Info on distances to objects in space

Question: Can a cheap telescope tell you how many miles/km away an object is that you focus on? (Via some digital readout onscreen, or perhaps via some calculating function that you can access in its computer.) Specifically - for objects within our solar system, and definitely within, say, the orbit of the moon. This would let you know whether a strange object was a planet or just something in near-earth orbit or space junk.
Comment: While most general telescope questions receive good answers in [Astronomy SE](https://astronomy.stackexchange.com/) yours question is interesting and different, as you are asking about measuring distances of objects in the solar system and especially *space junk and other objects in Low Earth Orbit*, which makes this definitely on-topic here.
Comment: which cheap telescopes have computers?
Comment: Plenty of consumer telescopes have computer-controlled tracking mounts. Image processing, I don't know.
Answer: If you ask can a distance to a selestial object be calculated from focal distance - the answer is No. Even closest satellites in low Earth orbit will be beyoud hyperfocal distance of a telescope.
But if you want to know "what object did I see in telescope" you can use special software like Stellarium for planets, nebulae, largest asteroids etc. For artifical satellites there are many online tools, my personal choice is www.heavens-above.com
Comment: It would be great if you could also at least estimate how far the hyperfocal distance of a cheap telescope, or even a fancy telescope really is. *"No" + link* is nice, but it doesn't really support your answer. How can the OP know your answer is correct? The link just has some ancient equations but no numbers.
Answer: There is an optical formula valid for all spherical lens and mirror telescopes.
For simplicty I am ignoring the hyperfocal distance.
1/f = 1/a + 1/b
f is the focal length, a is the distance of the object and b is the distance of the image.
I use a focal length of 1 m for an example. We solve the the formula above for b:
b = (a-1)/a
If the distance of the object a is 100 m, the distance of the image b is 0.99 m. For a = 1000 m we get b = 0.999 m.
If a is 100 km, b is 0.99999 m. For a = 1000 km we get b = 0.999999 m.
The difference between 0.99 m and 0.999 m is 9 mm, this is measurable with good precison. But the difference between 0.99999 m and 0.999999 m is only 9 µm, very close to the wavelength of visible light between 400 and 800 nm. Using a cheap telescope, it is impossible to measure 9 µm precisely. But you can't focus such a telescope so precisely anyway.
But the closest astronomical object, our Moon is about 400,000 km away. No way to distinguish between artifical satellites and natural astronomical objects
Answer: Let's try a different challenge; how well could the Hubble Space Telescope, outside of the Earth's atmosphere, tell if a small (unresolved) object was either 100 km away, or at infinity?
Let's assume perfect diffraction-limited optics and instead of a hard-edged Airy diffraction pattern, use a Gaussian beam with a 1/e^2 radius equal to the Hubble's mirror radius of 1.2 meters and a wavelength of 200 nm (it can go lower potentially).
From Wikipedia I get f/24 meaning f = 57.6 meters.
Using Gaussian Beam Optics, $w, z$ = 1.2, 57.6 meters, $\lambda$ = 0.2E-06 meters, I get
$$w_0= \frac{\lambda z}{\pi w} \approx 3 \mu m$$
That's the half-width of a Gaussian beam focussed to a Gaussian spot. The "Rayleigh Range" (a handy and significant parameter) $z_R$ is given by
$$ z_R = \frac{\pi w_0^2}{\lambda} \approx 144 \mu m$$
and the beam width at any point $z$ on either side of best focus is
$$w = w_0 \sqrt{1+\left(\frac{z}{z_R}\right)^2} $$
If the wavefront striking the mirror were not flat, but had a slight curvature because it was a spherical wave from far away, you can (somehow) use
$$R(z) =  z \sqrt{1+\left(\frac{z_R}{z}\right)^2} $$
to calculate the shift at the focal plane, but like most things in optics it ends up being the same as the classical behavior shown in @Uwe's answer, the "Lens Equation"
$$\frac{1}{f} =  \frac{1}{a}  + \frac{1}{b} $$
If $f$ = 57.6 meters and $a$ = 100,000 meters, then b is different from 57.6 meters by 1/100,000, or 10 $u$m.
If I draw that, it looks like the plot below. If you had a lot of time, and both the object of interest and a star (of roughly equal brightness) in the field of view, a very smooth focus mechanism that you could zoom back and forth by a few hundred microns, then eventually, after collecting tons of data, you might be able to notice the 10 micron shift in best focus, by looking at the offsets in slopes at around +/-$z_R$ the Rayleigh distance from best focus.
You can go back and do the Fourier optics for the Airy diffraction and hard circular aperture, and the results will be different but roughly the same. It is theoretically conceivable, but practically impossible, even with the HST above the atmosphere, to detect the difference between an object at infinity, and one at 100 km.
At 10 km, the curve would be shifted ten times farther, and that would be possibly detectable from a practical point of view, but you'd still spend a lot of time measuring very carefully. Below 10 km, for unresolved objects of identical brightness it might be possible routinely, but still a lot of work.
Plot axes are microns:
Answer: There won't be a readout, but you can do a quick estimate yourself: when you look at an object through a non-tracking telescope, it'll wander out of view after a while (how long this takes, depends on your field of view).
Close objects will move out of view a lot faster than stars. Satellites in low Earth orbit are difficult to track with a telescope (they'll cross the sky in a few minutes). Observe some of the planets to get an idea of how fast they're moving, that'll give you a base for comparisons.

Info on distances to objects in space


Accepting bank account information in a form

Question: What security concerns are there when accepting a user's bank account information (account number and routing number) via a form on a page that is using SSL, and posting it back to the server where I then curl off a HTTPS request to send that information to an ACH service like First ACH or ACH Direct via their API?
We wouldn't be saving the bank account information in our database.  I know another option is to use Paypal's Mass Pay API, but they think it's unprofessional (at least for their business) to require customers to have a paypal account to get paid.
Thoughts?
Answer: Lots of security concerns here.   Even if you aren't subject to regulatory statutes (depends on your business) I can think of quite a few.  I'll type till I'm tired of it.

Are your servers secure?  Locked in cages, secure site, two-person access, patched, audited and scanned with intrusion detection, and so on.  This is the most likely thing you've got to worry about.  If you think they are, they're probably not.
Is the network secure all the way down?  Even stupid things like a plan to make sure your domain name doesn't expire is important. Do you have automated controls for shutting down the site in case of intrusion?
Is your software secure?  Auditing by a third party for security problems is a must.  Don't skimp on this.  From the front end (CSS attacks) to the back end, everything needs to be locked solid.
Is your development cycle secure? Dual-person controls on the software.  Don't let your developers patch the servers, or deploy code without a code review.  Admins need monitoring too.
Everything needs to be encrypted.  Better to not store the account numbers at all.  If you do store them, encrypt them and keep the keys elsewhere.

... okay, I'm tired.
Not to rain on this parade, but if you have to ask about this kind of thing on SO, you're really not prepared to do this.  Buy a solution, hire a professional in the field, or spend a lot of time studying this problem and have someone audit what you're doing.
Comment: This all makes sense, and each point is a valid concern.  Assuming I have these important, but indirect security practices in place, what about the security of the actual process listed above.  Basically, SSL all the way through, no local storage of bank info, and encryption of customer ACH account numbers in db?
Comment: In short, building a terribly secure transport means squat if any of the nodes (process or system) are compromised.
Answer: You use SSL on both streams and you don't store. I'd say there are no concerns,
just keep your server well guarded and use strong crypto.
Should someone manage to get access on your server you could
be in trouble, but that's pretty much true for every kind of online
financial transaction.
Answer: Lyons offers an "account verification" API which might work: http://www.lyonsreg.com/products/account-verification.asp
Answer: Your design sounds relatively sound.  The main issue is that your web application has to be running in a secure environment, which I'm guessing you don't have a lot of control over if you're hosted.
Ask you hosting company if they are PCI-DSS compliant.  If they can hit that bar then you're probably in good hands.


Security of AES in CTR mode with fixed nonce (but different key)

Question: If I am using AES in CTR mode and use a unique key every time but the same nonce, is this secure?
If so, does it make a difference if the nonce is public?
Answer: Yes, if you use a different key for each message, then it's secure, and it's not a problem if you use a public (or fixed) nonce.
One example where this comes up is the Integrated Encryption Scheme, which uses a symmetric system to encrypt the actual message, and which internally generates a key that will be used only for that message.  Note that in IES, you'll also need to implement integrity protection as well (e.g. a MAC)
Answer: Yes, this is probably safe.  If you choose the keys randomly and/or using a secure KDF, with no chance of an attacker being able to directly influence them, I'd even upgrade that "probably" to "definitely" (assuming that AES itself is not broken, of course).
However, it's worth noting that in many common models of nonce-based encryption the attacker is assumed to be able to choose the nonce. If the attacker gets to choose part of the key instead, that potentially opens the cipher to related-key attacks.  AES is known to be theoretically vulnerable to such attacks, in the sense that there are known related-key attacks that, while still too slow to carry out in practice, can in theory break AES faster than pure brute force.
So while currently, as far as we know, AES is still practically unbreakable even if the attacker gets to choose arbitrary relations between the keys, that's definitely a known weak point that future attacks might conceivably weaken further.  As the old crypto maxim goes, "attacks only get better, not worse."  In any case, since related-key attacks are easy enough to avoid by using a good KDF, there's really no excuse to design a system that could be vulnerable to them.


