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# Q: How is it that Bell’s Theorem proves that there are no “hidden variables” in quantum mechanics? How do we know that God really does play dice with the universe?

Physicist: Bell’s theorem, and its philosophical fallout, is one of the most profound discoveries since relativity.

Bell’s theorem states (among other things) that the universe is fundamentally unpredictable, and that quantum mechanical things (for example: everything) are not actually in one state.  If a box could contain either a blue marble or a red marble, then when you open it you’ll see either one or the other.  In “reality” it was one color or the other before you open the box.  In QM, it can be both before you open the box (it’s actually still both afterwards, but moving on…).

Einstein (and most other physicists of the time) believed that if you knew everything about a system of particles (no matter how big) that you could theoretically predict what that system will be doing in the future, perfectly.  Homeboy thought that the only reason that the movement of air molecules seems to be random, is that we can’t perfectly measure that exact position and velocity of every single one.  So he thought that every particle truly is in some particular state, but that we merely don’t know for sure what that state is (the marble in the box has only one color, but we don’t know what it is).

The idea that randomness and unpredictability are caused by unknown (or unknowable) things is called “hidden variable theory” (The ‘Stein believed in this).  For example; 2, 2, 3, 6, 0, 6, 7, 9, 7, 7, 4, 9, 9, … is not random, but seems random.  It would be really hard to predict the next term (7) if you don’t know the hidden variable.  (BTW, the “hidden variable” is: this is the decimal expansion of $\sqrt{5}$)

Bell’s theorem essentially boils down to a proof that the result of an experiment doesn’t exist until the measurement is made (so it can’t be predicted).  Hidden variable theory presupposes that the particles involved are in definite states, which means that the result of a measurement already exists before the measurement is made.  For example: before you open a gift what you’ll see is already set in stone.  The gift is a set thing before you open the box.  This is not the case for most quantum mechanical systems.

Here’s one of the experiments that demonstrates Bell’s theorem, and two ways to look at it.

An entangled pair of photons is created and fired in opposite directions. En route the polarizers are randomly oriented, then the detectors measure whether or not the photons pass through. This is done hundreds of thousands of times to measure the relationship between 1) the difference in angles between the polarizers and 2) the probability of measuring the same result.

The experiment: Step 1: Create a pair of entangled photons and fire them in opposite directions.  Entangled particles always yield the same result when they are subjected to the same measurement, and are likely to yield the same result for similar measurements.

Step 2: Randomly orient the polarizers, after the entangled pair is created, but before either is detected (this is hard to time, and is really fast).  This is done so that the photons “don’t know what to expect” and “can’t compare notes”.  Information about polarizer A would have to travel faster than the speed of light to get to photon B before photon B hits it’s own polarizer.  So, without faster than light effects (which don’t exist for many, really good reasons) the photons are each acting independently.  The orientation is random so that the photons can’t “plan ahead”.

Step 3: Measure the polarization.  If the detector “clicks” then the photon made it through the polarizer, and therefore has the same polarization.  If the detector doesn’t click, then the photon had the opposite polarization and was stopped.

The probability of the measurements being the same (for an entangled pair) is $P = \cos^2{(\theta)}$, where $\theta$ is the difference in angles between the polarizers.  It is tricky to see why, but this probability is impossible if you assume that the result of a measurement exists before the measurement is made.  Here’s why.

The possible polarizations for polarizer A (red) and polarizer B (blue).

Algebraic approach: Restricting the possible angles of the polarizers to 0° and 45° for A, and 22.5° and 67.5° for B, run the experiment. Here’s what’s about to happen:

1) If you could predict the outcome of each version of the experiment, then you could find a definite value of L (see below).

2) For strictly (unarguable) mathematical reasons L = ±2.

3) Experimentally we find that the average value of L is 2√2.

4) But this is a contradiction, so we cannot actually make useful predictions.

Now it’s happening:

If polarizer A is at 0° and the detector clicks then you’d say “A0 = 1″, and if the detector doesn’t click then “A0 = -1″.  Similarly, you can define B67.5, A45, and B22.5.  Just for the hell of it, take a look at: L = A0B22.5 + A45B22.5 + A45B67.5 – A0B67.5 = (A0 + A45)B22.5 + (A45 – A0)B67.5

L = (A0 + A45)B22.5 + (A45 – A0)B67.5 = ±2, since either (A0 + A45) = ±2 and (A45 – A0) = 0, or (A0 + A45) = 0 and (A45 – A0) = ±2.  So if you could fill out each of these values (A0, A45, B22.5, B67.5), then L = A0B22.5 + A45B22.5 + A45B67.5 – A0B67.5 = ±2 which, notably, is less than or equal to 2.

However, you can’t make all of these measurements simultaneously, so you can’t actually get A0B22.5 + A45B22.5 + A45B67.5 – A0B67.5 for each run of the experiment.  The best you can do is find one of these four terms each time you run the experiment.  For example, if the polarizer A was randomly set to 45° and the detector clicked, and polarizer B was randomly set to 22.5° and the detector didn’t click, then you just found out that A45B22.5 = (1)(-1) = -1 for that run.

You can however find the expectation value by running the experiment over and over and keeping track of the results and polarizer orientation.

E[A0B22.5] + E[A45B22.5] + E[A45B67.5] – E[A0B67.5] = E[A0B22.5 + A45B22.5 + A45B67.5 – A0B67.5] ≤E[|A0B22.5 + A45B22.5 + A45B67.5 – A0B67.5|] = E[2] = 2.

So E[A0B22.5] + E[A45B22.5] + E[A45B67.5] – E[A0B67.5] ≤ 2.  This is one version of “Bell’s Inequality”, and it holds if each term (A0, A45, B22.5, B67.5) has a value.

Using the fact that the chance of getting the same result is $P = \cos^2{(\theta)}$, and that each term is 1 when the results are the same ((1)(1) or (-1)(-1)), and -1 when the results are different ((1)(-1) or (-1)(1)), you can calculate each term.  For example:

$E[A_0B_{22.5}]=P(same)-P(different)=\cos^2{(22.5)}-(1-\cos^2{(22.5)})=\frac{1}{\sqrt{2}}$

You’ll find that:

$E[A_0B_{22.5}]+E[A_{45}B_{22.5}]+E[A_{45}B_{67.5}]-E[A_0B_{67.5}]=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{-1}{\sqrt{2}}=2\sqrt{2}$

Holy crap!  $2\sqrt{2}>2$!  But that’s a violation of Bell’s inequality!  But the existence of each measurement (whether or not you actually do that measurement) is all you need for Bell’s inequality!  So if the inequality is false, then the result of those measurements don’t exist if the measurement isn’t made!

Maybe, if you’re clever and have ready access to a time machine, you could go back and do all the measurements you didn’t make the first time. Then all the results would have to exist! They’d just have to!

Me and my time machine vs. quantum mechanics: If the results exist, but you just didn’t happen to do all the measurements, why not get a time machine?  Then you could do one measurement, go back, do a different measurement, go back, do a different measurement, …  Then every possible result would be known.

However, once again that correlation probability ($P = \cos^2{(\theta)}$) screws things up.

So, for example, if the photon goes through at 50°, and then you go back in time, change the polarizer to 51°, and repeat the experiment, then there’s a 99.97% chance (cos2(1°) = 0.9997) that the photon will go through again.

One result from probability says that $P(x=z)\ge P(x=y)+P(y=z)-1$.  Do this twice and you get $P(w=z)\ge P(w=x)+P(x=z)-1\ge P(w=x)+P(x=y)+P(y=z)-2$.  So if you measure in the 0° direction to find A0, then go back and change the angle by 1° and repeat this until you’re measuring at 90°, then:

$P(A_0=A_{90})\ge P(A_0=A_1)+P(A_1=A_2)+\cdots+P(A_{89}=A_{90})-89$ $=90\cos^2{(1^o)}-89=0.9726$

So, if you go back and forth in time to measure whether or not the photon goes through at 1° increments, then there’s a 97% chance that by the time you get to 90° you’ll be getting the same result you did at 0°.  However, in reality $P(A_0=A_{90})=\cos^2{(90^o)}=0$.

But this is a contradiction.  So the results of each measurement (A0, A1, A2, …, A90) can’t all exist.

If I had to guess, every time you go back in time the experiment is completely reset, and the experiment becomes completely random again.  The reason (such as it is) is below this unsettling picture.

Wait. Wait… Why?

But why?!: It turns out that the reason that the results of a quantum event can’t be predicted, is that every possible result of that event plays out.  So if you ask “will I see the photon go through the polarizer?” the answer is “yes, some versions of you will see the photon go through” and an equally valid answer is “no, some versions of you will not”.

If different versions of you will see every possible result, then the result can’t be predicted, and doesn’t really exist one way or the other until after the measurement is done.  At that time the different versions of you will disagree on the result.  But don’t worry too much.  You’ll never meet you’re parallel-universe twins.

This entry was posted in -- By the Physicist, Philosophical, Physics, Probability, Quantum Theory. Bookmark the permalink.

### 23 Responses to Q: How is it that Bell’s Theorem proves that there are no “hidden variables” in quantum mechanics? How do we know that God really does play dice with the universe?

