As someone who once spent more than 10 years sick, I can tell you that most of what you know about healing is wrong. Now, who ends up doing the healing?  Is it the surgeon who removes the gallbladder that's gone bad? Or who does some kind of intestinal surgery because of some inflammation or some infection you have? Or is it the dietitian that gives you a little change in your diet and then the healing comes? Who actually does the healing? In this video, I thought I would share a very personal look at a bit of my own healing journey and share five of the lessons I learned being someone who was formerly sick for about 10 years. Hi guys, I'm Dr. Alex Hain, board licensed acupuncturist and doctor of acupuncture and traditional Chinese medicine and author of The Health Book, Master the Day. So let's jump in. Healing lesson number one, your illness is a gift. If you see it that way.   It was painful and I was losing money and time that I could have for my whole life going to illness and healing. And he said, listen, every little thing that you go through, if you view it through the lens of that silver lining, there is something to learn. And that thing that you have to learn maybe there's a way that you think about life. Maybe you feel like there's not enough time or you have to be driven because being driven means you're gonna be successful.  And so this root of I don't feel good enough pushes you to do something else that then sows the seeds of illness. Or maybe you realize you're really unhappy at your job and that's been the root of this sort of emotional eating going out and getting drinks after work every night that put on 30 pounds and cause that acid reflux.  Healing lesson number two, the body does not lie. You know, one particular day during my healing journey I was feeling particularly self-loathing or maybe just depressed because again, I was putting in all this time and money. My ego views myself as quite an intelligent person and someone who lives an overall healthy life but to be having all these symptoms felt like something was not right. It felt like there was a mismatch that I didn't deserve a dangerous word to have these symptoms.  I work out four days a week. I like my work. How could I be so sick was the thought that came through my mind? And I also thought, I feel like I'm really smart. Why can't I figure this out? And it was amazing this cognitive dissonance that was happening with me psychologically that I see with so many of my patients and usually the smarter they are or the more driven they are, the more that this is true. The body all day long was showing me signs and symptoms of in my case extreme stress, heart palpitations every day, insomnia almost every single day, acid reflux, indigestion. My body was showing me, my body's in panic, five alarm fire as my mentor said and yet I couldn't recognize that because my mind, mentally, I was like, I don't feel that stressed but my body was showing the different story, the real story. And so that's why I say to my patients, the body does not lie because while you may be thinking that you're fine, the body tells the true story because the body is not only in some ways the subconscious mind, the body is the receptacle of the damage of the nervous system, of all of the various rituals and habits, good or bad that you do, one of the ways they primarily manifest is in the body. So if you mentally think you're fine but your body is showing all kinds of signs and symptoms of dysfunctional disease, you might wanna check in there because there's a giant disconnect happening. Healing lesson number three, the canaries are always singing. You guys know this whole canary in the coal mine analogy or metaphor, right? That apparently these coal miners back in the day would bring canaries into the mines and if carbon monoxide was there the canaries would start singing and would die potentially if it was too much. So that was a signal to the miners they had to get out pronto. So there are these very, very subtle signs in the body that are trying to tell you that you're moving on the road to disease. That may be just a little bit of indigestion from time to time or it may be a full blown tumor growing off the side of your neck. But in general, the canaries are always there. Are you paying attention? A lot of the time we're too busy, we're driven by the mind, I need to succeed, I need to provide, I need to do this, I need to get things done because insert reason. But the canaries are always telling you to slow down and stop and listen. I'll never forget, as soon as I started feeling better in my healing journey the first let's say like week or two where I had normal energy and I didn't feel like I was chronically fatigued or I wasn't having heart palpitations or an arrhythmia, finally got back in the gym. I finally started staying out late a little bit. I got drinks with my friends and stayed out till two or three a.m. And within about a week those symptoms came crashing back full force. And it went to show that when people go through really serious long-term health crises, chronic fatigue, Lyme disease, cancer, you have to be extremely careful with your resources for literally years after prioritizing rest and good nutrition and not pushing in your life. You have to prioritize that because you can easily backslide. But when I say that canaries are always singing the balance is always this very delicate scale that just starts with 1% indigestion once and then it's indigestion a few times a week and then it's full-blown GERD, acid reflux daily and then it's potential risk of increased esophageal cancer or Barrett's esophagus, let's say. That little balance starts always the same way with just 1% before it becomes quite serious. Healing lesson number four, when you are your weakest look for your trusted advisors or mentors. You know, it is in these stages where if you are seriously ill or even sometimes if you're not, if you've spent $300 in supplements and it hasn't done anything, pay a specialist $300 who actually knows what they're talking about because while it seems like a lot of money, it isn't in terms of continuing to use supplements that don't most of the time do anything. When you are really seriously ill, this is the time to look for trusted advisors, people who are sort of these guides on this healing quest, this healing journey because sometimes when you're really seriously ill, that can be the difference between having faith that this could work and giving up. I'll never forget driving up to one of my mentor's homes. He lived about an hour outside of a city and it was this beautiful winding country drive up into the forest, up into the hills, up into these very moss and lichen covered trees. Going through this hour drive out of the city, up into this really, really rural farm unmarked even. It felt like I was a person on the Camino de Santiago or doing a hage or some Christian going to, let's say Fatima in Portugal. I really felt like I was on this quest to meet my healer and something about that when you're incredibly scared, if you're very, very ill, can give you so much hope and faith that that placebo on its own can be something that keeps you going for another day. For a lot of the patients that come to see me for cancer, they tell me that weekly visit is something that gives them a lot of hope and a lot of faith. Regardless of any healing that is happening with me, with our oncologist, or even if sometimes they come in and they are terminal, that ritual gives them purpose and gives them hope and that inner fortitude it provides is a powerful healing force for me. Healing lesson number five, there are some initial signs of imbalance that you should pay attention to the most. So within traditional Chinese medicine, we say, there's a balance between yin and yang. And these are both theoretical concepts as well as practical ones. Work and rest, yang and yin. If you alter that balance too much, your 80% of your day is all work, 20% percent's rest, you're working 14 hours a day and you're work sleeping five, you will 100% of the time develop health problems. Maybe anxiety, maybe depression, maybe GI shoes, maybe you'll have a heart attack or stroke, 100% of the time something will happen with enough time. But day to day, it's important to pay attention to the subtle signs first. And in my mind, the first signs of the balance going in the wrong direction are nervous system signs and symptoms, like mood, you're noticing yourself feeling really agitated a lot. You're noticing yourself feeling really sad a lot or depressed a lot. Those are subtle signs that the macro balance of your life is not in a way that is producing health, it is producing disease. And only you know what that is and only you know where that's coming from. You may have to introspect, but that is the first sign. The second one being sleep. Because in traditional Chinese medicine, we say, sleep is often a mirror of the emperor, the heart, like the big spirit. And in cultures all around the world, the heart is associated with happiness or depression. So keeping the heart happy and paying attention to internally, how am I feeling and how am I actually doing? Now there are many other healing rituals and practices that you can do to heal. I've put together four of them. In a free guide, it's the first link below this video. Four daily rituals that could potentially help you add years to your life with traditional Chinese medicine. That's a free download. And there's also information below this video on if you wanna book a visit with me in Los Angeles or virtually via telemedicine, you can see my clinic info. And a brand new excitement that's really cool. We just launched something called the Healing Library, which is a series of online programs on how you can heal with traditional Chinese medicine. I launched this to develop more specialized information for you, as well as to keep this channel sponsor-free. There's no shortage of wellness and doctors that are throwing supplements and throwing all kinds of things that realistically probably don't do anything. I know you've bought them, my patients tell me. And I figured, what could I do to keep this channel sponsor-free? Launch online programs to learn about healing with purely traditional Chinese medicine. So the Healing Library, there's a link right below this video to learn more about it. And you can check out if some of those courses connect with you. But before you guys go, there's one more related video right here on my healing journey that I think can help.


