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Sciencemadness Discussion Board » Special topics » Technochemistry » identification of unusual electronic component, large. Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues

Author: Subject: identification of unusual electronic component, large.
Panache
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identification of unusual electronic component, large.

Been meaning to ask about these two items for awhile, me being electronics inept. I believe they were used as part of a stepup transformer, they were in oil, the silver bases and top sheets are all highly polished aluminium, never felt anything so slippery and smooth.

Better question is what to do with them other than use them as chandeliers, which is their present task.

watson.fawkes
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The one with the zig-zag on it is a Cockcroft-Walton multiplier for high voltage. It's got ten stages, so that's the multiplication factor. Each of the diode assemblies (they look too large to have only a single junction) could well be rated 15 kV (that's 15 x 1 kV junctions in a single package). The whole multiplier looks like it might have operated at 100 - 150 kV. You didn't post of photo of the back side of the zig-zag, but I'd guess that's where the capacitors for the cascade are.

The other one looks like a stack of "doorknob" capacitors. There are six of them. A breakdown rating on these of 25 kV is typical; these could go higher.

All the polished and rounded metal is specifically engineered to suppress corona discharge. Submersion in oil is also standard for HV equipment; it further suppresses corona. Aside: oil doesn't always suppress corona, which is why self-quenching HV oils can be pricey.

So, it looks like you have the two main HV components of a very dangerous HV capacitor discharge system. As a trial calculation of its capacity, assuming operation at 150 kV and capacitors of 1200 pF each. Total energy storage is 1/2 CV2. For these values, that comes out to 81 Joules. That's enormous, and would be very instantly lethal, even for a smallish mistake. I have no irrational fear of HV, but this is no idle screwing-around gear.
Wizzard
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I believe a baseball traveling at 60MPH has about 8J of energy, so that's enormous!!!
Twospoons
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 Quote: Originally posted by watson.fawkes That's enormous, and would be very instantly lethal, even for a smallish mistake. I have no irrational fear of HV, but this is no idle screwing-around gear.

Especially since HV at that level will reach out and get you - you don't have to actually touch!

Helicopter: "helico" -> spiral, "pter" -> with wings
edgecase
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Sell it to Dr. Megavolt... I think his videos are on YouTube. Look for the guy wearing a large birdcage for a helmet, holding a wooden plank drawing a 10 foot arc.
Panache
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Ok so chandeliers they stay beautiful things though, thanks for your replies, I much prefer understanding the description. If anyone in Melbourne wants the oil that watson mentioned pm me to arrange it it's about forty liters. I'll take the extra photos including ones of the 'mace' and 'gun' sections which where the large bit sticking into the large round gaps in the devices.
Is there any possibility of any pgm's in any of this stuff and if so which bit.
dann2
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The mother of all Birkeland-Eyde reactors for making nitric acid, anyone?
watson.fawkes
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 Quote: Originally posted by Panache If anyone in Melbourne wants the oil that watson mentioned pm me to arrange it it's about forty liters. [...] Is there any possibility of any pgm's in any of this stuff and if so which bit.
The oil might be specially-made dielectric oil, or it might be relatively pure mineral oil. Both are used.

As for PGM content, unlikely. There might be some silver in the doorknob capacitors, but not a lot, since there are only two plates inside those. They're worth more whole, maybe 30 - 50 USD each (if I'm guessing right).

If you want to sell the parts, go on 4hv.org and post pictures. You'll get more salivation there than here. You could well get someone who wants this gear even after paying for international shipping. You should know that the rarest part is all the good machining and fabrication, not so much the electronic components themselves. Dimensions are matched though, so both are worth less if separated.
 Quote: Originally posted by dann2 The mother of all Birkeland-Eyde reactors for making nitric acid, anyone?
This is a pulse generator, able to fire every second or so; this is an educated guess (obviously). For the B-E process you need a continuous arc and a different power supply.
Panache
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Quote: Originally posted by watson.fawkes

 Quote: Originally posted by dann2 The mother of all Birkeland-Eyde reactors for making nitric acid, anyone?
This is a pulse generator, able to fire every second or so; this is an educated guess (obviously). For the B-E process you need a continuous arc and a different power supply.

what about one of these then (see attachment)?
My manual states the power supply is 280V Dc, 3- 5A, from a 240V supply (mains power), this feeds the high frequency power amplifier running 6 transistors.
from manual...'the driver system is controlled by the et51 gpp circuit via optocouplers. the small signals are amplified to the power necessary to drive the big transistors. This system is self-adjusting and gives exactly the basic-current which gives the optimal efficiency of the big transistors.'

goes on to talk about collector currents, impulse peak currents, resonance circuits, impulse generators, synchronized repition frequency (18kHz-35kHz depending upon which electroder you're using) and a whole lot of other stuff beyond me. It has a built in ozone extractor treatment unit.

and best of all it works! corona is a very pretty thing.

they retail at around AUS $125K, i got my one second hand with zero hours clocked (it was a second unit of two in a lab that never used it) for AUS$100
for that price i had to buy it really, i'm from a poly film background so i knew its use, however it owes me a further \$100 i spent on castors for it, its heavy. i should really try and sell it but i so loath cold calling i never try(seriously if someone in australia wants to broker this you can have ten percent of whatever you get, pm me)

Attachment: CP-Lab.pdf (103kB)

SM2
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I've found the best way to stumble upon such items is usually at something like a 3M or Imation auction. Also, certain government auctions. Turnkey, hand picked, matched, and ready to go (as Watson pointed out). Most the crap at military surplus retail level is, well, crap.
watson.fawkes
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 Quote: Originally posted by Panache what about one of these then (see attachment)?
Yes, a continuous output corona generator is certainly another application for such a multiplier.

I should also mention, however, that such devices can be, in certain ways, even more dangerous to work with than pulse discharge devices, because a corona generator may source more current than a generator that charges a capacitor bank. That is, these things can always deliver significant current, not just when their output capacitors are charged.

 Sciencemadness Discussion Board » Special topics » Technochemistry » identification of unusual electronic component, large. Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues


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# SdcaRegressionTrainer Class

## Definition

The IEstimator<TTransformer> for training a regression model using the stochastic dual coordinate ascent method.

public sealed class SdcaRegressionTrainer : Microsoft.ML.Trainers.SdcaTrainerBase<Microsoft.ML.Trainers.SdcaRegressionTrainer.Options,Microsoft.ML.Data.RegressionPredictionTransformer<Microsoft.ML.Trainers.LinearRegressionModelParameters>,Microsoft.ML.Trainers.LinearRegressionModelParameters>
type SdcaRegressionTrainer = class
inherit SdcaTrainerBase<SdcaRegressionTrainer.Options, RegressionPredictionTransformer<LinearRegressionModelParameters>, LinearRegressionModelParameters>
Public NotInheritable Class SdcaRegressionTrainer
Inherits SdcaTrainerBase(Of SdcaRegressionTrainer.Options, RegressionPredictionTransformer(Of LinearRegressionModelParameters), LinearRegressionModelParameters)
Inheritance

## Remarks

To create this trainer, use Sdca or Sdca(Options).

### Input and Output Columns

The input label column data must be Single. The input features column data must be a known-sized vector of Single.

This trainer outputs the following columns:

Output Column Name Column Type Description
Score Single The unbounded score that was predicted by the model.

### Trainer Characteristics

Is normalization required? Yes
Is caching required? No
Required NuGet in addition to Microsoft.ML None
Exportable to ONNX Yes

### Training Algorithm Details

This trainer is based on the Stochastic Dual Coordinate Ascent (SDCA) method, a state-of-the-art optimization technique for convex objective functions. The algorithm can be scaled because it's a streaming training algorithm as described in a KDD best paper.

