text:
# 37. A nurse administers the prescribed inhaled medication then auscultates the patient’s lung after the medication...

###### Question:

37. A nurse administers the prescribed inhaled medication then auscultates the patient’s lung after the medication has finished. In what phase of the nursing process would the nurse’s action of auscultating the lungs occur? A. Planning B. Assessment C. Evaluation D. Diagnosis

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format: plain


text:
# listOMLDataSets

0th

Percentile

##### List the first 5000 OpenML data sets.

The returned data.frame contains the data set id “data.id”, the “status” (“active”, “deactivated”, “in_preparation”) and describing data qualities. Note that by default only the first 5000 data sets will be returned (due to the argument “limit = 5000”).

##### Usage
listOMLDataSets(number.of.instances = NULL, number.of.features = NULL,
number.of.classes = NULL, number.of.missing.values = NULL, tag = NULL,
data.name = NULL, limit = 5000, offset = NULL, status = "active",
verbosity = NULL)
##### Arguments
number.of.instances

[numeric(1) | numeric(2)] If not NULL, subsets the entries with respect to the given values or, if a vector of length 2 is passed, the given ranges.

number.of.features

[numeric(1) | numeric(2)] If not NULL, it subsets the entries with respect to the given values or, if a vector of length 2 is passed, the given range.

number.of.classes

[numeric(1) | numeric(2)] If not NULL, subsets the entries with respect to the given values or, if a vector of length 2 is passed, the given ranges.

number.of.missing.values

[numeric(1) | numeric(2)] If not NULL, subsets the entries with respect to the given values or, if a vector of length 2 is passed, the given ranges.

tag

[character] If not NULL only entries with the corresponding tags are listed.

data.name

[character(1)] Name of the data set.

limit

[numeric(1)] Optional. The maximum number of entries to return. Without specifying offset, it returns the first 'limit' entries. Setting limit = NULL returns all available entries.

offset

[numeric(1)] Optional. The offset to start from. Should be indices starting from 0, which do not refer to IDs. Is ignored when no limit is given.

status

[character] Subsets the results according to the status. Possible values are {"active", "deactivated", "in_preparation"}. Default is "active".

verbosity

[integer(1)] Print verbose output on console? Possible values are: 0: normal output, 1: info output, 2: debug output. Default is set via setOMLConfig.

##### Value

[data.frame].

##### Note

This function is memoised. I.e., if you call this function twice in a running R session, the first call will query the server and store the results in memory while the second and all subsequent calls will return the cached results from the first call. You can reset the cache by calling forget on the function manually.

Other listing functions: chunkOMLlist, listOMLDataSetQualities, listOMLEstimationProcedures, listOMLEvaluationMeasures, listOMLFlows, listOMLRuns, listOMLSetup, listOMLStudies, listOMLTaskTypes, listOMLTasks

Other data set-related functions: OMLDataSetDescription, OMLDataSet, convertMlrTaskToOMLDataSet, convertOMLDataSetToMlr, deleteOMLObject, getOMLDataSet, tagOMLObject, uploadOMLDataSet

##### Aliases
• listOMLDataSets
##### Examples
# NOT RUN {
datasets = listOMLDataSets()
tail(datasets)
# }

Documentation reproduced from package OpenML, version 1.7, License: BSD_3_clause + file LICENSE

### Community examples

Looks like there are no examples yet.


text:
# 13

Decision TheoryEmbedded AgencyAI
Frontpage

This post is about following type signature, which I call the type of agency: . You can also think of it as consequentialism or doing things on purpose. This post will be a rant with a bunch of random thoughts related to this type signature, and it will likely not make sense. It will also be sloppy and will have type errors, but I think it is worth posting anyway.

