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20pcs Remy Tape 16inches Human Hair Extensions #1b

16" Natural Black(#1b) 20pcs Tape In Human Hair Extensions

Start With: $39.99 $124.95
* select Length:16 Inches
  • 16 ”
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* select color:Natural Black(#1b)

Quick Overview

Type: Tape On
Color: #1b
Length: 16 Inches
Texture: Straight
Qty: 1 Set-20pcs

Reg: $124.95

Now: $39.99

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16" Natural Black(#1b) 20pcs Tape In Human Hair Extensions
$ 39.99 New Available online
16" Natural Black(#1b) 20pcs Tape In Human Hair Extensions
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Hear the VOICES from our customers.You can TRUST us as they did.
Willow Parrish Loves COCO REMY HAIR.
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  1. taylor sabourin
    A++++ thanks s...
    Posted on 9/1/14
    A++++ thanks so much !
  2. sherry renee dusek
    Posted on 8/30/14
    Nice piece and color is really really really pretty with its ombre purple to pink, my teenage daughter loved it.
  3. Satenik
    Five Stars
    Posted on 8/30/14
    Very good quality actually, silky shiny hair
  4. Origami owl
    I just have a question....... Please answer
    Posted on 8/28/14
    I ordered light green to vanilla can u tell me if it looks nice I have dirty blonde/brown hair and colors like blue a,d green look good in my hair. I have thin hair and a sensitive scalp so do they weigh down your hair and itch like crazy!?!?? And does it look fake
  5. D. Carducci "Nadeya77"
    This is REAL hair, it's thin, but beautiful and color-matches perfectly. It's also. TON cheaper than anywhere else!!!
    Posted on 8/26/14
    I literally JUST came from my local beauty supply store and bought two packs of 18" Color #27 Remy Clips. When I say a "pack" I mean 2 individual clips in each pack, so a total of 4 INDIVIDUAL clips, not sewn together, or graduated down like this one on Amazon. Literally 4 individual clips I wanted to add to my pony tail for length. Even though this Amazon product says "7 pcs" I think they mean "7 layers" because it actually includes 16 individual pieces sewn into a graduated pattern of 7 rows, just like the picture shows.So here are some AMAZING things I've already discovered that I think anyone considering this purchase should know:1. This is a PERFECT color match for me. I HIGHLY recommend finding out if you have a local beauty store that sells this brand, and go COLOR-MATCH it FIRST. I was in between two colors and could have easily gone with either, or even mixed them together for a highlighted look.2. Amazon's price is amazingly lower than the beauty stores. For the 4 measly individual clip-ons I purchased at the store, I paid a whopping $32.68; but for $24 on AMAZON (with Free Super Saver 2-Day Shipping) you get 16 individual pieces, already sewn together into a full head piece. Pretty amazing savings. PRICE SAVINGS: If you do the math and divide out the cost of each individual clip, I paid $8.17 per individual clip from the beauty supply store, and only $1.50 per individual clip on Amazon!!! For the EXACT same product! Needless to say, I'm returning these 4 individual extensions tomorrow morning first thing. I've already ordered this 16 piece kit from Amazon tonight, and can't wait to get it on Tuesday. I will update my review then to let you know what I think.3. These extensions are NOT for full thick hair, or for those of you wanting the full, thick hair look. They are pencil strait, and super silky thin. The four individual clips in my ponytail do look a little "skinny", but they are believable, and my husband is sooooo thrilled (I chopped off all my mid-back length hair after our wedding into a bob, now I'm growing it back out, but its taking an eternity:-). So, keep that in mind if you don't already have thin hair, or if you are wanting super full extensions. Maybe two sets of these extensions would give you the fullness you want, but for me, it's about believability. I'm not competing for the title of: "Biggest Hair in Texas" any time soon:-)4. I've worn synthetic extensions as a ponytail before, so I know the difference between real and synthetic. Plus, the proof is in a simple test. Cut a tiny piece of hair and burn it with a match. If it has that distinctive burnt hair smell, then you have the real deal. I mean, it's not like they are giving you 5 pounds of extensions and calling it real hair. It is a reasonable amount of hair to be considered human hair in my opinion. Even if its not human hair, I could care less. For the price, and believability (ESPECIALLY for a girl living in the heart of L.A.:-) its worth every penny to give it a shot.5. Because its so thin, and presumably, real human hair, you will need to be GENTLE with taking care of it. Brushing through it like its growing out of your head is not smart, it will start falling out of the clips, even though it is sewn into each clip. Use a delicate, tangle free brush and gently brush downward, once, maybe twice. Over brushing will not make it grow like your real hair, so try and avoid this bad habit. I've already caught myself running my fingers through it several times, and did brush it a bunch before reading the packaging and more info from Remy online. The better you take care of it, the longer it will last.6. For all the negative reviewers out there, if you want a head full of real human hair, that's full and thick and that appeals to your every wish and desire, you're going to pay thousands of dollars for it. Yes, I said thousands. Real hair isn't cheap, (again, the reason why these clips are on the thin side, it's not cheap , and they aren't giving it away for free, so you get the amount of hair you pay for). If you still want huge, full hair without spending tons of money, maybe order the synthetic stuff and see how you like it. Some of it might actually not be so bad. Since I'm not looking for that kind of look, and because I already have pencil straight, thin hair, would rather have this product over anything else while I'm patiently growing out my own real hair:-).
  6. Cour
    i loved them
    Posted on 8/14/14
    loved them, they matcched my hair perfect and everyone kept saying how nice they looked, i would get them again
  7. Meg Jacobi
    Totally shocked at how great they are
    Posted on 8/11/14
    Totally shocked at how great they are. I have $600 Extentions I have put into my hair and I like these better😀
  8. Justin
    Great product, great price!
    Posted on 8/10/14
    I bought these for my wedding and they matched my hair color perfectly! They are true to color, easy to style and blend in with your natural hair. Would definitely buy from this seller again!
  9. DrG
    Very nice
    Posted on 8/9/14
    Ive had these extensions a month now and theyre still in great condition. I was worried before i recieved them because of all the negative reviews. I was very satisfied though, yes they are pretty thin but what do you expect for $20? other than being thin theyre the best extensions ive ever had, ive been buying 18" extensions from other stores for about 8 years and these have been great especially for the price.
  10. Pen Name
    first timer
    Posted on 8/8/14
    I've never gotten clip ins before but these are very soft, I could actually use my brush on them and hair didn't come out. They stayed in very nice and a little tattoo sick on and weave cap came in the order. I don't know as gifts but that's cool.
Color Natural Black(#1b)
Style straight
Length Range 16 Inches
Hair Extension Type tape in
Brand Hairextensionsale
Unit Weight(g) 30g
Quantity 20PCS
  • 10"
  • 12"
  • 14"
  • 16"
  • 18"
  • 20"
  • 22"
  • 24"
  • 26"
  • 10"
  • 12"
  • 14"
  • 16"
  • 18"
  • 20"
  • 22"
  • 24"
  • 26"
  • 10"
  • 12"
  • 14"
  • 16"
  • 18"
  • 20"
  • 22"
  • 24"
  • 26"
Length I need:
10" 12" 14" 16" 18" 20" 22" 24" 26"
Our hair length depends on the different hair textures you choose. The straight hair naturally falls longer because it lies completely flat when it's measured. We measure the length of our curly hair when it is in a straight position because it has more volume for the same weight. The length of our wavy hair texture lies in-between that of the straight and curly as you can see in the above picture.This guide will help you to determine where a length will fall when placed at the top of the head. The hair will fall even longer when placed at the nape area.
Our color ranges in all the popular hair colors listed below, which are all in stock and ready to ship in time. These colors list can accommodate 90% of our consumers. If you are still not sure about your hair color, you can order a color ring to see the entire range of colors we have for sure. We can also offer color customizing on all our products if you have special needs.

    Section your hair horizontally in line with the tips of your ears, making sure you start from the bottom of the neck. This will make it easier for you with how to apply tape hair extensions.

    Start by placing the first tape hair extension into your hair gently. Be careful not to apply the extension too close to the root of the hair. This should allow the tape extension to move and sit comfortably without pulling and causing too much friction between your scalp and the roots of your hair. With this method, you are allowing for the tape hair extensions to also be styled (ponytail or bun updo, tight curls or braids) without forcefully pulling on your roots too much. Otherwise, the tape hair extensions could potentially be uncomfortable and unable to style. This could also cause damage or inflammation to the scalp.

    Raise the first applied extension up, clipping it to your hair bundled on top. Some people are very concerned with how to apply tape hair extensions by this point! But do not stress too much, it is completely normal for your own hair to attach itself up to the tape hair extension.

    Take your second tape hair extension and apply it by pressing it tightly against your hair with the first tape extension that you have already clipped up. Your tape extensions are now neatly stuck together with your own hair sitting comfortably in between.

