The Naked Scientists

The Naked Scientists Forum

Author Topic: How significant is this?  (Read 7031 times)

Offline cheryl j

  • Neilep Level Member
  • ******
  • Posts: 1460
  • Thanked: 1 times
    • View Profile
How significant is this?
« on: 07/08/2014 02:13:04 »
I was just wondering if someone with a better understanding of physics could tell me how significant this finding is - whether they are "chipping away" at the uncertainty principle or not.

http://www.scientificamerican.com/article/particle-measurement-sidesteps-the-uncertainty-principle/

The article seems to hedge a bit.


 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #1 on: 07/08/2014 03:41:20 »
I was just wondering if someone with a better understanding of physics could tell me how significant this finding is - whether they are "chipping away" at the uncertainty principle or not.

http://www.scientificamerican.com/article/particle-measurement-sidesteps-the-uncertainty-principle/

The article seems to hedge a bit.
I've heard about articles like this for a while now. They're using the wrong definition of uncertainty. It's inconsistent with orthodox quantum mechanics. Therefore they ended up proving something else.

If you're sharp and read carefully and you're a good student then you'll notice a difference between how the term uncertainty is defined in undergraduate texts such as texts entitled Modern Physics and texts which are entitled Quantum Mechanics. The two classes of textbooks have a tendency to teach quantum mechanics a bit differently. They define uncertainty differently and when that's done there's no reason to expect to get the same results and meaning as when it's defined otherwise. I think that's what's going on. Here is what the Stanford Encyclopedia of Philosophy has to say about this
http://plato.stanford.edu/entries/qt-uncertainty/
Quote
The notion of ‘uncertainty’ occurs in several different meanings in the physical literature. It may refer to a lack of knowledge of a quantity by an observer, or to the experimental inaccuracy with which a quantity is measured, or to some ambiguity in the definition of a quantity, or to a statistical spread in an ensemble of similary prepared systems. Also, several different names are used for such uncertainties: inaccuracy, spread, imprecision, indefiniteness, indeterminateness, indeterminacy, latitude, etc. As we shall see, even Heisenberg and Bohr did not decide on a single terminology for quantum mechanical uncertainties. Forestalling a discussion about which name is the most appropriate one in quantum mechanics, we use the name ‘uncertainty principle’ simply because it is the most common one in the literature.
What they're talking about is something like this - In modern physics textbooks, basic physics textbooks and books written for the layman, authors tend to think of and define uncertainty as being the same thing as the accuracy in a measurement while in quantum mechanics texts it means the same thing as standard deviation. The former depends on the instruments and techniques used to make measurements while the later is calculated strictly from the wave function itself. Here are a few resources which talk about the different meanings of uncertainty.
http://statintquant.net/siq/siqse2.html
http://plato.stanford.edu/entries/qt-uncertainty/

I hope that helps. I doubt the author of that article is aware of this but you never know. I'll write to him and find out.
« Last Edit: 07/08/2014 04:22:05 by PmbPhy »
 

Offline chiralSPO

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 1872
  • Thanked: 143 times
    • View Profile
Re: How significant is this?
« Reply #2 on: 07/08/2014 03:54:59 »
The article closes by saying "physicists seem to have found a way to get more data with less measurement..." I'm not sure I agree. It would seem to me that they are getting the same amount of information, but now related to more parameters.

Because the uncertainty in momentum and uncertainty in position are inversely proportional, it makes sense that the less precisely the value of one parameter is measured, the more precision can be expected for the measurement of the other. The researchers found a cool trick to actually allow them to take advantage of this proportionality. So I don't think they are able to get more information this way, just more useful information. I am certain that if they were to make the mirrors substantially smaller, or use any other technique to refine the position measurement, there would be a corresponding increase in the distribution of measured momenta.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #3 on: 07/08/2014 04:31:51 »
Quote from: chiralSPO
The article closes by saying "physicists seem to have found a way to get more data with less measurement..." I'm not sure I agree. It would seem to me that they are getting the same amount of information, but now related to more parameters.

