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Dimensional analysis, and the notion that quantities with different units cannot be equal, is often used to justify very specific arguments, for example, you might use it to argue that a particular formula cannot possibly be the correct expression for a particular quantity. The usual approach to teaching this is to go "well kids, you can't add apples and oranges!" and then assume that the student will just find it obvious that you can't add meters and seconds.

I'm sorry, but... I don't. I'm not convinced. $5$ meters plus $10$ seconds is $15$! Screw your rules! What are the units? I don't know, I actually don't understand what that question means.

I'm specifically not convinced when this sort of thing is used to prove that certain formulae can't possibly be right. Maybe the speed of a comet is given by its period multiplied by its mass. Why not? It's a perfectly meaningful operation - just measure the quantities, multiply them, and I claim that the number you get will always equal the current speed of the comet. I don't see how "but it doesn't make sense to say mass times time is equal to distance divided by time" can be a valid counterargument, particularly because I don't really know what "mass times time" is, but that's a different issue.

If it's relevant, I'm a math student and know extremely little about physics.

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So, is 5 meters plus 10 microseconds also equal to 15? Now set our two equations equal, subtract the meters, and we've just proven that 1 microsecond = 1 second. Great job! –  chase Feb 7 at 18:39
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If you want a "pure math" example, what is $1 + (5\ 2\ 3)^T + \begin{pmatrix}a\ b\\c\ d\end{pmatrix}$? –  chase Feb 7 at 18:46
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Here is a link to Terrence Tao's blog, where he discusses mathematical underpinning of physical units. I personally guess that types might be a better approach. @KyleKanos: I don't see how what you answered clears things up at all. You just say "that's how we do it". The motivation, or why certain things "make sense" and others don't, is lost on me. I, for one, am a PhD student with a physics degree and I don't quite why physical units work either. –  NikolajK Feb 7 at 19:02
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As long as nobody here is actually trying to challenge the validity of dimensional analysis, I think this is a pretty good question. –  David Z Feb 7 at 20:30
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@GlenTheUdderboat, that is assuming some other conversion factor which converts the time to length. In your case, it's the speed of light. In a F1 race, it's the speed of the cars. End of the day, we use length to measure length, whatever magic you do to make it look like we're using time. –  Pranav Hosangadi Feb 7 at 21:28
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7 Answers 7

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Physics is independent of our choice of units

And for something like a length plus a time, there is no way to uniquely specify a result that does not depend on the units you choose for the length or for the time.

Any measurable quantity belongs to some set $\mathcal{M}$. Often, this measurable quantity comes with some notion of "addition" or "concatenation". For example, the length of a rod is $L \in \mathcal{L}$ a measurable quantity. You can define an addition operation $+$ on $\mathcal{L}$ by saying that $L_1 + L_2$ is the length of a the rod formed by sticking rods 1 and 2 end-to-end.

The fact that we attach a real number to it means that we have an isomorphism $$ u_{\mathcal{M}} \colon \mathcal{M} \to \mathbb{R}, $$ in which $$ u_{\mathcal{M}}(L_1 + L_2) = u_{\mathcal{M}}(L_1) + u_{\mathcal{M}}(L_2). $$ A choice of units is essentially a choice of this isomorphism. Recall that an isomorphism is invertible, so for any real number $x$ you have a possible measurement $u_{\mathcal{M}}^{-1}(x)$. I'm being fuzzy about whether $\mathbb{R}$ is the set of real numbers or just the positive numbers; i.e. whether these are groups, monoids, or something else. I don't think it matters a lot for this post and, more importantly, I haven't figured it all out.

Now, since physics should be independent of our choice of units, it should be independent of the particular isomorphisms $u_Q$, $u_R$, $u_S$, etc. that we use for our measurables $Q$, $R$, $S$, etc. A change of units is an automorphism of the real numbers; given two units $u_Q$ and $u'_Q$, the change of units is $$ \omega_{u,u'} \equiv u'_Q \circ u_Q^{-1}$$ or, equivalently, $$ \omega_{u,u'} \colon \mathbb{R} \to \mathbb{R} \ni \omega(x) = u'_Q(u_Q^{-1}(x)). $$ Therefore, $$ \omega(x+y) = u'_Q(u_Q^{-1}(x+y)) \\ = u'_Q(u_Q^{-1}(x)+u_Q^{-1}(y)) \\ = u'_Q(u_Q^{-1}(x)) + u'_Q(u_Q^{-1}(y)) \\ = \omega(x) + \omega(y). $$ So, since $\omega$ is an automorphism of the reals, it must be a rescaling $\omega(x) = \lambda x$ with some relative scale $\lambda$.

