Is using Photoshop a good habit in photography?

Question: I am an amateur. I have a Nikon L120 bridge camera. While taking snaps with that camera I am not able to take get the background out of focus. It is as in-focus as the subject I am trying to capture. I can use photoshop to make it out of focus, but is using PhotoShop to enhance photographs a good habit? If the outcome is good, can the photographer take the credit?
Comment: This isn't about ethics as much as it is about good/bad practices for a novice; it's a very different thing.
Answer: That can be answered with both "yes" and "no".
Almost all digital images go through some form of post-capture processing, whether that be during the "development" of RAW images on a computer or merely the conversion to JPEG using a picture style in the camera. These sorts of adjustments are very much analogous to what we would do in the old days when selecting a film to use, choosing the processing method, and perhaps making and developing the print. For these sorts of adjustments, Photoshop (or one of its immediate competitors) is an awkward, but thoroughly legitimate choice. (Something like Lightroom is a much easier environment to work in for these more global adjustments than, say, the Adobe Camera Raw window in Photoshop.)
After that, it really boils down to intent. In this case, you are trying to make up for the deficiencies of your camera. A purist approach would be to either accept the image as your camera provides it or, if your camera can't provide the image you intended, change your camera. And that purist approach is all well and good as far as it goes, but it's expensive and prevents you from taking the pictures you want until you've paid the entry fee to join the club. As you progress down the rabbit hole that is the photographic hobby, you'll probably begin to acquire gear that is more suitable to whatever style you develop. But if you can express your vision in the meantime only through manipulation, then that is certainly legitimate.
On the other hand, using something like Photoshop to make up for deficiencies in your photography on a regular basis would be a bad habit to get into. Sure, we've all blown the shot at some point and had to rescue it through heroic action, but that should be a fairly rare event. If you find that you're being careless with the camera because you can "fix it in post", then you've gotten into a bad habit. If most of your images need to be recomposed or need major exposure adjustments, then you're relying too much on the software, and should probably try to go Photoshop-free for a period (even if that means shooting camera JPEGs exclusively) just to break you out of that habit.
The object of the game here, no matter what the end product may be, is to get things as right as you can in camera before you proceed to manipulation. That applies to "straight" landscapes and photojournalism, but it also applies to complex composited creations that could never have been shot "straight".
On to the specific topic of bokeh in post: you'll probably find that a plugin such as Topaz Labs Lens Effects or Alien Skin Bokeh 2 will do a much better job than you can do with the native tools in Photoshop (or similar), since they'll allow you to create a depth map or depth gradient to realistically simulate shallow depth of field effects.
Answer: Whether using photoshop is or not a good practice for a novice, is arguable. Personally I think a novice photographer should make a reasonable effort to obtain the best results right out of the camera. A few arguments are:
Most non dslr cameras shot jpg only. Jpg is a good format, but it also looses a bit of image information. This limits how far a picture can be enhanced or recovered. Bad exposure, heavy white balance shifts, movement or out of focus blur, are often non recoverable.
Even though some effects can be obtainable through editing, they are an imitation of what physics can do, given your lens/sensor properties, and of course the subject and the light lying on them. This makes a true difference between a real "sharp subject, blurry background" and a post production blurred bg. Just to mention one: blurriness in the picture is a function of the distance of the real object from the lens, that is, nor al objects in the background will be equally blurred unless they all are in the same plane (same distance).
Other effects that can not be easily obtainable in post is lighting effects. For example "Rembrandt Lighting". If it's not done by correctly placing subject and light sources during capture, is extremely difficult to achieve in post because, the obtained image should have certain color and tone variations due to light incidence angle, reflections from other objects and that is obviously influenced by your subject's shape. To achieve the same in post, you'd want to add/remove shadows, but that has to be done with specific gradients that are not linearly shaped, tonal shifts, saturation shifts. You can end up deforming terribly a face...
If you spend more time editing your photos rather than preparing your capture, then you are being more an editing artist than a photographer. It's not a bad thing per se, but, "Basic Photography" and "Basic Photo Editing" are rather different courses (if it were about taking classes and class topics) and each has a very specific set of abilities with some overlapping knowledge areas.
A very good observator can distinguish a pre-produced photo from one with the effects applied after, specially with difficult to simulate effects like bokeh (out of focus) and lighting.
Finally, just in case: To achieve what you mention (out of focus background) try these tips:

If your camera has portrait mode, try it.
Place your subject far from the backgroud, don't make them stand right in front of a wall...
Step back and zoom in: this also lessens lens distortion on face and body
for a more balanced proportion. It avoid big noses or big foreheads.
The zooming in from a distance tends to improve bokeh for most cameras.
Answer: Photoshop has a lot of good uses. It has a lot of bad uses too. If you feel like you are using it to make up for a skill that could be better, then it would be better to go out and improve your skills. Photoshop can also be used to do things that are difficult to do otherwise- for example getting a picture of a single person on a street with more than one person. I take two snaps a couple paces (in time, same physical location) apart and I usually have enough data to get the image I imagined.
On the topic of focus, adding blur in Photoshop is difficult since focal blur is distance dependent. I have not tried doing this in a while, so maybe someone made a blur decoder that generates a z mask for you? Otherwise it is a time intensive task to do correctly.
Answer: I think you might need to think about if you want to be good at photography or good at photoshop.  There are no penalties for doing one over the other,  you can even get good at both over time.
People can be harsh about photoshop use but if the mobile application Instagram is any indication of where photography is headed I don't think its such an issue to do a lot of post processing.
To achieve the effect of blur in a camera you basically need a differnt lens, or use your lens differently.  You want to lock your lens at the largest aperture you lens will allow - inversely the lowest number 1.2 , 1.8 , 2 , 4.  Try messing around with the camera set to Aperture priority mode with the aperture fully open.  Try long shots and close ups.
Answer: I think that there are a variety of tools available to make the image you want.  Use what best helps you achieve your goal.  There is no right or wrong. I my world, "good" is what produces the results you want. You have to decide what you think is good.
Having said that, I think it would be non-controversial to say that the better the raw material, the better the result.
Answer: Photoshop has the advantage of being the defacto standard for photoediting. While there are many, many other tools for photoediting, including many that are more geared toward the beginner, Photoshop is essentially the standard.
Having started with 3 other programs before relenting and buying Photoshop, I know this from experience. When you are stuck, or need to understand how to use a feature, there is an incredible amount of online tutorials, blogs, videos, books, and classes that will instruct you on anything to do with Photoshop, much of it free or inexpensive. It is this ecosystem that makes the Photoshop purchase worthwhile, vs any other tools. Other tools have much less available.
As for ethics, ask yourself if it was ethical for Ansel Adams to dodge and burn his images, to adjust development time as he processed his photos. He did the version of photoshop from his day: wet darkroom development. Adams would never dream of sending his film off for development and printing, just as many photographers would never publish an image that they had not edited in Photoshop.
Sure, you can change or add things to an image with Photoshop, but you know what? You could do that before Photoshop or even computers.
Answer: As with anything in life you can over do it.  Practicing with moderation is key.   I am a professional instructor at a university in Graphic Design & Photography and I don't try to scare my students away from editing photos or touching them up.  However, getting a shot as best as you can in camera will only make it look even better when you take it into Photoshop! So really, there is not definitive answer for this besides practicing with moderation! Best of luck to you!
Answer: Digital editing of the photo is a tool, like any other. Being a novice means you need advice when to use which tool.
Given what you are trying to do, it would be more efficient to get this effect on the camera using Jahaziel's tips as that is faster.
Answer: The question is all about personal preference, in my opinion (pun intended). Photoshop is an excellent tool for photographers to enhance or alter their images.  Photoshop can also be used as a crutch for photographers who are either too lazy or not experienced and/or gifted enough to get the composition they wanted out of the shot before touching it up.
To me, there's a difference between "touching up a photo" during post-processessing, and changing the soul of the image.  It's a fine line to draw, and also can become a bad habit if the photographer isn't careful.  I use photoshop to make sure my color balance and tonal structure is calibrated properly, so I can get the printer to reproduce the photo exactly how it is supposed to look.  I also occasionally use it for cropping and other post-production techniques depending on what my client wants out of the photo.
As a general rule, I always try to get as close as possible to the image I am going for while im on the scene.  If I feel like I have a good bit of work to do post-processing that involves "cleaning up" my images (i.e. fixing mistakes) then it's clear to me I didn't do a good job capturing the image.  From there I use Photoshop as a good teaching tool because I can get very specific on the things I made mistakes on (i.e. compositional issues vs. exposure and metering).
There is no question Photoshop enhances our ability to provide a better image than we could without it.  But when I start to dive more into the realm of covering up mistakes as opposed to retouching already-present characteristics that I wanted, then I have to ask myself what I'm going after.


Moving Drives Between Macs

Question: I'm considering buying a new MacBook Pro after today's refresh, and I am wondering if OS X handles simply pulling a drive out of one MacBook Pro and putting it in another. I know that in the Windows world this is — most of the time — possible, but I have never tried it with OS X. Does the OS handle the sudden change in hardware, or does it 'sh*t bricks' and abort mission? In theory you'd think it could handle it, but anyone with definite experience would be good.
Answer: In general, this is a very safe, OS X boots other hardware in general situation.
However, when new hardware is released - often a special build is made and until the next revision or two of software, you can't boot that hardware so you'd need to take a hard drive from the specialized machine to an older one instead of the other way around.
It's far safer to migrate from a Time Machine or target mode or from running OS using file sharing / network. That way each machine has the correct OS and can flag apps or system files for non-transfer when there is a known problem migrating them.
Answer: Odds are, the new drive is larger than the old, no? Certainly younger, and harddrives have a failure rate that rises with age.
I suggest you use migration assistant and just move the data over to the new machine.

Moving Drives Between Macs


Problems while installing Fedora 19

Question: Currently I have Windows 7 installed in my machine with a 500GB hard disk.
C(52 GB) drive has the windows7 installed.
D Drive contains files which I do not wish to install.
There is 52 GB of unallocated memory.
I want to replace the C drive and use that partition along with the 52 GB of unallocated memory and install Fedora 19 there.
These are the steps that I've followed.

Boot from DVD.
Install to Hard Drive.
Selected English as language
Installation Destination by default has Automatic partitioning
selected. I clicked on it and it redirects to Installation
Destination page.
Click on Reclaim Space
Select C drive, click on delete click on reclaim.
Then I get this error Error checking storage configuration. Click for details. After I click on it, I get the following details:

you have not created a bootloader stage1 target device
sda5 must have one of the following disklabel types: gpt.

