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# 7.5: Geometric Series

Created by: CK-12

## Learning objectives

• Find the value of a finite geometric series.
• Determine if a geometric series converges or diverges.
• If an infinite series converges, find the sum.

## Introduction

In lesson 2 we began working with series (sums), and in lesson 4 we proved formulas for certain sums, such as the sum of terms in an arithmetic series. In this lesson we will focus on geometric series, the sum of terms from a geometric sequence. We will prove a formula that allows us to find the sum of a finite number of terms, and we will return to the issue of the sum of an infinite number of terms.

## Finite Geometric Series

A finite geometric series is simply a geometric series with a specific number of terms. For example, consider again the series introduced in lesson 2: 50+25+12.5+....The series is geometric: the first term is 50, and the common ration is (1/2).

The sum of the first two terms is 50 + 25 = 75. We can write this as S2=75
The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S3=87.5

To find the value of Sn in general, we could simply add together the first n terms in a series. However, this would obviously be tedious for a large value of n. Given the regular pattern in a geometric series - every term is (1/r) of the previous term, and the nth term is an = a1rn - 1 , we can use induction to prove a formula for Sn .

Claim: The sum of the first n terms in a geometric series is $S_n = \frac{a_1(1 - r^n)} {1 - r}$

Proof by induction:

1. If n = 1, the nth sum is the first sum, or a1 . Using the hypothesized equation, we get $S_1 = \frac{a_1(1 - r^1)} {1 - r} = \frac{a_1(1 - r)} {1 - r} = a_1$. This establishes the base case.
2. Assume that the sum of the first k terms in a geometric series is $S_k = \frac{a_1(1 - r^k)} {1 - r}$.
3. Show that the sum of the first k+1 terms in a geometric series is $S_{k + 1} = \frac{a_1(1 - r^{k + 1})} {1 - r}$.
$S_{k + 1} = S_k + a_{k + 1}$ The $k + 1$ sum is the $k^{th}$ sum, plus the $k + 1$ term
$= \frac{a_1(1 - r^k)} {1 - r} + a_1r^{k + 1 - 1}$ Substitute from step 2, and substitute the $k + 1$ term
$= \frac{a_1(1 - r^k)} {1 - r} + \frac{a_1r^k(1 - r)} {1- r}$ The common denominator is $1 - r$
$= \frac{a_1(1 - r^k) + a_1r^k(1 - r)} {1 - r}$ Simplify the fraction
$= \frac{a_1 \left[1 - r^k + r^k(1 - r) \right]} {1 - r}$
$= \frac{a_1 \left[1 - r^k + r^k - r^{k + 1} \right]} {1 - r}$
$= \frac{a_1 \left[1 - r^{k + 1} \right]} {1 - r}$ It is proven.

Therefore we have shown that $S_{n}=\frac{a_1(1 - r^n)} {1 - r}$ for a geometric series. Now we can use this equation to find any sum of a geometric series. For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:

$S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{50 \left(1 - (\frac{1} {2})^6 \right)} {1 - \frac{1} {2}} = \frac{50 \left(1 - \frac{1} {64} \right)} {\frac{1} {2}} = \frac{50 \left(\frac{63} {64} \right)} {\frac{1} {2}} = 50 \left(\frac{63} {64} \right) \left(\frac{2} {1} \right) = 98 \frac{7} {16}$

The figure below shows the same calculation on a TI-83/4 calculator:

We can use this formula, as long as the series in question is geometric.

Example 1: Find the sum of the first 10 terms of a geometric series with a1 = 3 and r = 5.

Solution: The sum is 58,593.

$S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{3(1 - 5^7)} {1 - 5} = \frac{3(1 - 78,125)} {-4} = \frac{3(-78,124)} {-4} = 58,593$
Notice that because the common ratio in this series is 5, the terms get larger and larger. This means that for increasing values of n the sums will also get larger and larger. In contrast, in the series with common ratio (1/2), the terms gets smaller and smaller. This situation implies something important about the sum.

### Infinite Geometric Series

Now let’s return to the situation in the introduction in lesson 2: you are walking a 100-mile charity walk. You walk halfway, 50 miles. Then you walk half of that, 25. If you keep walking half of the remaining distance, how will you ever finish the walk?

At the end of lesson 2 we represented the sum of the infinitely many pieces of this walk as: $\sum_{n = 1}^\infty 50 \left(\frac{1} {2} \right)^{n - 1}$ . We know that the pieces have to add up to 100 miles. But how is it possible for the sum of an infinite number of terms to be a finite number?

To find the sum of an infinite number of terms, we should consider some partial sums. Above we found three partial sums: $S_2 = 75 , S_3 = 87.5,$ and $S_6 = 98 \frac{7} {16}$ or $98.4375.$ Now let’s look at larger values of $n:$

$S_7$ $= \frac{50 \left(1 - (\frac{1} {2})^7 \right)} {1 - \frac{1} {2}} \approx 99.2$
$S_8$ $= \frac{50 \left(1 - (\frac{1} {2})^8 \right)} {1 - \frac{1} {2}} \approx 99.6$
$S_{10}$ $= \frac{50 \left(1 - (\frac{1} {2})^{10} \right)} {1 - \frac{1} {2}} \approx 99.9$

As n approaches infinity, the value of Sn seems to approach 100. In terms of the actual sums, what is happening is this: as n increases, the nth term gets smaller and smaller, and so the nth term contributes less and less to the value of Sn . We say that the series converges, and we can write this with a limit:

$\lim_{n \rightarrow \infty} S_n$ $= \lim_{n \rightarrow \infty} \left(\frac{50 \left(1 - (\frac{1} {2})^{n} \right)} {1 - \frac{1} {2}} \right )$
$= \lim_{n \rightarrow \infty} \left(\frac{50 \left(1 - (\frac{1} {2})^{n} \right)} {\frac{1} {2}} \right )$
$= \lim_{n \rightarrow \infty} \left(100 \left(1 - \left(\frac{1} {2} \right)^{n} \right ) \right)$

As n approaches infinity, the value of $\left(\frac{1} {2} \right)^{n}$ gets smaller and smaller. That is, the value of this expression approaches 0. Therefore the value of $1 - \left(\frac{1} {2} \right)^{n}$ approaches 1, and $100 \left(1 - \left(\frac{1} {2} \right)^n \right)$ approaches 100(1) = 100.

We can do the same analysis for the general case of a geometric series, as long as the terms are getting smaller and smaller. This means that the common ratio must be a number between -1 and 1: |r| < 1.

$\lim_{n \rightarrow \infty} S_n$ $= \lim_{n \rightarrow \infty} \left (\frac{a_1(1 - r^n)} {1 - r} \right)$
$=\frac{a_1} {1 - r},$ as $(1 - r^n) \rightarrow 1$

Therefore, we can find the sum of an infinite geometric series using the formula $S = \frac{a_1} {1 - r}$.

Example 2: Determine if the series converges. If it converges, find the sum.

a.$40 + -20 + 10 + -5 + ...$ b.$1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27}+ ...$ c.$3 + -6 +12 + -24 + ...$

Solution:

a. $40 + -20 + 10 + -5 + ...$ converges
The common ratio is $\frac{-1} {2}$. Therefore the sum converges to:
$\frac{40} {1 - \left (\frac{-1} {2} \right)} = \frac{40} {\frac{3} {2}} = 40 \left(\frac{2} {3} \right) = \frac{80} {3}$
b. $1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27} + ...$ converges.
The common ratio is (1/3) . Therefore the sum converges to:
$\frac{1} {1 - \frac{1} {3}} = \frac{1} {\frac{2} {3}} = \frac{3} {2}$
c. The series 3 + -6 + 12 + -24 + ... does not converge, as the common ratio is -2.

Remember that the idea of an infinite sum was introduced in the context of a realistic situation, albeit a paradoxical one. We can in fact use infinite geometric series to model other realistic situations. Here we will look at another example: the total vertical distance traveled by a bouncing ball.

### The Infinitely Bouncing Ball

Consider a bouncing ball that bounces to a particular percent or fraction of its previous height with each bounce. If the ball is left alone, theoretically, it will continue this pattern infinitely.

We can think of the total distance as the distance the ball travels down plus the distance the ball travels back up. The downward bounces form a geometric series, and the upward bounces form a geometric series. So we can think of the total distance as the sum of two series. (You can also develop one series, which you will do in the review questions). The example below shows how to find the total distance. Example 3: A bouncing ball

A ball is dropped from a height of 20 feet. Each time it bounces, it reaches 50% of its previous height. What is the total vertical distance the ball travels?

Solution:

We can think of the total distance as the distance the ball travels down + the distance the ball travels back up. The downward bounces form a geometric series:
20 + 10 + 5 +...
The upward bounces form the same series, except the first term is 10.
So the total distance is: $\sum_{n=1}^\infty 20 \left ( \frac{1}{2} \right )^{n-1}+\sum_{n=1}^\infty 10 \left ( \frac{1}{2} \right )^{n-1}$.
Each sum converges, as the common ratio is (1/2). Therefore the total distance is:
$\frac {20}{1-\frac {1}{2}}+\frac {10}{1-\frac {1}{2}}=\frac {20}{\frac {1}{2}}+\frac {10}{\frac {1}{2}}=40+20=60$
So the ball travels a total vertical distance of 60 feet.

## Lesson Summary

In this lesson we have found the sums of finite geometric series, and we have returned to the idea of an infinite series. A geometric series is either convergent or divergent. In this lesson you have learned to determine if a geometric series converges or diverges: a geometric series will converge if |r| < 1 , and the sum is $S=\frac{a_{1}}{1-r}$. It is important to note that other kinds of series converge besides geometric series. You will examine these other types of series when you study calculus.

## Points to Consider

1. How can infinitely many numbers add up to a finite number?
2. How does induction help us prove formulas that simplify large sums?