Okay, let's see if I can do this in one take. So one of the cool and nice but dangerous features that was added to OpenShift at 4.8 release was the ability to edit a deployment within your dashboard and it gives you a real slick form view and I'll show you what that means by that. So if I pick this application and then go to actions and edit deployment, it takes me over to edit the deployment. Now instead of just having the YAML view, which you would always have before, now you have a form view. So it kind of takes, it abstracts out the YAML and puts it in a nice, easy to understand format. The advantage being you don't need to know the format of the YAML. You don't need to do all the indentation and all that. You just throw out this form and it does it for you. This is really dangerous because if you use this in production, like you could change production. This is like patching code on the fly. This is really a dangerous thing, probably a really bad idea to do in production. So maybe you have a staging environment that mimics production as you do. Even that's dangerous because you say, well, we'll change it. We'll test this and then if it works, we'll go and edit the source code. That's all well and good but sometimes you'll forget to edit the source code. It's worth for humans that happens. What I like to do with this, the way I use it, is as a teaching tool. So for example, I want to change my deployment source code on my PC to include a secret value and put it into an environment variable in my code. So what I'm going to do in the form view is scroll down to here and I can see that I can add an environment variable from a config map or a secret. Now these are existing. So I'm going to click on that. I'm going to call it Foo so it's obvious to see and then I'm going to select a resource and I can select a config map that already exists or in this case, I'm going to use a secret called Red Hat Cloud Services Service Account. And then inside that secret, it obviously has two keys and so I can pick the client ID and there's a bug because you have to click it twice. So now it's going to update my YAML for me when I click save to reference that key in that secret and put it into the environment variable called Foo. So now that I've updated it, let's go back and look at the deployment and then we'll go back into the YAML view and what I'll do is I'll scroll down and I'll say, oh look, there it is, line 206 to 210. There's the values that it inserted in my YAML. And then what I'll do is I'll make note, you know, I'll see the indentation, it goes under environment variables and its reference is a value from and the secret key ref and things like that. I'll copy and paste that into my source code, you know, like for example here, this is a different deployment but it's the same idea. And then I'll then I'll know how to properly use it. So that's how to edit in the dashboard and be careful and use it wisely. Thanks.


Hello and good morning, my friends. I'm Atul Kedia. I'm a fourth year undergraduate student at IT Bombay. And I'm going to discuss some quantum mechanics problems with you. So let's begin with the first problem. This is the problem we are encountering right now. I'll give you a couple of minutes to attempt it, give it a thought, and then we'll discuss how to go about with the solution. So just to make sure that you are able to read the question, I'll just read it out loud. Show that the Bohr condition, Bohr's condition of quantization of angular momentum leads to the condition of formation of standing waves of electrons along the circumference of the Bohr model of hydrogen atom. So the Bohr's quantization condition comes from one of his postulates that the energy difference between transitions of electrons from one state to the other has to be an integral multiple of h. So let's go about with the problem. The condition is that the angular momentum has to be an integral multiple of the reduced Planck's constant. So taking this beyond, it'll be the angular momentum for an electron around an atom at a radius. Let's say this is a nucleus of the atom. And the electron is revolving at a distance of, let's say, around r, around the nucleus. So the angular momentum would be given by p into r. Precisely speaking, it will be r cross p. But then the radius and the direction of the momentum would be perpendicular to each other. So the cross product would equivalently result into r into p with some direction. So r into p will be equal to n into h over 2 pi. Now we know from De Broglie's hypothesis that the momentum and the wavelength of an electron are interrelated. The relation is given by the wavelength of a massive particle that is given by h over the momentum. So now we can make a substitution of the Planck's constant over here by lambda into the momentum. And then we get a very unique, very interesting result. The result comes out this way. p r into p is equal to n lambda into times p over 2 pi. Here, I've changed from reduced Planck's constant to the Planck's constant divided by 2 pi. Here, I have substituted the Planck's constant by lambda times p. And then, so from the Bohr's quantization condition, we have arrived at this relation. Here, we simply cancel out the momentum of the particle. And so this gives 2 pi into radius is equal to n times the wavelength of the electron. So this is a very profound implication. This, in one way, says that the electron has to have an integral multiple of the wavelength of the electron has to be. And the multiple of an electron's wavelength should be equal to the circumference of the electron's revolution circuit. So in way, you can see it this way. The relation was 2 pi into radius would be equal to n times lambda, where the radius is the radius of revolution. And the wavelength is the de Broglie wavelength of an electron. So now, assuming, picturing it in this way, that there's a nucleus in the center, the electron is revolving at a distance of r. From here, circularly. Now, this is the condition says that the circumference of this should be equal to n times the wavelength, which means, which implies that if the electrons wavelength, if the electrons, the waves crest starts over here, let's say, then the wavelength comes out. And since it's a multiple of the wavelength has to be equal to the circumference of the circle, there's a constructive interference between the wave that is coming back from the other side of the electron. So you can see this way. It comes out this way. And then, eventually, it will have a constructive interference. So you can think of it this way. That if this n were equal to 1, let's say, that means the circumference is exactly equal to the wavelength. So the length covered overall in the circumference is equal to the wavelength. And so after one revolution, the electron would have covered exactly one wavelength amount of distance. So there would have been such an interference. This being the crest over here, this being the trough, this being the crest, this being the trough. So this is one complete revolution. Similarly, if this were not true, then we would have observed, if we would have started from one point, come around, we would have reached with a phase difference, which would have led to a destructive interference amongst the two waves. The original going wave, the original wave that was leaving from here, and the wave that is coming from the other side. I hope that is clear. I can make another illustration of this. Starting from here, here, here, here. If we say one wavelength amount of length is covered in this much, let's say, if we take 2 pi r is equal to 4 times the wavelength, n being 4. Then this much circumference covers one wavelength, so like this. Then remaining covers another wavelength amount of distance. Next one covers another wavelength amount of distance. And the fourth one, again, covers the same amount. Had this multiple been n by 2 or something, there would have been a destructive interference after coming back over here. So in that case, the electron would not have been possible. It would have not been stable at all. So this is a very profound implication that the electron has to have a wavelength that is somewhat related to the circumference of the radius at which it is rotating. We can go to the next problem now. So I'll give you a couple of minutes to attempt this problem again. The dispersion relation for a lattice wave propagating in one-dimensional chain of atoms of mass m bound together by force constant beta is given by the following equation. Omega is given by omega 0 times sine of k a by 2, where omega 0 is equal to the square root of 4 beta by m. Here a is the distance between atoms and beta is the expression given. Show that in the long wavelength limit, the medium is not in dispersive. And then find the phase and group velocities at k equals pi by a. So the dispersion relation given here is this. Omega equals omega 0 times sine of k a by over 2, where omega 0 is given to be under root of 4 beta by m. So in the long wavelength limit, when the wavelength is extremely long, let's say wavelength is tending to infinity, we could simply say that k is related to wavelength by this relation as we know. In the long wavelength limit, k would be tending to 0. k would be a very small quantity. And so we can say that the dispersion relation could be simply written as omega is equal to omega 0. We can make a sine a small angle approximation, which is sine of let's say k a over 2 minus of k a over 2, the whole cube, 1 by 3 factorial plus higher order terms. In the small angle approximation, we could simply say that small k approximation. We could simply say that the higher order terms are negligible as compared to the first order term. So omega can simply be written as omega equals omega 0 times k a over 2. It is an approximation. So now we have a simpler dispersion relation for the system given. For this, if we try to calculate the group and the phase velocities, the phase velocity is given by, as Professor Shivprasad must have covered in class, that the phase velocity is given by v phase is equal to omega over k. And v group is given by del omega over del k. So in this case, the phase velocity comes out to be exactly equal to the group velocity. And both of them are equal to omega 0 a over 2. So now this is a very interesting result that we have obtained. Here we are getting the phase velocity and the group velocity are equal. So this, in turn, implies that any disturbance that is created in the wave would be propagating with a constant speed and would not be varying over time or space. These velocities are dependent absolutely on constants, omega 0 a and omega 0 n a. So we know that the wave and the wave propagates absolutely at a constant speed. This can also be seen as this thing. Any disturbance that is created on the wave, let's say this is the wave over here, the group velocity would be given this way. So I'm making a picture of how the wave actually looks. The wave would actually be a product of two factors. The phase velocity factor and the group velocity factor. Together, both of them would give us the total wave. The group velocity would individually give the bigger, would be comparable to the wave packet that we think about in quantum mechanics. The wave packet of a photon or any particle is what the group is supposed to be, whereas the velocity of phase is actually the actual space perturbations or actual space oscillations and velocities, so to say. So this can be thought of as the velocity of group would be as these, whereas the phases would be individual particles in between which would be oscillating with some frequency, which is probably different from the group phase, group frequency. But the velocities of both of them would be the same. All of these particles would be moving together. All of them would be moving together. And these internal crests and troughs would be maintained as time passes. So after some time, we