I then go to custom layout
Add mount point / 30 gb ext4
Add mount point /boot 2 gb ext4
Add mount point swap 6gb swap
Add mount point /home 65.99 gb ext4
Click on Done -> Accept Changes

Then I get the following error

you have not created a bootloader stage1 target device
sda2 must have one of the following disklabel types: gpt.

What am I doing wrong? Please help.
Answer: Switch your system to BIOS mode boot or else create a GPT partition table (gparted or gdisk).

Problems while installing Fedora 19


MapServer dealing with Huge Aerial Images

Question: I've been trying to evaluate MapServer.
Is there a way to handle the large size of Aerial image with any kind of Tiling or Caching?
Answer: Yes. We are using mapserver for our National aerial imagery service and other raster maps, and geowebcache for tiling and caching. There are other tiling cache servers around, and the newest one is mapcache, which is integrated with mapserver.
Answer: In my opinion the relevant issue is the image format. For a large aerial image it is recommended that you use a format that (a) achieves high a compression ratio and (b) features partial decompression.
As far as I know, the only formats that meet these two requirements are ECW, MrSID and GeoTIFF. Unfortunately ECW and MrSID are propietary. About GeoTIFF, there is a good Q & A at this website: "Should GDAL be set to produce GeoTIFF files with compression? Which algorithm should be used"
Despite it is a propietary format, it is posible to serve large aerial photos in ECW format with MapServer. I have done it and it works fine.


Is there any reason not to do laundry with bar soap?

Question: When traveling light, is there any reason not to use standard hotel-issue bar soap to wash clothes in a hotel sink?  Carrying powder laundry detergent (or worse, liquid!) seems like a big hassle if standard hand/bar soap will do an adequate job.  If I go this route, are there any common clothing materials that won't stand up to this sort of washing over time?
Comment: @rlesko: Mainly because the consequences if spilled are much higher than with dry soap. Air travel restrictions on liquid can be an issue, too, when not checking luggage. I realize there are ways to safely transport liquids, even in low-pressure situations found in an aircraft cargo hold, but it's just easier to transport dry substances, and easier yet to transport nothing at all--thus the question. :)
Comment: Why do you say `or worse, liquid!`? I regularly use liquid detergent for traveling since they have a screw on cap and are very practical for use. They come in travel size and are more effective than powder detergents.
Answer: In the olden days you did do laundry with bar soap - but it was laundry bar soap - harsher and unscented. You can still get it in Mexico and probably lots of places.
I guess "hand soap" as it is (or was) usually called differs mainly by being more expensive and probably has ingredients to be less harsh on the skin, and it's usually scented.
Whether any of these extra ingredients is bad for fabrics I doubt but you never know. But being gentler on skin you can be pretty sure it's going to be gentler on stains also, so less effective on clothes.
So if the convenience of carrying fewer kinds of soap outweighs the higher price and lesser cleaning strength then sure - wash your clothes with hand soap while you're on the road. But now and then you probably want to wash it with laundry soap too.
(If you've ever tried washing your hands with laundry soap you will notice it can badly dry your skin. I made this mistake recently where the hostel bathroom didn't have hand soap but did have laundry powder. After a week or so the skin only on my hands was raspy and I realized my mistake.)
Comment: Your answer got me to look for bar-form laundry soap, and came across the apparently popular [Octagon All-Purpose Soap bar](http://www.amazon.com/Octagon-Purpose-Laundry-Soap-Colgate/dp/B000GCOLQ6/ref=sr_1_1?ie=UTF8&qid=1332013762&sr=8-1) (and of course a number of competitors), which would make for a useful all-purpose travel soap I suspect. Not quite as convenient as hotel soap, but more reliable and more certain to be available.
Answer: Hand soap and laundry soap are two different things. What you want to do is to make the water "thinner" (destroying it's natural surface tension) so it gets the dirt out. You can get the same effect with pretty much any soap or dishwashing liquid. However most powder laundry detergents also have some whitener in it that make your clothes "appear" white and clean, and this is something you don't have in hand soap.
What I usually do is get some paper or magazine and an empty & dry 0.5L fruit juice bottle (or anything with a big bottleneck and a screw bottle cap) and then funnel some powder laundry detergent into the bottle using the rolled magazine or paper. This way it's sealed, quite compact and won't make a mess in my luggage when it gets thrown around on luggage belts. I do this mostly because a lot of hostels charge you $2 dollars for 1 small bag of powder (enough for 1 wash load), while you can get a small box of powder (~20 wash loads) for $4 in the supermarket.
Comment: ... and if the police asks you about the strange white powder in the juice bottle, be prepared to explain.
Answer: I am a permanent traveller, and I wash my laundry with hotel bar soap or shampoo. Some is hanging over my hotel balcony to dry now, in fact. I've never ruined anything with it, and I've been hand washing silks and knit wools regularly. In my opinion it is more important to be careful in not using too hot of water, wringing or hanging things that shouldn't be hung to dry. Though with bar soap in general you do want to be careful not to use too much (i.e. don't rub the bar directly on a thick knit) or it can become difficult to remove.
I mostly do this with my socks, underwear, tops and other easy to wash items. Then I plan to use a washing machine every few weeks for items I wear many times without a wash like denim and jackets. For this I carry these:
GrabGreen Laundry Detergent
I keep them in their plastic bag in my hardside suitcase, and have never had any troubles with breakage or mess. If I am going on a short side trip, I bring one "just in case" in a small plastic bag. When it is laundry day I toss one or two into my laundry bag to take to the washer.
Comment: Hi @Rebeka. If you can, try to put your links here behind some descriptive text, especially long ones like this one. Thanks (-:
Comment: I second this. I have some hand-wash-only clothes (underwear and special T-shirts), and I was them using liquid soap from Yves Rocher, with no problem. Actually, liquid soap seems to very well wash off even red wine, dark vinegar and motor oil (all have happened to me recently).
Comment: I use my hand soap I know I do not react bad on, not just any hotel soap bar.
Answer: I agree with Mouviciel about the validity of Aleppo soap as an all-purpose product, but I find it a bit too expensive (at least here in Italy it is normally marketed as something posh), so I would recommend Marseille soap. If you buy it pure, i.e. with no perfumes or other things added, you can really use it for everything, and it is less expensive than the other option.
Or else, if you happen to be travelling through Southern India, you can always find small bars of laundry soap, often marketed under familiar brandnames, which are extremely reliable to do the job, easy to carry around because of their size, and absolutely cheap (5 rupees or 10-15 cents).
Comment: Though I don't like Marseille soap for hair, I agree about its multipurpose usage (and price). I use it even for dish.
Comment: @mouviciel You're right about the hair, and I admit that, having long hair, I've never used it that way personally. I was indeed reporting the opinion of male friends who are balding... I'll try it for dishes, thank you for the tip.
Answer: The washing powder has some advantages:

You can't use soap in washing machines (it generates too much froth)
It's much less labour intensive to use washing powder even manually (tested with 2 kids last month, when both of them had all their clothes completely dirty every day)

but it obviously is a pain to carry. If you are just travelling on your own and you are not planning to use washing machines on campsites/hostels then probably soap will be fine.
Comment: Shampoo or liquid 'soap' for shower or bath works as well.
Answer: Although it is not as popular in the US, special bar "soap" specifically made for doing laundry does exist, and is often carried at common places like grocery stores--it only seems hard to find because it's so small and the demand is low so it may be on a top shelf where you can easily overlook it. Look for brands like Fels-Naptha, Octagon, Lirio, or Zote, or search for a resaler online: http://www.google.com/products?q=bar+laundry+soap
I would expect laundry bar-soap to work significantly better than hand-soap should bringing liquid or powder laundry detergent be non-desirable. You really want a detergent more than a "soap" per-se, glycerin isn't going to get clothes clean, so depending on the main ingredients in the soap it may not be effective for clothes.
Answer: When I travel light, I carry Aleppo soap and I use it for everything: skin, hair, shaving and laundry.
I never found something better for that many usages.
Answer: There is really no real reason why you can't use bar soap for laundry (at least for hand laundry.) It wouldn't be practical at the laundrette/laundromat because you'd need it either flaked or in liquid form to work with the machines (a little grater might help with this; grate into powder and either "melt" it in some hot water to make a liquid detergent or leave it in powder form and put directly into the allotted space in the machine.
Answer: Why not? They do a pretty good job too. I ran out of laundry detergent while away for a 2 month travel and unfortunately nearby shop ran out of laundry detergent (it was in an isolated area).
I substituted by using Lux bath soaps (they are all that's available). Leave my clothes smelling good and got rid of the stains too. I rub the soap on the wet clothes, scrub and scrunch them a little, leave them soaked in the sink for a while and rinse. My clothes smell great once they dry :D
Answer: You can buy sheets of laundry detergent. The ones in the supermarket usually have fabric softener in them too, which is ok if you're going to use a washing machine. For the sink you can use something like Travelon sheets. They appear to be scaled for sink-sized loads (too much soap makes rinising a big challenge) and packaged for traveling. Not the cheapest option, but handy.
The Amazon page is rich in links to other handy things too: sink stoppers, in-room laundry lines, inflatable hangers, ... who knew?
Comment: I know there are many alternatives; but this doesn't answer the question of whether there is a reason not to use bar soap.


Licence of audio samples

Question: I am not sure if this is the right place to ask, but is there a free-to-use licence for audio samples like drum-kit sounds? And what if I want to use the samples in a commercial application? And what if I produce the samples on my own, using a commercial software, will the samples be free then?
Comment: perhaps useful... http://avp.stackexchange.com/q/2091/2569
Answer: If I understand your question right then you are looking for samples to use in a commercial application.
There is a big difference between purchasing the samples and creating them yourself. As far as I know if you create any sound using commercial software then the end result is your own. Eg if I use Cubase to record and release a song then Steinberg have no claim on the song, and likewise if I use a particular synth or sample library to create the work. I know this gets curious when it comes down to for instance someone trying to re-sell on actual samples from a synth or library which I presume would be a breach of copyright, so you would have to estimate where you stand on this I suppose and get legal advice if necessary.
On the other hand if you want to use samples directly in a commercial application, like a game for instance,  then the usage would be covered by the licence agreement. In general I know more about other elements like photos and pieces of software but I suppose samples and clips are similar. In that regard there would be 2 licences, and this is the case for instance with Audio Jungle . The typical licence allows for normal usage then as far as I know the extended licence ( which is many times the price of the standard licence ) allows for you to include the file in something you are re-selling. But there will be limitations to this so you would really need to research it pretty well and to be sure to be sure you'd need to get the info direct from the licence holder.
In general if the licence is not clear and easy to follow ( and get clarification on ) then it's probably better to look elsewhere.
I would say you are probably in no danger of getting sued by any of the DAW manufacturers if you create your own sounds on their software though.
Answer: If they are 'free to use' like those samples that you can buy from loopmasters, producerloops etc. then they are exactly that unless stated otherwise. Most bought samples come with a license agreement that should explain how you can use them. It depends where you got the samples from as license agreements can differ between suppliers.

