Tag Info

Hot answers tagged

27

The thing is that $\mathrm{dm}$ is a single symbol, not a combination of two symbols. Yes, it can be understood in terms of a prefix and a base indicator, but it is still a single symbol. An analogy to the concatenation of variable is inappropriate. Reference to an authoritative statement: The grouping formed by a prefix symbol attached to a unit ...


13

If I saw the word "amp" written as such in a paper in my field (astrophysics) it would strike me as a bit informal. I would expect to see the full "ampere" written. That said, it is rare to actually write out the full name of a unit; usually it follows a number and is given its standard abbreviation. When abbreviated to e.g. "$5\ \mathrm{A}$", I would ...


11

Technically, apparently, your teacher is correct. BIPM and NIST In the official brochure from the Bureau international des poids et mesures (BIPM, the keepers of SI units) in §5.1 Unit symbols we find: It is not permissible to use abbreviations for unit symbols or unit names, such as sec (for either s or second), sq. mm (for either mm2 or ...


9

They're not used because it's ugly to read such texts with parentheses and it's time-consuming to write it down. A decimeter is indeed a "product" of "deci" and a meter, so the origin is analogous to the product of two real numbers $ab$. But once we define the new derived unit ${\rm dm}$, we treat it as a single object, so it really means what you would ...


6

What Goldstein means by $\nabla_iV_i$ is $$ \nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i $$ which is indeed a vector. Here, $\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$ is the position of the $i$th particle (with respect to the origin), ...


5

This is typical to see in situations where the potential is a function of the coordinates of more than one particle: $$ V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N). $$ The force produced by such a potential on the $i$th particle is the gradient of this function with respect to that particle's coordinates, while holding all the other ...


5

This is notation for the imaginary part of a complex number. It is a fraktur letter I, and its counterpart for the real part is a fraktur letter R. Thus, if $z=x+iy$ and $x,y$ are real, one writes $$ \mathfrak{R}\,z=x\ \ \text{ and }\ \ \mathfrak{I}\,z=y. $$ A good chart of the fraktur alphabet is in this Yale resource, which includes handwriting guidance, ...


4

$1\text{ dm}^3$ can be seen as a shorthand way to write $1\text{ dm}\cdot 1\text{ dm}\cdot1\text{ dm}$. The motivation behind this is that the quantity $1\text{ dm}^3$ represents the volume of a cube with each side of length $1\text{ dm}$, and the way to find the volume is to multiply the three sides: $LWH=(1\text{ dm})\cdot(1\text{ dm})\cdot(1\text{ ...


3

In your example, ab is implied to be a multiplication - a*b; but dm is a single indivisible token. Imagine second2 - that doesn't imply that the last letter should be squared, and it's the same with decimeters. Mathematical notation is often ambiguous, and leaves many things implied and underspecified with the expection that the reader will fill the gaps ...


3

The $E$ he is refering to is the Young's Modulus, which is the elastic modulus that tells you how the tensile stress for a wire relates to its extensional strain, i.e. $$ \text{stress} = E \text{ strain} $$ in a thin wire. In the experiment he is summarizing in the graph, you pull a thin wire until it breaks, while measuring the stress on each end. The ...


3

Yes, there are sometime different conventions for indicating vectors in hand-writing and printing. Yes, overset arrows in handwriting and boldface in printing is one of those conventions. No, it is not the only convention. Yes, you should familiarize yourself with the most common conventions in your sub-discipline. Yes, you should read the section on ...


2

If it is actually the imaginary part of a complex variable, then just write $Im[\cdot]$ instrade of the curly character.


2

This is just a convention and nothing more. We treat the $\mathrm{dm}$ as a single symbol. Note that there are much stranger notations of powers, such as squared sines: $$ \sin^2(x) = \big(\sin(x)\big)^2 $$ You can not treat this formally as written - it's not a square of the sine, but the square of the value it returned.


1

For an electromagnetic wave, this is the propagation constant. It can be expressed as the sum of two terms: the attenuation constant and the phase constant.


1

The algebraic rule that if $xy=0$ then $x=0$ or $y=0$ does hold for tensor products: if you had $a^x b_y = 0$ you could conclude that $a^x = 0$ or $b_y = 0$. But there's no such rule for contractions like $a^x b_x$. For example suppose $a$ and $b$ written out in components were $(1,0)$ and $(0,1)$ respectively. You can't just move a variable to the other ...


1

Often times, an operation is something that changes the state of the system, like a measurement. But it can be other things that will change the state of the system.


1

The unitary matrix as given represents an operation on the Hilbert space of states. It could be a measurement, it could be a symmetry, it could be time evolution, or something else. It is called "randomizing" here, because it transforms a state of definite spin (the $\lvert 0 \rangle = \left(\begin{matrix} 1 \\ 0 \end{matrix}\right) $ state) into a state of ...


1

All the possible states for your systems are encoded as rays in a Hilbert space. For expample, imagine that we are interested in the spin direction of an electron. The Hilbert space for this system is $\mathcal{H}=\mathbb{C}^2$. We can take an orthonormal basis, for example, the $z$ component of the spin, which it would be represented by the orthonormal ...


1

According to the Wikipedia page, amp is acceptable, but is not a correct SI unit. I think your instructor is being thorough and making certain that you know the correct term to use.


1

From math and the power rule: $\dfrac{d(x^2)}{dx} = 2x$ And we assume that L is a function of time: $\vec{L} = \vec{L(t)}$. To refresh you on the chain rule: if x were a function of time, then $\dfrac{d(f(x))}{dt} = \dfrac{d(f(x))}{dx} * \dfrac{dx}{dt}$. Back to math: $\dfrac{d(x^2)}{dx}$ is actually $2x\dfrac{dx}{dx}$ if you apply said chain rule. ...


1

It is the easiest to think of this problem in the component form. Let's say the vector $\vec L$ is in a 3 dimensional space (that is usually what we use). So in component form, i.e., written as a 3 by 1 matrix, $\vec L=(L_1,L_2,L_3)$. What is the left-hand-side of your expression? It is $\vec L\cdot\frac{\mathrm d\vec L}{\mathrm dt}=L_1\cdot\frac{\mathrm ...


1

For me in my own field (optics, where one most often comes across it in engineering considerations), "amps" is common spoken usage, particularly for compound words such as milliamp or microamp. For written usage, I'm afraid I like to see the full name - it is, after all, recalling a very great man of science André-Marie Ampère. Even so, curiously, the SI ...



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