    Repeat steps from 1-4, ensuring that the tape hair extensions are sitting side by side with a 1cm distance in between each extension, to also allow for some movement over time. Complete your first row of tape, and then move on to your second row (usually 1 inch up from first row) and work your way up on your head. Keep in mind that if you are placing your extensions too far onto the left and right sides of your head, your tape hair extensions may become visible. Ensure that your tape hair extensions stay covered by your own hair.
  • You have now learnt how to apply tape hair extensions!
  1. Keep your hair extensions clean and untangled by taking time to care for them every day.
  2. Do not comb or brush the hair extensions while hair is wet.
  3. Never sleep with wet hair extensions. Always make sure the hair is dry and pulled up or preferably put into a braid to keep hair from tangling at night.
  4. Brush the hair with a hair extension brush several times a day. Be sure to brush properly, holding the roots to prevent pulling on the bonds. Use a wide tooth comb to remove tangles if needed.
  5. Use high quality hair care products but don’t overdo it, too much hair care product can cause build up and tangling over time. So you want to be careful with the amount of product you use. Never use any products containing oil, alcohol or ethanol as this can damage the hair extensions and also wear down on the adhesive. Try your best to use sulfate free shampoo and conditioner as well as 100% natural hair products to keep hair extensions in good condition.
  6. When working out with hair extensions or swimming with hair extensions, be sure to remove excess sweat and chlorine after working out.
  7. When using hot tools be sure to use a heat protectant spray! The more you use heat on your hair, the shorter the life-span of the hair extensions.
  8. Do not use a flat iron directly on the tape tabs. This will weaken the bond and they will slip out.
  9. When using conditioner, condition from mid-shaft to ends only. Do not apply conditioner to your scalp when wearing tape in hair extensions.
  10. When using serums or oils avoid contact with the tape, use on mid length to ends only. We use natural products such as coconut oil on the ends of the extensions to keep the hair healthy and shiny over time.
  11. Every 6-10 weeks, your hair extensions will need to be reapplied to move the bonds closer to the scalp and to keep your hair healthy.
  12. Tape in extensions are not best for those who have lost a significant amount of hair due to any illnesses or for those that have extreme hair thinning. Consult with your stylist about another option.
  13. Clarifying shampoo BEFORE going to your stylist is KEY. The stylist should also use clarifying shampoo prior to installing the hair extensions. Conditioner should not be used during application of hair.
  14. Since a person’s average natural hair loss is between 50-100 strands per day you will see a small amount of naturally released hairs in between the tape bonds at time of removal. This is normal and should not be interpreted as hair loss due to wearing hair extensions.
At the end of the day, this hair is gorgeous and so are you! We want to keep it that way!
So take care of your hair and you will find out how long they last!

Payment methods

1.Pay via Credit Card / Visa, Mastercard, AMEX, JCB ect.

Note: Please choose one payment methods at your convenience. Remember to try another if you can not go through one. Don't Worry! we only charge you once. the order will be cancelled automatically if your payment didn't go through.

FREE SHIPPING for Orders Over $59

Standard Shipping(9-15 Business Days):

We give International standard delivery worldwide which comes with an online tracking number and need a signature. Orders can be tracked through
USA: USPS: http://www.ups.com
UK: Royal Mail: http://track.royalmail.com
Australia: Australia post: http://www.auspost.com.au

The postage for standard shipping is $6.95. If your order is over $59, you get Free shipping.

Express Shipping(3-4 Business Days):
If you require rush order, you can choose our Express Shipping option for $15.95. Orders can be tracked through

Note: Once your order is shipped out, you will receive a tracking number by Email from us. If you don't receive it by any chance, please contact us at: contact us

Q: What are differences between human hair & synthetic hair?

A: Human Hair offers the most natural look and feel. It is remarkably soft with a shine and movement that is not easily duplicated in synthetic hair. It is also extremely versatile. Human Hair can be cut and styled to suit your personal tastes. And while it can be a more expensive pick, with the proper care, it is also more durable and can last over a year.

With technological advances, synthetic hair has come a long way in recent years. In some cases it is difficult to tell the difference -- the denier and texture are such that they feel almost like human hair. As a matter of fact, with some of the higher quality synthetic pieces, you would never know they were not natural human hair if you didn’t have the human hair right next to it to compare!

Human Hair
● Ooffers the most natural look & feel
● Presents the most styling versatility
● Requires more styling - especially after washing
● Typically more expensive than synthetic hair
● Less durable than human hair — 4-6 months with proper care.(It is important to follow our care recommendations in order to get the optimal life span from your hair!)
● Premium & Remy Human Hair are the highest quality human hair available
● Hair cuticles kept running in uniform direction resulting in less tangling
● More careful collecting and processing techniques result in higher quality product and contribute to higher price
●The most durable hair type
Synthetic Hair
● higher quality synthetic hair offer a look and feel quite similar to human hair
● offers limited styling versatility (hot tools cannot be used)
● requires little maintenance & can usually be worn right out of the box
● typically more affordable than human hair
● very durable - can last over a year with proper care.(It is important to follow our care recommendations in order to get the optimal life span from your hair!)
● Heat-Friendly Synthetic Hair offers more styling versatility (hot tools may be used - max 350°)
● Can be more challenging to style than human hair
● Care must be taken to avoid tangling
● Colors may vary from non-heat-friendly synthetic hair
● Less durable than regular synthetic hair — 2-3 months with proper care(It is important to follow our care recommendations in order to get the optimal life span from your hair!)

Q:What are differences between human hair & synthetic hair?
A: Unfortunately, any heat styling runs the risk of damaging synthetic wigs. Wigs with synthetic hair can be purchased pre-curled or in any style you desire. Alternatively, you could curl the wig with a styling method that does not involve heat. Wigs with human hair will not be damaged by heat styling.

Q: What type of fiber is used for your synthetic wigs/hairpieces?
A:The fiber is made from the raw material Kanekalon, which is imported from Japan. It's commonly used for hair extensions, weaves, wigs and doll hair. Kanekalon is flame-retardant and comes in a variety of styles and textures, plus more than 120 colors.

Q: Can I color or straighten the hair in this wig?
A: Wigs made with human hair can be colored or styled as you would your own hair. Synthetic wigs can be colored, but heat styling of any kind could be damaging to the fibers.


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Tuesday, April 17, 2018

Fire Line Equipment

Your Central Pennsylvania E-ONE Dealer

Choose the Material that Works Best for Your Needs.

You have unique requirements. So E-ONE offers body design and fabrication processes, customized to your department’s specific needs, with one of the industry’s most durable frames. Our unique, all-aluminum, high-strength extrusions minimize the amount of an accident’s damage to the cab. Because of the modular design, repairs to the body can be made in sections. E-ONE’s all-aluminum, extruded body is lightweight, corrosion- resistant, and provides years of trouble-free service. The all-welded tongue and groove design is superior to a bolted design, and every E-ONE body is backed by our 10-year structural warranty.

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Divination Light

Download Divination Light for PC

By Charles Angel

  • Category: Reference
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Divination Light for PC Screenshots

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Divination Light includes Tarot Card and Horoscope Analysis. Tarot Card may predict your health, love and career which may help you to make right options by learning your past, present and future status. Horoscope Analysis may help you learn about your Zodiac signs which might reflects your life tracks by the ascent or descent of the stars. Features - Tarot Card : Health, Love and Career help you understand your current situation. - Zodiac Horoscope: Select your sign and show description of your horoscope, and characteristics. - Zodiac Sign Match: Select the zodiac sign, discover the love horoscope and zodiac compatibility in your relationship. Subscription pricing and terms： For unlocking all features and content, Premium Membership offers one auto-renewable subscription options: 1.49$/week,The first three days of subscription are free. End of trial and subscription renewal： - Payment will be charged to iTunes Account at confirmation of purchases. - Subscription automatically renews unless auto-renew is turned off at least 24 hours before the end of the current period. -Account will be charged for renewal within 24-hours prior to the end of the current period, and identify the cost of the renewal. -Subscriptions may be managed by the user and auto-renewal may be turned off by going to the user’s Account Settings after purchase. -Any unused portion of a free trial period, if offered, will be forfeited when the user purchases a subscription to that publication, where applicable. -Privacy Policy: http://z.mpq20.pro/irk/divinationlight/PrivacyPolicy.html -Terms of Service: http://z.mpq20.pro/irk/divinationlight/TermsofUse.html


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Olympocalypse! Canada loses in hockey and I make a video

I realize it’s probably a bit tacky to suddenly return after not posting for a long while to try to drive more traffic to a video I made, but hey I still have the most posts on Vancouver Metblogs so I figure it should be allowed.  Granted most of those posts were just a video of the sea otters at the Vancouver Aquarium floating around their tank holding hands, but it’s hard work making that seem new after the fourth hundred time.  And gosh aren’t they cute?

In exchange for housing during the Olympics an out-of-town friend of mine provided me with tickets to see the Men’s Olympic Ice Hockey game between Canada and the United States of America.  Inspiredby the righteously awesome “A Night At The Emirates” video [yt] I set out to make my own fan’s perspective video.  It’s hard to say how it turned out because I’ve spent too much time on it to see it clearly and my wife loves me just enough not to insult it to my face.

So enjoy the video as you prepare to watch Team Canada (hopefully) beat up on Germany tonight.  Looking on the bright side it was a valuable wake-up call early in the tournament that hopefully will let us adjust before facing tougher opposition like Russia; though to be fair that’s what the game against Switzerland should have been.

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Write a project proposal examples

They can be solicited or unsolicited. How long will it take? However, the DOE is undecided on where to put this high level radioactive waste. In this stage, aftershocks occur. Objectives I propose to review the available literature about using Yucca Mountain as a possible repository for spent nuclear fuel.

For that reason, much pressure exists to select Yucca Mountain as a repository site; otherwise, this money would have been wasted. In Tokyo, the damage from the earthquake was less, but the resulting fires were more devastating.

