Do absolute space and time exist?

Question: Einstein's theory proved that most properties of a body are relative to something and are not absolute. I am still wondering if physicists still think that there is a way that Newton's absolute space and time still exist?
If they do, how are the ways that we prove their existence? and how do they exert their influence (effects)?
Comment: Possible duplicate of [Is spacetime absolute?](https://physics.stackexchange.com/q/183577/)
Comment: I think you may be confused about what "relative" and "absolute" mean.  Newton's equations do not depend on any "absolute" spatial coordinate system.
Comment: The [principle of relativity](https://en.wikipedia.org/wiki/Galilean_invariance) has been known since Galileo's day.  Einstein's theory rescued the principle of relativity after Maxwell cast doubt on it with his [theory of electromagnetic waves](https://en.wikipedia.org/wiki/James_Clerk_Maxwell#Electromagnetism).
Comment: What does "proved to a great existence" mean?
Comment: Of course the answer depends on what you mean by the words.   There are cosmological models with a preferred global time coordinate.  Would you count these as models in which "absolute space and time exist"?
Answer: Generally, no, Newton's concept is now seen as a limit case of Relativity - you only get (exact) Newtonian mechanics when the velocity and masses of all objects involved reach zero. There have been various attempts to tweak Newtonian dynamics to make it match reality (known as MoND, Modified Newtonian Dynamics), but without success, there's always some observation that doesn't match.
Not that Newtonian dynamics isn't useful, it's a much simpler way to calculate as long as you're dealing with a situation where the deviation from Relativistic dynamics is negligible, but in the strictest sense it is always "wrong" to some extent.
Answer: Spacetime is a consequence of Einstein's false constant-speed-of-light postulate:
"Special relativity is based on the observation that the speed of light is always the same, independently of who measures it, or how fast the source of the light is moving with respect to the observer. Einstein demonstrated that as an immediate consequence, space and time can no longer be independent, but should rather be considered a new joint entity called "spacetime."  http://community.bowdoin.edu/news/2015/04/professor-baumgarte-describes-100-years-of-gravity/
So spacetime doesn't exist, space and time do. Einsteinians know that:
Nima Arkani-Hamed (06:09): "Almost all of us believe that space-time doesn't really exist, space-time is doomed and has to be replaced by some more primitive building blocks."  https://www.youtube.com/watch?v=U47kyV4TMnE
"Rethinking Einstein: The end of space-time. It was a speech that changed the way we think of space and time. The year was 1908, and the German mathematician Hermann Minkowski had been trying to make sense of Albert Einstein's hot new idea - what we now know as special relativity - describing how things shrink as they move faster and time becomes distorted. "Henceforth space by itself and time by itself are doomed to fade into the mere shadows," Minkowski proclaimed, "and only a union of the two will preserve an independent reality." And so space-time - the malleable fabric whose geometry can be changed by the gravity of stars, planets and matter - was born. It is a concept that has served us well, but if physicist Petr Horava is right, it may be no more than a mirage. Horava, who is at the University of California, Berkeley, wants to rip this fabric apart and set time and space free from one another in order to come up with a unified theory that reconciles the disparate worlds of quantum mechanics and gravity - one the most pressing challenges to modern physics."  https://www.newscientist.com/article/mg20727721-200-rethinking-einstein-the-end-of-space-time/

Do absolute space and time exist?


Is there a word or expression when you are thinking too much?

Question: When you are thinking too much and creating situations that will never happen when solving a problem, is there an expression or word for this? What is a better way to express this idea without using "thinking too much" or is this fine?
Comment: please give a context in which you would use the word or expression.
Answer: The word is "overthinking".

Overthink—to think too much about (something); to put too
much time into thinking about or analyzing (something) in a way
that is more harmful than helpful
— Merriam-Webster
Answer: You can use the world overthink.

Overthink (verb): to think about something too much, in a way that is not useful:
Eg: I overthink everything and worry too much.

[Cambridge English Dictionary].
You can also use the sentence "think too much" to express the same thing.
Answer: exaggerating things would be one option. Also, many of the OVER- verbs would also do the job. I personally like overdramatize because it reminds me of the funny expression "to be a drama queen".
Other idiomatic expressions you might like are

worry over nothing
make a big deal
press the panic button
blow things out of proportion


three way interaction in lmer

Question: I have two continuous variables (cfreq, and LanPro), and one catergorical variable (cond_aud, as shown in the picture). The summary of a lmer model shows a three way interaction of aud (EA and NoA) x LanPro x cfreq. I used emmip function to make an interaction plot, which shows that the difference between different levels of "LanPro" is smaller when "cfreq" is higher in EA than in NoA, but I am not sure whether my understanding is correct and don't know how to interpret this interaction and test the significance for the difference. Any advice would be much appreciated.
Answer: A few things are evident from the plot:

In all cases association of cfreq with the outcome is negative and linear
Increasing values of LanPro are associated with lower values of the outcome.
The differences between the levels of LanPro are lower at higher values os cfreq, however this is more pronounceed in cond_aud = EU than the other two groups.

The latter point seems to be the focus of the question. Looking at the 1st and 3rd plot, the differences between the levels of LanPro are lower at higher values os cfreq, however this appears to be very small, whereas in the middle plot it is far more pronounced. That is the two-way interaction between cfreq and LanProshould be greater at the EA level of cond_aud than the other two levels. That means we would be looking for a meaningful 3-way interaction term involving the EA level of cond_aud. Specifically, if CA is the reference level for cond_aud then the model would estimate two 3-way interaction terms: cfreq:LanPro:cond_audEA and cfreq:LanPro:cond_audNoA, and we would expect the former to be meaninfully larger than the latter
Assuming that you have repeated measures within subjects, or some other kind of clustering (otherwise why are you using lmer?), then you can explore this with the following model:
Y ~ cfreq * LanPro * cond_aud + (1|subject)

This will estimate, assuming CA is the reference level for cond_aud:

the overall intercept which will be the estimates mean response for CA when cfreq and LanPro are zero.

4 main effects (one each for cfreq and LanPro and 2 for each non-reference level of cond_aud) for each of the variables which will tell, for cfreq and LanPro the association of a 1 unit change with the outcome when LanPro is at it's reference level (ie the respective slopes), and for LanPro the asssociation of the difference between its reference level and the other two levels, with the outcome, when cfreq and LanPro are zero.

5 2-way interactions: cfreq:LanPro, cfreq:cond_audEA, cfreq:cond_audNoA, LanPro:cond_audEAand LanPro:cond_audNoA, which can be interpreted as the interaction involving those variables when the 3rd variable is zero (or for cfreq:LanPro when cond_aud is CA.

2 three way interactions which can be interpreted as above to answer the research question
Comment: You are welcome. It is conditional on `cond_au` being at its reference level (and `cfreq` being at 0). This interpretation of main effects holds when it is involved in an interaction. However, it is obvious from the plots that it also applies unconditionally. You don't need a test.
Comment: Hi would you be able to help with a similar issue here: https://stats.stackexchange.com/questions/529850/significant-interaction-but-no-significant-simple-effects-in-lmer? Really appreciate your expertise.
Comment: Your explanation is very helpful. Based on your model above (except that I use NoA as baseline), does a main effect of LanPro indicate that the outcome decreases when the values of LanPro increases regardless of the cfreq and the cold_aud conditions? Or does it only apply to the reference level (i.e., when conditions_aud is NoA)?

three way interaction in lmer


Form Always prompting to update

Question: I have a form (2007 filler form template - deployed on SharePoint 2010, opens in infopath) and when it opens I am always getting a prompt saying a 'new version is available and to update the form'. Does anyone know a possible cause and/or fix for this as I am clueless.
Answer: Have you tried clicking on the prompt to update the form? Give this a try. I am not entirely sure how InfoPath pulls forms down from the server but, it may have a cache that stores copies of these forms.
Comment: Yes, once I update the form updates and runs fine. The problem is that I get this everytime the form is opened and the form needs to be updated. I figured this would be a caching problem - would this be a server side cache or user sided?
Comment: My guess would be user. Could be an problem if you are using InfoPath Filler 2007 trying to commuincate with SP 2010

Form Always prompting to update


How to multiply UV coordinates?

Question: I have a plane simply UV mapped and using a texture for the albedo.
I want to have the same UV mapping used for the normal maps, but multiplied by 5 or 10, I don't know how much exactly yet.
Is it possible to not make another mapping on another channel, and just feed to the normal maps a UV coordinates multiplier?
Answer: Use a MAPPING vector node to manipulate the values that are fed to it with an INPUT UV MAP node, and in the SCALE fields change the scale of the texture: the values act as multipliers.
Below is the actual node setup.
Comment: Just for completeness, you can also use a Converter > Vector Math node. Or you can split the coordinates with convert > separate XYZ and then use a Convert > Math node for individual control on each of the vectors, and recombine using Convert > Combine XYZ .

How to multiply UV coordinates?


Backup jailbreak

Question: I had a jailbreak on my iPod 5 (6.1.2) but then I upgraded it to 6.1.3 and I had backed it up when it was jailbroken and when i had restored it it asked if I wanted to use a backup and I used the one I did while I was jailbroken and now I have battery percentage on my iPod 5. How can I make sure it's jailbroken or not. Thanks
Answer: Battery percentage is a normal feature. Based on what you described, you are not jailbroken. But the way to verify it is see if you have the Cydia app installed.
Comment: Battery percentage is not normal on an iPod touch. It's set in a .plist, though, so if you restore from a backup that had it, you'll still have battery percentage.