can just expect the entire wave to be traveling as it is the same picture and moving together along the time direction, time or space, either of those. So the second part asks us to find the group and phase velocities at k equals pi by a. So at k equals pi by a, the original dispersion relation was firstly omega 0 into sine of k a over 2. The phase velocity would be omega over k, which would be equal to omega 0 sine of k a over 2 over k. And this would be the phase velocity. And the group velocity would be del omega over del k, which is equal to omega 0 into derivative of sine k a over 2 with respect to k. That would be a over 2 cos of k a over 2. OK, so these evaluated at k equals pi by a. k equals pi by a, sine of k equals pi by a would be 1. So this would be omega 0 over k. And the group velocity, on the other hand, would be 0, because since cos of pi by 2 would be 0. All right, so like I said, in the earlier case, we got both the phase velocity and the group velocity as a constant. The constant was equal to, if you don't remember, it was equal to omega 0 a over 2 in the previous case, vg and v phase in the earlier case. But now, we are getting both of them are firstly not equal in this, when we don't do the approximation. And when we do a substitution of k equals pi by a, we get a difference in both of them. So this also implies that one of the wave packet does not move at all, because the phase velocity, the group velocity is given as 0. Whereas the phase velocity is non-zero, that means the internal waves, the waves in between each group would be moving and propagating through space and time. Whereas in the earlier case, we saw that both of them moved at the same speed. And so the entire wave moved together in space. All right, I think we can go to the next problem now, third problem. I'll give you a few minutes to give it a thought. So the question was, find the group and phase velocities of the matter wave associated with a free particle under the assumption the frequency is defined using the kinetic energy or the total relativistic energy. So we can keep the relativistic energy case hold on for a time, time being. We have to find the phase and group velocities for when the energy is related by the kinetic energy only of a particle. So the group and the phase velocities, when the particle's energy is related by only its kinetic energy. The particle's energy, let's try to write the dispersion relation firstly, is that the particle's energy E is given just by the kinetic energy, which is equal to the momentum squared over 2 times the mass. All right, the momentum given by the de Broglie hypothesis is related to the wavelength. The momentum is related to the wavelength by p is given by h over lambda, where lambda is the particle's de Broglie wavelength. We can find the relation between p and the wave vector k through this, because lambda is just equal to 2 pi over k. So p would be equal to h over hk over 2 pi, or simply h cross times k. So we've got a relation between. So if we substitute p into the energy's expression, where E is equal to p square over 2m, we would get a relation between the energy and k. That will be h cross squared k squared over 2 times the mass. Now we know that we can attribute this energy to the wave nature of the particle with a particular wavelength, the wavelength and frequency. The frequency of which would be given by the relation energy is equal to h cross times omega, or which is simply equal to h into nu. So this in turn reduces to omega is equal to h cross omega is equal to h cross squared k squared over 2m, which is equal to omega is equal to h cross k squared over 2 times the mass. Now this is a simple relation between omega and k, and we can just simply find the group and the phase velocities using this. The group velocity as we did before would be given by the derivative of omega with respect to k. And the phase velocity would be given by just by dividing omega by k. That would be h cross k over 2m. And group velocity would be h cross k over m since the derivative would take out the two. All right, so this is one relation that we have found out using the energy being related just by the kinetic energy over here. And then in putting the momentum as h cross into k, and the energy as h cross into omega, we found out the omega versus k relationship. And then we found out the phase and the group velocities. Now if we take this, if we try to attempt the same problem using the relativistic energies, we would get a slightly different result. Let's try this out. The second part of this question says this. Then if the energy is related by the total relativistic energy, then how does the phase and the group velocity turn out? So the relativistic case. So probably you might be knowing from before, although it has not been covered in class yet. But in the relativistic case, the energy is given by the famous Einstein's equation e squared is equal to p squared c squared plus m not squared c raised to 4, where m not is the rest mass of the particle. And the momentum, p is the momentum of the particle. c over here is just the speed of light. And this is the total energy. So it's not a linear relation anymore. It's the energy squared is related by the momentum squared and the mass squared in some way. There's a restriction this way. And the entire problem changes because of this relation. So now going beyond substituting p is equal to h cross time k. We get energy is equal to, we reduce the square root sign, h cross square k squared c squared plus m not squared c raised to 4. All right. Now we equate this to h cross into omega and solve it beyond. So h cross into omega is equal to squared, h cross squared k squared c squared plus m not squared c raised to 4 raised to half. And now just finding the phase and the group velocities would be simple. We have a omega and k relationship. You can just divide omega by k and find the phase velocity. And we can differentiate omega with respect to k and find the group velocity. So omega by k, which is equal to the phase velocity, is equal to this not so good looking expression, some k relationship plus some constant, some other constant over k. And the group velocity would be the derivative of this with respect to k. So differentiating this with respect to k would give us firstly 1 over h cross from over here. And so half of this 2k into h cross square c squared over this entire bracket. This entire bracket comes over here on differentiating. So again, now we see the expression has changed entirely. It is dependent on k over square root of k squared plus square root of something like k squared, some constants c not plus some c 1 over here also c not and some c 1. These relationships aren't exactly as we found out in the previous case, which were just linear. The phase velocity and the group velocity were linearly related to k earlier. But in this case, we see there's a difference. So that's how relativity changes the entire situation. The phase and group velocities change entirely due to this. All right, I think we can go to the next problem. It's a very interesting problem. So I'll read out a lot of the question once. A wave packet is constructed by superposing waves, their wavelengths varying continuously in the following way, y of x comma t, which is the wave, it is given by an integral of a of k, an amplitude, times cos of kx minus omega t times dk, where k is the wave vector. And now there are some conditions given on A. A k is just a constant between some range of k from k equals minus delta k by 2 to k not plus delta k by 2. And A is equal to 0 otherwise. Now we have to sketch the approximate by the actual wave and estimate the uncertainty in the position that we obtain from this wave given. So we can continue with the problem now. I'll go about with this. So firstly, we are given A of k is a constant A in the range of k between k not minus delta k over 2 to k not plus delta k over 2 and 0 otherwise. So I'll just make a plot to keep this in mind. So if this is k versus A of k, it would be 0 all throughout. And then there will be a box type relation and then 0 again. So this is a constant over here. A is a constant equals A naught. And the range is given by the condition given over here. This is k not minus delta k over 2 and this is k not plus delta k over 2. So now evaluating the actual wave given as the relation given over here, the displacements along the y direction are given as y of xk as the integral. Now we can reduce this integral by substituting the values of A in the ranges given. So A is equal to, so from the range 0 to k not minus delta k over 2, A will be just equal to 0. So the integral would be 0 times cos of kx minus omega t times dk plus in the range of k not minus delta k over 2 to k not plus delta k over 2, it will be just a constant times cos of something times dk. Plus the third term would be exactly as this term except that the integral would be from k not, I'll just write it down. The integral would be from k not plus delta k over 2 till infinity. And again the coefficient would be 0 times cos of something times dk. So again these two integrals would be 0 obviously because the coefficients are given as 0. We only have to evaluate this one integral over here in the range from k equals minus a k not minus delta k over 2 to k not plus delta k over 2. So the displacements y of x comma t would be equal to A times the integral of k not minus delta k over 2 to k not plus delta k over 2 of cos of kx minus omega t times dk. All right. So let's go ahead and evaluate this integral. The integral of cos would be just sine, sine divided by the coefficients. So we can write that down. Y of x comma t would be evaluated as A times, some A times. The integral would evaluate as a sine of kx minus omega t, sine of kx minus omega t over x within the range of k being from k not minus delta k over 2 to k not plus delta k over 2. All right. So now substituting, we make the substitutions and do some simplifications. This would be, we can just remove the x over here and group it with rA and this would turn out to be sine of k not plus delta k over 2 times x minus omega t minus the whole thing subtracted by sine of slightly tedious k not minus delta k over 2 times x minus omega t. So now we make some trigonometric simplifications. This is sine of A minus sine of B. We can write it as cos of A plus B by 2 into sine of A minus B by 2. 