Licence of audio samples


Could hydrogen replace oxygen in cellular respiration?

Question: I was wondering what oxygen actually does in the body. I have seen a few answers to other questions that involve the electron chain and I am really not sure what that is. So I was wondering what oxygen does and could hydrogen do the same thing as a substitute?
Comment: Thanks for the clarifying edit @MadScientist. I would rescind my vote to close the question if I could, but alas...http://meta.stackexchange.com/questions/915/can-we-have-the-ability-to-rescind-a-close-vote-before-it-closes.
Answer: No, hydrogen could not replace oxygen because it has entirely different characteristics. The most important one is probably its electronegativity - oxygen 'pulls' electrons much 'stronger' than hydrogen.
Basics: Reduction potential
Oxygen is the so-called terminal electron acceptor of the electron transport chain in eukaryotes. You can see "reduction potential" as a kind of stored "energy" which molecules have, similar to the power stored in batteries (very similar actually). To make this text a bit shorter I'll call it "RP" from now on.
One maybe confusing detail is that a substance with low RP has "more energy" than a substance with high RP, so it is the opposite way of thinking.
In very generalised terms, metabolism means that molecules with a low RP (glucose) are oxidised (burned) and turn into molecules with much higher RP (CO2). Coupled with this, a different molecule with very high RP (oxygen) is reduced and becomes a molecule with slightly lower RP (H2O). (You may have heard this before - it's called a redox reaction.) *
The important part is that the RP "released" by the oxidisation (burning) is larger than the RP "taken up" by the reduction. The surplus leaves as energy - heat and light if you just burn the glucose. This is a spontaneous process, meaning it will occur just by itself - even if it takes a long time if nobody drops a match on it.
The idea of metabolism is to let that process happen - but use as much of the energy it releases as possible. This works by not just letting it burn, but intercepting that burning process at different stages so that at each step a bit of the RP can be taken off and stored in something else. This "something else" is NAD which I'm sure you've encountered before. Each step that glucose is burned down, another bit of NADH is made, which then has a respectable reduction potential.
NADH (leaving out NADPH here which is a bit different) is channeled into a process called oxidative phosphorylation which retrieves the reduction potential in an actual form of energy.
Basics: Terminal electron acceptor
Finally, the reduction potential I've been talking about all the time is really just electrons, involved in bonds which are "happy" to react. Passing down the reduction potential as I've explained is really a passing down of electrons into more and more stable, less reactive bonds. That's why it's called "electron transport chain". At the end of oxidative phosphorylation, those electrons are dropped onto O2 and make it into H2O. That's why O2 is called the "terminal electron acceptor".
Why Hydrogen can't replace Oxygen
Now to come back to why hydrogen cannot perform oxygen's function in our body. We use glucose as our source of reduction power and oxygen as our terminal electron acceptor. O has a high electronegativity (3.5) so it pulls electrons strongly towards it. H's electronegativity is only 2.1, so it's much weaker. O as a terminal electron acceptor works because it pulls them much stronger than H when they bond, so an O-H bond is almost like giving oxygen an electron. In order for hydrogen gas (H2) to perform the same function, it would need to be possible to drop electrons onto hydrogen in a bond where it pulls them much stronger than the other partner. They do exist, and such compounds are called hydrides. But the catch is: unlike H2O, these are normally strong reducing agents, meaning that hydrogen would rather not be in that bond. This not a feasible option for cell respiration, at least in humans, because it requires a lot of RP input. Making oxygen into H2O does not require a lot, it's a very cheap electron acceptor.
I hope I was able to put this in understandable terms. Let me know if I need to clarify anything.
--
*It works with other molecules than glucose->CO2 / O2->H2O too, many prokaryotes do that and in fact that's how batteries work
Comment: There's a small mix-up in your otherwise very nice post. The reduction potential describes the ability of a substance to be reduced. Thus oxygen does not have a low RP. In fact it has a pretty high reduction potential - it is a pretty good oxidizer. Similarly, glucose has a pretty low reduction potential. It is hard to reduce, it would much rather be oxidized. :-) High reduction potential->Strong oxidizing agent. I find it a bit counter-intuitive too.
Comment: Hahaha, what a blunder! I managed to let my lecturer's graphics mislead me. It shows a vertical reduction potential scale - but I didn't realise it's inverted with the lowest (most negative) at the top. I'll correct that of course, thanks!
Comment: You're welcome, you could always vote up my comment :D
Answer: I would say that while you could not just replace oxygen with hydrogen and expect life to just go on normally, it's possible for living things to be energetically provided for by H2 if it were found in the environment in enough plenty.  This is just speculation as Armatus points out metabolism does not work this way in animals, but I think that its quite possible a living organism could live off of H2 as an energy source.
Extremophiles can metabolize sulfur in underwater vents rather than oxygen.
In the soil, anoxic metabolisms reduce nitrogn gas to ammonia.
Photosynthesis uses light to reduce CO2 to glucose.  Atmospheric molecular oxygen is provided entirely by photosynthesis, and at one point there was little or none of it in the atmosphere and living things were found everywhere.  The glucose metabolism came after O2 appeared.
The main point is that H2 stores a reasonable amount of energy even if free oxygen is not available to drive the formation of water from another oxide, or even more exotic chemistry. There's not thermodynamic reason that it couldn't work and biology on earth has proven to be versatile in getting chemical energy wherever it needs to.
Its xenobiology and speculation, but I feel that if microorganisms evolved on a gas giant with a primarily hydrogen atmosphere, they could use the redox energy from H2, which is quite reactive even without O2 present.


For all w, CDF of y at w is greater than the CDF of x at w. Prove that the probability ( y < x ) >= 0.5

Question: F y(w) >= F x(w), for all w.  Prove P{y<x} >= 0.5
I had this question on the exam and boy did it stump me.  It was simple to understand why it was true, but I tried several different paths and got nowhere.
Any ideas?
Comment: What distribution are we talking about here?
Comment: Are $X$ and $Y$ independent?
Comment: Any distribution, and I believe they were independent if it matters.
Answer: Here's a rough sketch of one approach:
\begin{align}
\mathbb{P}[Y\lt X]
&= \int_{-\infty}^{\infty} \mathbb{P}[Y\lt X|X=x] \cdot f_X(x) \,dx
\\
&= \int_{-\infty}^{\infty} F_Y(x) \cdot f_X(x) \,dx
\\
&\geq \int_{-\infty}^{\infty} F_X(x) \cdot f_X(x) \,dx
\\
&= \left[ \frac{1}{2} \left(F_X(x)\right)^2 \right]_{-\infty}^{\infty}
\\
&= \frac{1}{2}.
\end{align}
Answer: For definiteness, let us assume that $X$ and $Y$ are positive random variables taking values
in $[0, \infty)$,and $F_Y(w)\le F_X(w)$, where $F_X(x), F_Y(y)$ are the CDFs and the equalities
take place only at zero and in the limit $w \rightarrow \infty$. For simplicity, let us assume further that
both CDFs are strictly increasing. The latter condition implies that we can invert the x-CDF and substitute in the y-CDF
to obtain the function $F_Y(F_X)$. In this function, both independent ($F_X$) and dependent ($F_Y$)
variables range in $[0, 1]$ and the value of F_Y is always less than F_X (except at 0 and 1).
(It is easy to see that, if one plots the x-CDF and y-CDF on the same axes, and see that at every value of the x-CDF, the y-CDF lies to the left of the x-CDF.
Now, $P(Y\lt X)$ is equal to the area under the line of $F_Y(F_X)$ , which is less than the area of a right isosceles rectangle of unit sides which is equal to 0.5.
Edit:
Here is the proof (as asked by Eruditass) that the required probability can be computed from the area of under the $F_Y$ graph as a function of $F_X$.
Using the convolution formula to compute the probability density function of $Y-X$:
$f_{Y-X}(w) = \int_{0}^{\infty}f_X(x) f_Y(w+x) dx$
In terms of which, the required probability is given by:
$P(Y < X) = P(Y -X<0) = \int_{-\infty}^{0} f_{Y-X}(w)  dw$
Substituting the convolution formula:
$P(Y < X)  = \int_{-\infty}^{0} dw  \int_{0}^{\infty} dx f_X(x) f_Y(w+x) $
Changing the order of integration (the integration is over bounded functions) and performing a change of variables $ u = w +x$.
$P(Y < X)  = \int_{0}^{\infty} dx f_X(x)  \int_{-\infty}^{x} du   f_Y(u) $
Using the definition of the cumulative distribution function, the second integral is just the Y-CDF
$P(Y < X)  = \int_{0}^{\infty} dx f_X(x)  F_Y(x) $
Again using the definition of the CDF:
$f_X(x) dx = dF_X(x) $
Changing the integration variable from $x$ to $F_X$, we obtain the final result:
$P(Y < X)  = \int_{0}^{1} dF_X  F_Y$
Comment: Your conclusion is the opposite of the desired one. I think you mean that the value of $F_Y$ is always *greater* than $F_X$. Then the graph of $F_Y$ as a function of $F_X$ will lie *above* the line $F_Y=F_X$, and the area (which is $P(Y Fx(w), so it should logically work itself out.