Backup jailbreak


Problem 1 Chapter 11 Wald

Question: I'm currently trying to solve problem 1, Chapter 11 of Wald, General Relativity.
The request is to derive from the condition
$$
\tilde\nabla_a \tilde\nabla_b \Omega=0\text{ at }\mathscr I^+,
$$
where $\Omega$ is the conformal factor defining the compactification of spacetime, the fact that in the chart $(u, \Omega, \theta, \phi)$, the behavior of  $\tilde g_{uu}$, $\tilde g_{u\theta}$ and $\tilde g_{u\phi}$ is $O(\Omega^2)$ near $\mathscr I^+$.
We also know that at $\mathscr I^+$ the metric is
$$
ds^2=d\Omega du + d\theta^2 + \sin^2\theta d\phi^2.
$$
My try: since $\Omega$ is a coordinate we have $\partial_a \Omega=\delta^\Omega_a$; thus, at points of $\mathscr I^+$,
$$
0=\tilde\nabla_a \tilde\nabla_b \Omega=\partial_a \delta^\Omega_u-\tilde\Gamma^c_{ab}\delta^\Omega_c=0-\tilde\Gamma^\Omega_{ab}\implies \tilde\Gamma^\Omega_{ab}=0.
$$
Now, since by compatibility $0=\tilde\nabla_a\tilde g_{bc}=\partial_a\tilde g_{bc}-\tilde \Gamma^d_{ab}\tilde g_{dc}- \tilde \Gamma^d_{ac}\tilde g_{bd}$, we have (at $\mathscr I^+$).
$$
\partial_\Omega \tilde g_{uu}=2\tilde\Gamma^d_{\Omega u}\tilde g_{du}= 2\tilde \Gamma^\Omega_{u\Omega}=0
$$
This is fine, since $\partial_\Omega\tilde g_{uu}=0$ at $\mathscr I^+$ implies by Taylor expansion in $\Omega$
$$
\tilde g_{uu}(\Omega)=\tilde g_{uu}(0)+0 + O(\Omega^2)=O(\Omega^2)
$$
as requested. However, a similar trick does not work for $\tilde g_{u\theta}$ and $\tilde g_{u\phi}$, since
$$
\partial_\Omega \tilde g_{u\theta}=\tilde\Gamma^d_{\Omega u}\tilde g_{d\theta} + \tilde\Gamma^d_{\Omega \theta}\tilde g_{du}= \tilde\Gamma^\theta_{\Omega u}\tilde g_{\theta\theta}+\tilde \Gamma^\Omega_{\Omega\theta}= \tilde\Gamma^\theta_{\Omega u} + 0;
$$
which at first sight does not vanish, at $\mathscr I^+$. I also tried using the mixed version $\tilde\nabla_a \tilde\nabla^b \Omega=0$, but this avails nothing as we get
$$
\partial_a\tilde g^{b\Omega}=-\tilde \Gamma^b_{au}
$$
and one cannot extract further information from this as far as I get.
Please, lend a hand!
EDIT: Perhaps I have found a way out.
The metric is $u$-independent at $\mathscr I^+$, since it has the form given in the second formula above for any $u$, provided $\Omega=0$. Hence $\partial_u\tilde g_{ab}=0$ and the compatibility condition yields
$$
0=\tilde\nabla_u \tilde g_{ab}=0-\tilde \Gamma^c_{ua}\tilde g_{cb} - \tilde \Gamma^c_{ub}\tilde g_{ca} \implies \tilde\Gamma^c_{ua}\tilde g_{cb}=-\tilde\Gamma^c_{ub}\tilde g_{ca}, \text{ at }\mathscr I^+.
$$
Now, substituting in the above computation,
$$
\partial_\Omega\tilde g_{u\theta}=\tilde\Gamma^c_{u\Omega}\tilde g_{c\theta}=-\tilde\Gamma^c_{u\theta}\tilde g_{c\Omega}.
$$
Now, also $\partial_A \tilde g_{u\Omega}=0$ at $\mathscr I^+$, for any angle $A=\theta,\phi$ yielding
$$
\tilde\Gamma^c_{Au}\tilde g_{c\Omega}+\tilde\Gamma^c_{A\Omega}\tilde g_{cu}=0
$$
and hence
$$
\partial_\Omega\tilde g_{u\theta} = \tilde\Gamma^c_{\Omega\theta}\tilde g_{cu}= \tilde\Gamma^\Omega_{\Omega\theta}=0
$$
and similarly for $\phi$.
I am still a little perplexed though, since this strategy also applies to getting $\partial_\Omega\tilde g_{u\Omega}=0$, which is not written down by Wald.
EDIT 2: Silly me! Even if $\partial_\Omega\tilde g_{u\Omega}=0$ is true, but one cannot infer $\tilde g_{u\Omega}=O(\Omega^2)$ near $\mathscr I^+$ because $\tilde g_{u\Omega}(\Omega)=\tilde g_{u\Omega}(0)+O(\Omega^2)=1+O(\Omega^2)$.
Answer: Perhaps the most handy way is to proceed as follows: up to $\tilde\Gamma^\Omega_{ab}=0$ everything is fine. Now, write down this identity at $\mathscr I^+$ to get, since $\tilde g^{\Omega a}=\delta^c_u$ and $\partial_u\tilde g_{ab}=0$,
$$
\frac{1}{2}\left(\partial_a\tilde g_{ub}+\partial_b\tilde g_{ua}\right)=0.
$$
Now, fix $a=\Omega$ and pick $b=u,\theta,\phi$; then $\partial_b \tilde g_{u\Omega}=0$ by inspection and we have
$$
\partial_\Omega\tilde g_{ub}=0,\text {at }\mathscr I^+.
$$
Then, by Taylor expansion in $\Omega$
$$
\tilde g_{ub}(\Omega)=\tilde g_{ub}(0)+O(\Omega^2)=O(\Omega^2)
$$
for $b=u,\theta,\phi.$
Comment: For those reading this in the future, it might he helpful for you to explain how exactly this proves the claim, it's not immediately obvious.

Problem 1 Chapter 11 Wald


Create a digital signature hash via Gnosis/Safe Multisig Wallet

Question: I have a multisig wallet setup at https://safe.global/. I need to prepare a TX which calls a contract, and one of the parameters of the function I am calling needs a digital signature of the (tightly encoded) passed-in function parameters. How do I construct this digital signature hash parameter? If I would have access to the private key, I could construct it myself (programmatically, e.g. via hot wallet), but since I don't have access to the multisig wallet's private key, naturally, how can I do the same in this case?
Comment: Does the contract support ERC-1271? https://eips.ethereum.org/EIPS/eip-1271 If not then it's impossible
Answer: I dont think you can execute this function , because multisig dont have a private key , its a Smart Contract Wallet . So , it cant be done . One thing you can try , you can pass the signature of your metamask wallet which is the owner of the multisig wallet.
I hope this helps


Allowing cookies from a specific site in Safari

Question: I'm a new Mac user and I have a financial site which says I must allow cookies...and I'm okay with that, I just don't know how to do that. I've looked in my settings and preferences but it appears I have to allow cookies for everyone or no one at all. Would someone be able to help me? I'd rather not add additional programs if I can avoid it... OS X must be able to do this, right?
Comment: What version of OS X and Safari are you using?
Comment: Any solutions to this that work with the current macOS and Safari?  (The solution below doesn't work (and lacks detail too) as the Privacy tab has changed for the worse.  Extensions, I guess?
Answer: Go to Safari's settings and from there you can allow certain sites to leave cookies.  There are several other settings worth investigating here too.
To get to settings, from the Menu click Safari > Preferences... and go to the Privacy tab.
Comment: I realize it's been 6 years, but is there an updated version of this answer? It doesn't seem to work any more.
Comment: @OldBunny2800 If you're getting told that cookies must be allowed, the website should correct it. Most browsers are standardizing blocking 3rd party cookies, which was the original question here.
Comment: The website I'm trying to use is a two-factor authentication site. It has a checkbox for "remembering" my browser for 30 days, or basically storing a cookie on my system. This process works fine in Google Chrome, but doesn't work in Safari, meaning I have to use the 2FA every single time I log in.


Are Bayesian Networks parametric or non-parametric models?

Question: After searching Wikipedia, I found that there are both parametric Bayesian models and non-parametric Bayesian models. What about Bayesian Networks? When building up a Bayesian Network model, I don't need to give much prior information. Does that mean Bayesian Networks are non-parametric models?
Comment: "I don't need to give much prior information" Ironically this is the exact distinction you are missing. On what do you set your priors? A finite set of path weights and residual variances? You're looking parametric. On an infinite space of probability models? You're veering into non-parametric.
Comment: No. Bayes nets can be parametric. It only has to do with the models used to relate edges. Non-parametric Bayesian regression models to estimate paths in the graphical model make the Bayesnet a non-parametric Bayes net. If you just use linear regression (inverse gaussian prior on residual variance for regressions between nodes), it's a parametric Bayesnet procedure.
Answer: Based on those definitions:
Nonparametric:Algorithms that do not make strong assumptions about the form of the mapping function are called nonparametric machine learning algorithms. By not making assumptions, they are free to learn any functional form from the training data.
EX: k-Nearest Neighbors, Decision Trees
Parametric:Assumptions can greatly simplify the learning process, but can also limit what can be learned. Algorithms that simplify the function to a known form are called parametric machine learning algorithms.
EX: Logistic Regression, Linear Discriminant Analysis
And in my knowledge I can: Yes, Bayesian Belief Networks with discrete variables are indeed nonparametric, because they are probabilistic models based conditional dependencies between their variables. I worked with those models (discrete) in ecology.
BUT,Bayesian networks with continuous variables make strong assumptions about their data (I believe one of them is that the variables follow a Gaussian distribution). These models themselves are parametric.
Comment: Where are these definitions coming from? I don't think these are accurate taken either from stat or ML literature.