2 times of sine of, it will be 2 times cos of A plus B by 2 and sine of A, this minus this by 2. So the relation simplifies to 2 times A over x times cos of sum of these 2 divided by 2 which would be k not x minus omega t. This is a simpler relation and multiplied by sine of the difference of these 2 divided by 2. So we can write it as this, delta k into x minus omega t. This would be divided by 2 and this would remain as it is, alright. So we have simplified our original integral from our original integral of y of x, t equal to integral of A k into cos of k x minus omega t, we have arrived at this relationship between the wave displacements y of x, t and these 2 trigonometric terms. Now getting back to the question, what is the question, the question asked is that sketch this, we will do the sketching in a while. Before that we will just notice that one of these in the limit that k not is much greater than k delta k. This one would have, this sine function would have a very slow oscillation frequency since delta k is a very small quantity as compared to k not, I will just write it down, k not is given if in the limit k not is much greater than delta k. The oscillations with respect to sine function, the sine function would be much slower since much slower in space coordinate and so in like we saw in the earlier problem, these could be said as the group velocity, group frequency. This would correspond to the group frequency whereas these would correspond to individual particles oscillating within each group. So, you can see it this way, since delta k is a smaller quantity, let us say if this were the overall delta k factor oscillation, this is the sine delta k times x over 2 oscillation. Within each, within this much of length, there will be a large number of oscillations of the cos factor. So, I will just write it down at this as this. So, firstly if we say that at x equals 0, we can firstly ignore the omega t factor for a while and discuss this only at a constant time, time equals 0, t equals 0 let us say. So, at t equals 0, the sine factor, sine of kx minus delta k into x over 2 would be oscillating in this fashion again continuing and at x equals 0, it would be 0 and at x equals, x equals delta pi over 2 delta k, it would again be 0, ok. So, where we know delta k is a very small quantity, hence this length is a very large quantity. But within this length, each of these oscillations would happen. So, as for at any given x value, these would be oscillating since these are relatively larger quantities and we would observe oscillations like this as we had seen earlier in the previous problem. Again, this highlights on the group and the phase velocities and now, so the envelope would actually look like this, if we take it beyond it would be, it would look something like this where the number of oscillations within the, within each group would be different from how many are there which would be dependent entirely on the k naught and delta k naught how I have made right now, alright. So, the problem also asks us to verify the uncertainty in the position given by the envelopes, envelopes wavelength. So, firstly we notice that the envelopes, envelopes wavelength would be given by when this factor sign of 2 delta k x over 2 would be, would change by 2 pi, so that would, would make a delta k into the wavelength, let us say x prime over 2 would equal to 2 pi. So, that will be equal to, the wavelength would be equal to x prime is equal to 4 pi over delta k alright. So, to calculate the uncertainty in delta x, delta x, the uncertainty in delta x has to be taken as the difference between 1 maxima and 1 minima within the group velocity, group wave. So, that would correspond to a phase shift of just pi instead of pi by 2 and so we will get delta x would be equal to, so this can be taken as the wavelength delta x would be equal to 2 pi over delta k alright. So, now we have found a relationship between the delta x and the delta k, the relationship was delta x is equal to 2 pi over delta k. Now, we make some simplifications, now taking delta k to the other side delta k times delta x would be equal to 2 pi and noticing that p is given by k times h cross, so multiplying both sides by h cross and substituting k h cross into delta k as delta p. So, I will just write it down, multiplying both by h cross would give us h cross delta k delta x is equal to 2 pi times h cross alright. So, now this can be written as delta p since h cross is just a constant and so delta p would be given by h cross times delta k, I will just write it down delta p would be just h cross times delta k and so this is delta p times delta x is equal to 2 pi times h cross which is equal to which and now we have found out the uncertainty principle. We have obtained at the uncertainty principle through just by assuming that the oscillations of a particle are given by some sine functions and so this is a very profound result that we have obtained right now alright. I think we can move on to the next problem, so the information given in the problem is that the average value of momentum p x is 0 and the momentum in x direction is given as 0 and so the uncertainty the standard deviation in the momentum is just given by the square root of the average value the expectation value of p x squared. Now we are requested to obtain firstly the minimum kinetic energy that the proton and electron would have if they were confined to a nucleus of approximately approximate diameter of 10 to the minus 14 and we have to see that this argument results in the electron not being it turns out to be that this energy is very unphysical. The electron could not have this much amount of energy staying within the nucleus, so I think we will go about with the problem. So the delta, so we are given that delta problem 5 the expectation value of momentum or the average value is given as 0 and the standard deviation in the momentum is given by square root of the average value of p of x times squared. Had this not been 0 we would have had another term over here written as p x the average of p x the whole squared, but since it is given 0 we can just ignore this alright. So firstly we are to evaluate the kinetic energy using this. So the kinetic energy would be given by p squared over 2 m. So we are given that the electron or the proton are confined to a diameter of 10 to the minus 14. So we can say that delta x is equal to 10 to the minus 14 of meters alright. Using the uncertainty relation that delta x times delta p is equal to let us say anything h cross or h bar it does not really matter we will just be working by with the order of magnitudes this entire problem can be just needs us to do an order of magnitude analysis it does not need really to give a precise answer. So we can just take it as h cross usually the exact relation would be h cross over 2 but then this is this would work absolutely. So delta p would be given by h cross over delta x we make the substitution over here for the kinetic energy of the nucleus in the of the particle in the nucleus and it would be kinetic energy is equal to the h cross squared over 2 times mass times delta x the whole squared alright. So now keeping the mass as a variable. So in the part in the question it is asked that we can we have to solve this for electrons as well as protons we can keep we can substitute at the end of each for the electron and a proton separately and see how the results are for now we will just make the substitution this is equal to the h cross would be given by 10 to the minus 34 the whole squared over 2 times the mass we can ignore the 2 also but let us just keep it over here times delta x squared delta x squared would be minus 14 the whole squared alright. So this would turn out to be of the order 10 to the minus 40 over 2 times the mass. If we try to evaluate this in this is in Joules so if we try to evaluate this convert this into electron volts we will have to multiply and divide by electrons and the net result would be something like divide by electrons so it will be 10 to the minus 10 to the positive 119 electron volts divide by 1.6 I will just give you a couple of minutes to attempt this I think I am getting the number somewhere wrong I was hoping to get a positive and extremely positive number oh ok alright alright I think I have gotten it so yeah this is the relation right now it is the kinetic energy is given by this I have converted so till over here the kinetic energies were in Joules I will just write it down in Joules and over here it is in electron volts so I will just copy the relation in another sheet and explain kinetic energy is equal to 10 to the minus 40 over 2 times the mass multiplied by 10 raise to 19 over 1 over 1.6 electron volts alright. So simplifying this a little bit gives 10 to the minus 21 times some we can ignore this 3.2 times mass so now substituting for the mass of electron and proton separately for electron it would be the mass of the electron is around 9 into 10 to the minus 31 in SI units. So, this would this would be this would give us 10 to the minus 21 over some constant some this ok we can ignore the constants and write it this way 9 into 10 to the minus 31 and similarly this is for electron ok electron and similarly for the proton the same analysis would give us this Ke would be equal to 10 to the minus 21 over 3.2 into the mass which is equal to 10 to the minus 27 in this case and to some into some constant we can ignore that it is an approximation. So, this gives us so alright both of them are in electron volts this one gives us a total of this is approximately 10 to the 10 in order whereas, this one is over here of the order of 10 to the 6. So, kinetic energy of electron is about this much and of a proton is about 10 to the 6 whereas, often electron is of the order of 10 to the 10 electron volts which is which can be said as a almost to giga electron volts this is almost like a 10 giga electron volts this is extremely large a normal electrons the electrons rest mass is given by is about 0.5 mega electron volts whereas, this is in giga giga electron volts this is extremely large and unphysical for an electron to have whereas, for a proton on the other hand the same result gives us something like 10 to the 10 electron volts which is equal to a mega electron volt which is very physical and a protons I will write this down mass of an electron a protons mass is given is about 900 or 930 mega electron volts. So, this is very physical as compared to the kinetic energy of the proton that we have observed is very physical it makes a lot of sense for a proton to have such a kinetic energy and for electron to not have such a kinetic energy since the electron cannot be restricted in such a small area also another analysis of the momentum we could do using the uncertainty relation and the delta x we were given the delta x that was given was 10 to the minus 14 since I am running short of time I will just I will do a small analysis of the momentum of the particles of each of the particles in within that nucleus given this delta x we can find the delta v the uncertainty in the velocity would be given by h by h or h cross by delta x into m the mass alright. So, this would be of the order of 10 to the minus 34 over delta x is given as a 10 to the minus 14. So, we can simply write this as ok right we will write it down as 10 to the minus 14 and mass would be 10 to the minus for an electron it would be 31 and for a proton it would be minus 27. So, if we reduce this for an electron we get delta v for an electron would be about 10 to the 11 if we reduce this to this and same the same thing for a proton would be around 10 to the 7 since this turns to 10 to the minus 27 the entire thing the entire relation reduces to 10 to the 7. Now, from relativity in hindsight we know that the speed of an electron speed of any particle cannot be more than the speed of light over here we are observing that the speed of electron is coming the uncertainty in the in the electrons velocity or in some way the electrons speed is coming to be more than the speed of light which is entirely unphysical. Hence, we can easily say that the electron is cannot be bound within this much delta x which is to say the it cannot have been bound within a nucleus. So, moving on there are two other parts to this problem. So, the next part is pretty simple we can go ahead the next part is just says that in a square well potential ok. So, we have to find the ground state energy of a particle in an infinite square well potential again this is an approximate problem we are not we are not dealing with exact values the delta x over here can be taken simply as delta x is equal to for an infinite potential delta x can simply be taken as the length of the potential itself. Again, so now the energy ground state energy delta k the ground state energy or the energy so to say would be p squared over 2 m which would be again we can substitute p as h cross over L delta p is equal to h cross over delta x which is equal to delta h cross over the length. So, now we get h cross squared over 2 m L squared. If I am sure that you have already done this problem of a particle in an infinite well now we are seeing that the relation over here is gives that we are we have obtained is not exactly not entirely equal to what we have expected we would have expected the actual answer would have been pi times h cross squared over 2 m L squared for a ground state for the ground state energy whereas, we are observe we have calculated as h