First, interpret these arrows as causal arrows, but you can also think of them as function arrows. This is saying that the causal relationship from to causes to happen. Think of as an action and as the goal. The reason that happens is the fact that it has as a consequence. There are not normally exponential objects like this in Bayes' nets, but I think you can modify so that it makes sense. (I'm not sure that this works, but you have a Cartesian closed category with nodes that are the nodes in your Bayes net, and add small number of morphisms from product nodes to individual nodes, corresponding to the functions in the Bayes' net. The acyclicness of the Bayes' net roughly corresponds to this category being thin. Then you can consider having other types of morphisms that can keep the category thin.)

If you have a game between two agents with action nodes and , with utilities and . The game implements a pair of functions and . We can Curry these functions and think of them as and . Bringing in the agency of both players leads to cycle. This cycle does not make sense unless the agency arrows are lossy in some way, so as to not be able to create a contradiction.

Fortunately, there is another reason to think that these agency arrows will be lossy. Lawvere's Fixed Point Theorem says that in a Cartesian closed category, unless has the fixed point property, you cannot have a surjective function , in Set this is saying that if has more than one element, you cannot have an injection . i.e. The agency arrows have to be lossy.

Also, notice that Argmax, takes in a function from some set to , and returns an element of the domain, , so Argmax has type .

This one is a bit more of a stretch, but if you look at gradient descent, you have some space , you have a function . The gradient can be thought of as a function from infinitesimal changes in to infinitesimal changes in . Gradient descent works by converting this gradient into a change in . i.e. Gradient descent looks kind of like .

# 13

New Comment

It's interesting to notice that there's nothing with that type on hoogle (Haskell language search engine), so it's not the type of any common utility.

On the other hand, you can still say quite a bit on functions of that type, drawing from type and set theory.

First, let's name a generic function with that type . It's possible to show that k cannot be parametric in both types. If it were, would be valid, which is absurd ( has an element!). It' also possible to show that if k is not parametric in one type, it must have access to at least an element of that type (think about and ).

A simple cardinality argument also shows that k must be many-to-one (that is, non injective): unless B is 1 (the one element type),

There is an interesting operator that uses k, which I call interleave:

Trivially,

It's interesting because partially applying interleave to some k has the type , which is the type of continuations, and I suspect that this is what underlies the common usage of such operators.

I found a paper about this exact sort of thing. Escardo and Olivia call that type signature a "selection functional", and the type signature is called a "quantification functional", and there's several interesting things you can do with them, like combining multiple selection functionals into one in a way that looks reminiscent of game theory. (ie, if has type signature , and has type signature , then has type signature .


text:
# Open Source Friday

## Working on open projects

Posted by on August 16, 2018

## Reproducible Analysis

#### A pythoninchemistry cross-post

Open source Friday in an initiative started by GitHub, which pushes people and businesses to contribute to open source code once a week.

As part of a personal experiment, to grow my skills as a programmer while improving code that exists in the open-source community, I am going to try and spend (at least some of) my Fridays working on open source programming projects. Obviously as a PhD student I have a bit more freedom with my time than the typical programmer, so I understand that not everyone is able to do this.

I am going to try and (to some extent) document the open-source work I do, and maybe this will allow me to offer insight to contributing to open source code.

This started on last Friday (2018-08-10), and I am going to briefly write about one of my contributions.

### My first pull request (in mdanalysis)

mdanalysis is a python library for molecular dynamic simulation analysis. I wanted to contribute something after getting back in touch with one of the maintainers (@richardjgowers) whom I had met a conference a few years ago.

Richard suggested issue #1897 as a good place for a newbie to work on. This particular issue was related to the PDB file format, in particular the fact that if there are more than 99999 atoms in the PDB trajectory it starts to count in base36. This seemed initially to be an easy fix, simply if the number is greater than 99999 convert from hex using int(number, 36).