Collision of Rolling body

Question: When a rolling body collides with a wall elastically,its torque is zero.But direction of velocity of particle is reversed.So will the angular momentum be conserved?
Comment: I think it should remain conserved ,in intervals,before the ball.colloide,during ball.collide,after collision ,and direction angular momentum is along angular velocity vector,which has changed after collision
Comment: But mvr direction changes then how angular momentum be conserved
Comment: Angular velocity direction is same after collision but direction of v is changed then how in this situation angular momentum be conserved
Comment: Yes angular momentum will be conserved torque delta t=change in angular momentum since elastic collision k.e will be conserved, hence angular momentum will be conserved
Answer: If there is no friction between the body and the ground then the body will rebound with the same speed as it had before hitting the wall and will rotate with an unchanged angular velocity.
Even though there will be relative motion between the body and the ground as there is no friction no mechanical energy is dissipated at the point of contact so such motion continues without change.
If there is friction between the body and the ground then the frictional force on the body will be in such a direction as to reduce the velocity of the centre of mass of the body whilst also providing a torque which will reduce the angular velocity of the body and eventually reverse the direction of rotation such that the no slip condition is satisfied.
During this time the body loses mechanical energy and heat is produced.
—-
There are two components to the angular momentum of the body.
The first component is due to the rotation of the body about its centre of mass and it is sometimes called the spin angular momentum.
The spin angular momentum does not change during the collision.
The second component is due the translational motion of the centre of mass and this is sometimes called the orbital angular momentum.
This is the angular momentum whose value is $mvr$ about a point on the ground.
The the body hits the wall it experiences a horizontal force due to the wall which reverses the direction of the linear momentum $mv$.
That force exerted by the wall can also be thought of a torque about a point on the ground acting on the body.
This torque reverses the direction of the orbital angular momentum $mvr$ about a point on the ground.
Comment: Here friction is absent.But my doubt is that since direction of velocity reverses after collision wilk the direction of mvr changes?
Comment: @MuhammedYoosuf The angular momentum cannot change because, in the absence of friction, there is no torque applied to body.
Comment: My doubt is that when velocity reverses after collision mvr will change to -mvr? Is it wrong or correct
Comment: @MuhammedYoosuf The body will still be spinning in the same direction and with the same magnitude as before the collision.
Comment: Will the angular momentum produced due to translation to be conserved?
Comment: @MuhammedYoosuf In response to your comment I have added to my answer.
Answer: Basic definition for angular momentum to be conserved,there should be absence of sudden external impulsive force which can change the momentum,since there is no impulsive torque before ,during,after collision,since it is elastic collision hence k.e energy remain conserved hence velocity,remain same ,hence after all these argument ,it clearly describe that in given situation angular momentum remain conserved.
Answer: Linear moment isn't conserved when hitting a wall because the wall can't move and thus its momentum (which is zero) can't change.
Conversely, angular momentum isn't conserved, because angular momentum is just linear momentum at a distance. The interpretation of angular momentum conservation is that of conservation of the geometry (line in space) where linear momentum acts through.
In this case, before the impact, the center of rotation is at the contact point $R$ below the center of mass, and the momentum axis at a distance $\tfrac{I_c}{m r}$ above the center of mass (here $I_c$ is the mass moment of inertia of the object).
What happens next depends on if friction is considered or not. In the frictionless case, the ball is no longer rolling after the bounce. Its center of rotation is now $R$ above the center of mass, and the momentum axis $\tfrac{I_c}{m r}$ below the center of mass.
With friction, that situation is by far more complex, but suffice to say the angular momentum isn't conserved either.
There is a special coefficient of friction value of $\mu = \tfrac{I_c}{m r^2}$ which makes the ball instantly roll after the bounce, although it lifts up from the floor as friction causes a vertical impulse in addition to the regular bounce.

Collision of Rolling body


WWII documentary series with end credits that used a diorama

Question: I remember a series about World War II where the end credits were shown over a camera that panned over a diorama that show the progress of the war that started with normal scenes that slowly transitioned by the end of the credits to bombed out cities. I thought it might be the classic BBC The World at War, but the episodes I've seen on Youtube don't have the diorama.
Comment: The World at War does show a top-down view of a moving diorama that sounds like what you are describing.  https://youtu.be/m0mrOeJwXBI
Comment: At the end of The World at War, episode freezes the last frame of the historical footage they are using and shows the credits over it.The show I'm thinking of had a long miniature country side and towns like one might build for a train set that showed the passage of time as they panned front to back. Thank you I love that music and Lawrence Olivier's voice.
Answer: I think you might be thinking of People's Century https://www.youtube.com/results?search_query=People%27s+Century
Comment: I think you can see how I might confuse it with The World at War. Both these series were excellent. Thank You


How to make swirled edge circle

Question: I am trying to make a swirled edge circle like this for a logo in illustrator. What would be the best method to use?
Thank you!
Comment: It's not 1 circle. it's multiple circles/ellipses.
Answer: I guess you are not going to copy an existing logo, but need a little resembling circle

Draw a circle with no fill. Delete a part of the line.

Make rotated and moved copies. You can also draw some different curves like in your example.

Make width variations  with the width tool or apply an art brush. Your own example has complex profiles which need art brushes.

Have color variations to make the turns distinguishable. I have only 3 different grays, but there's no limit how rich colors are possible. You may need to expand the strokes to filled areas to apply complex effects.
Custom art brushes can be made easily. In the next image in the left there's 3 partial circles like above and the applied (simple) art brushes are shown in the right.
Comment: thank you! That is very helpful!

How to make swirled edge circle


deploying in salesforce from Dev Org to Production Org

Question: Can i create an outbound change set in a developer organization and deploy it to production ?
I have created my Project in my developer edition can i deploy it to production without using sandbox org is it possible or do i need sandbox edition to deploy my project to production ?
Answer: No, you can not able to do this.
To deploy change set, you need to have connection between two orgs.
In this case, your Developer Edition will not have connection with Production Org.
Deployment Connections

A deployment connection is required between two Salesforce orgs to
  send change sets from one org to another. You can't create deployment
  connections between arbitrary orgs. Instead, you create connections
  between all orgs affiliated with a production org. For example, if you
  have a production org and two sandboxes, a deployment connection is
  created between production and each sandbox. Also, a deployment
  connection is created between the two sandboxes.

If you don't have connection between two org then you use a Force.com IDE to deploy metadata.
Comment: So do i need sandbox org first and then i will be able to deploy to production right @Devendra ?
Comment: You can use a Eclipse IDE to deploy metadata from Developer Edition to Production Org. to use a change set you need to have connection between two orgs. Generally, we have multiple developer Sandboxes from which we do the deployment to Prodution org.
Comment: @ALEX_RILEY Yes. Once you have sandbox you can create a connection between Production and related Sandbox Org. And, then can able to deploy changesets.
Answer: No you cannot use a Change Set from a Developer Org to another org you would have to use a tool like the Salesforce Deployment tool or one of the IDEs that support this.
Options are explained here by Salesforce
Sandboxes are a separate type of Salesforce Org from the Developer environments which are on the test.salesforce.com url rather than salesforce.com.


I can't deploy my Upgradeable contract using truffle plugins

Question: I'm trying to deploy the following smart contract:
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.4;

import "@openzeppelin/contracts-upgradeable/token/ERC20/ERC20Upgradeable.sol";
import "@openzeppelin/contracts-upgradeable/proxy/utils/Initializable.sol";

contract MonstersOnTheWayUp is Initializable, ERC20Upgradeable {
    function initialize() initializer public {
        __ERC20_init("Promethium", "PRM");
    }
}

I'm using this script for migration:
// migrations/3_deploy_upgradeable_box.js
const { deployProxy } = require('@openzeppelin/truffle-upgrades');

const MonstersOnTheWayUp = artifacts.require("../contracts/MonstersOnTheWayUp");

module.exports = async function (deployer) {
  await deployProxy(MonstersOnTheWayUp, [""], { deployer, initializer: 'initialize' });
};

But when I try to run the command:
npx truffle deploy --reset --network rinkeby

I get the following error:
Error: The requested contract was not found. Make sure the source code is available for compilation

Why is this happening?
This is my project directory:
Comment: Does this helps: https://forum.openzeppelin.com/t/how-to-verify-a-contract-on-etherscan-bscscan-polygonscan/14225 ?
Answer: const MonstersOnTheWayUp = artifacts.require('MonstersOnTheWayUp.sol');


How do I do a clean upgrade of roundcube on Debian buster?

Question: I have a Debian buster machine on an AWS instance. /etc/apt/sources already has the backport url in it.
I have installed Roundcube already using the stable release. I want to upgrade this package to the testing release with as little trouble as possible so I'm reading over this: https://backports.debian.org/Instructions/
However, I'm confused by the language in the Sloppy distributions section:

To guarantee a clean upgrade path from a previously stable
distribution to the next stable it is not allowed to upload packages
from the current testing to that backport distribution. To allow some
newer packages for those systems we create so called sloppy
distributions.
You can use it exactly as the current backports distribution, just add
another line to your sources.list and install new packages via apt-get
-t stretch-backports install

I'm not sure if this applies to my situation or not. Or does this apply only after I upgrade from buster to bullseye?
And can do I just do apt-get -t buster-backports install roundcube? Or do I have to completely remove the old version of roundcube first?
Answer: The text you cited is only relevant for package distributors.
If you want to use the backport version of roundcube, you are good to go, by simply using your stated command, without the need to remove the installed version.


In a proof that is reliant on proven theorems, does one assume the reader's familiarity with said theorems, or explicitly include their logic?

Question: In composing a proof that is reliant on proven theorems, does one simply assume the reader's familiarity with said theorems, or does one explicitly include their logic in the new logic?
Comment: Do whatever enhances the clarity of your writing.
Comment: But do include a reference to a published proof of the known theorems that you use.  Preferably a page-number reference, rather than a whole book that has the result in it somewhere.
Comment: There's no need to prove known theorems. If it is not something well known, you can give a reference to the paper where it's proved.
Answer: It's a judgment call, depending, in part, on the audience you are addressing, and your purpose for writing a proof, or a paper with proofs, etc. And so, as Asaf answers, the answer to your question: "depends on the context."
If you are writing proofs for classes (assigned or recommended), for example, it is best to error on the side of caution and be more explicit rather than less, and include more rather than less. One rarely loses marks for including more information than needed, whereas it is common to lose marks for not including enough. So when justifying a statement in such a proof: you can refer to proofs/theorems in your class text provided they have been covered in class. You can often do so by referencing the number/letter used in the text or in class, or by referring to it by using a commonly-used name of a particular theorem. E.g., "By the Fundamental Theorem of Arithmetic, it follows that...".
It is often helpful, when writing proofs for classes, to also justify an assertion by referring to definition(s) that your assertion relies upon: e.g. "by the definition of congruence modulo n, we know that...", without needing to restate the entire definition, unless you are introducing an unfamiliar term that you plan to use in your proof.
These suggestions are just as much for your (present and future) benefit as they are for your audience, and/or for demonstrating mastery of the material relevant to the statement you are asked to prove. Of course, for proofs required in coursework or in an exam, I'd highly recommend that you consult your instructor on this matter. If preparing for preliminary examinations, you can access and work out some problems from past prelims, and discuss with your advisor whether your proofs/solutions are adequate: (What should I have included? What could I have excluded? etc).
Attachment I:
You might find the following exposition written by Dr. John M. Lee helpful:

Some remarks on writing mathematical proofs (pdf), also available for downloading from John Lee's website. It provides very sound advice on writing math proofs, discussing, among other things, considerations with respect to the intended audience, etc.

Internet Bibliography:
If interested, you might want to explore the links available at

How to Write Math Proofs.

