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# The rate of a certain chemical reaction is directly

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07 Aug 2005, 20:00
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The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B present is increased by 100 percent, which of the following is closest to the percent change in the the concentration of chemical A required to keep the reaction rate unchanged.

A. 100 % decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-rate-of-a-certain-chemical-reaction-is-directly-90119.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Dec 2013, 02:01, edited 1 time in total.
Added the OA.
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07 Aug 2005, 20:11
Let's say here is the formula of how A and B affects the reaction speed:

rate = A^2/B

After the concentration B is increased by 100%, the percentage of B
become 2B, to keep 'rate' the same, need to have A^2 being doubled.
Which means A increased by 2^(1/2) times.

In closest percentage, that is 40% increase

D
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08 Aug 2005, 06:01
But the question does say the relationship between A and B is multiplication.
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08 Aug 2005, 07:14
does inversely proportion suggest when B increase, the reaction speed decrease - in essense it is equivalent to a division.
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08 Aug 2005, 09:47
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In general and in full

r = k A^2 / B for some constant k<>0
(Question should also give B<>0)

So we have

r1 = k A1^2 / B1
r2 = k A2^2 / B2

r1 = r2
B2 = (1+100%)B1 = 2B1

k A1^2 / B1 = k A2^2 / B2
A1^2 / B1 = A2^2 / 2B1
2A1^2 = A2^2
(sqrt(2)A1)^2 = A2^2
A1,A2>0
A2 = sqrt(2) * A1

A2 = (1 + 41.4%) * A1

About 40% increase
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15 Aug 2005, 13:31
qpoo wrote:
Let's say here is the formula of how A and B affects the reaction speed:

rate = A^2/B

After the concentration B is increased by 100%, the percentage of B
become 2B, to keep 'rate' the same, need to have A^2 being doubled.
Which means A increased by 2^(1/2) times.

In closest percentage, that is 40% increase

D

plz QPoo could explain why doubled A^2 is to increase it by 2^(1/2) times.
thanks
regards
mandy
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The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B present is increased by 100 percent, which of the following is closest to the percent change in the the concentration of chemical A required to keep the reaction rate unchanged.
A. 100 % decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase

NOTE: Put directly proportional in nominator and inversely proportional in denominator.
$$RATE=\frac{A^2}{B}$$, (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).

We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means $$x^2=2$$ --> $$x\approx{1.41}$$, which is approximately 40% increase. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}$$

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-rate-of-a-certain-chemical-reaction-is-directly-90119.html
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Re: The rate of a certain chemical reaction is directly [#permalink]

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04 Dec 2013, 02:02
Bunuel wrote:
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B present is increased by 100 percent, which of the following is closest to the percent change in the the concentration of chemical A required to keep the reaction rate unchanged.
A. 100 % decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase

NOTE: Put directly proportional in nominator and inversely proportional in denominator.
$$RATE=\frac{A^2}{B}$$, (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).

We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means $$x^2=2$$ --> $$x\approx{1.41}$$, which is approximately 40% increase. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}$$

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-rate-of-a-certain-chemical-reaction-is-directly-90119.html

DS problems about this concept:
the-amount-of-coal-a-train-burns-each-mile-is-directly-93667.html
in-a-certain-business-production-index-p-is-directly-63570.html

PS problems about this concept:
a-is-directly-proportional-to-b-when-a-8-b-88971.html
in-a-certain-formula-p-is-directly-proportional-to-s-and-80941.html

Hope it helps.
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Re: The rate of a certain chemical reaction is directly   [#permalink] 04 Dec 2013, 02:02
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# Book:Richard S. Millman/Geometry: A Metric Approach with Models/Second Edition

## Richard S. Millman and George D. Parker: Geometry: A Metric Approach with Models (2nd Edition)

Published $\text {1991}$, Springer Verlag

ISBN 0-387-97412-1.

Part of the Undergraduate Texts in Mathematics series.

### Contents

Preface
Computers and Hyperbolic Geometry
CHAPTER 1: Preliminary Notions
1.1 Axioms and Models
1.2 Sets and Equivalence Relations
1.3 Functions
CHAPTER 2: Incidence and Metric Geometry
2.1 Definition and Models of Incidence Geometry
2.2 Metric Geometry
2.3 Special Coordinate Systems
CHAPTER 3: Betweenness and Elementary Figures
3.1 An Alternative Description of the Cartesian Plane
3.2 Betweenness
3.3 Line Segments and Rays
3.4 Angles and Triangles
CHAPTER 4: Plane Separation
4.1 The Plane Separation Axiom
4.2 PSA for the Euclidean and Poincaré Planes
4.3 Pasch Geometries
4.4 Interiors and the Crossbar Theorem
CHAPTER 5: Angle Measure
5.1 The Measure of an Angle
5.2 The Moulton Plane
5.3 Perpendicularity and Angle Congruence
5.4 Euclidean and Poincaré Angle Measure (optional)
CHAPTER 6: Neutral Geometry
6.1 The Side-Angle-Side Axiom
6.2 Basic Triangle Congruence Theorems
6.3 The Exterior Angle Theorem and Its Consequences
6.4 Right Triangles
6.5 Circles and Their Tangent Lines
6.6 The Two Circle Theorem (optional)
6.7 The Synthetic Approach
CHAPTER 7: The Theory of Parallels
7.1 The Existence of Parallel Lines
7.3 The Critical Function
CHAPTER 8: Hyperbolic Geometry
8.1 Asymptotic Rays and Triangles
8.2 Angle Sum and the Defect of a Triangle
8.3 The Distance Between Parallel Lines
CHAPTER 9: Euclidean Geometry
9.1 Equivalent Forms of EPP
9.2 Similarity Theory
9.3 Some Classical Theorems of Euclidean Geometry
CHAPTER 10: Area
10.1 The Area Function
10.2 The Existence of Euclidean Area
10.3 The Existence of Hyperbolic Area
10.4 Bolyai's Theorem
CHAPTER 11: The Theory of Isometries
11.1 Collineations and Isometries
11.2 The Klein and Poincaré Disk Models (optional)
11.3 Reflections and the Mirror Axiom
11.4 Pencils and Cycles
11.5 Double Reflections and Their Invariant Sets
11.6 The Classification of Isometries
11.7 The Isometry Group
11.8 The SAS Axiom in $\mathscr H$
11.9 The Isometry Groups of $\mathscr E$ and $\mathscr H$
Bibliography
Index

Next

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text:
# How do you find the average value of the function for f(x)=x^2-sqrtx, 0<=x<=3?

Oct 5, 2017

$3 - \frac{2 \sqrt{3}}{3}$

#### Explanation:

Recall the formula for determining the average value of a function $f \left(x\right)$ on an interval from $\left[a , b\right]$:

$\text{Avg. Value} = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

For this problem:

$A v g = \frac{1}{3 - 0} {\int}_{0}^{3} \left({x}^{2} - \sqrt{x}\right) \mathrm{dx}$

$= \frac{1}{3} {\int}_{0}^{3} \left({x}^{2} - {x}^{1 / 2}\right) \mathrm{dx}$

$= \frac{1}{3} {\left[{x}^{3} / 3 - \frac{2}{3} {x}^{3} / 2\right]}_{0}^{3}$

$= \frac{1}{3} \left({3}^{3} / 3 - \frac{2}{3} {3}^{3} / 2\right) = \frac{1}{3} \left(\frac{27}{3} - \frac{2}{3} \cdot 3 \sqrt{3}\right)$

$= \frac{1}{3} \left(9 - 2 \sqrt{3}\right) = 3 - \frac{2 \sqrt{3}}{3}$


text:
# fsst

Fourier synchrosqueezed transform

## Syntax

s = fsst(x)
[s,w,n] = fsst(x)
[s,f,t] = fsst(x,fs)
[s,f,t] = fsst(x,ts)
[___] = fsst(___,window)
fsst(___)
fsst(___,freqloc)

## Description

s = fsst(x) returns the Fourier Synchrosqueezed Transform of the input signal, x. Each column of s contains the synchrosqueezed spectrum of a windowed segment of x.

example

[s,w,n] = fsst(x) returns a vector of normalized frequencies, w, and a vector of sample numbers, n, at which the Fourier synchrosqueezed transform is computed. w corresponds to the columns of s and n corresponds to the rows of s.

example

[s,f,t] = fsst(x,fs) returns a vector of cyclical frequencies, f, and a vector of time instants, t, expressed in terms of the sample rate, fs.

example

[s,f,t] = fsst(x,ts) specifies the sample time, ts, as a duration scalar. t is in the same units as ts. The units of f are reciprocal to the units of ts.

example

[___] = fsst(___,window) uses window to divide the signal into segments and perform windowing. You can use any combination of input arguments from previous syntaxes to obtain the corresponding output arguments.
fsst(___) with no output arguments plots the synchrosqueezed transform in the current figure window.

example

fsst(___,freqloc) specifies the axis on which to plot the frequency.

## Examples

collapse all

Generate 1024 samples of a signal that consists of a sum of sinusoids embedded in white Gaussian noise. The normalized frequencies of the sinusoids are $2\pi /5$ rad/sample and $4\pi /5$ rad/sample. The higher frequency sinusoid has 3 times the amplitude of the other sinusoid.

N = 1024; n = 0:N-1; w0 = 2*pi/5; x = sin(w0*n)+3*sin(2*w0*n);

Compute the Fourier synchrosqueezed transform of the signal. Plot the result.

[s,w,n] = fsst(x); mesh(n,w/pi,abs(s)) axis tight view(2) colorbar

Compute the conventional short-time Fourier transform of the signal for comparison. Use the default values of spectrogram. Plot the result.

[s,w,n] = spectrogram(x); surf(n,w/pi,abs(s),'EdgeColor','none') axis tight view(2) colorbar

Generate a signal that consists of two chirps. The signal is sampled at 3 kHz for one second. The first chirp has an initial frequency of 400 Hz and reaches 800 Hz at the end of the sampling. The second chirp starts at 500 Hz and reaches 1000 Hz at the end. The second chirp has twice the amplitude of the first chirp.

fs = 3000; t = 0:1/fs:1-1/fs; x1 = chirp(t,400,t(end),800); x2 = 2*chirp(t,500,t(end),1000);

Compute and plot the Fourier synchrosqueezed transform of the signal.

fsst(x1+x2,fs,'yaxis')

Compare the synchrosqueezed transform with the short-time Fourier transform (STFT). Compute the STFT using the spectrogram function. Specify the default parameters used by fsst:

• A 256-point Kaiser window with β = 10 to window the signal

• An overlap of 255 samples between adjoining windowed segments

• An FFT length of 256

[stft,f,t] = spectrogram(x1+x2,kaiser(256,10),255,256,fs);

Plot the absolute value of the STFT.

mesh(t,f,abs(stft)) xlabel('Time (s)') ylabel('Frequency (Hz)') title('Short-Time Fourier Transform') axis tight view(2)

Compute and display the Fourier synchrosqueezed transform of a quadratic chirp that starts at 100 Hz and crosses 200 Hz at t = 1 s. Specify a sample rate of 1 kHz. Express the sample time as a duration scalar.

fs = 1000; t = 0:1/fs:2; ts = duration(0,0,1/fs); x = chirp(t,100,1,200,'quadratic'); fsst(x,ts,'yaxis') title('Quadratic Chirp')

The synchrosqueezing algorithm works under the assumption that the frequency of the signal varies slowly. Thus the spectrum is better concentrated at early times, where the rate of change of frequency is smaller.