cross squared over 2 m L squared. Now, this is to say we are we are dealing with just approximations we have taken delta x as just equal to L had more information information being given about the nature of uncertainty in the position we would have had a better estimation of what the energy would have been and hence we could have actually had obtained the correct relation of pi squared h cross squared over 2 m L squared. Alright, so we can go to the next problem the next part but next part is next part is slightly simple. So, I would leave that for you people to attempt that since we are running out of time we will attempt the sixth problem now. So, I will simply explain the problem what the problem states the problem states that in actual physical atoms the excited atoms usually the electrons in higher state not their ground state they usually emit some photons and come back to the ground state. Now, we are given over here that the lifetime of at which for which the electron stays in the highest state is of the is of around 10 to the minus 8 seconds. Now, we are expected to find the uncertainty in the energy in the B part and the uncertainty in the frequency using this information the information given is again the that the life the lifetime of an electron at a highest energy state is equal to 10 to the minus 8 seconds. So, delta t is given to be of 10 to the minus 8 seconds. Alright, we know the uncertainty principle for relating the uncertainty in energy and the uncertainty in the time as delta E times delta t would be equal to h cross simply to say h cross I will ignore any factors do or anything. So, the second part just simply tells us to evaluate this value of uncertainty in the energy that can be simply done by making the substitutions. In the first part rather we are asked to find the frequency the uncertainty in the frequency. So, we will evaluate we will go ahead and evaluate that that is the uncertainty in the energy would be the uncertainty h cross h into the uncertainty in the frequency times the delta t would be equal to h cross. Now, cancelling the h cross and h we would get delta nu delta t is equal to 1 over 2 pi. So, delta nu is simply equal to 1 over 2 2 pi times delta t this is equal to this is equal to 10 to the 8 over 2 pi. So, we can just simply say that it is around 10 to the 7 sorry this is 10 to the 7 hertz the uncertainty in the frequency is coming out to be 10 to the 7 hertz. Now, so this is the uncertainty we have observed. Now, we are asked to find out the uncertainty the fractional uncertainty the delta of nu by nu for the radiation of a wavelength given as 5000 angstroms. So, we are supposed to find delta nu over nu given lambda is equal to 5000 angstroms. Now, we can simply we can simply do the calculations and find the frequency frequency would be equal to C over the lambda which would turn out to be 3 over 5 times 10 to the 10 to the 8 plus 10 to the minus 7. So, that will be 10 to the 15. So, this is the frequency of light and so the fractional uncertainty in the particles frequency would be simply 10 to the minus 10 to the 7 over this factor where 10 to the 15 we can ignore the 3 by 5. So, delta nu over nu is given by this. So, we see that it is a very small factor it is a of the order of 10 to the minus 8 hertz it is of the order of 10 to the minus 8. So, so we can simply say that the so when we observe the spectrum of an electron excited of an electron de-exciting from an excited state we can simply say that the electron the uncertainty in the frequency is very less, but still if we resolve the frequency to of the order of 10 to the 8 then we could see the uncertainty we could see a width within the spectrum power spectrum observed over there. Alright. So, I think this is the last problem now we can go to some questions if anyone has doubts questions and regarding anything in quantum mechanics. Center 1 2 5 6 go ahead for your question. How group of those are generated when single particle is under motion. Okay your question is how a group is formed when we are dealing only with a single particle so to say right. Yeah obviously. Yeah alright so it is an interesting question. So firstly so the whole the entire discussion about quantum mechanics relies on the fact that a particle is not just one particle but there is one particle we can just we can say that it's a particle associated with a wave with it alright. So, there's a wave so to say for each particle if there's an interference of one particle with another particle we can say that the there's a interference of two waves and then there's a there's a there's this difference that comes in the phase and the group velocities. So you're right we are not the the phase and the group velocities does not come due to one particle but the due to the interference of multiple particles let's say two particles and then there's a so that will lead to some differences in the frequencies and which would be which would give us the a difference in the phase and the group velocities. Another question is there sir. Yeah. Then according to Dave Wembley's hypothesis lambda is equal to H by P. Yeah. Where P stands for L in the way. Yeah. Why didn't we use group or group velocity in that occasion. Okay so alright so I think Professor Shivprasad could answer this much better I'll give it an attempt this okay so the de Broglie's hypothesis is about one particle when de Broglie says that lambda is equal to H by the momentum the he said it is said only for one particle the so let's say one electron over there so we can when we make a substitution to find the lambda it's for one particle now the group and the phase velocities that you said are for multiple like I said for when two particles interfere there's a group there's a phase and a group velocity so hence there's a difference the the de Broglie's hypothesis does not apply to that an interfered particle if had we a better understanding of how the interfered particle can be thought of as one particle together then we would have said one part as one particle alone then we could have said that we could have applied the de Broglie's hypothesis to find one wavelength but since that is not true and we and I'm pretty sure that we wouldn't be finding that because okay so firstly because we as we saw the wave changes the waves amplitude firstly changes and its internal frequencies also internal wave also changes so there's a difference the this is not exactly a normal wave which which was assumed in the de Broglie's hypothesis so there's a difference I think Professor Shifrasad could explain this much better diagram the wave what you had drawn it looks like coming from a single particle that's I have an elementary doubt in that actually in the wave packet that I made to a crest and a trough and within within there are electr there are oscillations are increasing and then decreasing are you talking about that yes that that is my doubt yeah so that is about that is when more than one particle when more than one particles are there together so again again the k relationship that we saw so in problem number four we saw that the displacements were given as an integral over k all right so that means not only one k value are available there are multiple k values available for in the in the analysis whereas according to de Broglie's hypothesis there'll be only one part one k value all right so there's a difference so over here there are multiple particles whereas then the de Broglie's hypothesis deals with only with one particle any one particle center 1176 go ahead from karnataka sir in the second problem okay while solving the second problem we got vg is equal to zero yeah yeah that means particle will be a drift yes yes I think that in the contradict result I think okay let's just look at the problem so the dispersion relation was a relation was given as a omega over some sine function so and we were we found out the case in the group velocities for k equals pi by a so yeah I think it is possible that for this one k value k equals pi by a the phase the group velocity comes out to be zero but it in general that need not be the case so I would say that the particle as such is not moving but in in other cases of k values there the particle would be moving and we would get a non-zero group velocity and in the same problem yeah in the next part we got v phase greater than vg is that possible sir yeah v phase is equal to omega by k yeah will be h cross k divided by 2m and vg is equal to do b omega divided by dk is h cross k divided by m so that is so yeah vg is twice the v phase yes that's correct there's a there's a difference between the two so the phase velocity was given as h cross k by 2m and the group velocity was given as h cross k by m so physically we could think of this as this the group as such is moving with a slower rate but the oscillations within are moving much faster so let's say initially the when the group is over here the oscillations were within them but as time passes the oscillations move move beyond and they move into the next next group so the oscillations keep shifting from one group to the next one and travel beyond that's how it is next one three five go ahead from Tamil Nadu the first problem which we have solved that circumference of the circle is equal to an integral multiple of wavelength yes right yes which is applicable only for the hydrogen atom or any other atoms this would apply to any atom which has only one electron so in in the in physics we call it like as hydrogen like atoms these would be atoms that have only one electron we could say it is a helium plus ion or a lithium two plus ion so any atom that would have only one electron so firstly the deeper the boss boss model ball model applies only to atoms that are hydrogen like which have basic which basically means which has only one electron associated to the nucleus so there's one nucleus which could have any number of protons or neutrons but there's there should be only one electron associated with it so it could be lithium two plus ion or a helium plus ion or anything beyond that or a hydrogen atom so this relation this result that we obtained is true for all of them the phase velocity of the wave is greater than the velocity of light then what about the physical nature of the de Broglie wave the phase velocity of a particle of the particle is greater than the velocity if it turns out to be greater than velocity of light you're asking what would be the nature of de Broglie I mean how would de Broglie's okay so like I said earlier phase and the group velocities come as come into the picture when we are talking about multiple particles interacting together all right so in that case again applying the de Broglie's hypothesis does not make sense because that applies only to a single particle case and anyways if the phase velocities come out to be larger than speed of light then that's not entirely a problem because those aren't physical matter moving with the speed of something greater than light this actual physical particles are only the groups that are formed the groups are the actual physical particles that are moving that turns out to be larger than the speed of light then that is a problem not if the phase velocity comes out to be larger than the problem larger than the speed of light if you take the de Broglie relation lambda is equal to actually P for a particle of having a mass m moving at certain velocity and it will have only a unique momentum then the corresponding wavelength also should be unique can we say that the de Broglie wave is monochromatic no because an assumption that you made was the momentum is unique so like like we know there's an uncertainty in the momentum and the position of a particle if we okay let's take an example if we take the electrons wavelength to be a constant particles wavelength to be a constant one value that would mean the particle has one momentum and we can we can picture it as a wave moving in free