Consider a typical physical formula, e.g., $$ F \colon Q \times R \to S \ni F(q,r) = s, $$ where $Q$, $R$, and $S$ are all additive measurable in the sense defined above. Give all three of these measurables units. Then there is a function $$ f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R} $$ defined by $$ f(x,y) = u_S(F(u_Q^{-1}(x),u_R^{-1}(y)). $$

The requirement that physics must be independent of units means that if the units for $Q$ and $R$ are scaled by some amounts $\lambda_Q$ and $\lambda_R$, then there must be a rescaling of $S$, $\lambda_S$, such that

$$ f(\lambda_Q x, \lambda_R y) = \lambda_S f(x,y). $$

For example, imagine the momentum function taking a mass $m \in M$ and a velocity $v \in V$ to give a momentum $p \in P$. Choosing $\text{kg}$ for mass, $\text{m/s}$ for velocity, and $\text{kg}\,\text{m/s}$ for momentum, this equation is $$ p(m,v) = m*v. $$ Now, if the mass unit is changed to $\text{g}$, it is scaled by $1000$, and if the velocity is changed to $\text{cm/s}$, it is scaled by $100$. Unit dependence requires that there be a rescaling of momentum such that $$ p(1000m,100v) = \lambda p(m,v). $$ This is simple -- $10^5 mv = \lambda mv$ and so $\lambda = 10^5$. In other words, $$ p[\text{g} \, \text{cm/s}] = 10^5 p[\text{kg} \, \text{m/s}]. $$

Now, let's consider a hypothetical situation where we have a quantity called "length plus time", defined that when length is measured in meters and time in seconds, and "length plus time" in some hypothetical unit called "meter+second", the equation for "length plus time" is $$ f(l,t) = l + t. $$ This is what you've said - $10 \text{ m} + 5 \text{ s} = 15 \text{ "m+s"}$. Now, is this equation invariant under a change of units? Change the length scale by $\lambda_L$ and the time scale by $\lambda_T$. Is there a number $\Lambda$ such that $$ f(\lambda_L l, \lambda_T t) = \lambda_L l + \lambda_T t $$ is equal to $$ \Lambda f(l,t) = \Lambda(l+t) $$ for all lengths and times $l$ and $t$? No! Therefore, this equation $f = l + t$ cannot be a valid representation in real numbers of a physical formula.

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Regarding your edit (v2) "And for something like a length plus a time, there is no way to uniquely specify a result that does not depend on the units you choose for the length or for the time", I commented to the question that the definitions or metre and second are interdependent. My comment was intended to be mostly funny (I have a bad sense of humour), but it is not completely without serious intent. (I think it would be good to at least observe this definitional interdependence.) Not sure though. –  Glen The Udderboat Feb 7 at 21:14
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Hm, you mean since they are related through the speed of light? It is a serious point, but if we go too far down that route then we'll be telling Jack M to do whatever he wants with the values of physical quantities as long as they are in Planck units! A thoughtful discussion of why dimensional analysis takes different forms between classical physics, SR, QM, QFT, and "TOE" (i.e. turning on and off c=1, hbar=1, G=1 in various combinations) would be interesting. –  jwimberley Feb 7 at 21:32
    
I have to say this does make sense to me. Although, if I wanted to be stubborn I could question why we're so sure there couldn't be a formula that only happens to work in one set of units, but I admit that would be an unusual situation. –  Jack M Feb 8 at 10:48
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It may be worth noting that the base spacial unit is not a distance but a vector; likewise with force. Energy is not merely any force times any distance, but rather a force vector and a distance vector which is parallel to it. –  supercat Feb 8 at 20:29
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@jpmc26 Within the framework I've presented, I'd say that the problem with the Celsius and Fahrenheit scales is that they are linearly related to the abstract temperature, but not isomorphic to it. If 0 represents absolute 0, $u_K(0) = 0$, while $u_C(0) = -274$ and $u_F(0) = -460$. So, the latter two scales fail to be isomorphic to the real temperature, since e.g. $u_C(x) = u_C(x+0) \neq u_C(x) + u_C(0)$. On the other hand, any rescaling of the Kelvin temperature scale is allowed (e.g. the Rankine scale). –  jwimberley Feb 11 at 13:03
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All right, I'm cashing in my comments to provide an answer:

Let's start with an example that doesn't invoke dimensions, units, or physics at all. How do we evaulate the following expression? $$ 1 + \begin{pmatrix}5\\2\\-9\end{pmatrix} + \begin{pmatrix}a\ b\\c\ d\end{pmatrix} $$

The answer is, we don't. Not without defining some special convention, like that every quantity is multiplied by some invisible matrix such that they all end up being $3\times3$ matrices... that's just arbitrary, and inconsistent.