Crash in png_set_read_fn() on Windows 7

Question: I'm cross-compiling some C code for Windows 7 and linking to libpng12.dll found here. Unfortunately, the line below causes a crash (GDB says segmentation fault):
png_structp png = png_create_read_struct(PNG_LIBPNG_VER_STRING, NULL, NULL, NULL);

I'm not sure where to start to debug this--that function seems very straightforward and unrelated to the rest of the code. Top of the backtrace looks like this:
#0 0xabababab in ?? ()
#1 0x6cc556d1 in png_set_read_fn () from c:\...libpng12.dll
#2 0x6cc7a5d4 in png_libpng_ver () from c:\...libpng12.dll
#3 0x00000038 in ?? ()
#4 0x0028f928 in ?? ()

I guess #4 is png_create_read_struct(), since #5 is the function I call it from.
Answer: Please generate a core dump firstly by using ADplus or windbg etc.., and then look at the assembly code around the address of #0 (0xabababab ), also you can check the variable values with regards to those assembly. If you have the source code, you may have a good chance to understand the logic of why it crashes.
Answer: Solved (unfortunately by trial & error) but just in case anyone else experiences this issue, it's caused by an incorrect version of zlib (in my case, too old). You can check for this quite easily at runtime:
fprintf(stderr, "   Compiled with libpng %s; using libpng %s.\n",
        PNG_LIBPNG_VER_STRING, png_libpng_ver);
fprintf(stderr, "   Compiled with zlib %s; using zlib %s.\n",
        ZLIB_VERSION, zlib_version);

Crash in png_set_read_fn() on Windows 7


ASP.NET reportviewer running batch reports in background

Question: I have a asp.net 4.0 page using reportviewer control to deliver reports.There is option to render reports in the page or to pdf/word.
There is also an option to provide a set of (or range of) parameters and the reportviewer would render multiple reports(sometimes over 100) in PDF/WORD.
The way this operates is when the "run batch" button is pressed, the page posts back and the local report is rendered for the available parameters or parameter combinations(into PDF reports).
This obviously causes a Postback and the postback takes very long to return.Is there a way to make the rendering of reports happen in background so the page can return immediately,
In the web I have seen vague references to background threads etc but nothing seems to work. Could somebody provide valuable ideas???
Comment: You put "There is also an option to provide a set of (or range of) parameters and the reportviewer would render multiple reports(sometimes over 100) in PDF/WORD." Are You able to point me in the direction of an article which covers this? Thanks
Answer: maybe if you post your request to a new page (if you can), it can solve your problem.


Why do the system sounds keep turning on?

Question: Whenever I record with the voice recorder app on my Galaxy S5, the "system" sounds turn on— that is, the bloop sound whenever I click on something, the typewriter sounds when I type, the lock-click sound when the phone unlocks, etc. I would rather these sounds stay off permanently, and am tired of having to keep turning them off in settings.  How can I get these sounds to stay off?
Comment: No idea *why* it happens. To fix that you can try automation using [Macrodroid](https://play.google.com/store/apps/details?id=com.arlosoft.macrodroid). Create a macro with Trigger: Application launched > your recorder app. Action: volume change > set system sounds to zero. Constraints: blank. This should fix the problem or you may need to set other sounds to zero if it is not a system sound ( like notification sound). Let me know how it goes
Comment: Are you rooted?
Comment: It seems you keep the phone in silent mode. When you start audio recording app app, system automatically changes its mode from silent(or vibrate) to Normal mode. Check your audio Profiles and also turn off Audible touch tones, Audible selection etc in you sound settings.
Answer: I reckon this has to do with Samsung Galaxy S5 enhanced audio features typically Wise Voice
According to Samsung:

Wise Voice, a new feature of the Galaxy S5, lets the volume and
  quality of a voice call remain consistent, no matter how far or near
  you speak from the phone. When you speak too far away from the phone
  or the sound is too low, Wise Voice detects it and turns the volume
  up; when you speak too close to the phone, it automatically lowers the
  volume.

I believe this somewhat have effects on system sounds, as volume is auto-controlled by the system to suit the specific environment.
Additionally the Samsung Galaxy S5 has 2 microphones. As explained from this article:  https://forum.xda-developers.com/galaxy-s5/help/microphone-problem-t2807110

The secondary microphone is used for active (ambient) noise
  cancellation. This can be weighted as well when recording audio.
  Normal mode uses both mics omnidirectionally. Interview mode uses the
  secondary mic only in a directional beam pattern. And conversational
  mode takes directional top and bottom audio.

Your issue could be caused by a number of things from poor audio quality, poorly chosen presets or an obstructed  microphone triggering volume to up.
As a possible fix try adjusting the Samsung "wise voice" feature that adds microphone gain to compensate  audio recorded from a distant place or tweak the notifications or sound mode settings or perhaps automation to auto-off sounds when recording (e.g using Tasker):


Is any part of the Peace of Westphalia still valid today?

Question: The treaties comprising the Peace of Westphalia contain many clauses. Most were invalidated since then. Is any of the clauses still legally valid? If not, when was the last of them invalidated or replaced?
Comment: I wonder if this question might be a better fit for https://history.stackexchange.com. But before I migrate it I would be interested to read some more opinions.
Comment: @Philipp I agree. I think, the answer is the dissolution of the Holy Roman Empire of the German Nation in 1806. But that's a history question.
Comment: The Swiss Confederacy was accepted as independent of the HRE. Switzerland is still independent of either Germany or Austria. Does that count?
Comment: I see it personally as worthwhile in the Politics SE as the question does ask what clauses of the Peace of Westphalia are still in impact today (probably in the areas of international relations, diplomacy and economics, all of which relate to politics). However I imagine the History SE is the most likely to provide an answer for the person asking the question.
Comment: I did tag it international law. Not all the clauses relate to the empire.
Answer: You are putting it in a framework governed by law. However, the context of international relations is chaotic and not governed by any laws other than strength (in its various forms, cultural, economic, military).
International laws such as the Treaty of Westphalia are then just some form of contractual agreements valid until the agents decide to not honor them anymore. At a later stage, they may become part of doctrine and become precedents for the resolution of future disputes (in a mechanism similar to common law).
In the specific case of the Peace of Westphalia treaty, if you read the text, you can see that it is actually comprised by many agreements regarding small portions of sparse territories assigned to vassals, therefore among the years I think that technically they have taken full effect, and therefore the law can actually be considered valid.
Of course, if at a later stage other treaties affect these territories, while the law is still valid (it has taken its full effect), the new laws supersede the first one, marking the historical evolution of societal organization.
Answer: While not necessarily directly applicable, the Peace of Westphalia is frequently attributed as the source of a conceptual Copernican revolution in international affairs in which legal authority of divided between sovereign nation states that each have a particular territory which other sovereign nation states respect in the absence of good cause to do otherwise.
Before that time, authority over particular people and places was seen in more of a feudal and personal network of relationships of particular living leaders, rather than as a comprehensive system of sovereign states.
This concept, sometimes called Westphalian Sovereignty, is embedded deeply, sometimes explicitly and more often, reading between the lines, in almost all modern treaties and in international affairs more generally.


Missing workflow task

Question: A user is receiving an email for task that is overdue.  The document that it relates to has been deleted.  I have searched for the task and looked in the task history but I can not find it.  Any suggestions how to track it down or stop the email?
Answer: Please try to clear the config cache followed this knowledge base article, restart the SharePoint Timer and wait for the effect.
KB Article by MS
You can also run the following query against the Web Application content database to determine if there is an orphaned item in the table.
Select * from SecheduledWorkItems where ID='GUID'
If the item and associated workflow tasks are removed while the workflow is still running, it will leave out the related records in ScheduledWorkItems table which timer services checks regularly hence sending out the overdue notification emails at the time. It is because the workflows associated with the document were not yet completed at the time and they need to monitor the status of those tasks.
Sometimes this may not work. Once we had this same issue, we had a rule written in outlook for the particular user and routed it to the deleted items folder.

Missing workflow task


Number of binary words that form a group of Hamming weight at most d

Question: Consider binary words in {0,1}^n whose Hamming weight is at most some constant d. We want to select some of these words such that they form a group under addition. How many words can we choose at most?
Answer: Your structure is known as a group anticode with maximal distance $d$. The obvious construction is to take all vectors of the form $(x_1,\ldots,x_d,0,\ldots,0)$, obtaining a group of size $2^d$. Ahlswede, Another diametric theorem in Hamming spaces: optimal group anticodes showed that this is optimal, and furthermore, the solution I described is unique up to isomorphism, except when $d = 2$, in which case another solution is $(0,0,0,\bar{0}),(1,1,0,\bar{0}),(1,0,1,\bar{0}),(0,1,1,\bar{0})$.
Comment: Great help. Thanks.


Wrong resolution of viewport

Question: I'm usually use Blender with second monitor on my MacBook but when it's disconnected and I try to run blender it's starts in wrong resolution, like on screenshot below. To fix that I connect monitor and drag Blender window to second monitor workspace. So I can't start Blender without second monitor. Any suggestions?
Comment: After December update resolution of main window been fixed. But all new windows appear (preferences, render view) on new workspace with wrong scale.
Answer: It's really complicated because it involves hardware and software. But...
When your machine starts the operating system it sets one of the monitors as the main one, if you disconnect one of the monitors with your machine still on, the system still thinks the main monitor is turned on.
To avoid further problems, it is recommended not to disconnect the second monitor with your machine still connected! For this you should turn off your machine, disconnect the second monitor and then turn it on again so that the operating system recognizes the first monitor as the primary one.
Note: The final version of Blender 2.80 is still in development, many bugs can and will occur, this resolution problem may be part of that release.

Wrong resolution of viewport


How to plot a multicolored curve using a single plot command in matplotlib

Question: If I understand correctly, the function:
matplotlib.pyplot.plot(x, y)

plots len(x)-1 separate line segments - one going from (x[0], y[0]) to (x[1],y[1]), one going from (x[1],y[1]) to (x[2], y[2]), etc. In my application, I want to display a curve consisting of a series of line segments connecting data points in this way, but there is an extra piece of data (z) associated with the transition between each of these data points, that I want to represent by the color of the line segment. Clearly one way of doing this is the following:
for i in range(len(x)-1)):
    matplotlib.pyplot.plot(x[i:i+2],y[i:i+2], color=z[i])

but is there a way to do it that doesn't involve a separate call to matplotlib.pyplot.plot for each line segment?
Answer: You should be able to use matplotlib.collections.LineCollection which accepts a colors parameter (sequence of RGBA tuples).


What is a good browser for a publicly accessible computer?

Question: Im setting up a computer in a hotel lobby and it needs to not have any history or remembered information between browsing sessions. Chrome is the most compatible browser but it seems difficult (if even possible) to completely disable history. Firefox is ok but i really don't care for the newest version of it. Edit: I also need to be able to prevent users from creating bookmarks and installing addons. Is there a way to force Chrome and/or Firefox to always use Private Browsing? Is there another browser out there built specifically for this purpose?
Comment: On Linux firefox has a system-wide /etc/firefox wtih a profile that can force settings onto a user's profile and cannot be overriden.
Comment: Caution:  If the computer is physically accessible, it is hard to secure.  If hacked it could be set to appear to reboot, or even reboot, but not before sending details of the session somewhere.
Answer: Probably the best option would be to run a cut down Linux installation with Chrome/Chromium or Firefox that runs in RAM only, starts straight into the browser  and then reboots on exit.
It will be fast to operate and allow no changes to the settings, bookmarks, history, cookies, etc. to be preserved between sessions - you could even run it from a USB that has a physical switch for read-only mode to further ensure that nobody can make changes with any lasting impact.
An example of setting this sort of thing up can be found here.
Comment: Thanks, I like this idea a lot....I almost wonder if Chrome OS would be a good choice....I hate Chrome OS personally, but it could be ideal for a public terminal...depending on how annoying it is to use lol
Comment: @Braden'sPC's: Personally I would be tempted to use a RaspberryPy 3 with the same sort of OS changes.