space continuous wave moving from from minus infinity to plus infinity moving in free space this would be of one wavelength and the de Broglie's hypothesis would give us one momentum associated with it but then with this with this condition we have lost the uncertainty the certainty in the position of the position of the particle so when we when we say that the particle is a wave when we take a wave which is which is spread it out around the whole space we are saying that the uncertainty in the position is infinite like we can find the electron from minus infinity to plus infinity anywhere but then so so we have lost the uncertainty the the uncertainty principle has shown it shown its role over there now if if you if we make the particle a little bit constricted to some small range of space and it is finite texture yes it is not from minus infinity to flux plus infinity wave packet is a finite size yes that's what i'm saying so if we make it a finite sized particle that's what i was i was about to say that if we make it a finite size particle we would have to include more terms in the in the wavelength condition so so okay if we want to distribute if you if you want to find a Fourier transform of or the or the spectrum of or the dispersion relation for this particle which is localized with respect to the frequencies it would be it would have a lot of terms not just one term like like in the previous case so then again the uncertainty has shown its role and now we don't have a distinct we don't have a distinct wavelength associated with this this and and the momentum so again this is interrelated as you may see like that you see you see the uncertainty in the wavelength the which is somehow related to to the momentum by the de Broglie's hypothesis and we know that the momentum and we see that there's an uncertainty in the momentum because there's an uncertainty in the lambda in the wavelength so that gives us the uncertainty in principle back again the uncertainty in the x in the and the multiply the uncertainty in the momentum is a constant is greater than greater than a constant at any time in the fifth problem we solved today we proved that the delta v in the order of 10 to 11 and delta vp is in order of 10 to 7 and based on that how can we say that electron could not exist within the nucleus can you comment on this yeah so as I said then the so we said we found delta p delta vp to be of the order of 10 to the 11 and the other one was for so the uncertainty in the electrons velocity was of the order of more than speed of light and that of the proton was less than speed of light now in some way the uncertainty of the law of a particle gives us reflects about the particle's actual velocity so once we are saying that the uncertainty is greater than c then we can clearly say that this is not this is unphysical relativity implies that the absolute speed of speed of any particle could be equal to speed of light so as soon as in any argument if we see something which is something exceeding the speed of light we simply say that that's unphysical that's that's impossible so yeah so thank you very much it was interesting solving these problems today with you thank you and bye bye


You'll experience patient care that is specific to you with exceptional follow-up care that ensures your hearing and balance needs are being met. Audiology Associates at 1428 Northlake Drive in Prestonsburg. On Sunday, February 19th, a portion of Niki Road between Phelps and Stone had broken away from the mountain side due to heavy rainfall. The Kentucky Department of Transportation is working around the clock to make the necessary repairs and ensure the road will be passable once again. Public Information Officer Shantana Woodward shares more details. We've got a lot of rain here lately. And a lot of times when this rain happens and we've had a few really cold days, the rain will kind of get down in between the rocks and we'll have this freeze and thaw cycle where it will freeze. And then once it thaws, it'll leave gaps and stuff in those gaps. Well, it's what tends to lead to our embankment failures. And it'll cause the rock underneath the road to kind of sink down and sometimes even kind of shear off if it's close to a creek bank. You know, in this particular area in Niki Road, you know, it's along the ridge line that, you know, so it's not that close to the creek bank. But it has, you know, it's got several different breaks throughout there and we've deemed it unsafe for travel right now. According to a new update, there has been an additional slip in the road which has worsened the initial damage. Woodward advises motorists to seek an alternate route if commuting from Stone to Phelps or vice versa. There is no current estimation of when it will be fixed. We've had a bit of an update that we've actually had a bit of a slip that has happened directly underneath where this break is located. And because of that, the road has dropped down several more feet. And so what was originally a smaller repair has now turned into a more extensive long term type repair. Our guys are out there assessing the situation. We'll get started on it just as soon as, you know, we can. But in the meantime, Niki Road is closed for the foreseeable future. Our number one priority is making the roads in our district, our state, maintain roads safe for you and everyone else to travel. Reporting from Mountain Top News, I'm Nick Colum.


The results of this study suggest that dietary counseling has a positive impact on the diets of children with familial hypercholesterole. Children with FH who received dietary counseling were found to consume more low-fat dairy products, cheeses, and margarines than those without FH. Additionally, they also consumed less sugar and sweeteners than those without FH. These findings indicate that dietary counseling can help improve the dietary habits of children with FH, which may lead to better cholesterol control.


The final item of business today is the member's business debate on motion number 10420, in the name of Gordon MacDonald, on celebrating the contribution of independent retailers to the Scottish economy. This debate will be concluded without any questions being put, and I would be grateful if those members who wish to speak in the debate could press the request to speak buttons now, please. Before I begin, I would like to thank John Lee of the Scottish Grocers Federation for his assistance, and I also welcome those convenience store owners who are in the gallery. The independent convenience store sector has been a cornerstone of all of our local communities for generations. Wether it's a pint of milk in a newspaper in the morning or a few messages on your way home from work, they are open, ready to serve from early in the morning to late at night in every city, town, rural and island community across Scotland. Indeed, Scotland has more convenience stores per head of the population than any other part of the UK employing over 41,000 people. To put that into context, that is more people than the combined total employment provided in the agriculture, forestry and fishing, or indeed in the motor trade in Scotland. It's not just the direct employment that local shopkeepers provide but also their contractors and suppliers. The vast majority of convenience stores are owned and operated by small business owners, and because they are part of the local community, they tend to use local contractors and suppliers. Research by the Scottish Grocers Federation identified that a typical convenience store reinvests £1.25 million into the local economy by using local tradesmen, local produce suppliers, local shop fitters, local garages, as well as local legal and accountancy firms. As a result, convenience stores are also, I believe, one of the cornerstones of a community in the same way as the primary school, the community centre, the post office and the pub. Their presence in a village, town or in a neighbourhood helps to support that community. In addition, many corner shops provide a range of services from a cash machine, bill payment service, home delivery or even just a notice board. It all helps to sustain that local community. All is not well in the convenience store sector, especially among those family owned businesses, as there are two main issues affecting them. The growth of the big four supermarket convenience store network and the resultant over provision in grocery stores. The decision by the large supermarket chains to move into the convenience store sector in Scotland has put pressure on small independence. In the current economic climate, customers' income, in many cases, has not kept pace with inflation. The outcome is that the available spend in an area is spread more thinly across all the retailers, resulting in the viability of some stores to be questioned. The expansion by the big four supermarket chains into convenience stores is not just detrimental to small independent shopkeepers but other small businesses, as these large conglomerates tend to have national contracts for maintenance, servicing and international suppliers, resulting in the loss of local economic benefit generated by having a locally owned store. The other issue that I mentioned is over provision of grocery stores, and I will use an example from my constituency to highlight the potential problems. Along the stretch of the B701 from Oxcang's Road north to Collington Main's Drive, there are two large supermarkets, one multiple convenience store operator and three independently owned convenience stores in a distance of just over a mile in length. An area of my constituency that most people would consider as being well served by grocery stores has healthy competition and choice. However, the council has decided, along with its partner, to sell land previously occupied by a social work building in a church to a major grocery discount store operator. The local community are split. Many families would welcome the discount, while others have signed a petition opposing the building of yet another supermarket. The reason for doing so includes the effect of increased traffic congesting, resulting in additional traffic noise and deteriorating air quality, plus the effect on existing employment and ultimately possible store closures. There is also disappointment within the local community that the land has not been designated for housing as there is unmet demand for new housing in the area. Many members of the now closed St John's Church were under the impression that the land would be used for much-needed housing. Now they find a potential licence premise will be on the previous church ground in an area where there already exists eight licence premises. It is times like this that the city planners have to take a step back and ask themselves whether the location of another supermarket is in the best interest of the whole community. In doing so, council officials should examine the job creation claims and whether those employment opportunities are really new posts or simply displaced employment from other existing operators who need to downsize their operation to meet the reduced revenue. I study by the UK Retail Trade Association ACS titled Job creation claims and new supermarket retail developments found that in many instances the claims of jobs created simply represented a headcount. Part-time posts make up 49 per cent of all retail opportunities and therefore the actual full-time equivalent jobs will be in most cases substantially lower than that. If the new supermarket is giving the go-ahead, my concern is for the survival of all three small independent stores and possibly the reduced employment opportunities provided by the existing supermarket operators. The convenience store sector, unlike the large supermarket chains, does not have reserves that allow them to trade at a loss for a prolonged period of time. Since 2008, many small businesses experienced difficult trading