I have taken several classes related to nuclear waste, economics, and environmental write a project proposal examples. Our ability to engineer, construct, and operate facilities rounds out that objective. This research culminates in a formal report, which will be completed by December 5, Management Plan This section presents my schedule, costs, and qualifications for completing the proposed research.

Project Proposals contain key information about your project. I am a senior in the Engineering Physics Department at the University of Wisconsin at Madison, majoring in nuclear engineering and physics.

More than 30, metric tons of nuclear waste have arisen from U.

We started with only three people and now employ over employees. This discussion, drawing much from overview chapters in Earthquakes, Animals and Man [Deshpande, ] and California Quake [Meyer, ], will put into perspective how accurate, or inaccurate, the named methods are and what hurdles face engineers who try to predict earthquakes.

Proposals can be internal within your organization or external written from one independent organization to another. This document proposes a literature review of whether Yucca Mountain is a suitable site for a nuclear waste repository.

Because engineering students are the primary audience for my proposed research topic and may not be familiar with the history of nuclear waste, I will provide a background on past methods used for waste storage. Who will be doing it? You should also provide a brief setting and history behind the project.

Unfortunately, nuclear fission, the process used to create this large amount energy, creates significant amounts of high level radioactive waste. The stir caused by theory suggests that researchers have not explored all sides of the safety issue concerning potentially hazardous situations at Yucca Mountain.

The economics involved in selecting a site is another criterion. This document has proposed research to evaluate the possibility of using Yucca Mountain as a possible repository for this spent nuclear fuel. Should I require additional sources other than the ones I have, I will search for them in the library system at the University of Wisconsin.

A Project Proposal is not a contract. Also, because one of the most significant factors in determining the life span of a possible repository is how long the waste storage canisters will remain in tact, the waste site must be located in a dry climate to eliminate the moisture that can cause the waste canisters to corrode.

Since I already possess literature on the subject of Yucca Mountain as a nuclear waste site, most of my time will be spent sorting through the literature to find key results, and presenting those results to the audience.

Despite these successes, the Chinese failed to predict the earthquake that struck the city of Tangshan on July 27, ; this earthquake killedpeople and injuredmore [Bolt, ]. Justification of Proposed Review On the morning of April 18,the population of San Francisco was awakened by violent shaking and by the roar caused by the writhing and collapsing of buildings [Hodgson, ].

Rating Radioactive Waste Risks," Science, vol.

44+ Project Proposal Examples – PDF, Word

Two days later, a magnitude 4. To achieve these two goals, I will rely on three books that give an overview to earthquake prediction: The only costs, which will be minor, are for copying articles, printing the review, and spiral binding the review.

According to the Department of Energy DOEa repository for high-level radioactive waste must meet several criteria including safety, location, and economics [Roush, ]. The formal presentation will be on October 27, and the formal report will be completed by December 5. Even more disastrous was the Kwanto earthquake in Japan that devastated the cities of Yokohama and Tokyo on September 1, [Hodgson, ].

These earthquakes struck without warning and with disastrous results. As you sit down to write your own document keep in mind that although some details will necessarily change to accommodate your particular project type, scopeand industry, the general structure is more or less the same for any kind of project proposal.

Scientific American Library, Some project professionals even suggest writing the project summary last.Proposal Writing Basics Whether it’s a business, project, or a different type of proposal, the goal is the same: to convince the reader to make the choice you propose.

Therefore, your proposal has to. Sample proposals for the Writing Guidelines for Engineering and Science Students: guidelines to help students of science and engineering make their writing more efficient for others to read and to make the process of writing more efficient for them to perform.

This proposal includes my methods for gathering information, a schedule for. Looking for a project proposal example or a template? Check out the project proposal toolkit, with FREE to use template, samples, examples, guide and even video tutorials.

Example Guru Project Proposals One of the trickiest parts of securing a freelance job is knowing how to write a winning project proposal. Not only do you need to show off your strengths and experience; you need to write the proposal in a way that suits the Employer.

The following project proposal outline is in a generic format so it can be applied to several kinds of circumstances and projects.

As you sit down to write your own document keep in mind that although some details will necessarily change to accommodate your particular project type, scope, and industry, the general structure is more or less the same for any kind of project proposal. A Sample Research Proposal with Comments A research project or thesis will take at least two semesters to complete.

Prior to starting a research, i.e.

Write a project proposal examples
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Monday, 9 January 2017

Engine & Driveability Problems: 12 Engine Vacuum Leaks

engine intake vacuum gauge reading

Engine Vacuum Leaks

Copyright AA1Car
Have you ever tried to tune an engine only to find it won't idle or run right? Or have you ever been confronted with an engine that just doesn't seem to run right no matter what you've done or replaced? You may be dealing with an engine vacuum leak.
Sometimes a vacuum leak will whistle or hiss and make itself obvious. But oftentimes, a vacuum leak will disguise itself as an ignition or fuel problem that defies diagnosis. Either way, an engine vacuum leak is bad news because allows "unmetered" air to enter the engine and upset the air/fuel ratio.
So how do you know when a vacuum leak is causing a problem? If the engine is experiencing any of the following symptoms, a vacuum leak is probably responsible:

  • Too fast an idle speed. If an engine without computerized idle speed control is idling too fast and refuses to come down to a normal idle speed despite your best efforts to back off the carburetor idle speed screw or air bypass adjustment screw (fuel injection), air is getting past the throttle somewhere. Common leak paths include the carburetor and throttle body gaskets, carburetor insulator spacers, intake manifold gaskets, and of course, any of the engine's vacuum fittings, hoses and accessories. It is even possible that leaky O-rings around the fuel injectors are allowing air to leak past the seals. Another overlooked item can be a worn throttle shaft.
  • A rough idle or stalling. A performance cam with lots of valve overlap can give an engine a lopping idle, but so can a vacuum leak. A really serious leak can lean the air/fuel mixture out to such an extent that an engine won't idle at all. An EGR valve that is stuck open at idle can have the same effect as a vacuum leak. So too can the wrong PCV valve (one that flows too much air for the application), or a loose PCV hose. The rough idle in these cases is caused by "lean misfire." The fuel mixture is too lean to ignite reliably so it often misfires and fails to ignite at all. Lean misfire will show up as elevated hydrocarbon (HC) readings in the exhaust, enough, in fact, to cause a vehicle to fail an emissions test.
  • Hesitation or misfiring when accelerating. This may be due to a vacuum leak, but it can also be caused by a weak or inoperative accelerator pump in a carburetor, dirty injectors, or even ignition problems such as a cracked coil, worn spark plugs or incorrectly gapped plugs.
  • An idle mixture that defies adjustment. When setting the idle mixture adjustment screws on a carburetor, the idle speed should start to falter as the adjustment screws are turned in to lean out the mixture. If the screws seem to have little or no effect on idle, you have either got a carburetor problem or a vacuum leak.
The important thing to keep in mind about vacuum leaks is that they have the most noticeable effect at idle. At part and full throttle, there is so much air entering the engine that a little extra air from a vacuum leak has a negligible effect.
TIP: If you have a scan tool, look at the Short Term Fuel Trim (STFT) and Long Term Fuel Trim (LTFT) values. Normal range is plus or minus 8. If the numbers are +10 or higher for STFT and LTFT, the engine is running LEAN. If you rev the engine to 1500 to 2000 rpm and hold it for a minute or so, and the STFT value drops back down to a more normal reading, it confirms the engine has a vacuum leak at idle. If the STFT value does not change much, the lean fuel condition is more likely a fuel delivery problem (weak fuel pump, restricted fuel filter, dirty fuel injectors or a leaky fuel pressure regulator) than a vacuum leak.
For more information about using fuel trim to diagnose a lean fuel condition, see What Is Fuel Trim?, or view this article on Fuel Trim by Wells Manufacturing (PDF file).
Before we get into the various techniques of finding and fixing vacuum leaks, let's quickly review vacuum's role in fuel delivery.

Intake vacuum exists in the intake manifold as a result of the pumping action of the engine's pistons and the restriction created by the throttle valve. Were it not for the throttle choking off the flow of air into the engine, there would be little if any vacuum in the intake manifold (like a diesel). The downside of intake vacuum is that it creates pumping losses and reduces engine efficiency.
On older carbureted engines, vacuum is needed to pull fuel into the engine. Vacuum siphons fuel through the idle, main metering and power circuits. An engine with a vacuum leak, therefore, will likely be an engine that suffers from the symptoms of lean carburetion such as lean misfire, hesitation, stalling and rough idle. But the same symptoms can also be caused by a clogged catalytic converter or other exhaust restriction, a leaky EGR valve or valve timing problems (all of which reduce intake vacuum).
Engines with Multiport Fuel Injection and Gasoline Direct Injection don't need vacuum to pull fuel into the engine because it is sprayed in under pressure. Even so, most of these engines still have a throttle for regulating airflow and engine speed. And like the older carbureted engines, a throttle body also creates an airflow restriction that creates vacuum inside the intake manifold.