Convergence is underwritten by periodically enforcing synchronization between primal and dual variables in a separate thread. Several choices of loss functions are also provided such as hinge-loss and logistic loss. Depending on the loss used, the trained model can be, for example, support vector machine or logistic regression. The SDCA method combines several of the best properties such the ability to do streaming learning (without fitting the entire data set into your memory), reaching a reasonable result with a few scans of the whole data set (for example, see experiments in this paper), and spending no computation on zeros in sparse data sets.

Note that SDCA is a stochastic and streaming optimization algorithm. The result depends on the order of training data because the stopping tolerance is not tight enough. In strongly-convex optimization, the optimal solution is unique and therefore everyone eventually reaches the same place. Even in non-strongly-convex cases, you will get equally-good solutions from run to run. For reproducible results, it is recommended that one sets 'Shuffle' to False and 'NumThreads' to 1.

This class uses empirical risk minimization (i.e., ERM) to formulate the optimization problem built upon collected data. Note that empirical risk is usually measured by applying a loss function on the model's predictions on collected data points. If the training data does not contain enough data points (for example, to train a linear model in $n$-dimensional space, we need at least $n$ data points), overfitting may happen so that the model produced by ERM is good at describing training data but may fail to predict correct results in unseen events. Regularization is a common technique to alleviate such a phenomenon by penalizing the magnitude (usually measured by the norm function) of model parameters. This trainer supports elastic net regularization, which penalizes a linear combination of L1-norm (LASSO), $|| \textbf{w}_c ||_1$, and L2-norm (ridge), $|| \textbf{w}_c ||_2^2$ regularizations for $c=1,\dots,m$. L1-norm and L2-norm regularizations have different effects and uses that are complementary in certain respects.

Together with the implemented optimization algorithm, L1-norm regularization can increase the sparsity of the model weights, $\textbf{w}_1,\dots,\textbf{w}_m$. For high-dimensional and sparse data sets, if users carefully select the coefficient of L1-norm, it is possible to achieve a good prediction quality with a model that has only a few non-zero weights (e.g., 1% of total model weights) without affecting its prediction power. In contrast, L2-norm cannot increase the sparsity of the trained model but can still prevent overfitting by avoiding large parameter values. Sometimes, using L2-norm leads to a better prediction quality, so users may still want to try it and fine tune the coefficients of L1-norm and L2-norm. Note that conceptually, using L1-norm implies that the distribution of all model parameters is a Laplace distribution while L2-norm implies a Gaussian distribution for them.

An aggressive regularization (that is, assigning large coefficients to L1-norm or L2-norm regularization terms) can harm predictive capacity by excluding important variables from the model. For example, a very large L1-norm coefficient may force all parameters to be zeros and lead to a trivial model. Therefore, choosing the right regularization coefficients is important in practice.

## Fields

 The feature column that the trainer expects. (Inherited from TrainerEstimatorBase) The label column that the trainer expects. Can be null, which indicates that label is not used for training. (Inherited from TrainerEstimatorBase) The weight column that the trainer expects. Can be null, which indicates that weight is not used for training. (Inherited from TrainerEstimatorBase)

## Properties

 (Inherited from StochasticTrainerBase)

## Methods

 Trains and returns a ITransformer. (Inherited from TrainerEstimatorBase) (Inherited from TrainerEstimatorBase)

## Extension Methods

 Append a 'caching checkpoint' to the estimator chain. This will ensure that the downstream estimators will be trained against cached data. It is helpful to have a caching checkpoint before trainers that take multiple data passes. Given an estimator, return a wrapping object that will call a delegate once Fit(IDataView) is called. It is often important for an estimator to return information about what was fit, which is why the Fit(IDataView) method returns a specifically typed object, rather than just a general ITransformer. However, at the same time, IEstimator are often formed into pipelines with many objects, so we may need to build a chain of estimators via EstimatorChain where the estimator for which we want to get the transformer is buried somewhere in this chain. For that scenario, we can through this method attach a delegate that will be called once fit is called.

format: plain


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# Help recalling theorem

1. May 26, 2008

### Gokul43201

Staff Emeritus
This is something of an odd request, I guess.

I have a very foggy recollection of a theorem in some field of applied mathematics - probably information theory. I think it has to do with a seemingly surprising result about the information needed to describe any real in some closed interval compared to the information needed to describe all the reals in that interval.

I can't recall anything more definite about this, and what I've said above may itself be more wrong than right. Hopefully there's just enough correct stuff there to help ring a bell with someone. Does anyone know what I'm rambling about?

Last edited: May 26, 2008
2. May 27, 2008

### CRGreathouse

The information needed in both cases is uncountably infinite -- in particular, $$\beth_1:=2^{\aleph_0}.$$

3. May 28, 2008

### maze

Wouldnt you be able to describe a single real with a sequence that converges to it (which is countable)?

4. May 28, 2008

### CRGreathouse

There are $$\beth_1$$ sequences consisting of $$\aleph_0$$ rational numbers. So, you could describe a real number that way, but it wouldn't be countable.

5. May 28, 2008

### Gokul43201

Staff Emeritus
Thanks CRG. I was about to ask how you get beth_1 for both cases... but I ought to give it some thought first. Also, this tells me that I probably haven't recalled the theorem correctly.

6. May 28, 2008

### CRGreathouse

A closed real interval is of the form $\{x:a\le x\le b\}$ for some real a and b. Thus describing the interval requires only giving two real numbers. Two real numbers can be combined 'for the price of one' in many ways, like interleaving digits:

1.12345 (interleave) 2.24680 = 21.1224364850

Work from the decimal point out, since real numbers can't be infinite.

The second point is only that $$\aleph_0^{\aleph_0}=2^{\aleph_0}:=\beth_1.$$

I'm just trying to jog your memory.

7. May 28, 2008

### CRGreathouse

Of course an arbitrary subset (rather than interval) of the reals has cardinality $$\beth_2=2^{\beth_1}.$$

If the result was information theory, then we're probably looking at the wrong stuff. Information theory usually deals with finite numbers of bits, right? $$2^{2^{\aleph_0}}$$ doesn't strike me as particularly 'applied'.

8. May 29, 2008

### maze

Perhaps they have infinities sitting around that get renormalized away?


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# Section 2.2: Problem 8 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Assume that is a set of sentences such that for any sentence , either or . Assume that is a model of . Show that for any sentence , we have iff .
By assumption, either every model of is a model of or every model of is a model of . Then, using the fact that no can be both a model of and a model of , we conclude that every model of is a model of () iff any model of is a model of (). Note that does not actually require the assumption.


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# Question:Use The Method Of Completing The Square To Solve The Following Equation. Choose One Answer. A. B. C. Solve The Equation By Factoring Or By Using The Property Choose One Answer. A. B. C. D. D. Use The Quadratic Formula To Solve The Following Equation. Choose One Answer. A. B. C. D.

## This problem has been solved!

Use the method of completing the square to solve the following equation.

a.
b.
c.
Solve the equation by factoring or by using the property

 a. b. c. d.
d.

Use the quadratic formula to solve the following equation.


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# Relation between Interior Product, Inner Product, Exterior Product, Outer Product..