See also the previous post: What can a writer assume in a proof?.
Comment: I just realized that you have a habit of referencing me (to my answers/comments) and later removing the reference. Strange. :-)
Comment: @Asaf - I re-referenced you: I had edited my first sentence, and ended up omitting the reference, not intending to have done so...so I just re-edited the first sentence to re-insert the reference. Besides, I'm not altogether clear when others "want" for me to reference their answers, (as they may or may not "endorse" any content I add) - And as for my comments, I sometimes delete them later when it seems they no longer apply or if they add only noise, and such. :-)
Comment: Oh, I don't mind you not referencing my answers/comments/whatever. I just find it peculiar as it wasn't the first time I saw that you began by referring to something I'd said and then removed that reference.
Comment: @Asaf I like to credit where credit is due (trying not to "hijack" answers)...But I'm never really certain that references are always welcome by the referenced, so I'm a bit ambivalent about when to or not to reference, that's all. Any way, I didn't take you to be complaining, or as being critical - not at all.
Comment: I don't mind being cited! :-)
Answer: It depends on the context. My masters thesis, for example, began with saying that I expect the reader to be familiar with forcing. I am not expecting the reader to be familiar with other topics which are relatively common, though. These topics are fully explained in my thesis.
When writing something one can usually foresee who is going to read the text, and what the readers are expecting to see written.
Over-detailed writing is very hard to read; but under-detailed writing is very hard to understand. It takes time and experience to find the balance. Consult an advisor or referee regarding your actual text.
This may also be affected by where you are sending the text. When writing a very short article (say 3-4 pages) it's fine to just quote results, but when writing a book it's expected to prove them as well.
Answer: With all proofs (in whatever format), you must first identify your intended audience. That will allow you to decide what you can / cannot assume, and how to include in the right amount of detail.
Are they students in your field that are interested in understanding the buildup of the theory? Are they 'professionals' in related fields who believe in the truth of your statements and only want to understand implications of your results? Are they high schoolers who want to know why quintics can't be 'solved' and only know the basics of group theory?
Having additional proofs is preferable. You can helps others along by labeling them as Theorem 1, Proof (or even placed in the Appendix), so if readers are familiar, they can skip it.


Python Twisted - Server communication

Question: I'm having a bizarre issue. Basically, the problem I have right now is dealing with two different LineReceiver servers that are connected to each other. Essentially, if I were to input something into server A, then I want some output to appear in server B. And I would like to do this vice versa. I am running two servers on two different source files (also running them on different processes via & shellscript) ServerA.py and ServerB.py where the ports are (12650 and 12651) respectively. I am also connecting to each server using telnet.
from twisted.internet import protocol, reactor
from twisted.protocols.basic import LineReceiver

class ServerA(LineReceiver);
   def connectionMade(self):
      self.transport.write("Is Server A\n")
   def dataReceived(self, data):
      self.sendLine(data)
   def lineReceived(self, line):
      self.transport.write(line)

def main():
   client = protocol.ClientFactory()
   client.protocol = ServerA
   reactor.connectTCP("localhost", 12650, client)

   server = protocol.ServerFactory()
   server.protocol = ServerA
   reactor.listenTCP(12651, server)

   reactor.run()

if __name__ == '__main__':
   main()

My issue is the use of sendLine. When I try to do a sendLine call from serverA with some arbitrary string, serverA ends up spitting out the exact string instead of sending it down the connection which was done in main(). Exactly why is this happening? I've been looking around and tried each solution I came across and I can't seem to get it to work properly. The bizarre thing is my friend is essentially doing the same thing and he gets some working results but this is the simplest program I could think of to try to figure out the cause for this strange phenomenon.
In any case, the gist is, I'm expecting to get the input I put into serverA to appear in serverB.
Note: Server A and Server B have the exact same source code save for the class names and ports.
Answer: You have overridden dataReceived.  That means that lineReceived will never be called, because it is LineReceiver's dataReceived implementation that eventually calls lineReceived, and you're never calling up to it.
You should only need to override lineReceived and then things should work as you expect.
Comment: I think I found the solution. I had to keep track of what's connecting to the server such as clients = [] and then appending an instance of the client to that list every time a connection is made. I'm guessing my above code was actually just sending the data to itself.
Comment: Hey, I tried removing the overridden dataReceived but it still has the same behavior. Server B isn't receiving anything from server A with sendLine.

Python Twisted - Server communication


Accessing fine-grained volume control using external keyboard volume wheel

Question: In macOS, the Volume keys change the volume in full steps, but it is possible make fine-grained changes using Option-Shift-Volume{Up,Down}.
I have an external keyboard connected which provides a volume wheel. Moving this wheel modifies the volume very quickly, in full steps. Pressing Option-Shift while moving the wheel is the perfect precision for me.
Is it possible to configure macOS (or the external keyboard) so the internal keyboard behavior isn't affected but the VolumeUp/VolumeDown keys from the external keyboard act in small steps, without pressing Option-Shift?
Otherwise, is it possible to modify the entire behavior so volume control always acts in small steps? (regardless of whether it's the internal or external keyboard)
Answer: You should be able to use Karabiner-Elements to do this. It allows you to customize keyboard actions on a per-device basis. You could set it up to replace Volume Up / Volume Down with Option + Volume Up / Option + Volume Down.


disable second mouse click, that clicked immediately.

Question: My mouse is corrupt. So that sometimes when I click one time, it is clicked two times itself.
Double-click happen when I just clicked once.
How can I disable mouse click, for 0.5 second, that after every mouse click.
Comment: Mice doesn't cost that much, buy a new instead of trying to workaround it with a nasty hack that will end up annoying you. When I doubleclick I'm pretty sure the clicks are far less than half a second apart.
Comment: Does it happen with a different mouse?  I think you're trying to solve a hardware issue with a really bad software patch.
Comment: my touchpad works fine. and also I see this in another Linux desktop(laptop).
Comment: With the same mouse or another one?
Comment: with same mouse
Answer: The most common issue with a defective mouse is the clicking which works on and off or starts to click by itself or double-click when you click.
Any software patch will only waste your time as it would be very annoying to work with and the situation will continue to deteriorate.


Is it technically possible to configure two different SSL certificates for the same domain?

Question: Say I have these URLs:

https://example.org/
https://example.org/criticalpath

I want the first one to be served with a domain validated commercial SSL certificate and the second one with an extended validation SSL certificate. Is it technically possible? Is it supported by SSL/TLS (and HTTPS?) standards and browsers?
I'm not asking in any way if this is even a good configuration (I would go for the EV for all the domain), I just need to know if that configuration is supported. I would really appreciate references to the standards.
Answer: Yes and No.
It's possible to have multiple certs that all have the same domain name on it.  You could have a Comodo example.org cert and an Entrust example.org cert, both of them valid and official, no problem.  I believe some load balancers will rotate which one is used on a per-connection basis, but only in a round-robin fashion.
The No is that you're asking if you can select which cert to use based on the URI (/ vs. /criticalpath).  The problem is, the URI is only available after the SSL connection has been nailed up using one of the two certs.  So you can't really choose which cert to use based on information you can only have after you've chosen the cert.
Now, I'm not going to say 100% it's impossible.  With Apache, for example, you can set the SSLCipherSuite configuration on a per-directory basis, so I wouldn't be surprised if there was some way to force a renegotiation that would involve a new cert.  But I rather suspect it's practically unimplemented even if it is theoretically possible.  (Apache SSLCertificate* is per-server or per-vhost, not per-directory).

Appendix: Musings upon renegotiation
Disclaimer: I haven't done any of this, it's a layman's interpretation of the RFCs and other documents.  Dozens of people here are far more qualified to comment than I.
I'm going to look at TLS 1.2 since it's neatly described within RFC 5246.
Section 22.214.171.124, "The HelloRequest message MAY be sent by the server at any time."  That should trigger the Client to renegotiate the connection.  (The client MAY ignore that request, and the server MAY drop the connection if the client ignores the request for too long.)
A renegotiation looks very much like an initial negotiation, although it may pass some information from the previous negotiation forward to try and smooth the path (e.g., here's the cipher we agreed upon last time).  So the Client sends a ClientHello, then the server sends a ServerHello, then the server sends a Server Certificate (7.4.2).
So my read of the RFC is that, yes, a server can force renegotiation of a connection, including the selection of a different server certificate.
I'll be honest with you, I don't know that you'll find existing software that will do what you want to do.  I would suggest you start playing with Perl Crypto::OpenSSL or Python's ssl lib to see if you can test it out.
Comment: A Web server _can_ start a renegotiation at any time, and that's exactly how IIS handles client certificates: it starts with a "normal" SSL handshake, then, when it learns the target path (HTTP request received), it may decide that, for that specific path, it needs a client certificate, and thus starts a new handshake (a renegotiation) by sending a `HelloRequest` message. That second handshake also involves a certificate from the server (not necessarily the same). A good question, though, is which server certificate (the first or the second one) the browser will show to the human user.
Comment: I'm not sure on this. Wouldn't this (certificate change on renegotiation) be prevented by the [Triple Handshake](https://www.secure-resumption.com/) countermeasures (or at least by some of the browsers)?
Comment: I'm very insterested in the part of your answer where you say `if there was some way to force a renegotiation`, because after getting the actual URL path I could make that, it is forcing the renegotiation, but is it a SSL/TLS supported operation?
Comment: I _think_ so, and have updated the answer (added **Appendix**) to lay out why I think so.
Answer: Nope. No web server can support this feature; not now, not ever. Here's why:
HTTPS follows these steps in this order:

Client connects to the server
SSL/TLS negotiation (includes exchanging certificates, certificate verification, and encryption setup)
Client sends request (includes server hostname, path, cookies, etc)
Server sends back response (includes response headers, content, etc)

Note that step 2 above happens before step 3. That is, the whole encryption thing is totally finished before the browser tells the server what URL it's looking for. This means that when the server selects an SSL certificate to use, it does not yet know what the URL path will be. Since this information is not available at that stage, it can't factor in to the decision of which certificate to use.
Also note that the hostname is sent to the server in step 3, which is why each certificate is typically installed on unique IP addresses; the server uses the IP address to determine which certificate to present. Newer versions of SSL/TLS added extension called Server Name Identification to allow for the client to specify the hostname during TLS negotiation, but this only works for hostnames.
There is no provision to allow certificate selection based on the URL path, nor will there ever be, because that would imply that the URL path needs to be sent to the server unencrypted as part of certificate selection. And sending the URL unencrypted would be unacceptable for a secure page.
Comment: If anybody wnats to look it up: The feature allowing to provide the hostname during TLS handshake is called "Server Name Indication" (SNI)
Comment: I think that the feature to allow the client to specify the hostname before the server to present a certificate is actually really common, a simple SSL configured Apache named virtual host does it
Comment: could you point to me to the SSL/TLS (or HTTPS?) spec where the step 3 is described?
Comment: @jaime http://www.ietf.org/rfc/rfc2616.txt
Comment: @jaime yes, putting multiple certificates on the same IP is getting more common, but it won't help you with your original question.
Answer: I am thinking it may be possible using SSL offloading, deep packet inspection, and TLS renegotiation.  However, if you google TLS renegotiation, the top 20 hits will be about renegotiation vulnerabilities-- so if you do it, you may end up making your site less secure, not more.
On the other hand, if your scheme were like this:
https://example.org/
https://criticalpath.example.org/
it would be pretty trivial.  You'd just route those two domains to different internal IPs, and bind the IPs to different certs.
Comment: With different domain names you don't need different IP addresses, one server can support multiple vhosts with different certs using SNI (Server Name Indication) -- as stated in tylerl's answer 3 years earlier.