Columns not created

Question: I'm using Subsonic (SimpleRepository) in my new project and enjoy using it but...
With one, and only one of my table, it does not create all the columns and i don't understand why.
Here is the code :
<code>public class Rating
{
    public Rating()
    {
        UsernameVotant = "";
        UsernameEvaluate = "";
        Note = 0.0;
        NoteAccueil = 0.0;
        NotePedagogie = 0.0;
        NoteRapportQP = 0.0;
        NoteContenu = 0.0;
        Comment = "";
        stageId = 0;
        DateNote = DateTime.Now;
        isValidate = false;
    }
    [SubSonicPrimaryKey]
    public int ID { get; set; }
    public DateTime DateNote;
    public int stageId;
    public string UsernameVotant;
    public string UsernameEvaluate;
    public int Note;
    public int NoteAccueil;
    public double NotePedagogie;
    public double NoteRapportQP;
    public double NoteContenu;
    [SubSonicLongString]
    public string Comment { get; set; }
    public bool isValidate { get; set; }
}
</code>

Called like my other classes :
IRepository _repoRun = new SimpleRepository(Core.Config.ArticlesDB, SimpleRepositoryOptions.RunMigrations);
<code>   public bool AddRating(Rating p)
    {
        _repoRun.Add<Rating>(p);
        return true;
    }
</code>

The table Ratings created contains the columns :
ID, Comment, isValidate
Whatever what I try to add as default value, the 3 columns contains the value :
ID = 1 (2, 3, 4...) -> works
Comment = ""
isValidate = false

As i noticed a probleme where naming a column "Read", i tryed to rename the columns, rename the table (which was "Vote" [in french]) but the probleme is the same as with my original table "Votes"
Could you help me please.
Thanks in advance (and sorry for my english)
Answer: The only properties you're defining in that class are ID, Comment and isValidate so these are the only columns SubSonic will generate. Change your fields to properties and SubSonic should create columns for them:
<code>[SubSonicPrimaryKey]
public int ID { get; set; }
public DateTime DateNote { get; set; }
public int StageId { get; set; }
public string UsernameVotant { get; set; }
public string UsernameEvaluate { get; set; }
public int Note { get; set; }
public int NoteAccueil { get; set; }
public double NotePedagogie { get; set; }
public double NoteRapportQP { get; set; }
public double NoteContenu { get; set; }
[SubSonicLongString]
public string Comment { get; set; }
public bool IsValidate { get; set; }
</code>

Columns not created


Ubuntu 20.04 fprint update

Question: I installed Ubuntu 20.04 and wanted to update it. After restart of a system software update popped up and stated that software update is available. I saw that it suggests to update fprint in a very long list of available updates. Is it included in main repository? I do not want to install anything from universe.
I did not change anything just installed Ubuntu and restarted the computer.
Comment: Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
Comment: I cannot find anything definitive on this. fprint is not installed on my Ubuntu 22.04 system and it is updating properly. Did you install an App that requires fprint? That is all I can think of
Comment: I did install the Ubunru 20.04 LTS and did not change anything. After restart of the system I got a notification that updates are available. In the update list I see fprint.
Answer: You have found the
Fingerprint reader support :

The fprint project aims to add support for consumer fingerprint reader
devices, in Linux, as well as other free Unices.
Previously, Linux support for such devices has been scattered amongst
different projects (many incomplete) and inconsistent in that
application developers would have to implement support for each type
of fingerprint reader separately. For more information on where we
came from, see the project history page.
We're trying to change that by providing a central system to support
all the fingerprint readers we can get our hands on. The software is
free software and in the long term we're shooting for adoption by
distributions, integration into common desktop environments, etc.

For programmers, this unfortunate name may cause a confusion with the
fprint C-function.
Comment: Is it included in main repository for Ubuntu 20.04?
Comment: Your experience indicates that it's included.
Comment: If my answer was helpful, please consider marking it as accepted (click the ✔ sign).

Ubuntu 20.04 fprint update


Inner dimensions of Anchor Hocking glass measuring cups

Question: Hi especially for those who already have the Anchor Hocking glass measuring cups. I wonder what are the inner dimensions (diameter) and height of the Anchor Hocking glass measuring cups (250 ml and 500 ml). I saw several contradicting dimension measurements all over the Internet so I ask this question. Thanks.
Comment: What practical problem would this solve? They're truncated cones, so the diameter changes all the way up.
Comment: I just wanna know the inner upper diameter of the, at least, 1 cup sized Anchor Hocking glass measuring cup before buying one.
Answer: Anchor Hocking has been making glass measuring cups for 50 years (possibly as long as 70).  During that time, they have had multiple different designs for their glass measuring cups, some taller and narrower, some wider and shorter.  Even within the last 5 years they've changed the design slightly (taller spouts).
As such, unless you're specifically asking about the 2021 measuring cup model, you're going to get a variety of different measurements.
Comment: I'm talking about I think it is the new one with red letters on it. Any idea?


How can I remove a stubborn screw from a toilet seat?

Question: I would like to install on a one piece Kohler toilet a bidet. When I tried to unscrew the screw from the seat, it just didn't move, not with the screwdriver nor the drill. So I put wd40 but still doesn't move. Can u please help me.
Comment: You want a penetrating oil like Kroil.  WD-40 is not penetrating oil.  It is Water Displacer (hence: WD).  It was made to stop rust on missile tanks.
Comment: So the screw doesn't move with a drill at full power and also doesn't start to strip?  This may mean you need more muscle or a better drill.  Also you can use some pliers to get it started.
Comment: Thank you for answers, but this is a one piece Kohler toilet. I don't have reach to the bottom of the screw. We tried with pliers & an electric dril. Still not moving
Answer: I was away & finally found The Answer to your problem. Your problem is Kohler & their very bad design. They use "anchors" & it's quite ridiculous, here are a couple of videos to demonstrate handling of Kohler's anchors.
Video 1
Video 2
Comment: I can't remove the seat, the screw doesn't move
Comment: Sorry, I was suspended for a week. If you haven't had any luck & can't get to any underside nut there's only 2 options left. First & kind of tricky is to drill-out the screw, you'll have to guess & go to bigger drill bits until the screw head pops off if you can't see how big the screw is. SECOND & YOUR BEST OPTION, is to remove the whole toilet. The nut for the seat screws is inside & should let loose from there. That toilet is heavy, maybe 100-lbs, & get all the water out that you can. Or, call Roto-Rooter.
Answer: We have 3 Kohler 1-piece Rialto toilets. I replaced the seats on all 3 of them. The first two took about 2 minutes each. Unscrew bolt, replace seat, re-screw bolt. But the 3rd toilet had bolts that would not budge. I bought an impact screwdriver from Amazon and went to work. Using a small sledge hammer the first bolt finally loosened after about 30 blows. What a workout! The second bolt was more stubborn and the groove (flathead) at the top was beginning to widen causing the flathead bit to slip. I used layers of heavy gauze and plastic waterproof tape to narrow the groove and finally got the thing to unscrew. What a pain!
Answer: This problem is caused by a basic design error.  The bolt is stainless steel and the anchor is aluminum.  These metals are never supposed to be used together in a fastener due to galvanic / bimetallic corrosion. This is particularly true on a toilet seat where the anchor will be exposed to acidic / alkaline fluids and cleaning products, which promote galvanic corrosion.  The aluminum anchor will corrode and 'glue' itself to the bolt.
I just ordered a replacement set, hoping that they fixed this Engineering 101 problem.  Nope!  Same aluminum post with a stainless steel bolt.  See the photos below.
I notified Kohler and sent them a galvanic corrosion metals chart and photos.  No reply yet.  I did find one potential online source for stainless steel molly / sleeve expansion bolts.
Mark


Periodicity trick for Kerr Black Holes

Question: I am slightly confused concerning the euclidean section of a Kerr black hole. In page 5 of the following paper
https://arxiv.org/abs/hep-th/9908022
it is said that in order to get the euclidean section, we need to set $t \to i \tau$ and $a \to i a$. (They consider general Kerr-Newman-AdS black holes but I am simply interested in Kerr asymptotically flat.) This makes sense because we want to keep the $dt \otimes d\phi$ components of the euclidean metric real. What confuses me is that if we do the analysis of the conical singularities as they mention, we will get the following periodicity for $\tau$ and $\phi$
\begin{equation}
\tau \sim \tau +\beta
\end{equation}
\begin{equation}
\phi\ \sim \phi+i\beta\Omega_H
\end{equation}
with $\beta$ the inverse temperature and $\Omega_H$ the angular velocity of the event horizon, namely
\begin{equation}
\Omega_H=\frac{a}{r_{+}^2+a^2}
\end{equation}
where $r_{+}$ is the event horizon and $a$ is the rotation parameter of the black hole.  What is strange to me is that if we take $a \to 0$ in Boyer-Lindquist coordinates, we get that
\begin{equation}
\phi \sim \phi
\end{equation}
because $\Omega_H$ vanishes. This becomes a trivial identification and it does not tell us anything about the periodicity of the $\phi$ coordinate. However, we also know that if we take the $a \to 0$ limit, we get the Schwarzschild black hole in Schwarzschild coordinates. In Schwarzschild Euclidean, we should take the $\phi$ coordinate to have period
\begin{equation}
\phi \sim \phi+2\pi
\end{equation}
and even though the Boyer-Lindquist $\phi$ is different than the $\phi$ in Schwarzschild, they match in the limit I am considering $a \to 0$. What does this imply? Does this mean that even though Kerr goes to Schwarzschild in the limit $a \to 0$ as a lorentzian geometry, their euclidean sections are not connected continuously somehow?
Edit1: I also have the notion that in lorentzian Kerr, the $\phi$ coordinate has periodicity $2\pi$. When we go to Euclidean, we seem to get this other periodicity: but shouldn't the periodicity of $2\pi$ be preserved as well? At least that is what happens in Schwarzschild. So we would have both
\begin{equation}
\phi\ \sim \phi+i\beta\Omega_H
\end{equation}
\begin{equation}
\phi\ \sim \phi + 2\pi
\end{equation}
It also confuses me that this manipulations are usually done based on the coordinate systems and therefore it is harder to get a notion of what it means to 'euclideanize' in a coordinate invariant way. If someone has a coordinate invariant way to talk about this analytic continuation, I would like to hear it.
Edit2: If we see what really is the expression in the identification of $\phi$, we get
\begin{equation}
i\beta \Omega_H=i4\pi \frac{r_{+}a}{r_{+}^2\left(1-\frac{a^2}{r_{+}^2}\right)}
\end{equation}
By doing the analytic continuation $a \to ia$, we have
\begin{equation}
i\beta \Omega_H=-4\pi \frac{r_{+}a}{r_{+}^2\left(1+\frac{a^2}{r_{+}^2}\right)}
\end{equation}
we see that it is alway less then $2\pi$ because
\begin{equation}
r_{+}=a+\sqrt{2}a
\end{equation}
defines extremality assuming the fact that we set $a \to ia$. So it seems to make the $\phi$ direction smaller in general. But if I try to compute the action on-shell
\begin{equation}
I=\int_{\partial \mathcal{M}}K-K_0
\end{equation}
I have to integrate from $0$ to $2\pi$ along $\phi$ to get the right result mentioned in
https://doi.org/10.1103/PhysRevD.15.2752
because since we are sending the boundary to infinity only the leading order of $1/r$ matters which is the same as in Schwarzschild. So I am confused what kind of geometry we have along $\phi$.
Answer: There is a subtlety in expressing the periodicities of variables, because sometimes, there may be a "shift" or "twist" in one variable as we go along a cycle in the other. This is indeed what is happening here.