On most engines, intake vacuum should be steady between 16 and 22 inches Hg (Mercury). A lower reading usually indicates a vacuum leak, or one of the other problems just mentioned. A reading that gradually drops while the engine is idling almost always points to an exhaust restriction. An oscillating vacuum reading usually indicates a leaky valve or badly worn valve guides that leak vacuum.
Although fuel injected engines do not rely on intake vacuum to pull fuelinto the engine, vacuum leaks can upset the carefully balanced air/fuel ratio by allowing "unmetered" air to enter the engine. The result is the same kind of driveability symptoms as a vacuum leak on a carbureted engine (lean misfirehesitation when accelerating, rough idle and possibly even stalling). Common leak points include injector O-rings, intake manifold gaskets, idle air control circuit and the throttle shaft.
Fuel injected engines also rely on intake vacuum to regulate the fuel pressure behind the injectors. Fuel delivery cannot be accurately metered unless a fairly constant pressure differential is maintained. So the fuel pressure regulator diaphragm is connected to a source of intake vacuum. Vacuum working against a spring-loaded diaphragm inside the regulator opens a bypass that shunts fuel back to the tank through a return line. This causes the fuel pressure in the injector rail to rise when engine load increases (and vacuum drops). Thus, the regulator uses vacuum to maintain fuel pressure and the correct air/fuel ratio. A vacuum leak changes the equation by causing a drop in vacuum and a corresponding increase in line pressure.

Vacuum is measured with a vacuum gauge. Most are calibrated in inches Hg (Mercury), but you may also see some gauges that are also calibrated in inches of H20 (water) or kilopascals (kPa) or even bars. One inch of vacuum measured in inches Hg equals 13.570 inches in H20, 0.4898 psi (pounds per square inch) or 3.377 kPa.
Here's a conversion table you can use to convert units of measurement:
pressure conversion factors provided by

Okay, now that we have covered what a vacuum leaks do, how do you find components that leak vacuum? One way is to visually inspect all the vacuum hoses and connections. Look for disconnected, loose or cracked hoses, broken fittings, etc. Hey, you might get lucky and find the problem in a few minutes, or you might waste half the day trying to find the mysterious leak. Vacuum leaks are often the elusive needle in a haystack. And if it is not a hose leaking vacuum but something else such as a gasket, worn throttle shaft, injector O-rings, etc., you may never find it using this technique.
A faster technique for finding intake manifold vacuum leaks is to get a bottle of propane and attach a length of rubber hose to the gas valve. Open the valve so you have a steady flow of gas. Then hold the hose near suspected leak points while the engine is idling. If there is a leak, propane will be siphoned in through the leak. The resulting "correction" in the engine's air/fuel ratio should cause a noticeable change in idle speed and/or smoothness (Note: on engines with computerized idle speed control, disconnect the idle speed control motor first).
Aerosol carburetor cleaner can also be used the same way. CAUTION: Solvent is extremely flammable, so do not smoke or use it if there are any sparks in the vicinity (arcing plug wires, for example). Spray the solvent on suspected leak points while the engine is idling. If there is a leak, the solvent will be drawn into the engine and have the same effect as the propane. The idle speed will suddenly change and smooth out.
TIP: If you have a scan tool, look at the Short Term Fuel Trim (STFT) value while you are using carb cleaner or propane to check suspected vacuum leak points. If there is a leak and some of the cleaner or propane is sucked in through the leak, you will see a momentary drop in the STFT reading. This confirms you have found a leak (keep checking because there may be multiple leaks!).
smoke leak detection machine
A smoke machine with UV dye can reveal tiny vacuum leaks.
Low pressure smoke is fed into the intake manifold, then you look for the telltale dye to find the leak.

A much safer technique is to use a smoke machine. These machines feed artificial smoke into the intake manifold, The smoke may also be mixed with an ultraviolet dye to make leaks easier to see. You then look for smoke seeping out of hoses, gaskets or cracks in the manifold and/or use a UV light to find the leak. This type of equipment is often needed to find small air leaks in the EVAP (evaporative emissions) control system. Smoke machines can cost $600 to $2000 or more depending on the model and features, so they are primarily for use by professional technicians.

Propane can also be used in conjunction with an exhaust analyzer (do NOT use carburetor cleaner or you may damage your analyzer!). Engine vacuum leaks almost always cause fluctuating HC readings, so an infrared exhaust analyzer can (1) tell you if there is indeed a leak, and (2) where the leak is using the propane procedure.
Two types of vacuum leaks can be diagnosed with an analyzer. The first kind is a general vacuum leak (PCV hose, brake booster, etc.) that leans out the mixture and causes a very low CO reading and only a slightly higher fluctuating HC reading. The O2 reading will also be high. The second kind of vacuum leak is a "point" leak that affects only one or two cylinders (a leaky manifold gasket or a crack or porosity leak in one of the manifold runners). This will be indicated by a normal or low CO reading combined with high fluctuating HC readings. O2 will again be high.
To find a leak, feed propane at suspected leak points until you note an improvement in idle quality and/or a change in the HC/CO/O2 readings. When you have found the leak, the idle should smooth out, HC and O2 should drop and CO rise.
It is important to note that an overly lean idle mixture will also cause a fluctuating HC reading the same as a vacuum leak. To tell one from the other, there is a simple "trick" you can use. Momentarily enrich the idle mixture to 1.5 to 2.0% CO by placing a clean shop rag over the top of the carburetor. If the engine smooths out and HC drops and remains stable, the problem is a lean idle mixture adjustment. If HC still fluctuates, however, the engine is still too lean in one or more cylinders indicating a vacuum leak.

If you like gadgets, there are electronic tools designed to detect vacuum leaks. An electronic vacuum leak detector will beep or flash when it detects ultrasonic vibrations that are characteristic of a vacuum leak. These tools use a sensitive microphone to listen for certain noise frequencies. Though extremely sensitive, these tools sometimes react to tiny leaks that are not really causing a problem, or "false" leaks such as the noise created by arcing inside the distributor cap or normal bearing noise in the alternator.

Another way to find an elusive vacuum leak is to pressurize the intake manifold with about three lbs. of regulated air. This can be done by attaching a regulator to your shop air hose, then attaching the hose to a vacuum fitting or the PCV valve fitting on the intake manifold, carburetor or throttle body. Do not apply too much pressure or you may create new leaks! With the engine off and air flowing into the manifold, spray soapy water on suspected leaks. If you see bubbles, you have found the leak.
You can also use the opposite technique, which is to apply vacuum with a hand-pump to various vacuum hoses and circuits to see if they hold vacuum. But this technique means tracing the entire circuit to see where it ends, and disconnecting and plugging any parts of the circuit that do not "dead end" against a diaphragm or valve.

Okay, now you have found the leak. Here are some suggestions on how to fix it:

Leaky vacuum hoses Replace them. If the end of a hose is loose or cracked, cutting it off and sticking it back on may temporarily eliminate the leak. But if the hose is rotten or age hardened, it needs to be replaced. Shortening hoses may also create additional problems. The hose may chaff or rub against other components, or pull loose as a result of engine motion and vibration. Use the correct type of replacement hose (PVC hose or vacuum hose capable of withstanding fuel vapors and vacuum without collapsing). Also, be sure the replacement hose is the same diameter and length as the original.

Carburetor or throttle body base gasket vacuum leaks Try tightening down the carburetor or throttle body mounting bolts. If that doesn't stop the leak, replace the gasket under the carburetor or throttle body. If there is a heat insulator or adapter plate under the unit, it may also have to be replaced depending on its condition. While the carburetor or throttle body is off, use a straightedge to check the base for flatness (and the manifold, too). Warped surfaces can prevent a tight seal, so if you find any it calls for resurfacing or component replacement.

Carburetor or throttle body throttle shaft vacuum leaks Wear here can only be repaired by resleeving the throttle shaft, which for all practical purposes means replacing the carburetor or throttle body with a new or remanufactured unit.

Intake manifold gasket vacuum leaks Try re torquing the intake manifold bolts, working from the center out in the recommended tightening sequence. If that fails, the intake manifold will have to be removed and the intake gaskets replaced. Sometimes the mating surface of the intake manifold or the heads will not be flat (check both with a straightedge). If warped, the intake manifold and/or heads will have to be resurfaced on a milling machine. Another problem to watch out for here are heads that have been milled or resurfaced to raise compression. To maintain proper alignment between the manifold and heads, metal also needs to be machined off the bottom of the manifold where it mates with the block, otherwise it will sit to high and the ports and bolt holes won't align.

EGR valve leaks If the valve isn't closing all the way due to carbon deposits on the stem or valve seat, cleaning may be all that is needed to cure the problem. Otherwise, the engine will need a new EGR valve.

Leaky power brake booster Replace it. But first make sure it is the booster and not just the hose or check valve that is leaking.

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Random physics

Alberto Verga, research notebook

\(\newcommand{\I}{\mathrm{i}} \newcommand{\E}{\mathrm{e}} \newcommand{\D}{\mathop{}\!\mathrm{d}} \newcommand{\Di}[1]{\mathop{}\!\mathrm{d}#1\,} \newcommand{\Dd}[1]{\frac{\mathop{}\!\mathrm{d}}{\mathop{}\!\mathrm{d}#1}} \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\braket}[1]{\langle{#1}\rangle}\)

Lectures on statistical mechanics.

Non interacting bosons and fermions

In classical physics the thermodynamics of non interacting particles is essentially that of the ideal gas, described by the equation of state, \(PV = NT\), relating the intensive variables \(P\) pressure and \(T\) temperature to the extensive ones, \(N\) the number of particles and \(V\) the volume. Interesting new phenomena appear in the quantum realm, even in the non interacting case, as a result of one of the most fundamental laws of nature: the identity of elementary particles, and by extension, of elementary excitations in condensed matter.