Compute and display the Fourier synchrosqueezed transform of a linear chirp that starts at DC and crosses 150 Hz at t = 1 s. Use a 256-sample Hamming window.

x = chirp(t,0,1,150); fsst(x,ts,hamming(256),'yaxis') title('Linear Chirp')

Compute and display the Fourier synchrosqueezed transform of a logarithmic chirp. The chirp is sampled at 1 kHz, starts at 20 Hz, and crosses 60 Hz at t = 1 s. Use a 256-sample Kaiser window with β = 20.

x = chirp(t,20,1,60,'logarithmic'); [s,f,t] = fsst(x,fs,kaiser(256,20)); clf mesh(t,f,(abs(s))) title('Logarithmic Chirp') xlabel('Time (s)') ylabel('Frequency (Hz)') view(2)

Use a logarithmic scale for the frequency axis. The transform becomes a straight line.

ax = gca; ax.YScale = 'log'; axis tight

This example shows that the Fourier Synchrosqueezed Transform, despite the sharp ridges it produces, cannot resolve arbitrarily spaced sinusoids. The width of the window used by the transform dictates how closely spaced two sinusoids can be and still be detected as distinct.

Consider a sinusoid, $f\left(x\right)={e}^{j2\pi \nu x}$, windowed with a Gaussian window, $g\left(t\right)={e}^{-\pi {t}^{2}}$. The short-time transform is

${V}_{g}f\left(t,\eta \right)={e}^{j2\pi \nu t}{\int }_{-\infty }^{\infty }{e}^{-\pi \left(x-t{\right)}^{2}}{e}^{-j2\pi \left(x-t\right)\left(\eta -\nu \right)}\phantom{\rule{0.16666666666666666em}{0ex}}dx={e}^{-\pi \left(\eta -\nu {\right)}^{2}}{e}^{j2\pi \nu t}.$

When viewed as a function of frequency, the transform combines a constant (in time) oscillation at $\nu$ with Gaussian decay away from it. The synchrosqueezing estimate of the instantaneous frequency,

${\Omega }_{g}f\left(t,\eta \right)=\frac{1}{j2\pi }\frac{{e}^{-\pi \left(\eta -\nu {\right)}^{2}}\frac{\partial }{\partial t}{e}^{j2\pi \nu t}}{{e}^{-\pi \left(\eta -\nu {\right)}^{2}}{e}^{j2\pi \nu t}}=\nu ,$

equals the value obtained by using the standard definition,

${\nu }_{inst}=\frac{1}{2\pi }\frac{d}{dt}\mathrm{arg}f\left(t\right)=\frac{1}{2\pi }\frac{d}{dt}2\pi \nu t.$

For a superposition of sinusoids,

$f\left(x\right)=\sum _{k=1}^{K}{A}_{k}{e}^{j2\pi {\nu }_{k}x},$

the short-time transform becomes

${V}_{g}f\left(t,\eta \right)=\sum _{k=1}^{K}{A}_{k}{e}^{-\pi \left(\eta -{\nu }_{k}{\right)}^{2}}{e}^{j2\pi {\nu }_{k}t}.$

Create 1024 samples of a signal consisting of two sinusoids. One sinusoid has a normalized frequency of ${\omega }_{0}=\pi /5$ rad/sample. The other sinusoid has three times the frequency and three times the amplitude.

N = 1024; n = 0:N-1; w0 = pi/5; x = exp(1j*w0*n)+3*exp(1j*3*w0*n);

Use the spectrogram function to compute the short-time Fourier transform of the signal. Use a 256-sample Gaussian window with $\alpha =20$, 255 samples of overlap between adjoining sections, and 1024 DFT points. Plot the absolute value of the transform.

Nw = 256; nfft = 1024; alpha = 20; [s,w,t] = spectrogram(x,gausswin(Nw,alpha),Nw-1,nfft,"centered"); surf(t,w/pi,abs(s),EdgeColor="none") view(2) axis tight xlabel("Samples") ylabel("Normalized Frequency (\times\pi rad/sample)")

The Fourier synchrosqueezed transform, implemented in the fsst function, results in a sharper, better localized estimate of the spectrum.

[ss,sw,st] = fsst(x,[],gausswin(Nw,alpha)); fsst(x,"yaxis")

The sinusoids are visible as constant oscillations at the expected frequency values. To see that the decay away from the ridges is Gaussian, plot an instantaneous value of the transform and overlay two instances of a Gaussian. Express the Gaussian amplitude and standard deviation in terms of $\alpha$ and the window length. Recall that the standard deviation of the frequency-domain window is the reciprocal of the standard deviation of the time-domain window.

rstdev = (Nw-1)/(2*alpha); amp = rstdev*sqrt(2*pi); instransf = abs(s(:,128)); plot(w/pi,instransf) hold on plot(w/pi,[1 3]*amp.*exp(-rstdev^2/2*(w-[1 3]*w0).^2),"--") hold off xlabel("Normalized Frequency (\times\pi rad/sample)") legend(["Transform" "First sinusoid" "Second sinusoid"],Location="best")

The Fourier synchrosqueezed transform concentrates the energy content of the signal at the estimated instantaneous frequencies.

stem(sw/pi,abs(ss(:,128))) xlabel("Normalized Frequency (\times\pi rad/sample)") title("Synchrosqueezed Transform")

The synchrosqueezed estimates of instantaneous frequency are valid only if the sinusoids are separated by more than $2\Delta$, where

$\Delta =\frac{1}{\sigma }\sqrt{2\mathrm{log}2}$

for a Gaussian window and $\sigma$ is the standard deviation.

Repeat the previous calculation, but now specify that the second sinusoid has a normalized frequency of ${\omega }_{0}+1.9\Delta$ rad/sample.

D = sqrt(2*log(2))/rstdev; w1 = w0+1.9*D; x = exp(1j*w0*n)+3*exp(1j*w1*n); [s,w,t] = spectrogram(x,gausswin(Nw,alpha),Nw-1,nfft,"centered"); instransf = abs(s(:,20)); plot(w/pi,instransf) hold on plot(w/pi,[1 3]*amp.*exp(-rstdev^2/2*(w-[w0 w1]).^2),"--") hold off xlabel("Normalized Frequency (\times\pi rad/sample)") legend(["Transform" "First sinusoid" "Second sinusoid"],Location="best")

The Fourier synchrosqueezed transform cannot resolve the sinusoids well because $|{\omega }_{1}-{\omega }_{0}|<2\Delta$. The spectral estimates decrease significantly in value.

[ss,sw,st] = fsst(x,[],gausswin(Nw,alpha)); stem(sw/pi,abs(ss(:,128))) xlabel("Normalized Frequency (\times\pi rad/sample)") title("Synchrosqueezed Transform")

Load a speech signal sampled at ${F}_{s}=7418\phantom{\rule{0.2777777777777778em}{0ex}}Hz$. The file contains a recording of a female voice saying the word "MATLAB®."

load mtlb % To hear, type sound(mtlb,Fs)

Compute the synchrosqueezed transform of the signal. Use a Hann window of length 256. Display the time on the x-axis and the frequency on the y-axis.

fsst(mtlb,Fs,hann(256),'yaxis')

Use ifsst to invert the transform. Compare the original and reconstructed signals.

sst = fsst(mtlb,Fs,hann(256)); xrc = ifsst(sst,hann(256)); plot((0:length(mtlb)-1)/Fs,[mtlb xrc xrc-mtlb]) legend('Original','Reconstructed','Difference')

% To hear, type sound(xrc-mtlb,Fs)

## Input Arguments

collapse all

Input signal, specified as a vector.

Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise.

Data Types: single | double
Complex Number Support: Yes

Sample rate, specified as a positive scalar. The sample rate is the number of samples per unit time. If the unit of time is seconds, then the sample rate is in Hz.

Data Types: double | single

Sample time, specified as a duration scalar. The sample time is the time elapsed between consecutive samples of x.

Data Types: duration

Window used to divide the signal into segments, specified as an integer or as a row or column vector.

• If window is an integer, then fsst divides x into segments of length window and windows each segment with a Kaiser window of that length and β = 10. The overlap between adjacent segments is window – 1.

• If window is a vector, then fsst divides x into segments of the same length as the vector and windows each segment using window. The overlap between adjacent segments is length(window) – 1.

• If window is not specified, then fsst divides x into segments of length 256 and windows each segment with a 256-sample Kaiser window with β = 10. The overlap between adjacent segments is 255. If x has fewer than 256 samples, then the function uses a single Kaiser window with the same length as x and β = 10.

For a list of available windows, see Windows.

Example: hann(N+1) and (1-cos(2*pi*(0:N)'/N))/2 both specify a Hann window of length N + 1.

Data Types: double | single

Frequency display axis, specified as 'xaxis' or 'yaxis'.

• 'xaxis' — Displays frequency on the x-axis and time on the y-axis.

• 'yaxis' — Displays frequency on the y-axis and time on the x-axis.

This argument is ignored if you call fsst with output arguments.

## Output Arguments

collapse all

Fourier synchrosqueezed transform, returned as a matrix. The number of columns of s equals the length of x. Time increases across the columns of s and frequency increases down the rows of s, starting from zero. If x is real, then its synchrosqueezed spectrum is one-sided. If x is complex, then its synchrosqueezed spectrum is two-sided and centered.

Normalized frequencies, returned as a vector. The length of w equals the number of rows in s.

Sample numbers, returned as a vector. The length of n equals the number of columns in s. Each sample number in n is the midpoint of a windowed segment of x.

Cyclical frequencies, returned as a vector. The length of f equals the number of rows in  s.