## Review Questions

1. Find the sum of each series: a. The first term of a geometric series is 4, and the common ratio is 3. Find S8. b. The first term of a geometric series is 80, and the common ratio is (1/4). Find S7.
2. Find the sum: 5+10+20+...+640 (Hint: if an=640 , what is n?)
3. Use a geometric series to answer the question: In January, a company’s sales totaled \$11,000. It is predicted that the company’s sales will increase 5% each month for the next year. At this rate, what will be the total sales for the year?
4. For questions 4 – 6, determine if the series converges or diverges. If it converges, find the sum.
5. -3+6+-12+24+...
6. 240+60+15+...
7. 9+6+4+(8/3)+...
8. In this lesson, we proved the formula for the sum of a geometric series, $S_n=\frac {a_1(1-r^n)}{1-r}$, using induction. Here you will prove this formula without induction: (1) Let Sn=a1+a1r+a1r2+...+a1rn-1 (2) Multiply Sn by r to obtain a second equation (3) Subtract the equations and solve for Sn.
9. A ball is dropped from a height of 40 feet, and each time it bounces, it reaches 25% of its previous height. a. Find the total vertical distance the ball travels, using the method used in the lesson. b. Find the total vertical distance the ball travels using a single series. (Hint: write out several terms for each bounce. For example, the first bounce is: 40 feet down + 10 feet up = 50 feet traveled.)
10. Below are two infinite series that are not geometric. Use a graphing calculator to examine partial sums. Does either series converge? a. $1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+...$ b. $1+\frac {1}{4}+\frac {1}{9}+\frac {1}{16}+...$

1. a.$S_{8}=\frac {4(1-3^8)}{1-3}=13,120$ b.$S_{7}= \frac {80 \left ( {1- \left ( \frac{1}{4} \right )^7} \right )}{1-\frac {1}{4}} \approx 106.66$
2. $S_{8}=\frac {5(1-2^8)}{1-2}=1275$
3. $S_{12}=\frac {11000(1-1.05^{12})}{1-1.05}= \175,088.39$
4. The sum does not converge because r = -2.
5. The sum converges. S = 320.
6. The sum converges. S = 27.
8. (1) $S_n=a_1+a_1{r}+a_1{r^2}+...+a_1{r^{n-1}}$
(2) $r{S_n}=a_1{r}+a_1{r^2}+a_1{r^3}+...+a_1{r^n}$
(3) $S_n-rS_n=a_1-a_1{r^n}$
$\Rightarrow S_n(1-r)=a(1-r^n)$
$\Rightarrow S_n=\frac {a(1-r^n)}{(1-r)}$
9. a.$\sum_{n-1}^\infty 40 \left ( \frac{1}{4} \right )^{n-1}+\sum_{n-1}^\infty 20 \left ( \frac{1}{4} \right )^{n-1}=66\frac {2}{3}$ b.$\sum_{n-1}^\infty 50 \left ( \frac{1}{4} \right )^{n-1}=66\frac {2}{3}$
10. a.This series does not converge. b.This series converges around 1.65. (The actual sum is $\frac {\pi^2}{6}$.)

## Vocabulary

Converge
If the limit of the partial sums of series exists and is finite, the series converges.
Diverge
If the limit of the partial sums of series does not exist or is infinite, then the series diverges.
Geometric series
A geometric series is the sum of terms from a geometric sequence.

Feb 23, 2012

Dec 29, 2014


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# [texhax] Printing error using \widehat{} and \widetilde{} in the limits of \int^{}_{}

Herman Z. Bennett hzb at MIT.EDU
Mon Dec 27 04:55:59 CET 2004

Hello,

I am having trouble printing a pdf file generated with latex + dvi2pdf (from
Winedt’s dvi2pdf.edt I can see the pdf is produced with dvipdfm.bat). I would
very much appreciate any help on the matter.

The problem:

I write \int_{\widetilde{A}}^{\widetilde{A}}, then I do latex  + dvi2pdf, and
the pdf file on screen looks ok. I print the file and Acrobat gives me the
might not print correctly.” The printout is a page with no \widehat nor
\widetilde in the limits of the integral (the hat and the tilde disappear from
the limits of the integral and it looks like \int_{A}^{A}). This does not
happened if the \widetilde{A} or \widehat{A} is outside the limits of the
integral. Here is the exact code:

\documentclass[letterpaper, 12pt]{article}
\usepackage{calc}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\textwidth 6in \textheight 9in \footskip .65in \parskip .2in

\begin{document}

Hola $\widehat{A} \widetilde{A}$ (This one is ok)
$$\widetilde{A} \int_{\widetilde{A}}^{B}$$
(the first $\widetilde{A}$ is ok, but the one in the lower limit of $\int$ only

shows an $A$ without the tide).

\end{document}

1.- C:\localtexmf\miktex\config\updmap.cfg contains only the following lines:

And I did initexfm –u and  initexmf –mkmaps after the changes (DOS).

2.- C:\texmf\web2c\updmap.cfg contains the line

3.- I also tried “dvips, open GV and send to print” and it worked fine.
Similarly,  pdflatex works fine too. Due to unrelated issues, these two options
do not satisfy my needs and I would prefer to do latex + dvi2pdf instead (but I
still need to solve the issue discussed in this email).

4.- If I do the same example with \int_{A}^{B} there is no problem.

5.- In case it could be useful, the resulting pdf file has the following fonts
(from Doc Properties/Fonts):

CMR12, CMEX10, CMMI12, CMMI8, cmex8
(they all say: Type 1, Built-in, Embedded Subset, Type 1)

Thank you very much for any help on how to solve this problem,

Herman

PS: Sorry if you see this posting twice. I sent it to pdftex at tug.org, but I
realized afterwards that texhax at tug.org is a more appropiate list for this
question. Thanks in advance for any help!

format: plain


text:
# Test Video

Aller à : Navigation, rechercher

This is an example

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This is the same example, but I used the extension

$$a + b = \frac{1}{2}$$


text:
# Math Help - Help please

1. Mr. Adler and Mr. Bosch work for the same company making $40,000 and$38,000 a year,respectivly. The employer wishes to raise both there saleries for the next year so they will be making the same amount of money as each other. The employer will use \$12,000 for the total of their raises. By what % will Mr. Bosch salary be raised? Exspress ur answer to the nearest tenth.

2.Due to a bowling score of 204 in his last game, Remy raised his average from exactly 156 to 158. What score must eh bowl in the next game to raise his overall everage to exactly 159?

3. Of all four-digit positive integers contaning only digits from the set {2, 4, 6, 8} what fraction of them have at least one of their digits repeated? Exspress ur answer as a common fraction.

2. poke

3. 1)
40,000+40,000*x/100=38,000+38,000*y/100
the sum of their augmentation is 12,000 ,so we have

40,000*x/100+38,000*y/100=12,000
400x+380y=12,000
20x+19y=600

from the first equation,you will get
20x-19y=-100
you do combination and you will get
40x=500
4x=50
x=12.5

19y=600-20(12.5)
19y=600-250
19y=350
y=18.4
the first will get an augmentation of 12.5%
the second 18.4%


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# Band-pass filter vs. Serial Low-pass+High-pass filter - Is there a difference?

Is there a difference between a band-pass filter (BPF) and tying a low-pass filter's (LPF) output to the input of a high-pass filter (HPF)?

It seems to me that they should do the same thing---or perhaps they are algebraically equiveilent (are they?). If not, what is the benefit of using a BPF instead of a serial LPF+HPF?

For example:

Serial LPF+HPF:

Is that effectively the same as a BPF?

• Yes. There is a time where you'd chose one approach and a time where you'd chose another. I have another post elsewhere here on the topic. That may be enough for you. But if not, just say so and I'll write something up for you.
– jonk
Jul 9, 2020 at 4:14
• No. For radios you would not consider these equivalents due to the difference in input and output impedance. Jul 9, 2020 at 4:30

Your intuition is correct, the 2nd is derived from the first through frequency transformations applied to the transfer function. Usually, the first step is to calculate the lowpass prototype, which has unity, non-scaled frequency, and which has the form:

$$H(s)=\frac{\omega^2}{s^2+\frac{\omega}{Q}s+\omega^2}\tag{1}$$

This form is reached after calculating the roots of the filter. Transforming this into a bandpass is done by subsituting $$\s\$$ into the real-valued transfer function. For simplicity, let's consider all the terms $$\1\$$ ($$\BW=\omega_2-\omega_1\$$ is the bandwidth, $$\\omega_0=\sqrt{\omega_1\omega_2}\$$ is the center frequency):

\begin{align} H_{LP}(s)&=\frac{1}{s^2+s+1}\tag{2} \\ H_{BP}(s)&=H_{LP}(s)\biggr|_{s\to\frac{s^2+\omega_0^2}{BW\,s}} \\ &=\frac{1}{\left(\frac{s^2+\omega_0^2}{BW\,s}\right)^2+\left(\frac{s^2+\omega_0^2}{BW\,s}\right)+1} \\ &=\frac{BW^2s^2}{s^4+BW\,s^3+(BW^2+2\omega_0^2)s^2+BW\omega_0^2\,s+\omega_0^4}\tag{3} \end{align}

Equation $$\(3)\$$ gives a new set of roots which can be used to group $$\(3)\$$ into two transfer functions:

\begin{align} H_{BP}(s)&=H_{LP}(s)\cdot H_{HP}(s) \\ &=\frac{a_0}{s^2+b_1s+b_0}\cdot\frac{s^2}{s^2+c_1s+c_0} \end{align}

which are nothing but a lowpass and a highpass. The connection to the topology you are showing is that $$\s\$$ represents one reactive element (depends whether it's series or parallel, normally $$\s=L\$$ and $$\\frac1s=C\$$), while $$\\frac{s^2+\omega_0^2}{BW\,s}\$$ represents a resonant bandpass $$\LC\$$: if it's in series with the load then it's a series $$\LC\$$ (G1 and G3 in your 2nd picture), if it's shunt then it's a parallel $$\LC\$$ (G2).

For notch lowpass/highpass filters, the resultants will be the same combination of two lowpass/highpass filters, but each of them notch.

As @Tony Stewart shows, you cannot calculate the two resultant lowpass/highpass filters and build them as separate filters, as in your 1st picture, since 1) as @jonk links in the comments, there is a loading effect, and 2) these two resulant filters are meant to be used in their composed form (your 2nd picture).

