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2 votes
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Kramers-Kronig relations for a Gaussian function

I'm not sure that PSE is the best site to ask this question, but I'll answer it anyway. The main issue is that the KK relations do not apply on $f$. Intuitively, the Fourier transform of $f$ is again ...
LPZ's user avatar
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5 votes
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Hermiticity of a projection operator

Indeed, you missed that $\sigma_{00}^0=I$. \begin{equation} \hat{U}=I+\sigma_{00}(e^{i\omega}-1), \hspace{6 mm} \hat{U}^{\dagger}=I+\sigma_{00}(e^{-i\omega}-1), \end{equation} and it is now easy to ...
Ruben Campos Delgado's user avatar
2 votes
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Use of infinity in physics

You sweep the infinitely big error under an infinitely distant rug. You lose me a bit here - there is no error to sweep. Two sets have the same cardinality if and only if the elements of each can be ...
J. Murray's user avatar
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2 votes

Use of infinity in physics

Fourier analysis is a much more routine example: the Fourier transform is defined by an integral with infinite limits. It gets used all over physics, as well as in other sciences, and even more in ...
John Doty's user avatar
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0 votes

Inverse Square vs Exponential

As far as I understand, in exponential equation the variable (x) is in the exponent. In a power function the variable is the base. In inverse square law, the intensity is inversely related to the ...
Narayanan 's user avatar
1 vote

Character Table and Binary Basis

The final column is obtained by representing the abstract group as an isometry subgroup. Each set of quadratic polynomials (separated by a semicolon) span an irreducible representation whose type is ...
LPZ's user avatar
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1 vote
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Writing the most general form resistivity tensor

You need to write the vector equation $$\vec{E}=\frac{1}{ne}(\vec{j}\times\vec{B})+\frac{m}{ne^2\tau}\vec{j}$$ in components. This gives $$\begin{pmatrix}E_x\\E_y\\E_z\end{pmatrix} =\frac{1}{ne}\...
Thomas Fritsch's user avatar
0 votes

Lagrangian total time derivative - continues second-order differential

I think that the textbook 'Classical Dynamics of Particles and Systems' by Marion and Thornton has mentioned this restriction (Section 7.4 in the fifth edition): However, certain transformations that ...
user19833's user avatar
3 votes

Can we call numbers unidirectional vectors?

Good morning, what you're saying is interesting. Of course, real numbers are vectors, belonging to $(\mathbb{R},+,*)$, which is an $\mathbf{R}$-vector space. It's a one-dimensional version of the ...
Gorga's user avatar
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5 votes
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Why does $\oint_C \vec{E}\cdot d\vec{\ell}=0$ imply $\nabla\times \vec{E}=\vec{0}$?

The first equation is only for all closed loops, not for all contours. That’s why you can’t conclude $\vec{E}=0$. The only thing you can conclude is that $\vec{E}=\text{grad}(V)$ for some function $V$....
peek-a-boo's user avatar
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2 votes

Why does $\oint_C \vec{E}\cdot d\vec{\ell}=0$ imply $\nabla\times \vec{E}=\vec{0}$?

For the very reason you say at the end of your question: while the second relation holds for all surfaces, the first one only holds for all closed loops, and not any arbitrary line. Let's do some ...
basics's user avatar
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1 vote

How can a triangle have a sum exceeding 180 degrees in a curved space?

You can imagine putting mirrored square tiles on a disco ball. No matter how hard you try, you end up leaving awkward gaps between the tiles or having to cut parts off the tiles to completely cover ...
KDP's user avatar
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3 votes
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How can a triangle have a sum exceeding 180 degrees in a curved space?

Here are three diagrams to illustrate that the angle sum of a triangle can differ from $180^\circ$. The Wikipedia article Spherical Geometry might be of interest? It has a nice illustration relating ...
Farcher's user avatar
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2 votes

How can a triangle have a sum exceeding 180 degrees in a curved space?