However, after implementing some tests, which failed, I discovered that as with many things about the PDB file format, it wasn’t a very easy fix. It turns out, as discussed in the pull request, the PDB file format doesn’t fall back to standard base36, instead implementing a case-sensitive hybrid36. Basically the way that the PDB atom indexing is as follows:

• counts from 1 to 99999,
• once it reaches 100000, there are only 5 columns available for numbers so it goes to A0000, then A0001, …, A000Z, etc.,
• this then counts until ZZZZZ, which is equivalent to 43770015,
• then it uses the lower case versions, a0000,
• this continues to zzzzz, meaning that in 5 columns the pdb can count to 87440031.

So late Friday afternoon, after I included some tests that hopefully immortalised some of my office mates in the code, the pull request was accepted and I became an official contributor to mdanalysis.

Hopefully this is a series I will continue when I have time.


text:
## 1. Divisor sums

1. ### Moments of divisor functions in short intervals

#### Problem 1.1.

[S. Lester] Evaluate $$\int_X^{2X} \left| \sum_{x\leq n\leq x+H} d_k(n) \ - \ \mbox{main term}\right|^{2\ell}\,dx$$ for different ranges of $H$.
1. Remark. [K. Soundararajan] Find connections with the paper of Keating, Rodgers, Roditty-Gershon, and Rudnick.
• #### Problem 1.2.

[B. Rodgers] Some coefficients of the Ramanujan tau function appear as coefficients of the piecewise polynomial in the work of Keating, Rodgers, Roditty-Gershon, and Rudnick. Is this a coincidence or not?

Cite this as: AimPL: Moments of zeta and correlations of divisor sums, available at http://aimpl.org/zetamoments.


text:
# Number System Aptitude Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Number System Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. Find the cube root of the number $$13824$$?

1. $$23$$
2. $$24$$
3. $$27$$
4. $$26$$

Answer: (b) $$24$$

Solution: Cube root of the number $$13824$$,$$\sqrt[3]{13824} = \sqrt[3]{24 \times 24 \times 24}$$ $$= (24)^{3 \times \frac{1}{3}} = 24$$

1. What will come in place of question mark $$(?)$$ for the given expression $$4^{?} = 64 \times 64 \times 64$$?

1. $$4$$
2. $$8$$
3. $$9$$
4. $$5$$

Answer: (c) $$9$$

Solution: Given expression $$4^{?} = 64 \times 64 \times 64$$ $$4^{?} = (4)^3 \times (4)^3 \times (4)^3$$ $$4^{?} = 4^{3 + 3 + 3}$$ $$4^{?} = 4^9$$ $$? = 9$$

1. Solve the expression $$(11)^2 \times (11)^{108} \div (11)^{107}$$?

1. $$1331$$
2. $$1333$$
3. $$1330$$
4. $$1311$$

Answer: (a) $$1331$$

Solution: Given expression, $$(11)^2 \times (11)^{108} \div (11)^{107}$$ $$= (11)^2 \times (11)^{108 - 107}$$ $$= (11)^2 \times (11)^1 = (11)^{2 + 1} = (11)^3 = 1331$$

1. Find the square root of the number $$4225$$?

1. $$54$$
2. $$65$$
3. $$67$$
4. $$58$$

Answer: (b) $$65$$

Solution: Square root of the number,$$\sqrt[2]{4225}$$ $$= \sqrt[2]{65 \times 65} = 65$$

1. Solve the expression $$64 + [{4 \times 5 (3 + 2)}]$$?

1. $$165$$
2. $$164$$
3. $$163$$
4. $$161$$

Answer: (b) $$164$$

Solution: Given expression, $$64 + [{4 \times 5 (3 + 2)}]$$ $$64 + [{4 \times 5 \times 5}]$$ $$64 + [100] = 164$$

1. Find the remainder of the expression $$\frac{35 \times 45}{8}$$?

1. $$6$$
2. $$5$$
3. $$1$$
4. $$7$$

Answer: (d) $$7$$

Solution: According to basic remainder theorem,$$\frac{35 \times 45}{8}$$ $$= \frac{3 \times 5}{20} = \frac{15}{8} = 7 \ (Remainder)$$