The picture above is a torus in the complex plane where $z = x + i y$. The torus is defined by two equivelence relations,
\begin{align}
z \sim z + 1 \\
z \sim z + \tau.
\end{align}
If we write
$$
\tau = \tau_x + i \tau_y
$$
then we can write the equivalence relations as
\begin{align}
(x, y) &\sim (x + 1, y) \\
(x, y) &\sim (x + \tau_x, y + \tau_y).
\end{align}
There is a "shift" in $x$ as $y$ goes from $0$ to $\tau_y$. Note that the range of x is $1$ and the range of $y$ is $\tau_y$. $\tau_x$ gives the shift.
This is exactly what is happening in the Euclidean Kerr metric. If one defines the variable
$$
\Phi \equiv \phi - \Omega t
$$
where $t$ is euclidean time, then the equivalence relations are
\begin{align}
(t, \Phi) &\sim (t + \beta, \Phi) \\
(t, \Phi) &\sim (t, \Phi + 2 \pi ).
\end{align}
If one uses $\phi = \Phi + \Omega t$, then this becomes
\begin{align}
(t, \phi) &\sim (t + \beta, \phi + \beta \Omega) \\
(t, \phi) &\sim (t, \phi + 2 \pi ).
\end{align}
$\beta \Omega$ is therefore not giving the periodicity of $\phi$ but rather the "shift" in $\phi$ as one makes a full cycle in $t$.

Periodicity trick for Kerr Black Holes


The Quantum Hall Effect

Question: In the integral quantum Hall effect, one has that in regions where $R_{xy}$ (the Hall resistance) is a constant, $R_{xx}$ surprisingly goes to zero. Why does that happen? Do impurities in the material play a role in this?
Answer: $R_{xx}$ is the longitudinal resistance, i.e. the resistance along the direction of current flow -- to be clear, this is the direction where the voltage difference from the battery or power source is. So a vanishing $R_{xx}$ means that there is dissipationless flow, and is kinda the "big thing" about the IQHE (Integer Quantum Hall Effect) and why it relates to topological insulator.
The fact that $R_{xx}=0$ is not directly caused by impurities. Impurities guarantee that there are plaeatux in $R_{xy}$ and hence large regions where $R_{xx}=0$.
The reason why $R_{xx} = 0$ is that at that precise value for magnetic field strength $B$ and carrier number $N$, the Landau levels are completely filled. So there are literally no other states an electron could scatter into. If you had Ohmic (i.e. dissipationful) conduction, you'd have an electron being scattered around and hence moving at a constant velocity (Drude model). This would require it to change motional state at each scattering event. But if there are no states available, then it cannot scatter and it just keeps going along. Hence the carrier flow is dissipationless (and topologically protected, in that it does not matter whether you put a dent on the edge of the material, electrons will go around it and still move along the edge).
The presence of impurities just broadens the otherwise fixed-energy Landau levels, thereby guaranteeing that the dissipationless flow occurs for a range of $B$ (or $N$) and not just at the specific combination filling the Landau level.
Just to be clear, the plot we are talking about is ($R_{xy} =R_{\mathrm{H}}$ and $R_{xx} = R_x$):
Comment: So if I understand correctly, the Landau levels in this system are the various possible wave functions whose centers are at a fixed separation from each other. Also, a_xy=(ne^2/h), where a is the Hall conductance and n is the number of Landau levels that are filled. The characteristic step graph of Hall resistance "is" because Landau levels are getting filled one by one and not that all are filled from the onset. Then according to you, Rxx should be zero only when all the levels are filled, and not throughout the whole graph. Also, how would you explain the spikes in Rxx then?
Comment: The Landau level is just the energy level. The collection of the individual closed orbits (the ones you are talking about) is the just the *degeneracy* of the Landau level, i;e. how many particles there are per energy level. The spikes in $R_{xx}$ are when you are filling the *next* broadened Landau level. Now there are states that electrons can scatter into, so they do not have to go *only* forward. This causes usual Ohmic dissipation.
Comment: So when you say that Landau level is filled, you mean that all the possible degenerate states associated with that Landau level is filled?
Comment: Yes. All the states in a given energy level are filled.
Comment: Oh! Got it. Thanks a lot :)

The Quantum Hall Effect


Safari HTML for Planck Constant/Reduced Planck Constant

Question: I want to display the symbols commonly known as h and h-bar.  This is for Safari and iOS.  What do I enter into my html?
If there is a table of such things somewhere, a reference to it would be great.  Thanks!
Answer: You're looking for a list of unicode characters and an explanation of how to use them. Unicode characters should work in any browser regardless of platform.
The ℏ is written as ℏ and ℎ as ℎ in a html document.
If you're representing complex equations you might get on better with LATEX
Answer: Regarding information about such things in general, the Unicode standard is authoritative in matters of character codes. But it does not prescribe the use of characters. Various sources need to be consulted; in matters of mathematical and physical notations, you can consult the ISO 80000 series of standards (available from standards organizations for rather high prices).
The Planck constant can be viewed as letter h in italic, but the non-normative well-written advisory document Unicode in XML and other Markup Languages says: "to indicate Planck's constant one should use U+210E and not h" (the logic behind this is not simple, but logical). For the reduced Planck constant, there really is no choice. So use the characters U+210E PLANCK CONSTANT and U+210F PLANCK CONSTANT OVER TWO PI. The way to enter them depends on the authoring context.
In modern authoring, you should be using a Unicode-enabled editor and should be able to insert these characters directly. Failing this, the character references mentioned by toomanyairmiles can be used.
Then there's the character problem. These special characters have limited support in fonts. You should probably analyze their availability and suggest a list of fonts in CSS, as outlined in my Guide to using special characters in HTML.
Note that the reduced Planck constant character has varying shapes (glyphs), with either horizontal or slanted stroke over the letter h. The glyph in the relevant Unicode chart has a slanted stroke. This is not normative but suggestive, and it makes it clear that the character is different from h with stroke (ħ, a letter used in some natural languages).
Comment: You raise a very good point about font support, and in the fractured world of mobile that may be a very big issue. Thing is I'm not clear if you think my answer is right or wrong, or if there is a better approach?
Comment: @toomanyairmiles, your answer is correct, but instead of character references, real characters can be used, too, and additional markup may be needed (to enclose the characters into elements, so that their font can be set as different from normal copy text font, if needed).
Comment: Checking the characters on my Android, I saw both of them as small boxes, like I sort-of expected. So if only the Planck constant is needed, using `h` is more practical than the recommendation. If you also need the reduced Planck constant, you could consider using an embedded font, or using MathJax (which renders LaTeX notations with client-side JavaScript).


Fatal error in latex

Question: Hi there I am using texstudio version 2.12.6. I tried compiling a document (CV) but it keeps repeating the fatal error : "cannot-use-pdftex" The fontspec package requires either XeTeX or LuaTeX to function. You must change your typesetting engine to, e.g., "xelatex" or "lualatex" instead of plain "latex" or "pdflatex". For immediate help type H .
I tried to configured the default compiler from pdflatex to xelatex but it will not just work. It keeps returning to pdflatex when I go back to check.
Any help will be appreciated.
Comment: your question is just about the editor you are using, but you have not said which editor that is. On the command line, calling latex directly switching from pdflatex to xelatex simply means typing `xelatex file` instead of `pdflatex file`.
Answer: You can either go to the Tools -> Commands menu and choose xelatex, like this:

Or, change the default compiler, go to the Options -> Configure TeXstudio, choose the Build icon and change the default in the editing window:
Comment: It wouldn't still work. The error provided is as stated below. Undefined control sequence. \begin{document}
Missing number, treated as zero. \begin{document}
Font \fb=''[cmr10]'' not loadable: metric data not found or bad.
XLTXTRA IS TO BE USED ONLY UNDER XETEX. LOAD FONTSPEC DIRECTLY, INSTEAD.
Comment: Did you click on OK after you've changed the default compiler?
Comment: Yes I did that.
Comment: Which distribution do you have, on which system?

Fatal error in latex


What happens to the semiconductor in an LED when it is given voltage that it cannot handle?

Question: Let us suppose we have an LED with the allowed voltage range of 1.5V to 4.5V and we gave it somewhat 12V. Now we know that the LED will instantly stop forever. But I want to know what will happen to the material (the semiconductor, etc) when this happens.
Answer:
what will happen to the material?

The LED may experience an incident of Electrical Overstress (EOS).  This depends upon if the supply can source more current than the LED's maximum allowed current. And if the LED can tolerate the supply's maximum current whether the supply's voltage output will reduced itself to the LED's forward voltage. Over current and over voltage (i.e. over driven) will result in an EOS event.
EOS damage in LEDs can be catastrophic in that the
LED is permanently non-functional immediately after the EOS event; alternatively there may be just partial damage whereby a significant degradation in performance or the complete failure of the LED only occurs later on. Partial EOS damage in LEDs might manifest itself as reduced light output, poor thermal performance, and/or a shorter service life.
When an LED is overdriven the failures here include the thermal overstressing of the semiconductor or a fused wire bond.  Notice in the photo below the stress is greatest where the bond wire is attached to the semiconductor material.
Example of a fused wire bond
In this experiment, an LED with a 250 mA max was driven with 1000 mA, four times the data sheet
maximum current.   Failure occurred in about
10–20 seconds.