Purple garlic color change

Question: I bought a bag of peeled garlic cloves as a convenience(I know, shame on me) I stored the unused cloves in the fridge and the began to show purple areas on the cloves. I don't think this is anything harmful but my partner didn't like it and insisted I throw away what was left. Is this just another natural color change for garlic such as the blue/green or green colorations that can occur. Anyone have a definitive answer? I'm not trying to one-up her, just trying to be safe.
Answer: The purple is just the anthocyanins in the garlic reacting to conditions. It's perfectly safe. Some types of garlic actually turn purple before they are picked if they are fertilized close to maturity; I do this now by choice, because I've decided I like the purple. The purple will occur more frequently and to a greater degree if the garlic is cooked with or peeled cloves are stored in contact with something acidic.

Purple garlic color change


How to find the input variables for a classification problem?

Question: I am working on a  classification problem. I have 1000+ features in this dataset. I don't know how to select the right variables/ features that can actually contribute to predicting the output. What are the different methods through which I can identify the important variables that can be used out of these 1000+ variables.
Answer: Two approaches come to mind:

Use all of them and perform feature selection to identify indicative features. Algorithms include Information Gain and Mutual Information.
Hand-pick some features you deem indicative intuitively, or delete the ones you deem non-indicative.

Generally, you have some sort of hypothesis that you are testing where you attest the capability of (a subset of) your features (and the algorithm you'll use) to uniquely identify the label/class they belong to. The features you use are one of the most important things in the classification task. So it is good to spend time to intelligently find out which features contribute.
For this task, you do have a lot of possible features so the automatic feature selection approach might be better.
Comment: Also use all of them and reduce dimensionality. e.g. PCA extracts meaningful features according to the variance.
Comment: Exactly. You can even consider it as a small research project. You assume that there are some features that will work well based on your intuition, and pick those. Or, you assume nothing and input "everything" into a feature selection algorithm. The output for that will be the machine's way of telling you "These N features work the best for me to discriminate between the possible labels". Good luck!
Answer: The general approach in feature selection is to get a score of each feature in the data set and select top features. We can run algorithms like GBM or Random Forest on all the variables simply to get a ranking of variable importance. We can also use χ² (chi-squared) statistic with cross validation to select a user-specified percentile of features with the highest scoring. But, the disadvantage with these approaches is not detecting correlations between features.
We can also use backward elimination: features are tested one by one and the ones that are not statically significant are deleted from data set. In forward selection that starts with out any variable in dataset and then adds variables that are statically significant. Hope this helps.
Answer: You can perform logistic regression (if your dependent variable has two classes) that penalizes based on the L1 norm. You can choose the correct sparsity parameter (typically $\lambda$) that chooses how strongly to seek sparsity based on cross-validation. The model will force many non-informative features to be 0. This is a form of feature selection. See here: http://scikit-learn.org/stable/modules/linear_model.html#logistic-regression


Creating a custom control/action for a document library

Question: I am wondering if it is possible to add a custom control/action for files in a document library in SharePoint. For example, when I select a file or click the ellipsis for a file I get the options to preview, share, copy link, download ,etc... I want to know if it's possible to create my own custom option for this which will trigger a ms flow and pass the file properties to ms flow?
To be more specific, I already have a ms flow that automatically picks up files when they are uploaded to sharepoint and copies these files to a different environment. I just need to see if I am able to implement a custom option, when you click that option it will trigger the flow for that file and copy the file to the different environment or if it is even possible to add  custom options for files in a document library.
Answer: @Omi in a hellcat,
@Murilo Santana has provided a good solution. you may also have a look SPFx ListView Command Set Extension:

It can trigger an HTTP flow:

https://www.linkedin.com/pulse/download-sharepoint-list-item-pdf-using-microsoft-flow-maniyani/

BR
Answer: It looks like you need a flow with the "For a selected file" trigger.
That trigger allows you to select a file in a library to initiate a flow from there - you can also set input options when running the flow. Inside the flow you will then be able to use any other actions you'd like.

You will see this option on SharePoint


Mass-Energy relation

Question: Einstein mass- energy relation  states $E=mc^2$. It means if energy of a paricle increases then mass also increases or vice-versa.
 My question is that what is the actual meaning of the statement "mass increases"?  Is really the mass of the particle increasing or what?
Comment: Originally Einstein derived his relationship $\Delta E=\Delta mc^2$ for macroscopic bodies capable of absorbing and radiating electromagnetic waves. So he meant literally the body mass change. This formula is not applicable to "elementary" particles incapable to absorb/emit other species.
Comment: Related: http://physics.stackexchange.com/q/8610/2451 and http://physics.stackexchange.com/q/43813/2451
Answer: The rest mass of an object is, by definition, independent of the energy. But all other forms of mass are indeed increasing with the energy, as $E=mc^2$. With the relativistic interpretation of the kinetic energy, the total mass is
$$ m  = \frac{m_0}{\sqrt{1-v^2/c^2}}$$
Here, $m_0$ is the mass measured at rest, i.e. the rest mass. The corrected, total mass goes to infinity if $v\to c$ and it holds for the following interpretations of the mass:

the inertial mass, i.e. the resistance towards the acceleration, increases. For example, the protons at the LHC have mass about 4,000 times larger than the rest mass (the energy is 4 TeV), and that's the reason why it's so hard to accelerate them above their speed of 99.9999% of the speed of light and e.g. surpass the speed of light. It's impossible to surpass it because the object is increasingly heavy, as measured by the inertial mass
the conserved mass. If you believe that the total mass of all things is conserved, it's true but only if you interpret the "total mass" as the "total energy over $c^2$". In this conserved quantity, the fast objects near the speed of light indeed contribute much more than their rest mass. If you considered the total rest mass of objects, it wouldn't be conserved
the gravitational mass that enters Newton's force $Gm_1m_2/r^2$. If an object is moving back and forth, by a speed close to the speed of light, it produces a stronger gravitational field than the same object at rest. For example, if you fill a box with mirrors by lots of photons that carry some huge energy and therefore "total mass" $m=E/c^2$, they will increase the gravitational field of the box even though their rest mass is zero. Be careful, in general relativity, the pressure from the photons (or something else) creates a gravitational field (its independent component curved in a different way), too.

Despite this Yes Yes Yes answer to the question whether the total mass indeed increases, Crazy Buddy is totally right that especially particle physicists tend to reserve the term "mass" for the "rest mass" and they always prefer the word "energy" for the "total mass" times $c^2$.
Answer: The correct expression is
$$E=\frac{mc^2}{\sqrt{1 - {v^2}/{c^2}}}$$
where $v$ is the velocity of the particle. The mass $m$ of the particle is a constant and cannot increase [*]. The energy of a particle increases with its velocity. For a particle at rest $v=0$ and $E=mc^2$.
[*] This is the modern concept of mass used in physics. For instance the mass of an electron is $m_\mathrm{e} = 9.10938188 \times 10^{-31}$ kg.
Some answers here mention the old "relativistic mass" concept $m_\mathrm{rel} \equiv {m}/{\sqrt{1-v^2/c^2}}$ and mention a supposed "rest mass". The term "rest mass" is a misnomer, for instance $m_\mathrm{e}$ is not the mass of an electron at rest and this term is still more misleading for particles such as photons (which are never at rest). For additional criticism see the recent Okun paper Mass versus relativistic and rest masses
Answer: Yes. You're somewhat (not whole part) right. But, The actual terminology of mass-energy equivalence is that, It gives the energy associated with a "massive" particle or the total energy content obtainable from the system.
But Nowadays, relativistic mass $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ isn't considered as a good term by Physicists. Because, they're eager & more satisfied with the true rest mass $m_0$ and rest energy $m_0c^2$ they actually measure (with their goggles) rather than the apparent relativistic mass.
Moreover, there are satisfactory explanations that the rest mass does not vary with an object's inertial motion relative to the observer's frame of reference.
Answer: Yes, mass actually increases when energy increases. Though very slowly.
This is because the equation E=mc^2 says that a huge amount of energy  is only equal to a small amount of mass. However it is still there.
However there are times when this energy can increase so drastically even mass is increased significantly. Rest mass usually does not increase for the most part because rest mass is internal however even this can impact mass and energy. The rest mass of a neutron is less in side an atomic nucleus than alone. This rest mass is then lost as a lot of energy. So even rest mass subjected to E=mc^2. Just not as much as other types of mass. Most of a protons or neutrons mass is not really rest mass but the kinetic energy of the quarks that actually make them up. Even true rest mass(the rest mass of fundamental particles) comes from the energy of the interaction between particles and the Higgs field.
Answer: That's not a relation, that's more like the equivalence between mass and energy. The main concept under this formula is that mass is a property of any energy and energy is the property of any mass, connected with a constant - $c^2$.
It means that if you have a body at rest with a mass of $m$, it would have energy $E=mc^2$.
For example, the rest mass of electron is $m_e=0.511\mathrm{MeV}/c^2$. And when electron and positron annihilates the photons with total energy of $E = 1.022\mathrm{MeV}$ is radiated.
So, about the question: yes, heavier elementary particles would have more energy under annihilation. But you can't make electron with bigger mass.