Time instants, returned as a vector. The length of t equals the number of columns in s. Each time value in t is the midpoint of a windowed segment of x.

collapse all

### Fourier Synchrosqueezed Transform

Many real-world signals such as speech waveforms, machine vibrations, and physiologic signals can be expressed as a superposition of amplitude-modulated and frequency-modulated modes. For time-frequency analysis, it is convenient to express such signals as sums of analytic signals through

$f\left(t\right)=\sum _{k=1}^{K}{f}_{k}\left(t\right)=\sum _{k=1}^{K}{A}_{k}\left(t\right){e}^{j2\pi {\varphi }_{k}\left(t\right)}.$

The phases ϕk(t) have time derivatives k(t)/dt that correspond to instantaneous frequencies. When the exact phases are unknown, you can use the Fourier synchrosqueezed transform to estimate them.

The Fourier synchrosqueezed transform is based on the short-time Fourier transform implemented in the spectrogram function. For certain kinds of nonstationary signals, the synchrosqueezed transform resembles the reassigned spectrogram because it generates sharper time-frequency estimates than the conventional transform. The fsst function determines the short-time Fourier transform of a function, f, using a spectral window, g, and computing

${V}_{g}f\left(t,\eta \right)={\int }_{-\infty }^{\infty }f\left(x\right)g\left(x-t\right){e}^{-j2\pi \eta \left(x-t\right)}\text{\hspace{0.17em}}dx.$

Unlike the conventional definition, this definition has an extra factor of ej2πηt. The transform values are then “squeezed” so that they concentrate around curves of instantaneous frequency in the time-frequency plane. The resulting synchrosqueezed transform is of the form

${T}_{g}f\left(t,\omega \right)={\int }_{-\infty }^{\infty }{V}_{g}f\left(t,\eta \right)\text{\hspace{0.17em}}\delta \left(\omega -{\Omega }_{g}f\left(t,\eta \right)\right)\text{\hspace{0.17em}}d\eta ,$

where the instantaneous frequencies are estimated with the “phase transform”

${\Omega }_{g}f\left(t,\eta \right)=\frac{1}{j2\pi }\frac{\frac{\partial }{\partial t}{V}_{g}f\left(t,\eta \right)}{{V}_{g}f\left(t,\eta \right)}=\eta -\frac{1}{j2\pi }\frac{{V}_{\partial g/\partial t}f\left(t,\eta \right)}{{V}_{g}f\left(t,\eta \right)}.$

The transform in the denominator decreases the influence of the window. To see a simple example, refer to Detect Closely Spaced Sinusoids with the Fourier Synchrosqueezed Transform. The definition of Tgf(t,ω) differs by a factor of 1/g(0) from other expressions found in the literature. fsst incorporates the factor in the mode-reconstruction step.

Unlike the reassigned spectrogram, the synchrosqueezed transform is invertible and thus can reconstruct the individual modes that compose the signal. Invertibility imposes some constraints on the computation of the short-time Fourier transform:

• The number of DFT points is equal to the length of the specified window.

• The overlap between adjoining windowed segments is one less than the window length.

• The reassignment is performed only in frequency.

To find the modes, integrate the synchrosqueezed transform over a small frequency interval around Ωgf(t,η):

${f}_{k}\left(t\right)\approx \frac{1}{g\left(0\right)}{\int }_{|\omega -{\Omega }_{k}\left(t\right)|<\epsilon }{T}_{g}f\left(t,\omega \right)\text{\hspace{0.17em}}d\omega ,$

where ɛ is a small number.

The synchrosqueezed transform produces narrow ridges compared to the windowed short-time Fourier transform. However, the width of the short-time transform still affects the ability of the synchrosqueezed transform to separate modes. To be resolvable, the modes must obey these conditions:

1. For each mode, the frequency must be strictly greater than the rate of change of the amplitude: $\frac{d{\varphi }_{k}\left(t\right)}{dt}>\frac{d{A}_{k}\left(t\right)}{dt}$ for all k.

2. Distinct modes must be separated by at least the frequency bandwidth of the window. If the support of the window is the interval [–Δ,Δ], then $|\frac{d{\varphi }_{k}\left(t\right)}{dt}-\frac{d{\varphi }_{m}\left(t\right)}{dt}|>2\Delta$ for km.

For an illustration, refer to Detect Closely Spaced Sinusoids with the Fourier Synchrosqueezed Transform.

## References

[1] Auger, François, Patrick Flandrin, Yu-Ting Lin, Stephen McLaughlin, Sylvain Meignen, Thomas Oberlin, and Hau-Tieng Wu. "Time-Frequency Reassignment and Synchrosqueezing: An Overview." IEEE® Signal Processing Magazine. Vol. 30, November 2013, pp. 32–41.

[2] Oberlin, Thomas, Sylvain Meignen, and Valérie Perrier. "The Fourier-based Synchrosqueezing Transform." Proceedings of the 2014 IEEE International Conference on Acoustics, Speech, and Signal Processing (ICASSP), pp. 315–319.

[3] Thakur, Gaurav, and Hau-Tieng Wu. "Synchrosqueezing-based Recovery of Instantaneous Frequency from Nonuniform Samples." SIAM Journal of Mathematical Analysis. Vol. 43, 2011, pp. 2078–2095.

## Version History

Introduced in R2016b


text:
# Characterizing the image of a faithful left adjoint.

If there is an adjunction between a category $C$ and $D$ such that the left adjoint $$L:C \to D$$ is faithful (but not full), can one describe the image of $L$ in terms of the co-unit of the adjunction? If it helps, in my situation the right adjoint $R$ has a further right adjoint; I am concerned with the case actually that there is faithful functor $$f:E \to F$$ and I would like characterize the image of its prolongation $$f_{!}:Set^{E^{op}} \to Set^{F^{op}}$$ (which is again faithful).

-
The fixed points (i.e. $d$ such that the counit $LRd \to d$ is an isomorphism) are in the image, but in general the image is larger. Consider sets and pointed sets: $R : Set_∗ \to Set$ is the forgetful functor and $L : Set \to Set_*$ adds a formal point. The unit for a set $X$ is $X \hookrightarrow X + \*$, the counit for a pointed set $(Y,y)$ is $Y + \* \twoheadrightarrow Y$ where $\* \mapsto y$. The counit is never an isomorphism, but $L$ is essentially surjective and faithful as required. –  Martin Brandenburg Jan 14 '12 at 10:49
In another example I have in mind, the co-unit is epi on the image. I wonder if this is a characterization or not. –  David Carchedi Jan 14 '12 at 18:20
What do you mean by "epi on the image"? –  Martin Brandenburg Jan 15 '12 at 12:11
@Martin: In your example, the left adjoint $L$ is essentially surjective, so everything is in its image (up to iso) and the co-unit is always an epimorphism. In the case I was looking at, when I know I have something in the image, the counit happened to be epi. I was wondering if the converse were true, i.e. if the co-unit of an object was epi, then it was in the essential image. –  David Carchedi Jan 15 '12 at 14:54


text:
# Proof of equipartition theorem for $i$ not equal to $j$

The generalized equipartition theorem (where variables need not be quadratic) states that if $$x_i$$ is a canonical variable (position or momentum variable), then

$$\left\langle x_i \frac{∂H}{∂x_j}\right\rangle=δ_{ij}\cdot k_BT.$$

Wikipedia provides proofs of this for a microcanonical ensemble (for any $$i$$, $$j$$) and for a canonical ensemble (ostensibly for $$i=j$$ only). However, the case of a canonical ensemble with $$i$$ not equal to $$j$$ (i.e., $$\langle x_i \, ∂H/∂x_j\rangle =0$$) appears to be missing proof. Is it a simple extension?

Yes, the proof is almost identical to the one on the Wikipedia page, and can be found e.g. in Statistical Mechanics (3rd ed) by RK Pathria and PD Beale, p62. We start with $$\left\langle x_i \frac{\partial H}{\partial x_j} \right\rangle = \frac{\int x_i \frac{\partial H}{\partial x_j} e^{-\beta H} d\Gamma}{\int e^{-\beta H} d\Gamma} .$$ Integration of the numerator by parts with respect to $$x_j$$ gives $$\int x_i \frac{\partial H}{\partial x_j} e^{-\beta H} d\Gamma = \int \left[ \frac{1}{\beta} \int \frac{\partial x_i}{\partial x_j} e^{-\beta H} dx_j \right] d\Gamma'$$ where $$\Gamma'$$ stands for all the coordinates except for $$x_j$$, and a boundary term $$x_i e^{-\beta H}$$ evaluated at $$x_j=\pm\infty$$ has been omitted: it is assumed to vanish on the grounds that $$H$$ becomes infinite whenever $$x_j$$ becomes infinite (or is bounded by the container walls). The important point is that $$\partial x_i/\partial x_j = \delta_{ij}$$ which comes out of the integral. That gives us $$k_BT \, \delta_{ij} \int e^{-\beta H} d\Gamma$$ and the integral cancels with the denominator in the original expression.


text:
News: IMPORTANT UPDATE REGARDING SITE IN FORUM ANNOUNCEMENT FOLDER.

• May 25, 2018, 04:18:48 AM

### Author Topic: Cheapskate stories  (Read 625788 times)

0 Members and 1 Guest are viewing this topic.

#### Piratelvr1121

• Member
• Posts: 9123
##### Re: Cheapskate stories
« Reply #495 on: April 11, 2013, 08:08:47 AM »
I always kind of wondered what the point and purpose was to that.  So okay you have a huge great home to brag about but what are you going to do if people want to come see it? Sit them on shabby furniture? Feed them the cheap food you get for themselves?
Beyond a wholesome discipline, be gentle with yourself. You are a child of the universe, no less than the trees and the stars.  You have a right to be here. Be cheerful, strive to be happy. -Desiderata

#### Venus193

• Member
• Posts: 17103
• Backstage passes are wonderful things!
##### Re: Cheapskate stories
« Reply #496 on: April 11, 2013, 08:28:35 AM »
The resale value of the property?

Some people don't get that this is not a good way to live.

#### Piratelvr1121

• Member
• Posts: 9123
##### Re: Cheapskate stories
« Reply #497 on: April 11, 2013, 10:37:23 AM »
I guess so.  I mean if someone has a huge house and have the finances to properly fund not only the house but the utilities, decent furniture, and to feed themselves, fine, why not?

But as I said, I don't get buying a huge expensive house in a premium location just to sit on a hard wooden floor and stare at each other because they couldn't afford anything but bunny ears on their tv.
Beyond a wholesome discipline, be gentle with yourself. You are a child of the universe, no less than the trees and the stars.  You have a right to be here. Be cheerful, strive to be happy. -Desiderata

#### heartmug

• Member
• Posts: 2310
##### Re: Cheapskate stories
« Reply #498 on: April 11, 2013, 10:59:39 AM »
I lived in Monterey, CA for a while in the mid 80s and worked in a county office.  One of the men who worked there owned a home on 17 Mile Drive (very, very expensive real estate, at least back then there were several celebrities including Clint Eastwood who lived there).  Thinking back now, I can't remember the sacrifices this man and his wife made to be ablel to live there; I do remember him cutting coupons out of the office copy of the newspaper, and buying lunch at the grill at KMart on double hot dog days.  They basically poured every cent of their income into their house, which is certainly their choice, but it always seemed to me to be detrimental to their overall quality of life - sure, the house was great, but the furniture (I heard) was pretty sparse, they had one 15 year old car on its last legs, and of course there was no room at all for fun items in the budget.  Kind of a grim life, to me, but each to his own.