X-Ray of the above LED failure mode

In the image below the same LEDs were driven with a single pulse of 3000 mA for 300 ms. Due to the higher power load, the bond wire melted.

Source: The Basic Principles of Electrical Overstress (EOS) App Note
Answer: An LED can be modeled with a very complex circuit, but for the purposes of answering this question, it can be modeled as a voltage drop with a series resistor in parallel. The allowed voltage range is determined by keeping this series resistor in mind. More specifically, the power dissipated across this resistor should be less than what the physical package of the LED can dissipate into the environment. When the supply voltage  is more than the rated value, the resistor generates more heat than what the package can dissipate into the environment thus heating the LED gradually. Once a particular temperature is crossed, the junction can burn down in several different ways depending on the construction of the junction.
Comment: The magic smoke (that makes it work) leaks out.
Comment: An LED cannot be modeled as a resistor.  An LED can be damaged from over voltage and a resistor cannot.   You do not model a voltage drop, its the result of the modeling of a resistance.  An LED could be modeled as a variable resistor where the resistance is a function of the current flowing through it.
Comment: TL;DR: It melts.
Comment: ^that^ or the bondwire melts like in a fuse, or the semiconductor material evaporates, or a combination of all that is mentioned here.
Comment: Ohh. Well, that clears my doubt. Thanks everyone!
Comment: I have accidentally put 12 vdc across an LED.  It makes a "pop" sound, much like a glass fuse with sudden current overload.


Balun for impedance matching?

Question: I would like to match the output impedance of amplifier IC (1.5 + j1.7) with an antenna of impedance (50 or 75 Ohms) at around 950 MHz.
I thought of using a balun so that I could avoid the inductor and capacitor.There are 1:1 or 1:4 or 1:10 or 1:16 impedance conversion baluns. But here I have complex impedance at the source.
Could any one recommend how to deal with this ?
Answer: You could use a shunt capacitor to make the output impedance of the amplifier real, and then use a proper balun to achieve 50 or 75 ohms.
Comment: Moni, do you need any clarification?

Balun for impedance matching?


Does L'hopital work for one sided limits?

Question: Simple question, to which I don't know the answer. Does it work the same even if we are only interested in one-sided limits, and it won't cause problems that the actual limit doesn't exist?
Answer: One of the key assumptions of L'Hospital's theorem is that the functions $f,g$ be derivable over an open interval one of whose end points is $c$, the point to which $x$ tends:
$$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}.$$
The proof due to Taylor and presented on Wiki does operate only over this one sided interval of common differentiability.
So the L'Hospital rule is basically about one sided limits.
To be very rigorous, first, one has to examine the derivatives on both sides of $c$. It may turn out that the left and the right limits differ or only one of them exits.
Answer: I am not 100% sure what you mean, but let me state the following things and you tell me if that answers your question.
First, let me state l'hopital's rule:
Let $I=(\tilde{x}_0,x_0)$ be a non-empty open Interval and $ f,\, g \colon I\to\Bbb{R} $ differentiable functions, s.t. $x\nearrow x_0$ both functions tend to 0 OR both functions diverge ($-\infty$ or $+\infty$).
If $ g'(x) \neq 0$  $\forall x \in I$ holds and $\tfrac{f'(x)}{g'(x)}$ converges for $x\nearrow x_0$ to $c$ or diverges ($-\infty$ or $+\infty$), then so does $\tfrac{f(x)}{g(x)}$. Analogue for $x\searrow \tilde{x}_0$.
If $I$ is fully contained in an open intervall, where the above conditions hold, then especially the following holds$$\lim_{x\to x_0}\frac{f'(x)}{g'(x)}=c~\Rightarrow~\lim_{x\to x_0}\frac{f(x)}{g(x)}=c.$$
This also holds true for interval boundaries $x_0=\pm\infty$.
Secondly, recall that:
By definition the limit $\lim_{x\rightarrow x_0}f(x)$ exists if and only if $\lim_{x\searrow x_0}f(x)$ and $\lim_{x\nearrow x_0}f(x)$ exist.
Hence, for the statement you only need onesided limits, and the existence of the bothsided limit in the result implies the existence of the onesided ones.


How to show records for a particular month

Question:
Show the Sales, Profit and ShipDate fields (only) of any records from the month of January (any year) where Region and Province are both Ontario.
Comment: What datatype is ShipDate?
Answer: ....... AND Month(Shipdate) = 1
Answer: Does your shipdate in invoice table contain a date or a month only. If it's a date then you need to define an expression (a new field in your query as expression Month (shipdate) and then put your WHERE criteria January in newly defined expression. Hope this helps.


Is it possible for a shell script to contain both tcsh and bash commands in the same script?

Question: Say, I modify a lot of code that uses tcsh, but since a lot of people say that bash is more modern, I'd also like to learn bash along the way. Is it possible for me to use a script that contains both of these commands at the same time? Since this is for personal use, elegance is not an issue.
Answer: I wouldn't recommend mixing and matching shell types within a script.

You lose efficiency in launching a shell to execute a command and return a result.
One of the reasons to keep everything a script needs in-script is that that script has access to all of the variables declared and returned from functions, etc.  Passing an entire variable stack to another shell just to execute a couple commands and then return, seems a bit counter-intuitive.

I would say your best bet would be to copy a known script and change the calling shell, and then see what breaks, so you know where the differences in syntax are that have to be worked out.
But, in short, trying to mix and match within a script just seems like a bad idea.  (personally)
Answer: I agree with the other answers that it's not generally a great idea.
But you can invoke a shell from a script.  If you have a script currently written to use tcsh and you want to use a few bash-specific features, you can invoke bash explicitly.  For example, bash makes it much easier to redirect stdout and stderr independently.
#!/bin/tcsh -f

echo Hello, tcsh
bash -c 'echo Hello, bash 1>&2' # writes to stderr
echo Good-bye, tcsh

In the long run, though, it's probably better to translate the entire script to use bash.
Obligatory link: http://www.faqs.org/faqs/unix-faq/shell/csh-whynot/
Answer: No, it is not possible.
The reason is that the shell reads in chunks of the file as it executes it, and a chunk might contain shell code that you want to run in a different shell, but since its already been read by the first shell, the second shell cant read it if it were to take over execution of the script. (This could work if the shell had the ability to lseek() the file, but shell languages just arent that advanced)
However there is a dirty way of doing this; zsh.
zsh has a command called emulate which lets you switch shell languages on the fly. But using zsh just to emulate tcsh and bash is just plain dirty, not to mention overly complex.
.
Also I wouldnt say bash is more modern. It's more widespread sure, but it adopts features at a very slow pace. Bash is more focused on compatibility than anything.
Comment: Adopting features at a "very slow pace" isn't really a "bad thing".  Look at how many distros Debian has spawned, and some of those have then spawned countless others, but in so many cases, when people start running into unexpected behavior and want to get back to "doing things right", they take a pause, and start again from Debian as a base.


Why can't photons have a mass?