Now, how do we interpret $5 \text{meter} + 10 \text{seconds}$? The answer is, we don't. Again, not without defining some arbitrary, inconsistent, and meaningless convention. You proposed it's equal to 15; well I will define $5\text{m}+10\mu \text{s}=15$ as well, and now we've just proven that microseconds equal seconds.

The lesson is, there's not a meaningful way to perform addition on disparate types of quantities. In physics, we refer to the general types as dimensions. Examples of dimensions are: length, time, energy, mass, etc. Units are specific ways to represent dimensions, for example meters and feet are both units of dimension length.

So, how does this relate to your comet example? You've observed (correctly) that we can in fact multiply different dimensions. E.g. mass times velocity has dimensions of momentum. But that still doesn't mean you can compare disparate quantities. The condition for your calculation to be correct is: $$ (\text{comet mass})\times(\text{period}) - (\text{the true speed}) = 0 $$ But as we've already shown, this expression is meaningless since we cannot add (or subtract) quantities of differing dimensions.

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But I don't see how you did show that (your last sentence). My claim would be that that equation will always be numerically true, which is meaningful. –  Jack M Feb 7 at 19:22
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@JackM: It's not meaningful because it's not physically true. –  Kyle Kanos Feb 7 at 19:26
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@JackM: It's not meaningful because it's not consistent. You want to do numerical comparisons, but you've got to be comparing numbers then. If you gave me a method for turning (mass)x(period) into a single number that I can compare with the speed (also as a single number), that method would not work for any other comet. You'd have to change your magical (mass)x(period)->number function for each individual comet, and so you haven't learned anything useful about comets, you're just making up numbers. The point of physics is to unify many different observed phenomenon to a set of rules. –  chase Feb 7 at 19:28
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@JackM: No we establish that through math. Math requires you to compare or perform operations on quantities. If you do this in a way that is consistent, you get meaningful results. I invite you to consider my microseconds example more carefully; the point there is that you've got to come up with some way to combine different dimensions via addition/subtraction. And any such attempt will be inconsistent. –  chase Feb 7 at 19:33
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The expression you describe is perfectly legitimate working in a tensor algebra, by the way. –  Sean D Feb 8 at 18:01
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In physics, you're not allowed to ignore the units; they come along for the ride on every sub-step of every calculation. From a mathematics perspective, consider the units to be variables, so instead of 5 meters + 10 seconds, you have 5x + 10y. Unless you arbitrarily assign x = y = 1, there is no way you're getting 15 out of this; at the end of the day, you still have 5x + 10y, and physics is uninterested in complex numbers. At the end of a computation, you need one number, and "5x + 10y" is two numbers. And here's the rub in physics: you're not allowed to assign values to those variables. They are basic, irreducable units; you can't say "meters are 1".

On the other hand, you're allowed to multiply units like nobody's business. "Furlongs per fortnight" is a silly phrase, but you can use it and everyone will (after some conversion) understand exactly what you're saying:

$$13440\frac{\text{furlongs}}{\text{fortnight}} \times \frac{1 \; \text{mile}}{8 \; \text{furlongs}} = 1680\frac{\text{miles}}{\text{fortnight}}$$

$$1680 \frac{\text{miles}}{\text{fortnight}} \times \frac{1 \; \text{fortnight}}{336 \; \text{hours}} = 5 \frac{\text{miles}}{\text{hour}}$$

(Cue a horde of angry physicists descending upon me for my use of Imperial units.)

