Now, let me repeat this point about Lyapunov stability and the theorems in the simplest form in which we looked at it yesterday last time. If this is a critical point, say the origin is a critical point of a system and you have some neighborhood of this origin and I would like to understand the stability or otherwise of the system in this neighborhood. Then the statement was for the dynamical system given by x dot equal to f of x, one constructs or considers or pulls out of the hat an auxiliary function v of x called the Lyapunov function which has the following properties. The properties are if v of x is such that v of 0 is 0 and v of x is greater than 0 not equal to 0 anywhere but strictly positive for all x not equal to 0 in this neighborhood. So, let us call this neighborhood u then v of x is a positive definite function if it is less than 0 for all x not equal to 0 in u it is a negative definite function if it is greater than or equal to 0 you can find other 0s in the neighborhood u then it is semi definite and if it is less than or equal to 0 and you can find other 0s then it is negative semi definite with these statements with this with these definitions the statements we made were the following. So, this is positive then v is positive definite and so on. Now the statements made were the following the rate of change of this function v v dot turns out to be my very simple piece of algebra this turns out to be grad v dot f and the statements were if you have a positive definite Lyapunov function and v dot is strictly negative in this neighborhood namely its negative definite then the critical point is asymptotically stable on the other hand if you have a positive definite Lyapunov function and v dot is only negative semi definite in other words there exist other points other than the origin where v dot vanishes then what you can assert is that the critical point is stable you cannot assert that it is asymptotically stable and if you have a positive definite v and v dot happens to be positive also then it is unstable or if v is negative definite and v dot is negative then it is unstable this is for sure. Now in the examples we looked at we came across instances we came across one instance where in the neighborhood of the origin for a positive definite Lyapunov function we discovered that in the presence of damping v dot was minus gamma y squared where gamma was the damping coefficient and I commented that this only enables us to deduce that this particular critical point at the origin is not necessarily asymptotically stable but merely stable because we know that minus gamma y squared is vanishes everywhere on the x axis so we can assert that it is stable but we cannot go on to assert that it is asymptotically stable which is what the actual critical point was in the presence of damping. So it is not that now now it might turn out it might turn out that you have a better Lyapunov function not the one that we chose but a better Lyapunov function which is positive definite for which you can show that v dot is actually negative definite in which case you can assert that this is asymptotically stable as well it is not that this thing exhausts all possibilities it is just that you could choose a bad Lyapunov function in which case you cannot make any statement at all could change sign or you could choose one which turns out to be only negative or positive semi definite in which case you can assert something about stability but not about asymptotic stability but the point is if you can find a Lyapunov function which is such that it is positive definite and v dot is negative definite in the neighborhood then you can definitely assert that the critical point is asymptotically stable now we have seen there are critical points which are asymptotically stable but not stable and the critical points which are stable but not asymptotically stable and certainly there are critical points which are both stable as well as asymptotically stable and the corresponding Lyapunov functions would reflect these possibilities in various cases. So these are not mutually exclusive categories in that sense something can be stable as well as asymptotically stable but something which is stable does not have to be asymptotically stable and something which is asymptotically stable does not have to be stable it is true that you have proved something is asymptotically stable you made a very powerful statement you said that points going to fall into this a tractor eventually so it is a very strong statement in that sense stability is weaker than asymptotic stability it says things do not go away from the neighborhood even if you wait long enough once they have entered the neighborhood but it does not say they are going to fall into the critical point at all. So I hope this clarifies to some extent the question that you had so let us take an example now go back a little bit and look at a mechanical example of a bifurcation and I will work this problem out fully because it is a standard problem so a mechanical we will see there is a bifurcation and I am going to ask you at the end what sort of bifurcation this system displays and it is an extremely simple one but it is an illustrative one because it says something about Lagrangian mechanics as well which we have not really discussed in detail in this course at all but this would give us an opportunity to in a free capitulate or ideas about Lagrangian mechanics it is a following problem I have a hoop of wire which is say in the shape of a circle and I have a bead