The view!  He must have bought it for the view and probably the re-sale value.  I know that area.
One option in a tug of war with someone is just to drop the rope.

#### Piratelvr1121

• Member
• Posts: 9123
##### Re: Cheapskate stories
« Reply #499 on: April 11, 2013, 12:24:05 PM »
That kind of reminds me of when my husband and I sold our starter home and bought a much bigger house in a nicer neighborhood for ourselves and our two daughters, who were toddlers at the time.  I often say that if I could go back in time, I wouldn't have done it.   The new house had a lot more room for all of us, true, and we loved it (still do) - but it ate up every last cent we had for a long time and put us into severe debt.  It's now paid off (thanks in large part to some help from my parents), but I would never do that again.

To add insult to injury, when we sold our starter home, we got less than we'd originally paid for it.  When the buyer sold it himself a few years later, he got double what we'd got.  Sigh!

That's one thing I'm afraid of about moving from where we are.  True, it is a bit small for the 5 of us and will be even more when the two older ones are teens but then once they go off to college or whatever, it'll be less crowded and it'll just be DH, myself and our youngest.

That and many of the homes around here that we like are not gas heat like this one is, but rather require oil which can be pricey, and the bigger the house the more oil to heat. Though on the other hand, the big Victorians around here tend to go for around $200-$300 thousand.   Which isn't too bad when I think that where I used to live they'd be closer to about 600,000
Beyond a wholesome discipline, be gentle with yourself. You are a child of the universe, no less than the trees and the stars.  You have a right to be here. Be cheerful, strive to be happy. -Desiderata

#### Zenith

• Member
• Posts: 502
• Life's Little Lunatic
##### Re: Cheapskate stories
« Reply #500 on: April 11, 2013, 11:38:54 PM »
I used to buy a $15-20 purse at one of the discount stores 3-4 times a year. I then realized these purses weren't made well and I was tired of repairing or discarding them. I now buy one or two well made purses a year for$30-40 and they last. I just change them out if I get bored. Same with shoes, if they aren't made well then they aren't worth a dollar, if you have to keep replacing them.

My current purse is a beautiful, heavy and soft leather item that cost me $120. My most expensive bag ever... It's now 3 years old, and that's young for me and my leather bags. Most of them get used for 10+ years. I had a nice pair of good leather flats that were$105 new.  I wore them for 12 years.  I resoled them 5 times, and re heeled them another 3 or so on top of that.  I probably spent another $100 over the years, spiffing them up. But,$200 over 10 years, with no problems caused by ill fitting shoes, breaks down to $20 a year - and I have NEVER had a$20 pair of shoes that fit me or would last more than a few months.  There are quite a few things that are expensive up front, but will pay for themselves many times over in length of use time.

Doc Marten's boots. I used to LIVE in those things, and never managed to wear a pair out. At worst, a pair might be downgraded to use for painting & such where I don't want to wear shoes that I care about getting paint-splattered.

Careful with the new styles of Doc boots. My friend bought some black and white ones and the white flaked off almost instantly. Apparently the white had been painted on and not dyed and could not handle the natural creasing from walking. I bought a high heeled b & w womens version of the same thing and thankfully i got to treat them before any paint flecked off but they will need to be redone within a year or so. I also have a pair of bloodred traditional Doc which have no colour problems 5 years down the track. There is a problem with their white stuff though and it's not only Doc shoes with this problem. Still very hardy and comfortable boots but I have also noticed their soles are made of much softer rubber now then the ones from 10-15 years ago. I have more wear on my occasional wearing on my 2 year old red boots than my 20+ 4th hand black boots which have hardly any wear on the soles at all.
« Last Edit: April 11, 2013, 11:41:52 PM by Zenith »

#### jedikaiti

• Swiss Army Nerd
• Member
• Posts: 2908
• A pie in the hand is worth two in the mail.
##### Re: Cheapskate stories
« Reply #501 on: April 11, 2013, 11:52:29 PM »
That's too bad. Seems they may have succumbed to the lower quality = higher profits line of thinking.
What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science!

"The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture

#### nuit93

• Member
• Posts: 1094
##### Re: Cheapskate stories
« Reply #502 on: April 12, 2013, 12:05:12 AM »
I used to buy a $15-20 purse at one of the discount stores 3-4 times a year. I then realized these purses weren't made well and I was tired of repairing or discarding them. I now buy one or two well made purses a year for$30-40 and they last. I just change them out if I get bored. Same with shoes, if they aren't made well then they aren't worth a dollar, if you have to keep replacing them.

My current purse is a beautiful, heavy and soft leather item that cost me $120. My most expensive bag ever... It's now 3 years old, and that's young for me and my leather bags. Most of them get used for 10+ years. I had a nice pair of good leather flats that were$105 new.  I wore them for 12 years.  I resoled them 5 times, and re heeled them another 3 or so on top of that.  I probably spent another $100 over the years, spiffing them up. But,$200 over 10 years, with no problems caused by ill fitting shoes, breaks down to $20 a year - and I have NEVER had a$20 pair of shoes that fit me or would last more than a few months.  There are quite a few things that are expensive up front, but will pay for themselves many times over in length of use time.

Doc Marten's boots. I used to LIVE in those things, and never managed to wear a pair out. At worst, a pair might be downgraded to use for painting & such where I don't want to wear shoes that I care about getting paint-splattered.

Careful with the new styles of Doc boots. My friend bought some black and white ones and the white flaked off almost instantly. Apparently the white had been painted on and not dyed and could not handle the natural creasing from walking. I bought a high heeled b & w womens version of the same thing and thankfully i got to treat them before any paint flecked off but they will need to be redone within a year or so. I also have a pair of bloodred traditional Doc which have no colour problems 5 years down the track. There is a problem with their white stuff though and it's not only Doc shoes with this problem. Still very hardy and comfortable boots but I have also noticed their soles are made of much softer rubber now then the ones from 10-15 years ago. I have more wear on my occasional wearing on my 2 year old red boots than my 20+ 4th hand black boots which have hardly any wear on the soles at all.

Are you ordering yours directly from the UK or from a US store?

#### laud_shy_girl

• Member
• Posts: 285
##### Re: Cheapskate stories
« Reply #503 on: April 12, 2013, 06:49:18 AM »
I lived in Docs but they stopped making them the way they used to. The quality is just not there anymore.
“For too long, we've assumed that there is a single template for human nature, which is why we diagnose most deviations as disorders. But the reality is that there are many different kinds of minds. And that's a very good thing.” - Jonah Lehrer

#### Calistoga

• Member
• Posts: 391
##### Re: Cheapskate stories
« Reply #504 on: April 12, 2013, 10:54:37 AM »
Well. I'm guilty of a bit of my own cheapskateness I fear.

I have to wear khaki's to work. Without fail after about 6 months, I develop a tear in the crotch area of my pants. But instead of buying new ones, I take this one pair of old ones that I have, cut a patch, and hot glue it to the pants over the tear. Then I sew it later. I have two pairs of khakis that have 3 patches or more...the only reason I'll get a new pair is if I actually out grow the old ones.

BUT I think that some of DH's family is worse.

MIL in particular is obsessed with plastic take away cups. Not the thick, reasonably you could use this a few times kind, but the cheap clear plastic ones and even the paper ones. Woe on the person who tries to throw away a solo cup.

#### Carotte

• Member
• Posts: 1744
##### Re: Cheapskate stories
« Reply #505 on: April 12, 2013, 11:46:28 AM »
I have to wear khaki's to work. Without fail after about 6 months, I develop a tear in the crotch area of my pants. But instead of buying new ones, I take this one pair of old ones that I have, cut a patch, and hot glue it to the pants over the tear. Then I sew it later. I have two pairs of khakis that have 3 patches or more...the only reason I'll get a new pair is if I actually out grow the old ones.

You should try putting up a patch before you get a hole, maybe some fusible fabric? That way it will last longer.
I wouldn't say it's being a cheapstake, just like mending socks, sometimes clothes have a lot more life in them if you just patch a small hole or two.

#### alkira6

• Member
• Posts: 655
##### Re: Cheapskate stories
« Reply #506 on: April 12, 2013, 11:50:32 AM »
I have to wear khaki's to work. Without fail after about 6 months, I develop a tear in the crotch area of my pants. But instead of buying new ones, I take this one pair of old ones that I have, cut a patch, and hot glue it to the pants over the tear. Then I sew it later. I have two pairs of khakis that have 3 patches or more...the only reason I'll get a new pair is if I actually out grow the old ones.

You should try putting up a patch before you get a hole, maybe some fusible fabric? That way it will last longer.
I wouldn't say it's being a cheapstake, just like mending socks, sometimes clothes have a lot more life in them if you just patch a small hole or two.

This is what I've learned to do with jeans. I have "thunder thighs" that rub against the seams on the inside leg and they wear harder than the rest of the jeans. I have some old jeans that I have cut into strips that I use fusable webbing to patch in and then sew in so that the inside legs are re-enforced for hard wear. For some reason I do not have this same issue with any other pants, just jeans.

#### Calistoga

• Member
• Posts: 391
##### Re: Cheapskate stories
« Reply #507 on: April 12, 2013, 11:52:28 AM »
Humm... I'll have to try that.

I'm also debating just getting some khaki colored jeans, since it makes no sense to me to wear these thin flimsy pants to work in a kitchen.

#### alkira6

• Member
• Posts: 655
##### Re: Cheapskate stories
« Reply #508 on: April 12, 2013, 12:09:31 PM »
This is what I've learned to do with jeans. I have "thunder thighs" that rub against the seams on the inside leg and they wear harder than the rest of the jeans. I have some old jeans that I have cut into strips that I use fusable webbing to patch in and then sew in so that the inside legs are re-enforced for hard wear. For some reason I do not have this same issue with any other pants, just jeans.

As a lifelong fat person, I have become quite the expert at patching the inner thighs of pants.  Nearly every pair of my pants has patches there, but since I made the pants, the patches are out of the exact same material.   As for jeans.....I've patched jeans for friends, but I don't wear them.