Mass-Energy relation


Gmail Syntax Highlighter

Question: I sent my code to my friend over Gmail but as you guess, the code is not highlighted in Gmail. Is there solution to highlight code that is sent over Gmail?
Answer: Paste your code to an online (maybe offline) syntax highlighter, like http://tohtml.com/ . Then copy/paste it to gmail & send.
Comment: This is good for me but I wish there should be a gmail plugin in gmail labs. Thanks Ahmet...
Comment: It somehow seems to make the font size very small
Comment: Somehow it messes up highlighting for quotation marks in python.
Comment: @EricWalker same for SQL too
Comment: That doesn't seem to work well for me now. Using https://github.dev was the best I could find right now.
Answer: consider using gist.github.com
some benefits:

you can copy and paste the syntax highlighted code into gmail

(its not always perfect but gets the job done, see screenshot below)

snippet(s) can be shared with a hyperlink (also bookmarkable)
you don't need an account to share code snippets

code can still be kept private and is only shared with those have the secret link

you can specify the language or snippet type, or even multiple snippets on a single page

each lang can be specified with github flavored markdown

you can also access historical changes of the snippet making it easy to track revisions over time
Comment: Do not use this extension as it is removed from ChromeWebStore for a good reason. The reason it got banned is explained by @Drenai
Comment: I think Gist does not support on 2021, when I paste to Gmail, no syntax highlighted. I prefer write via markdown, then paste to Gmail, e.g. I write in Typora, it works for me.
Comment: In 2022, still seeing no syntax highlighting when copying from Gist to Gmail. However, using https://github.dev did work wonderfully, and looks the ~same.
Answer: Unless you absolutely have to have the code in the email, I would just put your code on Pastebin and link to the code.
If you must put your code in the email, you could use a GreaseMonkey script such as this C++ highlighter, however that will not work for the person you're sending it to, only you. Another solution is to export your code as HTML (with coloring), and send it to the person as HTML. Try Visual Studio's CopySourceAsHtml plugin if you're using Visual Studio.
Comment: I prefer to use internal application not external link. So generating highlighted html code for sourcecode is the best way.
Answer: If your language is not available at http://tohtml.com but you use GitHub you can also paste into a (private) gist there - http://gist.github.com/ - then copy&paste back to gmail.
Answer: Use markdown-here. You can write markdown in mails with the extension. The extension format mail content to html and highlight the code. I think this is an excellent extension.
An example
Comment: Note this has not been maintained for many years now.

It also does not support converting rendering, editing, and then rendering again, which was a dealbreaker for me.
Answer: If I have to send a snippet of Java code from Eclipse, I copy paste the code into Microsoft Word and then from Word into Gmail. That preserves the HTML formatting. If you use an email program such as Outlook, just copying and pasting from Eclipse into Outlook works.
Answer: You can use MSFT's monaco editor via github.dev for this in two ways:
1. Just copy nice code

Open https://github.dev
Delete the contents of the markdown file, and write your code in syntax highlighting blocks (eg, "```ts").
Copy code into gmail:

Personally, this is exactly what I was looking for.
2. Copy full markdown

Write your whole email as markdown in the README editor.
Press the "[Preview] README.md" tab.
Copy/paste the result into gmail:

It is still nicely editable within gmail (though of course it won't re-highlight your syntax if you edit code).
The syntax highlighting is different (and in my opinion, uglier) with this option. The markdown preview engine is also a little wonky (definitely not the same as Github's); see the source for the above screenshot for an example (note the need to over-indent the inner numbered list).
Note that when you close & reopen the tab, the last contents you had will still be there. IMO this is a tremendous advantage, but if you are writing something private, you should delete the contents before closing the github.dev tab.
Answer: Try http://jsbin.com/. Easy to write html/JavaScript and test. Just copy paste the code from jsbin to gmail.
Answer: mac-oriented solution with Xcode:
I convert the code into an RTF file (using Xcode's qlmanage -p source_file -o Command described in here http://hints.macworld.com/article.php?story=20091104002118164).
Then open it in Word (NOT in textedit, when copying from textedit color gets lost), and copy and paste it from Word into gmail.
Answer: For those of you looking to copy/paste from xCode to email, I simply copy from xCode and paste into Mac's default email client ("Mail").  It keeps all the formatting with colors.

Gmail Syntax Highlighter


What is the difference between the mass of 100g and the weight of 1newton?

Question: I was flicking through my physics textbook (as you do when you need to revise for a test that is going to decide your grade of the whole year), when a certain question caught my eye it read:
Explain the difference between a mass of 100g and a weight of 1newton.
Comment: possible duplicate of [What is the difference between weight and mass?](http://physics.stackexchange.com/q/43195/)
Answer: Well, the difference is that weight $W$ is a force and mass $m$ is... a mass.
$$W=mg$$
The difference is the coefficient $g$, roughly equal to 10.
Answer: The mass is an intrinsic property of an object, weight is the force with which that object pushes the ground (depends not only on the object but on the gravitational force in it's vicinity as well).
On the moon the mass of that object will stay the same, but it's new weight will be 1/6 of it's weight on earth because moon's gravity is much weaker.
Answer: Mass is a scalar.  Weight is a vector.
100 grams indicates the amount of matter in an object, and is a measure of the object's inertia, its resistance to acceleration.  At relativistic speeds, or when dealing with sub-atomic particles, mass also is a measure of the amount of energy in an object.
1 Newton is a measure of the force exerted on an object by a gravitational field, or equivalently by acceleration.  The weight vector of an object on Earth points toward the center of the Earth.  If weight is caused by acceleration, the weight vector points opposite from the direction of acceleration.
A mass of 100 grams has no directional component, whereas a weight of 1 Newton has both directional and mass components.  It is the directional force necessary to accelerate 1 kilogram of mass at the rate of 1 meter per second, per second (1Kg * 1m/sec^2).
These differences are important to avoid confusion in dimensional analysis.  If you say that 100 grams is one tenth of 1 Newton, that would be comparing apples to oranges.


Overshoot control

Question: Suppose I have the following step response:
$$y_{step}(t)=\frac{kP}{1+kP}(1-e^{\frac{-t}{\tau}})$$
where k is constant and P is the plant.
How may I determine the values of k for which there would be no overshoot?
Comment: Are there *any* values of \$k\$ for which there is overshoot?
Comment: But how may I check for these values? Given a certain function, how does one normally check for potential overshoots? Do I simply check whether y_step(t) exceeds 1?
Comment: For a time domain expression, if a function does *not* overshoot or undershoot but steadily approaches the 'final' value, what does that imply about the (time) derivative of the function?
Answer: The particular step response you show can't overshoot, as k and P are constants. The dynamics of the step response are determined by the exponential function only.
In general, you can check if a linear system will overshoot with a step input by analyzing its frequency response.
The maximum value of the Bode plot of magnitude is related to the overshoot you will have, given a step input, because the step input excites your system at every frequency.
If the maximum of the Bode plot of magnitude of a closed-loop system is 1 (or less, allowing a steady-state error as in the closed-loop system you brought as an example) you have no overshoot.
You could find the topic of critical damping interesting, too.
Comment: P might not be linear.  VERY true for a 1st order plant & the rest of this there should not be any overshoot. But a 2nd order plant with a high proportional aspect to it?
Comment: Actually, P(s)=s^2+a1*s+a0 and Y_step(s)=[kP(s)]/[1+kP(s)]. For what values of k will there be no overshoot?
Answer: If there is no overshoot, then, mathematically, it means that the function monotonically increases toward the limiting value at high values of t (which is just the fraction without the exponential factor). This means that its derivative of the function stays positive.
Overshoot means that the function rises and then falls. When it switches from rising to falling, its slope becomes zero very briefly and then negative.
You can analyze the derivative, looking for values of k which make it touch or cross zero.
Comment: You mean the \$y_\text{step}(t)\$ function isn't necessarily the right one? If it is the actual step function, then it represents the actual time-domain response, and it boils down to, for what values of the parameters does the function not show a "hump".
Comment: Is there any way to check for overshoots without having to use an inverse Laplace (I mean, how else could I determine the derivative wrt time)? My initial expression for y(t) is most likely incorrect.

Overshoot control


Adding a data dictionary to Trimble R1 reciever through ArcGIS Online

Question: I don't even know how to properly google this effectively, so maybe someone can help me on here. My company purchased a Trimble R1 GNSS Receiver. The data is managed through ArcGIS Online. I have a data dictionary that I need to use, but I can't figure out how I am supposed to upload it in ArcGIS Online. I have been creating Feature Layers to collect other data with generic data dictionaries - I use Field Notes Template currently.
Is there a way to upload a data dictionary to ArcGIS Online to use with ArcGIS Collector on my phone?
Answer: Check out this tech session from the user conference in 2018 talking about how to use ArcGIS Online, Collector and high accuracy GPS receivers: https://www.youtube.com/watch?v=9QkYNJtNoQE
Comment: That was super helpful, especially in regards to setting a transformation. However, unless I missed it, did it cover if you can upload a pre-existing data dictionary? I actually figured out that to edit attribute fields I just needed to switch from Overview to Data and edit the fields that way. But I still haven't figured out how to add a Trimble Data Dictionary file to a Feature Layer.
Comment: you will have to add the fields prior to publishing or after using the workflow you are currently using. Or you can do this through the REST api https://developers.arcgis.com/rest/services-reference/add-to-definition-feature-layer-.htm


Use child element of group to clip or cover other element

Question: I'm working on an SVG blueprint and I've run into a bit of an issue.
I believe the issue arises because I've had to group two, semi-translucent elements so that when they overlap it appears as if they are one solid unit.
Here is an MCVE, which should be viewed in Chrome since I didn't use "absolute" values for one element's transform-origin.
leftRoomFill.style.fill = '#fff';
leftRoomDoorFill.style.fill = '#fff';
leftRoomFill.style.opacity = '1';
leftRoomDoorFill.style.opacity = '1';
console.log(window.getComputedStyle(leftRoomFill).opacity);
console.log(window.getComputedStyle(leftRoomDoorFill).opacity);

The issue is that, even though it's logging to console that the elements' opacities are set to 1, it appears as if the elements' opacities don't change at all and are still 0.
I've tried the following which further cements my belief that the issue is an issue of opacity.
rightRoomFill.style.zIndex = '1';
leftRoomFill.style.zIndex = '2';
leftRoomDoorFill.style.zIndex = '2';