Identity of particles

The identity of particles is a fundamental fact of quantum physics: the spectrum of solar hydrogen is identical to the one recorded in the laboratory; the exchange of particles of an ideal gas, even if they are distinguishable in classical mechanics, do not increase the entropy (Gibbs paradox). The identity of photons can be experimentally tested, as for instance in the interference of two photons in a beam splitter, the Hong-Ou-Mandel effect.

In order to understand the meaning and devise the physical consequences of the particles identity, let us consider a simple system consisting of two particles. Its hilbert space is

$$\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_1\,.$$

where \(\mathcal{H}_1\) is the one particle hilbert space. A two particle state is then of the form,

$$\ket{s_1} \otimes \ket{s_2} = \ket{1,2}\,,$$

where we introduced the basis kets \(\ket{s}\) for the single particle states with \(s\equiv 1,2,\ldots\), the labels associated to the set of quantum numbers (i.e. momentum, spin,…): particle 1 (slot one in the ket) is in state \(s_1=1\) and particle 2 (slot two in the ket) is in state \(s_2 = 2\). Now, exchanging \(2\) and \(1\), \(\ket{2,1}\), that is putting particle 1 in state \(2\) and particle 2 in state \(1\), \(12\rightarrow21\), the physical properties of the system must be preserved. This exchange leads to the two possibilities:

$$\ket{1,2} = \ket{2,1},\; \text{or} \; \ket{1,2} = - \ket{2,1}\,,$$

since both are compatible with the fact that another exchange returns the system to its initial state, \(12\rightarrow 21 \rightarrow 12\). The first case corresponds to symmetric states and the second one to antisymmetric states, their are related to the spin of the particles (in fact, it is a quantum relativistic phenomenon): integer spin particles, called bosons, have symmetric states under permutation of particles; half integer particles, fermions, have antisymmetric states. Photons are bosons of spin \(1\), electrons are fermions of spin \(1/2\).

Instead of the hilbert space of \(N\) particles \(\mathcal{H}_N\), it is possible to describe a system of an arbitrary number of particles using a representation in terms of occupation numbers, the so called fock space \(\mathcal{F}\), which is a direct sum of symmetrized hilbert spaces:

$$\mathcal{F} = \bigoplus_{N=0}^\infty \mathrm{S}_\pm\mathcal{H}_N\,,$$

where \(\mathrm{S}_+\) is the symmetrization operator for bosons and \(\mathrm{S}_-\) is the fermion’s antisymmetrization operator, and \(N=0\) corresponds to the vacuum state \(\ket{0}\). The occupation numbers \(n_s\) are simply the number of particles in a given state \(s\) (we use the convention \(n_1 = n_{s_1}\), etc.).

A state in the fock space is defined by the corresponding set of occupation numbers \(\ket{n_1,n_2,\ldots}\). It can be build from the vacuum state applying successively creation and annihilation operators of particles at a given state (an abstract generalization of the harmonic oscillator operators \(a\) and \(a^\dagger\), respectively). For instance, the state \(\ket{1,2}\) of two bosons can be constructed as:

$$\ket{1,2} = a^\dagger_1 a^\dagger_2 \ket{0}= a^\dagger_2 a^\dagger_1 \ket{0}= \ket{2,1}$$

from which we obtain the commutation relations,

$$[a_s,a_r] = 0 = [a^\dagger_s, a^\dagger_r]$$

(using that \(a\) is the adjoint of \(a^\dagger\)). A similar calculation for fermions leads to anti-commutation relations

$$\{c_s,c_r\} = 0 = \{c^\dagger_s, c^\dagger_r\}$$

using \(c\) and \(c^\dagger\) for the fermionic operators. An immediate consequence of the anticommutation of fermions is that two fermions cannot be in the same state \(\ket{1,1}\) (because it it symmetric!): this is the content of the Pauli exclusion principle. The occupation numbers \(n_s\) are the eigenvalues of the operators \(a^\dagger_s a_s\) for bosons and \(c^\dagger_s c_s\) for fermions. The boson operator is similar to the harmonic oscillator, its spectrum are integers; the spectrum of fermion operator contains, in contrast, only the two values \(0,1\).

Therefore, a \(N\)-particles boson state can be written,

$$\ket{n_1,n_2,\ldots,n_{N_s}}\,,\quad \sum_{s=1}^{N_s} n_s = N\,,$$

where particles in a similar state are grouped: the \(N\)-state is thus given by the number \(n_s\) of particles in each different state \(s\). For a fermion state we must track the order of the occupation numbers \(n=0,1\),

$$\ket{n_1,\ldots,n_N}\,,\quad n_s = 0,1\,,$$

because the permutation of two occupation numbers change the state sign.

The boson occupation numbers can be any integer number \(n_s=0,1,2,\ldots\), and the fermion occupation numbers can only take on two values \(n_s=0,1\).

The fact that the number of particles is arbitrary, makes the fock space the natural framework for the grand canonical ensemble. The hamiltonian of free bosons has the general diagonal form,

$$H_B = \sum_s \epsilon_s a^\dagger_s a_s\,,$$

where the \(\epsilon_s\) are the energy levels and \(a^\dagger_s a_s\) is a number operator with, as we said above, eigenvalues the occupation numbers \(n_s \in \mathbb{N}\). A similar expression holds for the free fermions,

$$H_F = \sum_s \epsilon_s c^\dagger_s c_s\,,$$

where the number operator of state \(s\), \(c^\dagger_s c_s\) has a spectrum of occupation numbers \(n_s=0,1\) with two values. The respective grand partition functions are

$$Z_B(T,\mu) = \mathrm{Tr}\, \E^{-(H_B - \mu N_B)/T}\,,\quad N_B = \sum_s a^\dagger_s a_s\,,$$

$$Z_F(T,\mu) = \mathrm{Tr}\, \E^{-(H_F - \mu N_F)/T}\,,\quad N_F = \sum_s c^\dagger_s c_s\,.$$

Using the explicit form of \(H_B\) and \(N_B\), and the properties of the fock states to exchange the trace and the product (the basis states are product states of the occupation numbers), we can put the grand partition function in the form,

$$Z_B= \mathrm{Tr}\,\exp\left(-\sum_s \frac{\epsilon_s - \mu}{T}a^\dagger_s a_s\right) = \mathrm{Tr}\,\prod_s \exp\left(-\frac{\epsilon_s - \mu}{T} a^\dagger_s a_s\right) = \prod_s \sum_{n_s=0}^\infty \E^{-(\epsilon_s - \mu)n_s/T}$$

which can be easily summed up,

$$Z_B(T,\mu) = \prod_s \frac{1}{1-\E^{-(\epsilon_s - \mu)/T} }\,.$$

A similar calculation for the fermions, the only difference is in the values of \(n_s\), leads to,

$$Z_F(T,\mu) = \prod_s \sum_{n_s=0,1} \E^{-(\epsilon_s - \mu)n_s/T} = \prod_s \left(1 + \E^{-(\epsilon_s - \mu)/T} \right)\,.$$

We remark that, in the boson case, the condition for convergence is ensured if the chemical potential is negative \(\mu < 0\) (\(\epsilon > 0\) for free particles), or more generally when the energy is bounded from below, \(\mu < \min \epsilon_s\).

The grand potential of non interacting bosons and fermions is

$$\Phi(T, \mu, V) = \pm T \sum_s \ln \left( 1 \mp z \E^{-\epsilon_s/T} \right)\,,\quad z = \E^{\mu/T}$$

where \(\epsilon_s\) is the energy spectrum, \(z\) the fugacity, and the top sign is for bosons and the bottom one is for fermions. In the classical limit \(\E^{-(\epsilon_s-\mu)/T} \ll 1\), the grand potential reduces to

$$\Phi = -T \sum_s \E^{-(\epsilon_s - \mu)/T}$$

which corresponds to the Boltzmann distribution.

The expected values of the energy \(\braket{H}\) and the particle number \(\braket{\hat{N}}=N\) are given by the bosons and fermions occupation numbers \(\braket{a^\dagger_s a_s}\) and \(\braket{c^\dagger_s c_s}\), respectively:

$$n_s = \frac{1}{\E^{(\epsilon_s - \mu)/T} \mp 1}\,, \quad \sum_s n_s = \braket{\hat{N}}=N\,,$$

$$\braket{H} = \sum_s \epsilon_s n_s\,,$$

where \(n_s\) are the bose (upper sign) and fermi (lower sign) one particle distribution functions EX.

