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19

The symbol $\Delta$ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small. The symbols $d,\delta$ refer to infinitesimal variations or numerators and denominators of derivatives. The difference between $d$ and $\delta$ is that $dX$ is only used if $X$ without the $d$ is an actual quantity that may be ...


14

The wording of the question suggests that the electrons were the first objects or particles whose charge required the people to establish the sign convention. But that's obviously not the case. The electron was discovered by J. J. Thomson in 1897 but for much more than a century before that moment, people had already been studying electric (and magnetic) ...


11

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 ...


10

First, recall what a partition is. A partition of a set $X$ is a way to write $X$ as a disjoint union of subsets: $X=\coprod_i X_i$, $X_i\cap X_j=\emptyset$ for $i\neq j$. When the elements of the set $X$ are considered undistinguishable, what matters are the cardinals of the set only, and we have a partition of an integer number, $n=n_1+\ldots+n_k$. For ...


9

The relative sign is not just a convention. Once you decide that $E$ is represented by $i\hbar \partial/\partial t$, there must be a minus sign in the formula for $p$, namely $p=-i\hbar \partial / \partial x$. Or vice versa. First of all, there has to be $i$ or $-i$ in all the formulae because $\partial/\partial x$ is an anti-Hermitian operator (because of ...


8

The kilogram is defined by a prototype (the "International Prototype Kilogram", IPK) -- basically, a kilogram is by definition the mass of a metal cylinder sitting in a vault in Paris. People have made a bunch of other metal blocks with almost exactly the same mass (as near as they could get), called "sister copies". To measure a mass extremely accurately in ...


8

Yes, to some extent. Once you choose which of the electron or positron is to be considered the normal particle, then that fixes your choice for the other leptons, because of neutrino mixing. Similarly, choosing one quark to be the normal particle fixes the choice for the other flavors and colors of quarks. But I can't think of a reason within the standard ...


8

The clockwise direction is normally defined by the right hand grip rule. When your thumb is pointing away from you, your fingers are curled clockwise. So when you look at a clock the axis of rotation is away from you through the clock. I'd guess the downvotes are because people believe your question is not physics related, but in fact this rule is how ...


7

This is analogous to the definition of an empty product in mathematics. For a finite non-empty set $S=\{s_1,\ldots,s_n\}$, the product over $S$ can be defined as $$\prod_{s\in S}s=s_1\times \cdots\times s_n.$$ For such a product you'd want disjoint unions to map into products: if $R\cap S=\emptyset$, then you want $\prod_{x\in R\cup S}x=\left(\prod_{s\in ...


7

You can absolutely have negative pressure in solids or liquids. Think of an elastic solid being forced to expand due to adhesion to the walls of some chamber. That has negative pressure even if the comparison is a total vacuum. Depending on the bulk modulus of the material being stretched and the strength of the interaction with the walls of the chamber ...


7

Mass isn't always first. For example we write Newton's law for the force between two objects as: $$ F = \frac{Gm_1m_2}{r^2} $$ I don't think there are hard and fast rules. I suspect conventions have arisen over the years and we have all got used to what we learned at school, which was taught by teachers who are used to what they learned at school and so ...


6

I would add to John's answer that $a$ is not always constant. It represents the second derivative of motion, and thus is potentially a function of time. So, the overall conventional ordering in equations (in Mathematics as well as Physics) is, $$\mathrm{Constant \times Parameter \times Variable}$$ where I'm distinguishing between, say, $G$ which is ...


6

There are two separate issues here. (1) Why does it make sense to consider a dipole moment as a vector? (2) Given that it's a vector, why does it make sense to say that it points in this particular direction, rather than the opposite direction. Intuitively, it makes sense to define a dipole as a vector because when we put it in a field, it aligns itself ...


6

It is a matter of convention. The sign convention of Clausius and the sign convention of IUPAC are the two prevailing sign conventions. Both of these assign a sign to the work done differently. The former, used primarily in physics assign a positive sign to the work done by the system while the latter assigns positive sign to the work done on the system. ...


5

When radioactive element A decays to produce element B, the (infinitesimal) number of decayed elements A, $dN$, that occurs in a small time interval, $dt$, is proportional to the initial population of A, $N$: $$ -\frac{dN}{dt}\propto N $$ Assuming the proportionality is a constant, then the above becomes $$ -\frac{dN}{dt}=\lambda N $$ which has a known ...