1. wrf3 says:

Do you have any opinion on Joy Christian’s work: Disproof of Bell’s Theorem by Clifford Algebra Valued Local Variables (http://lanl.arxiv.org/pdf/quant-ph/0703179)?

2. Israel says:

But does the polarizer constitute a measurement? Because it is not as if the photon isn’t interacting with the crystal lattice of the polarizer or the metal and dielectric of a wire grid polarizer.

3. Physicist says:

There’s a subtle, but important, difference between “interaction” and “measurement”. For example, sugar water affects the right and left circular polarizations of light slightly differently, with the effect that rotates the linear polarization. So the sugar water interacts with the photons, but no measurement is ever made. The amount of knowledge and the lack of knowledge you have about the photon before and after passing through the sugar water stays the same.
In a polarizer, the light certainly needs to pass through it, and so will interact with the constituent matter (air, plastic, whatever). However, the “polarizerness”: that light with one polarization is blocked, while light with the opposite polarization passes through, is absolutely a measurement. Whether or not light passes through tells you something about the photon’s polarization that you may not have known before.

4. Johnny says:

This is so difficult to accept… From what I understand: it is not because we don’t know the “hidden variable” it is because it doesn’t exist at all!

Is that so?

So a particle doesn’t have momentum — if it has a position.

This just brings up more questions, like: Why does this not apply to bigger things, like humans? How big does a thing have to be before the theorem is true?

5. The Physicist says:

It always has both, just not a definite value of either. The uncertainty principle applies on all scales, but you’d never notice it on a large scale. The errors implied by the uncertainty principle are really, really, crazy small. But they’re not zero.

6. Johnny says:

Ah ok, I see.

7. dr.line says:

yeah i’m sure i will never meet my are parallel-universe twins either.

8. Andrew says:

“The errors implied by the uncertainty principle are really, really, crazy small. But they’re not zero.”

I just posted in another article of yours referring to basically the same thing. I’m not a physicist, which may be obvious, but as an outsider it is hard not to think that a lot of modern physics is what I would call “mistaking the map for the road.”

The Uncertainty Principle, for instance: just because we do not possess tools that can measure location and velocity perfectly at the same time is not proof that the particle actually sort-of-exists in a probability cloud. A probability cloud may be a great tool for describing the results we get when observing an electron, but that should not be taken to assume that it represents the ultimate reality of what is going on. Just because we cannot prove or know exactly where it is and at what velocity is not proof that it doesn’t have a true location and velocity.

The idea that light is both a particle and a wave (forgive me if this is no longer a relevant concern, but I know that it was at one time)…seriously, has it never occurred to anyone that it is neither and that particles and waves are artifacts of human perception, even (or perhaps especially) when that perception is filtered through millions of dollars of instruments?

This is not some magical case of reality being simultaneously in two contradictory states at one time–this is the result of our failure to fully grasp the ultimate reality, which is quite probably not self-contradictory if we could see it for what it is.

My guess it that the “hidden variables theory” is right on the money, and furthermore that its basic concept is more broadly applicable than was originally intended.

I have the feeling that a whole host of the things that have been mathematically “proven” in the last few decades will someday be seen to have no relation to the reality they are attempting to model.

9. The Physicist says:

What Bell’s Theorem shows (in the article above) is that “local hidden variables” are inconsistent with experiment (impossible). This means that either 1) thing conspire with each other in a very mysterious, faster-than-light way (this is non-local), or 2) things are not in a definite state (no hidden variables).
There’s a lot of very direct, straightforward experimental evidence for many-stateness (for example, in fact there’s even a post that talks about one example!), and also a lot of both theoretical and experimental evidence showing that faster than light stuff is impossible.
Physicists are pretty good about tying theory to physical experiment, and at the end of the day physical evidence always wins (even if it messes everything up, or makes you look bad). So, while quantum theory may be a pretty set of theories and equations, if they weren’t all thoroughly backed up by experiment after experiment, they would have been abandoned.
With regard to the uncertainty principle, it is a mathematically derived statement, but it’s based on very sound physical principles, and it’s stood the test of time (and also tests). It’s just one manifestation of “many-statedness”.

10. Fanciful Dope says:

It’s taken me a very long time to wrap my head around Bell’s Inequality, but I finally have, thanks to some simplified math I found off in some corner of the internet. (My math is… bad. I can’t math.). My request is: can you write an article that explains it in a way that could be understood with just a good grasp of algebra? It’s such an important cornerstone of physics, and it goes ignored/misunderstood mostly because of that, I think, when really, it’s a very simple thing.

Thanks, and I love the site.

11. EinsteinWasRight says:

After decades of trying, no experiment has been able to rule out Einstein’s local realism or establish QM’s predicted, non-local entanglement as an actual phenomenon.

Google “loopholes in Bell test experiments” and you will see. All experiments to date suffer from problems, generally either insufficient detection rates or cross-talk problems. The Bell’s Theorem inequalities, correctly accounting for real experiments, are not violated.

The best Bell’s Theorem experiment to date – I think it’s Ansmann’s – seems almost close to establishing “spooky action at a distance” but is compatible with and actually accurate enough to be in support of local realism.

12. chris says:

My god, you really did try hard to make this article completely incomprehensible. Well done.

13. Dale Stahl says:

I understand Bell’s Inequalilty and the modified inequality that takes account of detection inefficiencies. Surely potential detection inefficiencies affect the prediction of QM. However, I have never seem a modified QM prediction that also takes account of detection inefficiencies. Can you direct me to any such accounts?

14. The Physicist says:

They show up a lot in the experimental papers that actually verify the Bell inequalities. Errors are a pretty standard problem (they’re the problem), and dealing with them is such old hat that it generally gets glossed over.
Paul Kwiat and Anton Zeilinger have both done a number of (fairly dense) papers on this sort of thing.

15. Dave Hollomby says:

It’s already been mentioned, but it’s worth restating, that a violation of Bell’s Inequality does not disprove the possibility of hidden variables. Bell himself was a fan of Bohm’s theory, which is a hidden variables theory.

The theorem states that there is a difference between the predictions of theories based on local hidden variables and those of QM. The key word here is ‘local’. Violations of Bell’s Inequality suggest that at least one of the assumptions he made in proving his result (which he took from the Einstein-Podolsky-Rosen argument) must be false: either nature is non-local or hidden variables do not exist (or both).

But in the EPR argument, the orthodox interpretation of QM (non-real, no hidden variables) was shown to require non-locality to account for QM’s predicted correlations between measurements separated by a sufficient distance. Einstein simply would not countenance the idea that nature could be non-local, that there was no ‘spooky action at a distance’, and so he concluded that QM was incomplete because it had no other way of accounting for its predicted correlations.

So even if Bell’s Theorm is taken (wrongly) as demonstrating that there are no hidden variables (non-real), the EPR argument is that nature must still be non-local. So, what violations of Bell’s Inequality suggest is that nature is non-local.

I believe that since the above posts were made, there has been significant progress in closing the loopholes that would still permit a local hidden variable theory. It’s still possible nature does employ hidden variables, but if she does, she does so non-locally.

16. Max Graham says:

I am a Learner with a capital ‘L’ when it comes to this area so forgive me if my question/comment is ridiculous, but here goes.
Is one possible answer to this anomaly (and various others I’ve heard of) that the whole cosmos down to its smallest particle exists within another ‘reality’ that has its own and different set of laws governing the behavior of its constituent parts? (and that includes our cosmos which is within the greater reality). If that were so then when we get down to the quantum level we are looking at the interface between the physical reality that we know and experience and this other greater reality. This ‘other reality’ can only be discerned when we come across these ‘odd behavior’ incidents. An oversimplification perhaps, but the existence of one ‘other and greater reality’ would maybe do away with the need to imagine the multiplicity of realities that have been posited by some leaders in the field.


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# Removing Header from Chapter in a Thesis Written in LaTeX

I am writing a thesis using the LaTeX template files that seem to have been used at my department for a while. The problem is, when I compile the code, I get a header with chapter number and title on the page where chapter 2 starts.

It doesn't happen though on the page where chapter 1 begins.