1. Find the factors of the number $$150$$?

1. $$10$$
2. $$14$$
3. $$12$$
4. $$16$$

Answer: (c) $$12$$

Solution: $$150 = 2 \times 3 \times 5 \times 5$$ $$= 2^1 \times 3^1 \times 5^2$$ then factors of composite number $$= (1 + 1)(1 + 1)(2 + 1)$$ $$= 2 \times 2 \times 3 = 12$$

1. Solve the given expression $$(7)^{11} \div (7)^{10}$$?

1. $$5$$
2. $$8$$
3. $$7$$
4. $$2$$

Answer: (c) $$7$$

Solution: Given expression,$$(7)^{11} \div (7)^{10}$$ $$= (7)^{11 - 10} = (7)^1 = 7$$

1. What will come in the place of question mark $$(?)$$ for the given expression $$7^{?} = 49 \times 343 \times 2401 \div 343$$?

1. $$2$$
2. $$4$$
3. $$6$$
4. $$8$$

Answer: (c) $$6$$

Solution: Given expression $$7^{?} = 49 \times 343 \times 2401 \div 343$$ $$7^{?} = (7)^2 \times (7)^3 \times (7)^4 \div (7)^3$$ $$7^{?} = (7)^{2 + 3 + 4 - 3}$$ $$7^{?} = 7^6$$ $$? = 6$$

1. Solve the given expression $$(4 + 5 \times 2) \times {16 \div 2 (5 - 1)}$$?

1. $$25$$
2. $$20$$
3. $$28$$
4. $$21$$

Answer: (c) $$28$$

Solution: Given expression,$$(4 + 5 \times 2) \times {16 \div 2 (5 - 1)}$$ $$= 14 \times {16 \div 8} = 14 \times 2 = 28$$


text:
Chapter 11, Problem 13E

### Intermediate Accounting: Reporting...

3rd Edition
James M. Wahlen + 2 others
ISBN: 9781337788281

Chapter
Section

### Intermediate Accounting: Reporting...

3rd Edition
James M. Wahlen + 2 others
ISBN: 9781337788281
Textbook Problem
258 views

# Bailand Company purchased a building for $210,000 that had an estimated residual value of$10,000 and an estimated service life of 10 years. Bailand purchased the building 4 years ago and has used straight-line depreciation. At the beginning of the fifth year (before it records depreciation expense for the year), the following independent situations occur: 1. Bailand estimates that the asset has 8 years’ life remaining (for a total of 12 years). 2. Bailand changes to the sum-of-the -years’-digits method. 3. Bailand discovers that the estimated residual value has been ignored in the computation of depreciation expense. Required:For each of the independent situations, prepare all the journal entries relating to the building for the fifth year. Ignore income taxes.

To determine

Prepare necessary journal entries of Company B for each of the given situations.

Explanation

Depreciation expense: Depreciation expense is a non-cash expense, which is recorded on the income statement reflecting the consumption of economic benefits of long-term asset on account of its wear and tear or obsolesces.

Straight-line depreciation method: The depreciation method which assumes that the consumption of economic benefits of long-term asset could be distributed equally throughout the useful life of the asset is referred to as straight-line method.

Double-declining-balance method: The depreciation method which assumes that the consumption of economic benefits of long-term asset is high in the early years but gradually declines towards the end of its useful life is referred to as double-declining-balance method.

Prepare necessary journal entries of Company B for each of the given situations as follows:

Situation 1: Building has remaining useful life of 8 years.

 Date Account Title & Explanation Debit ($) Credit($) Depreciation expense (2) 15,000 Accumulated depreciation-Building 15,000 (To record the depreciation expense incurred at the end of the fifth accounting year)

Table (1)

• Depreciation expense is a component of stockholder’s equity. It decreases the value of stockholder’s equity by $15,000. Therefore, debit depreciation expense account with$15,000.
• Accumulated depreciation is a contra asset, and it decreases the value of asset by $15,000. Therefore, credit accumulated depreciation account with$15,000.