Why is P(y < x) equal to the line under Fy(Fx) ?  Why not Fy(fx)? And what do the functions mean without something feeding in like (w)?
Answer: The assertion is false, even for independent random variables $X$ and $Y$. Fix $u$ in $(0,\frac12)$ and consider i.i.d. Bernoulli random variables $X$ and $Y$ with
$$P(X=0)=P(Y=0)=1-u,\quad
P(X=1)=P(Y=1)=u.
$$
Then $[Y<X]=[Y=0,X=1]$, hence $P(Y<X)=u(1-u)$, which is at most $\frac14$ and can be as close to $0$ as desired. In particular $P(Y<X)<\frac12$.
Let us now consider the probability that $[Y\le X]$. Then the independence hypothesis is crucial. To see this, consider $u$ in $(0,1)$, $X$ uniformly distributed on the interval $(u,1+u)$ and $Y$ uniformly distributed on the interval $(0,1)$. Then $F_Y(x)\ge F_X(x)$ for every $x$. One can realize $(X,Y)$ in at least three ways.
(1) If $Y=X-u$, then $P(Y\le X)=1$ hence $P(Y\le X)\ge\frac12$.
(2) If $(X,Y)$ are independent, $P(Y\le X)=\frac12+u-\frac12u^2$ hence $P(Y\le X)\ge\frac12$.
(3) If $Y=\varphi(X)$ with $\varphi(x)=x+u$ if $x\le1-u$ and $\varphi(x)=x+u-1$ otherwise, one can check that $Y$ is uniformly distributed on $(0,1)$ and that $[Y\le X]=[X>1-u]$ hence $P(Y\le X)=2u$ and $P(Y\le X)$ is as close to $0$ as desired.
This shows that the result $P(Y\le X)\ge\frac12$ cannot hold in full generality.
Finally, for independent $X$ and $Y$, indeed $P(Y\le X)\ge\frac12$.
A proof which does not assume the existence of densities is as follows. Note that $$P(Y\le X)=E(P(Y\le X|X))=E(F_Y(X))\ge E(F_X(X))=P(X'\le X),
$$
where $X'$ and $X$ are i.i.d. Now, by symmetry $[X'\le X]$ and $[X\le X']$ have the same probability and $P(X'\le X)+P(X\le X')=1+P(X=X')\ge1$ hence $P(X'\le X)\ge\frac12$ and we are done.


Getting the IR Tower working with RCX and RobotC

Question:
Possible Duplicate:
Lego Mindstorms RCX RIS 2.0 IR tower Windows 7 / Windows XP driver

Trying to get RobotC to work with my old RCX 2.0. I'm running windows 7 64 bit. I've connected the IR tower via a Serial to USB cable. I get the error from RobotC when I go to send the firmware "RCX Controler Failed to Respond"
I'm assuming this is because the drivers are for 64 bit only. Has anyone had success running a WindowsXP virtual machine and trying the drivers? Or any luck getting RobotC and the IR tower running on a Win7 64 bit machine.
Answer: Please see this question: Lego Mindstorms RCX RIS 2.0 IR tower Windows 7 / Windows XP driver
Basically there are no 64bit drivers available and the OP on the question I linked to says he had no luck with an XP VM either.


Can I soak and rejuice the pulp that comes out of a juicer?

Question: If I'm juicing Oranges it takes quite a few oranges to get a decent amount of juice. Would it be possible to soak the pulp in water and send it through the juicer again? If so, what is the best way to go about that? What ratio of water to pulp would be optimal?
Answer: The water soluble parts of the orange are all already solved in its juice. So in mixing the pulp with water, you aren't leeching any new orange components. The result won't be orange juice by any stretch, just water which has washed the last drops of juice your juicer didn't get.
If you drink your juice diluted (I prefer it that way: less sweet, less fructose, better value), doing it might provide it with a better diluting liquid than pure water. But it won't be anywhere near real orange juice for drinking on its own.
If you want to get some more orange taste out of the pulp, you can try boiling the pulp, or using a different solvent like alcohol or a fat. All will produce something very different from true orange juice (a syrup for the first if you add sugar, liqueur with alcohol or an essential oil with fat), and the amount will be rather small. So probably not worth the work.
Answer: I'd think that the resulting juice would be bitter from the orange plup and not sweet like the first juice.
Answer: I'm not completely confident, but it strikes me as sure, you could add water to the pulp, and then you could squeeze water right back out of it. It'd have a very slight orange flavor.
If you want to get more orange flavor out of your oranges, zest them before juicing and then use the zest.
Answer: Turning your question around, we've found that with our twin-gear masticating juicer (an older version of this), we usually run the pulp through it a second time.  It comes out pretty moist the first time, but the second time it's pretty much squeezed dry.


Novell Netware Support/Recommendation?

Question: My previous job had used Novell products for almost 15 years.  From my experience, it seemed to work rather well over the years.  The last few years though have been met with bad support and long support times.  To my knowledge, it was a decent size support contract that I would think with the amount of money being spent; support would be top notch.  Well I heard recently that they will be replacing all Novell servers/software with windows servers on the main servers with Linux servers here and there.  Novell used to be a major competitor to windows and I am getting the feeling this is not so much the case anymore.  Every job interview I have went to has kind of chuckled over my mention of working with Novell at my previous employer and I kind of have to laugh back a little. I am hearing nothing but negative stories of Novell use and support now days. The only people I find using it are people that have been using it for a long time.  Am I the only one seeing Novell decline? Is there some area of Novell that has really thrived? Their conversion to SUSE Linux seemed to generate a lot of interest years ago, but that has since disappeared.  Anyone actually have any happy Novell support/usage stories?  Would anyone actualy use/recommend Novell services and server software for implementing a brand new network infrastructure?
Comment: I have, er, fond memories of Personal NetWare and getting their Windows Client32 to behave with legacy database locking.
Answer: Novell is still making money, but not for the reasons they were making it back around 2001. NetWare/Open Enterprise Server is a small piece of their "operating system" market, which otherwise contains their SLES and SLED products. GroupWise is still around and the community around it is vibrant. Zen for Desktops (now ZENWorks Configuration Management) is still around, but a small piece of the overall ZEN product line.
That said, they've been attempting to buy their way into the datacenter infrastructure market, most notably their purchase of PlateSpin a while back. Interestingly, they seem to not be rebranding Plate Spin as Novell Plate Spin, even though the big VM installations are very into Plate Spin due to it's P2V and V2V capabilities.
Their identity management product, Novell Identity Management, was the #1 identity management product for a number of years and may still hold that tile. There are now other competitors to it that are almost as featured. Novell was in the enviable position of being the party you have to be cheaper than in order to compete with for some years. They also earned a lot of consulting money with IDM deployments.
ZENWorks has expanded WELL beyond what it was in 2001. It now includes more than simple Desktop management. There is an Asset Management piece that allows more detailed tracking of both hardware and software assets, including licensing compliance. The ZENWorks Orchestrator is very useful in heterogeneous virtualization environments. They've recently added an 'Endpoint Security Management' piece, another acquisition IIRC, to get into the Network Access Controls market. And none of these require NDS or eDirectory, and can run in a pure Windows Active Directory environment.
Their support has gotten more annoying since they hey-day in 2002. But then, so has everyone's. Novell has offshored their helpdesk functions just as fast as everyone else in the industry. While they now allow unlimited call in, they now require a support contract before you do so; gone are the per-incident contracts of yester-year. Novell support used to be top of the industry in terms of experience, now they're just dead average.
This is not the Novell we all knew in 1999.
Would I use Novell stuff for a brand new install? In certain instances, absolutely.
If I were looking to do any kind of identity management and didn't have something already, or had a cobbled together rube-goldberg assemblage of db extracts, perl and powershell scripts, and nightly batch-runs, I'd be really interested in Novell Identity Manager. I'd try and get funds for it, since the environment is a lot cleaner to work in. In an environment like ours where we have several different identity databases, some of them application-specific, it would make maintainability much nicer.
If I were looking for workstation management more featured than simple GPO's and some SQL-server backed WMI scripts, their ZENworks stack has a lot of very nice features in it. Especially if I was in an industry subjected to compliance audits of any kind (HIPPA, SOX, FERPA, PCI, others).
If I had a need for a lot of enterprise managed Linux workstations, SUSE Linux Enterprise Desktop would be a very good choice. There isn't a lot of demand for that, but the product is there if you need it.
Comment: +1 for IDM. Essential in a public company (or one that might go public) or an audited university/higher-ed context these days.
Answer: My organization uses Novell products. Our mail system is Groupwise. We also use ZenWorks.
Your post was interesting to me because it marks literally the first time I've ever heard of another (specific) organization using Novell products.
I think they may have a larger customer base in Europe than in the U.S.
Comment: Yah, my old employer had all the main products in use (Groupwise, Zenworks, etc.) also.
Answer: Novell is dead.  Has been for years.  This isn't a recent thing.  There are countless articles dating back to 2000 talking of the decline, and impending death, of the Netware product and the Novell brand.
I have been a consultant for many years, and the last time I saw a Novell shop was last millennium.
Comment: I loved seeing windows get some competition, but you just cant beat the support behind windows.
Comment: If you still think of Novell in terms of Netware, you don't know much about Novell. They're 100% a Linux company these days, and a pretty active contributor to the Open Source community. They fund the development of the SuSE Linux distribution, and the Mono .Net framework, for instance.
Comment: Novell easily beats the support behind Windows. Most of it 'just works'. It's nice grabbing kids out of college who know Linux and just plugging them straight into OES. "Huh? Tree?" "It's just LDAP." "Oh!" -- and you've got a Novell admin. No MSCE, et. al.
Answer: We finished migrating off Groupwise about a year ago and that was the last reason we had NDS around.
In our experience, there was a decline in the quality of support and a marked rise in the cost while there seemed like a lack of a clear future for the platform and the email system.
A migration to Novell on SUSE to support Groupwise didn't make sense to us given the circumstances.  If we were going to undertake that kind up effort, it made more sense to us to scrap it altogether and ditch the ongoing cost and replace the apps and services at a lower cost on our standard platforms (Redhat and Windows).
Best case scenario for sticking with it would have been a migration to Novell services on SUSE to support Groupwise and then likely be faced sooner rather than later with another migration off Groupwise.
Seemed better to make a clean break with Novell & GW's uncertain future.
Comment: Groupwise on OES/Linux is remarkably stable and really nice compared to the Netware world. Migration was dead simple, and we can do all kinds of fun things from the OES2 (essentially, SLES 10sp2 or OpenSuSE 10) stack underneath. I'd say you really missed out.
Comment: What exactly did we miss out on?  We replaced a Groupwise system with Zimbra running on Red Hat at a much lower cost both in software and hardware.

Groupwise is a nice email system and can provide a good user experience.  It may be bulletproof on linux.  Our experience with increasingly expensive but unsatisfactory support from Novell and what seemed like an uncertain future for the product didn't balance against those factors.
Comment: I remember moving from novell 5 to 6 and then to 6.5 and every time had some wacky issues that took forever to resolve.

Novell Netware Support/Recommendation?


Why is $\int (dp/2\pi) |p \rangle\langle p| = 1 $?