on this hoop of wire a bead of mass m which moves frictionlessly on this ring of wire and let us say this hoop is in a vertical plane and the bead is constrained to move on this circle and let us choose for convenience coordinates such that the origin of coordinates is at the bottom position of this hoop which is in a vertical plane and let us suppose this plane is the YZ plane for instance and this hoop has a radius R what sort of motion does this constrained system have this point mass under gravity if I take it up here and let go it is going to oscillate without any dissipation back and forth about this point here and it is a constrained system because it is constrained to move on this hoop as it stands how many degrees of freedom does this particle have now how many independent degrees of freedom does it have well I fixed the plane of this hoop and that is the YZ plane so to start with this particle has only Y and Z coordinates the X coordinate is always 0 moreover I am saying that it moves on the circle so it is clear on the circle Z is a function of Y and therefore it has one independent degree of freedom which you can choose in many ways a convenient one would be to choose the angle it makes with the vertical axis its instantaneous position makes with the vertical axis that would be convenient degree of freedom but now I make the problem a little more interesting by saying that this hoop is set rotating about the Z axis with uniform angular speed omega so the entire hoop rotates about this diameter with uniform angular speed omega and so does the bead also rotate and the question is can we do the dynamics can we solve for the dynamics of this point particle of this mass with this time dependent constraint namely it is sitting on this hoop but the hoop itself is rotating at a constant angular speed omega so this is the problem to start with so what do we do we start by writing down the Lagrangian of the particle which is in this case the kinetic energy minus the potential energy of the particle and let us do this by saying L which would initially be a function of the XYZ coordinates of this particle because now that it is set rotating there is also an X axis coming out of the board you have a right handed coordinate system here and this L would be a function of all the coordinates as well as the generalized velocities corresponding velocities and what would this be well to start with it would be equal to one half m times the kinetic energy which I could write as X dot squared plus Y dot squared plus Z dot squared in Cartesian coordinates minus the potential energy and what is the potential energy of this hoop well it when it is here at Z equal to 0 potential energy could be taken to be 0 and then when it is at a height Z above the XY plane its potential energy is just mg Z so this is minus mg Z to start with but then the hoop is the bead is constrained to move on this hoop and what sort of coordinate system should I choose when I do that should I choose spherical polar coordinates there is an axial symmetry here it is rotating about the Z axis so what would be a good set of coordinates to choose cylindrical polar coordinates so let us choose cylindrical polar coordinates and what are these coordinates well the radial distance the axial distance of this particle from the Z axis which is equal to the square of that distance is X squared plus Y squared that would be one of them so let us choose rho there let us choose the azimuthal angle phi so X is rho cos phi and Y is rho sin phi a plane polar coordinates in the XY plane and of course Z itself these would be the cylindrical polar coordinates and what is the kinetic energy in such a case so L becomes equal to one half m what does X dot squared plus X dot squared plus Y dot squared become in cylindrical in polar coordinates rho and phi exactly there is an radial velocity rho dot squared plus yes indeed rho squared phi squared because that is what X dot squared plus Y dot squared become when you have when you go from X and Y to plane polar coordinates rho and phi plus Z dot squared is always present this is there minus mg Z this is what the Lagrangian would be but now we have to impose the constraint that Z is a function of the other coordinates because it has to remain on the circle and we need an equation for this circle itself at any instant of time now what is the equation to this circle the circle has a center at Z equal to R and rho equal to 0 and in the static case when it was not moving what would have been the equation to this circle it is a circle in the YZ plane with the center at Y equal to 0 Z equal to R and its radius is R so indeed this circle would have been for example it would have been Y squared plus Z minus R whole squared equal to R squared that would have been the equation of this circle when it is static when it is in the YZ plane the center of the circle is at Y equal to 0 and Z equal to R now when it starts rotating there is this circular action symmetry about the Z axis in the XY plane and therefore what would this equation become instead of Y squared you simply have to replace it by the actual distance of this particle from this axis it becomes rho squared indeed so this is the constraint therefore if you solve for Z it says Z minus R whole squared equal to R squared minus rho squared or Z equal to R plus or minus square root of R squared minus rho squared which root should I choose the two roots for Z and what do they correspond to here yeah if you fix a value of the other coordinate you could either