Question: Why can't photons have a mass? Could you explain this to me in a short and mathematical way?
Comment: This is on an entirely different question, does the photon have a mass. The question is different, the answers are different. In fact, the "duplicate" question has NO answer that tells why it is believed that photons are massless. And the question is one of some interest. In addition to the current literature (i.e. http://arxiv.org/abs/0809.1003 ), J. D. Jackson has a discussion of this question and the question here has attracted 5 answers, none from a less than 400 rep author, and with a total of 9,000 reputation. Reopen the question.
Comment: Jackson's book on Electrodynamics starts off in the introduction on the experimental evidence for why we believe a photon does not have a mass. It is not an axioma as far as I know.
Comment: Well, if photon has a zero mass, why they can't escape a black hole? Is this why any body with a mass can't escape a black hole?
Answer: The other answers explain that there's no paradox but they don't explain why the particular particle called photon is massless.
It's massless because it is the messenger particle responsible for electromagnetism which is a long-range force. Its range is infinite so the mass has to be zero. One may view the Coulomb potential as the zero-mass limit ($m\to 0$) of the Yukawa potential
$$V(r) = \frac{\exp(-mr)}{r} $$
OK, so why is it massless and why the range is infinite? It's because of the unbroken $U(1)$ gauge invariance for the electromagnetic field that acts on the electromagnetic gauge potential as
$$A_\mu\to A_\mu+\partial_\mu \lambda$$
The mass term (in the Lagrangian) for a gauge field would have the form $m^2 A_\mu A^\mu/2$ and it is not invariant under the gauge invariance above. The gauge invariance is needed to make the time-like mode $A_0$ unphysical - otherwise it would produce quanta with a negative norm (because of the opposite sign in the signature for the timelike direction) which would lead to negative probabilities.
However, gauge fields may consistently become massive via the Higgs mechanism - like the W-bosons and Z-bosons. Then they lead to short-range forces. Beta-decay is mediated by the W-bosons.
Comment: @LubošMotl This answer is wrong. If you add the mass term $\frac12 m^2 A^2$ to the QED Lagrangian, you break gauge invariance but there are no negative norm quanta. The theory (Proca Lagrangian) is well defined, unitary and finite (there are problems with naïve power-counting renormalisation, but as long as the current is conserved, all measurable quantities are finite). The most correct and general treatment of massive photons is through the Stückelberg mechanism (which becomes the Proca theory in the unitary gauge). The S. theory is perfectly well defined, unitary, covariant  and finite.
Comment: @MurodAbdukhakimov: That's not quite correct. Neutrinos always travel at some significant fraction of $c$ ;-)
Comment: Because the strong force is confining, so only color-neutral ("uncharged") particles are allowed to exist in isolation. Correspondingly, the force among such neutral particles decreases quickly with the distance. In fact, the decreases is faster than the power law because the gluons self-interact so they are confined, too. The mass of colored objects is a subtle thing - it depends on the RG scale and the masslessness is only relevant at very short distances, much shorter than the proton radius (QCD scale) where the confinement starts to matter.
Comment: @MurodAbdukhakimov: that nothing with mass can travel at the speed of light is an axiom of physics. Einstein's equations show that in order for mass to travel at the speed of light would take an infinite amount of energy. Also, results show that neutrinos ALWAYS travel sub-C. So, you asked Crazy Peanut how he knew that neutrinos did not travel at the speed of light. I am going to reverse that on you. How do you know that neutrinos travel at the speed of light?  Your profile lists theoretical particle physics as your specialty, so you must have some basis for your stance.
Comment: OK, I actually agree. I would still bet anything that the photon is exactly massless - and there are also reasons that will be understood beyond the QFT framework why it has to be so. I think it's not an accident that the light but massive gauge bosons we know get their mass from the Higgs mechanism, not via Stueckelberg, and in quantum gravity, the Stueckelberg masses have to be high enough - comparable to the fundamental scale - for consistency, for some reasons similar to the weak gravity conjecture if not exactly that one.
Comment: Neutrinos also travel at the speed of light but have non-zero masses.
Comment: @ϚѓăʑɏβµԂԃϔ: How do you know that?
Comment: @MurodAbdukhakimov: Well, I believe in *observations* ;-)
Comment: Why the strong force has a short-range but gluons are massless?
Comment: Can the answer be because it is impossible to accelerate them ?
Comment: Hi Professor Motl. Is that also related with the mass gap of Yang-Mills fields?
Comment: Hi @XiaoyiJing - most things are related in some way. The mass gap of Yang-Mills fields is just a different statement of the fact that there exist no massless - or arbitrarily light - allowed (color-neutral) bound states in QCD and similar theories. For QCD, the conclusion is the opposite one - a massless particle does not exist.
Comment: Luboš That the electrostatic force is a infinite force seems to be a postulate. Please pay attention to my answer [here](http://physics.stackexchange.com/questions/232787/do-we-know-why-the-field-produced-by-charge-extends-to-infinity/232856#232856).
Comment: The related elaboration shows how with a postulate about finite electrostatic fields the exchange between positive and negative charges happens without virtual photons.
Comment: And the EM radiation is not a force. But of course it's range is infinite because photons are indivisible particles, travelling until they hit something.
Answer: There is nothing special about the photon having zero mass. Although zero is the smallest mass any particle can have, it is as good as any other value. In this sense, there is no mathematical proof that the photon has to have zero mass, this is a purely experimental fact. And, to our best knowledge, the photon mass is consistent to zero.
If you want to describe a theory with a zero mass vector in a manifestly relativistic way, you have to have gauge invariance. This is a mathematical fact. As is the fact that if you force this symmetry to be quantum mechanically exact, the mass will not receive quantum corrections (perturbatively, at least). Gauge theories can be shown to have all sorts of other nice features (like IR finiteness, if you sum enough virtual and real diagrams) and that makes us believe that at low energies they are the right theories.
But one would be inverting the logical order within physics if one says that the mass of the photon is zero because EM is described by a gauge theory. EM is described by a gauge theory because the photon has zero mass. There would be no problem with special relativity either. The fact that the maximal velocity is the same as the velocity of light in the vacuum is, again, an experimental fact (equivalent to the one we are discussing here) but by no means necessary by any mathematical theorem.
Comment: Sure, any zero mass field equation will have this property, not only Maxwell's equation. I know that historically it was not done this way, but teaching in the historical sequence is usually the worst thing to do. Today, the way people understand is the following: you measure the mass and the spin of the particle. From these values and special relativity you can get a well define lagrangian with which you calculate all the rest.
Comment: If Maxwell's equations are to be relativistically covariant, then light must travel at some constant c' in all frames. If it wasn't the same as the universal limiting velocity c, you'd end up with a different velocity for a photon and hence light from the velocity addition formula. From this it must have zero rest mass, correct?
Comment: "If you want to describe a theory with a zero mass vector in a manifestly relativistic way, you have to have gauge invariance.". This is wrong. Maxwell equations allow for magnetic charges, in wich case no $U(1)$ gauge invariance is possible. But source-free EM waves are still possible, even with magnetic charges.
Comment: IR and EM? I am no familiar with these acronyms
Answer: According to the special theory of relativity, any particle with a finite mass requires an infinite amount of energy to reach the speed of light. Therefore no particles with any intrinsic mass can travel with the speed of light.
The energy required to attain a speed $v$ is given by $E$ = $\frac{mc^2}{\surd1-v^2/c2}$ - $mc^2$
As $v$ approaches $c$, $E$ approaches $\infty$.
Only massless particles are allowed to travel at the speed of light. Photon is massless, hence it can travel with the speed of light. The energy of a photon is given by $E = pc$ where p is the momentum of the photon.
Comment: It is a nice precise explanation but it's wrong, unless you make the circular definition "the c used in Maxwell's field equations is the same c used in special relativity". If there were some proof that it was impossible in the manner of 2+2=4 people wouldn't be getting paid to set limits on the mass of the photon as in Phys.Rev.D82:065026,2010, "Upper Bounds on the Photon Mass", Antonio Accioly, José Helayël-Neto, Eslley Scatena, http://arxiv.org/abs/1012.2717 Google Proca electrodynamics.
Comment: Good answer, just minor points: "are allowed" -> "are forced/required", "it can travel" -> "it has to travel". Because otherwise it seems like they can also decide not to :)
Comment: This doesn't explain why photons don't have (rest) mass - very small rest mass, yes.
Comment: @John McVirgo: Are you really voting on the basis of the merit of the answer? BTW your comment is absolute nonsense.
Comment: @sb1 "why can't photons have mass?" was the question. It could be that photons have a finite rest mass and so don't reach this limiting velocity. How can we know that the speed of a photon is this limiting velocity?
Comment: @john McVirgo: Photons are particles of the electromagnetic field. Electromagnetic field is described by Maxwell's field equations. The requirement, that these equations are invariant under inertial transformations forced us to accept a limiting velocity $c$ which is nothing but the velocity of the electromagnetic waves. Hence photons which are particles of em waves move with this limiting velocity $c$. Do you understand now why light moves with the speed of light? ;)
Comment: @sb1, that's a nice, precise explanation.
Comment: @Carl Brannen: Are you sure the above explanation is wrong? Are you sure you understand the context of the paper you cited? I am not too sure. Since you have gone to a more generalized context than Maxwell's electrodynamics, namely proca theory. By this you have brought QED quite unnecessarily. Yes, under proca theory a massive photon is allowed and in the appropriate limit the proca theory produces the classical Maxwell's theory. There is nothing circular about the above point. The c used in Maxwell's theory is indeed the same c in special relativity. I've already explained why and..
Comment: ..nothing can obscure this simple logic. You certainly didn't provide any logic except referring to a paper which, as anticipated, is based on proca theory. If you change the context of the discussion then many things can change. If a new TOE is discovered then many current ideas may change accordingly but within their appropriate context the ideas will remain safe for ever.
Comment: I am quite sure. I'm not going to argue the point, but you might consider the possibility that the editors at Phys. Rev. D, the authors of the paper, the reviewers, and me are right and you're wrong. You might try doing a further literature search, I gave you a 2010 reference but this is an old problem.
Answer: I think the central issue is the invariant interval and invariant mass.  A particle moving in spacetime has the interval $ds^2~=~dt^2~-~dx^idx_i$.  There is the corresponding invariant mass $m^2~=~E^2~-~p^2$, which is the momentum spacetime interval. So let us consider the plane wave $\psi~=~exp(-i{\vec k}\cdot{\vec x}~+~i\omega t)$ $=~exp(-ik^\mu x_\mu)$.  The Laplacian operator $\Delta~=~\nabla^2~-~\partial^2/\partial t^2$ applied to $\psi$ is
$$
(\nabla^2~-~\partial/\partial t)~=~(\omega^2~-~k^2)\psi~=~\hbar^{-2}(E^2~-~p^2)\psi.
$$
This is an eigenvalued problem with $\Delta\psi~=~\lambda\psi$.  If the particle has mass this eigenvalue is the mass squared.  This means there is dispersion as $|k|~=~\sqrt{\omega^2~-~m^2} $ and for $\omega~=~2\pi/\lambda$ we then have that
$$
|k|~=~c\sqrt{2\pi/\lambda^2~-~m^2}
$$
The velocity of a wave is then only exactly $c~=~1$, or $|k|~=~2\pi/\lambda$, with of course $kc~=~\omega$, for $m~=~0$, and otherwise we contradict our assumption of the wave existing on a null interval $ds^2~=~0$.  If the invariant interval in the spacetime is zero, then the corresponding invariant mass in the momentum-spacetime must be zero.
The above Laplacian in the case of a photon is predicted as the wave equation operator by Maxwell's equations.
Answer: In the context of special relativity, anything that travels at the speed of light can't have a nonzero rest mass. One way to see this is that the kinetic energy of an object of mass $m$ moving at speed $v$ is
$$
mc^2\left({1\over\sqrt{1-v^2/c^2}}-1\right),
$$
which tends to infinity as $v\to c$. Physically, this means that it would cost an infinite amount of energy to raise a massive particle up to speed $c$.
As far as special relativity is concerned, it's logically possible that photons do have mass and travel at speeds (slightly) less than $c$. (This would mean that the quantity $c$ that occurs in special relativity should not be called "the speed of light.") The experimental limits on this possibility are extremely severe, though.
I may have misguessed the level of your question and the sort of answer you're looking for. For instance, there are separate reasons for believing the photon to be strictly massless based on gauge invariance.
Answer: put simply - mass terms for photons break gauge invariance.
Comment: In its current form this is a very poor answer. Please look at some of the more highly voted answers on the site for a better model.
Comment: You need to elaborate or state WHY a mass term breaks guage invariance or this answer is useless.
Comment: useless is a tad strong - should i elaborate and state why gauge symmetries hold in standard model too?????? this was meant to be a 'short' and 'mathematical' answer as requested.

Why can't photons have a mass?


Do I need a hammer drill to install brackets for drapes on a brick wall?

Question: I want to put up brackets for drapes. I have a masonry bit, but are using a simple (and possible cheap) cordless drill.  I can seem to penetrate the brick.  Do I need a hammer drill to do the job.
Answer: Petty much going to need a hammer drill for brick. You could try going into the mortar, but it sounds like you already dulled your bit. Try a fresh bit going into the mortar joint but don't expect superior pull-out strength.
Answer: Depending on the brick, a drill bit for glass will go through brick faster, easier than standard carbide bits. Sometimes just with a regular drill, it is worth a try.