Because the uncertainty in momentum and uncertainty in position are inversely proportional, it makes sense that the less precisely the value of one parameter is measured, the more precision can be expected for the measurement of the other.
Nope. As I said, the definition that they're using for uncertainty is wrong. Uncertainty is the standard deviation of an ensemble of measurements. Didn't you know that? See
http://www.aip.org/history/heisenberg/p08a.htm

The value of the uncertainty of both position and momentum is determine from the wave function. Experiment can only verify what theory predicts. The only way to change the value of the uncertainty is to start with a new wave function.
Quote
The uncertainty relations involve the uncertainties in the measurements of these variables. The "uncertainty" -- sometimes called the "imprecision"--is related to the range of the results of repeated measurements taken for a given variable. For example, suppose you measure the length of a book with a meter stick. It turns out to be 23.6 cm, or 23 centimeters and 6 millimeters. But since the meter stick measures only to a maximum precision of 1 mm, another measurement of the book might yield 23.7cm or 23.5 cm. In fact, if you perform the measurement many times, you will get a "bell curve" of measurements centered on an average value, say 23.6 cm. The spread of the bell curve, or the "standard deviation," will be about 1 mm on each side of the average. This means that the "uncertainty" or the precision of the measurement is plus or minus 1 mm.
...
967878d1da852d4b07a961e3168b0fff.gifq is the uncertainty or imprecision (standard deviation) of the position measurement.

967878d1da852d4b07a961e3168b0fff.gifp  is the uncertainty of the momentum measurement in the q direction at the same time as the q measurement.
Do you see now that uncertainty is a standard deviation and has nothing to do with imprecision in measurements?
 

Offline alancalverd

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 4699
  • Thanked: 153 times
  • life is too short to drink instant coffee
    • View Profile
Re: How significant is this?
« Reply #4 on: 07/08/2014 06:53:07 »
A lot of confusion arises from the use of "uncertainty" (the result of human activity) when discussing
"indeterminacy" (Heisenberg's inherent property of nature). The latter term gives a much better picture of what's going on in quantum mechanics.   
 

Offline lightarrow

  • Neilep Level Member
  • ******
  • Posts: 4586
  • Thanked: 7 times
    • View Profile
Re: How significant is this?
« Reply #5 on: 07/08/2014 10:35:09 »
Nope. As I said, the definition that they're using for uncertainty is wrong.
...
Pete, can you help me finding the original article or something more detailed? That Scientific American page doesn't say much...

--
lightarrow
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #6 on: 07/08/2014 10:39:12 »
Nope. As I said, the definition that they're using for uncertainty is wrong.
...
Pete, can you help me finding the original article or something more detailed? That Scientific American page doesn't say much...

--
lightarrow
I think this is it - http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.112.253602
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
Re: How significant is this?
« Reply #7 on: 07/08/2014 14:25:09 »
The article closes by saying "physicists seem to have found a way to get more data with less measurement..." I'm not sure I agree. It would seem to me that they are getting the same amount of information, but now related to more parameters.

This is an example of hype getting ahead of the science.  I believe there are two things going on here.

The first is weak measurement, which often gets hyped as breaking the uncertainty principle.  It doesn't.  What happens is that one measures an observable, let's say position, of a particle very poorly.  The uncertainty principle says that a sufficiently poor position measurement won't effect the particle's momentum much.  No violation of the uncertainty principle.  Then take many identically created particles.  Repeat this poor measurement many times.  Each measurement will differ slightly and be equally poor on its own, but taken together, they tell you about the position of these particles.  Since the uncertainty principle only applies to a single measurement, not to averaging over many, it isn't violated. 

Compressive sensing, which is the other thing they use, is a way of taking many less measurements than you'd traditionally expect by employing what are called priors.  Let's say (as a silly example) you want to describe the contents of a box full of (6-sided) dice.  Your measurement is to roll them all and sum the result.  I do so and tell you that you get a value of "4".  From that single measurement, you know that you have many options: 1+1+1+1, 1+1+2, 2+2, 3+1, 4 (and that's not counting ordering of dice).  A prior would be if I told you that your answer uses the smallest possible number of dice.  Then you would know for certain that you only have 1 die and it's rolled a 4.  This is a silly example, but this type idea is extremely powerful in many real-world applications where we know the types of things we're measuring.  It's been very powerful in MRI/CT imaging where you take a bunch of measurements that look nothing like the final image and process them to get a 3D image of the body.  You need lots of measurements because traditional MRI/CT do not try to use the fact that we know the general structure of the body.  For all they care, you're measuring some random object each time and you need tons of data to figure out its internal structure.  If you instead you use some of what we know about anatomy, you can cut way down on the number of needed measurements. 