• They are not algebraically equivalent as shown. With no load the top reduces to a 5th order and the bottom to a 4th order polynomial, yet with load may be similar for s21with a 6th order polynomial. Jul 10, 2020 at 16:20
• Great answer, I'm still wrapping my head around the math. Thanks! Jul 10, 2020 at 18:40
• @TonyStewartSunnyskyguyEE75 That's part of the conclusion in the last paragraph. Jul 10, 2020 at 18:51
• It wasn’t clear from the math that they are not equivalent. But they can be loaded and chosen to get similar but not identical results. Jul 10, 2020 at 18:59
• @TonyStewartSunnyskyguyEE75 Sure, provided the loading effect is minimized (which loses power), but why would anyone complicate like that when the tried-and-tested method is OP's 2nd picture? :-) Jul 10, 2020 at 19:01

rough Simulation You need a load R to obtain an impedance ratio on the output and so looking at the s11 s22 impedance of the in,out , they are different, but similar but not the same s21 transfer function.

This configuration is sub-optimal for a BPF. Here a double tuned 3~8.5 MHz BPF.

The left and right images just switch the output response.

Below is a narrow BPF 8th order Cheb. Filter Chosen here for 50 Ohm input and load impedance using the Large passive Filter option.

• Thanks for the graphs, it is neat to see the response of the two filters in an example. Jul 10, 2020 at 18:40


text:
MathSciNet bibliographic data MR2465825 46L05 (47L30) Park, Efton; Trout, Jody On the nonexistence of nontrivial involutive \$n\$$n$-homomorphisms of \$C\sp *\$$C\sp *$-algebras. Trans. Amer. Math. Soc. 361 (2009), no. 4, 1949–1961. 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:
## # Square Root of a Fraction

It is really easy to find the square root of a fraction if the numerator and denominator are perfect squares. It is also pretty easy if you can reduce the fraction to perfect squares.  If the numerator and denominator are not perfect squares, it is a little bit harder because you have to rationalize the […]

## ###$$Reciprocals What is a Reciprocal? A reciprocal is a fraction that has been flipped upside-down. The word comes from the Latin root word “reciprocus” which means “alternating” or “going backward and forward”. Ancient mathematicians chose this word because you have to switch or “alternate” the numerator and denominator to find the reciprocal of a fraction. Examples […] ## ###$$ Denominators

What is a Denominator? A denominator is the bottom number of a fraction. The word comes from the Latin root word “denominaire” and it means “that which gives a name to”. Ancient mathematicians chose this word because the denominator of a fraction names the size of a fractional part.  The size of a fractional part […]

## ##$$Numerators What is a Numerator? A numerator is the top number of a fraction. The word comes from the Latin root word “numerare” and it means “the thing by which the number of something is denoted”. Ancient mathematicians chose this word because the numerator of a fraction tells you the number of fractional parts you are […] ## ####$$ Fractions to Decimals

It is really easy to convert fractions to decimals, especially if you use a calculator. On this page, I’ll show you how to convert fractions to decimals with step-by-step instructions and examples.  If you need to know how to convert decimals to fractions or how to convert fractions to percents, check out the linked pages. […]


text:
# Real Polynomials in inner product space projections?

Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define $$\langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1).$$

(i) Prove that $\langle\cdot,\cdot\rangle$ defines an inner product.

(ii) Find the angle between the polynomials given by $p(t)=-4+9t-6t^2+t^3$ and $q(t)=3-6t+3t^2$.

Out of curious from looking at this question:

Curiosity Question: (iii) If we define $\langle\cdot,\cdot\rangle$ on the space of all continuous functions I can see that it isn’t an inner product. right? Can someone explain to me why not though? I know we can probably still use the projection formula to project a continuous function onto the subspace V.

Can you show me how to project $\sqrt{t}$ onto $V$ to obtain a cubic polynomial that is “near” $\sqrt{t}$.

I saw this in a numerical analysis book which explained a little about these things a couple days ago. I think it was by Isaacson or Keller. This question came up. From reading, I know some things about this problem as you can probably tell from my curiousity question. I have come to the point where I am frustrated trying to solve this problem though. I am trying to self teach myself. Can someone please help me?

• Continuous functions aren't necessarily differentiable. – Pedro Tamaroff Nov 3 '13 at 23:19
• On continuous functions, this bilinear form is not defined. On functions that are three times differentiable at $1$, it is defined. But no longer definite. For instance $f(t)=(t-1)^4$ has $<f,f>=0$. – Julien Nov 3 '13 at 23:20
• There is some kind of analogue for weak differentiable, square integrable functions, the Sobolev spaces. – Christoph Nov 3 '13 at 23:39

$$\sqrt{t}=a+bt+ct^2+dt^3 \longrightarrow(*)$$
We need to find the coefficients $a,b,c,d$. This can be achieved by taking the inner product of both sides of $(*)$ with the functions $1,t,t^2,t^3$ will result in a system of four equations which solving them gives the desired coefficients. For instance,
$$<\sqrt{t},1>=< a+bt+ct^2+dt^3, 1 >.$$


text:
Quickstart¶

To launch a fault-tolerant job, run the following on all nodes.

torchrun
--nnodes=NUM_NODES
--nproc_per_node=TRAINERS_PER_NODE
--rdzv_id=JOB_ID
--rdzv_backend=c10d
YOUR_TRAINING_SCRIPT.py (--arg1 ... train script args...)

To launch an elastic job, run the following on at least MIN_SIZE nodes and at most MAX_SIZE nodes.

torchrun
--nnodes=MIN_SIZE:MAX_SIZE
--nproc_per_node=TRAINERS_PER_NODE
--rdzv_id=JOB_ID
--rdzv_backend=c10d
YOUR_TRAINING_SCRIPT.py (--arg1 ... train script args...)

HOST_NODE_ADDR, in form <host>[:<port>] (e.g. node1.example.com:29400), specifies the node and the port on which the C10d rendezvous backend should be instantiated and hosted. It can be any node in your training cluster, but ideally you should pick a node that has a high bandwidth.

Note

If no port number is specified HOST_NODE_ADDR defaults to 29400.

Note

The --standalone option can be passed to launch a single node job with a sidecar rendezvous backend. You don’t have to pass --rdzv_id, --rdzv_endpoint, and --rdzv_backend when the --standalone option is used.

Note

If torchrun does not meet your requirements you may use our APIs directly for more powerful customization. Start by taking a look at the elastic agent API.


text:
## Monday, July 22, 2013

### LL and LR Parsing Demystified

My first adventures in parsing theory came when I was doing an independent study of programming languages in college. When I got to the part about algorithms such as LL, LR, and their many variations (Strong-LL, SLR, LALR, etc), I was fascinated. I felt like I was peering at some deep and powerful incantations, the significance of which I could not yet appreciate, but I was sure that someday terms like "left-to-right, rightmost derivation" would make perfect sense, and I looked forward to achieving this enlightenment.

I can say now, 10 years and a whole shelf of parsing books later, that I understand these algorithms pretty well. But the way I think about them is quite different than any of the literature that I have found. I think more from an implementation perspective than a mathematical one, which could have something to do with it. In any case, I want to explain how I think of these algorithms; hopefully some people will find this perspective intuitive, as I do.

This article will only address the parser from a black-box perspective: what are its inputs/outputs and what are its constraints? A planned future article will break open the black box for more details about the inner workings of these algorithms.

## Parsing And Polish Notation

If you studied Computer Science at university, or ever owned an HP calculator, you likely came across Polish and Reverse Polish notation. These are ways of writing mathematical expressions that don't need parentheses or order-of-operations rules. We're used to writing expressions as infix, where the operators go in between the operands:
  1 + 2 * 3

In this case, how do you know what order to do the operations in? You have to follow the conventional rules (PEDMAS), and if you want a different order you have to use parentheses, like:
  (1 + 2) * 3

Polish and Reverse Polish notation let you write these expressions without needing arbitrary order-of-operations rules or parentheses to disambiguate. They work by putting the operators before (Polish) or after (Reverse Polish) the operands. These are also known as prefix and postfix notation.
  // First example:
1 + 2 * 3  // Infix
+ 1 * 2 3  // Polish (prefix)
1 2 3 * +  // Reverse Polish (postfix)

// Second example:
(1 + 2) * 3  // Infix
* + 1 2 3    // Polish (prefix)
1 2 + 3 *    // Reverse Polish (postfix)

Besides not needing parentheses or a defined order of operations, Polish and Reverse Polish are much easier to write evaluators for (maybe the HP calculator's designers decided to use Reverse Polish so they could spend a week in the Bahamas). Here is a simple Reverse Polish evaluator in Python.
# Functions that define the operators and how to evaluate them.
# This example assumes binary operators, but this is easy to extend.
ops = {
"+": (lambda a, b: a + b),
"-": (lambda a, b: a - b)
}

def eval(tokens):
stack = []

for token in tokens:
if token in ops:
arg2 = stack.pop()
arg1 = stack.pop()
result = ops[token](arg1, arg2)
stack.append(result)
else:
stack.append(int(token))

return stack.pop()

print "Result:",  eval("7 2 3 + -".split())

Polish and Reverse Polish notation, as they are usually described, do require that all operators have a known arity. Arity is just the number of operands that the operator takes. This means, for example, that unary minus needs to be a different operator than subtraction. Otherwise we don't know how many operands to pop from the stack when we see an operator.

A similar formulation that avoids this problem is one like Lisp's s-expressions. S-expressions (and similar encodings like XML) avoid the need for fixed operator arity by explicitly marking both the beginning and end of each expression:
  ; Lisp-style prefix notation; the same operator can be used
; for different numbers of arguments.
(+ 1 2)
(+ 1 2 3 4 5)

; Lisp equivalents of our first two examples:
; Prefix: + 1 * 2 3
(+ 1 (* 2 3))

; Prefix: * + 1 2 3
(* (+ 1 2) 3)

This variant has slightly different tradeoffs (we traded fixed arity for required parentheses) but the underlying algorithms for parsing/processing these are quite similar, so generally we'll consider this a minor variant of prefix notation.

It might seem like I've veered off-topic, but I've been sneakily talking about LL and LR the whole time. Polish and Reverse Polish notation directly correspond, in my view, to LL and LR parsing, respectively. But to fully explore this idea we need to first describe what kind of output we expect from a parser.