The angles are the same in 2d and 3d! Your example of the sphere is perfect. If you take the North pole and two points on the equator then a triangle with 3 x 90 degrees is formed. If, for this ...
Jos Bergervoet's user avatar
1 vote

How can a triangle have a sum exceeding 180 degrees in a curved space?

So in essence a straight line for us (in respect to a 3D space) is in fact a curved line on a 2D space, this essentially means that the lines are "stretched" over the curve of our spherical ...
user394147's user avatar
0 votes

Consistency of perturbative theory when some of the first-order terms are smaller than second-order terms?

A perturbation series is never generally valid. Rather, there is a range in the parameter space where it is assumed to hold. A typical example is asymptotic freedom in QCD. You can be sure that your ...
Jon's user avatar
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0 votes

In real life, we can have a pencil of length 2 cm. Can we have pencil of length $\sqrt{2}$ cm?

It is philosophical in nature. Here is my intuitive opinionated answer. (1) We can easily define a certain Pencil to have certain length. When we say it is $4$ cm then we will get $2$ cm when we break ...
Prem's user avatar
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1 vote

Consistency of perturbative theory when some of the first-order terms are smaller than second-order terms?

There is no general guarantee that second-order terms will always be smaller than all first-order terms. The validity of the perturbation expansion and the relative importance of each order depend on ...
Testina's user avatar
  • 171
1 vote

Wave functions and Schwartz spaces (QM)

It is possible for a function to go to zero at infinity without "getting all flat". For example, imagine a function which oscillates in such a way that its amplitude goes to zero at infinity,...
Albertus Magnus's user avatar
3 votes

Can measurements in physics fundamentally only yield rational numbers?

There are several ways to answer your question: First, as you yourself noted, any practical measurement will always have a finite precision, i.e., finite number of decimal digits, so will always be a ...
Nadav Har'El's user avatar
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4 votes

Can measurements in physics fundamentally only yield rational numbers?

Any measured value has a finite precision, so, by definition, it must be a rational number. If the "true" value of an observable were an irrational number - $\sqrt 2$ for example - then we ...
gandalf61's user avatar
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4 votes
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A confusion about the inner product keeps disturbing me

But very often, I hear that people [cannot give a reference, :-(] say it is an operation between a vector from $\mathbb{V}$ and a dual vector from $\mathbb{V}^*$? If somebody could explain this simply ...
peek-a-boo's user avatar
  • 6,100
1 vote

A confusion about the inner product keeps disturbing me

Note: this is not a math answer, go to math.stackexchange.com for that. This is a physics answer. So my absolute favorite definition of the inner product is from a Kip Thorne lecture on "...
JEB's user avatar
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1 vote

A confusion about the inner product keeps disturbing me

There are differences in real vectors and imaginary vectors. In particular, real vectors in a measure space (the dot product establishes a measure) are combined in a dot product by multiplication of ...
Whit3rd's user avatar
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0 votes

Can the different differentiation notations be equated and do they have an integral definition?

In physics the most common notations for differentiation are, $\frac{da}{dt} = \dot{a} = a'$. The notation $\frac{da}{dt}$ is due to Leibniz, the notation $\dot a$ is due to Newton, and the notation $...
Gnome Chomsky's user avatar
2 votes

Can the different differentiation notations be equated and do they have an integral definition?

The equality you write in your post is most certainly not true. Arguably the only pair for which equality holds is: $$\dot{a} = \frac{\mathrm{d}a}{\mathrm{d}t}$$ Which is true by definition of the &...
Cody Payne's user avatar
1 vote

Dirac Delta applied to the gradient of a function

The supplementary section of a paper I am reading uses the "substitution" property of the Dirac delta function : $$\mathbf{v}\left(\mathbf{x}_j\right)\delta\left(\mathbf{x}-\mathbf{x}_j\...
hft's user avatar
  • 19k
0 votes

Another Fourier transformation but with a $\sqrt{\mathbf{q}^2 + m^2}$ term now

So, I have been tackling this integral for awhile, and the main issue is the square root term that makes the integral highly divergent at $q\rightarrow \infty$. The way to get around this is to look ...
MathZilla's user avatar
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