circumstances and it was the introduction of the Scottish Government's small business bonus scheme that assisted them. Across Scotland, 92,000 small businesses, many of them owners of local convenience stores have had their rates abolished or substantially reduced. However, it is not just about existing businesses. We need to make our shopping areas more attractive, offering a wider and more diverse mix of high street businesses. We need to encourage younger entrepreneurs to create new businesses, making use of the premises that are currently vacant in our towns and centres, town centres and neighbourhood shopping areas. A more diverse high street or shopping area will generate a higher footfall that should benefit all businesses in the area. The Carnegie Trust has developed a number of retail initiatives from pop-up retail festivals to a town's characteristic online toolkit. It stated in its briefing that, for many towns, the contribution of independent retailers is a crucial factor in the long-term sustainability, diversity and vibrancy of high streets. The sustainable diversity and vibrancy can only be achieved if everyone involved in developing our towns and cities recognises the importance of the contribution that is made by the independent convenience store sector to our economy and understands the importance of sustaining it. I was expecting a couple more minutes as I write down my notes because I just decided at the very last minute to step in. Independent stores actually play a very, very important part in the local economy throughout the country. I was thinking about it. People at my dad retired people. My dad used to go along to the local shop for his newspapers, his milk, his pension and also became a social event because he got to know personally that retailer. When you think about it, other people in the same circumstances as my dad, that might have been the only person that that person spoke to the whole day, so it's really, really important that retailers really play a very, very important part in social communication with individuals. Also, from the economy point of view for retailers or independent retailers, they're under a lot of pressure just now because of the economy and they have been struggling to survive. Things like parking can actually hinder their sales as well, which is quite a problem. Also, as I said before, they're struggling with the big multinational chains, which do provide a service but don't provide a personal service, which is very, very important. The local retailers, coming back to my previous life as a trainer, provide employment for a lot of young people as well. They were at the forefront and at the forefront of delivering the modern apprenticeship programmes and take an active part in the modern apprenticeship programmes and see the benefit of it. That's really important as well. I'm also the convener of the Towns and Town Centre Cross Party Group. I'm hoping that the Government will look at more initiatives and local authorities, look at what they can do to preserve those businesses and help them to survive. So, in winding up my very short speech, hopefully I'll ask the Scottish Government just now what can they actually do to help those local businesses. We're looking for more support for them in the way of training and also to help to look at how can you tackle the red tape that hinders these businesses to actually survive and develop their businesses. Thank you, Presiding Officer. Many thanks. I now call Chick Brody to be followed by Gavin Brown. Thank you, Presiding Officer. May I thank Gordon MacDonald for bringing his debate to Parliament this evening. I echo his sentiments, read the contribution made by our food clothing and services retailers. It's right that we celebrate the contribution of our independence convenience source retailers, often the Cinderella of our focused sectors as part of retail in our national economic strategy. Strong independent businesses have a vital role to play in the future, not just of our high streets, as Gordon mentioned, but if they're strong and well managed, they can attract new business and grow into more medium-sized operations. As a whole, the retail sector employs 255,000 jobs, 41,000 in this particular area. It is a very large private sector employer employing 14 per cent of a non-government workforce. Our retailers have had a difficult few years with the recession. Small and medium retail businesses, particularly the independent convenience stores, have seen a decline in sales volumes from 2008 until the first half of 2012, and there has been some limited growth since 2012. On a wider scale, the success and the feed-through in the leakage from the spending of recent Commonwealth games has reflected, I believe, in the fruit fall figures and the flow of business that embraces small independent operations. I want to praise the efforts of the retail sector that has tempted shoppers with not just keen prices and promotions, but with providing best value and customer service quality. The challenge, of course, is to maintain growth and ensure that we gain long-term benefits from the sector. There has also been the challenge to the convenience stores of the out-of-town supermarket developments in the past few years. However, I am told, in fact, in discussions that I have had with Lee Sparks, that the supermarkets seem to be re-appraising their investment strategies with supermarkets looking at town centre investment. That will provide the independence with new challenges to be confronted. The Scottish Government has recognised the importance of our retail in its town centre review, which included sectors to experts from all over the country, including some smaller representation of smaller independent units. We are looking at sharing resources and services, how we develop creativity and enterprise, and generally contributing to the health of our town centres. All of that is key. It is stated in that review that the Scottish Government and local authorities needed to recognise and prioritise the importance of town centres for sustainable economic growth. The light blood of that flows naturally from the independent convenience stores. We need to ensure that the retail sector, in all its dimensions, is seen to offer a good career path. To do so, we need to work with all the schools, development providers, our colleges and our employers, securing an aspiration for those who would like to be involved in retail and start at the bottom, if you like, working with independent convenience stores through the creation of the industry. The appropriate apprenticeships. Our independent retail plays a vital role in the success of our high streets. The town centre review group has made a great start in supporting the sector. The challenges remain. We must embrace those who know how to run independent convenience stores, and there are great opportunities to grow that particular sector. I thank the Scottish Grocer Federation, the National Federation of Retail News Agents and the Carnegie Trust for their very helpful briefings, which many members will have taken advantage of in preparing for this debate. We are right to talk about the enormous contribution made to the economy by independent retailers. Some figures have been put out there already, but they are worth repeating. 41,000 people across Scotland employed in the sector, 5,500 stores. Most importantly, there is a multiplier effect from those stores. We heard about the stats of investment in stores, but that leads on to local trades, shop fitters and suppliers around the local area. It is a matter of fact that a far higher percentage of each pound spent by any local retailer circulates back to the local economy when you compare it to an out-of-town centre or to online. Margaret McCulloch touched on that point too, but I do not know whether that is by accident or by design, but it is remarkable that the number of people who get their very first job, their very first opportunity on the employment ladder in some kind of independent retailer—again, that is something that ought to be recognised and applauded. They contribute nationally and locally, but they are also more than that. They are also a local community hub wherever they happen to be located. We heard in the briefings about the in-store post offices, the ATMs and the bill payment services, and some of Gordon MacDonald touched on the fact that they become, in many cases, the main source of advertising for the local community. If you want to find out when the gala day is and what is happening, if you want to find out who are the best local suppliers to speak to for whatever trade, if you want to find out what classes or training events are on, your local retailer is very often the first port of call and also the best port of call. Many of them might know that they do this for free, but even those that charge do so at very reasonable and very, very fair rates. Also, while I have no scientific evidence to back this up—only personal experience—those who own and operate independent retailers in their community are far more likely, in my experience, to either be involved in or, indeed, to run their local community council. So, when you wrap all that together, they genuinely give back in a way that many other businesses either do not or do so far less. Some of the challenges that the sector faces have been touched upon, but there are others that have not been mentioned yet, and I will take time to do that just now. In terms of the challenges that are mentioned by businesses, business rates are very near the top of the pile. The Scottish Government should be applauded for the small business bonus. We supported it from day one, we encouraged it to accelerate it and it has, I think, been a resounding success. The question for the minister, the question for all of us now, is where next with the small business bonus? Can we raise the thresholds so that more local retailers come in underneath the threshold and actually benefit from it? Can we have more of a stepped system so that if you are just above it, you do not miss out entirely? You still get some benefit from the small business bonus. Is there something that we can do to slow down the rises in business rates? I do not offer all the answers, but I think that there are questions that all of us have to think about. In terms of regulation, Deputy Presiding Officer, I know that this is something that the minister has formed on, particularly in his previous role, but they have been hit retailers by tobacco, alcohol and environmental burdens. Some businesses are hit by one of them. Many independent retailers have been hit by all three over a fairly short space of time. Is there something that we can do to reduce the burden of regulation on them? It is more difficult to apply to regulation or to comply with regulations as a smaller business, as compared to a far larger one. Can we do more in closing, Deputy Presiding Officer, to help them on the issue of illicit trade? Something where the Scottish Grosses Federation hosted an excellent session within the Parliament just a couple of months ago. Illicit trade damages legitimate businesses, harms consumers and helps those who are trying to get round the rules. Any of those issues that can be touched upon by the minister will be gratefully received. I would like to congratulate my colleague, Gordon MacDonald, on securing today's debate. The benefits of convenience stores in Scotland are many, as we have heard already in the debate. The boost employment allows people to start up and develop their own business, enhance the Scottish economy and serve local communities. In Scotland now has more convenience stores per head of population than in the other part of the UK, with one shop per 949 people. They are indeed the lifeblood of many of our communities, and convenience stores employ approximately 1.6 per cent of the Scottish working population. The benefit of independent stores