Question: In quantum mechanics, why is $\int (dp/2\pi) |p \rangle\langle p| = 1 $ where $|p \rangle$ represents momentum eigenstate?
Comment: Are you sure about the two pi? A family of orthonormal vectors spans the whole space (is 'complete') is the sum of all the projectors onto each single basis vector gives the identity operator. A projector onto the normalized state |p> is |p><p|$, and the identity operator is just 1
Comment: It's from A.Zee's Quantum Field Theory page 11.. So it is wrong?
Comment: It's not wrong, it's just a way to normalize states that is a bit unconventional. Maybe it has advantages elsewhere. I'll check Zee later
Answer: The factor of $2\pi$ is a convention Zee uses which normalizes the $|p\rangle$ states so that they are not quite orthonormal:
$$ \langle p | p' \rangle =  2\pi \delta(p-p') $$
which means that Zee's $p$ states are different from Ahmed's $p$ states by a factor of $\sqrt{2\pi}$. Let me call these unit normalized $p$ states $||p\rangle$, then
$$ \sqrt{2\pi} || p \rangle = |p\rangle $$
This is something Dirac does also, and it becomes essential for relativistic states, when you want
$$ \int {d^3 p \over (2\pi)^3 2\omega_p} |p\rangle\langle p| = 1 , $$
because then all the parts are manifestly invariant andso the integral over $p$ is an integral over the mass-shell hyperboloid:
$$ \int {d^4 p \over (2\pi)^4} (2\pi) \delta(p_\mu p^\mu - m ^2) = \int {d^3 p\over (2\pi)^3 (2\omega_p)} , $$
where the delta function projects you onto the mass shell in a relativistically invariant way. But for this to be true, you need
$$ |p\rangle = (\sqrt{2\pi})^3 \sqrt{2\omega_p} || p \rangle $$
where $\omega_p = \sqrt{p^2 + m^2} $. This is called a relativistically normalized state. The relativistically normalized states are natural, and define relativistically normalized creation operators,
$$ \alpha(k) |0\rangle = |k\rangle , $$
which are related to the usual nonrelativistically normalized creation operators as follows:
$$ \alpha(k) = (\sqrt{2\pi})^3 \sqrt{2\omega_k} a_k . $$
In terms of the $\alpha(k)$'s and their conjugates, the field Fourier expansion is manifestly invariant:
$$ \phi(x) = \int e^{ikx} \alpha(k) + e^{-ikx} \alpha^\dagger(x) {d^3p \over (2\pi)^3 2\omega_k} . $$
This looks trivial, but it makes everything in scattering theory and mode expansions relativistically consistent. So you can make trivial all the first chapters of any old quantum field theory books with mode expansions which look opaque, because they are noncovariant.
Answer: I don't think this question has been answered. The original poster asked: why is
\begin{align}
\int dp |p\rangle \langle p| = 1,
\end{align}
up to unimportant normalization issues that have been discussed extensively (but are off the point).
The above statement is equivalent to the question of completeness of a set of states on an Hilbert space (which ahmed addressed but did not answer the question why the LHS of the original poster's equation is equal to the RHS).
Completeness is, heuristically, the statement that a set of vectors (functions on a Hilbert space) spans the space of possible (piecewise continuous and some degree of smoothness (differentiable)) functions on that space. In other words, just about any "well-behaved" function can be expanded in a basis of states that is complete.
A proof of completeness of eigenfunctions of differential equations is given in Chapter VI, Section 3, p. 424 of Courant & Hilbert's Methods of Mathematical Physics, Vol. 1, first English edition. It's too lengthy to reproduce here.
But motivation (not a "proof") for the completeness relation can be given as follows. We consider the expansion of the Dirac $\delta$ function in a particular basis, say the position basis. We assume that it can be written as an expansion over momentum eigenstates in the position basis, $e^{-ipx'}$:
\begin{align}
\delta(x-x') &= \int_{-\infty}^\infty\!dp\,c_p(x)\frac{e^{-ipx'}}{\sqrt{2\pi}}.
\end{align}
Clearly the coefficient, $c_p$ in the expansion must depend on $x$ since the $\delta$ function on the LHS does.
The completeness relation is determined as follows. Mulitply the above equation by $e^{ip'x'}/\sqrt{2\pi}$, interchange the order of the two integrals, and integrate over $x'$ on the interval $(-\infty,\infty)$. Use orthonormality of the plane waves, $\int_{-\infty}^\infty \frac{dx'}{2\pi} e^{i(p'-p)x'}=\delta(p'-p)$ to get
\begin{align}
c_{p'}(x) &= \frac{1}{\sqrt{2\pi}} e^{ip'x}.
\end{align}
Substitution back into the expression for $\delta(x-x')$ gives
\begin{align}
\delta(x-x') = \int_{-\infty}^\infty \frac{dp}{2\pi}e^{ip(x-x')}.
\end{align}
Rewriting this
\begin{align}
\delta(x-x') &= \langle x|x' \rangle = \langle x|1|x' \rangle \\
&= \int_{-\infty}^\infty\! dp\, \langle x|p\rangle\langle p|x'\rangle \\
\langle x|1|x' \rangle &= \langle x| \cdot\Big[\int_{-\infty}^\infty\! dp\, |p\rangle\langle p|\Big] \cdot |x'\rangle.
\end{align}
And completeness follows from the arbitrariness of the values of $x,x'$.
Nota Bene: This is not a proof as we have made several questionable steps (like reversing the orders of integration $dx' \leftrightarrow dp$) and essentially assumed the result we wanted (by using orthonormality).
What this does, though, is to make plausible the idea that if you need to expand a function in a complete set of states, you need a relation like $\int\! dp\, |p\rangle\langle p| =1$ to hold.
Comment: -1: OP was asking about why the $2\pi$ factors appear in some books and not others, not about why basis states are supposed to be orthonormal.
Comment: I was here when OP asked the question, the issue was the 2pi factor, which confused OP, because it is not orthonormal, but a different convention from other books. I understand what was being asked, really, I am not offtopic. I shouldn't have downvoted you, I suppose, you didn't say wrong things, just things OP already knows.
Comment: You need to read the question again, Maimon: "In quantum mechanics, why is ∫(dp/2π)|p⟩⟨p|=1 where |p⟩ represents momentum eigenstate?" This is asking about completeness not normalization. -1 for being off-topic.
Answer: The composition of any state in terms of the momentum eigenstates is  $\lvert\psi\rangle = k\int \mathrm{d}^{3}p\lvert p\rangle\langle\psi \vert p\rangle$. The set of momentum eigenstates form a complete set of orthogonal state vectors in Hilbert space (this set is uncountably infinite). The components of this decomposition are the $\langle\psi \vert p\rangle$ for each $p$ (a complex number).
To simplify things, $\int \mathrm{d}^{3}p\lvert p\rangle \langle p\rvert$ is an operator. Let $\int \mathrm{d}^{3}p\lvert p\rangle\langle p\rvert = A$, then $A\lvert\psi\rangle = \lvert\psi\rangle$ for an arbitrary $\lvert\psi\rangle$ then $A$ must be the identity operator.


Conditional Probability

Question: could you help with this:
Let $x_{1},x_{2}$ ~ Geo(p)
i'd like to know:
i  got:
P($x_{1} = k$ | $x_{1} + x_{2}$) = $\frac{P(x_{1} =k;x_{2} = n-k)}{P(x_{1}+x_{2} = n)}$ =  $\frac{P(x_{1}=k) \times P(x_{2} = n-k)}{P(x_{1}+x_{2} = n)}$ = $\frac{(1-p)^kp(1-p)^{n-k}p}{\sum_{k=0}^{n} (1-p)^kp(1-p)^{n-k}p}$ = 1
but accoring my equations that is to be equals 1. it is correct (i don't think so)? how to solve it ?
Thanks Guys.
Comment: You have a ratio with a sum of multiple terms on the bottom, and just _one_ of those terms on top. Why _ever_ would you think that is equal to $1$? You wouldn't write $3/(2+3+4+5)=1,$ would you?
Comment: It helps to specify *which* Geometric Distribution you are using.  It looks like "count of failures before first success" or $x_1\in\{0,1,\ldots\}$.
Comment: David, following the equation and if we simplify we get something like a / a (a!=0).
I agree with you , Graham, but how to solve this conditional probability ? any tip ?
Comment: On the contrary, you do not get anything at all like $a/a.$. If you had shown the intermediate steps of your simplification in your question, it would have been simple to point out the error. But the correct calculation has been given in an answer, so perhaps you can see it yourself.
Comment: Oh... Yes, on answear below i saw my error, and you are correct, it is not a a/a. Thanks Guys
Answer: Yes there is cancelation, but you have a series in the denominator.
$$ f_{x_1\mid x_1+x_2}(k\mid n)~{=~\dfrac{(1-p)^k p\cdot (1-p)^{n-k}p}{\sum_{j=0}^n [(1-p)^j p\cdot (1-p)^{n-j}p]}\\=~\dfrac{(1-p)^{n}p^2}{\sum_{j=0}^n [(1-p)^n p^2]}\\=~\dfrac{1}{\sum_{j=0}^n 1} \\=~\dfrac 1{n+1}}$$
Also note that the support for this function is: $\{(k,n)\in\Bbb Z^2: 0\leq k\leq n\}$
Now, can you identify the distribution family?
Comment: PS: Some errors of confusion can be avoided by *not reusing* the same variable in the numerator for the series' iterator in the denominator.
Comment: this seems like a uniforme distribution is not it ?
Comment: Thanks so much man, but i didnt see the distribution family, is it a regular/known distribution ?
Comment: Very much so,@RobertMota  What well-known distribution do you know has the form $~p_{\small Y}(y) = \dfrac{1}{n+1}~\big[y\in\{0,\ldots, n\}\big]~$?
Comment: @RobertMota The conditional distribution of $x_1$ given $x_1+x_2$ is a *Uniform Discrete Distribution*, to be exact.

Conditional Probability


create a front end post page template withTinyMCE,

Question: there are alot of front end posting suggestion with wp insert post but is it posible to create a front end post submision with TinyMCE, and similiar editor
Answer: If plugin is okay with you than you can try WP Quick Post Widget. It has Editor you are looking for and also supporting all Custom Meta Boxes. So you can use that too for front end.
Check this.
http://wordpress.org/extend/plugins/quick-post-widget/
Answer: If you got the rest of the form all set and ready you should check out nicedit.com.   That Editor would allow you to turn your textarea into a fully functional editor.
i have to this solution while build a wordpress based article website and found that using the TinyMCE is either impossible or i havent found the right explenation..
Anyway... NiceEdit would provide a toolbar similer to wordpress one and the data will be sent along with it (if your form is build in the correct way) when submitting a post.
Hope this helps :)
p.s:
if anyone know of a way to integrate wordpress TinyMCE i would also love to know


What is an example of a topological space that is not homotopy equivalent to a CW-complex?