The pants that I make for myself already have a V shaped insert that is made of two pieces of fabric fused together and sewn into each leg.  I find that I have much less frustration doing it from the front end. That or making sundresses. Over the past few years the medium weight dress up/dress down sundress has become my friend.

#### Zenith

• Member
• Posts: 502
• Life's Little Lunatic
##### Re: Cheapskate stories
« Reply #509 on: April 12, 2013, 10:48:41 PM »
I used to buy a $15-20 purse at one of the discount stores 3-4 times a year. I then realized these purses weren't made well and I was tired of repairing or discarding them. I now buy one or two well made purses a year for$30-40 and they last. I just change them out if I get bored. Same with shoes, if they aren't made well then they aren't worth a dollar, if you have to keep replacing them.

My current purse is a beautiful, heavy and soft leather item that cost me $120. My most expensive bag ever... It's now 3 years old, and that's young for me and my leather bags. Most of them get used for 10+ years. I had a nice pair of good leather flats that were$105 new.  I wore them for 12 years.  I resoled them 5 times, and re heeled them another 3 or so on top of that.  I probably spent another $100 over the years, spiffing them up. But,$200 over 10 years, with no problems caused by ill fitting shoes, breaks down to $20 a year - and I have NEVER had a$20 pair of shoes that fit me or would last more than a few months.  There are quite a few things that are expensive up front, but will pay for themselves many times over in length of use time.

Doc Marten's boots. I used to LIVE in those things, and never managed to wear a pair out. At worst, a pair might be downgraded to use for painting & such where I don't want to wear shoes that I care about getting paint-splattered.

Careful with the new styles of Doc boots. My friend bought some black and white ones and the white flaked off almost instantly. Apparently the white had been painted on and not dyed and could not handle the natural creasing from walking. I bought a high heeled b & w womens version of the same thing and thankfully i got to treat them before any paint flecked off but they will need to be redone within a year or so. I also have a pair of bloodred traditional Doc which have no colour problems 5 years down the track. There is a problem with their white stuff though and it's not only Doc shoes with this problem. Still very hardy and comfortable boots but I have also noticed their soles are made of much softer rubber now then the ones from 10-15 years ago. I have more wear on my occasional wearing on my 2 year old red boots than my 20+ 4th hand black boots which have hardly any wear on the soles at all.

Are you ordering yours directly from the UK or from a US store?

Sorry about the quote tree. I ordered from the US store but his pair of b&w shoes came from the UK and mine from the US. They were shipped from whoever had them in stock really. Mind you the pricing was fantastic. If I'd bought my shoes here in Australia they would have cost me over $300 and ordering from overseas cost me$130 inc shipping (on sale). I've been told painted white is a tricky thing to adhere properly to leather in any shoe brand. Any teeny imperfection can ruin the whole thing. Friend isn't too bothered by his 'mouldy' looking shoes and he couldn't be bothered sending them back for an obvious manufacturing defect. Shrug, his loss. If my white comes off like his has I'll just recover in fabric.

Just be aware of painted shoes vs dyed shoes (which is which isn't often mentioned). Dyed will hold up to creasing much better, painted often doesn't. Never knew that until I talked to a cobbler.


text:
### Practice Session for each note on this lesson

 Characteristics of a Group Take Practice Test
 Characteristics of a Period Take Practice Test
 Mendeleev's Periodic Table Take Practice Test
 Modern Periodic Table Take Practice Test
 Reactivity of Metals and Non- Metals Take Practice Test


text:
MathSciNet bibliographic data MR1040388 (91i:32026) 32H35 (32F20 32H10 35N15) Bell, S. Mapping problems in complex analysis and the $\overline\partial$$\overline\partial$-problem. Bull. Amer. Math. Soc. (N.S.) 22 (1990), no. 2, 233–259. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# 3.11: Wave Propagation

Wave motion typically involves a packet of waves encompassing a finite number of wave cycles. Information in a wave only can be transmitted by starting, stopping, or modulating the amplitude of a wave train, which is equivalent to forming a wave packet. For example, a musician will play a note for a finite time, and this wave train propagates out as a wave packet of finite length. You have no information as to the frequency and amplitude of the sound prior to the wave packet reaching you, or after the wave packet has passed you. The velocity of the wavelets contained within the wave packet is called the phase velocity. For a dispersive system the phase velocity of the wavelets contained within the wave packet is frequency dependent and the shape of the wave packet travels at the group velocity which usually differs from the phase velocity. If the shape of the wave packet is time dependent, then neither the phase velocity, which is the velocity of the wavelets, nor the group velocity, which is the velocity of an instantaneous point fixed to the shape of the wave packet envelope, represent the actual velocity of the overall wavepacket.

A third wavepacket velocity, the signal velocity, is defined to be the velocity of the leading edge of the energy distribution, and corresponding information content, of the wave packet. For most linear systems the shape of the wave packet is not time dependent and then the group and signal velocities are identical. However, the group and signal velocities can be very different for non-linear systems as discussed in chapter $$4.7$$. Note that even when the phase velocity of the waves within the wave packet travels faster than the group velocity of the shape, or the signal velocity of the energy content of the envelope of the wave packet, the information contained in a wave packet is only manifest when the wave packet envelope reaches the detector and this energy and information travel at the signal velocity.

The modern ideas of wave propagation, including Hamilton’s concept of group velocity, were developed by Lord Rayleigh when applied to the theory of sound[Ray1887]. The concept of phase, group, and signal velocities played a major role in discussion of electromagnetic waves as well as de Broglie’s development of the concept of wave-particle duality and the development of wave mechanics by Schrödinger.

## Phase, group, and signal velocities of wave packets

The concepts of wave packets, as well as their phase, group, and signal velocities, are of considerable importance for propagation of information and other manifestations of wave motion in science and engineering which warrants further discussion at this juncture.

Consider a particular $$k, \omega$$, component of a one-dimensional wave,

$q(x,t) = Ee^{i(kx\pm \omega t)} \label{3.113}$

The argument of the exponential is called the phase $$\phi$$ of the wave where

$\phi \equiv kx - \omega t \label{3.114}$

If we move along the $$x$$ axis at a velocity such that the phase is constant then we perceive a stationary wave. The velocity of this wave is called the phase velocity. To ensure constant phase we require that $$\phi$$ is constant or, assuming real $$k$$ and $$\omega$$

$\omega dt = k dx \label{3.115}$

Therefore the phase velocity is defined to be

$v_{phase} = \frac{\omega}{k} \label{3.116}$

The velocity we have used so far is just the phase velocity of the individual wavelets at the carrier frequency. If $$k$$ or $$\omega$$ are complex then one must take the real parts to ensure that the velocity is real.

If the phase velocity of a wave is dependent on the wavelength, that is, $$v_{phase} (k)$$, then the system is said to be dispersive in that the wave is dispersed according the wavelength. The simplest illustration of dispersion is the refraction of light in glass prism which leads to dispersion of the light into the spectrum of wavelengths. Dispersion leads to development of wave packets that travel at group and signal velocities that usually differ from the phase velocity. To illustrate this consider two equal amplitude travelling waves having slightly different wave number $$k$$ and angular frequency $$\omega$$. Superposition of these waves gives

\begin{align} q(x, t) & = A ( e^{i[k x-\omega t]} + e^{i[(k+\Delta k) x-(\omega+\Delta \omega) t]} ) \nonumber \\ & = A e^{i [ ( k+\frac{\Delta k}{2} ) x- ( \omega+\frac{\Delta w}{2} ) t ]} \cdot \{e^{-i [\frac{\Delta k}{2} x-\frac{\Delta \omega}{2} t ]} + e^{i [\frac{\Delta k}{2} x-\frac{\Delta \omega}{2} t ]} \} \nonumber \\ & = 2 A e^{i [ ( k+\frac{\Delta k}{2} ) x- ( \omega+\frac{\Delta \omega}{2} ) t ]} \cos [\frac{\Delta k}{2} x-\frac{\Delta \omega}{2} t ] \end{align} \label{3.117}

This corresponds to a wave with the average carrier frequency modulated by the cosine term which has a wavenumber of $$\frac{\Delta k}{2}$$ and angular frequency $$\frac{\Delta \omega}{2}$$, that is, this is the usual example of beats. The cosine term modulates the average wave producing wave packets as shown in figure (3.9.1). The velocity of these wave packets is called the group velocity given by requiring that the phase of the modulating term is constant, that is

$\frac{\Delta k}{2} dx = \frac{\Delta \omega}{2} dt \label{3.118}$

Thus the group velocity is given by

$v_{group} = \frac{dx}{dt} = \frac{\Delta \omega}{ \Delta k} \label{3.119}$

If dispersion is present then the group velocity $$v_{group} = \frac{\Delta \omega}{\Delta k}$$ does not equal the phase velocity $$v_{phase} = \frac{ \omega}{k}$$.

Expanding the above example to superposition of $$n$$ waves gives

$q(x,t) = \sum^{n}_{r = 1} A_r e^{i ( k_r x \pm \omega_r t)} \label{3.120}$

In the event that $$n \rightarrow \infty$$ and the frequencies are continuously distributed, then the summation is replaced by an integral

$q(x,t) = \int^{\infty}_{-\infty} A(k) e^{i(kx \pm \omega t)} dk \label{3.121}$

where the factor $$A (k)$$ represents the distribution amplitudes of the component waves, that is the spectral decomposition of the wave. This is the usual Fourier decomposition of the spatial distribution of the wave.

Consider an extension of the linear superposition of two waves to a well defined wave packet where the amplitude is nonzero only for a small range of wavenumbers $$k_0 \pm \Delta k$$.

$q(x,t) = \int^{k_0 + \Delta k}_{k_0−\Delta k} A(k) e^{i(kx - \omega t)} dk \label{3.122}$

This functional shape is called a wave packet which only has meaning if $$\Delta k << k_0$$. The angular frequency can be expressed by making a Taylor expansion around $$k_0$$

$\omega (k) = \omega (k_0) + \left(\frac{d \omega}{dk} \right)_{k_0} (k − k_0) + \dots \label{3.123}$

For a linear system the phase then reduces to

$kx − \omega t = (k_0 x − \omega_0 t) + (k − k_0) x − \left( \frac{d\omega}{dk}\right)_{k_0} (k - k_0) t \label{3.124}$

The summation of terms in the exponent given by \ref{3.125} leads to the amplitude \ref{3.123} having the form of a product where the integral becomes

$q(x,t) = e^{i(k_0 x - \omega_0 t)} \int^{k_0+\Delta k}_{k_0 − \Delta k} A(k) e^{i(k - k_0)[x-(\frac{d\omega}{dk})_{k_0} t]} dk \label{3.125}$

The integral term modulates the $$e^{i(k_0 x - \omega_0 t)}$$ first term.

