Tag Info

Hot answers tagged dimensional-analysis

82

It doesn't matter where the equation came from - a fit to experimental data or a deep string theoretic construction - or who made the equation - Albert Einstein or your next-door neighbour - if the dimensions don't agree on the left- and right-hand sides, it's nonsense. Consider e.g. my new theory that the mass of an electron equals the speed of light. It's ...

76

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 ...

74

The answers are no and no. Being dimensionless or having the same dimension is a necessary condition for quantities to be "compatible", it is not a sufficient one. What one is trying to avoid is called category error. There is analogous situation in computer programming: one wishes to avoid putting values of some data type into places reserved for a ...

70

Whenever I think about this problem I go back to one of Joel Spolsky's articles, "Making Wrong Code Look Wrong", which talks about Hungarian notation. Not only the useless kind of Hungarian notation, where variables are named in a way that describes their types (f_pos for a float, d_pos for a double, etc.) - this is "Systems Hungarian" in the article - but ...

67

first of all, the question you are asking is very important and you may master it completely. Dimensionful constants are those that have units - like $c, \hbar, G$, or even $k_{\rm Boltzmann}$ or $\epsilon_0$ in SI. The units - such as meter; kilogram; second; Ampere; kelvin - have been chosen partially arbitrarily. They're results of random cultural ...

33

A standard argument to deny possibility of inserting dimensionful quantities into transcendental functions is the following expression for Taylor expansion of e.g. $\exp(\cdot)$: $$e^x = \sum_n \frac{x^{n}}{n!} = 1 + x +\frac{x^2}{2} + \dots\,.\tag1$$ Here we'd add quantities with different dimensions, which you have already accepted makes no sense. ...

33

It depends what you mean by "unit". If you mean something like "seconds", then no. Counterexample: 1 minute = 60 seconds has different units on both sides, but they're both representing a duration, so they can still be equal. If you mean something like "time", then yes. An equation means two things are equal, i.e. the same. For that to be true, they have ...

29

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 ...

26

Yes, logarithms always give dimensionless numbers, but no, it's not physical to take the logarithm of anything with units. Instead, there is always some standard unit. For your example, the standard is the kilometer. Then 20 km, under the log transformation, becomes $\ln(20\;\textrm{km}\;/\;\textrm{km}\;)$. Similarly, the log of 10 cm, with this scale is ...

26

You are absolutely right about the dimensional analysis. The use of $\ln T$ etc. is always a shorthand for $\ln \left(\frac{T}{T_0}\right)$ which is okay to use if for some reason you don't care about $T_0$, i.e. because it cancels out or you are interested in the asymptotic behaviour only. In any expression where you have to take derivatives to get ...

25

The expression $(\hbar G/c^3)^{1/2}$ is the unique product of powers of $\hbar, G,c$, three most universal dimensionful constants, that has the unit of length. Because the constants $\hbar, G,c$ describe the fundamental processes of quantum mechanics, gravity, and special relativity, respectively, the length scale obtained in this way expresses the typical ...

23

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 ...

23

As ACuriousMind has already noted, you can geometrically interpret the length of the cross product of two vectors as the area of the parallelogram (or as twice the area of the triangle) spanned by them, and (the absolute values of) its components as the areas of the projections of that parallelogram onto the coordinate planes. As for the dot product of two ...

22


21

Here's one "mathematical" but highly unphysical answer. Using that $km\cdot km = (km)^2$ etc, we can formally define arithmetic of numbers with units over a graded algebra $A = \oplus_{k\in \mathbb{N}} V_k$ where $V_k = \otimes^k V$ where $V$ is treated as a one-dimensional real vector space ($V_0$ is the scalar $\mathbb{R}$). The choice of unit is the ...

19

Lubos Motl's answer is completely right, but I'll add my perspective anyway. For many compound units, you shouldn't try to "visualize" the meaning of the unit, but you should think of it as reminding you about relationships between that quantity and others. Why are the units of Newton's constant $G$ ${\rm N\ m^2/kg^2}$? It's because $G$'s "purpose in life" ...

18

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 ...