Question: It would also be nice if someone can explain this comment appearing on the Wikipedia page on CW-complexes:
"The homotopy category of CW complexes is, in the opinion of some experts, the best if not the only candidate for the homotopy category."
Answer: My favorite example of a space which is not homotopy equivalent to a CW complex is the Long Line. All it's homotopy groups vanish (exercise 1) but the long line is not contractible (exercise 2). It's too long!
Answer: A complete explanation of the Wikipedia comment would require delving into the theory of model categories in more detail than would be reasonable here. Unfortunately, there's a paucity of good resources on model category theory at the moment. The standard references are the books by Hovey and Hirschhorn.
A model category is an abstraction of the foundations of classical homotopy theory. It provides a set of structures we can use to pass from a category of objects we wish to study to an associated homotopy category. In a nutshell, the idea of constructing a "good" homotopy category (of some category, which in this case is the category of spaces) is to do two things. First, we want to regard certain classes of maps (namely, weak homotopy equivalences) as being invertible. This is done through the notion of localization of a category, which is analogous to (in fact, a generalization of) the notion of the localization of a ring. Secondly, we want to replace every object by an object that behaves nicely with respect to homotopical constructions. These are the "fibrant and cofibrant objects." For example, in homological algebra, we replace chain complexes by quasi-isomorphic injective resolutions so that they behave nicely with respect to left-exact functors. In our setting, CW complexes play this role; namely, we can replace any space by a weakly equivalent CW complex.
The role of actual CW complexes (as opposed to general fibrant and cofibrant objects, which may for instance only have the homotopy type of a CW complex) is not immediately obvious from the basic definitions. It comes from a more general construction in model category theory in which the class of cofibrations can be reconstructed from small sets of generating cofibrations. The inclusions of spheres as the boundaries of n-balls play this role in the Quillen model structure on spaces, and the cofibrations one constructs from this generating set are retracts of relative cell complexes.
As Reid points out in his answer, the homotopy category of CW complexes is only "right" up to equivalence; the fact that there are different equivalent "homotopy categories" of spaces stems from the fact that there are different (but in some sense equivalent) model structures that one can put on the category of spaces. Consequently, it's at least a little important to be careful about what we mean by "the" homotopy category.
Answer: The `quasi-circle" of Hatcher's Algebraic Topology, Exercise 10 of 4.1, is a nice example. The point is that is has trivial homotopy groups but is not homotopy equivalent to a point, contradicting Whitehead's theorem (I think that is the name of the theorem).
Answer: The Cantor set, or any space X with no nontrivial path which is not discrete.  If X did have the homotopy type of a CW-complex, the identity map from X with the discrete topology to X would be a homotopy equivalence by Whitehead's theorem.  But it's easy to see that the only possible homotopy inverse is the identity map, which is not continuous.
Answer: Some Google-fu turned up an example here.
Answer: I hope what that comment on Wikipedia means is that the homotopy category of CW complexes is the "right" homotopy category up to equivalence.  There are many other ways to produce the homotopy category, but they all yield equivalent categories.  The category of all spaces and homotopy classes of maps is not equivalent to the homotopy category (at least by the obvious functor) as these examples demonstrate.


Do singularities have a "real" as opposed to mathematical or idealized existence?

Question: I was thinking of, for example a Schwarzchild metric at r=0, i.e. the gravitational singularity, a point of infinite density. I realise that there are different types of singularities--timelike, spacelike, co-ordinate singularities etc. In a short discussion with Lubos, I was a bit surprised when I assumed they are idealized and I believe he feels they exist. I am not a string theorist, so am not familiar with how singularities are dealt with in it.  In GR, I know the Penrose-Hawking singularity theorems, but I also know that Hawking has introduced his no-boundary, imaginary time model for the Big Bang, eliminating the need for that singularity. Are cosmic strings and other topological defects singularities or approximations of them (if they exist).  In what sense does a singularity exist in our universe? --as a real entity, as a mathematical or asymptotic idealization, as a pathology in equations to be renormalized or otherwise ignored, as not real as in LQG, or as real in Max Tegmark's over-the-top "all mathematical structures are real"?
Comment: Not sure I can answer this question very well, but let me say this: singularities in physical theories always indicate that something is wrong in the particular domain to which you are applying them. No *observable variables* can be infinite in value in physics.
Comment: Anyway, I thought string theorists were intent on eliminating black hole singularities by infesting the interior of the event horizon with fuzzballs :)
Answer: Dear Gordon, I hope that other QG people will write their answers, but let me write mine, anyway.
Indeed, you need to distinguish the types of singularities because their character and fate is very different, depending on the type. You rightfully mentioned timelike, spacelike, and coordinate singularities. I will divide the text accordingly.
Coordinate singularities
Coordinate singularities depend on the choice of coordinates and they go away if one uses more well-behaved coordinates. So for example, there seems to be a singularity on the event horizon in the Schwarzschild coordinates - because $g_{00}$ goes to zero, and so on. However, this singularity is fake. It's just the artifact of using coordinates that differ from the "natural ones" - where the solution is smooth - by a singular coordinate transformation.
As long as the diffeomorphism symmetry is preserved, one is always allowed to perform any coordinate transformation. For a singular one, any configuration may start to look singular. This was case in classical general relativity and it is the case for any theory that respects the symmetry structure of general relativity.
The conclusion is that coordinate singularities can never go away. One is always free to choose or end up with coordinate systems where these fake singularities appear. And some of these coordinate systems are useful - and will remain useful: for example, the Schwarzschild coordinates are great because they make it manifest that the black hole solution is static. Physics will never stop using such singularities. What about the other types of the singularities?
Spacelike singularities
Most famously, these include the singularity inside the Schwarzschild black hole and the initial Big Bang singularity.
Despite lots of efforts by quantum cosmologists (meaning string theorists working on cosmology), especially since 1999 or so, the spacelike singularities remain badly understood. It's mainly because they inevitably break all supersymmetry. The existence of supersymmetry implies the existence of time-translational symmetry - generated by a Hamiltonian, the anticommutator of two supercharges. However, this symmetry is brutally broken by a spacelike singularity.
So physics as of 2011 doesn't really know what's happening near the very singular center of the Schwarzschild black hole; and near the initial Big Bang singularity. We don't even know whether these questions may be sharply defined - and many people guess that the answer is No. The latter problem - the initial Big Bang singularity - is almost certainly linked to the important topics of the vacuum selection. The eternal inflation answers that nothing special is happening near the initial point. A new Universe may emerge out of its parent; one should quickly skip the initial point because nothing interesting is going on at this singular place, and try to evolve the Universe. The inflationary era will make the initial conditions "largely" irrelevant, anyway. However, no well-defined framework to calculate in what state (the probabilities...) the new Universe is created is available at this moment.
You mentioned the no-boundary initial conditions. I am a big fan of it but it is not a part of the mainstream description of the initial singularity as of 2011 - which is eternal inflation. In eternal inflation, the initial point is indeed as singular as it can get - surely the curvatures can get Planckian and maybe arbitrarily higher - however, it's believed by the eternal inflationary cosmologists that the Universe cannot really start at this point, so they think it's incorrect to imagine that the boundary conditions are smooth near this point in any sense, especially in the Hartle-Hawking sense.
The Schwarzschild singularity is different - because it is the "final" spacelike singularity, not an initial condition - and it's why no one has been talking about smooth boundary conditions over there. Well, there's a paper about the "black hole final state" but even this paper has to assume that the final state is extremely convoluted, otherwise one would macroscopically violate the predictions of general relativity and the arrow of time near the singularity.
While the spacelike singularities remain badly understood, there exists no solid evidence that they are completely avoided in Nature. What quantum gravity really has to do is to preserve the consistency and predictivity of the physical theory. But it is not true that a "visible" suppression of the singularities is the only possible way to do so - even though this is what people used to believe in the naive times (and people unfamiliar with theoretical physics of the last 20 years still believe so).
Timelike singularities
The timelike singularities are the best understood ones because they may be viewed as "classical static objects" and many of them are compatible with supersymmetry which allowed the physicists to study them very accurately, using the protection that supersymmetry offers.
And again, it's true that most of them, at least in the limit of unbroken supersymmetry and from the viewpoint of various probes, remained very real. The most accurate description of their geometry is singular - the spacetime fails to be a manifold, i.e. diffeomorphic to an open set near these singularities. However, this fact doesn't lead to any loss of predictivity or any inconsistency.
The simplest examples are orbifold singularities. Locally, the space looks like $R^d/\Gamma$ where $\Gamma$ is a discrete group. It's clear by now that such loci in spacetime are not only allowed in string theory but they're omnipresent and very important in the scheme of things. The very "vacuum configuration" typically makes spacetime literally equal to the $R^d/\Gamma$ (locally) and there are no corrections to the shape, not even close to the orbifold point. Again, this fact leads to no physical problems, divergences, or inconsistencies.
Some of the string vacua compactified on spaces with orbifold singularities are equivalent - dual - to other string/M-theory vacua on smooth manifolds. For example, type IIA string theory or M-theory on a singular K3 manifold is equivalent to heterotic strings on tori with Wilson lines added. The latter is non-singular throughout the moduli space - and this fact proves that the K3 compactifications are also non-singular from a physics viewpoint - they're equivalent to another well-defined theory - even at places of the moduli spaces where the spacetime becomes geometrically singular.
The same discussion applies to the conifold singularities; in fact, orbifold points are a simple special example of cones. Conifolds are singular manifolds that include points whose vicinity is geometrically a cone, usually something like a cone whose base is $S^2\times S^3$. Many components of the Riemann curvature tensor diverge. Nevertheless, physics near this point on the moduli space that exhibits a singular spacetime manifold - and physics near the singularity on the "manifold" itself - remains totally well-defined.
This fact is most strikingly seen using mirror symmetry. Mirror symmetry transforms one Calabi-Yau manifold into another. Type IIA string theory on the first is equivalent to type IIB string theory on the second. One of them may have a conifold singularity but the other one is smooth. The two vacua are totally equivalent, proving that there is absolutely nothing physically wrong about the geometrically singular compactification. We may be living on one. The equivalence of the singular compactifications and non-singular compactifications may be interpreted as a generalized type of a "coordinate singularity" except that we have to use new coordinates on the whole "configuration space" of the physical theory (those related by the duality) and not just new spacetime coordinates.
It's very clear by now that some singularities will certainly stay with us and that the old notion that all singularities have to be "disappeared" from physics was just naive and wrong. Singularities as a concept will survive and singular points at various moduli spaces of possibilities will remain there and will remain important. Physics has many ways to keep itself consistent than to ban all points that look singular. That's surely one of the lessons physics has learned in the duality revolution started in the mid 1990s. Whenever physics near/of a singularity is understood, we may interpret the singularity type as a generalization of the coordinate singularities.
At this point, one should discuss lots of exciting physics that was found near singularities - especially new massless particles and extended objects (that help to make singularities innocent while preserving their singular geometry) or world sheet instantons wrapped on singularities (that usually modify them and make them smooth). All these insights - that are cute and very important - contradict the belief that there's no "valid physics near singularities because singularities don't exist". Spacetime manifolds with singularities do exist in the configuration space of quantum gravity, they are important, and they lead to new, interesting, and internally consistent phenomena and alternative dual descriptions of other compactifications that may be geometrically non-singular.
Comment: Many thanks for the detailed answer, Lubos. It was a pleasure to read.
Comment: Aren't there corrections to the geometry in the vicinity of an orbifold fixed point due to radiative corrections? The radiative corrections differ from the unorbifolded case because in a quantum field theory over orbifolds, we have to eliminate zero-point modes not respecting the orbifold projections.
Comment: Dear @QGR, there are corrections to various physics phenomena near the orbifold singularity - because of the projection on the spectrum as well as the existence of a twisted sector with completely new states - but the metric itself, as seen e.g. by propagating gravitons etc., remains fully uncorrected - due to a supersymmetry non-renormalization theorem. In general, cancellations implied by supersymmetry make things *more consistent* rather than *less consistent*. In this case, SUSY also cancels corrections to a singularity so it remains singular: no problems arise because of that.
Comment: š: Right, I was thinking of the nonsupersymmetric case.
Answer: The singularity in a Schwarzschild metric is a three dimensional space where the Weyl curvature diverges.  There are reasons why we might think this is "quantized," for otherwise the quantum evaporation would present the outside world with a singularity.
An observer that follows a string into a black hole will observe the tidal forces on the string increase.  It is extended along the radial direction and its energy increases.  The extension is along the $X^+$ direction, and the gauge for the system is set on the $X^-$ direction.  This is opposite the gauge condition the exterior observer employs in observing the string fill up the stretched horizon of the black hole.  The analysis for the string on the horizon can be found in Susskind & Lindsey "Black Holes, Information and the String Revolution"  If it is desired I can work out the case of the string which falls into the black hole.
The string falling towards the singularity becomes highly excited and reaches the singularity in some state.  It probably ends as an open string on a brane dual to the NS5-brane.  The string ends in this state of affairs, or it reaches its upper temperature limit and becomes something we are not well acquainted with.
The density of states for a string with respect to modes n is
$$
\eta(n)~\sim~ exp(4\pi n \sqrt{\alpha'})
$$
that defines a partition function $Z~=~ \int \eta(n)exp(-n/T)dn$.  The temperature is computed by $1/T~=~\partial Z/\partial n$ and the path integral diverges for a temperature greater than
$$
T_H~=~4\pi \sqrt{\alpha'}
$$
which is the Hagedorn temperature.  This is proportional to the reciprocal of the string length. The entropy of the system is the logarithm of the density of states the $S~\sim~ 1/nT_H$, which in the large n limit is zero.  The modes number is given by $n~=~1/(\sqrt{d}M_s)$, for d the number of degrees of freedom and $M_s$ the string mass.
These physical conditions occur before the Planck energy is reached.  Consequently the Schwarzschild singularity is valenced by quantum mechanics.  The singularity is valenced by, or really replaced with, a D-brane with a gas of II strings or a D-brane with a gas of D0-brane solitons or … .
Answer: Of course, not. Singularities are prohibited by the Cosmic Censorship principle.
Even more: no matter can ever move under the so-called "event horizon" because as one approaches the horizon or any other potential well, the time slows, and on the horizon it slows infinitely. Thus if the horizon existed, nobody could reach it in finite time, and the black hole existence time is finite, so any black hole would evaporate before the falling observer reaches the event horizon.
The singularity and event horizon terminology is useful only in theory which does not account for quantum effects and thermodynamics.