Working note (1):

Calculate the accumulated depreciation expense for four years of building under straight line method.

Accumulated depreciation expense  = Cost of acquisaition Salvage valueEstimated useful life ×4 years=$210,000$10,00010 years×4=$80,000 Working note (2): Calculate the annual depreciation expense for remaining life. Depreciation expense = (Cost of acquisaitionAccumualted depreciation (1)) Salvage valueRemaining useful life =($210,000$80,000)$10,0008 years=$15,000 Situation 2: Company B has changed the deprecation method (sum of the year’s digit).  Date Account Title & Explanation Debit ($) Credit($) Depreciation expense (5) 34,286 Accumulated depreciation-Building 34,286 (To record the depreciation expense incurred at the end of the accounting year) Table (2) • Depreciation expense is a component of stockholder’s equity. It decreases the value of stockholder’s equity by$34,286. Therefore, debit depreciation expense account with $34,286. • Accumulated depreciation is a contra asset, and it decreases the value of asset by$34,286. Therefore, credit accumulated depreciation account with $34,286. Working note (3): Calculate the depreciable base of asset. Depreciable base = [Purchase priceAccumulated depreciaiton] –Residual value =($210,000$80,000(1)) –$10,000=\$120,000

Working note (4):

Calculate the denominator of the fraction for sum-of-the-year’s digit.

Sum-of-the-digits = 6×(n+1)2=6×(6+1)2=6×72=21

Note: In this case, the remaining useful life (6 years) of the asset is considered as “n”(10 years4 years)  to calculate the numerator of the fraction

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text:
# Arduino Mega 2560 like board but with more ram

I am making a POV RGB LED Globe that plays animations on its surface. Currently I am using a Arduino Mega 2560 with 256k of Flash Memory. My program and the boot loader take up 20k of memory leaving me with 236k.

There are 64 RGB LEDs using 8 bits of brightness that change color 64 times a revolution. $\left( 64 \cdot 64 \cdot 3 \cdot 8 \right) \div 1024 = 12\mbox{ kB}$ == 12288 BYTES

Each frame of animation takes up 12288 BYTES (12k) of memory. That only leaves me with 19 Frames of animation. Good but not great, at 10 frames a sec thats only 2 secs of animations. I was hoping for ~64 frames of animation or 768k RAM. (768+20=788k total)

I have tried a few load 'on the fly' and steaming methods (SD, Wireless) but they can't keep up with the data required while streaming the data to the LEDs. I am going to try using a simple compression method next but I don't have high hopes that the system will be able to uncompress the data while steaming the color information out to the LEDs.

I have currently tested 10 frames of animation using the Arduino Mega 2560 and it works. Kind of, there are still some bugs. But I don't really want to switch off an Arduino board if I can help it.

My questions are:

1. Are there pre-built Arduino boards with more ram?
2. Can I add RAM to an existing Arduino board?
3. Suggest a different Dev board thats is VERY similar to the Arduino with a c++ compiler.

Edit: I am using the "PROGMEM" keyword to store my animations in the Flash memory not the SRAM. http://www.arduino.cc/playground/Learning/Memory