I want to remove the header from the start page of chapter 2 but despite trying a lot, I could not figure out what is causing it to appear there (and not on start page of chapter 1) and how to get rid of it. Will appreciate any help. Here is the MWE for the main LaTeX file:

\documentclass[11pt,a4paper,twoside]{report}
\usepackage[top=3cm, bottom=3cm, left=2.5cm, right=2.5cm]{geometry}
\usepackage[latin1]{inputenc}

% Page number must be in the upper righthand corner
\usepackage{fancyhdr}
\fancypagestyle{phdthesis}{
\fancyhf{} %clear all headers and footers fields
\rhead{\thepage} %prints the page number on the right side of the header
}
\fancypagestyle{plain}{\pagestyle{phdthesis}}

% to remove page numbers on empty pages (before chapter start)
\makeatletter
\def\cleardoublepage{\clearpage\if@twoside \ifodd\c@page\else
\hbox{}
\thispagestyle{empty}
\newpage
\if@twocolumn\hbox{}\newpage\fi\fi\fi}
\makeatother

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

%% TITLE PAGE AND DECLARATION
\pagestyle{empty}
\begin{titlepage}
\end{titlepage}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% CONTENTS
\tableofcontents

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% CHAPTERS
\newpage
\pagenumbering{arabic}

%% INCLUDE TEXT OF CHAPTER 1 IN THE FILE chp1.tex
\include{chp1}

%% INCLUDE TEXT OF CHAPTER 2 IN THE FILE chp2.tex
\include{chp2}

%% INCLUDE TEXT OF CHAPTER 3 IN THE FILE chp3.tex
\include{chp3}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% BIBLIOGRAPHY
\bibliography{mybib}

\end{document}

Here is the code for all the 3 chapters which are saved separately as chp1.tex, chp2.tex and chp3.tex in the same directory:

\chapter{{\it Chapter 1 Title}}\label{chp1}
{\it Text for Chapter \ref{chp1}}

\chapter{{\it Chapter 2 Title}}\label{chp2}
{\it Text for Chapter \ref{chp2}}

\chapter{{\it Chapter 3 Title}}\label{chp3}
{\it Text for Chapter \ref{chp3}}

• I can not reproduce the issue using your code. Maybe there is a \pagestyle{fancy} before the chapters. Remove \thispagestyle{myheadings} after the \chapter commands. Then pagestyle plain is used on the chapter pages. Off topic: \it is deprecated for a long time and do not use font commands inside the argument of sectioning commands. – esdd May 19 at 11:33
• Thanks a lot for taking time to look at my problem. Commenting out \thispagestyle{myheadings} after \chapter command worked for the header on chapter 2 start page but there was no \pagestyle{fancy} before the chapters. I don't understand but I very much want to, why it was this way? When I comment out \thispagestyle{myheadings} from chapter 2 LaTeX file, the header goes away but the same command appears on LaTex files for chapter 1 and 3 too and there no header appears whether or not I comment out this command. There's really no significant chunk of code that I didn't show here. – V Kahn May 19 at 12:06
• @esdd, you said you can't reproduce the issue using my code. Can I do something to make it more reproducible? All I want to do is it to understand why \thispagestyle{myheadings} command on chapter 2 start page creates a header but the same command does not do it for chapter 1 and 3. – V Kahn May 19 at 13:19
• Test your code example in a new folder. Then you can see what we get. There could also be a \pagestyle{phdthesis} or \pagestyle{plain} in your real code. Note that your code redefines plain using \fancypagestyle. They all activate the automatic setting of a \leftmark by \chapter and cleaning \rightmark. I am sure that your issue does not depend on the chapter number, but on the page number. You will get header entries only on even chapter pages, but not on odd chapter pages. – esdd May 19 at 18:33
• Thanks a lot @esdd. You were right, I tested the code with 4 chapters and this time I got headers on chapters 2 and 4 (which start on even pages), but not on chapters 1 and 3 (which begin on odd pages). I checked my complete code and found that just before table of contents, there is a page for disclaimer and there \pagestyle{phdthesis} does appear. So you are are right in all cases. I work almost all the time with article class. Here, why does this \pagestyle{phdthesis} class produce headers on odd pages? I looked at how it was defined in the document preamble but could not figure it. – V Kahn May 19 at 18:44

Remove the \thispagestyle{myheadings} after the \chapter commands.
Then your own pagestyle phdthesis will be used on chapter pages, because of
\fancypagestyle{plain}{\pagestyle{phdthesis}}


text:
# Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times,

Question:

Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.

Solution:

Length of the three measuring rods are 64 cm, 80 cm and 96 cm, respectively.
∴ Length of cloth that can be measured an exact number of times = LCM (64, 80, 96)
Prime factorisation:

$64=2^{6}$

$80=2^{4} \times 5$

$96=2^{5} \times 3$

$\therefore$ LCM $=$ Product of greatest power of each prime factor involved in the numbers $=2^{6} \times 3 \times 5=960 \mathrm{~cm}=9.6 \mathrm{~m}$

Hence, the required length of cloth is 9.6 m.


text:
# Wave equation and wave speed

1. Nov 6, 2013

### amiras

The wave on the string could be described with wave equation.

Wave equation has a factor v^2 = Tension/linear density.

It has dimensions of speed, but from where exactly does it follow that this is actually speed of propagation of the wave?

2. Nov 6, 2013

### Staff: Mentor

v^2 has dimensions of speed squared. You can calculate the forces in the string based on the deflection in the string, and get this formula as a result. I'm sure there are textbooks where this derivation is done.

3. Nov 6, 2013

### HallsofIvy

Staff Emeritus
That can be interpreted as either (or both) of two questions:
1) How do we know that
$$\dfrac{\partial^2 y}{\partial x^2}= \frac{1}{c^2}\dfrac{\partial^2y}{\partial t^2}$$
(the "wave equation) does describe wave motion with wave speed "c"?

2) How do we show that the movement of a string, under tension T and with linear density $\delta$ satisfies the wave equation with $c= T/\delta$?

To answer the first one, suppose that f(x, t) satisfies the wave equation. Let g(x, t)= f(x- ct_0, t- t_0). Show that g(x, t) also satisfies the wave equation. Of course, g(x, t) is just f(x, t) with time shifted by $t_0$ and distance shifted by $c t_0$ so that the difference in distances is $ct_0$ for a change in time $t_0$ so that the speed is $ct_0/t_0= c$.

Before answering the second, I have to admit it isn't exactly true! It is an approximation but a very good approximation for small waves. If you were to pull on the string or wire really hard, you might well permanently stretch or even break it!

Imagine a small part of the string between $(x_0, y(x_0))$ and $x_0+ \Delta x, y(x_0+\Delta x))$. Let the angle the string makes at $x_0$ be $\theta_{x_0}$. With tension, T, the upward force at $x_0$ is $T sin(\theta_{x_0}$. Let the angle at $x_0+ \Delta x$ be $\theta_{x_0+ \Delta x}$. The upward force at $x_0+ \Delta x$ is $T sin(\theta_{x_0+ \Delta x}$ so the net force is $T(sin(\theta_{x_0+ \Delta x})- sin(\theta_{x_0}))$. For small angles sine is approximately tangent so we can approximate that by $T(tan(\theta_{x_0+ \Delta x}- tan(\theta_{x_0}))$.

The tangent of the angle of a curve, at any point, is the derivative with respect to x so can write that as
$$T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}$$

If that section of string has length s and density $\delta$ then it has mass $\delta \Delta s$. For small angles, we can approximate $\Delta s$ by $\Delta x$. "Mass times acceleration" is
$$\delta \Delta x \dfrac{\partial y^2}{\partial t^2}$$
so that "force equals mass times acceration" becomes
$$T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0})= \delta \Delta x \dfrac{\partial^2 y}{\partial t^2}$$

Dividing both sides by $T \Delta x$, that becomes
$$\dfrac{\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}}{\Delta x}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}$$

Now we can recognize that fraction on the left as a "difference quotient" which, in the limit as $\Delta x$ goes to 0, becomes the derivative with respect to x. Since the function in the difference quotient is the first derivative of y with respect to x, the limit gives the second derivative:
$$\dfrac{\partial^2y}{\partial x^2}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}$$

the wave equation with $1/c^2= \delta/T$ so that $c^2= T/\delta$.

Last edited: Nov 6, 2013
4. Nov 6, 2013

### The_Duck

Convince yourself that for an arbitrary function $f(x)$ the function $y(x, t) = f(x - vt)$ solves the wave equation. Then convince yourself that this solution describes a wave whose shape is given by $f$ and which propagates in the positive $x$ direction at a speed $v$.


text:
# Bipartite state purity

1. Nov 12, 2015

### ma18

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

Consider the bipartite state:

|q> = a/sqrt (2) (|1_A 1_B> +|0_A 0_B>)+sqrt((1-a^2)/2) (|0_A 1_B>+|1_A 0_B>)

where a is between or equal to 0 and 1

a) compute the state of the subsystem p_b

b) compute the purity of p_b as a function of a

c) for what values of a is the purity of p_b at a minimum/maximum

d) Compute the entanglement entropy of the bipartite state, for what value of a is it at a min/max

3. The attempt at a solution

I have done a) and found p_b to be :

p_b = |q_b><q_b|

where |q_b> = (a+sqrt(1-a^2))/sqrt(2) (|0_B>+|1_B>)

the computation for this is long and I don't want to replicate it here...

b)

I simply applied:

p (0_B,0_B) = <0_B|p_b|0_b> for the 4 combinations of |0_b> and |1_b> and got the matrix, then taking the trace of this for the diagonal terms I get 1/2 and 1/2 so the purity was equal to 1 and thus not dependent on a

c) It is unrelated

d) I know that the equation is

S (p_AB) = S (Tr_B p_AB)

but I don't know how to proceed from here

Any help such as how to proceed or checking my previous steps would be greatly appreciated!

2. Nov 15, 2015

### ma18

Could anybody help with this please

3. Nov 16, 2015

### Staff: Mentor

How is purity defined?

4. Nov 16, 2015

### ma18

In my notes purity is defined as the square of the trace of a state's density operator, it is between 1/d and 1 where d is the dimension of the Hilbert space. P = 1 for pure states. The states with maximal classical uncertainty (maximally mixed states) have the minimum possible purity P = 1/d, we say that the state is 'maximally mixed'.