Free particles: ideal boson and fermion gases

The state \(s = (s, \boldsymbol p)\) of free particles is characterized by their spin \(s\) and momentum \(\boldsymbol p\). The energy spectrum,

$$\epsilon_p = \frac{p^2}{2m} \,,$$

possesses a \(g_s= 2s + 1\) spin degeneracy (\(m\) is the mass). Moreover, in a cubic box of volume \(V\), the momentum is

$$\boldsymbol p = \frac{2\pi \hbar}{V^{1/3}} \boldsymbol n\,,\quad \boldsymbol n \in \mathbb{Z}^3\,,$$

which, in the limit \(V \rightarrow \infty\) tends to a continuum, allowing the replacement of the sum over states by an integral:

$$ \sum_{(s,\boldsymbol p)} \ldots \rightarrow g_s \frac{V}{(2\pi \hbar)^3} \int_{\mathbb{R}^3} \D \boldsymbol p \, \dots$$

We note that in this limit, the grand potential \(\Phi \sim V\) is manifestly extensive. Therefore, the problem of the statistical properties of an ideal quantum gas reduces to the calculation of the integral:

$$\Phi(T,z,V) = \pm \frac{g_sVT}{(2\pi)^3} \int_{\mathbb{R}^3} \D \boldsymbol p \, \ln \left( 1 \mp z \E^{-\epsilon_{\boldsymbol p}/T} \right) = \pm \frac{g_sVT}{2\pi^2} \int_0^\infty p^2 \D p \, \ln \left( 1 \mp z \E^{-p^2/2T} \right)$$

where we introduced units such that \(m=\hbar=1\). After a change of variables \(p^2/2T \rightarrow x\) and one integration by parts, we obtain,

\begin{equation} \label{e:phi} \Phi(T,z,V) = \mp \frac{g_s V T}{\lambda^3} \mathrm{Li}_{5/2}(\pm z) \end{equation}

where \(\lambda = (2\pi/T)^{1/2}\) is the de Broglie thermal length, and \(\mathrm{Li}_\nu(z)\) is the polylogarithm function EX:

$$\mathrm{Li}_{\nu}(z) = \frac{1}{\Gamma(\nu)} \int_0^\infty \D x \frac{x^{\nu-1}}{z^{-1}\E^x - 1}\,, \quad \mathrm{Li}_\nu(z) = \sum_{n=1}^\infty \frac{z^n}{n^\nu}\,,$$

here \(\Gamma(\nu)\) is the Euler gamma function,

$$\Gamma(\nu) = \int_0^\infty \D x \, x^{\nu-1}\E^{-x}\,,$$

see the notes below. Usually, \(\pm\mathrm{Li}_{\nu+1}(\pm z)\) are called bose and fermi integrals, respectively. (See the note below.)1

Thermodynamic quantities are then obtained form the grand potential:

$$E = -T^2 \frac{\partial}{\partial T} \frac{\Phi(T,z)}{T}\,, \quad N = -z \frac{\partial}{\partial z} \frac{\Phi(T,z)}{T}\,, $$

for the energy and the number of particles, where, when deriving with respect to \(T\) one considers \(V=\text{const.}\) and \(z=\text{const.}\), and when deriving with respect to \(z\), \(T=\text{const}\); and

$$S = \frac{E - \mu N - \Phi}{T} = -\left. \frac{\partial\Phi}{\partial T}\right|_z + \left. \frac{\mu z}{T^2} \frac{\partial \Phi}{\partial z}\right|_T = -\left. \frac{\partial \Phi(T,\mu)}{\partial T}\right|_{V,\mu}$$

for the entropy. An immediate consequence of these formulas is that the ideal gas relation,

$$E = \frac{3}{2} PV$$

is satisfied for the free quantum gases EX.

We can immediatly compute the density,

$$\frac{N}{V} = -\frac{z}{T} \frac{\partial \Phi}{\partial z} = \frac{g_s}{\lambda^3} \mathrm{Li}_{3/2}(z) \approx \frac{g_s}{\lambda^3} \left( z \pm \frac{z^2}{2^{3/2}} \right)$$

we used in the last equation the approximation for \(z\ll 1\) (the few first terms of the series defining the \(\mathrm{Li}_\nu\) function), near the classical limit of high temperature (\(T \gg \mu\)); iteratively inverting this last approximation we obtain the fugacity as a power series of the density \(N/V=n\),

$$z \approx \frac{\lambda^3 n}{g_s} \left( 1 \mp \frac{1}{2^{3/2}} \frac{\lambda^3n}{g_s} \right)$$

from which we derive an approximated expression for the pressure \(P = -\Phi/V\),

$$P \approx nT \left(1 \mp \frac{1}{2^{5/3}} \frac{\lambda^3 n}{g_s} \right)$$

We observe that the effect of the exchange of identical particles is to decrease the pressure of the boson gas with respect to the classical one, and to increase the pressure of the fermion gas. These effectives attraction and repulsion, are not related to particle interactions (of course, the particles are free), they are the result of an entropic effect, the identity of particles increases the disorder, or in other terms, it decreases the quantity of information with respect to a system of distinguishable particles. Indeed, the entropy of a system of bosons or fermions is (EX):

$$ S = -\sum_s \left[ n_s \ln n_s \mp (1 \pm n_s) \ln (1 \pm n_s)\right]\,,$$

using the bose and fermi distributions: the first term corresponds to the classical entropy, to which a second term is added, increasing the value of \(S\). The classical limit, valid for bosons and fermions, is recovered when \(n_s \ll 1\), when the temperature is large with respect to any microscopic energy.

Black body and bose degenerate gas

In the limit of low temperature or high density, when the energy per particle is larger than the thermal one, new quantum effects appear in non interacting systems. One of such effects, the black body radiation, is even at the origin of quantum mechanics.

The photon, first described by Einstein (1905) in his work on the photoelectric effect, is a massless boson of spin \(1\), it has hence only two internal degrees of freedom or polarizations (which are perpendicular to its momentum), it is the quanta of the electromagnetic field. The photon energy \(\hbar \omega\) is proportional to the photon momentum \(\boldsymbol p = \hbar \boldsymbol k\), where \(\omega\) and \(\boldsymbol k\) are the frequency (multiplied by \(2\pi\)) and the wavevector, respectively:

$$\omega = c |\boldsymbol k|$$

with \(c = 299\, 792\, 458 \, \mathrm{m\,s^{-1}}\) the speed of light (exact value in SI units). The interaction of the electromagnetic field with matter realizes, in particular, through the emission and absorption of photons by the matter atoms; the balance of these processes, controled exclusively by the temperature, determines the thermodynamical equilibrium: this specific state of matter-radiation equilibrium, is called a “black body”. As a consequence of the exchange of photons with matter, the number of photons in a cavity maintained at a fixed temperature is not conserved, and the equilibrium condition leads to the constraint of zero chemical potential \(\mu = 0\). Under such constraint the canonical and grand canonical ensembles are identical.

Non interacting massive bosons also show an interesting quantum phenomenon when the energy per particle is much larger than the thermal energy: the ground state energy level \(\epsilon_0\), tends to populate at a macroscopic scale. This contrast with the behavior at high temperature where the levels occupation follows the Boltzmann law \(n_s \sim \E^{-\epsilon_s/T}\), and equipartition of the energy over all degrees of freedom dominates. This phenomenon, the Bose-Einstein condensation, was observed in the laboratory for the first time at the end of the twenty century, 70 years after their theoretical prediction.

Black body radiation

The partition function (it is convenient to introduce units such that \(k_{\mathrm{B}}=\hbar=c=1\)) is,

$$\ln Z = 2 \sum_{\boldsymbol k \ne 0} \ln (1 - \E^{\omega_{\boldsymbol k}/T} ) = \frac{2V}{(2\pi)^3} \int_0^\infty 4\pi k^2 \D k\, \ln(1 - \E^{-\omega_{\boldsymbol k}/T} )$$

(the factor \(2\) accounts for the two polarizations) which, after a change of variable from the wavenumber to the frequency \(k\rightarrow x = \omega/T\), and an integration by parts, gives EX,

$$\ln Z = \frac{VT^3}{3\pi^2} \int_0^\infty \D x \frac{x^3}{\E^{x}-1}= \frac{2V}{\pi^2}T^3 \mathrm{Li}_4(1) = \frac{\pi^2V}{45}T^3\,.$$

The free energy is therefore,

$$F = - \frac{\pi^2V}{45}T^4 = -\frac{\pi^2V}{45(\hbar c)^3}T^4\,,$$

from which one obtains the entropy and energy:

$$\frac{S}{V} = \frac{16\sigma}{3c} T^3 \,, \quad \frac{E}{V} = \frac{4\sigma}{c} T^4 \, \quad \sigma = \frac{\pi^2 c k_{\mathrm{B}}^4}{60(\hbar c)^3}\,,$$

where the expression for the energy is the Stefan law and \(\sigma\) the Stefan constant.

To compute the energy distribution of the radiation, we need to count the number of electromagnetic modes in a cavity of volume \(V\); this is naturally given by the wavenumber, \(k/2\pi \sim n \in \mathbb{N}\):

$$\text{modes } \sim V \times 2 \times \frac{\D \boldsymbol k}{(2\pi)^3} \;\rightarrow\; V \times \,2 \times 4\pi k^2 \frac{\D k}{(2\pi)^3}\,,$$

where the factor \(2\) comes from the two possible polarizations, and the last expression make use of the isotropy of the radiation field by counting the number of modes in a spherical shell of width \(\D k\). Note that the number of modes corresponds to the expression we used to replace the summation over the discrete wavenumbers to the integral: \(V \D \boldsymbol p/(2\pi \hbar)^3\).

