. This is done by noticing that: $$x=t, a=\frac{1}{2}a, \, b=v_0, \, c=d_0-d$$

In this sense, the quadratic's coefficients now have physical meanings. Before, they were simply symbols of abstraction. A simpler example would be the equation for linear thermal expansion: $$L=L_0+L_0\alpha \Delta T$$ This is simply a linear equation of the form $y=mx+b$ where

$$y=L, \, x=L_0, m=(1+\alpha \Delta T), b=0 $$

Just like before, the coefficients now take on a meaning, rather than just being an abstraction. Also, note that the coefficients--like in the abstraction--are things we know whilst the $x$ is something we don't know. (In most cases of this simple problem, $\alpha$ and $\Delta T$ are known quantities.)

Yet another example would be:

$$v=v_0+at$$

where this equation takes the form of, again, $y=mx+b$ with

$$m=a,\, b=v_0,\, x=t$$

$a$, acceleration, is usually taken to be constant and the intial velocity, $v_0$, is always constant. I hope this has illustrated the point better. :P

(Some of my remarks here are debatable because these equations are not the more complex and Calculus-esqe equations that describe the same thing, so take them a bit lightly. There are assumptions in all of these equations, I think.)

Is there some kind of list of coefficients according to what they do?

There is this: http://en.wikipedia.org/wiki/Physical_constant

Constants are just coefficients that don't vary and occur a lot. For example, acceleration due to gravity (on the Earth) is $9.8$ m/s$^2$ and is a particular value of the coefficient $a$ (when $a$ is taken to mean acceleration, of course). Here is another good reference: http://en.wikipedia.org/wiki/Variables_commonly_used_in_physics

I also use this a lot to understand some of the basic facts of the variables in Physics, a lot of them being coefficients: http://en.wikipedia.org/wiki/List_of_physical_quantities

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Thanks for great answer. This helps me understand a couple of things that were unclear. First, I am more specifically interested in G, the gravitational constant. From your answer I would say that G is not the coefficient but GMm is the coefficient of the variable 1/rr. Correct? Second, you say that coefficients are things we know. So, g in the kinematics equation is a number with a unit that we know independently, we cannot measure its value where it occurs as a coefficient. Correct? And is this also true for G? We need an independent value of G in order to call it a “coefficient” in F=GMm/rr –  Zeynel Jan 15 '12 at 0:11
    
I was going to provide the equation for gravitational force as another example, but it actually confuses me. For me, it depends entirely on what you know. It is 100% correct to say that $G$ is a constant. (Its value, to a reasonable amount of digits, is $6.67 \cdot 10^{-11}$ m$^2$ kg$^{-1}$ s$^{-2}$.) If $m_1$ and $m_2$ are known values, then it seems (to me) to be correct to say that $Gm_1m_2$ is the coefficient in the equation $F=G\frac{m_1m_2}{r^2}$. (I'm just using my own version of the exact same equation you stated, btw.) Now, this might confuse you, but follow me here: $g$, as in,. . . –  000 Jan 15 '12 at 6:59
    
$9.8$ m s$^{-2}$ is a constant and is considered a coefficient. However. . . These constants can be derived by silly algebra, if necessary. That's how most of them were derived, which is the funny part. People figured out the equations and that there is a constant of proportionality between (thing 1) and (thing 2) and measured them using algebra. To be specific, $g$ can be derived based on (in one case) the equation of motion: $d=d_0+v_0t+\frac{1}{2}at^2$ Assume that this is the motion of an object in free fall from an initial velocity of 0. And that no distance has been covered yet. . . . –  000 Jan 15 '12 at 7:06
    
Thus, we have: $d=\frac{1}{2}at^2$. If we know $d$ and $t$, then we can calculate the acceleration via algebra: $a=\frac{2d}{t^2}$ Now, if we are on Earth, then $a$ is $g$. That is, $a$ will reduce to $9.8$ m s$^{-2}$. In this sense, we can measure our constants and coefficients. However, our measurements will only be as precise as our measurements of the variables (in this case, $d$ and $t$). So, it is true that $g$ has an established value and that $G$ does as well, but it is a bit misleading to say, "we cannot measure its value where it occurs as a coefficient." –  000 Jan 15 '12 at 7:13
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The comment doesn't mean that physical coefficients must have a unit; it just means that some do.

There is no real difference, as far as I know, in the definition of coefficient for math or physics. The difference is that, in mathematics, units are rarely assigned to the quantities used, whereas in physics, units are fundamentally important to keep track of. This complicates the description of equations with coefficients slightly; consider, for instance, the kinematics equation

$$r = \frac{1}{2} a t^2 + v_0 t + r_0$$

To a mathematician, this equation describes $r$ as a simple quadratic in $t$, given a few other parameters, and we don't have to do any verification up front that it makes any sense; it is a certainly a proper family of functions. To a physicist, this equation can only make sense after we have verified that the units of the terms being added and equated match up. This means that the coefficient of each term must adjust to ensure that the product has the same unit as all of the other terms, as they do, in this case.

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Ok, thanks, now I realize that I am more interested in the relation of the coefficient to the term it qualifies (rather than units). So coefficient must be a number which is constant as in the kinematics equation. But in some examples that they give in that page, a coefficient is a property of the object, for instance, the lift coefficient is the property of the airfoil, or ballistic coefficient also. In the kinematics equation acceleration is not a property of t. I don't know if this makes sense. Is there some kind of list of coefficients according to what they do? –  Zeynel Jan 14 '12 at 3:47
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I think user22144's answer elaborates a bit better. Coefficients may be fundamental constants of the universe, properties of certain objects, or just features of a particular event ($a$ in the kinematics equation falls in the third category), and there may be more categories I haven't thought of. I don't know of a list of coefficients according that sort of categorization, which I think is what you're asking for? –  Dustan Levenstein Jan 14 '12 at 3:54
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