be here or here and we are interested in oscillations about this point here so we choose the plus root or the minus root we choose the minus root so this is the root we choose that is the physical root because when rho is 0 this is 0 Z is 0 sitting here so let us get rid of this plus and write minus for the branch we are interested in and now we are ready to eliminate out here this is phi dot squared by the way that is got to have dimensions of angle of velocity squared so that is the kinetic energy term and when I rotate it then what does phi become what does phi dot become if I rotate it at constant angular speed omega phi dot is omega itself and phi dot is identically equal to omega therefore we are now ready to write our Lagrangian down but we need to use this in order to eliminate Z dot from the problem and Z all together so if Z is this then it implies that Z dot is equal to minus 1 over twice square root of r squared minus rho squared it is rho that changes with time r is a constant and I am differentiating this with respect to time so you differentiate this and then you differentiate minus rho squared which is equal to minus 2 rho rho dot so that gives us equal to rho rho dot divided by square root of r squared minus rho squared we can put all that in here and we finally have our Lagrangian which is a function of rho and rho dot alone just the radial coordinate rho and it is corresponding generalized velocity rho dot and this becomes equal to 1 half m rho dot squared plus rho squared omega squared because I get rid of phi dot squared plus Z dot squared which as you can see is rho squared rho dot squared divided by r squared minus rho squared this term minus mg times Z which is mg r plus mg times square root of r squared minus rho squared and we can simplify this a little bit and get a Lagrangian which is equal to 1 half m we combine these two terms here and it is clear that you get just m r squared rho dot squared divided by r squared minus rho squared out here plus 1 half m rho squared omega squared that is this term plus mg square root of r squared minus rho squared minus mg r this is just a constant in a relevant constant that is how Lagrangian with 1 degree of freedom independent degree of freedom rho and it is a function of rho and the corresponding velocity rho dot the constraint has been taken into account automatically and notice that in the so-called kinetic energy term you have a coefficient which depends on the coordinate this is typical of a constraint problem you eliminated the constraint and it has become a function it is true that it is quadratic in the generalized velocity but it is got a coefficient which depends on the coordinate itself and there is this extra term which has emerged this looks like part of the potential what would this be due to physically what is happening what is that going to give us this gives us the effect of the quote unquote centrifugal acceleration that is exactly what it is as you can see because it is a rotating coordinate you end up automatically with the term which will give us this pseudo force once you write the equation of motion down and that is one of the advantages of the Lagrangian formalism where constraints can be taken into account easily and non inertial forces are automatically included it is emerged automatically once we wrote down what this thing was the kinetic energy was now what is the equation of motion we know that in a Lagrangian system of this kind the equation of motion if you have L as a function of q and q dot then delta L over delta rho the partial derivative of L with respect to the coordinate is equal to d over dt delta L over delta rho dot this is the Euler Lagrange equation for this simple system that is the equation of motion which is supposed to give you the correct counterpart of Newton's equation of motion well all we need to do is to apply that here so this would imply if I differentiate I get delta L over delta rho there is a term here it sits here differentiate this you have to differentiate this you have to differentiate this and equate it to d over dt delta L by delta rho dot which means twice or whatever rho dot divided by this and you get the equation of motion but we are interested in writing things down in the Hamiltonian framework because we would like to write everything as first order equations in this case and so we make a change from the Lagrangian to the Hamiltonian formalism and if you recall if you start with an L which is a function of q and q dot then the way to go to the Hamiltonian formalism is to go from this function to a Hamiltonian which is a function of q and p by making the following transformation you first define a generalized momentum which is delta L over delta q and that is the definition of the generalized momentum given a Lagrangian the momentum conjugate to a particular coordinate q is the partial derivative of L with respect to q then once you have done that with respect to q dot sorry with respect to the velocity then this Hamiltonian is equal to p q dot minus L and it is a transformation which takes you from the Lagrangian to the Hamiltonian. However you have to be careful that you can that you invert this equation this in general gives you p as a function of q and q dot because L is a function of q and q dot but in going to the Hamiltonian which is a function of q and p you must eliminate q dot and you do that by solving this equation for q dot as a function of q and p so this gives you p as a function of q and q dot you must now write q dot as a function