So they're basically using what they know about the possible class of solutions to cut way back on the data required compared to the case where they know nothing about the class of solutions.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #8 on: 07/08/2014 15:21:00 »
Quote from: JP
This is an example of hype getting ahead of the science.  I believe there are two things going on here.
The first is weak measurement, which often gets hyped as breaking the uncertainty principle.  It doesn't.  What happens is that one measures an observable, let's say position, of a particle very poorly.  The uncertainty principle says that a sufficiently poor position measurement won't effect the particle's momentum much.
I believe that’s a misunderstanding of the uncertainty principle. Uncertainty has nothing to do with the quality.accuracy of a measurement. You can determine the uncertainty without making any measurements whatsoever and those will be the uncertainty for that observable for that system for that wave function. The uncertainty being the standard deviation of the observable is determined entirely by the wave function. Given the same wave function one calculates both <x> and <x^2> and uses those to determine the standard deviation dx as being dx = sqrt(<x^2> - <x>^2). Likewise you can use the wave function to get dp = sqrt(<p^2> - <p>^2). Note that nowhere has a measurement been made. It can be shown that dp dx >= hbar/2.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #9 on: 07/08/2014 15:43:54 »
I was just wondering if someone with a better understanding of physics could tell me how significant this finding is - whether they are "chipping away" at the uncertainty principle or not.

http://www.scientificamerican.com/article/particle-measurement-sidesteps-the-uncertainty-principle/

The article seems to hedge a bit.
Cheryl - Did you read where the scientist who did the experiment wrote
Quote
"We do not violate the uncertainty principle,” Howland says. “We just use it in a clever way.”
So no. They didn't chip it away. Whew! Safe for another day. There's something I don't like about it though. The wrote
Quote
The filter provided a way of measuring a particle’s position without knowing exactly where it was—without collapsing its wavefunction.
In quantum mechanics I don't think that calling the inference of where the photon is can be said to be a measurement of where it is.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
Re: How significant is this?
« Reply #10 on: 07/08/2014 15:59:40 »
Quote from: JP
This is an example of hype getting ahead of the science.  I believe there are two things going on here.
The first is weak measurement, which often gets hyped as breaking the uncertainty principle.  It doesn't.  What happens is that one measures an observable, let's say position, of a particle very poorly.  The uncertainty principle says that a sufficiently poor position measurement won't effect the particle's momentum much.
I believe that’s a misunderstanding of the uncertainty principle. Uncertainty has nothing to do with the quality.accuracy of a measurement. You can determine the uncertainty without making any measurements whatsoever and those will be the uncertainty for that observable for that system for that wave function. The uncertainty being the standard deviation of the observable is determined entirely by the wave function. Given the same wave function one calculates both <x> and <x^2> and uses those to determine the standard deviation dx as being dx = sqrt(<x^2> - <x>^2). Likewise you can use the wave function to get dp = sqrt(<p^2> - <p>^2). Note that nowhere has a measurement been made. It can be shown that dp dx >= hbar/2.

I think we're saying the same thing in different ways.  A quantum "particle" always has wavelike properties.  The uncertainty principle is a statement about waves.  A wave can't be arbitrarily narrow in two Fourier conjugate variables.  The limit is precisely the uncertainty principle limit.  This is true of all waves: light, sound, water waves.  It is also true of the wave nature of quantum particles. 

People seem to imbue the quantum version with some magical properties simply because it has the word "quantum" in front of it, but it's a very well understood property of all waves.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
Re: How significant is this?
« Reply #11 on: 07/08/2014 16:14:18 »
Quote
The filter provided a way of measuring a particle’s position without knowing exactly where it was—without collapsing its wavefunction.
In quantum mechanics I don't think that calling the inference of where the photon is can be said to be a measurement of where it is.

Ok--looking at the PRL paper, this is applying something that's well known in classical wave theory to the fact that photons are also waves.  What's important to remember is that a "measurement" just means that you interact with the wavefunction to change it somehow and extract information from that.  For example, if you put a single pinhole in the photon's path and a camera behind the pinhole, you can tell if the photon went through that pinhole or not by looking for a "click" at the detector.  If instead you put a random array of pinholes (which is basically what they did), you now only know that the photon went through those open pinholes and didn't strike the mask.  In the wave formalism of quantum mechanics, both force the wavefunction to be zero everywhere on the obstructing screen and allow it to remain the same in the open pinholes.  The "weak" part of the measurement is that the single pinhole squeezes the wavefunction down to a tiny point, forcing it to spread in momentum whereas using a mask of tons of pinholes does not squeeze it nearly so much, so momentum doesn't change much.

The lens after the pinhole mask and the screen is just a way of ensuring that particle momentum is measured at each detector pixel.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #12 on: 08/08/2014 03:26:18 »
But if you have two pin holes you can't tell which one it went through and when it went through either of them. If you can't tell when something was at a place then you can't really have said to have measured its position.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
Re: How significant is this?
« Reply #13 on: 08/08/2014 13:48:01 »
But if you have two pin holes you can't tell which one it went through and when it went through either of them. If you can't tell when something was at a place then you can't really have said to have measured its position.