On every step of these conversions, the units never went away; they stayed with the numbers. When you're computing something in physics, the numbers you play with are real things; they represent a quantity of something, and that something doesn't go away just because you think it's inconvenient. So if an asteroid goes 10 meters in 5 seconds, it looks like this:

$$10 \; \text{meters} \times \frac{1}{5 \; \text{seconds}} = 2 \frac{\text{meters}}{\text{second}}$$

You can multiply and divide quantities however you like, and you'll end up with some funky units, but your final number will be valid — though they may be difficult to relate to everything else. (For instance, the viscosity of a fluid is measured in (kilograms per (meter * second)), which isn't particularly intuitive but is useful in the particular places that viscosity is used.)

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+1 for "consider the units to be variables." –  Ilmari Karonen Feb 8 at 16:03
    
+1 Perhaps the OP would better see units as variables by considering "5 meters" as "5 times the length of the reference sample of length" and "10 seconds" as "10 times the duration of the sample unit of time". This should make it clear that meters and seconds are like orange and apples. Well, at least until he discovers einsteinian relativity. –  Peltio Feb 9 at 11:11
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Most of the answers seem to be reiterating that you're not allowed to add lengths and times just because it doesn't make sense. Here's why it doesn't make sense.

If two objects have the same temperature in Celsius, then they have the same temperature in Fahrenheit.

If two objects have the same speed in meters per second, then they have the same speed in miles per hour.

But if two objects have the same "length-time sum" in meters + seconds, they may not have the same "length-time sum" in other units. For example, $$60\,\text{meters}+0\,\text{seconds}\ “=”\ 0\,\text{meters}+60\,\text{seconds}$$ but $$60\,\text{meters}+0\,\text{minutes}\ “\ne”\ 0\,\text{meters}+1\,\text{minute}.$$ So it doesn't make sense to think of the length-time sum as a real physical quantity instead of a numerical coincidence caused by our choice of units.

This is basically the last part of jwimberley's answer, but I thought it would be helpful to have an explicit example spelled out so that the point is blindingly obvious.

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Jack, I will explain the problem here first in a mathematical rather than physical way. The mathematical issue at play here is that the operation you are proposing is not well-defined at the level of basic physics. Let's take a look at some situations in math where this type of problem crops up that have nothing to do with physical units.

In calculus, we have $\int_a^b f(x)\,dx = F(b) - F(a)$ where $F(x)$ is any antiderivative of $f(x)$. What if someone came along and asked if a new operation $I(f,a,b) = F(b) + F(a)$ has any useful meaning in terms of the original function $f(x)$ and interval $[a,b]$. It does not, because if you change the antiderivative you change the answer. For any two antiderivatives $F(x)$ and $G(x)$ of $f(x)$, they differ by a constant, say $G(x) = F(x) + C$. This means that a difference of antiderivatives of $f(x)$ at $a$ and $b$ is independent of the choice of antiderivatives but a sum of antiderivatives of $f(x)$ at $a$ and $b$ is not: $$ G(b) - G(a) = (F(b) + C) - (F(a) + C) = F(b) - F(a) $$ while $$ G(b) + G(a) = (F(b) + C) + (F(b) + C) = F(b) + F(a) + 2C, $$ which is not $F(b) + F(a)$ unless $C = 0$ (i.e., $G(x) = F(x)$). So a difference of values of an antiderivative of $f(x)$ is a well-defined number in terms of the original function $f(x)$, but a sum of values of an antiderivative is not. If you want to provide a definite real number from an antiderivative of $f(x)$, and have that definite number be determined solely by $f(x)$ and not the choice of antiderivarive, differences make sense but sums do not. By the way, this has a role in physics: potential energy is only defined up to an overal additive constant, which explains why a difference of potential energy values has a physical meaning but a sum of potential energy values does not.