For a fun exercise, try implementing an algorithm to convert Polish to Reverse Polish notation. See if you can do it without building the entire expression into a tree first; you can do it with a stack alone. Now say you wanted to do the opposite (Reverse Polish to Polish) -- you can just run the same algorithm on the input, but backwards! Of course you can build an intermediate tree too, but this takes O(input length) space, whereas the stack-only solution takes only O(tree depth) space. How about going infix to postfix? There is a very clever and efficient algorithm for that called the Shunting Yard Algorithm.

## A Parser And Its Output

We can all agree that the input to a parser is a stream of tokens (which probably came from a lexer, but we can talk about that part another day). But what is a parser's output? You might be inclined to say "a parse tree," and while you can certainly use a parser to build a parse tree, it's also possible to consume a parser's output without ever actually building a tree at all. For example, this Bison example evaluates arithmetic expressions in-line with the parse. Every time a subexpression is recognized, it is immediately evaluated until the final result is just a single number. No parse tree is ever explicitly built.

For this reason, saying that the output of a parser is a parse tree is not general enough. Instead, I claim that the output of a parser, at least for the cases of LL and LR which we are discussing today, is a parse tree traversal.

Apologies if I've set off anyone's nonsense detectors. I can hear people protest that a tree traversal is an algorithm; an operation you perform on a tree. How can I say that a parser outputs a tree traversal? The answer lies in thinking back to Polish and Reverse Polish notation. These are usually thought of as just a notation for mathematical expressions, but we can think of them more generally as flat (serialized) encodings of tree traversals.

Think back to our first example of 1 + 2 * 3. Here is that expression written as a tree:

There are three ways of traversing a binary tree, as explained on Wikipedia: in-order, pre-order, and post-order. They differ in whether you visit the parent node before (pre-order), after (post-order), or in between (in-order) the children of that parent. These three correspond exactly to infix, Polish, and Reverse Polish notation:
  1 + 2 * 3  // Infix expression; in-order traversal.
+ 1 * 2 3  // Polish (prefix) expression; pre-order traversal.
1 2 3 * +  // Reverse Polish (postfix) expression; post-order traversal.

So Polish and Reverse Polish notation fully encode a tree structure and the steps you would take to traverse it. The main difference between these encodings and an actual parse tree is that Polish and Reverse Polish encodings are not random access. With a real tree you can choose to follow an interior node to its right subtree, its left subtree, or even (for many trees) its parent. With these linear encodings there is no such flexibility: you have to follow the traversal as it is already encoded.

But on the plus side, this allows the parser's output to be a stream that can be consumed while the parse is happening. This is how the Bison example from before could manage to evaluate the arithmetic expression without ever actually building a tree. If a fully-fledged tree is actually required, the linear tree traversal can easily build one. But in cases where it is not, the cost of building one can be avoided.

This brings us to a key point:
The primary difference between how LL and LR parsers operate is that an LL parser outputs a pre-order traversal of the parse tree and an LR parser outputs a post-order traversal.
This is equivalent to these more traditional but (in my view) more confusing and less intuitive explanations of the distinction:
• "LL parsers produce a leftmost derivation, while LR parsers produce a reversed rightmost derivation."
• "LL parsers build the tree from the top-down, while LR parsers build the tree from the bottom-up."
• LL parsers are often called "predictive parsers," while LR parsers are often called "shift-reduce parsers."

## The Shape of a Parse Tree

Our arithmetic expression tree that we've been using so far isn't truly a parse tree, because it doesn't correspond to a grammar. To examine actual parse trees, we'll need a real grammar. Unfortunately writing grammars for infix arithmetic expressions isn't as simple or elegant as you might expect. Encoding the precedence and associativity rules into a grammar that is unambiguous (and can be handled by LL and LR parsers) is pretty ugly and nonintuitive. This is one reason why LL and LR parsers are often extended with capabilities that let you specify operator precedence; for example, see the precedence features of Bison. But for the purposes of this article we want to discuss pure LL and LR.

So we need to ditch our arithmetic expressions example in favor of something that is easier to write a grammar for. We'll use JSON since it is very simple but just complex enough to be interesting.
object → '{' pairs '}'

pairs → pair pairs_tail | ε
pair → STRING ':' value
pairs_tail → ',' pairs | ε

value → STRING | NUMBER | 'true' | 'false' | 'null' | object | array
array → '[' elements ']'

elements → value elements_tail | ε
elements_tail → ',' elements | ε

I've used single-quoted strings for literal tokens and upper-case names like STRING for tokens whose spelling can vary (for example, "abc" and "" are both valid STRING tokens). All lower-case names are grammar rules (also known as "nonterminals").

You might wonder why I am doing this pairs_tail and elements_tail business instead of using a repetition construct, which many parser generators like ANTLR support. They let you write something like:
elements → value (',' value)*

While this is more convenient and leads to simpler grammars, it makes parse trees conceptually a bit more complicated, because the number of children for a given grammar rule can vary. Also, LR parsers can't support repetition operators (for example, Bison doesn't support them), whereas the grammar I wrote above can be used with both LL and LR parsers. So we'll use this slightly more complicated grammar for now.

Now that we have a grammar, we can look at an example of a stream of tokens and the resulting parse tree.
{"message":"Hello, World!"}

The token stream for this text is:
{ STRING : STRING }

And the parse tree, according to our grammar, is:

Notice that the leaf nodes (colored in green) are all tokens, and correspond exactly to the sequence of tokens that was the input to our parser. (I cheated slightly by making ε into a leaf node, but this turns out to be pretty clean and principled as we will see).

I claimed earlier that LL parsers output a pre-order traversal and LR parsers a post-order traversal. From this we can say what output we would expect to get from an LL and LR parser given this input:
// LL output: pre-order traversal.
object '{' pairs(1) pair STRING ':' value(1) STRING pairs_tail(2) '}'

// LR output: post-order traversal.
'{' STRING ':' STRING value(1) pair pairs_tail(2) pairs(1) '}' object

Since leaf nodes are always exactly the input tokens themselves, in exactly the order of the input, all the parser is really doing is inserting the interior nodes in the appropriate places. Another way of looking at it is that a parse tree is just a bunch of structure that is defined on top of the sequence of input tokens. We can see this more clearly if we rearrange our previous diagram slightly:

We are converging on a very simple model for how LL and LR parsers operate. Both read a stream of input tokens and output that same token stream, inserting rules in the appropriate places to achieve a pre-order (LL) or post-order (LR) traversal of the parse tree.

Here we see another advantage of thinking of a parser's output in terms of Polish and Reverse Polish Notation. It lets us model the parser's input and output both as simple, flat streams. This is a lot simpler than thinking of a parser's intermediate output state as a partially-built tree -- it is unclear how you could meaningfully consume or inspect that.

## Lookahead

LL and LR parsers are "on-line," meaning that they can start producing output while they are still consuming input. But in many cases the tokens prior to a stream position do not contain enough information for the parser to know whether it should insert a rule or not (or if so, which rule it should insert). So the parser will "look ahead" in the stream to see what the next token(s) are before it makes a decision. When you see designations like LL(1), LR(0), etc. the number in parentheses is the number of tokens of lookahead.

Note that the lookahead is relative to where the rule should be inserted, which (as you will remember) is before that rule's tokens for LL parsers or after that rule's tokens for LR parsers. That means that LL lookahead counts from the beginning of the rule's tokens, whereas LR lookahead counts from the end. This gives LR parsers a huge advantage, because they get to see all of the rule's tokens (and maybe some lookahead) before they have to commit to a decision, whereas LL parsers only get to see the first few tokens of the rule.

This is why there is such a thing as an LR(0) parser, whereas an LL(0) parser would be impossible; it would have no information with which to know what rule to use for the following tokens!

## Consequences

With this understanding of LL vs LR parsing, we can draw a number of very significant conclusions about why certain things are the way they are. These illustrate a lot of the pros/cons of LL vs LR parsing.

### LR parsers can handle more grammars

This follows from the previous section about lookahead. Since LR lookahead starts from the end of a rule, a LR(1) parser has strictly more information available to it when making a decision than an LL(1) parser. It follows that LR(1) parsers can parse strictly more grammars than LL(1) (modulo LL-only grammar extensions; see below). LR parsers can also handle left recursion, which LL parsers cannot.

Advantage: LR

On the other hand, since LL parsers commit to what rule they are parsing before they parse that rule's tokens, and LL parser knows the context of what it is parsing whenever it parses a token. While that is more difficult job (since they have less information to go on), it also leads to some important advantages.

### LL parsers can support regex-like operators in grammars

Knowing the parsing context makes it possible to extend grammars with rich regex-like operators like repetition, alternation anywhere (not just at the top level), etc. Basically each rule can form a DFA. This is possible with top-down parsing because the parser knows what rule it is in and can step through that rule's state machine as it is parsing it. I don't believe this is possible with bottom-up parsing (even if you could somehow get the parsing tables to do the right thing, the "reduce" step counts on the reduction having a fixed arity). This is a really nice advantage of LL, because grammars are often more readable with these rich grammar extensions. In practice this helps mitigate the more restrictive grammar rules of LL, since many things that you'd want left-recursion for you can use repetition operators instead.
// LR Grammar: nothing fancy allowed, alternation only allowed
// at the top-level.
//
// This is only allowed because it is equivalent to:
// pairs → pair pairs_tail
// pairs → ε
pairs → pair pairs_tail | ε

// Extended LL grammar; possible because we can build each
// rule into a DFA.
pairs → (pair (',' pair)*)?

The latter rule could be built into a DFA like (green states are "accept" states):

Knowing the context also allows for mid-rule actions (custom code that runs in-between any two elements of the rule). Bison supports this, but only by rewriting the grammar internally which makes it appear more complicated in any kind of visualization.

Advantage: LL

### LL parsers support inherited attributes

Knowing the context also allows LL-based applications to pass attributes/metadata down the tree as it is being built (this is sometimes referred to as "inherited attributes." (Both LL and LR parsers can support "synthesized attributes" which are passed up the tree).