5. Nov 17, 2015

### Staff: Mentor

You have a small mistake there. $\mathrm{Tr}(\rho)$ gives you the normalization of $\rho$, so you will always get $\mathrm{Tr}(\rho)^2 = 1$. To calculate the purity, you need to take the trace of the square of the density operator, $\mathrm{Tr}(\rho^2)$.

6. Nov 17, 2015

### ma18

Ah, that makes it work!

Alright so the only part I am confused about then is part d) and how to compute the entanglement entropy mathamatically

7. Nov 17, 2015

### Staff: Mentor

Your missing an equation for calculating the entropy of a density matrix.

8. Nov 17, 2015

### ma18

Yes I know. That is what I am searching for, I do not have it in my notes and there is no textbook :(

Do you know this equation?

9. Nov 17, 2015

### Staff: Mentor

10. Nov 17, 2015

### ma18

Thank you for the equation and the link!

Hmm I am getting a negative entropy for some values of a

How come you posted it as log 2 but it simply says log, presumably log 10 on the site?

11. Nov 17, 2015

### Staff: Mentor

I put the 2 there to be explicit, as I guessed this would cause confusion. In this field, log always mean the logarithm in base 2.


text:
Suggested languages for you:

Americas

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Q.9

Expert-verified
Found in: Page 473

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Rewrite in terms of sine and cosine functions: $\mathrm{tan}\left(\theta \right)csc\left(\theta \right)$

In terms of sine and cosine function we get,

$\frac{1}{\mathrm{cos}\theta }$

See the step by step solution

## Step 1: Given information

We are given an trigonometric function as $\mathrm{tan}\left(\theta \right)c\mathrm{sc}\left(\theta \right)$

## Step 2: We use trigonometric formulastan(θ)=sinθcosθcscθ=1sinθ

We get,

$\mathrm{tan}\left(\theta \right)c\mathrm{sc}\left(\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}\frac{1}{\mathrm{sin}\left(\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{cos}\left(\theta \right)}\phantom{\rule{0ex}{0ex}}=sec\theta$

## Step 3: Conclusion

The function in terms of sine and cosine can be given as $\frac{1}{\mathrm{cos}\theta }$


text: I am interested in magic tricks whose explanation requires deep mathematics. The trick should be one that would actually appeal to a layman. An example is the following: the magician asks Alice to choose two integers between 1 and 50 and add them. Then add the largest two of the three integers at hand. Then add the largest two again. Repeat this around ten times. Alice tell tells the magician her final number $n$. The magician then tells Alice the next number. This is done by computing $(1.61803398\cdots) n$ and rounding to the nearest integer. The explanation is beyond the comprehension of a random mathematical layman, but for a mathematician it is not very deep. Can anyone do better?


text:
# A number theory problem by Alejandro Castillo

$\large \dfrac1m + \dfrac1n - \dfrac1{mn^2} = \dfrac34$

Let all the solutions of the pairs of integers $$(m,n)$$ that satisfy the equation above be denoted by $$(m_1, n_1), (m_2, n_2), \ldots, (m_k, n_k )$$. Find $$\displaystyle \sum_{i=1}^k (m_i + n_i)$$.

×


text:
# Renate Loll video lectures on quantum gravity (Perimeter Scholars series)

1. Jan 27, 2010

### marcus

Lecture 1: http://pirsa.org/10010094/ (25 January)
Lecture 2: http://pirsa.org/10010095/ (26 January)
Lecture 3: http://pirsa.org/10010096/ (27 January)
Lecture 4: http://pirsa.org/10010097/ (28 January)

This is planned as a three-week series of one-hour lectures.

She does not use slides. She writes on the blackboard as she talks.
Lecture 1 gave the outline for the course. Loll is well-organized.

She plans to cover nonperturbative QG: Both the canonical (e.g. Loop) and the
covariant or path integral (e.g. triangulations) ways of doing nonpert. QG.
I just watched the first hour lecture. My impression is she has her act together
and the series will prove a valuable video resource.

Last edited: Jan 28, 2010
2. Jan 28, 2010

### torquil

Agreed!

I skimmed through the the first 2.5, and watched the rest of the third one in detail. She has arrived at an expression for a graviton field operator, based on considerations about gravitational waves in linearized gravity. The lectures seem promising.

EDIT: Lecture 4 was mainly an introduction to second quantization, and introduction of path integrals.

Torquil

Last edited: Jan 29, 2010
3. Jan 30, 2010

### Finbar

Watched the first two. Will watch the third today.

They seem good. But as she points out its a hard subject to lecture.

4. Jan 30, 2010

### conway

I watched the first one and found it to be pretty pointless. I wasn't going to watch any more but the topic of Lecture 4 piqued my interest. It seems her motivation was that she whipped through a bunch of material in Lecture 3 about gravitons and afterwards from the students questions she realized that they didn't understand the "basics" of second quantization. So she did a "review" of this in the first 15 minutes of Lecture 4. I have to say she didn't do all that bad a job of it except for the fact that she seemed really embarrassed about the fact that she was even doing it. I didn't think it was kid stuff.

Then she more or less derived the path integral formulation which I followed most of the way through with some lapses before losing her at the end. But I'm still going to ask one question about her closing comments, regarding how the effective contributions to the path are dominated by paths which are "nowhere differentiable". (I think I'm capturing the gist of what she said.) She elaborates by describing these paths as everywhere zigzagging.

My question is that in the complex domain (because she talks about integrating along paths in complex time) the definition of differentiability is very different from our intuition about smooth being equal to differentiable. All kinds of smooth functions in the complex domain are not "differentiable", only analytic ones are. For example, the function that stretches the complex domain as a rubber sheet mapping x to 2x and y to 3y is not "differentiable", but it is obviously smooth.

So my question is: is she making a mistake by describing the paths in question as somehow jagged?

5. Jan 30, 2010

### torquil

I think I can add something to this: After the rotation, the path is still considered as a function of only one real variable, not a complex variable. This is despite the fact that the justification for the rotation is that the path is an analytic function in the complex plane of rotation, and something about the Lagrangian also being analytic there (I'm not sure).

So the differentiation in question is with regard to this real variable which determines the position on the imaginary axis, so it is differentiation in the "real sense". It corresponds to differentiating the value of the path's position with respect to the length along the imaginary axis.

Because of the assumption about the path being an analytic function, the derivatives along the imaginary axis and the real axis are related, right? Allthough this seems like a contradiction, since it cannot be analytic if it is nowhere differentiable in the real sense along the imaginary axis

Torquil

6. Jan 30, 2010

### Physics Monkey

As torquil points out, one definitely should not think of any of the "typical" paths as analytic objects. They are jagged stochastic objects, continuous but nowhere differentiable in the real variable sense. For example, the "Wick rotation" or "analytic continuation" of the free particle path integral in quantum mechanics is formally identical to the problem of diffusion (where we reinterpret imaginary quantum mechanics time as real diffusion time). Here by diffusion I mean mathematical Brownian motion most rigorously described as a stochastic differential equation. This differential equation can be thought of as describing a random walk on a discrete lattice in a particular limit of vanishing step size and step time. You must take the limit of step size going to zero in such a way that the step size $$a$$ and step time $$\tau$$ satisfy $$a^2 /\tau \rightarrow \text{constant}$$. This formula is enough to see that the velocity $$v \sim a/\tau$$ is typically going to diverge in this limit.

The more sensible meaning of analytic continuation in these contexts comes from the correlation functions. These objects, formally defined using the path integral, should be analytic in their arguments (at least in some domain, depending on which correlation functions you study). This is the more rigorous meaning of analytic continuation in the path integral.

Last edited: Jan 30, 2010
7. Feb 2, 2010

### marcus

Lecture 1: http://pirsa.org/10010094/ (25 January)
Lecture 2: http://pirsa.org/10010095/ (26 January)
Lecture 3: http://pirsa.org/10010096/ (27 January)
Lecture 4: http://pirsa.org/10010097/ (28 January)
Lecture 5: http://pirsa.org/10010098/ (29 January)
Lecture 6: http://pirsa.org/10020060/ (1 February)
Lecture 7: http://pirsa.org/10020061/ (2 February)

Evidently planned as a three-week series of fifteen one-hour lectures, pitched at Masters student level, which I suppose might become the core of a textbook. (Basically an introduction to the main approaches in 4D quantum gravity research.) Have to look at Claus Kiefer's QG book to see if there is a niche for a Loll book.
Kiefer's Oxford monograph on QG came out in 2004 and costs about \$175. The second edition appeared in 2007. Here's the TOC:

1: Why quantum gravity?
2: Covariant approaches to quantum gravity
3: Parameterised and relational systems
4: Hamiltonian formulation of general relativity
5: Quantum geometrodynamics
6: Canonical quantum gravity with connections and loops
7: Quantisation of black holes
8: Quantum cosmology
9: String theory
10: Quantum gravity and the interpretation of quantum theory
11: References

There could be a market niche. Loll's triangulations approach only became widely visible in 2004 when they got a result for 4D, and in 2005 when they showed dimensional reduction at small scale and the link to Asymptotic Safety. The field does look substantially different now in 2010. Loll is doing a really good job with these lectures and shows she has the command and communication chops to write a successful intro textbook to QG.

