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44

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 ...


24

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 ...


22

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 ...


19

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 ...


16

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 ...


15

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 ...


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 ...


14

$G$ 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 ...


14

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" ...


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 ...


12

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 ...


11

Particle physicists indeed do like to express temperature in the units of energy, usually in their most favorite unit of energy, a gigaelectronvolt. Kelvins are used for historical reasons as well as for the sake of having reasonable numbers in everyday conditions. Before the temperature-energy relationship $$ E \sim kT$$ was realized in the late 19th ...


11

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 ...


10

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 ...


10

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 ...


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 ...


9

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 ...


9

Regardless of the context and the meaning of the symbols, both sides of the equation have perfectly the same units: they are dimensionless. The integral has units $js$ as you write, using your notation, but the functional derivative has the compensating units $1/(js)$ so the units cancel. To see that dimension of the functional derivative is $1/(js)$, one ...


9

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 ...


8

The best way to think about it is that a number like 1 km consists of a dimensionless 1 multiplied by a unit, km. When you take the log of a product, you get the sum of the logs, so log(1 km) is the same as log(1)+log(km). This shows that the log of 1 km is neither a dimensionless quantity nor a dimensionful one. If it was dimensionless, then it would be ...


8

Using fundamental physical constants try to construct expression which unit is legth. So using dimensional analysis, we have a data of: $G = m^3 \cdot kg^{-1} \cdot s^{-2}$, $c = m \cdot s^{-1}$ 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 ...


8

Yes. The delta function always has the same dimensions as the inverse of its argument. You can read this from its definition, your first equation. So in one dimension $\delta (x)$ has dimensions of inverse of length, in three spatial dimensions $\delta^{3}(\vec x)$ or simply $\delta(\vec x)$ has dimension of inverse of volume, and in n spatial dimension ...


8

As the other answers (and dmckee's comments) note, yes, if you take the square root of a dimensional quantity then you need to take the square root of the units too: $$ \sqrt{4\;{\rm kg}} = 2\;{\rm kg}^{\frac12} $$ And no, I can't think of any meaningful physical interpretation for the unit ${\rm kg}^{\frac12}$ either. However, in the comments you say ...


8

Many questions here. Well, units have a formal definition in the sense that every physical quantity has two parts: A quantity and a unit. Take "distance". How would you describe distance using just a number? That's impossible. Stating just a number would immediately lead to the question "... of what?". Think about it like this: Everything you measure is ...


8

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 ...


8

But that's exactly the deeper meaning! Setting things up so that all coordinates are in the same units (besides being a reasonable requirement for $x^\mu$ to be considered a four-vector) is a constant reminder that time is really not that different from space. In fact, if it weren't for that sign in the metric, spacetime would be completely symmetric in its ...


7

I have at least an answer : Does this depend on the size of the funnel ? Yes, as it was expected (for a very large funnel, all sand falls in the same time, whereas for a very small funnel it doesn't fall at all. More precisely ... the complete answer is probably extremely complex, as one has to take into account the shape of the hourglass, the dynamics ...


7

Your last couple of comments, about units, are incredibly important. It only makes sense to compare two things if they have compatible units. So, to use your example, it doesn't really make sense to talk about the size of Planck's constant relative to the electron charge, but it does make sense to talk about the mass of the muon relative to the mass of the ...


7

The formula is obtained by dimensional analysis. Up to a constant dimensionless factor, the given expression is the only one of dimension length that one can make of the fundamental constants $\hbar$, $c$, and $G$. Discussions about the physical significance of the Planck length have no experimental (and too little theoretical) support, so that your second ...


7

1) The best answer here might be a little of both. This kind of question gets at the notion of what time is. You can define a system of measuring time by using light. The time between events is then the distance that light would travel in the duration between those events. Then by definition, the speed of light is $1$ and dimensionless, as we measure time ...



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