Following my previous question Relation between cross-product and outer product where I learnt that the Exterior Product generalises the Cross Product whereas the Inner Product generalises the Dot Product, I was wondering if the simple map that I drew below is at all an accurate representation of the links between these different products? Vertical lines denote generalisation-specification, horizontal lines denote "in opposition to". I'm just trying to get a quick overview before I dive in. Thanks

-
What does "interior product" mean? What does "outer product" mean? What do the lines on your diagram signify? – Chris Eagle Aug 13 '12 at 11:26
Hi Chris, the context and motivation is in the link in my question. Sorry for being unclear. – Ryan Aug 13 '12 at 12:07
The terms "exterior" and "outer" product are overloaded, and you're going to get clashing answers unless you are specific about what you mean. Actually I have never heard of an interior product... – rschwieb Aug 13 '12 at 12:25
Inner product generalizes dot product. Outer product is a particular case of tensor product and not related to scalar product. Cross product is a particular case of exterior product. Interior product is a particular case of exterior product. It is not well known and quite unnatural, because it depends on the choice of some vector. – Norbert Aug 13 '12 at 12:30
@Nobert Thanks for clarifying. The wikipedia page on Interior Product which implies that IP is opposed to EP but it must've meant just the naming. Anyway. – Ryan Aug 13 '12 at 13:51

I'm not sure what you're reading, but it certainly contains a lot of terms talked about in Clifford algebra (or "geometric algebra").

Clifford algebras were born of a synthesis of inner product spaces and Grassmann's exterior algebras, both of which have geometric applications.

A Clifford algebra is constructed from an inner product space $(V,Q)$ by generating an associative algebra (whose product is a descendant of the tensor product in the tensor algebra for $V$). These are compatible in a sense made clear in the Wiki.

Suppose we are in a real inner product space, and let's denote the algebra multiplication with $\otimes$ and the inner product with $\cdot$. One identity that holds for parallel vectors $v,w$ is $v\otimes w=v\cdot w$. Part of the reason this is true is that $v\otimes w=w\otimes v$ when $v,w$ are parallel.

When they are not parallel, then there is a skew component to $v\otimes w$. This can be retrieved explicitly by computing $\frac{1}{2}(v\otimes w-w\otimes v):=v\wedge w$. With this notation, $v\otimes w=v\cdot w+v\wedge w$ for all vectors $v,w$. This $\wedge$ is called the exterior or outer product in this algebra.

They key thing to know is that this inner and this outer product don't uniquely extend to the whole algebra (well the outer product has a slightly more natural extension than the inner product). They are not "natural" to the algebra really: they are mostly just a notational convenience with some algebraic properties that make them easy to view as products.

The cross product arises too in the discussion of Clifford algebras, but it seems to be regarded as second class to the outer product. I've seen this attributed to the cross-product somehow not carrying the correct information for physical applications.

I went and read a little and found out that "interior product" looks a lot like an important extension of the inner product in a Clifford algebra which is sometimes called the contraction.

In summary, I think these various products might become less mysterious if you read some of these linked articles and see where they are used. It's going to be especially enlightening when you find out where they are all defined (in the vector space $V$ or in an algebra containing $V$.)

-
Thanks for your input! – Ryan Aug 13 '12 at 12:30

The relation between the inner product of vectors and the interior product is that if you have a metric tensor (and thus a canonical relation between vectors and covectors = $1$-forms), the inner product of two vectors is the interior product of one of the vectors and the $1$-form associated with the other one. That is, if $g$ is the metric tensor, then the inner product of the vectors $v$ and $w$ is $g(v,w)$, and the $1$-form $\omega$ associated with $v$ is defined by $\omega(w)=g(v,w)$. Then it is obvious that $\iota_w(\omega) = \omega(w) = g(v,w)$.

The outer product is, as noted in answer to the other question you referred to, related to the tensor product. Indeed, if we associate row vectors with $1$-forms and column vectors with vectors, then we can write (using Einstein summation convention) the outer product of the vector $w = w^ie_i$ and the $1$-form $\omega = \omega_i\,e^i$ as the $(1,1)$ tensor $M = w^i\omega_j e_i\otimes e^j$ which describes an object that maps vectors to vectors. Its relation to the inner product is that you get the inner product of $w$ and $\omega$ by contracting the two indices of $M$ (which in the language of matrices corresponds to the trace of $M$).

The exterior product is related to the tensor product in that the exterior product of two forms (a form is a skew-symmetric tensor of type $(0,p)$) is just the antisymmetrization of the tensor product.

The cross product is a speciality of the three-dimensional space; here the space of $2$-forms has the same dimension as the space of $1$-forms; indeed, given a metric, the hodge star maps between them. Since the metric also allows to associate vectors and $1$-forms, you can define the cross product of $v$ and $w$ by the following procedure: Determine the $1$-forms corresponding to $v$ and $w$, calculate their exterior product (which is a $2$-form), apply the Hodge star to the result (which, given that we are in three dimensions, again results in a $1$-form), and finally determine the vector corresponding to that $1$-form.

-


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## SBT Prize at the end of chapter 7 – Math 7 SBT Horizon – Math Book

SBT Prize End of Chapter 7 – Math 7 SBT Horizon
==========

### Solution 1 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Let $$B = x{y^3} + 4xy – 2{x^2} + 3$$. Calculate the value of the expression $$B$$ when $$x = – 1$$, $$y = 2$$.

Detailed instructions for solving Lesson 1

Solution method

Substitute the values of the variable and calculate.

Detailed explanation

Substituting $$x = – 1$$, $$y = 2$$ into $$B = x{y^3} + 4xy – 2{x^2} + 3$$ we have

$$B = \left( { – 1} \right){.2^3} + 4\left( { – 1} \right).2 – 2. {\left( { – 1} \right)^2 } + 3 = – 15$$.

–>

— *****

### Solution 2 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Which of the following expressions is a monomial of one variable?

a) $$2y$$;

b) $$3x + 5$$;

c) $$12$$;

d) $$\frac{1}{3}{t^2}$$.

Detailed instructions for solving Lesson 2

Solution method

A monomial of a variable is an algebraic expression consisting of only a number, or a variable, or a product between numbers and the variable.

Detailed explanation

The one-variable monomial is a; c; d.

–>

— *****

### Solution 3 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Which of the following expressions is a one-variable polynomial?

$$5 – 2x$$; $$6{x^2} + 8{x^3} + 3x – 2$$;

$$\frac{2}{{x – 1}}$$; $$\frac{1}{4}t – 5$$.

Detailed instructions for solving Lesson 3

Solution method

Understand the concept of one-variable monomial, one-variable polynomial to determine.

A monomial of a variable is an algebraic expression consisting of only a number, or a variable, or a product between numbers and the variable.

A one-variable polynomial is the sum of all mononomials of the same variable.

Detailed explanation

The one-variable polynomials are “ $$5 – 2x;\,\,6{x^2} + 8{x^3} + 3x – 2$$; $$\frac{1}{4}t – 5$$.

–>

— *****

### Solution 4 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Write a quaternary one-variable polynomial with 5 terms.

Detailed instructions for solving Lesson 4

Solution method

A quaternary one-variable polynomial means that the degree of the largest variable is 4.

Detailed explanation

A quaternary one-variable polynomial with 5 terms is:

A(x) = x4 – 2x3 + 3x2 – 4x + 5.

Comment: This post has many answers.

–>

— *****

### Solve problem 5 page 33 SBT Math 7 Creative horizon episode 2 – CTST

State the degrees of the following polynomials:

$$A = 5{x^2} – 2{x^4} + 7$$;

$$B = 17$$;

$$C = 3x – 4{x^3} + 2{x^2} + 1$$.

Detailed instructions for solving Lesson 5

Solution method

The degree of the polynomial is the maximum exponent of the variable after reducing the polynomial.

Detailed explanation

The degree of $$A = 5{x^2} – 2{x^4} + 7$$ is 2.

The degree of $$B = 17$$ is 0.

The degree of $$C = 3x – 4{x^3} + 2{x^2} + 1$$ is 3.