18

I had an extensive look around, and I turned up four conventions. This included a short poll of google, other questions on this and other sites, and multiple standards documents. (I make no claim of exhaustiveness or infallibility, by the way.) Using $[q]$ to denote commensurability as an equivalence relation. That is, if $q$ and $p$ have the same ...

17

There is no reason why you should be "imagining" a squared second. Most quantities in physics don't have any canonical "geometric" visualization and there is no reason why they should have. What matters is that you should be able to calculate with it. For example, the gravitational acceleration on Earth is $9.81\,\,{\rm m/s}^2$. This simply means that the ...

17

Temperature is nothing else than energy per degree of freedom. It is purely for historical reasons that energy per degree of freedom is measured in Kelvin, and not in, say, micro-eV. It is just that these systems of units got fixed and became widely used before the statistical meaning of temperature became clear. For the same reason, mass measured in kg ...

17

The dimensional units in an equation must balance. Sometimes a dimensionless "unit" may appear on one side and not be obvious (or even present) on the other side. For example, consider the kinetic energy of a spinning object: $$K_s = \frac{1}{2}\mathcal{I}\omega^2.$$ A comparison of SI units yields the following: $$[J]=[kg\cdot m^2]\frac{[rad]^2}{[s]^2}$$ ...

16

This is a fun question. I have a hard time getting a good grip on the transformation that is $ln$ so I'll write things in terms of exponents. $$value = \ln(10\ \mathrm{ km})$$ $$e^{value} = 10\ \mathrm{ km}$$ The number $e$ is, of course, unit-less. If I raise a number to a power, what are the permissible units of the power? If I write $x^2$, I have an ...

16

Seems valid to me... I think the interpretation of your example could be that you are doing something with some joules, for some period of time, and that the product of the number of joules and the number of seconds has a certain value, and the units would be... joule-seconds! I can picture the operator of an energy storage facility quoting a price for ...

15

It's a side effect of the unreasonable effectiveness of mathematics. You are in good company thinking it is a little strange. Many quantities in physics can be related to each other by a few lines of algebra. These tend to be the models that we think of as "pretty." Terms manipulated by pure algebra tend to pick up integer factors, or factors that are ...

15

$G$ is just a constant of proportionality to get the units right (so that when $m_1$ and $m_2$ are in kilograms and $r$ is in meters you get a force in Newtons rather than wingdingalings or something really weird). Indeed cosmologists like to work in a system of units where $G = c = 1 \text{ (dimensionless)}$, and particle physicists like to work in units ...

15

Using fundamental physical constants, try to construct an expression which has a length unit. So using dimensional analysis, we have: $G = m^3 \cdot kg^{-1} \cdot s^{-2}$ $c = m \cdot s^{-1}$ and $\hbar = J \cdot s = kg \cdot m^2 \cdot s^{-1}$. Than we are to construct length $l = m$ in the following way: $$l = G^a c^b \hbar^d = m^{3a + b+d} \cdot ... 15 No. All equations have the same dimension on both sides. Dimensions are mass, distance, time, speed, acceleration, force, power, electric current, electric charge etc. As long as you work with symbolic relations, you only care about dimensions. The equation$$v = \frac{s}{t} (velocity = distance / time) works with any units as long as they are units for ...

15

The length of the cross product of two vectors is the area of the parallelogram spanned by them, so the square-meters are the correct unit as well as geometrically meaningful - it's really an area. The $x$-component is the area of the projection of the parallelogram onto the $y$-$z$-plane, the $y$-component the area of the projection onto the $z$-$x$-plane ...

15

Your friend's question is perceptive but not at odds with your earlier answer. When you compare the length of something with a unit (1 meter), the ratio is indeed a unitless number. But then all numbers (1.5, $\pi$, 42) are unitless. When you want to determine speed you divide displacement by time - each of which has units. But what you enter into you ...

14

Yes. If you square a variable, its unit of measurement is also squared, in the case of speed $v$ in $m/s$ ($ms^{-1}$), then $v^2$ is expressed in $m^2s^{-2}$. This is true for all physical variables (or constants).

Only top voted, non community-wiki answers of a minimum length are eligible