Path independence: conservative field vs non-conservative field

Question: Why is it so that if we have a smooth, conservative vector field, we can choose any two points within its domain and join them with any type of smooth primitive curve, we will get the same result no matter what form of the curve do we choose. For example, we can choose a straight line, parabola, circle, etc... all of them will yield the same result. Now, if the field is not conservative the result might be different for each curve that we chose. Can someone explain why this happens? In other words if the field is conservative it is path independent, why it is so?
Comment: What's your definition of conservative?
Comment: It has a potential function: $F = \nabla \varphi $
Answer: One definition of conservative vector field is that the integral over any closed contour (in the relevant domain) vanishes (meaning, equals zero).
A path integral's value switches sign if you switch the orientation of the path. Now, if you take two paths from one point to another and switch one of their orientations, you get a loop. Of course you can also argue the other way, that if path integrals only depend on endpoints, then the vector field integrated over any loop vanishes and so it is conservative.
Another definition is that $F=\nabla U$ for some scalar function $U$. The gradient theorem states that if a vector field is a gradient, then path integrals depend only on endpoints. Conversely, if path integrals depend only on endpoints then we can fix $a$ in the domain and define $U$ using path integrals as $U(x)=\int_a^x F(r)\cdot dr$ and verify that $\nabla U=F$ (this is done at Wikipedia).
Answer: Apply Stokes' theorem.  Over a surface $M$ which has a bounding curve $\partial M$, a vector field $F$ that is a gradient $\nabla \varphi$ obeys
$$\int_{\partial M} F \cdot d\ell = \int_M (\nabla \times F) \cdot dA = \int_M (\nabla \times \nabla \varphi) \cdot dA$$
But for any gradient, $\nabla \times \nabla \varphi = 0$.  Hence this integral is zero.
If $F$ is a gradient everywhere, then we can consider any closed curve $\partial M$--or given two points, any combination of paths between those points--and still get zero, allowing us to conclude that the line integral is path independent.


Some questions about NAT

Question:
I know that NAT is used to map Private IP addresses to Public IP addresses. But can we also map Public IP addresses to Public IP addresses? And does this serve any useful purpose?
Can the Router's WAN Port have Private IP when we use Double NAT?
Can we have Multi Level NAT? And can this, at least in theory, solve the problem of IP address exhaustion?
Comment: Did any answer help you? if so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer.
Answer:
Yes, you can NAT any IP to any IP (as long as you own the IP or it's a private block).  I've NAT public to public before to deal with weird asymmetric routing issues as an alternative to policy routing.
Yes, this type of scenario (where the router has private IPs but NATs public ones) typically occurs in large enterprise and campus networks, where you route a public IP block to a router and the router NATs these blocks without having an IP address on them.  It works, but it's very confusing for people unfamiliar with it or if it's poorly documented.
Yes, carrier-grade NAT (CGNAT) but it will break applications which depend on consistent source port mappings as well as make it extremely difficult to implement destination PAT.  No, this won't solve IP exhaustion, and it will make everyone's (developers and infrastructure) life difficult.  IPv6 is the only answer to IP address exhaustion.
Answer: "And does this serve any useful purpose?"

Yes, if you bought 3 static ip's from your IP, you could use 1 for a router to do NAT and act like a gateway to the internet for most of your network. You could use the other 2 to give them to computers to make them public on the internet. Like a web server for example.
This isn't double nat, but yes you can assign an IP to the device so you can access it's configuration options.
Yes ISP could use a much smaller portion of the Ip's they currently use by giving out private address's and using CGNAT. Like Quailman pointed out, you wouldn't be able to run any services from inside your network to the Internet, it would however help with IP exhaustion.

Some questions about NAT


How often should I have my bike checked?

Question: How often should I have my bike checked out? My dealer said once a month. If that's the case, how much should I expect to pay?
Comment: motorbikes are still user-maintainable. You should be able to do the basics yourself with standard tools.
Comment: What type (manufacture/model/engine) of bike is it? Is it brand new?
Comment: that's a 2009 Honda Aero
Comment: If it is anything like my 2001 Shadow ACE, taking it in that much is overkill.  However, if it is chain drive, then you should check to make sure it is tight (every 500 miles?).  I would stick to what the book says on maintenance and then adapt from there.
Answer: Normally each bike has different maintenance schedules. You can find info about recommended schedules in the user manual of your bike.
Keep in mind that depending on the type of bike you can have very different maintenance. For example, the maintenance required on a 125ccm 4 strokes bike is very low and could be done on your own, while maintenance on a racing enduro bike can be much more frequent and harder!
As for the costs,  I can Tell you a scheduled check in my country (italy) ranges from 70 to 200 euros maximum, depending on the bike and if you go to a specialized/authorized dealer (usually more expansive). Other maintenance (for example chain cleaning,  brake pads etc) depends on the bike and how you ride and your actual effort in maintaining the bike.


Is grounding the subpanel not required some times

Question: Is the sub panel properly grounded, only Load and Neutral wiring were located at the sub panel as shown below.  This Sub panel is at a single floor Condo, built 1970, Updated with Newer Electrical Appliances, all functional 3 prong receptacles.  Subject Condo in a multi Condo-building, maybe 30 Condos.
As a realtor in Southern California' s area, I see some inspectors' reports that reflect NO grounding bar at the sub panel, similar to the attached photos.  Similar conditions are found in condos type building, typically with several Condos, maybe 20-50 Condos at the building, typically built in the early 70' s and thru mid 80' s.
So I am trying to determine the following; Were this kind of wiring and / or sub panel installation commonly done back in the days, some 30-40 years ago, is it considered properly grounded done today?  Thank you!
Comment: As a sidenote: I hope the inspector called out the mess of paint/mud in the breaker box...
Answer: That work is in metal conduit.
You don't have any ground wires because the grounding function is being handled by the metal conduit pipes.  All my buildings have the same setup.   Nary a ground wire in sight.
The dead giveaway of conduit is the non-"standard" wire colors; it isn't all black-white with the occasional red.   You will also note writing on the wires itself, not something you'll see with cable installations.
Metal conduit is typically seen in commercial installations.  Note that large condo and apartment complexes, particularly high-rises, tend to be done to that standard.
Comment: And to finish the point, no ground wires means you don't need a ground bar. Home Inspectors are not fully versed in all aspects of what is allowed etc., so this kind of notation is not that uncommon, but it is irrelevant. Now, if someone were to ADD wires and those circuits were NOT in metal conduit, then you would need to have a ground wire add a ground bar to the panel.
Comment: @JRaef -- well, you'd need a ground *bar* then :) but even with a ground bar present, the feeder EGC can still be a metal conduit (you might have a conduit feeder but traditional MC for the branch circuits, for example)


Is it wise to put CV on homepage

Question: I'm unemployed and looking for a job. I have applied to all workplaces near me that I can imagine my education would be suitable. Now I'm at the point that I can't imagine who could recruit me. I'm wondering how useful would it be to learn basics of WWW-programming to make a homepage where I put my CV and tell that I'm willing to work. Would anyone take a contact via homepage?
Comment: "I have applied to all workplaces near me" - maybe you need to expand your target radius. Are you open to commuting/relocation?
Comment: What makes you think your CV on a homepage would make people call you? I can certainly see that working for web designers, but I have a hard time imagining that your local butcher finds his employees that way. Maybe a bit more detail would help us to help you to a better answer.
Comment: All workplaces? How many exactly? What type of job? Did you go there in person? Even if you apply online, you should go there in person with a paper resume the same day you apply online so they put a face to the resume. Have you shown your resume to others? Are you willing to compromise on the type of job you might be willing to accept? You've really told us nothing about yourself. And no, I don't think a web page would help. It's not because you have a web page that employers will come flocking to you. If anything, you came up with this scheme because it's one more way to procrastinate.
Answer: "basics of WWW-programming" involves learning HTML, CSS, Javascript, JQuery, back end programming in a language of your choice, and possibly (most likely) database functionality. This is not a trivial task.
You could potentially set up a Wordpress site which only requires minimal setup and little to no technical knowledge, however.
At the end of the day it really depends what you're trying to accomplish. Setting up a website is like opening a market stall in the middle of the forest: no one is going to know it's there unless you tell them about it, and even then, will they actually visit it? How are you going to attract traffic?
For most situations you're far better off simply creating a profile on professional networking sites such as Linkedin which are already insanely popular with recruiters and employers.
Research whether your particular industry uses one site over the other. Research what makes a profile stand out on that site. Get help rewriting your resume/cover letter, and maybe run some practice interviews with a friend or staff at a hiring agency.
If your approach isn't paying off, change it!
Comment: What? No the basics of WWW-Programming doesn't. A bit of HTML and CSS, on a webserver (Eg Apache) is all it takes for a personal home page. Depending on what your area wants from a personal website this alone might be enough for traffic (common in Academia for example).
Only reason to add a database is if you are blogging and hosting comments etc (and then you might as well move to a platform like wordpress)
Answer: If there's good stuff on your homepage that makes you worth looking up for companies, and you make sure that searching for your kind of specialization ends up at your homepage, yes, certainly. This is how a lot of IT-professionals get found by companies. They have a homepage with stuff they've made, blog posts with valuable information, and some contact info and a CV. IT-recruiters will find those and will reach out to them.
But if you need to learn web programming just to make a site to put your CV on? I wouldn't count on it too much. You're more likely to get lucky putting your CV on established networking websites like LinkedIn.

Is it wise to put CV on homepage


How do I start designing portlets in Eclipse?

Question: Is there any plugins for Eclipse for portlet design and deployment? I have never designed portlets so are there any good tutorials on the web as well?
Answer: You do have a tutorial about portlet development (not portlet deployment though, that would be through an appropriate eclipse-maven configuration)
Portal eclipse can also help to develop portlet in eclipse.
Answer: You can try http://sourceforge.net/projects/portlet-eclipse/
Also, you can try IBM Rational Application Developer, which is IBM's Eclipse based IDE. It's Portlet and portal wizards are good... but R.A.D is not opensource.


Impedance Matching using Transmission line

Question:
Hi! I was looking at this research paper for Energy Harvesting through Rectanna and I stuck with this thing. The author, first, calculated the Rzin of the rectifier circuit. He had a source resistance of 50 ohms. Then he calculated the length and width of the transmission line which he was using for impedance matching.
My question is how he calculated this? is there any formula or relation for finding these parameters? In the paper, there is nothing mentioned about this. Please let me know if anybody has knowledge about this. Thank you.
Comment: Lots of resources out there on this topic.  Search for microstrip (or stripline) impedance calculator.
Answer: Impedance matching goes in several stages.
First, you have to select a general method. Options are transformer, lumped component network, or transmission line network. These would be applicable in increasing order of frequency.
Assuming you chose a transmission line topology, then you would need to compute the electrical lengths and impedances to use. One option is to use the Smith Chart, which is a graphical method of doing the complex number calculations. It's a bit like a slide rule, so unknown to most youngsters, but very fast and intuitive in the hands of an expert. These days you would tend to use a simulator to iterate to a solution, or an optimiser to design the network for you.
Finally, turn the impedances into a physical realisation. There are many calculators, many of which are free online, to allow you to get from target impedance and substrate material to dimensions. To give you one easy to remember data point, line width = twice substrate thickness gives you roughly 50 Ω on FR4 microstrip, when the substrate thickness is much much greater than that of the metalisation.
Comment: Thank you so much for your guidance. Actually, I used a simulator and got electrical lengths and impedance. Now I have to convert them to transmission line parameters. How can I do this?
Comment: @MuhammadAwaisQasim It's possibly easier to take you through a worked example than simply tell you how to do it. What are your lengths and impedances, and what sort of substrate do you want to realise them on?
Comment: This information is very helpful. But if you do me a favor? I am editing the question and putting another picture there. There I will mention the whole circuit including the impedance matching network using transmission lines. Could you please see that and give me more detail about how I can achieve my length and width of impedance matching network using transmission lines.
Comment: @MuhammadAwaisQasim The author didn't go from Rzin directly to lengths and widths, he used an intermediate step where he calculated electrical lengths and impedances. You have to do the work, there is no magic. Either learn how to use a Smith chart, or to use a simulator, where you can iterate the transmission line parameters and observe the matching. Once you have electrical lengths and inpedance, then physical lengths and widths are trivial to get.
Answer: There are many tools as @SteveSh suggested, to measure impedance on stripline topology.
I would suggest pcb toolkit, Saturn pcb design, since it has many other calculations other then impedance calculation.