Advantage: LL

## Conclusion

I have described a alternate model for LL and LR parsers that is equivalent to, but more intuitive (for me at least), than most of the literature. We can consider a parser a black box that inputs and outputs streams of tokens and rules according to pre-order or post-order notation. So far we have not explored the inner workings of these parsers at all; we have just considered them black boxes and we have no idea how they work internally. We have not explored issues of what grammars they can handle and which they can't. We also have not explored the variants of LL and LR (Strong-LL, SLR, LALR, etc). I hope to explore these issues more fully, including example code, in a follow-up article.

#### 7 comments:

1. Thanks for the article. I'm looking forward to the next one, especially since I've been beating the "parser-output-is-traversal" drum for years.

2. I think that's the only explanation of LL and LR that will stick in my head. Thanks!

3. An addition with PEG parsing would be nice, also what happened to gazelle?!

4. i agree with ale, i have been mystified about LL and LR up to this point,....its starting to make some sense now :)

5. Really well written and much clearer than the competing explanations. I'm writing a compiler in my spare time and have been fiddling with expression parsing for a while now (hello, Shunting Yard). Thanks a lot for posting.

6. Personally, I strongly believe that the output of a parser should be the object graph that it represents, and by "object graph" I'm NOT talking about a parse tree (AST). Rather, think of a parser as a deserializer. Also, I think there's no reason to choose LL, LR, etc only - especially not in these days of advanced languages with embedded DSLs (such as LINQ in C#). Instead, pick and choose what works best for each subset of your language. Most importantly, though, I think it should be possible to extend the parsed language by simply creating new classes for new language features and "plugging" them in without the need to update a formal grammar and re-generate your parser. This article explains further, and includes full sources for a C# parser implemented using these ideas:
http://www.codeproject.com/Articles/492466/Object-Oriented-Parsing-Breaking-With-Tradition-Pa

7. Really nice post. Looking forward for the next one.


text:
# Semiconductors

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text:
Learn to click the perfect selfie Learn to click the perfect selfie

# CBSE Class 12 Physics Notes Chapter 14 Semiconductor Electronics - Materials, Devices and Simple Circuits

## What are Semiconductors?

Semiconductors are the core fundamental materials which are used in solid-state electronic devices such as transistors, diodes etc. The material’s atomic structure decides whether the material will turn out to be a metal, semiconductor or insulator. Semiconductors could also be elements such as Ge, Si or compounds such as CdS or GaAs.

Related Topics:

CBSE Class 12 Physics Notes

NCERT Solutions for Class 12 Physics Chapter 14

NCERT Exemplar for Class 12 Physics Chapter 14

Important Questions on Semiconductors

### Intrinsic Semiconductors

These are the type of semiconductors devices which are pure and have charged electrons and holes which are achieved as a result of thermal excitation. The hole’s number is found equal to that of the number of the electrons.

### Types of semiconductors

There are two types of semiconductors:

• N-type- in here,
$$\begin{array}{l}n_{e}> > n_{h}\end{array}$$
• P-type- in here,
$$\begin{array}{l}n_{h}> >n_{e}\end{array}$$

For metals,

$$\begin{array}{l}E_{g}\approx 0\end{array}$$
and for semiconductors
$$\begin{array}{l}E_{g}\end{array}$$
ranges from 0.2 eV to 3 eV and in the case of insulators
$$\begin{array}{l}E_{g}\end{array}$$
>3eV

### Diode

Diodes are being for the purpose of AC voltage rectification which means restricting the voltage to follow one direction only using a capacitor or a filter, a dc voltage can be achieved. The types of diodes are:

• Zener Diode- This is used in places where voltage regulation is needed.
• P-N junction diode- It is used in photonic or optoelectronic devices and the entity is the photon. Examples are solar cells, light-emitting diodes etc

### Types of Digital Circuits

There are special logic operational digital circuits. They are:

• NOR gate
• NAND
• NOT
• AND
• OR

### Important Questions

1. When a p-n junction is set to forward-biased,
1. we could witness a rise in the potential barrier.
2. we could see a reduction in the majority carrier current to the value zero.
3. there is a decrease in the potential barrier.
4. None

2. In transistor action, identify the correct statements.

1. The doping concentrations and size of the base, emitter, and collector are similar
2. The base region is found out to be thin and lightly doped
3. The collector junction and the emitter junction is reversed biased and forward biased respectively
4. Both the collector junction and the emitter junction are forward biased.

## Frequently asked Questions on CBSE Class 12 Physics Notes Chapter 14: Semiconductor Electronics Materials Devices and Simple Circuits

### Who discovered the transistor?

John Bardeen, Walter Brattain and William Shockley invented the first working transistors at Bell Labs, the point-contact transistor in 1947.

### What are ‘Semiconductors’?

Semiconductors are substances with properties somewhere between them. ICs(integrated circuits) and electronic discrete components such as diodes and transistors are made of semiconductors.

### What is a ‘Bipolar junction’?

A Bipolar Junction Transistor is a solid-state device in which the current flow between two terminals (collector and emitter) is controlled by the amount of current that flows through a third terminal (base).


text:
Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
 Already signed up? New to KöMaL?

# Problem P. 4555. (September 2013)

P. 4555. There is a 4-m long, 1.2 kg chain, wound small, at the bottom of a 2-m deep lake. At least how much work is to be performed while the chain is pulled out totally from the lake, by holding one of its end, and moving it vertically? The density of the material of the chain is 3000 kg/m3.

(4 pont)

Deadline expired on October 10, 2013.

Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat. $\displaystyle W=39{,}2~$J.

### Statistics:

 164 students sent a solution. 4 points: 101 students. 3 points: 11 students. 2 points: 33 students. 1 point: 12 students. 0 point: 7 students.

Problems in Physics of KöMaL, September 2013


text:
# Explosive Particle and momentum

#### TFM

1. The problem statement, all variables and given/known data

A stationary mass of 10 kg disintegrates explosively, with the evolution of $$4.2.10^5$$ J, into three parts, two having equal masses of 3 kg. These two masses move at right angles to each other. Calculate the velocity of the third part.

2. Relevant equations

$$E_k = \frac{1}{2}mv^2$$

Momentum = mv

3. The attempt at a solution

Okay so I know that this is a momentum question. And it needs to be constant, therefore

$$mv_{before} = mv_{after}$$

the initial momentum before is 0, so the total final miomentum also equals 0. this means that since the two known partiles are travelling perpendicular, the third, which mus have mass 4kg, is travelling at an angle opposite, probably 45. However, I am slightly uncertain, since they don't give the speed, although they give with the evolution of $$4.2.10^5$$ J. I am slightly unsure what this is, do I need to calculate the velocities of the two 3kg particles using the kinetic energy formula and the energy given above?

Related Introductory Physics Homework News on Phys.org

#### LowlyPion

Homework Helper
1. The problem statement, all variables and given/known data

A stationary mass of 10 kg disintegrates explosively, with the evolution of $$4.2.10^5$$ J, into three parts, two having equal masses of 3 kg. These two masses move at right angles to each other. Calculate the velocity of the third part.

2. Relevant equations

$$E_k = \frac{1}{2}mv^2$$

Momentum = mv

3. The attempt at a solution

Okay so I know that this is a momentum question. And it needs to be constant, therefore

$$mv_{before} = mv_{after}$$

the initial momentum before is 0, so the total final miomentum also equals 0. this means that since the two known partiles are travelling perpendicular, the third, which mus have mass 4kg, is travelling at an angle opposite, probably 45. However, I am slightly uncertain, since they don't give the speed, although they give with the evolution of $$4.2.10^5$$ J. I am slightly unsure what this is, do I need to calculate the velocities of the two 3kg particles using the kinetic energy formula and the energy given above?
Initially at rest looks like it means the center of mass will be preserved at no velocity.

So ∑ mv = 0

Since you have 3V1 i + 3V2 j for the 2 pieces, then you have 4V3(i,j) equals the sum of those 2.

3V1 i + 3V2 j = 4V3x i + 4V3y j

Where V3x2 + V3y2 = V32

The energy also tells you that 1/2(3)V12 + 1/2(3)V22 + 1/2(4)V32 = 4.2*105

#### TFM

Okay, so would I now I find V1 in terms of the energy and V 2 like so:

$$4v_3 = 3v_1 + 3v_2$$

$$(4v_3)^2 = (3v_1)^2 + (3v_2)^2$$

$$(v_3)^2 = \frac{(3v_1)^2}{16} + \frac{(3v_2)^2}{16}$$

$$\frac{1}{2}3v_1^2 + \frac{1}{2}3v_2^2 + \frac{1}{2}4v_3^2$$

$$\frac{1}{2}3v_1^2 + \frac{1}{2}3v_2^2 + \frac{1}{2}4\frac{(3v_1)^2}{16} + \frac{(3v_2)^2}{16}$$

Rearrangeing to give:

$$\frac{21}{8}v_1^2 + \frac{21}{8}v_2^2 = E$$

$$v_1^2 + v_2^2 = \frac{8}{21}E$$

$$v_1^2 = \frac{8}{21}E - v_2^2$$

#### LowlyPion

Homework Helper
That's sort of where I was heading, but I was intending that

1.5V12 + 1.5V12 + 2V32 = 4.2*105

Recognizing that V1 = -4/3V3x and V2 = -4/3V3y

1.5(-4/3V3x)2 + 1.5*(-4/3V3y)2 + 2V32 = 4.2*105

2*V3x2 + 2*V3y2 +2V32 = 4.2*105

But V3x2 + V3y2 = V32

Substituting I get 2*V32 + 2V32 = 4V32 = 4.2*105

#### TFM

Okay, I am not quite getting the same answer.

$$v_1 = \frac{4}{3}v_{3(x)}$$

$$v_2 = \frac{4}{3}v_{3(y)}$$

inserting:

$$\frac{3}{2}v_1^2 + \frac{3}{2}v_2^2 + 2v_3^2 = E$$

$$\frac{3}{2}(\frac{4}{3}v_{3(x)})^2 + \frac{3}{2}(\frac{4}{3}v_{3(x)})^2 + 2v_3^2 = E$$

$$\frac{3}{2}\frac{16}{9}v_{3(x)}^2 + \frac{3}{2}\frac{16}{9}v_{3(x)}^2 + 2v_3^2 = E$$

giving:

$$\frac{48}{18}v_{3(x)}^2 + \frac{48}{18}v_{3(x)}^2 + 2v_3^2 = E$$

?