Sure you can, but you haven't localized it with the same precision as if you used a single pinhole.  Even if you don't like calling that a measurement, and I can understand why it might not be the best term for it, it's the standard terminology in the field.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #14 on: 08/08/2014 14:47:52 »
Quote from: JP
Sure you can, but you haven't localized it with the same precision as if you used a single pinhole.  Even if you don't like calling that a measurement, and I can understand why it might not be the best term for it, it's the standard terminology in the field.
A car leaves Boston MA at 8:00 am travels to and arrives in Haverhill MA 45 minute later. You're telling me that given this information you can state that you've measured the cars position? If so then I'd have to strongly disagree. Especially for a quantum particle which can't even be said to have a position until it's been measured. For all you know it tunneled through the entire space and skipped the region between Boston and Haverhill altogether.
 

Offline chiralSPO

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 1872
  • Thanked: 143 times
    • View Profile
Re: How significant is this?
« Reply #15 on: 08/08/2014 17:17:49 »
"measurement" and "observation" are just as poorly defined as "uncertainty."

I would consider any use of an operator on the wavefunction as an "observation." As long as the two (or more) operators don't commute, there is going to be Uncertainty depending on which observation is made first, and the accuracy of that measurement.

From the point of view of an experimenter, there are many different sources of uncertainty, ranging from instrumentation to variability in experimental setup, but their measurements are ultimately limited by the Uncertainty Principle. Other particles can be the "observer" too (I hate it when people confuse epistomology with quantum mechanics--Schrödinger's Cat is not about what people can and can't know--it's about superposition, and as far as I'm concerned, whatever "detects" the decay of the atom and releases the poison is what does the "observation" thereby collapsing the superposition, but I digress)

Like JP, I think that thinking about this things as waves simplifies the intuition.

Standard deviation is kind of useless if we are talking about a single measurement (this is why they measure large numbers of particles), but that doesn't mean that the phenomena they describe are necessarily only observed in populations, it still applies on a particle-by-particle basis.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
Re: How significant is this?
« Reply #16 on: 08/08/2014 17:51:33 »
Quote from: JP
Sure you can, but you haven't localized it with the same precision as if you used a single pinhole.  Even if you don't like calling that a measurement, and I can understand why it might not be the best term for it, it's the standard terminology in the field.
A car leaves Boston MA at 8:00 am travels to and arrives in Haverhill MA 45 minute later. You're telling me that given this information you can state that you've measured the cars position? If so then I'd have to strongly disagree. Especially for a quantum particle which can't even be said to have a position until it's been measured. For all you know it tunneled through the entire space and skipped the region between Boston and Haverhill altogether.

Of course I'd say I'd measured it.  But the precision of my measurement would be low.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #17 on: 08/08/2014 17:53:56 »
Quote from: chiralSPO
"measurement" and "observation" are just as poorly defined as "uncertainty."
Uncertainty is a very precisely defined quantity. It's nothing more and nothing less the standard deviation of an observable. An observable is the eigenvalue of an operator corresponding to a physical quantity. E.g. to make a measurement all you do is record what it is that you're interested in. For example; take out your digital camera and take a picture. For a moment light will enter the iris of the aperture an hit the Charged Coupled Device (CCD). When light hits the screen its basically photons hitting the pixels on the CCD. When that happens the device allows a record of which pixel was struck and what energy of the photon/light struck it. That's how energy and position is measured in this case. Position and energy commute so that you can measure them simultaneously and exactly.

When you run an experiment, say, 100,000 times starting out in the same state and measuring the same observable and record the measurements then you can calculate the standard deviation which is what uncertainty is.

Quote from: chiralSPO
I would consider any use of an operator on the wavefunction as an "observation." As long as the two (or more) operators don't commute, there is going to be Uncertainty depending on which observation is made first, and the accuracy of that measurement.
The uncertainty has nothing to do with accuracy. That's a common misconception.

Quote from: chiralSPO
From the point of view of an experimenter, there are many different sources of uncertainty, ..
Nope. That's not true. The only thing that the uncertainty is a function of is the wave function. Please read the Wikipedia page on uncertainty. See http://en.wikipedia.org/wiki/Uncertainty_principle

Uncertainty is defined as (Note: I don't know how to use Latex enough to put the hat over the variables so just keep in mind that they're there)

6a3915937903b01569f826c9bfa8abcb.gif

186db476ba51c37d19f3e1391cd6b6e8.gif

(sorry about the Latex. I did it exactly as http://oeis.org/wiki/List_of_LaTeX_mathematical_symbols says and it still came out wrong)