–>

— *****

### Solution 6 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial $$P\left( x \right) = {x^3} + 64$$. Find the root of the polynomial in the set $$\left\{ {0;4; – 4} \right\}$$

Detailed instructions for solving Lesson 6

Solution method

Replace $$x = {x_0}$$ into $$P\left( x \right)$$ if $$P\left( {{x_0}} \right) = 0$$ then $$x = {x_0}$$ is the solution of $$P\left( x \right)$$.

Detailed explanation

With $$x = 0$$ we have $$P\left( 0 \right) = {0^3} + 64 = 64 \ne 0$$, so $$0$$ is not a solution of $$P \left( x \right)$$.

With $$x = 4$$ we have $$P\left( 4 \right) = {4^3} + 64 = 128 \ne 0$$, so $$4$$ is not a solution of $$P \left( x \right)$$.

With $$x = – 4$$ we have $$P\left( { – 4} \right) = {\left( { – 4} \right)^3} + 64 = 0$$, deduce $$– 4$$ is the solution of $$P\left( x \right)$$.

So $$– 4$$ is the solution of $$P\left( x \right)$$.

–>

— *****

### Solution 7 page 33 SBT Math 7 Creative horizon episode 2 – CTST

The length of the triangle is $$3y + 2$$; $$6y – 4$$ and the perimeter is $$23y – 5$$. Find the unknown side in that triangle.

Detailed instructions for solving Lesson 7

Solution method

The perimeter of a triangle is equal to the sum of its three sides. So, if we want to find the unknown edge, we subtract the known edges from the perimeter.

Detailed explanation

The unknown side in that triangle is $$23y – 5 – \left( {3y + 2} \right) – \left( {6y – 4} \right) = 14y – 3$$.

–>

— *****

### Solve problem 8 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial $$M\left( x \right) = 3{x^5} – 4{x^3} + 9x + 2$$. Find the polynomials $$N\left( x \right)$$, $$Q\left( x \right)$$ such that $$N\left( x \right) – M\left( x \right) = – 5{x^4} – 4{x^3} + 2{x^2} + 8x$$ and $$Q\left( x \right) + M\left( x \right) = 3{x ^4} – 2{x^3} + 9{x^2} – 7$$

Detailed instructions for solving Lesson 8

Solution method

Step 1: Add and subtract mononomials of the same variable to reduce the given polynomial.

Step 2: Sort the monomials by the descending power of the variable.

Step 3: Perform the calculation horizontally or vertically.

Detailed explanation

$$N\left( x \right) – M\left( x \right) = – 5{x^4} – 4{x^3} + 2{x^2} + 8x \Rightarrow N\left( x \right) = – 5{x^4} – 4{x^3} + 2{x^2} + 8x + M\left( x \right)$$

$$– 5{x^4} – 4{x^3} + 2{x^2} + 8x + 3{x^5} – 4{x^3} + 9x + 2 = 3{x^5} – 5{x^4} – 8{x^3} + 2{x^2} + 17x + 2$$

So $$N\left( x \right) = 3{x^5} – 5{x^4} – 8{x^3} + 2{x^2} + 17x + 2$$.

$$Q\left( x \right) + M\left( x \right) = 3{x^4} – 2{x^3} + 9{x^2} – 7 \Rightarrow Q\left( x \ right) = 3{x^4} – 2{x^3} + 9{x^2} – 7 – M\left( x \right)$$

$$3{x^4} – 2{x^3} + 9{x^2} – 7 – \left( {3{x^5} – 4{x^3} + 9x + 2} \right) = 3{x^4} – 2{x^3} + 9{x^2} – 7 – 3{x^5} + 4{x^3} – 9x – 2 = – 3{x^5} + 3{x^4} + 2{x^3} + 9{x^2} – 9x – 9$$

–>

— *****

### Solve lesson 9 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform multiplication.

a) $$\left( {4x – 5} \right)\left( {3x + 4} \right)$$;

b) $$\left( {2{x^2} – 3x + 5} \right)\left( {4x + 3} \right)$$

Detailed instructions for solving Lesson 9

Solution method

Understand the rule of one-variable polynomials: To multiply a polynomial by a polynomial, we multiply each term of one polynomial by each term of the other polynomial and then add the products together.

Detailed explanation

a) $$\left( {4x – 5} \right)\left( {3x + 4} \right) = 4x.3x + 4x.4 – 5.3x – 5.4 = 12{x^2} + x – 20$$;

b) $$\left( {2{x^2} – 3x + 5} \right)\left( {4x + 3} \right) = 2{x^2}.4x + 2{x^2}. 3 – 3x.4x – 3x.3 + 5.4x + 5.3 = 8{x^3} – 6{x^2} + 11x + 15$$

–>

— *****

### Solution 10 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform division.

a) $$\left( {64{y^2} – 16{y^4} + 8{y^5}} \right):4y$$

b) $$\left( {5{t^2} – 8t + 3} \right):\left( {t – 1} \right)$$

Detailed instructions for solving Lesson 10

Solution method

Detailed explanation

–>

— *****

### Solve lesson 11 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform division.

a) $$\left( {{x^4} + 6{x^2} + 8} \right):\left( {{x^2} + 2} \right)$$

b) $$\left( {3{x^3} – 2{x^2} + 3x – 2} \right):\left( {{x^2} + 1} \right)$$

Detailed instructions for solving Lesson 11

Solution method

Divide polynomials divide polynomials by setting the division calculation.

Detailed explanation

a)

So $$\left( {{x^4} + 6{x^2} + 8} \right):\left( {{x^2} + 2} \right) = {x^2} + 4\ ) b) So \(\left( {3{x^3} – 2{x^2} + 3x – 2} \right):\left( {{x^2} + 1} \right) = 3x – 2$$ .

–>

— *****

### Solution 12 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform division.

a) $$\left( {2{x^2} – 7x + 4} \right):\left( {x – 2} \right)$$

b) $$\left( {2{x^3} + 3{x^2} + 3x + 4} \right):\left( {{x^2} + 2} \right)$$

Detailed instructions for solving Lesson 12

Solution method

Divide polynomials divide polynomials by setting the division calculation.

Detailed explanation

a)

So $$\frac{{2{x^2} – 7x + 4}}{{x – 2}} = 2x – 3 – \frac{2}{{x – 2}}$$.

b)

So $$\frac{{2{x^3} + 3{x^2} + 3x + 4}}{{{x^2} + 2}} = 2x + 3 – \frac{{x + 2} }{{{x^2} + 2}}$$

–>

— *****


text:
## Elementary Geometry for College Students (5th Edition)

Line $\overline{AB}$ is where planes M and N intersect.


text:
# Picture Frame¶

## Introduction¶

While I was working with a friend mathematician, I showed him a couple ways physicists use math in a strange way.

One of the things that I showed him was the usage of commutators in quantum mechanics. They look like this:

That is zero if and commute, that is . If they do not do that, like general square matrices or operators, then the commutator gives you the way they do not commute.

## The Problem¶

He told me that they actually come from group theory and that there is a really neat problem that can be solved with them:

You want to put a framed picture onto the wall with a cord to the picture frame. The cord is a single one, and both ends are attached to the frame. How do you need to put the cord around nails, so that the picture can be taken down by taking out any of the nails?

For that, he told me, that I should write the commutator like that:

In the previous revision of this article, prior to 2013-08-18, I used the commutators with the implicit . Writing the commutator like he said is cleaner, since there is only one operation on this group of windings, namely performing them after each other.

I have seen a website where they give you the solution, but I would like to show the derivation he gave me. He told me, that the solution for is simply . Later on, I tried to understand what that actually means. This is what I came up with.