Another example is in geometry. We add angles but we don't ever multiply angles. Is there a mathematical problem with multiplying angles? Yes: angle measurement is intrinsically only determined up to an integer multiple of $2\pi$, and this property is respected by addition but not by multiplication. If $\theta_2 = \theta_1 + 2\pi{k}$ and $\varphi_2 = \varphi_1 + 2\pi{\ell}$ for some integers $k$ and $\ell$, then $$ \theta_2 + \varphi_2 = \theta_1 + \varphi_1 + 2\pi(k+\ell), $$ so the two sums $\theta_2 + \varphi_2$ and $\theta_1 + \varphi_1$ are again equal up to an integer multiple of $2\pi$. However, $$ \theta_2\varphi_2 = \theta_1\varphi_1 + 2\pi(k\varphi_1 + \ell\theta_1 + 2\pi{k}\ell) $$ and $k\varphi_1 + \ell\theta_1 + 2\pi{k}\ell$ is not an integer all the time. (If you have had abstract algebra, I could say the "problem" here is that $2\pi{\mathbf Z}$ is a subgroup of ${\mathbf R}$ but not an ideal in ${\mathbf R}$, so the quotient ${\mathbf R}/2\pi{\mathbf Z}$ can be given the structure of an additive group but not a ring.) If you want to say "but I can talk about $\sin(xy)$ for any numbers $x$ and $y$, and that is multiplying angles", I'd say it is not: in the expression $\sin(xy)$ with real variables $x$ and $y$, the numbers $x$ and $y$ must be regarded as real numbers, not angles. The story is different with $\sin(2x)$, which is well-defined when $x$ is thought of as an angle (a number up to addition by an integral multiple of $2\pi$). This distinction is why the $x$ in a Fourier series $$ f(x) = \sum_{n \in {\mathbf Z}} c_ne^{2\pi{i}nx} $$ can be thought of as lying on a circle if you wish, but the $x$ in a Fourier transform $$ ({\mathcal F}f)(x) = \int_{{\mathbf R}} f(y)e^{2\pi{i}xy}\,dy $$ can not and must be thought of on the real line: the Fourier transform is not a $2\pi$-periodic function of $x$, so it is not well-defined to regard the Fourier transform as a function on the unit circle.

In linear algebra, the trace of a linear operator $A \colon V \rightarrow V$ on a finite-dimensional vector space is defined to be $\sum_{i} a_{ii}$ where $(a_{ij})$ is a matrix representation of $A$ in a basis of $V$. It is crucial that that this sum is independent of the choice of basis. We used a basis to compute the trace, but if you want the trace to be a function purely of the operator $A$ then it has to have the same value no matter what basis you use on $V$. In a linear algebra course you learn that the trace is independent of the basis used to compute it. On the other hand, the "anti-trace" $\sum_{i} a_{i,n+1-i}$ (sum on the antidiagonal) or "border trace" (sum around the boundary of a matrix representation of $A$) are not well-defined because if you change the basis then the new matrix representation has a different value for its anti-trace or border trace. That's why you never hear anyone talk about such sums in linear algebra, since they are not well-defined functions of the original operator: they depend on the choice of basis. To the extent you agree that geometric concepts should not depend on your choice of coordinate system, you'll agree that useful concepts in linear algebra should be independent of the choice of basis.

In algebraic geometry, polynomials are not well-defined functions on projective space since their values change if the homogeneous coordinates change. But ratios of homogeneous polynomials do give the same answer for all homogeneous coordinates of a point, and that is why ratios of homogeneous polynomials are the natural functions on projective space.

In grade school math, addition of fractions is not $(a/b) + (c/d) = (a+c)/(b+d)$, since this operation is not well-defined: although 1/2 = 5/10 and 3/4 = 6/8, this fake way of combining fractions by adding numerators and denominators doesn't lead to the same answer when you change the way you write the fraction: $(1+3)/(2+4) = 4/6$ and $(5+6)/(10+8) = 11/18 \not= 4/6$. If you were to fix a preferred representation of fractions, such as the representation using relatively prime numerator and denominator with a positive denominator, then this "add the numerators and add the denominators" is a well-defined operation, but it would be very awkward to use because it would depend on the way you write the fractions. This fake addition does have an interesting application, which you'll learn if you read about Farey fractions; it just doesn't correspond to addition, so we shouldn't denote it as +, and it does not generalize to fractions where the numerator and denominator are in a ring that lacks unique factorization (and a preferred choice of unit multiple of each nonzero element).

If you don't think having operations be well-defined is important in math then you're going to be in for a mountain of trouble when you learn algebra (quotient groups) or differential geometry (manifolds), where you regularly have to define functions by making a choice and then check the answer is independent of the choice that was made (a choice could mean a coset representative or a choice of coordinate system near a point).