The group velocity is defined to be that for which the phase of the exponential term in the integral is constant. Thus

$v_{group} = \left(\frac{d\omega}{dk}\right)_{k_0} \label{3.126}$

Since $$\omega = kv_{phase}$$ then

$v_{group} = v_{phase} + k\frac{\partial v_{phase}}{\partial k} \label{3.127}$

For non-dispersive systems the phase velocity is independent of the wave number $$k$$ or angular frequency $$\omega$$ and thus $$v_{group} = v_{phase}$$. The case discussed earlier, equation (3.9.3), for beating of two waves gives the same relation in the limit that $$\Delta \omega$$ and $$\Delta k$$ are infinitessimal.

The group velocity of a wave packet is of physical significance for dispersive media where $$v_{group} = (\frac{d\omega}{dk})_{k_0} \neq \frac{\omega}{k} = v_{phase}$$. Every wave train has a finite extent and thus we usually observe the motion of a group of waves rather than the wavelets moving within the wave packet. In general, for non-linear dispersive systems the derivative $$\frac{\partial v_{phase}}{\partial k}$$ can be either positive or negative and thus in principle the group velocity can either be greater than, or less than, the phase velocity. Moreover, if the group velocity is frequency dependent, that is, when group velocity dispersion occurs, then the overall shape of the wave packet is time dependent and thus the speed of a specific relative location defined by the shape of the envelope of the wave packet does not represent the signal velocity of the wave packet. Brillouin showed that the distribution of the energy, and corresponding information content, in any wave packet travels at the signal velocity which can be different from the group velocity if the shape of the envelope of the wave packet is time dependent. For electromagnetic waves one has the possibility that the group velocity $$v_{group} > v_{phase} = c$$. In 1914 Brillouin[Bri14][Bri60] showed that the signal velocity of electromagnetic waves, defined by the leading edge of the time-dependent envelope of the wave packet, never exceeds $$c$$ even though the group velocity corresponding to the velocity of the instantaneous shape of the wave packet may exceed $$c$$. Thus, there is no violation of Einstein’s fundamental principle of relativity that the velocity of an electromagnetic wave cannot exceed $$c$$.

Example $$\PageIndex{1}$$: Water waves breaking on a beach

The concepts of phase and group velocity are illustrated by the example of water waves moving at velocity $$v$$ incident upon a straight beach at an angle $$\alpha$$ to the shoreline. Consider that the wavepacket comprises many wavelengths of wavelength $$\lambda$$. During the time it takes the wave to travel a distance $$\lambda$$, the point where the crest of one wave breaks on the beach travels a distance $$\frac{\lambda}{\cos \alpha}$$ along beach. Thus the phase velocity of the crest of the one wavelet in the wave packet is

$v_{phase} = \frac{v}{\cos \alpha} \nonumber$

The velocity of the wave packet along the beach equals

$v_{group} = v \cos \alpha \nonumber$

Note that for the wave moving parallel to the beach $$\alpha = 0$$ and $$v_{phase} = v_{group} = v$$. However, for $$\alpha = \frac{\pi}{2} v_{phase} \rightarrow \infty$$ and $$v_{group} \rightarrow 0$$. In general for waves breaking on the beach

$v_{phase}v_{group} = v^2 \nonumber$

The same behavior is exhibited by surface waves bouncing off the sides of the Erie canal, sound waves in a trombone, and electromagnetic waves transmitted down a rectangular wave guide. In the latter case the phase velocity exceeds the velocity of light $$c$$ in apparent violation of Einstein’s theory of relativity. However, the information travels at the signal velocity which is less than $$c$$.

Example $$\PageIndex{2}$$: Surface waves for deep water

In the "Theory of Sound" Rayleigh discusses the example of surface waves for water where he derives a dispersion relation for the phase velocity $$v_{phase}$$ and wavenumber $$k$$ which are related to the density $$\rho$$, depth $$l$$, gravity $$g$$, and surface tension $$T$$, by

$\omega^2 = gk + \frac{Tk^3}{\rho} \tanh (kl) \nonumber$

For deep water where the wavelength is short compared with the depth, that is $$kl >> 1$$, then $$\tanh (kl) \rightarrow 1$$ and the dispersion relation is given approximately by

$\omega^2 = gk + \frac{Tk^3}{\rho} \nonumber$

For long surface waves for deep water, that is, small $$k$$, then the gravitational first term in the dispersion relation dominates and the group velocity is given by

$v_{group} = \left( \frac{d\omega}{dk} \right) = \frac{1}{2} \sqrt{\frac{g}{k}} = \frac{1}{2} \frac{\omega}{k} = \frac{ v_{phase}}{2} \nonumber$

That is, the group velocity is half of the phase velocity. Here the wavelets are building at the back of the wave packet, progress through the wave packet and dissipate at the front. This can be demonstrated by dropping a pebble into a calm lake. It will be seen that the surface disturbance comprises a wave packet moving outwards at the group velocity with the individual waves within the wave packet expanding at twice the group velocity of the wavepacket, that is, they appear at the inner radius of the wave packet and disappear at the outer radius of the wave packet.

For small wavelength ripples, where $$k$$ is large, then the surface tension term dominates and the dispersion relation is approximately given by

$\omega^2 \simeq \frac{Tk^3}{\rho} \nonumber$

leading to a group velocity of

$v_{group} = \left( \frac{d\omega}{dk}\right) = \frac{3}{2} v_{phase} \nonumber$

Here the group velocity exceeds the phase velocity and wavelets are building at the front of the wave packet and dissipate at the back. Note that for this linear system the Brillion signal velocity equals the group velocity for both gravity and surface tension waves for deep water.

Example $$\PageIndex{3}$$: Electromagnetic waves in ionosphere

The response to radio waves of the free electron plasma in the ionosphere provides an excellent example that involves cut-off frequency, complex wavenumber $$k$$, as well as the phase, group, and signal velocities.

Maxwell’s equations give the most general wave equation for electromagnetic waves to be

$\nabla^2 \mathbf{E} − \varepsilon \mu \frac{ \partial^2\mathbf{E}}{\partial t^2} = \mu \frac{ \partial \mathbf{j}_{free}}{\partial t} + \nabla \cdot \left(\frac{\rho_{free}}{\varepsilon} \right) \nonumber$

$\nabla^2\mathbf{H} − \mu \varepsilon \frac{ \partial^2 \mathbf{H}}{\partial t^2} = −\nabla \times \mathbf{j}_{free} \nonumber$

where $$\rho_{free}$$ and $$\mathbf{j}_{free}$$ are the unbound charge and current densities. The effect of the bound charges and currents are absorbed into $$\varepsilon$$ and $$\mu$$. Ohm’s Law can be written in terms of the electrical conductivity $$\sigma$$ which is a constant

$\mathbf{j} = \sigma \mathbf{E} \nonumber$

Assuming Ohm’s Law plus assuming $$\rho_{free} = 0$$, in the plasma gives the relations

$\nabla^2 \mathbf{E} − \varepsilon \mu \frac{\partial^2\mathbf{E}}{\partial t^2} − \sigma \mu \frac{\partial \mathbf{E}}{\partial t} = 0 \nonumber$

$\nabla^2 \mathbf{H} − \mu \varepsilon \frac{\partial^2 \mathbf{H}}{\partial t^2} − \sigma \mu \frac{\partial \mathbf{H}}{\partial t} = 0 \nonumber$

The third term in both of these wave equations is a damping term that leads to a damped solution of an electromagnetic wave in a good conductor.

The solution of these damped wave equations can be solved by considering an incident wave

$\mathbf{E} = E_o \mathbf{\hat{x}}e^{i(\omega t− kz)} \nonumber$

Substituting for $$\mathbf{E}$$ in the first damped wave equation gives

$−k^2 + \omega^2 \varepsilon \mu − i \omega \sigma \mu = 0 \nonumber$

That is

$k^2 = \omega^2 \varepsilon \mu \left[ 1 − \frac{i\sigma}{\omega \varepsilon} \right] \nonumber$

In general $$k$$ is complex, that is, it has real $$k_R$$ and imaginary $$k_1$$ parts that lead to a solution of the form

$\mathbf{E} = E_o e^{-k_I z} e^{i(\omega t - k_R z)} \nonumber$

The first exponential term is an exponential damping term while the second exponential term is the oscillating term.

Consider that the plasma involves the motion of a bound damped electron, of charge $$q$$ of mass $$m$$, bound in a one dimensional atom or lattice subject to an oscillatory electric field of frequency $$\omega$$. Assume that the electromagnetic wave is travelling in the $$\hat{z}$$ direction with the transverse electric field in the $$\hat{x}$$ direction. The equation of motion of an electron can be written as

$\mathbf{\ddot{x}} + \Gamma \mathbf{\dot{x}} + \omega^2_0 x = \mathbf{\hat{x}} q E_0 e^{i(\omega t − kz)} \nonumber$

where $$\Gamma$$ is the damping factor. The instantaneous displacement of the oscillating charge equals

$\mathbf{x} = \frac{q}{m} \frac{1}{(\omega^2_0 − \omega^2) + i\Gamma \omega} \mathbf{\hat{x}} E_0 e^{i(\omega t − kz)} \nonumber$

and the velocity is

$\mathbf{\dot{x}} = \frac{q}{m} \frac{i \omega}{(\omega^2_0 − \omega^2) + i\Gamma \omega} \mathbf{\hat{x}} E_0 e^{i(\omega t − kz)} \nonumber$

Thus the instantaneous current density is given by

$\mathbf{j} = Nq\mathbf{\dot{x}} = \frac{Nq^2}{m} \frac{i \omega}{(\omega^2_0 − \omega^2) + i\Gamma \omega} \mathbf{\hat{x}} E_0 e^{i(\omega t − kz)} \nonumber$

therefore the electrical conductivity is given by

$\sigma = \frac{Nq^2}{m} \frac{i \omega}{(\omega^2_0 − \omega^2) + i\Gamma \omega} \nonumber$

Let us consider only unbound charges in the plasma, that is let $$\omega_0 = 0$$. Then the conductivity is given by

$\sigma = \frac{Nq^2}{m} \frac{i\omega}{i\Gamma \omega - \omega^2} \nonumber$

For a low density ionized plasma $$\omega >> \Gamma$$ thus the conductivity is given approximately by

$\sigma \approx −i \frac{Nq^2}{m\omega} \nonumber$

Since $$\sigma$$ is pure imaginary, then $$\mathbf{j}$$ and $$\mathbf{E}$$ have a phase difference of $$\frac{\pi}{2}$$ which implies that the average of the Joule heating over a complete period is $$\langle \mathbf{j} \cdot \mathbf{E} \rangle = 0$$. Thus there is no energy loss due to Joule heating implying that the electromagnetic energy is conserved.