Using the results of the grand canonical ensemble, we find the the average occupation number of photons of energy \(\omega\) is,

$$n_\omega = \frac{1}{\E^{\omega/T} - 1},$$

which gives the energy distribution (changing \(k\) by \(\omega/T\) in the formula for the number of modes):

$$E(\omega) \D \omega = \frac{VT}{\pi^2} \frac{\omega^3 \D \omega}{\E^{\omega/T} - 1}\,,$$

the Planck radiation law (1900). Note that, after integration, one obtains the Stefan law EX,

$$E = \frac{V}{\pi^2} \int_0^\infty \D \omega\, \frac{\omega^3}{\E^{\omega/T} - 1} = \frac{6V}{\pi^2} T^4 \mathrm{Li}_4(1) = \frac{\pi^2 V}{15} T^4\,,$$

or in Si units:

$$E = \frac{\pi^2 V k_\mathrm{B}}{15 (\hbar c)^3} T^4\,,$$

in accordance with the previous result. (See the note below, on the choice of units.)2

Bose-Einstein condensation

The density \(n = N/V\) of a free boson gas obeys the formula we derived previously using (\ref{e:phi}),

$$n = \left(\frac{T}{2\pi}\right)^{3/2} \mathrm{Li}_{3/2}(\E^{\mu/T}) = \left(\frac{T}{2\pi}\right)^{3/2} \sum_{k=1}^\infty \frac{\E^{\mu/T}}{k^{3/2}}$$

(for spinless particles, \(g_s = 1\)) that fails to converge for chemical potentials above \(\mu = 0\). At the threshold the temperature as a function of the density is given by:

$$T_c = \frac{2\pi}{\zeta(3/2)^{2/3}} n^{2/3} = 3.3125 n^{2/3}\,,$$

lowering further the temperature, at fixed density, would make the chemical potential positive (\(\zeta_{3/2}= \zeta(3/2)\) is the Riemann zeta function). The problem is that for \(T<T_c\), the approximation consisting in the replacement of the sum over the energy levels by an integral is no longer justified, because a fraction of the particles condenses into the ground state (\(\epsilon_{\boldsymbol p} = 0\)), the first term of the sum. Indeed, the number of particle residing in the ground state is,

$$N_0= \braket{a^\dagger_0 a_0} = \frac{1}{\E^{\mu/T}-1} = \frac{z}{1-z} \,,$$

which increases to infinity, or becomes negative, at \(\mu=0\). Therefore, to avoid the divergences we split the sum into a ground energy term and the rest, the excited states that can be approximated by an integral, to obtain the fraction of particles in the condensate EX:

$$\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2} \,, \quad T \le T_c\,.$$

The Bose-Einstein condensation is just the appearance of a macroscopic fraction of bosons occupying the zero energy ground state. One interesting prediction is that there is a sharp value of the temperature (\(=T_c\)) at which the transition between the normal state and the condensate happens; in addition, this critical temperature depends on the particle density.

In experiments, boson atoms are trapped in an external potential created by lasers. In a first approximation this potential is harmonic, and thus the energy levels turn to be of the form,

$$\epsilon_{\boldsymbol n} = \hbar \omega\, (n_x + n_y + n_z + 3/2)\,,\quad \boldsymbol n \in \mathbb{N}^3\,,$$

in three dimensions, where the characteristic frequency \(\omega\) depends on the curvature of the potential minimum. This expression differs notably with the free particle energy, which is quadratic in the momentum. However, the qualitative picture depicted above, remains true; even if important quantitative differences appear. Let us calculate the thermodynamics of a system of \(N\) bosons in a harmonic trap, with energies given by \(\epsilon_{\boldsymbol n}\).

The relation between the chemical potential \(\mu\) and the number of bosons \(N\) is obtained from the summation of the occupation numbers \(n_\epsilon\),

$$N = \sum_{\boldsymbol n} \frac{1}{\E^{-\mu/T}\E^{\omega(n_x + n_y + n_z + 3/2)/T} - 1}$$

(as usual we have \(k_{\mathrm{B}} = \hbar = 1\)). The critical value of the chemical potential is, instead of zero,

$$\mu_c = 3\omega/2\,,$$

for chemical potentials larger than this one, the number of particle would be negative. The number of particles in the ground state \(N_0\), \(\epsilon_{\boldsymbol n}=\epsilon_0\), is then determined by,

$$ N - N_0 = \sum_{\boldsymbol n \ne 0} \frac{1}{\E^{\omega(n_x + n_y + n_z)/T} - 1}\,,$$

which is the number of particles in the excited states. This sum, being convergent, can safely be approximated by an integral in the limit of large \(N\rightarrow\infty\):

$$N-N_0 = \int_0^\infty \frac{\D \boldsymbol n}{\E^{\omega(n_x + n_y + n_z)/T} - 1}\,.$$

The integral can be done transforming the integrand into a power series EX; however, it is more useful to calculate the density of states \(\nu(\epsilon)\) in order to change the \(\boldsymbol n\) variable by the energy \(\epsilon_{\boldsymbol n}\). In the continuous limit, which is our present approximation, the number of energy levels smaller than \(n = \epsilon/\omega\) is the volume \(\Omega(n)\) defined by:

$$\Omega(n) = \{n_x \ge 0,\; n_y \ge 0,\; n_z \ge 0,\;\text{with }\, n_x + n_y + n_z \le n\}\,,$$

that is the volume of a pyramid of base a triangle of area,

$$\frac{1}{2}\sqrt{2} \frac{\sqrt{2}}{2} n^2$$

and height \(n\), giving a volume \(V(n) = \text{area}\times \text{height}/3\),

the sixth of the cube of edge \(n\); using that

$$d\Omega(n) = \frac{n^2}{2} \D n = \frac{\epsilon^2}{2\omega^2} \frac{\D \epsilon}{\omega} = \nu(\epsilon) \D \epsilon\,,$$

we find the density of states of a three dimensional harmonic oscillator,

$$\nu(\epsilon) = \frac{\epsilon^2}{2\omega^3}$$

up to subdominant terms \(\mathcal{O(\epsilon/\omega)}\) (our approximation is consistent with the limit \(\epsilon/\omega \gg 1\)). Therefore, the integral over \(\boldsymbol n\) becomes,

$$N-N_0 = \int_0^\infty \D \epsilon\, \nu(\epsilon)n_\epsilon = \int_0^\infty \frac{\D \epsilon}{2\omega^3} \frac{\epsilon^2}{\E^{\epsilon/T} - 1} = \frac{\zeta(3)}{\omega^3} T^3\,.$$

The critical temperature corresponds to the threshold below which the ground state is populated with a macroscopic number of particles. It is then the temperature at which \(N_0=0\), or,

$$T_c = \frac{ \hbar \omega}{k_\mathrm{B}} \left( \frac{N}{\zeta(3)} \right)^{1/3} \,,$$

similar to the free case with, instead of the power \(2/3\), we obtain a \(1/3\) exponent. The main qualitative difference is that now the critical temperature depens on the number of trapped particles and not, as before, on the particle density.

An analogous computation gives the energy for \(T \le T_c\) (\(\mu = 3\omega/2\)):

$$E = \int_0^\infty \D \epsilon\, \nu(\epsilon)n_\epsilon \epsilon = \int_0^\infty \frac{\D \epsilon}{2\omega^3} \frac{\epsilon^3}{\E^{\epsilon/T} - 1} = \frac{3\zeta(4)}{\omega^3} T^4$$

which obviously determined by exclusively by the excited states \(\epsilon > 0\). Using these two last equations, we can write the energy of the condensate as,

$$\frac{E}{NT_c} = \frac{3\zeta(4)}{\zeta(3)} \left( \frac{T}{T_c}\right)^4\,.$$

Note that for free bosons the exponent of the temperature is \(5/2\), rather than \(4\) for the bosons in a trap. It is also worth mentioning that the harmonic trap makes the bosons behave much like a relativistic degenerate gas, for which the energy is linear in the momentum: \(\epsilon_{\boldsymbol p} \sim |\boldsymbol p|\) (compare with the black body expression of the thermodynamic energy, also proportional to the power four of the temperature \(E \sim T^4\)).

Degenerate fermions \(z \rightarrow \infty\)

The occupation number of a fermi gas at energy \(\epsilon\)

$$n_\epsilon = \frac{1}{\E^{(\epsilon - \mu)/T} + 1}$$

and at low temperature \(T \ll \mu\), tends to zero if \(\epsilon > \mu\) and to one if \(\epsilon < 0\):

$$ n_\epsilon = \theta(\epsilon_F - \epsilon),\quad \epsilon_F = \left.\mu\right|_{T=0}$$

at \(T=0\), where we defined the fermi energy \(\epsilon_F\) as the chemical potential separating the low energy region of occupied levels, to the high energy region of unoccupied levels. The existence of this energy is a result of the exclusion principle: adding fermions to a system a zero temperature, must be done by occupying progressively higher energy unoccupied states. This is in contrast to the bosons, for which increasing the population of the fundamental level do not contribute to the energy.