And if you don't think issues of "units" occur in math, you're mistaken. They are just hidden enough that you may not notice them. To measure angles we prefer to use radians. If you wanted to use other systems of measuring angles then the familiar derivative formulas for trigonometric functions would change: while $\sin'(x) = \cos(x)$ when $x$ is an angle in radians, if you change $x$ to degrees then $\sin'(x) = (\pi/180)\cos(x)$. We prefer radians because they lead to the simplest calculus formulas, without any weird factors like $\pi/180$ showing up. In Fourier analysis, some prefer to define the Fourier transform using $e^{ixy}$ instead of $e^{2\pi{i}xy}$, and then factors of $2\pi$ or $\sqrt{2\pi}$ start showing up in other formulas from Fourier analysis such as Parseval's formula. In linear algebra, we prefer to take as the "natural" isomorphism from a finite-dimensional vector space $V$ to its double dual space the mapping $v \mapsto {\rm ev}_v$, where ${\rm ev}_v(\varphi) = \varphi(v)$ for all linear functionals $\varphi$ on $V$, but there are other possibilities, namely $v \mapsto c\cdot{\rm ev}_v$ for any nonzero element $c$ of the underlying scalar field. Category-theoretic arguments show that these are essentially the only possible natural isomorphisms to the double-dual space.

Now I'll turn to physical measurements. If you want to add a length and a time together, you need to recognize that there is no natural standard for measuring either of these quantities: any two systems of measuring length differ by a scaling factor, and any two systems of measuring time differ by a scaling factor. Even if everyone on our planet used the metric system, it doesn't make that system physically profound. At some point in the past someone picked a length and declared it one meter, but that human convention doesn't have any physical importance. (If you think metric units are actually an essential part of the fabric of nature, then something has gone badly wrong in your education. Maybe the "radius of the electron" or the Planck length could be considered a physically fundamental length, but your question is on a much more elementary level than that.) The link between different measurements of the same physical quantity is not always just a scaling factor (temperature is the best example of that, where $F = (9/5)C + 32$), but for simplicity let's stick to conversions between different systems of measurement as being just scaling factors.

Because of the physical "fact" that different systems of measuring the same physical concept differ by a scaling factor, a physical measurement can be thought of as a real-valued function defined up to an overall positive scaling factor. If $f$ and $g$ are two ways of measuring the same physical quantity, then $g = cf$ for some positive $c$. For instance, if we are measuring length ($L$) and write $f_L$ for the meter-function and $g_L$ for the feet-function, then $c = 3.28$: $g_L(x) = 3.28f_L(x)$ (that is, to convert from meters to feet, multiply the meters value by 3.28). If we are measuring time ($T$), with $f_T$ for the second-function and $g_T$ for the minute-function, then $c = .016$: $g_T(y) = .016f_T(y)$ (to convert from seconds to minutes, multiply the second value by .016). Now ask yourself: if a function is defined up to an overall scaling factor, and another function is defined up to an overall scaling factor, what can I do with them and keep the result defined up to an overall scaling factor? You can multiply them or divide them, but you can't add them.

For example, if $g_L = 3.28f_L$ and $g_T = .016f_T$, then $g_L/g_T = 205f_L/f_T$. Recalling what these functions meant above, this last equation says if you want to convert from meters per second to feet per minute, multiply by 205. And $g_Lg_T = .05248f_Lf_T$, so to convert from meters-seconds (whatever that means) to feet-minutes, multiply by .05248.

Let's finally try addition: if $g_L = 3.28f_L$ and $g_T = .016f_T$, is $g_L + g_T = c(f_L+f_T)$ for some $c > 0$? This is the test for whether addition of measurements is well-defined. Since $g_L + g_T = 3.28f_L + .016f_T$, you want $3.28f_L + .016f_T = c(f_L + f_T)$, so you need $(3.28-c)f_L = (c-.016)f_T$, and therefore $f_L = ((c-.016)/(3.28-c))f_T$. In other words, you need to be able to convert between length and time: length and time have to be different ways of measuring the same thing. Are they? At an elementary level they are not, and that is why you can't add length and time physically.

To add two physical measurements in a well-defined way, we have seen (for the examples of length and time) that the two quantities you're measuring have to be convertible into each other. In relativity, we learn that the speed of light is a fundamental physical speed, and if we decide it is truly fundamental we can use it to convert between length and time. In general relativity it is convenient to declare the speed of light to be 1, which sets a definite conversion between meters and seconds, or feet and seconds, or any preferred choice of measuring length and time so that the value of the speed of light using those systems of measurement turns out to be 1. (It's like our preference for radians to degrees because in calculus the use of radians makes certain coefficients in derivative formulas equal to 1.) Once you have a standard for turning length into time then you can add length and time, and then all you're doing is adding length and length. Google the term "Planck units" to see how fundamental physical theories lead to a way of converting between mass, length, and time.