of p and q substitute that here and wherever q dot appears in L substitute for q dot in terms of p and q and you have a Hamiltonian and you are guaranteed after that Hamilton's equations of motion. So this is the procedure for going to a Hamiltonian let us do that here let us apply that here so what we need to do is to first find the momentum conjugate to the generalized coordinate row I should really write p sub row it is a radial kind of momentum but since there is only one degree of freedom let me just call it p and this is equal to delta L over delta rho dot and what is that equal to it says differentiate this pretending everything else is a constant except rho dot the two cancels and you get m r squared rho dot divided by r squared minus rho squared please notice now that we are not turning out it is not turning out that the momentum is some mass times a velocity not at all it is a function of the coordinates times the corresponding generalized velocity this again is typical of a problem which involves constraints but that is the definition of the canonical momentum conjugate to rho once you have this of course you can invert this trivial in this case this implies that rho dot is p times r squared minus rho squared divided by m r squared and we can write down therefore the Hamiltonian as a function of rho and p which is equal to p rho dot minus L we are ready to write down then the Hamiltonian therefore h of p, rho is equal to p times rho dot and rho dot is given by this so it is p squared times r squared minus rho squared divided by m r squared minus the Lagrangian but in the Lagrangian you must replace rho dot in favor of p since it must be a function of rho and p alone if you did that you get p times rho dot squared so it is going to give you when I square this I get an m r squared whole squared that cancels against this and then r squared minus rho squared the whole squared one power cancels against this and as a factor a half so it is clear we have one of this minus half of the same therefore it is just this that takes care of the kinetic energy part minus one half m rho squared omega squared that is this quantity minus mg square root of r squared minus rho squared plus mg r I hope you kept track of all the minus signs because if not we are going to get wrong answers so let us see what happens that is our Hamiltonian so we can get rid of this now we are ready to write down Hamilton's equation some motion now it is a non-linear Hamiltonian it certainly does not look like simple harmonic motion or anything like that and notice the signs of these terms now what are the equations of motion well we know that rho dot is delta H over delta rho that is the first of Hamilton's equations and that of course is going to give us something we already knew because delta H over delta p and this is something we already knew because if I differentiate this with respect to p it gives p times r squared minus rho squared over m r squared but we already knew that p was rho dot times m r squared over r squared minus rho squared that is how we got p so this is not telling us very much nothing new in the case it is just the old relation between the generalized momentum and the generalized velocity but the other equation says something interesting and that is p dot is minus delta H over delta rho and what is this equal to now we have to differentiate everything I differentiate with respect to rho and I get minus p squared rho divided by m r squared it is a plus sign because I already have a minus sign here and I have another one here so it is a plus sign and then plus m rho omega squared this term here minus mg divided by twice square root r squared minus rho squared I am not very happy with what happened I hope we did not leave out any minus sign anywhere in between because if we have then we are going to have an unphysical result here so I need to differentiate this and it came from minus should be a plus should that be a plus we started off by writing L equal to the kinetic energy minus mg z that was the potential energy but if you recall z itself was r minus square root r squared minus rho squared and so it gives us this is equal to various terms plus mg square root of r squared minus rho squared in the Lagrangian so when I wrote the Hamiltonian I put a minus sign there so that gives me a minus similarly similarly the original term in the Lagrangian was of the form one half m rho squared omega squared the phi dot squared term in the Lagrangian with a plus sign and when we write a minus for the Hamiltonian it is pq dot minus L that to appears with a minus sign so we are okay all right now let us differentiate this so we were in the middle of that and then I need to differentiate it and change the sign so there is a minus here and a minus here both these things go away and you get a plus and then I have to differentiate this with respect to rho so that is equal to a minus twice rho and therefore the two goes away and the minus turns up and let us take the rho out of the bracket and the m as well so these are the two equations of motion they are the Hamilton equations of motion again you see they are badly non-linear equations because you have all kinds of powers here you have a cube here and then you have the third power here and then you have this one over the square root and so on so it is really a non-linear problem in this sense now the question is where are the critical points of the system they happen when the right hand side is equal to 0 of course you put r equal to rho you end up