If anyone wants to check out Kiefer's book in more detail here is the OUP page:
http://www.oup.com/us/catalog/general/subject/Physics/QuantumPhysics/?view=usa&ci=9780198506874 [Broken]

Last edited by a moderator: May 4, 2017
8. Feb 4, 2010

### Finbar

Watching the 6th lecture now. Its really interesting seeing how Renate thinks. It really gives good insight into CDT I think. Her explanation of good and bad gauge fixing conditions reminds me of the conditions she applies to the CDT path integral.

9. Feb 4, 2010

### atyy

10. Feb 5, 2010

### torquil

She hasn't gotten that far yet. I'm currently watching the Loll's Seventh now.

EDIT: I think something is wrong with the Flash video at 25m0s in the seventh lecture. Thankfully, there are other formats available.

Lecture 8 is all about repetition of symplectic/Hamiltonian dynamics with constraints.
Lecture 9 is about canonical formulation of gravity.

Torquil

Last edited: Feb 5, 2010
11. Feb 13, 2010

### marcus

Loll's 3-week course is now complete, and available video online.
Lecture 1: http://pirsa.org/10010094/ (25 January)
Lecture 2: http://pirsa.org/10010095/ (26 January)
Lecture 3: http://pirsa.org/10010096/ (27 January)
Lecture 4: http://pirsa.org/10010097/ (28 January)
Lecture 5: http://pirsa.org/10010098/ (29 January)
Lecture 6: http://pirsa.org/10020060/ (1 February)
Lecture 7: http://pirsa.org/10020061/ (2 February)
Lecture 8: http://pirsa.org/10020062/ (3 February)
Lecture 9: http://pirsa.org/10020063/ (4 February)
Lecture 10: http://pirsa.org/10020064/ (5 February)
Lecture 11: http://pirsa.org/10020065/ (8 February)
Lecture 12: http://pirsa.org/10020066/ (9 February)
Lecture 13: http://pirsa.org/10020067/ (10 February)
Lecture 14: http://pirsa.org/10020068/ (11 February)
Lecture 15: http://pirsa.org/10020069/ (12 February)

Incidentally, Prof. Jorge Pullin is currently teaching a one-semester introductory LQG course for undergrads. The second problem set is due 15 February, in a couple of days. Since there is no undergraduate-level textbook, the students are getting xerox write-ups of the lectures. Here is a brief course description and list of topics:
http://www.phys.lsu.edu/classes/spring2010/phys4750/
==quote==
PHYS 4750: Introduction to loop quantum gravity
...
This course will be an introduction for undergraduates to loop quantum gravity. To our knowledge, this is the first time such a course is offered in the world. We will assume minimal prerequisites: some knowledge of Lagrangian mechanics, some quantum mechanics and special relativity.

Topics:

1. Introduction: quantum gravity, why, what?
2. Special relativity and electromagnetism.
3. Some elements of general relativity.
4. Hamiltonian mechanics including constraints and fields.
5. Quantum mechanics and elements of quantum field theory.
6. Yang-Mills theories.
7. General relativity in terms of Ashtekar’s new variables.
8. Loop representation for general relativity.
9. An application: loop quantum cosmology.
10. Further developments.
11. Open issues and controversies.
==endquote==

Last edited: Feb 13, 2010
12. Feb 13, 2010

### torquil

I have now watched all the lectures. She reached CDT in the last one. She didn't mention anything about asymptotic safety in that context, but did mention the results in the article you are quoting. I.e. the de Sitter limit and results for the spectral dimension.

Very nice lectures, I would say. A good introduction to a few different QG efforts.

Torquil

13. Aug 2, 2012

### alemsalem

sorry to post here, but I'm being economic.
I'm watching the lectures now and I'm studying from Kiefer's QG book.
can I find problem sets that go along with these two??


text:
The SIMID consortium prepared a new technical note (v2021-08-18) containing the model estimates of hospital and ICU admissions and load by a short term prediction model and a stochastic dynamic transmission model using observational data up to December 2021.

The full document is available here.

Conclusions

• The short-term prediction model depicts a further increase in new hospitalizations and ICU load, driven by the current trends in positivity ratio. The model predicts between 78 and 136 new hospital admissions and between 242 and 282 patients in ICU on August 28.

• A further increase in hospital admissions is predicted in all regions (Brussels, Flanders and Wallonia), with a doubling of the number of new hospitalizations from August 16 to August 28 in both Brussels and Wallonia, and an increase of 70% in Flanders.

• Social mixing and thus risk behaviour still drives the projected burden of disease. An increase of +50% of the behaviour we estimate for August 2021, shows hospital admission levels in line with the projections using the September 2020’’ behaviour which would result in a high pressure on hospital capacity. If the increase in risk behaviour is only +30% of the August 2021 situation, we project on average only half of the daily hospital admissions and ICU load.

We wrote also a report with additional results for the Brussels Captial Region, which is available here.


text:
# What is the fastest algorithm for multiplication of two n-digit numbers?

I want to know which algorithm is fastest for multiplication of two n-digit numbers? Space complexity can be relaxed here!

• Are you interested in the theoretical question or in the practical question? – Yuval Filmus Oct 19 '13 at 20:03
• Both, but more inclined towards practical one! – Andy Oct 20 '13 at 6:49
• For the practical question, I recommend using GMP. If you're curious what they use, look at the documentation or the source code. – Yuval Filmus Oct 20 '13 at 8:04
• Nobody knows. We haven't found it yet. – JeffE Oct 21 '13 at 2:55

As of now Fürer's algorithm by Martin Fürer has a time complexity of $n \log(n)2^{Θ(log*(n))}$ which uses Fourier transforms over complex numbers. His algorithm is actually based on Schönhage and Strassen's algorithm which has a time complexity of $Θ(n\log(n)\log(\log(n)))$

Other algorithms which are faster than Grade School Multiplication algorithm are Karatsuba multiplication which has a time complexity of $O(n^{\log_{2}3})$ ≈ $O(n^{1.585})$ and Toom 3 algorithm which has a time complexity of $Θ(n^{1.465})$

Note that these are the fast algorithms. Finding fastest algorithm for multiplication is an open problem in Computer Science.

References :

Note that the FFT algorithms listed by avi add a large constant, making them impractical for numbers less than thousands+ bits.

In addition to that list, there are some other interesting algorithms, and open questions:

• Linear time multiplication on a RAM model (with precomputation)
• Multiplication by a Constant is Sublinear (PDF) - this means a sublinear number of additions which gets for a total of $\mathcal{O}\left(\frac {n^2} {\log n} \right)$ bit complexity. This is essentially equivalent to long multiplication (where you shift/add based on the number of $1$s in the lower number), which is $\mathcal{O}\left({n^2} \right)$, but with an $\mathcal{O}\left(\log n\right)$ speedup.
• Residue number system and other representations of numbers; multiplication is almost linear time. The downside is, the multiplication is modular and {overflow detection, parity, magnitude comparison} are all as hard or almost as hard as converting the number back to binary or similar representation and doing the traditional comparison; this conversion is at least as bad as traditional multiplication (at the moment, AFAIK).
• Other Representations:
• [Logarithmic representation]: multiplication is addition of the logarithmic representation. Example: $$16 \times 32 = 2^{\log_2 16 + \log_2 32} = 2^{4+5} = 2^{9}$$
• Downside is conversion to and from logarithmic representation can be as hard as multiplication or harder, the representation can also be fractional/irrational/approximate etc. Other operations (addition?) are likely more difficult.
• Canonical representation: represent the numbers as the exponents of the prime factorization. Multiplication is addition of the exponents. Example: $$36 \times 48 = 3^2\cdot 5^1\times 2^{2}\cdot 3^1\cdot 4^1 = {2^2}\cdot {3^2} \cdot 4^1 \cdot 5^1$$
• Downside is, requires factors, or factorization, a much harder problem than multiplication. Other operations such as addition are likely very difficult.
• I believe a residue/Chinese Remainder Theorem-based approach with the right moduli can lead to speedups over traditional multiplication even with the conversion back; at some point this was in chapter 4 of TAOCP, at least as a footnote. (It still doesn't get near the FFT-based methods, but it's an interesting historical note) – Steven Stadnicki Oct 20 '13 at 17:15
• @StevenStadnicki oh cool, I need to look at that then; do you happen to know the complexity? – Realz Slaw Oct 20 '13 at 23:54


text:
# Will future deep space optical communications “ground stations” actually be in space, or on the ground?

Deep space communications ground stations are on the ground because their transmit and receive electronics alone is bulky and heavy, not to mention their 34 and 70 meter dish antennas!

But those constraints change when it can be a 1 meter or smaller telescope with a laser diode or photodiode (or superconducting thingy) at the focal plane, exploiting the much smaller theoretical diffraction limit. For example right now in DSN Now I see DSS-14 a 70 meter dish receiving signals from Juno 8.4 GHz. That's a wavelength of 3.6 centimeters, so $$(D/\lambda)^2 \approx \text{3.8} \times \text{10}^6$$.