I'll leave it to you to decide what this viewpoint has to say about the physical possibility of adding meters and feet. Hint: be careful about whether you're dealing with functions of the same object or different objects.

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an antenna engineer does multiply angles: the directivity of an antenna is roughly $ 4\pi/(\phi \theta)$ where $\phi$ and $\theta$ are 3db beamwidths in elevation and azimuth, respectively. –  user31748 Feb 8 at 23:59
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I had not known about antenna directivity before. The formula you write is an approximation ("roughly", as you say) based on the use of small angles somewhere. Therefore it is essential in this approximation to measure angles not with any radian measurement, but so that a small angle is measured as a small number (and not, say, as a number very close to $2\pi$). –  KCd Feb 9 at 0:14
    
The formula is an approximation because it assumes a smooth and relatively narrow beam. I deliberately left out an efficiency factor that implies the amount of energy radiated in the sidelobes, and yes the angles are measured in radians. The formula is really nothing more than saying that an isotropic radiator radiates uniformly in $4\pi$ while a directional one radiates in $\phi \theta$ so their ratio is the directivity by definition. If you multiply that ratio by say 0.5 to 0.8 you get an amazingly good estimate of most aperture and practical dishes (Cassegrain or otherwise)... –  user31748 Feb 9 at 0:40
    
I'm tempted to give this answer a +1 merely on the basis of its length. –  shortstheory Feb 9 at 12:46
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@shortstheory: or, equivalently, on the basis of the time it took to write? –  KCd Apr 18 at 8:05
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Physics is concerned with finding a mathematical model to describe and predict reality. Units are a convenient and useful way for us to distinguish between quantities that describe different facets of a system, and adding or subtracting quantities from these facets does not help us describe or predict reality. It is undefined just like adding vectors and scalars; it doesn't help us build a useful theory just like adding scalars and vectors doesn't help mathematicians build a useful theory. The rules are the product of comparing experiment with the model; they are not arbitrary.

There might be some equation in which you can use the product of mass and period to describe a comet's motion; I couldn't say off-hand. But our definition of "speed" is just labeling some particular concept in our mathematical model. (It could also be described as the rate of change in distance.) The choice of "speed" is just semantics. The name itself is arbitrary aside from the fact that listeners understand it to convey a certain meaning, but now I'm describing how language works instead of physics.

So if you can find an experimentally verifiable formula that involves adding numbers of different units, then we would consider doing it and figuring out more about how it works. (You would also be famous among scientists, I imagine.) But until that happens, we'll continue using this rule of thumb.

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Surely you can add or multiply any two numbers, those are indeed simple mathematical operations. But physics is not math, it is physics. We attach units to values to represent something physical and therefore units have meaning.

While a meter-second could have some use (via multiplication), it's impossible to add two differing units. Can you really move 5 meters and 10 seconds? Sure you can move 5 meters in 10 seconds (division), but going 5 meters and 10 seconds is a meaningless statement.

Lastly, speed is given in units of length per unit time (e.g. meters per second or miles per hour). Mass is given in kilograms and period in seconds, multiplying them you get kilogram-seconds, which isn't anything like meters per second, so you cannot get a velocity from multiplying mass and time, not matter how much you might try.

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As an aside, you probably shouldn't post with such an accusatory tone. I'd also strongly recommend taking at least the first semester physics course offered at your university. The insight you can gain will probably clear it up much more than my post (or someone else's post). –  Kyle Kanos Feb 7 at 18:42
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There is no accusatory tone. What there is is a humorous tone (which I'm honestly a bit surprised has gone so completely lost, but sorry about that). In any case, your argument still just seems to boil down to "it wouldn't make sense", which doesn't seem conclusive to me. –  Jack M Feb 7 at 18:45
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Screw your rules! is rather accusatory. –  Kyle Kanos Feb 7 at 18:46
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Honestly now, who would use a phrase like that seriously when talking about something like dimensional analysis? –  Jack M Feb 7 at 18:47
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@KyleKanos: Well, if it's arguing to you then I can understand - otherwise one has the chance to learn. And I don't think people think enough about units to understand them, mostly because the practical approach suffices most of the time. When I say I don't understand physical units I mean in the sense that mathematicians don't understand prime numbers. –  NikolajK Feb 7 at 19:38
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protected by Qmechanic Feb 8 at 23:18

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