Does anybody know why the code above has seemingly no effect whatsoever?
What I'm trying to do is use the door fill element to crop the fill of the room into which the door extends.
I know I could simply ungroup the fill elements and use colors without an opacity, but I'd really like to maintain the flexibility of being able to use semi-translucent fills.
Any ideas?
Comment: You'll likely be in trouble again when Chrome fixes their transform-origin bug then.
Comment: @RobertLongson ??
Answer: It's unclear what you are asking. Why do you think the room opacities are 0? They are becoming visible (green) for me in Chrome. So a value of "1" seems correct to me. What is supposed to happen that isn't?
Re your zIndex question: You'll probably have to use a different approach.  z-index doesn't currently work on SVG elements.  It has been added to SVG 2, but I don't think any browser supports it yet.
Comment: the opacity of each fill is set to `0` in CSS, and when you hover a room `that particular room's` fill is then set to `0.5` or `1` or whatever. what i wanted to do was set the door fill (which belongs to the left room) to `opacity: 1` and `z-index: 99` when you hover the right room, so that the door fill would obscure that part of the right room's fill. it doesn't even matter anymore, i just sacrificed the opacity
Answer: The zIndex property does not rearrange SVG elements (at least not yet).
It seems like you want to make the door fill from a left-room group "cover" the right-room floor upon mousing over the right room floor. However, you want to do so only having a single copy of the door fill which in this case is grouped together with the floor that that door opens out of, i.e. the left floor. It seems that you do not want to create a master door fill in a <defs> section and then copy it multiple times with <use> elements. While I'm not sure this is the best strategy, if you really want to do this, one way is to move the floor that is currently visible (i.e. the one moused over) to the visible bottom of the stack of SVG floors, i.e. to the code-top of the floor's parent node. You can do this with the following code that is run whenever a particular floor is moused over: myFloorNode.parentNode.insertBefore(myFloorNode, myFloorNode.parentNode.firstChild);. Then any other door fills that overlap that floor will "cover" it, "cutting out" that section of the floor.
The thing that I find a little messy about this approach is that it requires constantly shuffling your floor elements around. That's not necessarily lethal. However it does make me a little uneasy, as it might create some confusion if the re-shuffling ever gets any more complicated. It also requires you to set that door fill to be completely opaque which may or may not be acceptable. In any case, that's just my gut feeling, and will depend on exactly how you implement it.
You can do a simple implementation of this strategy by modifying your code as follows:
function setFillOpacity(){
  this.style.opacity = '0.5';

  // *** add the following line:
  this.parentNode.insertBefore(this, this.parentNode.firstChild);

  if (this.id === 'right-room-fill'){...

Note, however, that with just this one change, sometimes your door fill seems to be in the "closed" position even if the door stroke is in the "open" position. It seems to eventually work itself out, by "open"ing by the end of a dynamic animation of a door closing and then re-opening. However, I think you'll have to work on the code to ensure that the door fill starts in the correct position. I'll leave working through that problem to you.
Comment: i simply sacrificed opacity and ungrouped the door fills with the room fills. also, i've been told using deeply embedded `` with animations is really intensive on processing power, especially when you have half a hundred different elements, which is why i didn't bother doing that.


Can i chant the Ashtottara nama of Maa Lakshmi for better finanacial situation?

Question: my family's financial condition has been quite bad since a very long time(10+ years), we are in huge debts and are in a hand to mouth situation living by just by the grace of god. can i chant the 108 names of Maa Lakshmi to benefit a bit financially? and i searched the names on the internet but i wanted to make sure they're non-tantrik or do not contain any rahasaya naams, since i am uniniated hence i cant do those. if possible could you share the non tantrik ones? thank you all, i appreciate your help...
Comment: Yes you can chant .. do not worry about if the names from Puranas or Tantras. There are many websites from where you can find the Ashtottara stotras.
Answer: Most sadhakas dismay Artha, but to to grihastha dharmanirvaaha, devapitrukarya, etc one needs wealth. Thus there's nothing wrong in seeking for Artha, till the time it is earned by dharma and used for dharma.
but i wanted to make sure they're non-tantrik or do not contain any rahasaya naams

Asnwering main question here, the Ashtotram available online are mostly from Laxmi tantra. But since you want to avoid tantric stotrams, then rather go for Ashtotram from Narada upapurana.
The Phalashruti says,

श्रियं वर्चसमारोग्यं शोभनं धान्यसम्पदः ।
पशूनां बहुपुत्राणां लाभश्च सम्भावेद्ध्रुवम् ॥
With the blessings of Srì one shall get health, grains, wealth, animals, and with be benefitted with birth of alot sons.

if possible could you share the non tantrik ones?

Well, I have even jotted a few additional stotras too with fhalashruti (to boost faith in stotra),
Laxminarayan samhita Shree Laxmi namavali,

ऐतानि लक्ष्मीनामानि सदा प्रातः पठेध्दियः ।
स पुत्रपौत्रादियुक्तः श्रियमाप्नोत्यनाशिनीम् ||
One who reads these names of Laxmi daily at dawn, with be benefitted with Sons, grandsons, and along those with his Srì (wealth) shall be indestructible.

Agastyamunikrutha Laxmi stotram,

यः पठेत्प्रातरुत्थाय श्रद्धा-भक्तिसमन्वितः ।
गृहे तस्य सदा स्थास्ये नित्यं श्रीपतिना सह ॥
सुखसौभाग्यसम्पन्नो मनस्वी बुद्धिमान् भवेत् ।
पुत्रवान् गुणवान् श्रेष्ठो भोगभोक्ता च मानवः ||
If one reads this at dawn after waking with faith and devotion, in his house Srì along with her consort Srì pati (Vishnu) will be stationed permanently. He shall be blessed by happiness, fortune, and become intellectual. Will have sons, himself be a great virtuous man, and will enjoy all materialistic joy.

Devi bhagwatam, Indrakruta Laxmi stotram,

इदं स्तोत्रं महापुण्यं त्रिसन्ध्यं यः पठेन्न्रः ।
कुबेरतुल्यः स भवेत् राजराजेश्वरो महान् ॥
If one reads this stotram in three sandhyas (dawn, afternoon, dusk) shall himself become like Kubera (wealthy) and thus become a great emperor.

Padma Puran, Indrakruta Mahalaxmi stotram,

एककाले पठेन्नित्यं महापापविनाशनं ।
द्विकालं यः पठेन्नित्यं धनधान्य समन्वितः ॥
त्रिकालं यः पठेन्नित्यं महाशत्रूविनाशनं ।
महालक्ष्मीर्भवेन्नित्यं प्रसन्ना वरदा शुभा ||
Those who chant this once will get rid of sins, those who read this twice will be blessed with wealth and grains. Who who reads thrice will be blessed with destruction of enemies (internal & external) and Mahalaxmi will always be pleased thus giving thr best boons.

we are in huge debts

I would say stick to one stotra, but since you have debts you must chant Rinmochana Ganapati Stotram, aided with any of the aforementioned stotrams.
Even Kanakadhara Stuti by Bhagwadpada helps.
Not just these, there are stotras of Ganapati, Nataraja, Chandika, Narayana, Shiva, etc deities too, to give artha. All it takes is, a stout faith in your ishta devta. Do any stotram but with whole devout, the results would ripe sooner.
But, would like to add a short disclaimer, once Srì favours the person, let not the greed bite the person. Donate a fraction of wealth timely for devta, pitru, sadhu and jeeva seva. That way the wealth not only increases manifold times, but also gives the merits thus taking one near real goal of Moksha
श्रीकालिकार्पणमस्तु |


What will happen if Pinocchio says: My nose is going to grow

Question:
If his nose is not growing, he is telling a lie and his nose will grow but then he is telling the truth and it can't happen.
If his nose is growing, he is telling the truth, so it can't happen.
If his nose will grow, he will be telling the truth, but his nose grows if he lies so it can't happen.
If his nose will not grow, he is lying and it will grow but then he would be telling the truth so it can't happen.

What will happen? The universe will explode?
Comment: What happens next is up to the story-teller, because Pinocchio is a fictional character whose world is not constrained to any particular laws of physics (hence the quite literally *fabled* properties of his nose) nor logical consistency (which must perforce go out the window if Pinocchio succeeds in implementing the Liar's paradox). Or more precisely: the logic which is best suited for effective reasoning in Pinocchio's world depends on what the story-teller decides happens, and that logic needn't be classical logic.
Comment: Be careful. Saying that a falsehood is true does not a liar make. If you don't know what you are saying is false, then not a liar. You're just wrong.
Comment: Well, he *could* get plastic surgery to enlarge his nose... I mean, wooden surgery.
Comment: @NieldeBeaudrap: Well played, like the guy, who, told to measure the height of skyscraper with a barometer, dropped it off the top and timed it.
Comment: If he is a little boy than chances is that as he grows older his nose will indeed grow.
Comment: @Mitch: the analogy would perhaps be complete if the barometer had no markings on it. One might as well ask a physics forum, how Superman avoids getting third degree burns from air friction.
Comment: @NieldeBeaudrap: it's all just playing the game.
Comment: Pinocchio's nose grows when he lies. Lying and saying something that isn't true is not the same thing.
Answer: That looks like a paradox, but it isn't.
The assertion is so vague that it is true even if Pinocchio's nose doesn't grow immediately.
Even if he says: "My nose will grow immediately" that's a vague assertion, when does "immediately" start and end as a time lapse?
If he is more specific and states: "My nose will grow in the next five minutes" then the nose can make a suspension of disbelief, if he tells a lie in those five minutes then the nose will grow and the first assertion will be true, if he doesn't then the first assertion will be proved as false and if falsehood is considered as a lie then the nose will grow after the five minutes.
Everything that is false is a lie? That depends on some definitions, but if Pinocchio says something that is false and a mistake then the nose may (or may not) consider that Pinocchio is implicitly stating to be sure about something that he is not sure and actually wrong, hence a lie. That may be a question for Geppetto.
Nothing is true or false until it is.
If Pinocchio succeeds in creating a paradox (either the liar's paradox or some other paradox) then the nose will not react to that. A paradox is something that cannot be true and cannot be false and the nose grows on falsehood, right?
No matter which logic and assumptions (OWA and CWA for instance) that nose follows, in any logic where there can be paradoxes it is clear that the middle cannot be excluded.
Comment: While I want to give you credit for this answer, I still have to give you a –1, because your treatment of the paradox that the author was really getting at seems quite weak. The paradox is the crux, and you just hand-wave it away.
Comment: no worries about the -1. WRT "the paradox is the crux", there is no paradox, maybe the question could be reworded or formalized so that there is some paradox, but there is no paradox so far. Anyway, if Pinocchio succeeded in creating a paradox: "then the nose will not react to that." Therefore, if the question was fixed and a paradox included, it would be already be answered.
Comment: Actually, I thought you did a good job clarifying such that it would become a paradox? One can call it nit-picking, but in philosophy discussions, nitpicking is often a good thing! :-) I just think you ended up ensuring there was a paradox, and then left me hanging!
Comment: There is no paradox, but if there ever was one, the nose would not react to that, because it does not react to *paradoxes*, it reacts to *lies* (as I pointed in the original answer). It's a tale to tell children not to *lie*, but nobody cared to say not to *troll* or not to *hack*.
Comment: I suppose you could make a play on 'lie' vs. 'utter falsehood', but that seems to escape on a technicality. It just seems like you're taking the easy way out, and not addressing the issue head-on. But I could be wrong.
Comment: What's "the issue" in your opinion?
Comment: PD: I've just seen your answer. You can analyse it like that, which means the nose will crash with a "not enough heap space" exception and it will not grow. I'd hope that someone with enough knowledge to make a nose that grows (creating matter out of nothing, big deal) would be able to detect and prevent this before raising an exception. In any case, the nose doesn't grow.
Answer: The premise that Pinocchio's nose grows if and only if he claims any falsehood is given as:
<code>∀x: G ↔ C(x) ∧ ¬x
</code>
If Pinocchio's claim is that his nose will grow then this is given as:
<code>G ↔ C(G) ∧ ¬G
</code>
This is a contradiction. Therefore the premise that Pinocchio's nose grows if and only if he claims any falsehood is necessarily false:
<code>∃x: ¬(G ↔ C(x) ∧ ¬x)
</code>
Comment: I don't understand how "If anything can be proven true, anything can be proven false." relates to my answer. Also, re. your first comment, are you claiming that I am accepting a contradiction? Because I'm not. I'm arguing that the premises lead to a contradiction and so one of the premises is false. Reductio ad absurdum; proof by contradiction.
Comment: `∀x: C(x) ∧ ¬x ↔ G` isn't an axiom. It's a premise which leads to a contradiction. Therefore it is false.
Comment: "An axiom, or postulate, is a premise or starting point of reasoning. A self-evident principle or one that is accepted as true without proof as the basis for argument; a postulate. As classically conceived, an axiom is a premise so evident as to be accepted as true without controversy." Pinocchio's nose growing when he claims any falsehood doesn't quite meet the criteria of being self-evident. It's not an axiom. It's just a necessarily false premise. Recognizing it as false is the solution to the paradox it gives rise to.
Comment: Once you accept a contradiction in bivalent logic, anything can be proven true.
Comment: Then don't accept it.
Comment: If anything can be proven true, anything can be proven false. The conclusion of your argument is not the only logical conclusion.
Comment: If a set of axioms leads to contradiction, we can say that one of those axioms is 'bad', but why did you single out a specific axiom?
Comment: If the logical combination of two axioms produces a contradiction, at least one of those axioms is generally considered 'bad'.
Comment: Right. But again, the premise that Pinocchio's nose grows if and only if he claims any falsehood is hardly an axiom. Although as with your comment regarding axioms, that it is shown to lead to a contradiction is that it is proven to be false, à la the Exception paradox.
Comment: Why is it "hardly an axiom"?
Answer: your question is the question of is the statement