If instead a 0.85 meter telescope was used at 850 nm (a random, typical AlGaAs laser wavelength) we'd have $$(D/\lambda)^2 \approx \text{10}^{12}$$

Remember that the far-higher gain is also available at the other end of the link as well, so this analysis is a gross underestimate of the total improvement in link budget going optical, but that's okay because I haven't included some of the challenges.

A few watt laser is comparable to a few watt deep space probe transmitter, and only four orders of magnitude lower than a DSN transmitter (except for things lke Has DSS-43 ever been used in high power mode (>>20 kW) for an emergency situation?) so our circa-one-meter telescope "ground station" beats a 70 meter dish if it can be pointed steadily and around any intervening clouds

Hubble can be pointed steadily and around any intervening clouds, and there are several high altitude locations around the world where optical telescopes can be pointed steadily and usually around intervening clouds, often using adaptive optics. (see also the surprising answers to Why aren't ground-based observatories using adaptive optics for visible wavelengths?)

Question: Optical deep space transceiving stations of circa one meter diameter could be Earth orbit or on the ground in several places, with adaptive optics if necessary. What are the most important technical tradeoffs that will determine where they will most likely end up being deployed?

Below this answer I'd summarized the following collection of Q&A on optical communications for deep space:

"interplanetary radio communication" will go away and be replaced by optical in the not-so-distant future because a 20 kW transmitter or liquid helium-cooled front-end receiver at the end of 34 meter dish on Earth can be replaced by a few watt laser diode and an avalanche photodiode or one of those superconducting thingies at the end of a 20 or 50 cm diameter telescope. (see also)

A very heavy ground station, very likely to remain on the ground for the foreseeable future:

DSS-43 from here from NASA.

Summary of DSN diameters and transmit powers in this answer

• It's notable that as far as has been publicly discussed, Starlink will have optical interlinks only for satellite-to-satellite communications. For talking to the ground, it uses radio. – Christopher James Huff Nov 17 '20 at 5:00
• If the optical deep space transceiving station is placed in an area with (almost) no clouds like the Atacama desert, it should work. – Uwe Nov 17 '20 at 10:19
• @uhoh That sounds really interesting. Can you expand on what you mean by intermittent sampling and give an example? – ChrisR Nov 18 '20 at 4:34
• SpaceX is already deploying laser interlinks on Starlink satellites, which certainly qualify as small spacecraft. They need only equip a few satellites in one or two planes with some additional links for interplanetary relay satellites, and they'll have massively redundant downlink capability through the phased array antennas used to provide internet service. – Christopher James Huff Nov 21 '20 at 15:59
• @uhoh the standard Starlink sats are equipped with enough laser links to connect to neighboring Starlinks (likely previous and next in the same plane, and some more for crossing planes), and I was mainly thinking that they'd need additional links to handle duties outside of that...though they might just redirect one of the normal links instead. I wouldn't expect the low-altitude Starlinks to directly handle interplanetary comms, but I can see the standard inter-Starlink link hardware being sufficient for reaching a relay satellite in a higher orbit. – Christopher James Huff Nov 21 '20 at 23:17


text:
# (Why) Is orbital angular momentum conserved for point masses?

The introduction of the angular momentum as $$\vec l = \vec r \times \vec p$$ is also true for point particles. So $$\vec l$$ must refer to the orbital angular momentum (and not the "spin") in such a case.

If I have a particle moving straight with constant velocity and then I apply a force $$\vec F$$ perpendicular to the moving direction, I created a torque $$\vec M = \vec{r} \times \vec{F}$$ that creates the orbital angular momentum $$\vec l \neq 0$$. Now conservation of angular momentum says that $$\vec l$$ stays the same when the torque vanishes.

But wouldn't this mean that the point mass has to fly an endless circle without any force? This is in total contradiction to Newton's laws that a particle moves a straight line without force. Why isn't angular momentum conserved in this case? What actually is angular momentum, then?

• Have you tried calculating the angular momentum for a point particle moving at constant velocity along a straight line, which does not pass through the origin of your coordinate system? Doing so may allow you to answer your own question. May 3 at 15:11
• Hint: angular momentum may also be calculated and conserved for straight line motion of particles. Once your torque vanishes, so does the curvilinear path. May 3 at 16:53
• @KshitijKumar But why is it "conserved" then? It's obviously not conserved if it vanishes. If I turn off a force the momentum stays like it is at the momemt of turning off the force. This is called conservation of momentum. So conservation of angular momentum should mean "if I turn off the torque, then the angular momentum should stay like it is in the moment of turning off" - doesn't this mean the particle must fly on a curve? That's exactly my question. May 3 at 18:16

Imagine you had a particle traveling at constant $$\mathbf{v_0} = \dot{z_0}\mathbf{\hat{k}}$$ upwards. At time $$t = 0$$, your particle is at $$\mathbf{r} = z_0 \mathbf{\hat{k}}$$ and you apply an horizontal force $$\mathbf{F} = -F\mathbf{\hat{i}}$$, constant $$F$$, for time $$\Delta t$$.

The particle was subect to a force, so its velocity was at every instant of application of the force: $$\mathbf{v} = \mathbf{v}_0 + \Delta \mathbf{v} = \mathbf{v}_0 + \int_0^t\frac{d \mathbf{v}}{dt} dt = \mathbf{v}_0 + \int_0^t \mathbf{a} dt = \mathbf{v}_0 + \int_0^t \frac{\mathbf{F}}{m} dt$$ but the last integral really is just Newton's second law and: $$\int_0^t \frac{\mathbf{F}}{m} dt = -\frac{F}{m} t \mathbf{\hat{i}}$$ and the final speed: $$\mathbf{v}(t) = \dot{z_0}\mathbf{\hat{k}} -\frac{F}{m} t \mathbf{\hat{i}} \Longrightarrow \mathbf{v}(\Delta t) = \dot{z_0}\mathbf{\hat{k}} -\frac{F}{m} \Delta t \mathbf{\hat{i}} = \mathbf{v}'.$$ Its position is also given by integrating the last equation: $$\mathbf{r}(t) = \mathbf{r_0} + \Delta \mathbf{ r} = (z_0 + \dot{z}_0 t)\mathbf{\hat{k}} - \frac{1}{2}\frac{F}{m}t^2 \mathbf{\hat{i}} \Longrightarrow \mathbf{r}(\Delta t) = (z_0 + \dot{z}_0 \Delta t)\mathbf{\hat{k}} - \frac{1}{2}\frac{F}{m}(\Delta t)^2 \mathbf{\hat{i}} = \mathbf{r'}$$ After the torque/force ends, it stops accelerating and therefore just travels rectilinearly through space with $$\mathbf{v}'$$. In the same coordinate system, you can check that angular momentum is, after calculating speed and position (reset the clock, so we can start at $$t = 0$$ again, that's licit, we "stored" everything that happened before on the $$\Delta t$$s, $$\mathbf{r}'$$ and $$\mathbf{v}'$$): $$\mathbf{r}(t) = \mathbf{r}' + \mathbf{v}' t$$ $$\mathbf{v}(t) = \mathbf{v}'$$ $$\mathbf{L}(t) = \mathbf{r}(t) \times \mathbf{p}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) = m(\mathbf{r}' + \mathbf{v}' t) \times \mathbf{v}' = m\mathbf{r}' \times \mathbf{v}'$$ notice that the primed vectors are constant, so this new angular momentum also is. So you really get rectilinear motion and constant angular momentum in the end.

Angular momentum(and torque) does not necessarily mean that there are rotations in place. you can define angular momentum for straight lines, however, that isn't very useful to solve any problem.

If you have a particle moving in a straight line and apply a force perpendicular to it, it won't rotate unless there is a centripetal force somewhere. In your exemple, if you only apply a perpendicular force, the particle will simply move in both x and y directions in a parabola.

What will happen is that, say, you have a particle moving in a straight line in the x direction. The angular momentum is: $$r \times p_x$$. Then you apply the perpendicular force, adding a y component into the velocity. Since the force is perpendicular only, then the linear momentum is conserved in the x direction at all times, thus

$$\frac{dp_x}{dt} = 0$$

The angular momentum while applying the force is, thus: $$r \times (p_x + p_y)$$. Where $$p_x$$ is constant and $$p_y$$ is varying due to the force, the torque is: $$r \times \frac{dp_y}{dt})$$

Once the torque vanishes, no forces act in the y direction, thus

$$\frac{dp_y}{dt} = 0$$

And then the particle continues in a straight line.Take the time derivative of angular momentum after the torque vanishes: $$\frac{dL}{dt} = v \times (p_x + p_y) + r \times (\frac{dp_y}{dt} + \frac{dp_y}{dt})$$

The first term vanishes because the velocity vector is always parallel to the linear momentum, then you are left with the second, but both time derivatives are zero, thus:

$$\frac{dL}{dt} = 0$$

What does this mean? Nothing, because there is no rotations, however it's still a true result to say that angular momentum is conserved in absence of torques.

For a particle of mass $$m$$ with velocity $$\vec v$$, with respect to a point $$O$$ the angular momentum is $$(1) \vec l = \vec r \times m\vec v$$ where $$\vec r$$ is the vector distance from $$O$$ to the particle. The torque due to a force $$\vec F$$ on the particle is $$(2) \vec N = \vec r \times \vec F$$ The change in the angular momentum is $$(3) {d\vec l\over dt} = \vec N$$ The angular momentum and torque depend on the point taken as $$O$$. For a particle with no applied torque, the angular momentum is constant.