Look at the first equation in the Wikipedia page for uncertainty. The uncertainty relationship is

d9aa95b837cb0cbfe252001dc47b21f4.gif

Quote from: chiralSPO
--Schrödinger's Cat is not about what people can and can't know--it's about superposition, and as far as I'm concerned, whatever "detects" the decay of the atom and releases the poison is what does the "observation" thereby collapsing the superposition,...
As Griffiths explains about the cat in the box experiment in his QM text
Quote
Schrodinger regarded this as patent nonsense, and I think that most physicists would agree with him. There is something absurd about the very idea of a macroscopic object being a linear combination of two palpably different states. An electron can be in a linear combination of two palpably different states, but a cat cannot be in a linear combination of alive and dead.
You can read more about this in
http://bookzz.org/book/2031469/ca8981

Quote from: chiralSPO
Standard deviation is kind of useless if we are talking about a single measurement ...
That's why both uncertainty and probability has nothing to do with single measurements. It only has meaning for an ensemble of measurements or systems.

re - this is why they measure large numbers of particles - Exactly! :)

Quote from: chiralSPO
but that doesn't mean that the phenomena they describe are necessarily only observed in populations, it still applies on a particle-by-particle basis.
Not uncertainty. All quantum mechanics can do is tell what can happen statistically.
« Last Edit: 08/08/2014 18:03:30 by PmbPhy »
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
Re: How significant is this?
« Reply #18 on: 09/08/2014 16:48:15 »
The uncertainty principle (and uncertainty) can be useful outside the context of physically taking many measurements.  For example, if I know a priori what my wavefunction is, I can characterize the standard deviation of the set of possible measurements.  This is the wave interpretation of uncertainty relationships--a single particle's wavefunction will satisfy an uncertainty relationship, even if defining the standard deviation of a measurement of a single particle is nonsensical.  In this sense, it is reasonable to talk about the uncertainty in an observable of a single particle insofar as this describes the possible outcomes of experiments.

As for what describes a measurement, there's no strict line between a measurement and an interaction.  Imagine putting a circular hole in an opaque screen and firing photons at it.  If we have a detector after the screen, and we get a "click" on the detector, we know a photon went through the hole and we can say we've "measured" its position.  But there's no sharp line that says "if the hole is below X size, this is a position measurement, otherwise it isn't."  We have just as much right to claim that a 500 nanometer hole made a measurement as we do that a 1 meter hole made a measurement.  We just have a very poor precision in our 1 meter measurement (which we should disclose) if we want to get the exact position.  You may say that we could come up with a thought experiment where the pinhole becomes infinitesimally small and we do have perfect measurement in position, but once you do that, you have to account for quantum effects in the pinhole itself and so you still don't get perfect localization of the particle.
 

Online jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3914
  • Thanked: 53 times
  • The graviton sucks
    • View Profile
Re: How significant is this?
« Reply #19 on: 09/08/2014 18:08:58 »
Pete

You stated earlier in a previous post "The value of the uncertainty of both position and momentum is determine from the wave function. Experiment can only verify what theory predicts. The only way to change the value of the uncertainty is to start with a new wave function." Again you have brought up a very important point which needs to be highlighted. When you say "The only way to change the value of the uncertainty is to start with a new wave function" is so important to grasp with respect to uncertainty that it can't be overstated.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2760
  • Thanked: 38 times
    • View Profile
Re: How significant is this?
« Reply #20 on: 10/08/2014 13:05:50 »
Quote from: JP
For example, if I know a priori what my wavefunction is, I can characterize the standard deviation of the set of possible measurements.
I'm sorry JP but I don't understand this. What do you mean by characterize the standard deviation?

Quote from: JP
In this sense, it is reasonable to talk about the uncertainty in an observable of a single particle insofar as this describes the possible outcomes of experiments.
I don't see that at all. E.g. I measure the position of a particle in a box of width 4 units to be 3.2. What does that tell me about uncertainty?

Quote from: JP
As for what describes a measurement, there's no strict line between a measurement and an interaction.
I'd say that they're the same thing, wouldn't you?

Quote from: JP
We have just as much right to claim that a 500 nanometer hole made a measurement as we do that a 1 meter hole made a measurement.
I disagree. The hole doesn't make the measurement. The photon detector that goes
"click" made the measurement. The larger the surface area of the detector the less the precision in the measurement. That still tells you nothing about the uncertainty. When you consider an array of such detectors and collect data on a large number of "click"s and how they're spread out then you can determine the uncertainty from the spread through the standard deviation.
 

The Naked Scientists Forum

Re: How significant is this?
« Reply #20 on: 10/08/2014 13:05:50 »

 

SMF 2.0.10 | SMF © 2015, Simple Machines
SMFAds for Free Forums