#### LowlyPion

Homework Helper
Sorry, I jotted down the 4/3 without squaring. You have the right equation. I was apparently more focused on the method.

#### TFM

Well, it gives me a nice answer of 300m/s. This seems rather big, though...

#### LowlyPion

Homework Helper
Well, it gives me a nice answer of 300m/s. This seems rather big, though...
Roughing it out V1 = V2 ~ V3

So roughly speaking the masses are moving with similar velocities, which suggests that

420,000 J ~ 1/2*Σ m * V2

V2 ~ 84,000 or

V ~ 290 m/s

This isn't so far from the calculated 239, 319, 319 now is it?

#### TFM

Nope, not too bad at all.

Many Thanks, LowlyPion

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text:
Previous: Abstract Structure Up: Chu Spaces from the Next: Linear Logic

# Representing general structures with Chu spaces

A natural question to ask is, how general are the ideas of the preceding section? We have given two answers to this question elsewhere, one measuring the generality of Chu spaces in terms of arbitrary relational structures and their homomorphisms, of which the foregoing are examples, the other in terms of arbitrary small categories, for which a forgetful functor may or may not be given.

For all these situations we have shown [Pra93,Pra96], that the objects in question can be represented by Chu spaces in such a way that the continuous functions between the representing Chu spaces are exactly the homomorphisms of the corresponding represented objects. More precisely, in each case the category of interest embeds fully and concretely in the category of Chu spaces over an appropriate alphabet.

Proposition 5   The relational structures (A,R) of arity n embed fully and concretely in .

Proof:   (Outline) We first give the representation of (A,R) in the case when R consists of exactly one n-tuple . Take (A,r,X) to be the Chu space such that consists of those n-tuples of subsets such that there exists for which , and .

We then generalize this representation to arbitrary R by intersecting the X's that arise as above for each n-tuple in R, and restricting r to A times that intersection. One can then show that a function from (A,R) to (B,S) is a homomorphism if and only if it is a continuous function between the respective representing Chu spaces [Pra93].

The construction generalizes to multiple relations by combining them into one super-relation with arity the sum of its constituent arities. It further generalizes to multiple sorts by combining them into one sort with disjoint union and indicating the sorts with one unary relation per sort.

Structures so representable include groups, transforming standardly by group homomorphisms ( because the group multiplication relation ab=c is ternary), directed graphs (or binary relations) transforming by graph homomorphisms ( ), small categories transforming by functors ( because we need a ternary relation for composition and also a unary relation to mark the identities), and so on.

The corresponding theorem for categories is as follows.

Proposition 6   Every small category C embeds fully and concretely in the category of Chu spaces over the set ar(C) of arrows of C.

(In the absence of an explicitly given forgetful functor we take that given by the Yoneda embedding. On the one hand this is the standard way to make any small category concrete, on the other it is not a very economical way so the result is not as useful as the theorem following this.)

Proof:   Represent each object b of C as the Chu space F(b) whose points are the morphisms from any object a of C, whose states are the morphisms to any object c of C, and for which r(f,h) is defined as the composite f;h (= ). Represent each morphism of C as the continuous function F(g) = (which maps points of F(b) to points of F(b')), with adjoint (which maps states of F(b') to states of F(b).

That this is indeed the adjoint follows immediately from (f;g);h=f;(g;h), associativity of composition. That F is full and faithful is a nice exercise (or see [Pra96]). That F is concrete (commutes with the forgetful functor to ) holds when we take as the forgetful functor for small categories the functor that sends each object to the set of all morphisms to that object, which is just the covariant half of our Chu representation.

A category admitting a full embedding of any small category is called universal [PT80,Trn66,HL69]. Unlike all previous universal categories however, Chu is concretely universal in the sense that the embedding preserves the carrier. This is the case both for this representation of objects of an arbitrary small category and for the preceding representation of relational structures (which normally form large categories).

A variant of this categorical embedding holds for small concrete categories (C,U) where is a given faithful functor. The advantage of this theorem over the preceding one is that it is concrete with respect to the given forgetful functor, not with respect to one we make up for the occasion. For it to go through we require one minor additional condition on (C,U), namely that if then for all objects b, there must exist a morphism , necessarily just one since U is faithful.

Denote by Elt(C,U) the disjoint union of the U(a)'s over all objects a of C, itself a set since C is small. This is the set of all occurrences of elements in the underlying sets of the objects of C.

Proposition 7   Every small honest concrete category (C,U) embeds fully, faithfully, and concretely in .

Proof:   (Outline) Represent object a of C by the Chu space (U(a),r,X) where X is the set of all morphisms of C with domain a, and for each element u of U(a), r(u,x) = U(x)(u), the image of element u under the representation of x. It is then straightforward to show that this representation is faithful, full, and concrete.

Previous: Abstract Structure Up: Chu Spaces from the Next: Linear Logic
Vaughan Pratt
1998-03-14


text:
# I'm learning about relativity and, by extension, the classic

1. Nov 1, 2009

### fhisicsstudnt

I'm learning about relativity and, by extension, the classic Michelson-Morley setup. I cannot see why a "fringe effect" was expected and could use some discussion. Here is my reasoning.

Firstly we imagine the light signal to propagate from the SM (silvered mirror, in the "center") to M1 and M2 and back at c (distance for each of 2*x), arriving at the detector simultaneously. This is taking the whole thing to be "stationary". The data is recorded in the interferometer's own reference frame, the scientist is in that frame, and so the whole thing considers itself to be at rest consistent with both "new" and "old" mechanics. There is no "fringe effect".

Next we imagine the whole thing to be in uniform translation at velocity v. The only way to accomplish this is to set up a 2nd reference frame which we consider stationary, and which we consider the interferometer to be in uniform translation with respect to. For simplicity we imagine the interferometer to be translating along the axis of one of the arms, call it the a axis.

With respect to the stationary frame the light signal propagates at v1a = c+v one way and v2a = c-v the other. The distance for the signal to travel relative to the stationary frame is x-v*t1a one way and x+v*t2a the other. t1a = (x+v*t1a)/(c+v) = x/c and t2a = (x-v*t2a)/(c-v) = x/c. Total transit time is Ta = 2*x/c

The other signal propagates at vb= (c2+v2)1/2 both ways relative to the stationary frame. The distance for the signal to travel, relative to the stationary frame, is xb = (x2+v2*tb2)1/2 both ways. The transit times are both tb = xb/vb. Tota transit time is Tb = 2*tb = 2*x/c.

So I don't get it. Even without doing the math out, all we are doing is taking the same situation and analyzing it with everything in uniform translation. Whether we are thinking "classically" or in the context of SR, uniformly moving frames and stationary frames are indiscernible from each other. If we consider the interferometer as stationary and the two light beams return to the silvered mirror simultaneously, then any uniformly translating frame will consider the two light beams to return simultaneously also, this is true either classically or relativistically. The only difference between classical and relativistic is that the two beams will classically arrive at M1 and M2 simultaneously in any moving frame whereas relativistically they will not necessarily appear to arrive at M1 and M2 simultaneously, depending on the frame.

2. Nov 1, 2009

### DrGreg

Re: Mmi

Welcome to Physics Forums, fhisicsstudnt!

You've got this the wrong way round. The assumption (which the experiment disproved) was that the speed of light was c relative to the "stationary" aether frame and not c relative to the apparatus.

You can do the calculation two ways: in the supposed aether frame with a speed of c and two modified distances, or in the apparatus frame with a modified speed and a distance of x. You've modified both the distance and the speed!

3. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Hey, thanks DrGreg, that was really fast!

If we were to take a hypothetical stationary frame in which something moves at c within it and translate it vt relative to another frame, then whatever is moving within the translated frame gains vt. This is just basic classical kinematics isn't it? This is what I learned.

We can grant the velocity v either to the apparatus frame or to the other frame. Since the aether was supposed stationary, it makes sense to grant v to the apparatus frame. Everything in the apparatus frame is then moving at +v relative to the stationary frame.

The problem seems even simpler than this, though. In classical kinematics, two events that are simultaneous in one frame are simultaneous in another uniformly translating frame. So if the two pulses of light strike the silvered mirror simultaneously in the apparatus frame then they must also do so if the apparatus is in uniform translation. Classical and relativistic kinematics agree on this conceptually. What they disagree on is the simultaneity of non colocal events, namely the events wherein the pulses of light strike M1 and M2. If in the apparatus frame the pulses strike M1 and M2 simultaneously then, in relativistic kinematics, the pulses will not strike M1 and M2 simultaneously in a uniformly translating frame. However, since the event wherein the 2 pulses strike the silvered mirror is both colocal and cotemporal, no kinematics that I know of can describe them as non simultaneous in any unaccelerated frame.

Moving frames, viewed from another frame, appear to have different distances and speeds. I modified both because that's how I was taught kinematics. The result, that uniform translation does not alter the outcome, is also consistent with everything I've learned about classical kinematics. Uniformly translating frames are not different, none are special, all should get the "same result" regarding simultaneity, ordering of events, etc.

4. Nov 1, 2009

### JustinLevy

Re: Mmi

Let us make this even simpler.

Consider the same setup with sound waves and the medium as air. The velocity of the sound is $v_{sound}$ with respect to the air. Go into the 'moving frame', and the velocity isn't increased proportionally so that it still appears to be $v_{sound}$. Heck, there is actually a speed in which the sound waves would appear to be stationary in the 'moving frame'. There is a real and physical difference for sound propagation in these two frames.

You keep wanting to remove the medium, which is the very thing the MM experiment was trying to measure.

5. Nov 1, 2009

### JustinLevy

Re: Mmi

My comments were referring to you latest post. Rereading your first post, it seems like this is just an issue with coordinate changes.

Here are the two situations:
the setup as you described, then either
speed of light = c (a constant independent of direction)
OR
speed of light = c(1 - v cos theta), where theta is the angle from the direction of travel in the 'medium'

So for the first case, the time to travel one leg to the mirror and back is: t = x/c + x/c = 2x/c
And it is the same for the other leg.