with an infinity here so the critical point would occur at p equal to 0 and let us write that down p equal to 0 and certainly rho equal to 0 is a critical point because if I put rho equal to 0 this vanishes and this vanishes as well rho equal to 0 that is always a critical point that corresponds to this point the system at rest here which we know is an equilibrium point whether it is stable or not we do not know is there any other critical point in the system so you said p equal to 0 and this term goes away and you see this could vanish this bracket here could vanish when does that happen so this would vanish with provided omega squared minus g over r squared minus rho squared is 0 or omega squared was equal to this or r squared minus rho squared is equal to g squared over omega 4 that is the root here or rho squared is r squared minus g squared over omega to the power 4 provided this is positive provided this quantity is positive otherwise it is not going to happen it is not a real root what would this imply when is this going to be positive so another solution non-trivial solution for rho at the square root of this value provided r squared was greater than g squared over omega 4 or since omega is the quantity that you have in control r and g are constants this implies that you have a second solution I need this equation so I am going to ask you to dictate it to me afterwards so you have another critical point provided omega to the 4 is greater than g squared over r squared or omega is greater than omega critical which is equal to square root of g over r then you have another solution it gives you something else altogether we need to compute what that is so that root is given by this let us write it in reasonable form so you have a second root at rho squared another cp at p equal to 0 and rho equal to the square root of this so let us take out an r completely r times square root of 1 minus g squared over r squared over omega 4 but g squared over r squared is omega critical to the power 4 right so this is equal to omega critical squared omega critical 4 omega c is some constant its square root of g over r now what kind of stability do we have here what can we say about the stability of these solutions so we have to go back and write the equations of motion down and see what it does for linear stability and if you recall we had p we had rho dot is equal to m r squared p over r squared my was this right what was what was rho dot p over mr squared that was the equation and the other equation was p dot is equal to now you have to tell me what it was I believe there was a p squared rho over mr squared plus m rho omega squared plus minus m g rho root of r squared minus rho squared and that was it so what we did was to take the rho out and the m out so we wrote it in this form so you could rewrite this as in terms of rho over r and the factor g over r you could write in terms of omega c but anyway let's leave it in this fashion so we have one critical point here and another one there and let's look at the stability of these critical points so let's linearize rho equal to 0 p equal to 0 and then the equations of motion are rho dot is approximately equal to the r squared cancels p over m which is what you'd expect because you'd expect very close to this for small oscillations you'd expect that rho dot is just p over m or p is m rho dot as usual the nonlinearity doesn't play a role there and then this equation this term is already third order we're going to look at it near rho equal to 0 this can be got rid of and then what you get p dot is approximately equal to so this can be dropped because it's second order in rho this can be dropped because it's got third order terms and then you have omega squared minus g over r times rho but what's g over r omega c squared so you have m times omega squared minus omega c squared times rho what kind of critical point is that well the linearized matrix l is of the form 0 1 over m m times omega squared minus omega c squared and then a 0 here what kind of critical point is that it depends on the value of omega whether omega is greater than omega c or less than omega c it depends on that completely this determines now the stability at the origin so let's look at the case omega less than omega c first so you start rotating at a small speed first and see what happens then so I think we need to retain this so let's do this therefore omega less than omega c the eigenvalues of l well what are the eigenvalues of this matrix omega is less than omega c plus or minus with an i there right because omega is less than omega c so what kind of roots are these pure imaginary so it's center absolutely so you have a center which is stable at the origin this means that if you rotate this hoop at sufficiently slow speed when as it rotates this point is a stable center so if you start here it oscillates about this point while rotating so it comes back and you have undamped oscillations about the bottom that's your critical point which is stable and you can show simultaneously that the other critical point is unstable on the other hand what happens if omega is greater than omega c so this is stable and if omega is greater than omega c then the eigenvalues of l plus or minus square root of omega squared minus omega c squared what kind of critical point is this it's a saddle point and therefore unstable the origin becomes unstable you would have to do the same thing about this point to discover that this becomes stable so this implies yes you're right second critical point