Start with two nails, A and B. There is a basic loop around each nail, and respectively:

If you do and then directly, and wiggle on the cord a little, you will get the following, which I will call :

If you go through one of the basic loops in reverse, it will count as its inverse, say .

## The Solution¶

The solution is supposed to be:

That means doing , then , then ( backwards) and then ( backwards). After some wiggling, I get this:

Look closely. If you pull out nail A, the loop that is tight around A will become free, go around B and will finally release the picture frame. If you release B, the loop that goes below B will become free. The loop above B will just fall down. That will also release the frame. So this is the solution to that problem.

## Higher Orders¶

The really interesting part about the group theory approach is that it scales to higher orders fairly easily. So what about three nails? The definition for the commutator of higher orders is the following:

With the basic loops , and , the solution to the three nail problem should be the following:

System Message: WARNING/2 (\begin{align*} \left[[A, B], C\right] &= [ A B A^{-1} B^{-1}, C ] \\ &= A B A^{-1} B^{-1} C (A B A^{-1} B^{-1})^{-1} C^{-1} \\ &= A B A^{-1} B^{-1} C (B A B^{-1} A^{-1}) C^{-1} \\ &= A B A^{-1} B^{-1} C B A B^{-1} A^{-1} C^{-1} \end{align*})

latex exited with error [stdout] This is pdfTeX, Version 3.14159265-2.6-1.40.18 (TeX Live 2017) (preloaded format=latex) restricted \write18 enabled. entering extended mode (./math.tex LaTeX2e <2017-04-15> Babel <3.9r> and hyphenation patterns for 43 language(s) loaded. (/usr/share/texlive/texmf-dist/tex/latex/base/article.cls Document Class: article 2014/09/29 v1.4h Standard LaTeX document class (/usr/share/texlive/texmf-dist/tex/latex/base/size12.clo)) (/usr/share/texlive/texmf-dist/tex/latex/base/inputenc.sty (/usr/share/texlive/texmf-dist/tex/latex/ucs/utf8x.def)) (/usr/share/texlive/texmf-dist/tex/latex/ucs/ucs.sty (/usr/share/texlive/texmf-dist/tex/latex/ucs/data/uni-global.def)) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsmath.sty For additional information on amsmath, use the `?' option. (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amstext.sty (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsgen.sty)) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsbsy.sty) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsopn.sty)) (/usr/share/texlive/texmf-dist/tex/latex/amscls/amsthm.sty) (/usr/share/texlive/texmf-dist/tex/latex/amsfonts/amssymb.sty (/usr/share/texlive/texmf-dist/tex/latex/amsfonts/amsfonts.sty)) (/usr/share/texlive/texmf-dist/tex/latex/anyfontsize/anyfontsize.sty) (/usr/share/texlive/texmf-dist/tex/latex/tools/bm.sty) No file math.aux. (/usr/share/texlive/texmf-dist/tex/latex/ucs/ucsencs.def) (/usr/share/texlive/texmf-dist/tex/latex/amsfonts/umsa.fd) (/usr/share/texlive/texmf-dist/tex/latex/amsfonts/umsb.fd) ! Misplaced \omit. \math@cr@@@ ...@ \@ne \add@amps \maxfields@ \omit \kern -\alignsep@ \iftag@ ... l.20 \end{align*}\end{split} ! Misplaced \omit. \math@cr@@@ ...@ \@ne \add@amps \maxfields@ \omit \kern -\alignsep@ \iftag@ ... l.20 \end{align*}\end{split} ! Misplaced \omit. \math@cr@@@ ...@ \@ne \add@amps \maxfields@ \omit \kern -\alignsep@ \iftag@ ... l.20 \end{align*}\end{split} ! Misplaced \omit. \math@cr@@@ ...@ \@ne \add@amps \maxfields@ \omit \kern -\alignsep@ \iftag@ ... l.20 \end{align*}\end{split} ! Misplaced \omit. \math@cr@@@ ...@ \@ne \add@amps \maxfields@ \omit \kern -\alignsep@ \iftag@ ... l.20 \end{align*}\end{split} ! Misplaced \omit. \math@cr@@@ ...@ \@ne \add@amps \maxfields@ \omit \kern -\alignsep@ \iftag@ ... l.20 \end{align*}\end{split} [1] (./math.aux) ) (see the transcript file for additional information) Output written on math.dvi (1 page, 840 bytes). Transcript written on math.log.

I tried to create a picture with this. It is not really easy to see what is going on there.

As far as I checked, the picture falls down if one removes either A, B or C, so the answer seems to be correct. In principle, it should be possible to do this with four and more nails, but I think that it will be almost impossible to present this in a clear way.


text:
# Q: Do physicists really believe in true randomness?

Physicist: With very few exceptions, yes.  What we normally call “random” is not truly random, but only appears so.  The randomness is a reflection of our ignorance about the thing being observed, rather than something inherent to it.

For example: If you know everything about a craps table, and everything about the dice being thrown, and everything about the air around the table, then you will be able to predict the outcome.

Not actually random.

If, on the other hand, you try to predict something like the moment that a radioactive atom will radioact, then you’ll find yourself at the corner of Poo Creek and No.  Einstein and many others believed that the randomness of things like radioactive decay, photons going through polarizers, and other bizarre quantum effects could be explained and predicted if only we knew the “hidden variables” involved.  Not surprisingly, this became known as “hidden variable theory”, and it turns out to be wrong.

If outcomes can be determined (by hidden variables or whatever), then any experiment will have a result.  More importantly, any experiment will have a result whether or not you choose to do that experiment, because the result is written into the hidden variables before the experiment is even done.  Like the dice, if you know all the variables in advance, then you don’t need to do the experiment (roll the dice, turn on the accelerator, etc.).  The idea that every experiment has an outcome, regardless of whether or not you choose to do that experiment is called “the reality assumption”, and it should make a lot of sense.  If you flip a coin, but don’t look at it, then it’ll land either heads or tails (this is an unobserved result) and it doesn’t make any difference if you look at it or not.  In this case the hidden variable is “heads” or “tails”, and it’s only hidden because you haven’t looked at it.

It took a while, but hidden variable theory was eventually disproved by John Bell, who showed that there are lots of experiments that cannot have unmeasured results.  Thus the results cannot be determined ahead of time, so there are no hidden variables, and the results are truly random.  That is, if it is physically and mathematically impossible to predict the results, then the results are truly, fundamentally random.

What follows is answer gravy: a description of one of the experiments that demonstrates Bell’s inequality and shows that the reality assumption is false.  If you’re already satisfied that true randomness exists, then there’s no reason to read on.  Here’s the experiment:

The set up: A photon is fired at a down-converter, which converts it into two entangled photons. These photons then go through polarizers that are set at two different angles. Finally, photo-detectors measure whether a photon passes through their polarizer or not.

1) Generate a pair of entangled photons (you can do this with a down converter, which splits one photon into an entangled pair of photons).

2) Fire them at two polarizers.

3) Randomly change the angle of the polarizers after the photons are emitted.  This prevents information about one measurement to affect the other, since that would require that the information travels faster than light.

4) Measure both photons (do they go through the polarizers (1) or not (0)?) and record the results.

The amazing thing about entangled photons is that they always give the same result when you measure them at the same angle.  Entangled particles are in fact in a single state shared between the two particles.  So by making a measurement with the polarizers at different angles we can measure what one photon would do at two different angles.