Substitution of $$\sigma$$ into the relation for $$k^2$$

$k^2 = \omega^2 \varepsilon \mu \left[ 1 − \frac{i\sigma}{\omega \varepsilon} \right] = \omega^2 \varepsilon \mu \left[ 1 − \frac{Nq^2}{ \varepsilon m\omega^2} \right] \nonumber$

Define the Plasma oscillation frequency $$\omega_P$$ to be

$\omega_P \equiv \sqrt{\frac{Nq^2}{\varepsilon m}} \nonumber$

then $$k^2$$ can be written as

$k^2 = \omega^2 \varepsilon \mu \left[ 1 − \left(\frac{\omega_P}{\omega}\right)^2 \right] \tag{$$\alpha$$}$

For a low density plasma the dielectric constant $$\kappa_E \simeq 1$$ and the relative permeability $$\kappa_B \simeq 1$$ and thus $$\varepsilon = \kappa_E \varepsilon_0 \simeq \varepsilon_0$$ and $$\mu = \kappa_B \mu_0 \simeq \mu_0$$. The velocity of light in vacuum $$c = \frac{1}{\sqrt{\varepsilon_0\mu_0}}$$. Thus for low density equation $$\alpha$$ can be written as

$\omega^2 = \omega^2_p + c^2k^2 \tag{$$\beta$$}$

Differentiation of equation $$\beta$$ with respect to $$k$$ gives $$2\omega \frac{d\omega}{dk} = 2c^2 k$$. That is, $$v_{phase}v_{group} = c^2$$ and the phase velocity is

$v_{phase} = \sqrt{c^2 + \frac{\omega^2_p}{k^2}} \nonumber$

There are three cases to consider.

1) $$\omega > \omega_P$$: For this case $$\left[ 1 − ( \frac{\omega_P}{\omega})^2\right] > 1$$ and thus $$k$$ is a pure real number. Therefore the electromagnetic wave is transmitted with a phase velocity that exceeds $$c$$ while the group velocity is less than $$c$$.

2) $$\omega < \omega_P$$: For this case $$\left[ 1 − ( \frac{\omega_P}{\omega})^2\right] < 1$$ and thus $$k$$ is a pure imaginary number. Therefore the electromagnetic wave is not transmitted and in the ionosphere it is attenuated rapidly as $$e^{−( \frac{\omega_P}{c})z}$$. However, since there are no Joule heating losses then the electromagnetic wave must be complete reflected. Thus the Plasma oscillation frequency serves as a cut-off frequency. For this example the signal and group velocities are identical.

For the ionosphere $$N = 10^{−11}$$ electrons/$$m^3$$, which corresponds to a Plasma oscillation frequency of $$v = \omega_P / 2\pi = 3$$ $$MHz$$. Thus electromagnetic waves in the AM waveband ( < 1.6 $$MHz$$) are totally reflected by the ionosphere and bounce repeatedly around the Earth, whereas for VHF frequencies above 3 $$MHz$$, the waves are transmitted and refracted passing through the atmosphere. Thus light is transmitted by the ionosphere. By contrast, for a good conductor like silver, the Plasma oscillation frequency is around $$10^{16}$$ $$Hz$$ which is in the far ultraviolet part of the spectrum. Thus, all lower frequencies, such as light, are totally reflected by such a good conductor, whereas X-rays have frequencies above the Plasma oscillation frequency and are transmitted.

## Fourier transform of wave packets

The relation between the time distribution and the corresponding frequency distribution, or equivalently, the spatial distribution and the corresponding wave-number distribution, are of considerable importance in discussion of wave packets and signal processing. It directly relates to the uncertainty principle that is a characteristic of all forms of wave motion. The relation between the time and corresponding frequency distribution is given via the Fourier transform discussed in appendix $$19.9$$. The following are two examples of the Fourier transforms of typical but rather different wavepacket shapes that are encountered frequently in science and engineering.

Example $$\PageIndex{4}$$: Fourier transform of a Gaussian wave packet

Assuming that the amplitude of the wave is a Gaussian wave packet shown in the adjacent figure where

$G (\omega ) = ce^{− \frac{(\omega −\omega_0)^2}{ 2\sigma^2_{\omega}} }\nonumber$

This leads to the Fourier transform

$f(t) = c \sqrt{2\pi} \sigma_{\omega} e^{− \frac{\sigma^2_{\omega} t^2}{2}} \cos (\omega_0 t) \nonumber$

Note that the wavepacket has a standard deviation for the amplitude of the wavepacket of $$\sigma_t = \frac{1}{\sigma_{\omega}}$$, that is $$\sigma_t \cdot \sigma_{\omega} = 1$$. The Gaussian wavepacket results in the minimum product of the standard deviations of the frequency and time representations for a wavepacket. This has profound importance for all wave phenomena, and especially to quantum mechanics. Because matter exhibits wave-like behavior, the above property of wave packet leads to Heisenberg’s Uncertainty Principle. For signal processing, it shows that if you truncate a wavepacket you will broaden the frequency distribution.

Example $$\PageIndex{5}$$: Fourier transform of a rectangular wave packet

Assume unity amplitude of the frequency distribution between $$\omega_0 − \Delta \omega \leq \omega \leq \omega_0 + \Delta \omega$$, that is, a single isolated square pulse of width $$\tau$$ that is described by the rectangular function $$\prod$$ defined as

$\prod (\omega ) = \begin{cases} 1 && |\omega − \omega_0| < \Delta \omega \\ 0 && |\omega − \omega_0| > \Delta \omega \end{cases} \nonumber$

Then the Fourier transform us given by

$f(t) = \left[\frac{\sin \Delta \omega t}{\Delta \omega t} \right] \cos \omega_0 t \nonumber$

That is, the transform of a rectangular wavepacket gives a cosine wave modulated by an unnormalized sinc function which is a nice example of a simple wave packet. That is, on the right hand side we have a wavepacket $$\Delta t = \pm \frac{2\pi}{\Delta \omega}$$ wide. Note that the product of the two measures of the widths $$\Delta \omega \cdot \Delta t = \pm \pi$$. Example $$I.2$$ considers a rectangular pulse of unity amplitude between $$−\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$$ which resulted in a Fourier transform $$G (\omega ) = \tau \left( \frac{\sin \frac{\omega \tau}{2}}{\frac{\omega \tau}{2}}\right)$$. That is, for a pulse of width $$\Delta t = \pm \frac{\tau}{2}$$ the frequency envelope has the first zero at $$\Delta \omega = \pm \frac{\pi}{\tau}$$. Note that this is the complementary system to the one considered here which has $$\Delta \omega \cdot \Delta t = \pm \pi$$ illustrating the symmetry of the Fourier transform and its inverse.

## Wave-packet Uncertainty Principle

The Uncertainty Principle states that for all types of wave motion there is a minimum product of the uncertainty in the width of a wave packet and the distribution width of the frequency decomposition of the wave packet. This was illustrated by the Fourier transforms of wave packets discussed above where it was shown the product of the widths is minimized for a Gaussian-shaped wave packet. The Uncertainty Principle implies that to make a precise measurement of the frequency of a sinusoidal wave requires that the wave packet be infinitely long. If the length of the wave packet is reduced then the frequency distribution broadens. Then the crucial aspect needed for this discussion, is that, for the amplitudes of any wavepacket, the standard deviations $$\sigma (t) = \sqrt{ \langle t^2 \rangle - \langle t \rangle^2}$$ characterizing the width of the spectral distribution in the angular frequency domain, $$\sigma_A (\omega )$$, and the width in time $$\sigma_A (t)$$ are related:

$\sigma_A (t) \cdot \sigma_A (\omega ) \geqslant 1 \tag{Relation between amplitude uncertainties.}$

This product of the standard deviations equals unity only for the special case of Gaussian-shaped spectral distributions, and is greater than unity for all other shaped spectral distributions.

The intensity of the wave is the square of the amplitude leading to standard deviation widths for a Gaussian distribution where $$\sigma_I (t)^2 = \frac{1}{2} \sigma_A (t)^2$$, that is, $$\sigma_I (t) = \frac{\sigma_A (t)}{\sqrt{2}}$$. Thus the standard deviations for the spectral distribution and width of the intensity of the wavepacket are related by:

$\sigma_I (t) \cdot \sigma_I (\omega ) \geqslant \frac{1}{2} \tag{Uncertainty principle for frequency-time intensities}$

This states that the uncertainties with which you can simultaneously measure the time and frequency for the intensity of a given wavepacket are related. If you try to measure the frequency within a short time interval $$\sigma_I (t)$$ then the uncertainty in the frequency measurement $$\sigma_I (\omega ) \geqslant \frac{1}{2\sigma_I (t)}$$. Accurate measurement of the frequency requires measurement times that encompass many cycles of oscillation, that is, a long wavepacket.

Exactly the same relations exist between the spectral distribution as a function of wavenumber $$k_x$$ and the spatial dependence of a wave $$x$$ which are conjugate representations. Thus the spectral distribution plotted versus $$k_x$$ is directly related to the amplitude as a function of position $$x$$; the spectral distribution versus $$k_y$$ is related to the amplitude as a function of $$y$$; and the $$k_z$$ spectral distribution is related to the spatial dependence on $$z$$. Following the same arguments discussed above, the standard deviation, $$\sigma_I (k_x)$$ characterizing the width of the spectral intensity distribution of $$k_x$$, and the standard deviation $$\sigma_I (x)$$, characterizing the spatial width of the wave packet intensity as a function of $$x$$, are related by the Uncertainty Principle for position-wavenumber. Thus in summary the uncertainty principle for the intensity of wave motion is,

$\sigma_I (t) \cdot \sigma_I (\omega ) \geqslant \frac{1}{2} \label{3.128} \\ \sigma_I (x) \cdot \sigma_I (k_x) \geqslant \frac{1}{2} \quad \sigma_I (y) \cdot \sigma_I (k_y) \geqslant \frac{1}{2} \quad \sigma_I (z) \cdot \sigma_I (k_z) \geqslant \frac{1}{2}$

This applies to all forms of wave motion, be they, sound waves, water waves, electromagnetic waves, or matter waves.