What section I would be in (13yo, D#2 - A5, E6)?

Question: I can hit a D#2. I am 13. I am kind of confused on what section I would be (i.e. bass, baritone etc.) I can go up to an A5 note with my chest voice. In my head voice I can hit a E6 comfortably. Can someone please explain what section I should be in?
Comment: Your voice will most likely change - a lot - in the next few years. If you're interested in order to join a choir, then abide by the choirmaster's decision - for now. Otherwise, any answer will be about right for maybe a few months.
Answer: Below is a common reference chart of vocal ranges, but keep in mind that what matters is not simply the absolute lowest and highest notes you can hit, what really matters is the range you can use effortlessly and without straining.
Additionally, being 13, expect that range and other characteristics of your voice are going to keep changing, sometimes slowly and sometimes suddenly, over the next few years.
https://en.wikipedia.org/wiki/Vocal_range#Vocal_range_and_voice_classification
Comment: I have already hit puberty and I am in a choir but the director won't put me in a specific section.
Comment: I can't speak for the director's choices, but perhaps you can take this as a unique opportunity to gain experience doing a wider variety of parts and see how your still evolving voice handles all that. And by the way, it's not just puberty itself, but the still growing body, developing muscles, changes in proportions and shapes -- your voice will go on evolving for at least another 10 years before settling into a somewhat stable form.
Comment: Do you know that song that is in frozen 2? "Into the unknown"? I can hit all of those notes on 3 different octaves. My choir instructor is making me kind of nervous by not giving me a specific section. I don't know if that makes sense to anyone but yeah.
Comment: If your interest in singing is more than a hobby, never mind what the director says, take things in your own hands and become more aware of the many possibilities in front of you. Have you seen the movie "Maria by Callas"? https://www.imdb.com/title/tt7364566/ especially the parts about her early practice years (around your age too). You can learn a lot from it. Also look for interviews with great singers who talk about their early years, there's much to learn.


Why is the throughput (performance) of the Slotted Aloha protocol ≈ 0.36

Question: I've read that the throughput - that is, the amount of good useful messages relative to capacity - of the Slotted ALOHA protocol for communication networks is roughly 0.36%.
But its not clear how we arrived at that number.
Can someone explain it?
Answer: Unlike in the regular ALOHA protocol, where other nodes can send messages that interfere with ours at any time, in the Slotted ALOHA protocol, the only other time a message can be sent to interfere with ours is if it's sent at the exact time ours is sent (since in Slotted ALOHA messages can only be sent at specific intervals, like every 5 seconds for example)
Therefore, the probability that our message will be the only one, and have no interference is the probability (in a Poisson Process) that only 1 message on the system (ours!) is sent on the mark
$$P[(N(t + 1) - N(t)) = 1]  = \frac{{e^{ - \lambda\tau } (\lambda\tau) ^ 1 }}{{1!}}  = \lambda\tau e^{ - \lambda\tau} = \lambda e^{ - \lambda}$$
This gives us a function of the throughputs for all $\lambda$s.
To find the $\lambda$ with the highest throughput (that is, what expected rate gives us the highest throughput) we take the derivative and set to zero:
$$e^{-\lambda} - \lambda e^{-\lambda} = e^{-\lambda} (1 - \lambda) = 0  $$
$$\therefore \lambda = 1$$
In other words, when $\lambda$ is $1$, our throughput is highest. How high exactly? Just plug it back into the function we came up with, $\lambda e^{ - \lambda}$ and get:
$$ 1 * e^-1 = \frac{1}{e} = 0.36787....$$
Here's nice graph comparing the two ALOHAs, made by Reuven Cohen.
Answer: The answer to your question given is quoted directly for the most part from 'Computer Networking: A top down approach' by Kurose and Ross 4th Edition.
" we assume the following:
• All frames consist of exactly L bits.
• Time is divided into slots of size L/R seconds (that is, a slot equals the time to
transmit one frame).
• Nodes start to transmit frames only at the beginnings of slots.
• The nodes are synchronized so that each node knows when the slots begin.
• If two or more frames collide in a slot, then all the nodes detect the collision
event before the slot ends.
Let p be a probability, that is, a number between 0 and 1. The operation of slotted
ALOHA in each node is simple:
• When the node has a fresh frame to send, it waits until the beginning of the next
slot and transmits the entire frame in the slot.
• If there isn't a collision, the node has successfully transmitted its frame and thus
need not consider retransmitting the frame. (The node can prepare a new frame
for transmission, if it has one.)
• If there is a collision, the node detects the collision before the end of the slot. The
node retransmits its frame in each subsequent slot with probability p until the
frame is transmitted without a collision."
We make the following definition:
"A slot in which exactly one node transmits is said to be a successful slot. The efficiency of a slotted multiple access protocol is defined to be the long-run fraction of successful slots in the case when there are a large number of active nodes, each always having a large number of frames to send."
The derivation of the maximum efficiency - the answer to your question is given.
"To keep this derivation simple, let's modify the protocol a little and assume that each node attempts to transmit a frame in each slot with probability p (That is, we assume that each node always has a frame to send and that the node transmits with probability p for a fresh frame as well as for a frame that has already suffered a collision.) Suppose there are N nodes. Then the probability that a given slot is a successful slot is the probability that one of the nodes transmits and that the remaining N – 1 nodes do not transmit. The probability that a given node transmits is p; the probability that the remaining nodes do not transmit is (1 – p)^N-1. Therefore the probability a given node has a success is p(1 – p)N-1. Because there are N nodes, the probability that any one of the N nodes has a success is Np(1 – p)^N-1."
To obtain the maximum efficiency we find a p* that maximizes the expression.
Also, to find maximum efficiency for a large number of nodes, we have to let N tend to infinity in Np*(1 – p*)^N-1. After performing the necessary calculation, you will get the maximum efficiency to be 0.37.


Does anyone know an excerpt of writing which demonstrates the stress and emotions of an anticipated clash with a bully?

Question: The protagonist has difficulty with his emotions and social coping mechanisms, and is being bullied by a secondary character. I'd like to dissect the emotions and reactions in him as he learns that a confrontation is inevitable. Are there any published works I could read which does a good job of this already?
Answer: Yes; see Worm which is a Web Serial.  The initial arc deals with bullying of a 15-year-old girl by her classmates, and I found it haunting.
Answer: Other good examples, which come from books (so, not readily available), include Tangerine by Edward Bloor, and Speak by Laurie Halse Anderson.  In the first book many of the worst bullies in the Main Character's life are members of his family. In the second the Main Character spends the school year attempt to recover from rape and avoiding the rapist. It's a matter of degree and characterization for how your character deals with trauma.


How do I receive the payment?

Question: And the payment request, if i have not been paid.
Do what it is to have not come still pay?
How do I receive the payment?
Comment: This question is unintelligible. If someone understands it, please feel free to fix it.
Answer: If you have not received the Bitcoin payment you are expecting, contact the sender and ask for the transaction ID.
You can copy the transaction ID into a blockchain explorer like blockchain.info or https://www.blocktrail.com/ to track the payment. If you cannot see the transaction ID, the send never sucessfully broadcast the transaction to the Bitcoin network.
If you can see the payment but it has not confirmed you will be able to tell if the problem is related to the transaction fee being too low, the BTC being sent to the wrong address or something else.
Comment: The wallet Is the same even in the case of Bitcoin core?
It is being read  now wallet.
Comment: Why pay waiting a long time?
Comment: @Hiro any wallet is fine for checking the status of your transaction. Just making the blockchain is fully synced.

How do I receive the payment?


Is it legally ok to show concepts for existing companies?

Question: I sometimes do rebranding exercises for companies that already exist (eg. a new design for their website, a new logo, etc) but they're uncommissioned and done without asking for permission. Is it okay for me to show these rebranding concepts online or would there be copyright issues? I make no money from it.
Comment: I think this _may_ be better asked on [law.se]. I'm interested in answers though.
Comment: Depends on where you are and who you show them to. Best not play this game.
Comment: @jooja I don't think your answer helps much, it'd be best if you gave some examples, like how it depends where you are? Any examples of copyright laws regarding this in some major countries? Plus OP says it's intended to go online and make no money off of it, how would that be a problem? I'm not saying there wouldn't be one, just asking how.
Comment: @MrMerrick Im not your lawyer, trademark law is tricky. In most jurisdictions where both parties pay its always a problem to open oneself up to legal action.  If the re is no need to do it, dont! Its a risk not worth taking. What countries are major? China go ahead, USA thread with caution...
Comment: I think you'd be a lot safer if you changed a few letters in the name making it a spoof/parody. As long as the logo isn't the main focus of the project, I don't think it would detract from your concept. I.e. instead of 'Pepsi' you could do 'Papsi' or 'Popsi'. It would probably be even easier with companies that abbreviate like GSK - instead of 'Glaxo Smith Klein' it could be 'Gecko Smoke Krack', as long as the full name was referenced well. :)
Answer: What you do on your own is on your own - but the moment you start showing it, it than enters the public realm. And once it enters the public realm, legal frameworks are applied. At this point, what's legal and what's not becomes a negotiation between you, the public, the government, and other entities that may have a vested business interest in protecting their image and depiction in the public sphere. So now the question becomes "Do I have the resources (time, money, legal representation) to protect my creative work, even if it's somewhat derivative of others work, inside the legal framework of the region I work in?"
And to answer that question may require the trained skills of a legal representative.