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72

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 ...


31

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. ...


28

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

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 ...


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 ...


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 ...


20

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 ...


20

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 ...


20

$\newcommand{\t}{[\text{time}]}\newcommand{\e}{[\text{energy}]}\newcommand{\a}{[\text{angle}]}\newcommand{\l}{[\text{length}]}\newcommand{\d}[1]{\;\mathrm{d} #1}$Dimensions vs Units: I want to take an educational guess as to why angles are considered to be dimensionless whilst doing an dimensional analysis. Before doing that you should note that the angles ...


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 ...


16

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 ...


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

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 ...


15

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 ...


15

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

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).


14

The infinitesimal length interval between two events in spacetime $ds$ is defined by $$ds^2=c^2 dt^2 - dx^2 - dy^2 - dz^2$$ The creature is dimensionally consistent, because time is multiplied with a speed. You can think of $(t,x,y,z)$ as the four coordinates of spacetime $(x^0,x^1,x^2,x^3)$ and $c$ appears naturally in the equations. However, the usual ...


14

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 ...


13

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 ...


12

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 ...


12

When the electrostatic force was originally being studied, force, mass, distance and time were all fairly well understood, but the electrostatic force and electric charge were new and exotic. In the cgs system, the charge was defined in relation to the resulting electrostatic force (it's called a Franklin (Fr) an "electrostatic unit" (esu or) sometimes a ...


12

Reciprocal cartons for storing eggs might make little sense, reciprocal seconds (Hz) do. Joules times seconds (J.s) is equivalent to Joules per Hertz (J/Hz). It denotes angular momentum, the amount of energy per unit of rotational frequency. Similarly, reciprocal meters (wave numbers) also make perfect sense. So units like Newton times meter (N.m) are ...


12

OK, time to burn thru all my mod points: for (j in 0:1e6 cm,by=1Angstrom) { print("Hi, magic fairy, the pencil is" ,j, "cm long") } Sooner or later you'll get it right and off you go. (With apologies to the world-famous "If you tell me the height of the building I'll give you this barometer" story) EDIT: for those who never ran across it, the ...


12

The angle $\theta$ might have units of degrees, yes. It could also have been in e.g. radians. But putting it through the sine function removes the unit. The $\sin(\theta)$ is unitless. The sine and cosine functions are defined as the "distance" vertically and horizontally, respectively, to the point on the unit circle. That is, a distance per unit lenght. ...


12

Dimensional analysis can help to "guesstimate" the form of many important results but it can, for instance, not produce general solutions to equations of motion. It's an invaluable tool to understand the structure of physical theory, including quantum mechanics and relativity, and to check results for consistency, but it can rarely replace complex ...


11

The reason $c$ is important is not because it is the speed of light. It's important because it is a universal conversion factor between time and distance. If you have a certain amount of time $t$, you can calculate the corresponding amount of distance by multiplying it by $c$. Note: I'm not talking about the distance any particular object travels in the ...


11

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 ...


10

The metric tensor is unitless. That can be seen from the fact that $g_{\mu\nu}v^\mu v^\nu$ gives the square of the four-vector length of $v$, and thus has the unit of $v^2$. The scalar curvature is a contraction of the Ricci tensor. A contraction doesn't change the units. Also the Ricci tensor is a contraction of the Riemann tensor. The Riemann tensor is ...



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