Now for the second case, the time to travel the 'theta=0' leg is t = x/(v-c) + x/(v+c)
the time to travel the 'theta=90 degrees' leg is t = x/c + x/c = 2x/c

Compare that to what you wrote:
The problem is that you were doing t_(frame b) = x_(frame a) / v_(frame b)

That is why it was not working.
I'm sorry if my previous post confused things because I didn't read through all your math first.

6. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Comparing something that isn't sound to sound wouldn't, I think, make this simpler. The simplest place to start is, in my opinion, the basic result of classical kinematics:

Events simultaneous in a stationary frame are also simultaneous in any uniformly translating frame.

I don't think this is true. The "medium" is just a hypothetical stationary reference frame. I have invoked such a frame.

Just to be clear, my labels "a" and "b" are not to indicate different frames but rather different pulses. The "a" pulse is the one traveling in the direction of v and the "b" pulse is traveling perpendicular to v. All my quantities are wrt the stationary frame. The distance-traveled for b is just the pythagorean distance (x2+v2*tb2)1/2 and the velocity is just the pythagorean velocity (c2+v2)1/2.

The fundamental issue is whether events judged simultaneous in one frame are also judged simultaneous in a comoving frame, classically. I thought yes, which would rule out any chance of a positive result.

7. Nov 1, 2009

### Staff: Mentor

Re: Mmi

You have to put yourself in the same frame of mind as Michelson, Morley and just about every other physicist of their era. To them, the "medium" (the lumimiferous aether) was a real physical entity, albeit with rather strange properties.

8. Nov 1, 2009

### Cleonis

Re: Mmi

Let me concentrate on the question whether according to pre-relativistic theory of light-propagation frames are discernable.
In pre-relativistic theories there was a supposition of a luminiferous ether (a light bearing medium) and the light was supposed to propagate with a particular velocity with respect to that immovable ether, with no supposition of time dilation or length contraction effects.

I attach two diagrams, each is 256x256 pixels in size (I don't know how speedily they will be approved).
Both are illustrations http://www.cleonis.nl/physics/phys256/special.php" [Broken]

The red lines represent worldlines of clocks that are co-moving. In the diagram newtonian_stationary the red worldlines represent zero velocity reletive to the ether. The yellow lines represent the worldlines of pulses of light. The yellow worldlines are at an angle of 45 degrees, to represent the speed of light relative to the ether. The total procedure takes 10 units of time.

In the diagram newtonian_moving the red worldlines are at an angle, representing that the clocks have a velocity with respect to the immovable ether.
The diagram is a representation of the moving trains thought experiment as considered from a pre-relativistic point of view. The yellow worldlines (the pulses of light) are at an angle of 45 degrees in the diagram. From halfway along the train one lightpulse is sent to the front and the other lightpulse is sent to the rear. Both pulses are returned and they arrive back at halfway the train simultaneously.
The important thing: if the train has a velocity relative to the luminiferous ether then according to a pre-relativistic theory the total time to complete the procedure will be longer than 10 units of time. And the larger the train's velocity relative to the ether, the more the duration will exceed 10 units of time.

This illustrates why according to a pre-relativistic concept of light-propagation frames ought to be discernable.

According to pre-relativistic theory: If there would be multiple trains, all with exactly the same length, moving with a velocity relative to each other, then only onboard the train that is stationary with respect to the ether will the procedure take 10 units of time; on the other trains the procedure will take longer to complete.
It is only in terms of relativistic physics that the prediction is that on all trains the duration will be 10 units of time.

Cleonis

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9. Nov 1, 2009

### JustinLevy

Re: Mmi

You are misunderstanding something very fundamental. The problem is that you are asking the wrong question. We're trying to show you that, but since we are failing to explain it to you, we end up arguing about two different things. Please keep this in mind, when I try to explain again.

(A quick aside to prevent other accidental disagreements:
First, some terminology. Usually in modern physics, classical just means non-quantum. So GR is a classical theory. It is obvious what you mean by it here, and it is convenient, so I will use your terminology here: in this thread, by "classical" I mean "Newtonian"/"pre-relativity".)

YES, you are correct that two events that are simultaneous in one inertial frame will be simultaneous in all inertial frames classically.

But that is NOT the question the MM experiment was looking at. They rotated the experiment to check things, but to make the discussion simpler, let's look at it like this:

1] do the experiment with the interferometer at rest with respect to the 'aether'
then
2] do the experiment with the interferometer moving with respect to the 'aether'

Are the results the same?
The answer, classically with a luminous aether, is NO.
Therefore, classically, one could detect the aether.

Instead, you are doing only one experiment (#1), then just re-analyzing that same experiment in another coordinate system. Of course the results of a single experiment are independent of the coordinate system we use to analyze it.

It took me awhile to realize that is what you were doing. The thread, and ensuing confusion, make much more sense now.

Last edited: Nov 1, 2009
10. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Hey Cleonis,

Nice site. I saw the figures you referenced.

The scientist on board each train will consider himself in an unmoving frame and measure 10 units of time on any of the trains. Since the scientist considers himself unmoving on the train any results he obtains are valid in any other situation in which he considers himself to be in an unmoving frame. If the pulse moves at c wrt the aether and the train moves at v wrt the aether then each pulse goes c+/-v on the way out and c-/+v on the way in. If the train moves at 2*v then the pulses go c+/-2*v out and c-/+2*v in. In my original post you can replace all the v's with 2*v or 3*v and the transit time remains 2*x/c.The data collected in the MM setup was all collected in the apparatus frame (i.e. within the train itself), so again there seems to be no possibility of a positive result.

In the first diagram we are in the apparatus frame. The pulses all travel identical distances and identical velocities, arriving back simultaneously.

In the 2nd diagram we are in the aether frame, in which the apparatus/train is translating at v. In going to the 2nd diagram you change the angle of the red lines to signify the motion of the apparatus (or train) of v wrt the aether but do not change the angle of the yellow lines. You have to translate the entire frame don't you? You went from the train frame to the aether frame, everything in the train frame has to be translated, not just some things, right? Again that's how I was taught kinematics. If we had cars or baseballs flying around in a cage/garage, and suddenly the whole thing started flying away from me (uniformly), I would have to consider the baseballs and cars (and photons) to be translating also (they'd have some angle in the 2nd diagram). Transit times from a point back to that point would be the same whether it was going at v, 2*v, 3*v, etc.

I still don't understand how. Moving, unmoving, the interferometer itself always considers itself to be "motionless". A thing doesn't move wrt itself. In one experiment the interferometer is motionless wrt the aether and the pulses are simultaneous at the silvered mirror. In another experment the interferometer is translating wrt the aether and, since events simultaneous in one frame are simultaneous in a uniformly translating frame, the pulses are again simultaneous at the silvered mirror.

But what is the aether, except another coordinate system? The data is always collected in the apparatus frame, whether the apparatus is moving wrt the aether or not. The data is "moving" with the apparatus, so it can't "know" if the apparatus is moving wrt the aether or not.

Note that I'm not considering the additional layer of complication involved in the stationary frame actually collecting data from the moving apparatus frame. This would involve accounting for doppler shifts etc. which isn't really pertinent.

I just can't understand how, considering pre-relativistic kinematics, a positive result could possibly have occurred.

11. Nov 1, 2009

### Staff: Mentor

Re: Mmi

At the time when M&M did their experiment, physicists considered the luminiferous aether to be a real physical medium for light waves, just as air is a real physical medium for sound waves, or water for water waves.

In that picture, if the laboratory and the aether are moving with respect to each other, the effect on light waves should be exactly like the effect on sound waves if the laboratory and the air are moving with respect to each other. In the laboratory frame, there is a "wind" of aether or air "blowing" through the laboratory. Light and sound should travel faster when going "downwind" and slower when going "upwind."

12. Nov 1, 2009

### Cleonis

Re: Mmi

I have formed a new hypothesis as to how you arrive at your opinions. I now think you believe that emission of light is in its physics characteristics similar to emission of a bullet from a gun (with velocity composition according to the rules of pre-relativistic mechanics).

Your postings have demonstrated the following two things: you are unaware of how light-propagation is conceptualized (both the pre-relativistic point of view and the relativistic point of view), and most unfortunately: you are immune to explanation. I'm sorry to say that; it's a harsh assessment, but the time has come to make that call.

The irony of it all is that relativistic physics does have a profound bearing on that fundamental question: how does emission and propagation of light relate to mechanics?

Cleonis

13. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Agreed, if the lab is moving at v wrt the aether then light will move slower one way (c-v upwind) and faster the other way (c+v downwind). I took this into account in my original post. When you do this there is no net effect.

Are you telling me, then, that pre-SR something (*anything*) moving at c in one frame is also moving at c in a relatively moving frame with some velocity v? How can you justify this? It contradicts everything I know of. You are telling me that the water waves on the beach, moving toward the shore at c relative to the water/earth, also move at c relative to the speed boat. You're telling me that when I'm on a moving train and I shout to someone, the sound has velocity c relative to the ground/air and also the train? That the velocity is not c+/-v and the distance-traveled c+/-v*t? How can you justify slanting the red lines but not the yellow lines? When you translate a frame you translate everything moving within it.

Please, help me conceptualize how light propagates (in the pre-relativistic view for now). I doubt I am "immune to explanation" since I have made significant progress already through a physics program and have been a successful student in general.

14. Nov 2, 2009

### JustinLevy

Re: Mmi

The aether is not a coordinate system. It is the MEDIUM in which light propagates.

Is the "air" a coordinate system? No. But there may be an inertial frame in which the air is at rest.

So while the aether is not a coordinate system, there is an inertial frame in which it is at rest. Pre-relativity, this is a special frame in which the speed of light is isotropic and constant. In any inertial coordinate system moving relative to this rest frame, the speed of light would be anisotropic.

Do you understand this?
You need to understand the concept of the aether as a medium before we can continue. You seem to understand (for example you use c+v, and c-v, and your upstream vs downstream comments), but as soon as it comes to calculating anything you incorrectly mix the coordinate systems.

Alright, if you can understand that the aether is a medium, please reread my previous post and think about it for a bit. Then I'll work the math out for you as an example.

...