doesn't exist at all if omega is less than omega c because this is pure imaginary absolutely right yeah so it says wherever you are it's going to oscillate about the bottom most position and these are stable oscillations on the other hand the moment you have omega crossing the value omega c the origin becomes unstable and a new stable critical point emerges so this is going to help us find what kind of bifurcation this is and this is the interesting question but before we do that let's see what happens let's draw this bifurcation diagram and see what happens so here's the bifurcation diagram so it's not an exchange of stability transition at all it's not an exchange of stability bifurcation because this one doesn't exist only if omega is greater than omega c okay bifurcation diagram is as follows we plot rho equilibrium the equilibrium distance radial distance axial distance from the z axis as a function of omega which takes on only positive values and here's omega c p is always 0 so I don't plot that it comes out of this axis but p is always 0 at equilibrium and then it says this is your stable root okay and we know it becomes a saddle point as soon as omega exceeds omega c so the rest of this diagram should really be a dotted line and the new root that comes out is this what's the graph here what kind of behavior does it have at omega equal to omega c it's 0 it's bifurcating out of this point that's clear what happens as omega tends to infinity what happens to that equilibrium value of rho it tends to r so it's very clear physically that this is what's happening it's clear that this hoop as you rotate it faster and faster with some angular frequency omega this is the furthest it can actually go it can't go beyond that it's constrained to remain on this hoop so once it zooms up to this point it has to remain there and this is a stable oscillatory point so here if this is an equilibrium value then if you displace the hoop about it as it's rotating it would actually do this about this point so it gets shot out a certain distance from the bottom as long as you didn't start exactly at this point then of course it remains at that saddle point it's an unstable equilibrium point but a little bit of perturbation puts you away and it zooms up crawls up to this point and oscillates about this point so this goes off and asymptotically hits the value r as omega becomes very large now what kind of behavior does it have at this threshold would the slope be infinite would the slope be finite would it do this would it do this would it do this what would it do that's crucial to know yes we indeed we need to know exactly what the behavior is like and that's easily seen because you can see that this row equilibrium is in fact equal to r times square root of omega 4 minus omega c 4 divided by omega squared take it out of the square root and near omega equal to omega c it's clear you can write this as omega squared plus omega c squared times omega squared minus omega c squared and near omega equal to omega c you can replace this with omega c that with omega c and you basically have this minus that which in turn you could write as omega minus omega c times omega plus omega c and this harmless factor you could write as twice omega c so the whole thing goes like the square root of omega minus omega c so what's the shape of the curve here what's the slope what's the slope of y equal to square root of x at x equal to 0 it's infinite so this is in fact the thing like this goes off and this thing here in this neighborhood it goes like omega minus omega c to the power of that's the exponent and this is the stable root what kind of bifurcation is this is it an exchange of stability bifurcation no because there isn't a critical point at all to start with the other one doesn't exist for omega less than omega c is it a saddle node bifurcation that means a spare a stable unstable pair of critical point emerges as you cross the bifurcation value but that's not true because you already had one here so it's not saddle node it's not exchange of stability it must be a pitchfork bifurcation but then a pitchfork bifurcation is one where a stable point bifurcates into an unstable point and a pair of stable equilibrium points where is the other pair one where's the other member of this pair pardon me why is it not physically realistic because we because row can't be negative we plotted a coordinate where it can't be negative but it's indeed a pitchfork bifurcation because there are indeed two possible states of this suppose you color this part of the hoop green and this part of the hoop red certainly something that sticks here and something that sticks here can be distinguished between depending on what kind of infinitesimal perturbation you had here it would either zoom up to the right hand side or the left hand side the green side or the red side and therefore it is in fact the situation where you have two different equilibrium points one on one side of the hoop and the other on the other side of the hoop and you don't see that in this picture because this is just the actual distance indeed there are two critical points that have emerged from a single one so this is very much a pitchfork bifurcation you could use this to measure little g to measure everything else and you're sure that you don't have friction you could use this actually to measure little g because you could measure where the equilibrium point is and this will tell you directly in terms of r and g and omega c tells you so that's one possible thing you could