Using (2), the magnitude of the torque is $$(4)rFsin(\theta)$$ where $$\theta$$ is the angle between $$\vec r$$ and $$\vec F$$. With no force there is no torque; for example a particle with constant $$\vec v$$. A force parallel to $$\vec r$$ produces no torque; for example a particle in uniform circular motion under a centripetal force.

For a particle moving in a straight line under no force, the velocity $$\vec v$$ is constant. There is also no torque on the particle, so the angular momentum is also constant. Specifically, from any point $$O$$, the angular momentum is constant, since $$\vec l = \vec r \times m\vec v$$ has constant magnitude $$r_{perpen}mv$$ where $$r_{perpen}$$ is the constant component of $$\vec r$$ perpendicular to $$\vec v$$.

For a particle moving at constant speed in a circle, there is a force, the centripetal force (provided by gravity, or tension in a string, etc.) The direction of the centripetal force is radially inward towards $$O$$ at the center of the circle. The velocity $$\vec v$$ is always tangential to the circle. Although there is a force, there is no torque about $$O$$ using (4), since $$\vec v$$ and $$\vec r$$ are always anti-parallel and $$sin(\theta) = 0$$. The angular momentum is constant, equal to $$rmv = mr^2\omega$$ where $$\omega$$ is the constant angular speed.


text:
# Synopsis: Unequal parts

New experiments studying nuclear fission yield unexpected results.

Traditional models of nuclear fission predict that heavy nuclei break into unequally sized (asymmetric) parts, which is consistent with the findings of most experiments in low-energy fission. This is naturally explained by the fact that one of the fragments tends to be in the vicinity of doubly magic tin-$132$, which is highly stable.

The vast majority of fission experiments, however, have focused on heavy nuclides ranging from thorium to fermium. Now, new experiments performed at the ISOLDE facility in CERN and presented in Physical Review Letters (Andrei Andreyev et al.), probe a different corner of the nuclear chart. The team studies mercury-$180$ (with $80$ protons and $100$ neutrons) and finds that the fission products are also asymmetric. However, in this particular case, the outcome is counterintuitive, since a symmetric decay of the nucleus into two copies of zirconium-$90$ (with $40$ protons and $50$ neutrons) would have produced exceptionally stable nuclei.

The ISOLDE team’s puzzling result hints that a very subtle interplay between macroscopic and microscopic interactions plays a deeper role in the fission process than expected and is likely to inspire detailed theoretical studies and further experiment. – Abhishek Agarwal

More Features »

### Announcements

More Announcements »

Nuclear Physics

## Previous Synopsis

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### Viewpoint: Watching the Hoyle State Fall Apart

Two experiments provide the most precise picture to date of how an excited state of carbon decays into three helium nuclei. Read More »

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### Synopsis: Strong Force Calculations for Weak Force Reactions

Theorists have used lattice-QCD calculations to predict two weak-force-driven reactions—proton fusion and tritium decay. Read More »

Particles and Fields

### Synopsis: Proton Loses Weight

The most precise measurement to date of the proton mass finds a value that is 3 standard deviations lower than previous estimates. Read More »


text:
# how to generate 50hz frequency from PIC16F84A or 877A

Status
Not open for further replies.

#### babatundeawe

##### Full Member level 5
highest period of pwm in pic16f

i am working on a project to construct an inverter using PIC16F84A or 877A. the micro is to control the whole operation of the inverter and also to generate the frequency and also using pulse width modulation regulation. pls any idea, circuit diagram and source code will be appreciated. thanks in anticipation.

#### deepu_ttc

##### Member level 5
howto generate a frequency with pic

y dont u use 2 timer( in pic ) to generate a pwm signal and generate another square wave with other timer use a ckt that have a logic such that it has 2 i/p's and 2 o/p's and the behaviour will be if square wave is 1 then o/p-1 will be pwm and o/p-2 will be 0 and when square wave is 0 then o/p-1 will be 0 and o/p-2 will be pwm then connect these 2 o/p's to 2 bjts connected in series so u get required frequency

#### VVV

how to program a servo with 877a

You can use the CCP modules in the PIC16F877A, together with TMR2. Both can be used to produce PWM pulses.
What kind of inverter are you going to design? There are newer parts that can handle the bridge. For example PIC16F690. Check the datasheet.

### babatundeawe

Points: 2

#### progart

##### Newbie level 4
how to generate 50hz ac

You can use MCU's interrupt (from TMR)function for your project and get stronger results.
Calculate your TMR prescaler values which MCU's oscillator and write basic interrupt program...

#### xorcise

##### Full Member level 2
configure ccp module as 50hz pwm output

I have written tons of 50Hz PWM code for servos. The following might be helpful for some of your own ideas. It runs 8 servos at super high resolution using mikroBasic.

SERVO8_10000 occupies less than 200 ROM and can individually position 8 servos to within 1/10000th of an 180 degree arc +/- 0.1%...if you have such a type of 50Hz servo. Add USART easily and without jitter and interruptions. This code example will work on P16's running @ 20Mhz. It is easily ported to P18's and uses NO INLINE ASSEMBLER.

Code:
Code:
program SERVO8_10000
'*********************************************************************************
' SERVO8_10000 by Warren Schroeder on October 3, 2006
' Program successfully compiled using mikroBASIC 5.0.0.0
' Tested on 16F877A @ 20MHz

' Servo8_10000 is a demonstration of managing 8 servos every 20ms with more
' than 13-bit position resolution without using inline assembler code.
' You can theoretically achieve 10,000 servo positions per servo if using
' 500us to 2500us range servos and a 20MHz oscillator. Any servo timing range
' is possible up to maximum of 2500us.
'
' Theory of operation is to give each servo an exact 2500us frame to do its
' position duty-cycle.  The servo turns ON at the start of the 2500us frame,
' completes its duty-cycle, turns OFF, and then waits for the next frame. At
' the next 2500us frame a new servo is then turned ON and the cycle is repeated.
' Each of the 8 servos occupies a 2500us slot and refreshed sequentially.  All
' 8 servos thus require 20ms to complete.  At this point, we go back to the first
' servo and repeat this whole process. This method creates exact 20ms periods
' between same servo time frames, which is required for 50Hz servos.
'
' The example below assumes using servos with a timing range of 900us to 2100us
' Timer1 is setup for 0.2us per count (no prescaler @ 20MHz).  Therefore, the min
' and max position values for this servo would be 4500 to 10,500.  To acheive 90
' degrees of motion, or 1500us, the formula is 1500 x 5 = 7500.
'
' To use RX with USART I recommend using the mB USART library commands.  These
' commands do not implement the USART Interrupts and are ideal for this program.
' Put the USART_DATA_READY and USART_READ function in the two wait loops in the
' MAIN program.  USART has a 2-byte buffer which will not fill up and is not
' affected by the Timer interrupts as long as you take advantage of reading it
' regularly.  Using the two wait loops is the best time to check the RX buffer.
'**********************************************************************************

symbol T1_Switch = T1CON.TMR1ON
symbol T1_Int_Flag = PIR1.TMR1IF
symbol T1_Int_En = PIE1.TMR1IE
symbol T0_Int_Flag = INTCON.TMR0IF
symbol T0_Int_En = INTCON.TMR0IE
const IsEnabled = 1
const IsDisabled = 0
const IsSet = 1
const IsRst = 0
const IsOn = 1
const IsOff = 0
dim intflag as boolean
dim servo as byte
dim T1 as word absolute $0E dim T0 as byte absolute$01
dim pos as word[8]

sub procedure interrupt
If T0_Int_Flag = IsSet Then      ' if Timer0 interrupt
T0 = -196                     ' load Frame 196 x 12.8us = 2508us interrupt cycle
Clearbit(PORTB, servo)        ' force servo off...in case value was wrong
servo = inc(servo) And 7      ' next servo
Setbit(PORTB, servo)          ' turn on next servo
T1 = 0 - pos[servo]           ' load servo position...each count is 0.2us
T1_Switch = IsOn              ' turn Timer1 On
T1_Int_Flag = IsRst           ' reset Timer1 Interrupt Flag
T0_Int_Flag = IsRst           ' reset Timer0 Interrupt Flag
Else                             ' if Timer1 interrupt
Clearbit(PORTB, servo)        ' turn off servo
T1_Switch = IsOff             ' turn Timer1 Off
End If
intflag = not(intflag)           ' interrupt flag status
end sub

main:
TRISB = 0
OPTION_REG = $85 ' Timer0 prescaler = 64 = 12.8us PORTB = 0 INTCON =$E4                     ' GIE, PEIE, Timer0 Enabled; Timer0 Flag Set
T1CON = 0                        ' Timer1 Off; No prescaler
For servo = 0 to 7
pos[servo] = 7500             ' set for 1500us mid-point
Next servo
intflag = False
servo = 0
T1 = 0 - 4500                    ' load Timer1
T1_Int_Flag = IsRst              ' reset Timer1 Interrupt Flag
T1_Int_En = IsEnabled            ' Timer1 Interrupt Enabled
T1_Switch = IsOn                 ' Timer1 On
T0 = -196                        ' load TMR0 for 2500us Interrupt Cycle Count
T0_Int_En = IsEnabled            ' reset Timer0 Interrupt Flag
Setbit(PORTB, 0)                 ' turn on first servo

While 1=1
While intflag = False         ' WAIT until Timer1 Interrupt
nop                          ' place USART Read routines in here
Wend
While intflag = True          ' WAIT until Timer0 Interrupt
nop                          ' place USART Read routines in here
Wend
Wend

end.