It has been experimentally verified that if the polarizers are set at angles $\theta$ and $\phi$, then the chance that the measurements are the same is: $C(\theta, \phi) = \cos^2{(\theta-\phi)}$.  This is only true for entangled photons.  If they are not entangled, then $C = .5 = 50\%$, since the results are random.  Now, notice that if $C(a,b) = x$ and $C(b,c) = y$, then $C(a,c) \ge x+y-1$.  This is because:

$\begin{array}{l}P(a=c)\\= P(a=b \cap b=c) + P(a \ne b \cap b \ne c)\\\ge P(a=b \cap b=c)\\= P(a=b) + P(b=c) - P(a=b \cup b=c)\\\ge P(a=b) + P(b=c) - 1\end{array}$

We can do two experiments at 0°, 22.5°, 45°, 67.5°, and 90°.  The reality assumption says that the results of all of these experiments exist, but unfortunately we can only do two at a time.  So C(0°, 22.5°) = C(22.5°, 45°) = C(45°, 67.5°) = C(67.5°, 90°) = cos2(22.5°) = 0.85.  Now based only on this, and the reality assumption, we know that if we were to do all of these experiments (instead of only two) then:

C(0°, 22.5°) = 0.85

C(0°, 45°) ≥ C(0°, 22.5°) + C(22.5°, 45°) -1 = 0.70

C(0°, 67.5°) ≥ C(0°, 45°) + C(45°, 67.5°) -1 = 0.55

C(0°, 90°) ≥ C(0°, 67.5°) + C(67.5°, 90°) – 1 = 0.40

That is, if we could hypothetically do all of the experiments at the same time we would find that the measurement at 0° and the measurement at 90° are the same at least 40% of the time.  However, we find that C(0°, 90°) = cos2(90°) = 0 (they never give the same result).

Therefore, the result of an experiment only exists if the experiment is actually done.

Therefore, you can’t predict the result of the experiment before it’s done.

Therefore, true randomness exists.

As an aside, it turns out that the absolute randomness comes from the fact that every result of every interaction is expressed in parallel universes (you can’t predict two or more mutually exclusive, yet simultaneous results).  “Parallel universes” are not nearly as exciting as they sound.  Things are defined to be in different universes if they can’t coexist or interact.  For example: in the double slit experiment a single photon goes through two slits.  These two versions of the same photon exist in different universes from their own points of view (since they are mutually exclusive), but they are in the same universe from our perspective (since we can’t tell which slit they went through, and probably don’t care).  Don’t worry about it too much all at once.  You gotta pace your swearing.

As another aside, Bell’s Inequality only proves that the reality assumption and locality (nothing can travel faster then light) can’t both be true.  However, locality (and relativity) work perfectly, and there are almost no physicists who are willing to give it up.  Except for Bohm, who’s an ass.

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

### 46 Responses to Q: Do physicists really believe in true randomness?

1. Scott says:

Your gravy is lumpy. Don’t use so much flour.

2. Marvin Frandsen says:

The answer did not include examples of chaos (nonlinear dynamics) which, while technically deterministic, provide ‘random’ results regardless of how well initial conditions are known.

Still, nice answer, especially with regard to quantum theory.

I think the swearing at Bohm is uncalled for. Skeptics of a standard paradigm should exist and continually be looking for a break in the standard paradigm. Also, I believe Bohm did most of his very original and interesting work before Bell derived his inequalities. If being wrong is a crime then most theorists would be in the klink.

3. Physicist says:

Thanks kindly, and fair enough. In a chaotic system the predictions do get better, the better your measurements are (reducible randomness). The irreducible component comes in when your measurements run up against Heisenberg uncertainty, or when the evolution of your system is (eventually) overwhelmed by quantum effects. But this is just another way of talking about the quantum randomness in the post.
I mostly think that Bohm is an ass based on his stubbornness and his personal creepiness. He’s only a criminal according to McCarthy.

4. pdf23ds says:

I think a more useful way to talk about randomness is “uncertainty”. There are different kinds of uncertainty, and thus different kinds of randomness. Uncertainty about air currents and temperature and the exact velocity of coins and dice are what make those things random. Under controlled conditions, that uncertainty can be eliminated, and so those things aren’t *always* random. A lot of binary quantum events (like polarizers) are more about indexical uncertainty (from the perspective of MWI)–the uncertainty we have about what part of the universe we’re in. So “uncertainty” doesn’t work to explain quantum randomness in the Copenhagen interpretation.

I think this is a pretty good way to explain it to a general audience.

5. Sonoran says:

Uncertainty and randomness are not the same thing. Probablistic events have uncertainty associated with them e.g.: No one can predict exactly when a specific radioactive nuclide will decay (uncertainty) but the decay of these nuclides is a strictly governed probablistic function not a random one. There’s exactly a 50% chance it will decay within it’s half-life period for example.

A random event can’t be assigned any probability. A truly random event wouldn’t have any greater or lesser chance of occurring by any parameter (time, conditions, interactions etc).

I don’t know of any natural process that’s truly random (unbounded and completely indeterminate).

6. The Physicist says:

Don’t expect to find one!
It’s not possible to create a function (probability distribution or otherwise) that is uniform over all numbers. If the definition of random is: “absolutely no tendency to happen more at any time or place”, then nothing can be random.

7. FunkyHitRecords says:

Well, boy I feel dumb here:
“It’s not possible to create a function…that is uniform over all numbers”

but isn’t f(x) = 5 uniform over all values of x — all numbers? I agree for probability distributions because uniform probabilities in a distribution would approach 0 as the number of outcomes approached infinity, but *any* function? Am I not understanding the term “uniform” ?

8. The Physicist says:

That’s my bad. You’re exactly right. Your understanding of “uniform” is perfect.
I’ve been spending too much time only considering L1 functions.

9. Niu says:

Hi, I think this demonstration has multiple issues.

1. To quote Wikipedia, “Experimental results have demonstrated that effects due to entanglement travel at least thousands of times faster than the speed of light.” This conflicts with the presumption made in step 3.

2. The impossible chance of 40% only exists when you do impossible experiments.

3. To prove to cause a random effect in the real world, one must affect deterministic objects and change their ways randomly. Otherwise the world is perfectly deterministic with some state-unknown particles. And when these particles come in contact with the world, all results and therefore the world are deterministic, just like what happens in the experiment.

10. The Physicist says:

1. I was a little surprised to read that in Wikipedia. It’s at best misleading and at worst wrong. Entanglement is a fancy form of correlation, so if you’d like to say that the fact that two coins are already in the same state is an “effect that travels” then it definitely travels infinitely fast.
2. The 40% thing is just a tool to emphasize that the correlation found in entanglement experiments is classically impossible. It can’t happen if the particles are in a particular state.
3. The crux of Bell’s theorem is that a probability distribution for a thing being in a single (even unknown), state can’t exist. That is, you can show that it doesn’t make sense to ask “what is the probability that this particle is in state 1?” for a particle in an entangled pair. There’s a better proof of that here.

You could make the argument that the universe is deterministic, but only from a fairly weird point of view.

11. Niu says:

What I meant in 3 is: If one small universe only consists of entangled photons and fixed polarizers, in this case you “know everything about” the universe, and you can predict the chances. So there’s no randomness (i.e. unpredictability) in the universe when you “know everything about” it. No need to introduce multiverse.

12. The Physicist says:

The statistics of entangled particles and the idea of “one universe” (a universe with one state) are inconsistent. The “multiverse” is just a way of talking about a universe with many states (both large and small scale). The language here is a little weak: we need more words than just “universe”.
Even if you know everything about a state, the best you can do is find the probability of measuring it in a particular (different) state. For example, if I know that a photon is definitely diagonally polarized, then I can say that the chance of it being measured in a horizontal state is 50%. But, unfortunately, there’s no way to bump that to 0% or 100%.