to ethnic minorities is greater than for any other group, as 60 per cent of Chinese and 70 per cent of Pakistanis live in Scotland work in small businesses. Asians make up only 2.7 per cent of the Scottish population but are a vital asset to the independent convenience store sector. Many independent convenience stores are family businesses, and of those, almost two thirds employ at least one other family member. Independent convenience stores, through hard work of the owners and staff, contribute to economic growth. Industry body IGD has said that the value of the convenience sector across the UK will grow from £35.6 billion in 2013 to £46.2 billion by 2013. That is also reinforced by the fact that, despite an increase in the number of other retail sector shop closures, convenience stores have not followed the same trend with growth in the number of shop openings during January to March 2014 of 10 per cent. Convenience stores contribute to local economy. It is important to reiterate what has already been said by previous speakers, including Gavin Brown and Mr MacDonald. They contribute to local economy by using local tradesmen, produce suppliers, shop fitters, garages, legal and accountancy firms. Added services and facilities may be a crucial reason as to why the sector is growing. They are very fleet on foot on their feet. With developments such as the introduction of click and collect, courier delivery and online shopping return services and secure ETMs, footfall will hopefully increase. Independent convenience stores are run primarily by their owners. One-third of women, one in 10, are under 30, whilst one per cent are over 60. One in five have been in business for less than half a decade. One in five of those owners are more than one shop. Clearly, a vast range of people are able to operate and develop their own store, thus demonstrating business opportunities in the sector. More than three quarters of owners are the first generation of their family to own or own a business, creating sustainable jobs for generations of families as well as others in their community. It is, of course, no easy life, and many hours of service have to be dedicated to ensuring customer access for the optimum number of hours and the reasonable footfalls essential to any kind of living, let alone a good one. Many consumers appreciate the importance of supporting their local businesses, not just because of proximity. Consumer trust in local shops continues to be consistently higher than, for example, in supermarkets. A survey of more than 7,800 shops run by independent and corporate retailers reported that, despite competition among major supermarkets, local shops are as important to the community as the bobby on the beat. Customers become familiar with the staff, as has been pointed out by Margaret McCulloch. Often people also meet there for a wee bit of anata and a gossip, and many people are attracted by the range and flexibility of the stock in which local convenience stores can get in. The aid development growth in areas such as employment and economy have a very positive impact. They are a vital community asset and enhance community resilience, social and economic wellbeing, the social value of enterprise and innovation and the importance to the town and village shops. The fabric of Scotland cannot be underestimated, but, as we have heard, there are storm crowds on the horizon. Profits are being squeezed. We have increased costs at this time, from energy costs to the cost of basic produce. Challenging times for all people in society can be very difficult for shop owners to pass on all increased costs to customers. I believe, as Mr Brown has said, that we should look further where possible to reduce the rates burden. The small business bonus scheme has been a tremendous success. We know that the Small Business Federation has said that one in six shops went down the stank during the recession without it, but we look to see how we can move on from there. I reiterate what Gavin has said with regard to the issue of smuggling. We need to take more action to ensure that this does not impact more on the profitability of our small business store sector. I now invite Fergus Ewing to respond to the debate minister in seven minutes, if possible. I am very grateful to Gordon MacDonald for bringing this debate before us this evening, covering a matter of huge importance in every part of Scotland. I am very grateful for the contributions of all members across the chamber and different parties, recognising the significant role that independent retailers play in the economy. That is a very large role, and the retail sector has been quite resilient during the recession. It did not use the technical expression that Mr Gibson brought to the proceedings that go down the stank. It managed to overcome the buffeting of the recession very well. I am quite sure that that is in part because of the hard work, the effort that managers' owners of independent businesses brought to bear to ensure that they were offering an efficient, friendly, effective service, providing goods, fresh food and a variety of grocery products that people want throughout the country, giving people what they want and running the businesses well on a family basis.  As someone who was a small business person, I remember the pleasures and anxieties of running a business. Being responsible for the livelihood of a great many other people is a burden to bear. However, one that the independent retailers do extremely well in Scotland, as the figures show and the employment in the sector, as has been pointed out, is substantial indeed. Of course, there are challenges in facing all businesses in Scotland, and the burden of taxation is one of them. Various members, starting off with Gordon MacDonald, mentioned business rates, and Gavin Brown particularly mentioned that. We are pleased that, first of all, we reinstated the uniform business rate when we came into office in 2007 and we did that in 2008. That ended an extra overtaxation of up to eight or nine per cent, I believe the figure was, and that made a contribution since then. I am very pleased that we have had a measure of cross-party support for our introduction of the small business bonus, which now assists over 92,000 businesses, two in five premises. Mr Brown puts his points very reasonably, suggesting that we would like to do more. We would like to do more. There is no question about that. That applies across the board, because the regret many businesses for whom business rates is a significant burden. It has to be paid irrespective of turnover. It is a fixed sum based on the notional rental or the rateable value. That is a large and looming burden in running a business. I should say, out of fairness and particularly since we know that there are some of the independent retailers in the gallery, that I would be somewhat pleasantly surprised if the independent retailers were eligible for the small business bonus to be quite candid about that. That is because they are for the smallest businesses and 92,002 out of five receive that benefit. Mr Brown said, could we raise the thresholds and what about stepping? As far as the thresholds is concerned, on 11 December, John Swinney announced an expansion of the business bonus by increasing the upper threshold for businesses with multiple properties from 25,000 to 35,000. That was estimated to extend the benefit to over 4,000 additional eligible properties for the lifetime of this Parliament. We also have more generous provision with regard to empty relief than south of the border. Mr Mackay has introduced a new incentive fresh start relief, 50 per cent for 12 months, when some long-term empty property shops and offices become newly occupied. That would affect perhaps a small number of businesses but is there nonetheless to encourage and promote additional economic activity in the retail sector. Reference was made to regulation. Red tape was one of the references that the Scottish Government brought in. I was the pilot of this, the regulatory reform Scotland bill, which will make all public bodies subject to the duty of considering the economic impact of what they do. That general principle is one whose introduction was long overdue. It is not yet in force but we expect regulators in particular to act as though it were in force right now. That means that when taking decisions, public bodies must take account of the economic impact, how they affect existing businesses. We must actively seek to ascertain how decisions they take will affect businesses. In town centre planning, my colleague Janet Mackay in July with COSLA announced the town centre first principle. Perhaps reflecting some of the remarks that Mr MacDonald made about supermarkets, the recognition of the town centre first principle acknowledges that in some town centres things are very difficult indeed. Empty shops, a great deal of charity shops and perhaps a lack of choice to individuals in many towns throughout the country. I am not going to name them all but I think that this is a tendency that we have noticed and therefore the town centre first principle I think is a recognition and principle that we need to redress the balance against perhaps the dominance of the supermarkets as was mentioned by a great many speakers. Of course, my colleague Mackay mentioned new initiatives and we have had the small business bonus, the empty relief, the fresh start relief, we have got the business improvement districts, the town centre first principles. In one sense I suspect that for those in business here most of those are in practical day-to-day terms probably not of much relevance. And also those in business do not really expect or want government to come along with an occasional check abung a grant. That's not really what businesses want. They don't want initiatives where there are handouts. They expect to run profitably, to run the show themselves and to offer a good service to their customers and thereby make a decent living out of that and to look after their staff well and provide appropriate training. They're not looking for initiatives in the sense of handouts and I'm sure that wasn't what was suggested. But of course we are very keen in the Scottish government to work with the representative bodies as reference was made to the grossers federation representatives of news agents and also the Scottish retail consortium and I therefore do with my colleagues seek to have close relations with them and that is something that I will continue to do. And to the independent retailers and their sector in general I would say that I have worked with individual businesses not least in my constituency to help them get access to other services to become a post office to retain their lottery outlet for example and other matters of that sort. But if there are general matters or initiatives or measures which are creating barriers to success that are identified by independent retailers then I would be extremely happy to work with their representatives to that end and I'm grateful that Mr McDonald has given me the opportunity clearly to make that pledge. I also represent the initiatives that have been taken by the Scottish grossers federation recently in terms of healthy living and healthy food which we very much welcome. We've covered quite a wide range of topics this evening but I think that most important of all we've all of us across the different parties expressed our support, our appreciation for the economic, the practical and also the social, as Mary McCulloch said, function that smaller shops, independent shops do competing against megaliths, dominant supermarkets who have different practices and purchasing practices. And therefore find it from time to time difficult to compete in that way but nonetheless by their personal, local, effective, high quality service have proven to be very resilient in hard times and the message from all of us is long may that continue.