# Tag Info

21

It's not a mistake, and conventional current is not wrong or backwards. The labeling of one polarity of charge as "positive" and the other as "negative" is totally arbitrary. It could be done either way and everything would still work out the same. Franklin didn't choose wrong; he just chose. Labeling protons as negative and electrons as positive wouldn't ...

21

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 ...

21

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

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) ...

14

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 ...

13

Well, that equation for the force due to electric charges is only true for a very special choice for the unit of the electric charge. Typically, you would write down Coulomb's law as $k\frac{q_{1}q_{2}}{r^{2}}$, where $k$ is a constant of proportionality chosen to make the units work out. IN the SI system, the unit of charge is the Coulomb (C) and the ...

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 ...

9

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 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

(Someone resurrected this oldie in the queue, so just to be a contrary voice...) Ben Franklin did get it wrong. He had just developed a remarkable new theory of electricity in which positive (+) and negative (-) had specific and accurate meanings, and he was unable to apply the two labels in the way he intended. In Franklin's time electricity was thought ...

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

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 ...

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 ... 7 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. ... 7 The factor of$1/2\pi$is an artifact of the normalization convention being used for the momentum eigenstates. To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state$|\psi\rangle$in a given ... 7 To quote Stephen Gasciorowicz, Before evaluating these quantities to obtain an idea of their magnitude, we will introduce some notations that will be very useful. First, it is$h/2\pi$rather than$hthat appears in most formulas in quantum mechanics. We therefore define $$\hbar=\frac{h}{2\pi}=1.0546\times10^{-34}\,{\rm J\cdot s}$$ So basically it's ... 6 Intensity has units of watts per area: $$\left[I\right]=\rm\frac{W}{ m^2}$$ where the area is the surface area of the emitting source (in this case, the sun). This tells you the total amount of radiation present (over all wavelengths). The extra factor of 1/nm in your plot gives the spectral irradiance: $$\left[\mathcal E\right]=\rm \frac{W}{m^2\,nm}$$ ... 6 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 "=$" signs for a reason - a tensor is not equal to a single, ... 6 I would add to John's answer that$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 ... 5 I suppose you talk about the standard$2\pi$that appears in the rules for Fourier transform. The factor of$2\pi$or$1/2\pi$or two factors of$1/\sqrt{2\pi}$have to appear "somewhere" in the Fourier transform rules because this is what the mathematics implies. At any rate, if this is your question, it is a mathematical question and you may learn it in ... 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 Using your metric signature, the metric is: $$\eta_{\mu\nu}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$ The generic position vector is defined as (in Cartesian) $$x^{\mu}=\begin{pmatrix}t\\x\\y\\z\end{pmatrix}$$ And the quantity$x_{\mu}=\eta_{\mu\nu}x^{\nu}$. So you can see that$x_\mu$... 5 The first bullet would be read "$A$dot$B$" or "The dot product of$A$and$B$" The second bullet would be read "$A$cross$B$" or "The cross product of$A$and$B$" 5 The integrand$\vec E \cdot d\vec r$is$E\,dr$, not$-E\,dr$. The evaluation of the dot product is sort of done for you when you specify the curve on which you are integrating (i.e., your limits of integration in this case). You've double-accounted for the relative directions of$\vec E$and$d\vec r$. I suspect the underlying confusion is that you are ... 5 If the particle moves from the point$x$to$x+dx$, and assume$dx\gt 0$for simplicity, then its potential energy increases by $$dU = \frac{dU}{dx}dx$$ Well, it increases if$dU$is positive and decreases if$dU\$ is negative. So far I have only used the definition of the derivative – pure mathematics. However, the total energy is conserved. The sum of ...

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 ...

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