#### babatundeawe

##### Full Member level 5
pic 50 hz inverter source code

VVV said:
You can use the CCP modules in the PIC16F877A, together with TMR2. Both can be used to produce PWM pulses.
What kind of inverter are you going to design? There are newer parts that can handle the bridge. For example PIC16F690. Check the datasheet.

thanks for your post. i really appreciate the way you put it pls can you be more explanatory the way you did with the sg3525. pls i will appreciate if i can see a little source code and scematics. because i ve not tried anything on ccp and pwm with any micro. thanks in anticipation

#### xorcise

##### Full Member level 2
how to generate 50hz sine wave for 100 points

VVV said:
You can use the CCP modules in the PIC16F877A, together with TMR2. Both can be used to produce PWM pulses.
What kind of inverter are you going to design? There are newer parts that can handle the bridge. For example PIC16F690. Check the datasheet.

You cannot get 50Hz (20ms period) from the CCP PWM module, unless you run the PIC with less than 1MHz oscillator.

#### VVV

pic16f84 pwm code

xorcise is right about the oscillator frequency for 50Hz output.
But I don't think the inverter will be running at 50Hz. That is the output frequency, which is really just a modulation for the HF signal.

I will try to dig up some schematics.

#### mpopovics

##### Junior Member level 1
generate a pulse signal with pic16f84

Is there any way to generate sine wave?
Means, real sinewave.

#### VVV

servo8_10000 by warren schroeder

babatundeawe,

Can you give more details on this inverter? Input/ output voltages, power and desired waveform?

#### Mike K8LH

##### Full Member level 2
pwm pic 50hertz

xorcise said:
I have written tons of 50Hz PWM code for servos. The following might be helpful for some of your own ideas. It runs 8 servos at super high resolution using mikroBasic.

SERVO8_10000 occupies less than 200 ROM and can individually position 8 servos to within 1/10000th of an 180 degree arc +/- 0.1%...if you have such a type of 50Hz servo. Add USART easily and without jitter and interruptions. This code example will work on P16's running @ 20Mhz. It is easily ported to P18's and uses NO INLINE ASSEMBLER.
Nice example and method.

Here's another 8 channel (port b) method that uses the CCP "compare" mode with a 4 MHz clock for 1 usec pulse width resolution and 20 msec Servo period;

Code:
static unsigned char n = 0;
static unsigned int Pulse [] = { 1500, 1500, 1500, 1500,
1500, 1500, 1500, 1500,
20000 };
static unsigned char Servo = 1;

void isr_hi ()
{
/****************************************************************
*  K8LH Crazy-8 Hi-Rez Soft 8-channel (PORTB) Servo Algorithm  *
****************************************************************/

if (PIR1bits.CCP1IF == 1)     // if CCP1 "compare" interrupt
{ LATB = Servo;               // output new Servo pulse and
CCPR1 += Pulse[n];          // setup next "compare" value

PIR1bits.CCP1IF = 0;        // clear CCP1 interrupt flag

Pulse[8] -= Pulse[n++];     // adjust end-of-cycle off time
Servo <<= 1;                // and prep for next channel

if (n == 9)                 // if end of a 20-msec cycle
{ n = 0;                    // reset array index
Pulse[8] = 20000;         // reset the 20.0-msec period
Servo++;                  // reset Servo shadow to Servo 1
}
}
}

xorcise said:
VVV said:
You can use the CCP modules in the PIC16F877A, together with TMR2. Both can be used to produce PWM pulses.
What kind of inverter are you going to design? There are newer parts that can handle the bridge. For example PIC16F690. Check the datasheet.
You cannot get 50Hz (20ms period) from the CCP PWM module, unless you run the PIC with less than 1MHz oscillator.
Actually you can. Sort of (grin).

The secret is to break up the 20 msec Servo period into smaller PWM "frames" and stuff the PWM duty cycle register during the period match interrupt with the value for the next "frame".

Here's a 4-pin 8 channel example using a 4 MHz clock and a 74HC238 decoder IC which is modulated by the PWM CCP1 pin (this is not the code for the 12F683 design shown below);

Code:
unsigned char select = 0;       // Servo channel number, 0..7
unsigned char frame = 10;       //
unsigned int pulse1;

/*                                                               *
*  Servo[] array elements contain Servo pulse width values in   *
*  the range of 3000..9000 (250-nsec 'ticks') which represent   *
*  extended Servo pulse width ranges of 750 to 2250-usecs       *
*                                                               */

static unsigned int Servo [] = { 6000, 6000, 6000, 6000,
6000, 6000, 6000, 6000 };

/*****************************************************************
*  K8LH 8-Channel 74HC238 Hi-Rez 'PWM' Servo Algorithm Example  *
*****************************************************************/
#pragma interrupt isr_hi

void isr_hi ()
{ if(PIR1bits.TMR2IF)
{ PIR1bits.TMR2IF = 0;            // clear TMR2 interrupt flag

/**************************************************************
*  setup 74HC238 address lines and our work variables for    *
*  the next servo channel interval during the final (10th)   *
*  frame of the current servo channel interval (the final    *
*  frame in each interval will always have a 0% duty cycle   *
*  and all 74HC238 outputs are 'off' so it will be safe to   *
*  change the address lines)                                 *
*                                                            */
if(frame == 10)             // 10 PWM frames/Servo interval
{ frame = 0;                // reset PWM frame number
select %= 8;              // 8 Servo intervals/20-msec cycle
pulse1 = Servo[select];   // setup 'pulse' work variable

LATB = (LATB & 0b11111000) | select;  // set address lines

select++;                 // preset for next interval
}
frame++;                    // bump for next interrupt

/**************************************************************
*  setup CCP1 PWM duty cycle for next 250-usec PWM 'frame'   *
*                                                            */

if(pulse1 > 1000)           // if pulse1 > 250-usecs
{ CCPR1L = 250;             // do a 100% duty cycle frame
pulse1 -= 1000;           // subtract 250-usecs
}
else                        // do a variable or 0% frame
{ CCP1CONbits.CCP1Y = (pulse1 & 1);
CCP1CONbits.CCP1X = (pulse1 & 2);
CCPR1L = (pulse1 >>= 2);
pulse1 = 0;               // remaining frames are %0
}
}
}

#### xorcise

##### Full Member level 2
pic16f84a frequency tmr0

You certainly have a talent for this. I hadn't considered the CCP module in the first application... very nice. I've been programming PIC's for a couple years now and still learn something new every day.

#### ldanielrosa

##### Member level 4
pic16f84 square wave

I suppose I jumped in a bit late.

My approach would be similar to what progart said. Whatever the crystal frequency (I will assume 20MHz), I would assign a timed interrupt for 1024 Tosc (256 Tins) which is the highest frequency that would allow the full resolution of the PWM. This would give me 390.625 PWM periods (51.2 uS) per AC output cycle.

The '877 has two PWM outputs that share the same period, but have independent duty cycles. I would use them in tandem- one controlling the positive half cycles and the other for the negative, only one would be active at a time.

The means of determining the instantaneous period would be by adding a constant to a rolling sum. In this case a 24 bit sum should be accurate to about 0.001Hz. At each timed interrupt I would add 42,950 (0x00A7C6) to the sum, use the most significant 9-11 bits (depending on the resolution you want) to determine which phase (PWM output) and fetch a value from a lookup table for the PWM period.

MUSTAFA19

Points: 2


text:
# Ionized Gas Kinematics of Nearby Low-Mass Galaxies

### Abstract

Starbursting dwarf galaxies are thought to be the prime candidate for exhibiting out- flow of material: they are undergoing massive star formation periods and have masses small enough for the material to escape the small gravity well. However, the outflow rates for most dwarf galaxies, starbursting or post-starbursting, are still relatively un- constrained. This project sets out to observe and catalog the possible outflow of the average dwarf galaxy. Eighteen dwarf galaxies were observed during three observing runs from 2015 to 2017 using the SparsePak spectrometer at the WIYN telescope. Data reductions were made and maps of the H$\alpha$ flux, rotational velocity, and velocity dispersion were made, as well as maps for the [SII]/H$\alpha$ and [SII] ratios. Evidence for outflow was observed for few galaxies by comparing the velocity dispersion and Hα flux maps. For those that had outflow events, morphological features were identified as well in a majority of them.

Publication
Presented at AAS Meeting #231
##### Ray Garner
###### CWRU Astronomy PhD Candidate

I’m a budding scientist and amateur photographer raised in Georgia and living in Ohio. My research interests include galaxy evolution, star formation, satellite galaxies, and nebular diagnostics.