13. Andy says:

I didn’t read through the whole article just skimmed bits but I did pick up one part where the experiment is described.

3) Randomly change the angle of the polarizers after the photons are emitted.

But how do you randomly change the angle of the polarizers. In the future Scientists may be able to scan your brain and again take in all other atmospheric/physical variables and predict exactly where you will move the polarizer too. So it’s not truly random, only random if you believe that you are moving the polarizer to a random location. My argument may be a bit off here as I didn’t read that much into everything.

I don’t believe there is such a thing as true randomness. Only at this point in time it may appear as though some things are. If we had any true randomness then i think the universe would collapse from a knock-on-effect.

14. The Physicist says:

Any and every manner by which you determine how to randomly orient the polarizers works. This is an addressed concern.
That said, you have a point. It could be that, somehow, the universe is conspiring against us at every turn to make it seem as though things are random. That is, it may be that all of the results of the experiment that demonstrate the fundamental randomness of things may have been written in the stars (so to speak) at the beginning of time. However, predictions based on the quantum mechanics (which involves, among other things, fundamental randomness) seem to hold up.
Fate or not!

15. This blew my mind. Before reading this, I thought there was always a slight chance that one answer of another would come up. Damn, was I wrong!

16. gherkin says:

I’m not a physicist, but reading this makes me question things:

1) The Reality Assumption implies choice, which has inherent randomness in it, which results in it being disproven. Choosing to do the test or not as well as choosing to look at the results or not. Why is the choice of the tester not included in this experiment? If the tester actually “chose” then there is randomness, regardless of the test outcome. Instead, wouldn’t the particles that make up the state of the tester and the experiment also part of the experiment? Deciding to run that test seems like it should be the experiment, not the experiment itself.

2) “However, predictions based on the quantum mechanics (which involves, among other things, fundamental randomness) seem to hold up.” – We are a part of the system and therefore interact with it. The randomness is caused by our measuring it and interacting with it, which may indeed be no random choice to do so. Wouldn’t it be safer to say “We do not have the capacity to measure and record all required variables without influencing them, and therefore must account for the apparent randomness our lack of information provides” ? The absence of data does not imply absence of data.

17. Miske50 says:

After 20 years of studying the chaos theory I am deeply convinced that true randomness and stohasticity do not exist in our universe. Common sense dictates that in causal universe it is not possible. What appears to be random/stohastic is only a very very complex behaviour of a non-linear system. Classical example is coin flipping. It is not random at all – it is just a nonlinear system with large movements consisting of levers made of muscles, tendons and bones, i.e. a classical chaotic sytem, but deteministic.

18. Ernesto Zoffmann Rodríguez says:

Hey , i need to understand this experiment for a schoo project, could someone please tell me what
a
b
c
C
represent?
Thanks!

19. Andrew says:

“That is, if it is physically and mathematically impossible to predict the results, then the results are truly, fundamentally random.”

The idea behind hidden variable theory is that everything is causal, even if we yet fail to understand how. Indeed, the march of time has shown that every single phenomenon we previously thought to be random was actually causal.

If something is mathematically impossible to predict, that doesn’t logically mean that it is not the result of causal interactions that we cannot yet know (which is what Einstein was trying to say in the first place).

For all its complexity, this equation by Bell fails simple logical analysis. What is the reality he is proposing exactly, that in certain situations, things suddenly occur with no relation to the rest of the universe? To me, all he seems to prove is that it is not possible to know everything, which really has nothing to do with whether everything that occurs is causal or not.

20. Dave says:

Here’s the one reason I am not inclined to believe that true randomness exists:

If true randomness existed at the quantum level, how could anything beyond the quantum level be deterministic? If things were truly random, we would never be able to predict anything. And yet, I can predict where a ball will be when it is thrown, I can predict the time it will take to bring an of a given size to temperature T given other information.

If the microuniverse isn’t deterministic, then the macrouniverse would be just as random.

21. The Physicist says:

Beautiful point!
But as unintuitive as it is, while individual things may be random, an aggregate of them can be pretty predictable (when taken together), which is exactly what the macrouniverse is. There’s even an old post that talks about this.

22. Jim Nazium says:

It seems to me that there is something wrong with this argument. When you say “the chance the measurements of the entagled photons are the same”, it seems equivalent to “the chance the two polarizers are less than 90 degrees apart”. You then calculate a positive value for C(0,90), which is clearly wrong, therefore some premise is wrong, therefore rejcet the reality assumption. I’d be more inclined to reject C(a,c) >= x+y-1, which doesn’t seem to follow from the argument you presented. Also, this argument doesn’t seem to require any experimental evidence, it’s just a mathematical derivation. What did I miss?

23. The Physicist says:

Having the polarizers less than 90° apart doesn’t mean that they’ll measure the same result. The experimental evidence is what lead to the fact that $C(x,y) = \cos^2{(x-y)}$, which includes the fact that orthogonal polarizers always yield opposite results ($C(0,90) = \cos^2{(90)} = 0$). The $P(a=c)\ge P(a=b) + P(b=c) - 1$ is mathematically derivable (derived in the post in fact!).

24. Jim Nazium says:

I’m still missing something. Suppose I have a square with area = 1, and I throw darts at it so they are uniformly distributed. Then for any two numbers a and b that are both between 0 and 90 (degrees), suppose I mark a sub-square whose sides have length cos(b-a). The sub-square has area cos^2(b-a). Define C(a,b) as the probability that a dart lands in this sub-square, so C(a,b) = cos^2(b-a). Doesn’t this satisfy all the premises of this argument? That is, we should have C(0,45) >= C(0,22.5) + C(22.5,45) -1, but in fact we have C(0,45) = 0.5 while C(0,22.5) + C(22.5,45) -1 = 0.7. That’s a contradiction. We didn’t have to experiment with entangled photons, or assume counterfactual definiteness … ?

25. The Physicist says:

What you’ve rediscovered is that C(a,b) = cos^2(b-a) is a classically impossible correlation. That is, there is no simple probability distribution on the two particles that supports it, in much the same way that there’s no probability distribution on the two coordinates for where the darts hit. And yet, when we actually do the experiments on entangled pairs this is exactly the correlation we always get.
That’s why counterfactual definiteness is dropped; entangled photons can’t be described in terms of being in a particular state, the way a dart must be in a particular state, even if that state is unknown!

26. loopantenna says:

It seems the proof of the randomness of the universe is in the quantum entaglement phenomenon, right? Because for other ordinary quantum phenomena the appearent randomness comes from the unertainty principle that precludes the exact knowledge of each variable in the mathematical model. Basically each ordinary particle is in a state that we can never measure precisely hence the description as a super position of states. Whereas for entagled particles there is no precise state as if the pair hops randomly from one possible state to another in an actual super position of states. Does this interpretation makes sense?

27. R9 says:

I thought entangled photons were polarised at 90 degrees to each other? Wouldn’t that mean C(0°, 90°) = 1?

28. Diem says:

The gravy experiment can be easily disproved with some logic. It is called a circular argument, and you stated that you had to change the angle of both polarizers to two RANDOM angles, and this proves randomness in the end.

An example: say I have a box. I can prove I have a box because I have an object and the object is the box. You are proving something with the same thing you are trying to prove, and that doesn’t work out, no matter how much math to stuff in there.

29. Slobodan Mitrovic says:

A typical example of randomness and stochasticity is coin flipping by hand and thumb.
But, this is, actually a non-linear system of levers (bones) and tendons and muscles with large (like pendulum) displacements and therefore it can behave chaotically. It is per se very sensitive to initial conditions. Hence, in my mind, it is actually a truly deterministic chaotic system. I am 100% convinced that true randomness cannot exist in a causal universe, as randomness inherently assume that the things are happening for no apparent reason!