5

I endorse Kyle's answer. Just two short comments. The number 36.8% is literally $$ 36.8 \approx 100 \exp(-1) =\frac{100}{2.71828\dots} $$ Moreover, it is right to call this quantity "average lifetime" or just "lifetime" because it is literally the average value of the time for which a nucleus (or something else) from the ensemble lives. If the initial ...


5

In general, when replacing a free index with a specific one, no signs ever get introduced: \begin{align} \partial_a & \to (\partial_0, \partial_i) \\ \partial^a & \to (\partial^0, \partial^i). \end{align} This holds for all signatures. (On a side note, I'm being pedantic about not using "$=$" signs for a reason - a tensor is not equal to a single, ...


5

It's a matter of history. When George Stoney developed Stoney units in 1881, or when Robert Millikan performed the oil drop experiment in 1909, it wasn't yet known that it was possible for anything to have a charge smaller in magnitude than the charge of an electron. By the time the quark model was proposed, in 1964, the use of the "elementary charge" being ...


5

Continuous Fourier analysis, which contains both the Fourier transform and the Fourier series, and which is used in e.g. signal processing, naturally picks the average value of the left and right limits, cf. the Dini-Dirichlet criterion. For the Heaviside step function, this means that $$\tag{1} H(0)~=~\frac{1}{2} \left(\lim_{x\to 0^-} H(x)+ \lim_{x\to 0^+} ...


5

First off, please don't use units with $c\ne 1$ in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like $\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that $u^\mu=(1,0,0,0)$, not $(-1,0,0,0)$. This is ...


5

I'd say it's form latin tempus - http://la.wikipedia.org/wiki/Frequentia


5

The number $1$ may be linguistically described as "unity". This very number is the original source of various words in the terminology, like the "unit matrix" (a matrix behaving like the number $1$). It is a convention to write down that dimensionless quantities like the Mach number have units $1$ because the multiplication by $1$ changes nothing about the ...


5

Your confusion arises because the term (anti)clockwise, when used by itself, is ambiguous, and should always be used with a statement like "as seen from the top" (unless that is absolutely obvious$^1$). The reason for this is that "clockwise" defines a direction of rotation within a plane, but does not specify which side the plane is observed from. (In more ...


5

Yes. You are missing the fact that he is using the convention $$ \nabla = (\partial_1, \partial_2, \partial_3) $$ as opposed to $$ \nabla = (\partial^1, \partial^2, \partial^3) $$ The first convention is by far the most common in my experience.


5

Your second equation, $P(\nu,T) = \frac{2 h {\nu}^3}{c^2}$ $\frac{1}{\exp\bigl(\frac{h \nu}{kT}\bigr) - 1}$ is what is commonly referred to as Planck's law for radiation, although a more standard symbol used is $B_\nu(T)$. This is the energy radiated per time, per area, per frequency interval, per steradian. It is a formula for the 'specific intensity' of a ...


4

You're right, absolute pressure can't be negative. Of course, you can easily have a $20\: \mathrm{PSI}$ pressure differential (although not without pressure above $1\: \mathrm{ATM}$ since that's $14.22\: \mathrm{PSI}$ at sea level). Check out Wikipedia on the zero-reference: Absolute pressure is zero-referenced against a perfect vacuum, so it is ...


4

It started with conservation of quantum numbers, from baryon number when we did not know about quarks, to lepton number, when we discovered the positron.For the neutrino momentum and energy conservation played a role too, since it is only seen as a missing mass. In time the symmetries in the assignments of the quantum numbers became more and more evident ...


4

As commenters have pointed out, it's German Strecke. Note that $s$ is for displacement, whereas $d$ is for distance. Distance is the distance along the path traveled by a body, whereas displacement is the birds-eye distance traveled. Displacement can also be negative in 1-D, depending upon your reference positive direction. For some reason, Strecke ...


4

To formalize the comments as an answer: The difference between requiring $$(\alpha u,v)=\alpha(u,v)\quad\text{ (mathematician's definition)}$$ and $$\langle u, \alpha v\rangle=\alpha\langle u,v\rangle\qquad\quad\,\,\text{ (physicist's definition)}$$ is purely one of convention, and the two definitions are equivalent as $(u,v)=\langle v,u\rangle$. There's no ...


4

I) Consider an arbitrary coordinate transformation $$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$ Let $$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$ denote the corresponding Jacobian. Traditionally in physics, a scalar $\sigma$ transforms as $$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma, $$ a pseudo-scalar ...



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