In the experiment where the aether is at rest with respect to the interferometer, we have:
t(0 degrees) = x/c + x/c = 2x/c
t(90 degrees) = x/c + x/c = 2x/c

Now in the DIFFERENT experiment (not just a different coordinate system, but a DIFFERENT EXPERIMENT) where the aether is moving at speed v with respect to the interferometer, we have:
t(0 degrees) = x/(c-v) + x/(c+v)
t(90 degrees) = x/c + x/c = 2x/c

As you can see, they are different. That is why, if the pre-relativity aether idea was correct, it would be possible to measure the velocity of the aether.

Again, the problem with your math is that you are mixing in coordinates from different frames. Instead of
t_moving = x_moving / velocity of light_moving,
you are using
t_moving = x_aether rest frame / velocity of light_moving

Since in pre-relativity, all inertial frames will agree on the time, you can calculate the time in either coordinate system. But you cannot mix and match the distance and velocity like that. Choose one coordinate system and stick with it through-out the calculation.

Last edited: Nov 2, 2009
15. Nov 2, 2009

### A.T.

Re: Mmi

This is how light behaves in reality and is modeled in SR, not pre-SR.

Physics doesn't make or justify the laws of nature, it just describes them.

Not everything, just the Galilean coordinate transformation, which is replaced by the Lorentz coordinate transformation.

Water waves are not light

In the pre-relativistic view it behaves like water waves. But that was ruled out by the MM experiment.

Last edited: Nov 2, 2009
16. Nov 2, 2009

### Al68

Re: Mmi

For the same reason a positive result will occur for sound waves with the same setup apparatus in motion with respect to the air.

They assumed the aether was the medium for light in the same way air is a medium for sound, not just a coordinate system. And they assumed the speed of light was constant relative to the aether in the same way the speed of sound is constant relative to its medium (air).

17. Nov 2, 2009

### fhisicsstudnt

Re: Mmi

Thank you for the patience Justin.

Yes, mathematically the air can be considered a coordinate system. In fact, to treat it properly we must consider it as a coordinate system/frame. If we consider it at rest then other things move wrt to it (or not).

Okay, the physical aether itself is not a coordinate system. However mathematically we treat it as a stationary frame, a stationary coordinate system.

Yes, I understand the aether as a medium. In math it becomes the stationary frame.

...
Agreed.

I think it is bad setup/form to consider the aether to be moving, since it is physically defined as at rest. To be consistent with the physical hyp (stationary aether) the interferometer should be considered to be moving at v, however of course we can do it this way.

For 0 degrees, the light does indeed now approach the mirror at c-v and return at c+v in the lab frame if, before the lab frame was moving, light was approaching/returning at c. Also, if before the lab was moving light pulses had to traverse the distance x, they now have to traverse the distance x+ct1 and x-ct2. The distance an object must travel, as well as its velocity, are altered by moving the frame. It wouldn't make sense to alter one without the other. If a train moves away from me (and the ground) across a km of track at v, then the track sprouts rockets and the tracks fly away at w, I now consider the train to be movng v+w relative to me and to be traversing more than 1 km (relative to me/the ground). The train now moves much further than 1km across the ground (lab) when it moves 1 km across the track (aether). The time for the train to traverse the km of track is of course identical to before.

Consider:
A conveyor at the airport d long. Jack eschews the conveyor and walks beside it at w relative to the ground (stationary aether) and the lab. He is a light pulse in experiment one. I get on the conveyor (moving aether), which is "conveying" (aether blowing) at v relative to the ground (laboratory), and I walk at w relative to the conveyor/aether, v+w relative to the ground/laboratory. I am a light pulse in experiment 2. When I step onto the conveyor I make a footprint indicating my exact starting location. Jack makes a similar footprint to indicate his exact starting location. This marks the silvered mirror location in the actual experiment. Both of us walk to the end of the conveyor, turn around, and walk back. We will both land on our original footprints simultaneously, we will arrive back at the silvered mirror at the same time. No surprise there, this is just the parallel arm.

Consider a square conveyor moving perpendicular to the other one. I step on the conveyor, making my initial footprint to indicate the exact silvered mirror location, and approach the opposite side at w relative to the ground/lab, while being dragged perpendicularly at v relative to the ground/lab. I will reach the end of the conveyor (M1) after d/w units of time. That's a velocity of (w2+v2)1/2 relative to the moving aether and a distance of (d2+v2*tb2)1/2 relative to the moving aether.

I'll try to anticipate a potential misguided objection that I am granting myself w relative to the lab here, whereas I gave myself w relative to the aether before. The light pulse is moving at w relative to the aether and is then reflected through 90 degrees. All the velocity it originally had in the +y direction is now in the +z direction. It is as if I were walking along at w relative to the conveyor(v+w relative to the ground) and then pivoted 90 degrees to the side and continued my previous stride. I now have velocity w along the +z axis relative to the ground. I certainly cannot possibly retain my previous velocity relative to the aether after pivoting a quarter turn. I must perform an elementary vector analysis to get my new resultant velocity. The distance I travel is also greater relative to the aether as I am both moving the distance d across the conveyor and the distance vtb to the right.

Jack eschews the conveyor and approaches the opposite side at w relative to the stationary aether (and lab). We both reach the opposite side simultaneously (M2), turn around, and come back to the original side Again we each land our foot back onto our original footprint simultaneously. We both come right back to the silvered mirror at the same time.

No matter what we do, we will land right back on our respective starting footprints. We both go and then retrace our own exact steps.

This thought experiment is representative of the MMI.

I believe you are mixing coordinate systems. If light pulses move at c wrt the stationary aether when the lab is stationary, then the lab is moving through the aether at v the pulse in the perpendicular arm traverses a longer distance through the aether at a faster speed, there and back. This is as I illustrated with the conveyors.

Whether sound or light, this setup does not appear capable of returning a positive result. Has anyone done such an experiment with sound?

JustinLevy and Al68:

You are telling me that a frame in uniform translation with respect to the air will lead to different results with sound than a frame that is stationary with respect to the air in a MM setup?

Last edited: Nov 2, 2009
18. Nov 2, 2009

### JustinLevy

Re: Mmi

I just realized I was using the wrong form for the "anisotropic" speed of light for 'pre-relativity'. I think I was essentially still keeping around time dilation. Regardless of the specifics, I made a mistake in the math, which most likely made it difficult for you to understand the qualitative explanations as well.

So let me try again.
x' = x - vt
y' = y
t' = t

For something moving at speed c at angle 0 (with respect to the x axis) in the unprimed frame, it is moving at angle 0 in the primed frame as well, but with speed c-v.

For something moving at speed c at angle theta in the unprimed frame, such that it was moving at angle 90 degrees in the primed frame, then it's speed is:

c'_x = cos(theta) c - v = 0
c'_y = sin(theta) c

so cos(theta) = v/c, which gives sin(theta) = sqrt(1 - (v/c)^2)
so we have:
c_x = v
c_y = c sqrt(1-(v/c)^2)
c'_x = 0
c'_y = c sqrt(1-(v/c)^2)

=======================
Okay, now let's work out the prediction for the experiments again.
You can use whatever coordinate system you want. But choose one and stick with it through out the calculation.

In the experiment where the aether is at rest with respect to the interferometer, and using the rest frame of the interferometer (which is also the rest frame of the aether), we have:
t(0 degrees) = x/c + x/c = 2x/c
t(90 degrees) = x/c + x/c = 2x/c

Now in the DIFFERENT experiment (not just a different coordinate system, but a DIFFERENT EXPERIMENT) where the aether is moving at speed v with respect to the interferometer, we have:

In the rest frame of the aether (so the interferometer is moving with velocity v)
and since we are using the rest frame of the aether the speed of light is isotropic in this frame
t(90 degrees) = x/ (c sqrt(1-(v/c)^2)) + x/(c sqrt(1-(v/c)^2)) = (2x/c) * (1/sqrt(1-(v/c)^2))

t(0 degrees) = t_away + t_back
t_away = (x + v t_away)/c
so, t_away = x/(c - v)
t_back = (x - v t_away)/c
so, t_back = x/(c + v)
so, t(0 degrees) = x/(c - v) + x/(c + v)

In the rest frame of the interferometer (so the aether is moving with velocity v according to this frame) and since the aether is moving in this frame the speed of light is anisotropic in this frame
t(90 degrees) = x/(c sqrt(1-(v/c)^2)) + x/(c sqrt(1-(v/c)^2)) = (2x/c) * (1/sqrt(1-(v/c)^2))
t(0 degrees) = x/(c-v) + x/(c+v)
Which of course matches the results obtained when calculating this same experiment in the aether rest frame.

Notice however, that the result is DIFFERENT from the experiment where the interferometer was at rest with respect to the aether.
Hopefully you can see more clearly now how you were mixing the different coordinate systems incorrectly. (If you are using the distances according to the aether rest frame, then you must use the velocities according to that frame. Instead you used the velocities according to the interferometer rest frame.)

19. Nov 2, 2009

### yuiop

Re: Mmi

That would be right.

Here is a setup that might be easier to analyse or visualise. Imagine a large L shaped device being towed across a calm lake. The arms of the "L" shape are the same length as each other, as in the MM experiment. There are 2 swimmers. One has to swim from the fulcrum of the L alongside the arm that is parallel to the direction the L is being towed in, and then back to the fulcrum. The other swimmer has to swim a diagonal path to get from the fulcrum of the L to the end of the arm that is orthogonal to the direction the L is being towed in and then back to the fulcrum again. Even though the two swimmers swim at exactly the same speed as each other relative to the water and both start at the same time, the swimmmer that swims the diagonal path returns to the fulcrum before the other swimmer. In pre relativistic times, photons were thought to move in a manner somewhat similar to the way the swimmers move in the water and that is why they expected the light signals in the MM experiment to return at different times and cause an interferance pattern.

There is a nice flash animation of what I have just described here: http://www.upscale.utoronto.ca/PVB/Harrison/SpecRel/Flash/MichelsonMorley/MichelsonMorley.html

20. Nov 2, 2009

### fhisicsstudnt

Re: Mmi

Hey guys, I see it now. Thank you so much, this has been quite a learning experience for me. The math and the animation all helped.

I even did it out on graph paper. I redrew the cross displaced half a box over repeatedly with the dots always displaced by one box length. Quite a surprising result!

You guys rock.