Wizard quest entry

Question: I'm an elf wizard with good stats, lots of spells, resistances and intrinsics and level 15. I'm new to nethack but have been spoiling quite a bit. Does having teleport control interfere with the teleport for the quest or am I not high enough level? I got the telepathic message but the only teleporter I can find triggered teleport control.
Answer: The entrance to the quest is a magic portal, not a teleporter, so anything to do with teleportation doesn't have any effect whatsoever.
You can locate the portal with a crystal ball, liberal use of Detect Unseen or a wand of secret door detection, a cursed scroll of gold detection, or a non-cursed confused scroll of gold detection.
But what I usually do is just walk on every tile on the level with shift-move until I hit it.
Answer: No, you can still use the quest teleporter. IIRC, it is not the usual colour of teleportation traps. You can use that to make sure that what you've find is not the right one.
Answer: To enter the quest level, you need to be experience level 14 or higher. Teleport control does not interfere with entering the quest level. Presumably you have not found the actual portal, just a teleport trap that happens to be on the same level.

Wizard quest entry


What kind of pills was Tommy Riordan taking?

Question: In the movie 'Warrior', we see Tommy taking pills and he is later forced to give them up by his father after requesting his help in the training for the tournament 'Sparta'.
So, what are these pills he is taking and are they medicinal or recreational?
Comment: I believe they are steroids as his father says "You sounded like a rock coming through the door."
Answer: From an interview with Gavin O'Connor, the writer and director of the film:

The Tommy character reminded me in some ways of Mark Kerr, who was the subject of one of your documentaries, The Smashing Machine. How much did your experience from that production inform the sensibility of Warrior?
Smashing Machine was my introduction to the sport and the fighters in the sport. But since that film, I've met so many fighters and have been to so many fights and watched so many fights that that was just part of the DNA of what I was doing. I'm sure there's a page out of a lot of even boxers' lives, and people I know, that have sort of infiltrated the script.
If you're referring to the pills ... now that you mention it, probably unconsciously. The whole intent with Tommy was, he was coming home to see his father -- but he was coming home to see the father that he knew when he left, which was a man who was a drunk.

In the documentary (linked above), the fighter, Mark Kerr, is linked to prescription painkillers (opiates). While the reference does not tell us plainly that Tommy is using painkillers, it's a reasonably good pointer in that direction. All the familial issues that Tommy has had as well as his experience in the marines might explain why/how he became addicted to them. It's useful to note that opiates, besides dulling pain, also promote a sense of well-being and euphoria.
Prescription painkiller abuse is quite rampant nowadays in all strata of society.
Answer: Since there is no hard-evidence in the movie that would suggest either, nothing can be said definitively. But, I would say that those pills would most likely be anti-psychotic (neuroleptic) or anti-depressant of some kind.
The reason I say this is because Tommy (Hardy) is shown to have a tough childhood (mother's chronic illness and death, father's drinking issues) and later on his buddy in the Marines dies in friendly-fire, that eventually leads in Tommy going AWOL from active duty. Now I reckon, such scenarios are enough for any sane person to develop psychological issues.
NOTE: A friend suggested that he might be taking steroids, but that seemed a little off to me since Tommy was in the Marines (where I suppose steroids that are not prescribed would not be allowed) and frankly I doubt Tom Hardy's physique (though mammoth) is a result of steroid popping!
Answer: Most likely (from military perspective)
Seroquel (antipsychotic)
Trazodone (sleeping/antidepressant)
Zoloft (antianxiety)
Military doctors work on sliding scale. If one medication doesn't work, up the dosage. if higher dosage doesn't work, change medication, ad nauseam. These are most often prescribed for said conditions
Answer: Painkillers. Probably opiate based. They numb mental / emotional pain. They also alleviate the symptoms of post-traumatic stress.
When Tommy's Dad refers to Tommy sounding like a 'maraca' walking in here means he rattled as he walked because his pockets were full of pills.
Comment: Do you have any evidence to back this up?
Answer: The pills are painkillers. On the side of the pill bottle, it says Darvocet.


Configurations for D&D dungeon tiles?

Question: I'm looking to pick up some of the D&D dungeon tiles sets.  I'm wondering if the sets come with recommended designs for the tiles, or are their sites online where people have provided possible configurations or even specific adventures/encounters that make use of the tiles.  Or do I just need to do the designing myself?
Answer: The three 'starter' boxed sets (Dungeon, Wilderness, City) do not.
All of the six-sheet packs (the original DTx and DUx series, and current DNx) have a couple of suggested designs on the interior covers. (In a few cases, one or two of the suggested designs require two copies of the set - particularly "Arcane Towers", which was designed around buying two sets, but in general with some of the 'building interior' and transition sets).
I strongly recommend PyMapper as a mapping tool.  It has a simple drag-and-drop tile interface, prints well, and can keep track of the tiles in your collection and warn you when you use them up - taking into account that the tiles are double-sided.
Also, the published Living Forgotten Realms adventures are a great resource for map ideas, as most contain two or three maps made with the tiles.
(Also, all the tiles are extremely easy to design with; "drop them on table and arrange" will go a long way in a pinch.)
Answer: I can only speak for the three Master Sets (Dungeon, Wilderness and City), which do not.
I believe most (if not all) of the published adventures picture tile configurations for their dungeons, though.

Configurations for D&D dungeon tiles?


Why is the string theory graviton spin-2?

Question: In string theory, the first excited level of the bosonic string can be decomposed into irreducible representations of the transverse rotation group, $SO(D-2)$. We then claim that the symmetric traceless part (i.e. the 35 rep) is the spin-2 graviton - but isn't the label "spin-2" intrinsically 3+1 dimensional? I.e. it labels the representation under the little group $SU(2)$?
Answer: It is traditional to label massless (and some massive) states in higher dimension by their 3-d "spin" counterparts, even thought he label is completely inaccurate, as you say. All antisymmetric forms are "spin-1", the symmetric two-index object is "spin-2", a fundamental spinor is "spin 1/2" and a vector of spinors is "spin 3/2". These labels refer to the maximum helicity of the associated massless particle, although the number of components is completely different than in 4d. For learning the higher dimensional rotation group, there is an article by Scherk from the 1970s.
Answer: I don't know anything about string theory, but the graviton has been described as a spin-2 particle well before string theory. In his "Lectures on Gravitation" Feynman explains why the graviton must be integer spin, then explains why it can't be 0 or 1, then proceeds to attempt to build the quantum field theory of a spin-2 graviton, simply because it's the next simplest thing that could work. (It doesn't, Feynman abandoned this line of research.)
Comment: ?? Feynman ***only*** talks about 4 space-time dimensions, but this question is evidently about 10, by the looks of the rep he is adducing.