• (64 * 64 * 3 ) / 1024 != 12288. Also, we have gorgeous LaTeX support through MathJax,, so that could be $\left( 64 \cdot 64 \cdot 3 \right) \div 1024 = 12\mbox{ kB}$ (right click to view source) – Kevin Vermeer May 10 '11 at 2:31
• By my calculations, your data requirements are only about 40 kB/second, which should be pretty easy with on-the-fly loading, SD cards, or (some) wireless networking options, as well as serial Flash memory (I think you want Flash, not RAM, for this). Could you explain how you ascertained that they can't keep up? – Kevin Vermeer May 10 '11 at 3:03
• 12kB a frame and 10 frames a sec for a total of 120 kB a sec. \left( 12 \cdot 10 \right) = 120\mbox{ kB} But you are right that does not make sense. The SD card should be able to keep up at that rate. I will edit the question with my experience with the SD card. – Steven smethurst May 11 '11 at 4:41
• @Steven - Surround math with \$ signs to invoke LaTeX formatting. However, I'm missing the point where you need to multiply by three - Do you have 8 bits for each color of the RGB LEDs? – Kevin Vermeer May 11 '11 at 12:36
• @reemrevnivek - Yes, 8 bits of brightness per color on the RGB LEDs. Sorry I should have been more clear. In other words it takes 3 BYTES or 24 bits for every single RGB LED. – Steven smethurst May 12 '11 at 23:43

Try using an I2C SRAM chip.

Maybe a 256 kbit N256S0830HDA @ 3V from AMI fits the bill: http://www.modtronix.com/products/components/ram/n256s0830hda.pdf

See http://arduino_related.livejournal.com/1414.html for a tutorial.

• Thats fantastic thank you, I think this is exactly what I am looking for. – Steven smethurst May 11 '11 at 4:46

Data compression may be an option. You probably don't NEED every iteration of the 255/255/255 data set. Even sticking with 256 colors would be more than enough for what you are trying to do.

Then each LED only needs 1 byte of information to store it's brightness setting, increasing the amount of frames available to 57, which would be pretty close to the 64 you were looking for.

If there's enough I/O pins left, you may try to connect the SRAM chip and manually control it. You could try to use extra latches to store the address, to reuse the same set of pins for both data and address busses (e.g. MCS-51 based systems used to use 74HCT373 to latch the lower part of the address bus, see here). This approach should allow for faster data transfers compared to I2C SRAM chips.

Note, that the memory of the size you need will most likely come in a TSOP package or similar.

• The requirement looks to be actually for flash rather than RAM, and there are plenty of those in 8-pin SOIC packages which will be easy to solder - just only solder one corner pin and verify alignment before soldering any others. And after doing a few of those, TSOP will not be intimidating. – Chris Stratton Nov 12 '14 at 16:44

You can't really expand the addressable RAM; the chip has whatever it has inside it and doesn't present a memory bus outside. You could add an SD shield such as this one and use it as a solid-state disk, much as you're doing now with the on-board flash.

Some people use several Arduinos (or other processor), each one running a small section of a LED display.

For example, perhaps you could use 8 Arduinos, each one responsible for 8 RGB LEDs of your total of 64 RGB LEDs. (While more Arduino Megas would work, this also opens up the option of using smaller, lighter, lower-cost Arduinos, such as the Arduino Pro Mini, that have far fewer pins than the Arduino Mega).

This allows you to scale up your display to any number of LEDs, far more than any one Arduino could handle.

• This is still not cost effective. Microcontroller on-chip storage is orders of magnitude more expensive than external memory chips made with a fabrication process optimized for storage. So while there is indeed a point at which adding additional processors makes sense (or where it makes sense to have a universal product manufactured at scale you can sometimes use more than one of), it makes sense to load each board up with additional storage before you add more boards. – Chris Stratton Nov 12 '14 at 16:42
• @ChrisStratton: I agree that "there is indeed a point at which adding additional processors makes sense". Feel free to edit this answer if there is a better way to express that statement. – davidcary Nov 12 '14 at 17:16


text:
### On Einstein’s resolution of the twin clock paradox

On Einstein’s resolution of the twin clock paradox
C. S. Unnikrishnan

Gravitation Group, Tata Institute of Fundamental Research, Mumbai, India

In: Current Science, Vol. 89, No. 12, p. 2008 (2005)