also put this in a ball in a spherical ball and rotate it rotate the ball you'd still have this problem but this is the way a constrained mechanical problem is simplified when you use the Lagrangian framework and its stability is easily analyzed when you use the Hamiltonian framework in this fashion you could now ask what happens if I had different shapes I don't necessarily have to have a circular hoop so let's see what happens quickly see what happens if you have a slightly different shape of hoop one possibility is to write is to use for instance a parabolic hoop so here's z and once I set it rotating in this fashion a parabolic hoop if this z is half ky squared for example then on this parabola z is equal to one half k rho squared because that's why squared is replaced by x squared plus y squared once you set it rotating by cylindrical symmetry and then you see that z dot is equal to k rho dot in this fashion we run through our steps quickly then the Lagrangian becomes equal to one half m they certainly a rho dot squared plus this is z dot squared but the z dot squared is that k squared rho squared if I pull the rho dot squared out this is one plus this times rho dot squared in this fashion plus in this case one half m rho squared omega squared that came from the rho squared phi dot squared term in the kinetic energy minus mg times z but this is over 2 k rho squared this is what the Lagrangian is and what's the Hamiltonian well to run through these steps quickly the conjugate momentum p is delta l over delta rho dot and that if I differentiate here is m times 1 plus k squared rho squared rho dot or rho dot is p over m 1 plus k squared rho squared that's the conjugate momentum again it depends on the coordinate so what's the Hamiltonian this is equal to p rho dot minus l p rho dot gives you p squared over m times 1 plus k squared rho squared here first portion minus l therefore you got to subtract this minus this term times the square of this so you have a p squared over m squared that gives you 2m and therefore it's clear that you just get twice as before plus one half m rho squared omega squared minus thank you minus this plus mg over twice k rho squared like so so we can simplify this a little bit take out the m rho squared over 2 and then write this as omega squared minus gk that's it and what are the equations of motion so again we have rho dot is delta h over delta p that's not going to give us anything new except this whole thing p over m into 1 plus k squared rho squared but then p dot is minus delta h over delta rho and that gives you something non trivial this is equal to minus p squared over m 1 plus k squared rho squared whole squared multiplied by the derivative of this quantity which is equal to k squared rho this term I differentiate this and then I differentiate that and that gives me minus m rho omega squared I need a minus out here by differentiate so this becomes a plus what do you conclude what kind of critical points do you have and what's your conclusion you have a bifurcation here at all what happens now clearly p equal to 0 rho equal to 0 is a critical point so the linear matrix about that point so near so what happens well I mean it's clear this is going to this is a linear term and it's going to change sign depending on omega is less than square root of gk or greater than square root of gk so the linear term the linear linearized problem is p dot equal to p over m to first order and rho p rho dot sorry rho dot is this and raise this problem write it properly rho dot and p dot is equal to m times omega squared minus gk times rho so what do you conclude from this so l is 0 1 over m m times omega squared minus gk and a 0 here yes if omega squared if omega is less than so center for omega less than square root of gk you could have guessed that by dimensional arguments square root of gk is the only quantity of dimensions frequencies so other than omega itself so for omega less than square root of gk you end up with a center which means you have stable oscillations what happens if omega is greater than gk the square root of gk becomes a saddle becomes unstable but you don't have an alternative after that and what does the system do goes off to infinity it goes off to infinity exactly this critical point just becomes unstable that's it it was because you had the curvature in the circle was appropriate that you ended up with another stable point so it deeply depends on the function that you had it depends on whatever was multiplying it it depended on the shape of the wire the constraint which gave you which depended on the shape of the wire told you whether you had another solution or not yes if omega square equal to if omega equal to root gk yes well if omega equal to root gk this goes away the problem is not linearizable anymore it's intrinsically nonlinear as you can see so that's the marginal point at which the center is losing its stability at this point infinitesimally greater value of omega and you're off it goes off so you can only have stable oscillations about the center as long as omega is less than square root of gk I leave you to play with this and figure out what happens if instead of this kind of situation you had for example you took a stick and you put it on a stick just a rigid rod you put a bead on that and you rotated this about the vertical axis so that's just a linear function so the circle is special of course there are other functions for which this would happen but the circle was a simplest case and we didn't look at what would happen