Local to Global principle for reductive groups

Question: Let $G$ be a reductive group over an algebraic number field $k$. Denote with $k_v$ a local field and with $A$  the ring of its adeles, let $G_k$, $G_{k_v}$ resp. $G_A$ be the group of its $k$- resp. $A$- points.  What are necessary and sufficient conditions for a local representation $\pi_v$ of $G_{k_v}$ to appear as $\otimes_v$ $\pi_v$ in the right regular representation of $G_k \backslash G_A$? What is a good reference to study this local to global process?
Comment: The special case of this for $\mathrm{GL}_2 / \mathbf{Q}$ was asked a little while ago, here: http://mathoverflow.net/questions/42728/fourier-coefficient-of-a-modular-form
Answer: If $\pi_v$ is supercuspidal, then (after making a twist if necessary) it should be possible to find an automorphic $\pi$ with $\pi_v$ as the local factor at $v$.  This kind of result is usually proved (although their are sometimes other possibilities) by an application of the simple trace formula.  This method will also give good (although perhaps not complete) control of the ramification at other places.  If you look at the paper of Deligne, Kazhdan, and Vigneras  on Jacquet--Langlands for $GL_n$, I think you will find an expose of the technique.
If $\pi_v$ is not supercuspidal (or if one is not willing to make a twist), then this is not generally possible, just for cardinality reasons.  The link that David Loeffler provides to the discussion of the $GL_2$ case is somewhat indicative of the situation.
Comment: I think you mean Deligne-Kazhdan-Vigneras.  Rogawski also came up with an independent proof, and personally I find Rogawski's paper easier to read.
Answer: There is also a theorem of I think Hakim (my apologies if this is not Jeff's theorem), generalized by Prasad and Schulze-Pillot, that allows you to globalize representations distinguished with respect to a subgroup. This one uses a simple relative trace formula.
Answer: Also, if $\pi_v$ is square-integrable and $G=Sp$ or $SO$, Arthur proves this in his upcoming book. See my reply to the question embedding of local representation into automorphic representation


How do you wander about with your Handheld Recorder?

Question: This might be a stupid question but I have a hard time dealing with it though, I use a Zoom H4n with a dead kitten windscreen I want to take it with me everywhere I go but I find it a little to big to fit in my pocket. During the winter time it was alright cause I had my hoodie on most of the time and it would fit in those pockets, So I'm wondering how do you guys carry around your handheld recorders?
Answer: I've got my Sony PCM-D50 in a small zipper pouch with an emergency pair of earbuds, spare rechargeable batteries, and a couple of tiny cables (and room for a telephone coil pickup or contact mic). This stays in my daily carry-everything shoulder bag. That's for everyday life.
When I'm hiking, traveling, or vacationing, I strip down to the recorder, earbuds, and spare batteries and all that lives in my camera bag (holster-style, worn over the shoulder), which I never go anywhere without or have stored securely elsewhere.
Answer: I used to carry my Olympus in my pocket, since it was so small, but I've recently gotten a Sony D50 - much bigger. I'm currently exploring possibilities myself now, but here are a few solutions I've considered:
-I generally have a messenger bag with me, so I could keep it in there most of the time.
-Perhaps a small camera bag with a belt loop or shoulder strap.
-Maybe get a bag for it and connect it to yourself with a carabiner.
-Perhaps have some sort of large "belt clip" system, not dissimilar to a phone holster
-Speaking of holsters, you could buy a pistol holster large enough to fit your recorder and carry it in that.
-Cargo pants / shorts?
Just some ideas. Hopefully one of them works for you or sparks an idea! Good luck!
Comment: The only thing about the pistol holster is that it may be close to impossible to do stealth recording. Get down! He's got an H4n!
Comment: I've fit the D50 into my cargo pants pocket before. You'll probably be able to pull that off. Not the best solution, but it works in certain situations.
Comment: lol I could walk into a grocery store and threaten the cashier to give me all their money with it, like that guy who tried with a PS3 controller.
Comment: @Stephen - considering how that went for him, you might not want to try ;-)
Answer: One of the nice features of looking like I do is that no one really notices WHAT I'm doing.  Still, just use it as you want -- hold it like a gun if that's your thing, or set it on a tripod and just watch.  People will generally only hassle you if you look like you're not supposed to be doing what you're doing, and you're hardly breaking any law recording sound.
I read once of someone carrying business cards that said something along the lines of:
"(Please read silently)
I'm recording sound, not video.
Please refrain from asking me questions.  Thank you."
That may help, too.
Answer: I've got a Sony PCM-D50 and just found the perfect case for it. It's a Lowepro Apex PV AW
It's got a belt loop, a shoulder strap and also comes with a very handy built in rain cover. It's designed for Portable Photo Viewers, but seeing they are about the same size as the D50 I though I would try one out, i'm very glad I did.
The D50 fits in without any problems with a windjammer on, it is a snug fit, but thats what you want. It's relatively light and compact and as it looks like a camera case it goes completely un-noticed. Inside the case is also a small pouch for memory cards etc, and also has just enough space for a spare set of batteries and some earbuds.
I'm awaiting a new Rycote windjammer for the D50 and once I have it I will post some pictures.
Answer: I carry my sony m10 in my pocket, wrapped in a sock which i use as wind protection as well, but its considerably smaller than the zoom...
When i only used a zoom h4 i just had it in a backpack, which i carried all the time anyway
Answer: I bought an H1 for this specific purpose.  It's a tiny little guy that lives quite happily in the front pockets of all my jackets/hoodies/sweaters.  It's got the same mic setup as the H4N, which is nice, and because of it I've caught some very cool stuff that I have sorely missed.
Answer: I carry a messenger bag with me most days. My M10 fits perfectly in the cushioned cell phone pocket inside. If you're against having to keep track of a bag full of crap, maybe just a cell holster would fit?
Answer: This is one of my most frustrating situations, How to have to recorder ready to go at all times without it becoming a burden and start dragging down your "normal" day to day life.  I have the Sony D50 and ideally I try to throw it in a bag so it is nearby where ever I go, thats great for work days when I have a bag with me already filled with hard drives and a laptop as I go to the studio or screenings around town.  But the problem is that the perfect sound tends to happen when you are just wondering killing time on the weekend and have nothing with you, and it drives me crazy when that sound gets away.  I got a ZoomH2 because it is small enough to pop in most pockets and I thought that would be my solution.  Even though its quality is not the best it is better to get the sound on a so-so recorder then not get it at all - right?  But my problem with the H2 is it seems to take forever to boot up, and by the time it does sometimes the sound is lost even though I had the recorder with me.
The same thing has happened in my house where something is happening out side and then by the time I run to the basement grab my recording rig and point the mic I have missed the event entirely.  So as of a month ago between recording gigs, I have been keeping my stereo shotgun mic in its blimp on a mic stand cabled up to my field recorder in my living room, ready to go when something great presents itself.  Now my real problem is convincing my wife this is a good long term idea, as this is not her version of interior decoration.
Comment: @AzimuthAudio Get an H1, the boot time is half that of the H2.  It's also bell/whistle free, which is actually quite refreshing.  I know that Yamaha and and a couple other companies also make super small and simple stereo pocket recorders.  Mine lives in my breast pocket with a RedHead little Kryptonite on it.  Stylin'.
Comment: @AzimuthAudio I read on a post earlier for the H4n that the size of the ssd card helps determine the boot time? I've got a 16GB in my H4n and it can take a while to boot, this happened the other day with a roaring plane fly-by that I missed the best bits of because it took so long :[ I'm yet to test the theory myself (it's on the to-do list) but perhaps try keeping a smaller card in it? If you have one handy and a spare second compare the boot times and see if it helps at all. Hope that helps man
Answer: I'm usually carrying around something in my messenger bag anyway, so I always throw an H4n in as well.
Paul
Answer: I put mine in my pocket and record when theres stuff to record.  I wear hoodies too and it pokes out a little bit so i always keep my hand on it incase I hit a bump or something.
Answer: sony m10 here, in a little pouch. whatever bag i carry, there's always a pocket for it. no headphones at all, or i use the headphones from my ipod - sennheiser earbuds, and have a pair of binaural mics go straight in my ears, in case it's a tricky situation. when i feel more inspired i carry it in my pocket. great little thing, i've been raving about it for ages.
Answer: I have an H4n and i found an old CD player carrier does the trick perfectly - the zip bit at the front is perfect for carrying around the accessories such as the mains adapter, windshield, handle etc and the main compartment is padded and fits the H4n very well.  It also has a hole at the bottom (which i presume was for the headphones) but its great to run cables from if you're using external mics into the xlr inputs.  It also has a handy strap for hands free operation! It all folds up into a small bag so you can chuck it in with the rest of your gear.
Not sure how likely you are to get a portable cd player holder in this day and age, but you might find something similar.
Hope this helps!
Gillian
Comment: Rocking it old school! Sounds great.
Answer: Hi all, I'm the new kid on the block :)
I kept coming to this site whenever i was googling something i was wondering about and
i'm so glad to have found this place. What a great resource!
I'm sure i'll be asking a lot of stupid questions in the future, but i'll start with a tip:
When i don't want to bring anything else but my H4n i put in theDuet Carrying Case
Bought it for my traveling sound card when i'm playing live, but found it to be perfect
for the recorder too. There is also enough space for extra memory, batteries and the Rycote's portable Recorder Audio Kit etc. To small for headphones though...
Check it out,
Tommy
Comment: Welcome, thanks for the link looks great and to be the right size.
Answer: I use a zoom h1 most of the time because I can carry it everywhere and it fits perfectly in the rode videomic pro dead catwhich  covers the whole recorder and no one can tell what's in it, I also bought two plant mic baggies, search eBay for those, they're basically zippered linen fabric cases not designed to protect but to conceal, they fit the H1 and there's a larger for the h4n and you can just leave in the bags recording incognito, on a desk table outide etc, lots of tiny holes in the bags being linen, great stealth, check eBay for "plant mic baggie" for zoom h1, i have both the h1 and h4n about 12 bucks but perfect or discreet recordings.
Answer: I bought a carry case for my Roland R26. Its perfect as you can wear it over your shoulder and the case opens up to expose the screen and mics so you can still record while wearing it. Maybe Zoom make something similar?
Answer: I have my Zoom H2 with me all the time, I keep it in a small padded pouch attached to one of the straps on my rucksack.
Answer: I'm having a hard time finding a solution that protects the recorder and that remains practical, so I designed my own. You can learn more here : http://rgalix.com/audio/
Answer: If you have an iPhone, the Tascam iM2X is worth checking out. It basically clips into your phone and turns it into a recorder. The mics are the same as used in the Tascam DR series and it's has it's own internal A/D converter to minimise noise from the phone's preamp. A bit of a ballache to get it out, clip into your phone and boot the recording app but no more than booting up your Zoom. I've had pretty good experience with it for on the fly ambient recordings and it's super small:
http://tascam.com/product/im2x/