As discussed in chapter $$18$$, the transition to quantum mechanics involves relating the matter-wave properties to the energy and momentum of the corresponding particle. That is, in the case of matter waves, multiplying both sides of Equation \ref{3.128} by $$\hbar$$ and using the de Broglie relations gives that the particle energy is related to the angular frequency by $$E = \hbar \omega$$ and the particle momentum is related to the wavenumber, that is $$\mathbf{p} = \hbar \mathbf{k}$$. These lead to the Heisenberg Uncertainty Principle:

$\sigma_I (t) \cdot \sigma_I (E) \geqslant \frac{\hbar}{2} \label{3.129} \\ \sigma_I (x) \cdot \sigma_I (p_x) \geqslant \frac{\hbar}{2} \quad \sigma_I (y) \cdot \sigma_I (p_y) \geqslant \frac{\hbar}{2} \quad \sigma_I (z) \cdot \sigma_I (p_z) \geqslant \frac{\hbar}{2}$

This uncertainty principle applies equally to the wavefunction of the electron in the hydrogen atom, proton in a nucleus, as well as to a wavepacket describing a particle wave moving along some trajectory. Thus, this implies that, for a particle of given momentum, the wavefunction is spread out spatially. Planck’s constant $$\hbar = 1.05410^{−34} J \cdot s = 6.58210^{−16} eV \cdot s$$ is extremely small compared with energies and times encountered in normal life, and thus the effects due to the Uncertainty Principle are not manifest for macroscopic dimensions.

Confinement of a particle, of mass $$m$$, within $$\pm \sigma (x)$$ of a fixed location implies that there is a corresponding uncertainty in the momentum

$\sigma (p_x) \geq \frac{\hbar}{2\sigma (x)} \label{3.130}$

Now the variance in momentum $$\mathbf{p}$$ is given by the difference in the average of the square $$\left\langle( \mathbf{p} \cdot \mathbf{p})^2 \right\rangle$$, and the square of the average of $$\langle \mathbf{p} \rangle^2$$. That is

$\sigma (\mathbf{p})^2 = \left\langle (\mathbf{p} \cdot \mathbf{p})^2 \right\rangle − \langle \mathbf{p} \rangle^2 \label{3.131}$

Assuming a fixed average location implies that $$\langle \mathbf{p} \rangle = 0$$, then

$\left\langle ( \mathbf{p} \cdot \mathbf{p})^2 \right\rangle = \sigma (p)^2 \geq \left( \frac{\hbar}{2\sigma (r)} \right)^2 \label{3.132}$

Since the kinetic energy is given by:

$\text{Kinetic energy } = \frac{p^2}{2m} \geq \frac{\hbar^2}{8m\sigma (r)^2} \tag{Zero-point energy}$

This zero-point energy is the minimum kinetic energy that a particle of mass $$m$$ can have if confined within a distance $$\pm \sigma (r)$$. This zero-point energy is a consequence of wave-particle duality and the uncertainty between the size and wavenumber for any wave packet. It is a quantal effect in that the classical limit has $$\hbar \rightarrow 0$$ for which the zero-point energy $$\rightarrow 0$$.

Inserting numbers for the zero-point energy gives that an electron confined to the radius of the atom, that is $$\sigma (x) = 10^{−10}m$$, has a zero-point kinetic energy of $$\sim 1 \ eV$$. Confining this electron to $$3 \times 10^{−15}m$$, the size of a nucleus, gives a zero-point energy of $$10^9 \ eV (1 \ GeV )$$. Confining a proton to the size of the nucleus gives a zero-point energy of 0.5 $$MeV$$. These values are typical of the level spacing observed in atomic and nuclear physics. If $$\hbar$$ was a large number, then a billiard ball confined to a billiard table would be a blur as it oscillated with the minimum zero-point kinetic energy. The smaller the spatial region that the ball was confined, the larger would be its zero-point energy and momentum causing it to rattle back and forth between the boundaries of the confined region. Life would be dramatically different if $$\hbar$$ was a large number.

In summary, Heisenberg’s Uncertainty Principle is a well-known and crucially important aspect of quantum physics. What is less well known, is that the Uncertainty Principle exists for all forms of wave motion, that is, it is not restricted to matter waves. The following three examples illustrate application of the Uncertainty Principle to acoustics, the nuclear Mössbauer effect, and quantum mechanics.

Example $$\PageIndex{6}$$: Acoustic Wave Packet

A violinist plays the note middle C (261.625 $$Hz$$) with constant intensity for precisely 2 seconds. Using the fact that the velocity of sound in air is 343.2 $$m/s$$ calculate the following:

1. The wavelength of the sound wave in air: $$\lambda$$ = 343.2/261.625 = 1.312 $$m$$.
2. The length of the wavepacket in air: Wavepacket length = 343.2 $$\times$$ 2 = 686.4 $$m$$
3. The fractional frequency width of the note: Since the wave packet has a square pulse shape of length $$\tau = 2s$$, then the Fourier transform is a sinc function having the first zeros when $$\sin \frac{\omega \tau}{2} = 0$$, that is, $$\Delta \nu = \frac{1}{\tau}$$.

Therefore the fractional width is $$\frac{\Delta \nu}{\nu} = \frac{1}{\nu \tau}$$ = 0.0019. Note that to achieve a purity of $$\frac{\Delta \nu}{\nu} = 10^{−6}$$ the violinist would have to play the note for 1.06 hours.

Example $$\PageIndex{7}$$: Gravitational Red Shift

The Mössbauer effect in nuclear physics provides a wave packet that has an exceptionally small fractional width in frequency. For example, the $$^{57}$$Fe nucleus emits a 14.4 $$keV$$ deexcitation-energy photon which corresponds to $$\omega \approx 2 \times 10^{25}$$ $$rad/s$$ that has a decay time of $$\tau \approx 10^{−7}$$ $$s$$. Thus the fractional width is $$\frac{\Delta \omega}{\omega} \approx 3 \times 10^{−18}$$. In 1959 Pound and Rebka used this to test Einstein’s general theory of relativity by measurement of the gravitational red shift between the attic and basement of the 22.5 $$m$$ high physics building at Harvard. The magnitude of the predicted relativistic red shift is $$\frac{\Delta E}{E} = 2.5\times 10^{−15}$$ which is what was observed with a fractional precision of about 1%.

Example $$\PageIndex{8}$$: Quantum Baseball

George Gamow, in his book ”Mr. Tompkins in Wonderland”, describes the strange world that would exist if $$\hbar$$ was a large number. As an example, consider you play baseball in a universe where $$\hbar$$ is a large number. The pitcher throws a 150 $$g$$ ball 20 $$m$$ to the batter at a speed of 40 $$m/s$$. For a strike to be thrown, the ball’s position must be pitched within the 30 $$cm$$ radius of the strike zone, that is, it is required that $$\Delta x \leq 0.3$$ $$m$$. The uncertainty relation tells us that the transverse velocity of the ball cannot be less than $$\Delta v = \frac{\hbar}{2 m \Delta x}$$. The time of flight of the ball from the mound to batter is $$t = 0.5 \ s$$. Because of the transverse velocity uncertainty, $$\Delta v$$, the ball will deviate $$t \Delta v$$ transversely from the strike zone. This also must not exceed the size of the strike zone, that is;

$t \Delta v = \frac{\hbar t}{2 m \Delta x} \leq 0.3 m \tag{Due to transverse velocity uncertainty}$

Combining both of these requirements gives

$\hbar \leq \frac{2m \Delta x^2}{t} = 5.4 \ 10^{−2} J \cdot s. \nonumber$

This is 32 orders of magnitude larger than $$\hbar$$ so quantal effects are negligible. However, if $$\hbar$$ exceeded the above value, then the pitcher would have difficulty throwing a reliable strike.

This page titled 3.11: Wave Propagation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


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# Dominant synoptic patterns associated with the decay process of PM 2.5 pollution episodes around Beijing

The variation in the concentrations of ambient inline-formulaPM2.5 (particles with an aerodynamic diameter less than 2.5 inline-formulaµm) generally forms a continuous sawtooth cycle with a recurring smooth increase followed by a sharp decrease. The episode of abrupt decay of pollution is mostly meteorological in origin and is controlled by the passage of synoptic systems. One affordable and effective measure for quickly reducing inline-formulaPM2.5 concentrations in northern China is to wait for a strong wind to arrive. However, it is still unclear how strong the wind needs to be and exactly what kind of synoptic system most effectively results in the rapid decay of air pollution episodes. inline-formulaPM2.5 variations over the 28 pollution channel cities of the Beijing region are investigated to determine the mechanisms by which synoptic patterns affect the decay processes of pollution episodes. This work shows more obvious day-to-day variations in inline-formulaPM2.5 concentration in winter than in summer, which implies that wintertime inline-formulaPM2.5 variations are more sensitive to meteorological factors. There were 365 decay processes from January 2014 to March 2020, and 97 of them were related to the effective wet deposition. In total, inline-formula26 %inline-formula43 % of inline-formulaPM2.5 pollutant is removed by the wet deposition in different seasons. Two dominant circulation patterns are identified in summer. All the other three seasons have three circulation types (CTs), respectively. The three CTs in spring show the same patterns as those in autumn and winter. The circulation patterns beneficial to the decay processes all exhibit a higher-than-normal surface wind speed, a negative relative humidity anomaly and net outflow of inline-formulaPM2.5 from the domain. In addition, CT1 in spring, autumn and winter is controlled by northeasterly wind and features the most significant horizontal net outflow of air pollutants and effective upward spread of air pollutants to the free atmosphere. CT2 is the most frequent CT in autumn and winter, with the highest wind speed from the northwest, highest boundary layer height (BLH) and lowest relative humidity among the three CTs, all of which are favorable for the reduction of inline-formulaPM2.5 concentrations. In CT3, strong vertical wind shear within the boundary layer enhances the mixing of surface air pollutants, which is the extra cleaning mechanism besides dry and clean air mass inflow. inline-formulaPM2.5 concentrations show significant decreases of more than 37 %, 41 % and 27 % after the passage of CT1, CT2 and CT3, respectively. A dry airflow with a positive BLH anomaly and the effective horizontal outflow of air pollutants are the main reasons for the abrupt decay phase in summer. inline-formulaPM2.5 concentrations after the decay process show a significant decreasing trend from 2014 to 2020, reflecting successful emission mitigation. Emission reductions have led to a 4.3–5.7 inline-formula $M16inlinescrollmathmlunit\mathrm{normal µ}\mathrm{normal g}\phantom{\rule{0ex}{0ex}}{\mathrm{normal m}}^{-normal 3}\phantom{\rule{0ex}{0ex}}{\mathrm{normal yr}}^{-normal 1}$ 56pt15ptsvg-formulamathimgcfeda80ea327d109f9a60e50f897425b acp-21-2491-2021-ie00001.svg56pt15ptacp-21-2491-2021-ie00001.png decrease in inline-formulaPM2.5 concentrations in the 28 pollution channel cities of the Beijing region.

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Wang, Xiaoyan / Zhang, Renhe / Tan, Yanke / et al: Dominant synoptic patterns associated with the decay process of PM2.5 pollution episodes around Beijing. 2021. Copernicus Publications.

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