Abstract: Einstein addressed the twin paradox in special relativity in a relatively unknown, unusual and rarely cited paper written in 1918, in the form of a dialogue between a critic and a relativist. Contrary to most textbook versions of the resolution,
Einstein admitted that the special relativistic time dilation was symmetric for the twins, and he had to invoke, asymmetrically, the general relativistic gravitational time dilation during the brief periods of acceleration to justify the asymmetrical aging. Notably, Einstein did not use any argument related to simultaneity or Doppler shift in his analysis. I discuss Einstein’s resolution and several conceptual issues that arise. It is concluded that Einstein’s resolution using gravitational time dilation suffers from logical and physical flaws, and gives incorrect answers in a general setting. The counter examples imply the need to reconsider many issues related to the comparison of transported clocks. The failure of the accepted views and resolutions is traced to the fact that the special relativity principle formulated originally for physics in empty space is not valid in the matter-filled universe.

… den ganzen Artikel lesen

———————————-

Siehe auch YouTube-Video vom Autor: Galileo, Time and Space: Gravity affecting clocks

.

### 3 Antworten zu “On Einstein’s resolution of the twin clock paradox”

1. Frank Wappler

C.S. Unnikrishnan wrote („On Einstein’s resolution of the twin clock paradox“, Curr. Sci. 89, 2009; 2005):

> […]
> Counter-examples
> […]
> Consider an atomic clock B than is synchronized with another clock A during an initial encounter. Since the relative velocity between the clocks is known

… provided it is specified and agreed how to determine values of „velocity between clocks“ in the first place (see below) …

> and since the velocity of light is a constant […]

… a statement for which the notion „velocity“ appears prerequisite, too.

> […] Now consider the situation when A sends a signal at his time tA to B to freeze the clock and then decelerate, on receiving the signal. Since A can estimate the time at which the signal will reach B, he can also freeze his clock at that time […]

Such an estimate would surely have to be consistent with the relation

(1a)
speed at which B moved away from A
/
speed of light
==
duration of A from indicating the statement of the signal, „tA„, to indicating the freeze
/
duration of A from indicating the encounter with B to indicating the freeze;

or

(1b)
1 –
(speed at which B moved away from A
/
speed of light)
==
duration of A from indicating the encounter with B to indicating the statement of the signal, „tA
/
duration of A from indicating the encounter with B to indicating the freeze.

(Note that there is a particular relation between A’s indication of freeze and B’s indication of freeze which can be described in greater detail, involving the notion of „simultaneity“.)

> The readings can be compared after B comes to rest relative to A

… how this might be accomplished would first need to be specified and agreed, of course …

> or after a round trip […]

The „frozen readings“ of A and B may be compared when they meet again, too; just as directly as they were „synchronized“ at their initial encounter.

Of course, comparing incidental „readings“ should be especially interesting if the corresponding durations could be compared as well, obtaining the value of ratio

(2)
duration of B from indicating the encounter with A to indicating the freeze
/
duration of A from indicating the encounter with B to indicating the freeze.

> In fact, the possibility of switching off or freezing the clocks makes all the standard resolutions inadequate.

This may well be debatable; but it surely doesn’t stand in the way of further analyzing the stated (counter-)example:

As well as the ratio

speed at which B moved away from A
/
speed of light

from (1a) or (1b) above, the ratio

speed at which A moved away from B
/
speed of light

may be measured as well.

If both were found equal throughout the trial under consideration, as may be assumed in the following, then this value is abbreviated as „\beta;“. Thus:

(3)
speed at which B moved away from A
/
speed of light
==
speed at which A moved away from B
/
speed of light
==
\beta;.

Further, trying to obtain relations corresponding to (1a) or (1b), but involving duruations of B, one may arrive at

(4a)
(speed at which B moved away from A
/
speed of light)
==
duration of B from some particular indication, „tB„, to indicating the freeze
/
duration of B from indicating the encounter with A to indication „tB„;

or

(4b)
1 +
(speed at which B moved away from A
/
speed of light)
==
duration of B from indicating the encounter with A to indicating the freeze
/
duration of B from indicating the encounter with B to indication „tB„.