Using the free particle \(\epsilon = p^2/2m\) density of states (take \(m=1\)) EX,

$$\nu(\epsilon) = \frac{V\sqrt{2}}{\pi^2} \epsilon^{1/2} \,,$$

we can calculate the number of particles below the fermi level \(\epsilon_F\)

$$N = \int_0^\infty \D \epsilon\, \nu(\epsilon) n_\epsilon = \frac{V\sqrt{2}}{\pi^2} \int_0^\infty \D \epsilon \, \epsilon^{1/2} \theta(\epsilon_F - \epsilon) = \frac{2}{3} V \frac{\sqrt{2}}{\pi^2} \epsilon_F^{3/2} \,,\quad (T=0)$$

at zero temperature. This formula relates the gas density to the fermi energy:

$$\epsilon_F = \frac{(3\pi^2)^{2/3}}{2} n^{2/3} \; [\hbar^2/m] = T_F \, [k_\mathrm{B}]\,,$$

where we indicate in square brackets the dimensional factor, and defined the fermi temperature \(T_F\) (in kelvin). It is also useful to write the density of state in the form,

$$\nu(\epsilon) = \frac{3N}{2} \frac{\epsilon^{1/2}}{\epsilon_F^{3/2}}\,,$$

where we replaced the volume by the number of particles:

$$V = \frac{3\pi^2}{2^{3/2}} \frac{N}{\epsilon_F^{3/2}} \, [\hbar^2/m]^{3/2}\,.$$

In a similar way we calculate the energy,

$$E = \int_0^\infty \D \epsilon\, \nu(\epsilon) n_\epsilon \epsilon = \frac{V\sqrt{2}}{\pi^2} \int_0^{\epsilon_F} \D \epsilon \, \epsilon^{3/2} = \frac{2}{5} V \frac{\sqrt{2}}{\pi^2} \epsilon_F^{5/2}\,, \quad (T=0)$$

or, using the expression of \(N\),

$$\frac{E}{N} = \frac{3}{5} \epsilon_F \,.$$

These expressions are valid at \(T=0\), the behavior of the degenerated fermi gas at low temperature can be obtained using the asymptotic expansion of the polylogarithm function. In fact, keeping two terms of this expansion,1

$$-\mathrm{Li}_s(-z) = \frac{(\ln z)^s}{\Gamma(s+1)} + \frac{\pi^2}{6} \frac{(\ln z)^{s-2}}{\Gamma(s-1)} \,,$$

is enough to estimate the corrections to the zero temperature calculations. The exact formula for the energy is,

$$E = \frac{3}{2}PV = -\frac{3}{2}\Phi = \frac{3}{2} \frac{N}{\epsilon_F^{3/2}} T^{5/2} \int_0^\infty \D x \frac{x^{3/2}}{z^{-1}\E^x + 1}\,,$$

where we substituted the volume \(V\) by its expression in terms of the fermi energy,

$$V = \frac{3}{2} \frac{\pi^2}{\sqrt{2}} \frac{N}{\epsilon_F^{3/2}}\,,$$

and the integral is \(-\Gamma(5/2) \mathrm{Li}_{5/2}(-z)\). Therefore,

$$E = \frac{3}{2} \frac{N}{\epsilon_F^{3/2}} \left( \frac{2}{5} \mu^{5/2} + \frac{\pi^2}{4} T^2 \mu^{1/2} \right)\,.$$

The grand potential in the gran canonical natural variables is then,

$$\Phi(T,\mu,V) = -\frac{2\sqrt{2}}{3\pi^2} V \left( \frac{2}{5} \mu^{5/2} + \frac{\pi^2}{4} T^2 \mu^{1/2} \right)\,.$$

Now, the equation of \(N\) at finite temperature gives an implicit formula for the chemical potential:

$$N= \frac{3}{2} \frac{N}{\epsilon_F^{3/2}} T^{3/2} \int_0^\infty \D x \frac{x^{1/2}}{z^{-1}\E^x + 1} = -\frac{3}{2} \frac{N}{\epsilon_F^{3/2}} T^{3/2} \mathrm{Li}_{3/2}(-z)\,,$$

or, substituting the asymptotic expansion,

$$1 = \frac{\mu^{3/2}}{\epsilon_F^{3/2}} \left[ 1 + \frac{\pi^2}{8} \left( \frac{T}{\mu}\right)^2\right]\,,$$

which is readily inverted to obtain,

$$\mu = \epsilon_F \left[ 1 - \frac{\pi^2}{12} \left( \frac{T}{\epsilon_F}\right)^2\right]\,,$$

in agreement with the definition \(\mu = \epsilon_F\) at \(T=0\), plus a quadratic correction in the temperature. Note that to obtain this relation \(N=N(\mu\)) we also could use more straightforwardly \(N=-\partial \Phi/\partial \mu\).

Putting the value of \(\mu\) in the formula for \(E\), we finally get an explicit expression of the energy in the range \(T \ll \epsilon_F\),

$$E = \frac{3}{5} N \epsilon_F \left[ 1 + \frac{5\pi^2}{12} \left( \frac{T}{\epsilon_F}\right)^2\right]\,.$$
$$\frac{C_V}{N} = \frac{\pi^2}{2} \frac{T}{\epsilon_F}\,,$$

linear in \(T\), and similarly for the entropy,

$$S = -\frac{\partial \Phi}{\partial T} = N \frac{\pi^2}{2} \frac{T}{\epsilon_F}\,,$$

This is an important result, showing that the specific heat and the entropy at low temperatures of a fermi gas, tend to zero in accordance with the Nernst principle.

  1. Polylogarithmic function

    The polylogarithmic function is defined be the series,

    $$\mathrm{Li}_\nu(z) = \sum_{n=1}^\infty \frac{z^n}{n^\nu}\,,$$
    which can be viewed as a generalization of the Riemann zeta function,
    $$\zeta(\nu) = \mathrm{Li}_\nu(1) = \sum_{n=1}^\infty \frac{1}{n^\nu}\,.$$
    Analytic continuation gives the integral formula,

    $$\mathrm{Li}_{\nu}(z) = \frac{1}{\Gamma(\nu)} \int_0^\infty \D x \frac{x^{\nu-1}}{z^{-1}\E^x - 1}\,.$$

    The asymptotic expansion for large \(z\) of the polylogarithm is,

    $$\mathrm{Li}_s(-z) = -2 \sum_{k=0}^\infty \eta(2k) \frac{(\ln z)^{s-2k}}{\Gamma(s-2k+1)}\,$$

    where the Dirichlet function \(\eta\) is related to the Riemann zeta:

    $$\eta(s) = (1-2^{1-s}) \zeta(s)\,,\quad \eta(1) = \ln 2$$

    $$\zeta(2k) = -\frac{1}{2} (2\pi)^{2k} \frac{(-1)^{k} B_{2k}}{(2k)!}, \quad \zeta(2) = \frac{\pi^2}{6},\; \zeta(4) = \frac{\pi^4}{90}$$

    where \(B_n\) are the Bernoulli numbers. This asymptotic expansion can be obtained using the Sommerfeld method EX.

    The two first terms of the asymptotic expansion,

    $$-\mathrm{Li}_s(-z) = \frac{(\ln z)^s}{\Gamma(s+1)} + \frac{\pi^2}{6} \frac{(\ln z)^{s-2}}{\Gamma(s-1)} \,,$$

    The derivative of the polylogarithmic function of order \(\nu\) depends on the function of order \(\nu-1\):

    $$z \frac{\partial}{\partial z} \mathrm{Li}_\nu(z) = \mathrm{Li}_{\nu-1}(z)\,.$$

    Explicit values of the Gamma function,

    $$n! = \Gamma(n+1)\,, \quad \Gamma(n+ 1/2) = \frac{(2n)!}{4^nn!}\sqrt{\pi}$$

  2. Note on the choice of units

    With the convention \(k_\mathrm{B} = \hbar = c = 1\), one may apply the conversion factors:

    $$1\,[\mathrm{m}] = \frac{1}{c}\,[\mathrm{s}], \quad 1 [\mathrm{m^{-1}}] = \frac{\hbar}{c}\,[\mathrm{kg}] = \hbar c\,[\mathrm{J}] = \frac{\hbar c}{k_{\mathrm{B}}} \, [\mathrm{K}]$$

    using the SI numerical values of the physical constants. For instance, to recover the Stefan constant from its value, one introduces the dimensional units \(\mathrm{J/s\, m^2\, K^4}\) using the above factors:

    $$\frac{k_\mathrm{B}^4}{c^2\hbar^3} = 3.44716123286048\,10^{-7} \frac{\mathrm{J}}{\mathrm{s\, m^2\, K^4}}$$

    from which one obtains

    $$\sigma = \frac{\pi^2}{60} \frac{k_\mathrm{B}^4}{c^2\hbar^3} = 5.67035294585\,10^{-8} \, \mathrm{J/s\, m^2\, K^4}$$


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theanyspacewhatever at the Guggenheim Museum New York

With “theanyspacewhatever”, the Guggenheim Museum New York presents a group exhibition of individual installations for the Rotunda of Frank Lloyd Wright’s landmark building. The show was organized by the Guggenheim Museum’s Chief Curator, Nancy Spector, in close collaboration with the artists. The exhibition brings together ten artists, whose work reaches beyond the visual arts. Theanyspacewhatever features Angela Bulloch, Maurizio Cattelan, Liam Gillick, Dominique Gonzalez-Foerster, Douglas Gordon, Carsten Höller, Pierre Huyghe, Jorge Pardo, Philippe Parreno, and Rirkrit Tiravanija.
Among the works are Carsten Höller’s “Revolving Hotel Room” (see also VTV’s video “Carsten Höller: Carrousel / Kunsthaus Bregenz, Austria“), Angela Bulloch’s LED-installation “Firmamental Night Sky: Oculus 12”, and Liam Gillick’s Theanyspacewhatever Signage System. Philippe Parreno has installed a site-specific, illuminated marquee on the facade of the Guggenheim, Douglas Gordon is exhibiting a compilation of text pieces, and Liam Gillick intervenes in the Guggenheim’s operational systems with hanging aluminum signs.
Exhibition walk-through at the occasion of the press preview on October 23, 2008.

> Right-click (Mac: ctrl-click) this link to download Quicktime video file.

Posted in: art, New York, no comment, VernissageTV


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