30. mananjay says:

If there is no randomness… then I don’t belive life can exist.. then we are just a complex system .. what is a will it i belive is a substantial example of randomness..
If will do not exist then life is just acomplex robot.

31. Robin says:

Another way of looking at it, is that indeed everything in the end is causal, but every possible causal line of events will happen on a grander scale, in a giant network. We just cannot know ahead which nodes in the network seem be to be experienced by us (our local universe/time line), because of the lack of hidden variables to tell us so (a concept which after 60 years still holds up).
Using the same logic, there are then instances of ourselves in other universes/time lines that also undergo all other causal nodes. From that vantage point, arguing about randomness seems a moot point, because, which one is the “real you”?
Any thoughts?

32. Brian says:

Could it not be said that what we label as random are merely phenomena for which we don’t yet understand all of the laws which govern their behavior? Therefore, they may be regarded as random from a practical standpoint insofar as our ability to predict behavior, but not truly random as their behavior is in fact governed by properties and laws which we have yet to discover.

To say that something is truly random would be to say that its behavior is not subject to physical law.

33. Slobodan Mitrovic says:

Excellent comment!
In a universe that we comprehend today, there can be no true randomness as it is not subject to physical law, i.e. things would happen for no reason. If there is a reason for an event then it is not truly random as the result can be predicted, as a consequence, all in accordance with physical laws. If it cannot be predicted then it is not subject to physical laws as we know them hence it cannot exist in our universe.

34. David Peters says:

If you don’t believe that the best theory we have for how things exist at the quantum scale predicts true randomness than you have misunderstood the argument. If you can setup an experiment based on quantum probability such as the decay of a uranium atom then you cannot predict when that will happen with a certainty of 100% regardless of how much you know about the universe. The inability of 100% prediction does not come from a lack of instruments or accuracy. Just as the world is round and the sun shines, these are observations with a testable outcomes that fit the model. We cant measure or predict with any certainty because the certainty does not exist at any level.

If on the other hand we were to take a thousand or maybe a million particles, measure their decay and plot the times on a graph we will see that there is a ‘tendency’ for a lot of the times to occur close to a specific value. In a bell curve possibly. This still highlights the fact that you can predict where most of them will end up, but not all. And the not all is the key here because whilst the chances may be high of the time being in the middle of the curve, there is still no way that you can look at an individual particle and predict its decay with 100% accuracy. I will restate that: it is impossible to accurately predict the outcome and the impossibility is an inherent part of the universe and provides for a model that represents experimental outcome at a level where no other model can.

The outcome we see in a bounced basketball is because there are so many particles and most of them will be close to the centre of the bell curve in terms of measured position or velocity. This means that any particles dramatically removed from the centre of the curve are so few that there effect is practically unmeasurable. How ever this does not mean that there is absolutely no spooky quantumm stuff happening in your basketball. It is simply swamped by the probability curve.

35. jason roberts says:

okay, some of this was good and all , and thanks for that . but , true randomness does not exist except for in our minds . the term “random” is quintessentially a misnomer . how is that not completely obvious ?

36. Barry Simon Detector says:

Your “proof” is utterly specious. By confusing measurements with angles you invalidly apply to the correlation function a well-known mathematical inequality and unsurprisingly obtain a contradiction, then blame it on the assumption that more than two experiments can be performed. When Jim Nazium demonstrates that the inequality fails to hold, and hence the contradiction follows, even in the case of just two experiments (with angular differences of 22.5 and 45 degrees), you blame this on the correlation being “classically impossible,” yet there was no assumption of “classicality” and his objection holds in the non-classical case, so perhaps you’ve disproved the negation of the reality assumption instead.

37. Nick says:

Bell’s theorem depends on the assumption of free will.

You may find interesting to read these two posts by G. t’ Hooft, as well as his papers.
http://physics.stackexchange.com/questions/34217/why-do-people-categorically-dismiss-some-simple-quantum-models
http://physics.stackexchange.com/questions/30065/why-do-people-rule-out-local-hidden-variables

38. Pingback: Random observations | Peoria Pundit

39. Denny says:

I like the clarity of your explanation of the reality hypothesis. Unfortunately, your overall conclusion is misleading. You say that the test of Bell’s Theorem shows the reality hypothesis to be false and then, as a footnote, point out that locality and realism can’t both be true and that the majority choose locality. OK, let’s be forthright about it. I choose reality and a good argument can be made that relativity is compatible with both. Incidentally, the usual narrative is that Hidden Variables was demonstrated to be wrong by the experiment. In fact, Bohm’s non-local Hidden Variable theory, formulated long before the experiment was done, was vindicated right along with the usual interpretation of quantum theory.

A word about Bohm. All too many physicists have failed to examine the implications of quantum theory and Bohm grappled with it, first writing a book supporting the usual probabilistic interpretation, and then developing a mathematically consistent hidden variable theory when many argued it couldn’t be done. Like Einstein and others, he did not like the rank empiricism of quantum theory and for good reason. While almost the whole physics community considered EPR an empty philosophical question, it was literally the work of 3 guys, Einstein, Bohm and Bell, all fierce critics of the usual interpretation, that actually interrogated the issue – first raising the question and then coming up with a theorem, which lead to the understanding of entanglement we have now. Finally, big props for Bohm for refusing to testify before McCarthy. The creeps are those inside and outside the physics community who did.

40. Slobodan Mitrovic says:

Bouncing basket ball is an archetypal example of a DETERMINISTIC CHAOTIC SYSTEM!
I strongly support Jason’s excellent comment that true randomness can only exists in our minds and can be, as such, treated with abstract math (mathematical statistics) but this math DOES NOT EXPLAIN REALITY. This does not tell WHY!
To me, cosmos/universe is only an infinitely complex phenomenon, but deterministic. Complexity grows infinitely if not stopped by some external force but it always remain deterministic. After Big Bang, at which cosmos was a simple point, cosmos expanded (as we see now CHAOTICALLY) and is still expanding at an accelerated rate. Seemingly, nothing to stop its ever growing complexity and eventually, the birth of life (and intelligence) as its most complex form. But it seems that the dark energy would prevail, it is the force that will simplify the ever growing complexity by an eternal accelerated expansion and eventually lead us all into simple big freeze.
Things that come out of the blue for no apparent reason or cause (i.e. randomly) could not be subject to physical laws, but we see that cosmos was.

41. ybot says:

There can be many hidden physical variables that could affect a “random” experiment.
Doing the same experimente, some of us pay attention an know some conditions and ignore others.
My first impression is that we often know few compared to what we should to be 99% confidente of an outcome.
Probability calculus helps to test trials and define the chance to be certain of a statement.

42. Alexander says:

I was looking for an authoritative but readable article on randomness to link to on my family-friendly blog. This one fits the bill – with one small but crucial exception: the profanity in paragraph 3. Would appreciate it if you would edit with a family-friendly synonym. Thanks.

43. Alexander says:

Thank you!


text:
# Figure 2

 AMPT model predictions are shown as hollow symbols in the (top) and (middle) panels. (top) Comparison of observables SC(4,2) (red filled squares) and SC(3,2) (blue filled circles) to theoretical model from. Solid lines indicate the predictions with constant $\eta/s$, while the dashed lines indicate predictions for different parameterizations of $\eta/s$ temperature dependence (labeled in the same way as in Fig.1 in). (middle) Results divided by $\left< v_m^2\right>\left< v_n^2\right>$. (bottom) Comparison to MC-Glauber using wounded nucleon (WN) and binary collisions (BC) weights.