this statement is false

true or false?
however there are more things to consider now that we are discussing Pinocchio none that i see include the world exploding but here are some facts to consider

the ambiguity of my nose will grow as trylks pointed out
the fact that this statement must have more context an "if... then my nose will grow" is obviously case-dependent each with a yes/no answer (unless it is "if i say my nose will grow then my nose will grow" - which is the proper paradoxical statement)
does his nose grow if he is unaware that he is lying (and if no what about if he is unsure)
Answer: Bad question. Stated differently, you have constructed a paradoxical world. Check out self-reference. I will take your question to be equivalent to the liar paradox:

This sentence is not true.

Something tricky is going on here: recursion:

This sentence is not true.

Let me do a substitution, "This sentence" → "This sentence is not true.":

"This sentence is not true." is not true.

But wait, what is being pointed to by "This sentence"? The substitution did not bring any clarity! The problem here is that there is an infinite regress, caused by self-reference. This kind of paradox shows up elsewhere, like Russell's paradox, which attempts to construct the set R, with membership criterion: "all sets which do not contain themselves as a member".

If R contains itself, it is not a member of R.
If R does not contain itself, it is a member of R.

This might take a while to completely wrap your head around, but it is a deep result of what is now called naive set theory. In order to circumvent this problem, axiomatic set theories where developed which could not produce such paradoxes. However, they lose something: they are axiomatic, and thus not ambiguous like natural language. And yet, there seems to be something deep to this ambiguity. I won't go into that now, but it would make a good separate question.
There is a theory of computation aspect to self-reference, which shows up as Turing machines being able to print out their own description. This gets at the idea of self knowledge. And yet, Thomas Breuer's The Impossibility of Accurate State Self-Measurements questions this whole perfect self-knowledge enterprise. This Turing machine self-reference thing is very important; it shows up in the Halting Problem, which presents a huge obstacle to provability of Turing complete systems, which means we cannot guarantee properties we'd like to guarantee (like that your phone won't crash).
Douglas Hofstadter introduced the idea of strange loops in his Gödel, Escher Bach. The book is a layman's introduction to some neat theory of computation issues. I do not pretend to understand this 'strange loop' idea, but it definitely has to do with self-reference. It may be that consciousness itself has to do with self-reference; indeed, it is hard not to. So there's a lot to this liar paradox!


Image manipulation in hybrid mobile apps

Question: i'd like to ask if there is possibility to write sth like instagram app but not as native application, just hybrid app (titanium or phonegap with ui frameworks)?
Most of functionality for me is possible to implement, but i wonder about image manipulation.
I have experience in appcelerator titanium and icenium(cordova with kendoui) and my idea is to create app in titanium with webview where i will use some JS manipulation scripts, i've found pixastic.com and CamanJS.com(camanjs is not possible to implement i thing in my situation) so pixastic seems the best.. or any other ideas?
What do you think about performance of this solution? Any experiences about this issue? Any suggestions?
Answer: Check out this post right here: Create Instagram-esque filters with javascript? sounds like it could direct you in the right way.


Is upgrading to Utopic worth it? Will I encounter lots of bugs?

Question: I have a newer, touchscreen laptop/tablet (a 2-in-1? but it doesn't come apart, it folds back and becomes a tablet), and I'd like the fullest support possible for my newer device. When trying to upgrade by computer to Ubuntu Gnome 14.10 from 14.04 recently, I noticed that the release notes state that Utopic is a devolopment release, not intended for use on production machines. However, I've yet to see a way to install either gnome 3.14, or 3.16 on 14.04. However, I can update the Gnome DE to version 3.14 on Utopic. I was wondering if anyone on here could please tell me, exacly how buggy Utopic is. With a name like Utopic Unicorn, one would think that the software is fully mature. Will I likely encounter a lot of bugs? Is it worth updating past the LTS release to get support for my newer device?
Answer: 14.10 Utopic has been released (in October last year) and is stable, but is newer than 14.04 LTS Trusty (which could be 'more stable') - 15.04 Vivid is in development and will be released in April. See more here
As for GNOME, Ubuntu sort of misses out on newer versions due to the timing of GNOME's release and Ubuntu's. So if you want to install a newer version (e.g. 3.14 on Utopic), you need to use PPA's to update to it:
<code> sudo apt-add-repository ppa:gnome3-team/gnome3
 sudo apt-add-repository ppa:gnome3-team/gnome3-staging
 sudo apt-add-repository ppa:ricotz/testing
 sudo apt-get update
 sudo apt-get dist-upgrade
</code>
Note this is experimental, and may break other desktop environments including Unity. Source.
(I will add that Fedora and Arch, different distributions of Linux, often have the newer versions of GNOME more easily available by default)
Comment: Yeah, I thought so, but I read that after 14.04, the rest of the 6-month releases would be buggy and experimental, until the next LTS release. Maybe they changed their minds?
Comment: Also, the release notes, when I used the update-manager -d command, said that it was a development release


How do I download Minecraft 1.6.4?

Question: I tried connecting to a server this morning but it said "outdated client". I understand that that means I have to get Minecraft 1.6.4 but I have no clue how to do it. I tried getting it on Minecraft.net but I couldn't find it. Please tell me how to download it and use it.
Answer: The launcher will automatically update Minecraft to the latest version when you click Play… unless you have accidentally told the launcher not to. If the button says Play and bottom-right of the launcher says

Welcome, [username]
  Ready to play Minecraft 1.6.2

Then you have somehow told Minecraft to never update past 1.6.2. Fortunately, this is easy to fix.
Why is this happening?
The launcher has a Use version setting in your profile which it uses to decide what version of Minecraft to download and launch. By default it is set to "Use Latest Version", which means it will always check for updates and install them if it finds any. If it is instead set to a specific version of Minecraft, it will never try to update to anything else.
Somehow, your profile has this option set to 1.6.2, and the launcher is dutifully following the launch orders it is being given.
This can also happen if you have modded Minecraft and have set up the profile to launch the modded version. (Forge's installer sets up such a profile for you automatically.) in this case, not updating is because "a modded version" is a specific version of Minecraft. Forge sets up the profile this way on purpose since an update will wipe out Forge and disable your mods, which is not usually something you want the launcher to do to your carefully-modded game when Mojang releases an update.
Updating to the latest version
You can change the Use version option by clicking the Edit Profile button in the bottom left of the launcher. Set it to "Use Latest Version" and then click Save Profile. The profile editing window will close and now in the bottom right the launcher should say

Welcome [username]
  Ready to download & play Minecraft 1.6.4

If this doesn't happen or the big button says Play Offline you may not be online or Mojang's servers may be having trouble. Make sure you have a working Internet connection and double-check that Mojang's servers all show a healthy "green" status. If their servers aren't working right, you'll just have to wait until they are back online and can serve your update request properly again.
I see "release 1.6.4" in that option. Why not choose that?
You can also set this to "release 1.6.4" to accomplish the same thing... for now. You'll get the same problem next time Minecraft updates, though. Unless you know what you're doing, don't set this to anything other than "Use Latest Version".
Comment: it wasn't set to use latest version because I was using forge. that might be what was wrong. I should probably download the latest version of forge.
Comment: If you're using Forge it will not update. That's working as intended, since an update will remove Forge, which is why Forge adds a new Forge-only profile. Note that you will have to switch to the non-Forge profile and update to 1.6.4 first before the newest Forge will install, since it needs 1.6.4 as a base.
Answer: If you're having trouble obtaining the latest version of Minecraft, I'd recommend starting off with a fresh launcher. I've personally ran into an issue or two launching/updating Minecraft, and grabbing the latest launcher has solved it each time.

Go to http://minecraft.net/download and click the link Minecraft.exe (for Windows) or Show All Platforms if you're running another OS.
Save the file in a preferred location and run it.
Once you open it, everything should update automatically.

How do I download Minecraft 1.6.4?