There, the indication „tB“ of clock B was surely (in B’s judgement) after B’s indication of the encounter with A, and before B’s freeze.

Also, surely, this indication „tB“ of clock B should have some particular relation to A’s indication „tA“ of stating the signal.

This relation can be described in greater detail, involving the notion of „simultaneity“; thereby also motivating the following relation (5) between the durations appearing in (1b) and (4b):

(5)
duration of A from indicating the encounter with B to indicating the statement of the signal, „tA
/
duration of B from indicating the encounter with B to indication „tB
==
duration of B from indicating the encounter with A to indicating the freeze
/
duration of A from indicating the encounter with B to indicating the freeze.

(Short of spelling out the motivation of (5) completely, note that defining its denominators involved explicit appeal to the notion of „simultaneity“; while defining its numerators did not.)

Finally inserting (1b) and (4b), with abbreviation (3):

(6a)
duration of B from indicating the encounter with A to indicating the freeze
/
duration of A from indicating the encounter with B to indicating the freeze
==
(
(1 – β) *
duration of A from indicating the encounter with B to indicating the freeze
)
/
(
duration of B from indicating the encounter with A to indicating the freeze
/
(1 + β)
);

or

(6b)
duration of B from indicating the encounter with A to indicating the freeze
/
duration of A from indicating the encounter with B to indicating the freeze
==
Sqrt[ 1 – β^2 ]
==
duration of A from indicating the encounter with B to indicating the statement of the signal, „tA
/
duration of B from indicating the encounter with B to indication „tB„.

If comparison of the „readings“ mentioned above corresponds to this ratio of durations then it is said that the two clocks had kept (sufficiently) equal „design“ throughout the trial; or otherwise not.

Also note that the quantity

(7)
(1 – Sqrt[ 1 – β^2 ]) *
duration of A from indicating the encounter with B to indicating the freeze

may by some be referred to as „net proper time difference“.

2. Frank Wappler

p.s.
Besides my reply above (#1, 29. Januar 2013 um 02:28) there is already a certain thread of correspondence directly arising from C.S. Unnikrishnan’s article on record in Current Science, and publicly available from the same website; by

Ø.G. Grøn („Relativistic resolutions of the twin paradox“, Curr. Sci. 92, 416; 2007),

K. Hazra („On the resolutions of the twin paradox“, Curr. Sci. 95, 706; 2008), and

C.S. Unnikrishnan („Response“, Curr. Sci. 95, 707; 2008).

In my humble opinion, together they exhibit all-too-common flaws:
Grøn being not (quite) pedagogical enough to appear relevant;
Hazra and Unnikrishnan being not (quite) inquisitive enough to be conclusive.

3. Mark Lofts

Der Schwerpunkt auf der Auflösung der Zwillingsparadox ist vom Einsteins Aufsatz, um die Simpel zu verwirren und Kritiker mit pedantischen unwichtigen bezüglich einiger ‘durch Gravitation’ Beschleunigung Einzelheiten zu verblüffen, abgelenkt und gedeckt. Warum unwichtig? Wegen eine entscheidende Tatsache. Die auf gleichbleibender Geschwindigkeit Reise bloß in einen STR-Effekt sich verwickelt, im Gegensatz zu übrigen der Fahrt Teilen, innerhalb welcher außerdem auch ein von Beschleunigung und Langsamerwerden mit Umkehrung bzw. ‘Gravitation’ ATR-Effekt erzeugt wäre.

Also die beide Effekte sind unabhängig voneinander, weil die auf gleichbleibender Geschwindigkeit Entfernung von Gravitation oder Trägheit nicht unterworfen sein kann, außer wenn die Simpel von irreführender mathematischer Argumentation, woran die Zahlen sich aufgehoben werden, irreführt sind und dementsprechend beschloßen!

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