in the hoop problem if you started with something not on the lower half but up here and I leave you to analyze that it's fairly straightforward the only thing is in the constraint equation you have to write Z is R plus the square root of R squared minus rho squared and go through the algebra once again to see what would happen to a particle sitting here when you start rotating but the result we got was completely physical it said that initially you have for small velocities small angular velocities you have a stable oscillations about this origin and once you exceed a critical point then this gets flung out and you have oscillations about a new point and the largest value that point can have is in fact when you write here at this stage which happens when you have an unbounded angular speed and then it gets flung out as far as it can go here and we also found out what was the nature of this bifurcation diagram we found out what kind of exponent you had and we found that there was a kind of square root behavior the function of square root of omega minus omega c was way the way the new function the new solution took off so it's a fairly complete analysis although we didn't solve the equation of motion in general that's complicated because a motion in general could be very complicated depending on the initial conditions and you could translate it back to the original Cartesian coordinates but there isn't much percentage in doing that in this problem it's obviously of interest to stay with the cylindrical coordinates that we had yes pardon me yes you have yes you are now we didn't draw the phase diagram because we didn't really write down we didn't really draw the phase diagram at all because we didn't write the solutions down we just looked at what happens in the critical points and so on but you're right there's obviously a hyperbolic point at a certain stage and then there would be separators is coming out of it and so on so I leave this as an exercise to try and generate this but in practice although because the expression is fairly complicated it might be messy to do this but you can do this numerically can write these solutions down can solve this yes it's clear that once the critical point becomes unstable and it becomes a saddle point then there is no question of oscillation so absolutely so things would diverge yes we didn't look at the time period of small oscillations about this point we didn't do that we didn't write down what would happen but that's straightforward because for sufficiently sufficiently close to the origin for small oscillations it was like a simple harmonic oscillator problem for which you can straight away write down the solution you can write down the time period yes would it diverge yeah presumably this would end up diverging yes because it becomes unstable beyond that point but one should check this out explicitly we didn't do the large oscillations we kept things very close to the origin and that was it but now you make the amplitude larger and larger and then you begin to see what would happen your question is what happens to the actual time period of small oscillations itself in the linear problem yes indeed it diverges that you can see directly because let's go back his question is what happens to the time period of oscillations as omega hits omega C in the previous problem the circular hoop if you recall in the parabolic case yes indeed it becomes infinite as you can see immediately because here's our problem here's our here's our linear matrix so what are the eigenvalues lambda 1, 2 is equal to what are the eigenvalues of this matrix right yes okay now we're going to look at small values of omega to start with and therefore omega is less than the square root of this in which case let's write it as I times square root of GK minus omega squared and let's call omega C squared equal to GK right therefore this is equal to plus or minus I times square root of omega C squared minus omega squared those are the eigenvalues which means that the small oscillations will have e to the power plus or minus these eigenvalues multiplied by T and the time period of oscillation T is therefore proportional to 1 over square root of omega C squared minus omega square which diverges as omega hits omega C it is exactly his conjecture so it's clear that as you hit this critical point the time period diverges it's about to take off and beyond that it becomes unstable you could also have deduce this by looking at the shape of the potential does this correspond to what kind of thing does this correspond to when I put this in here and you have an equation which says rho dot is P over M and P dot approximately equal to and let's write this as omega C squared and keep omega less than omega C squared so minus M omega C squared minus omega squared times rho what kind of potential are we talking about it's like a simple harmonic oscillator with a potential this is an equation of motion so you have to integrate this so you get a potential effective potential we effective which is proportional to one half M omega C squared minus omega squared rho squared isn't it if I integrate this and change the sign right this is like the force f f of rho so V is equal to integral f of rho d rho with a minus sign since f is minus d V over d rho this is certainly true what kind of shape is this as a function of rho because we are looking at positive rho this is a parabola upwards provided omega smaller than omega C but it's getting flatter and flatter as omega approaches omega C and therefore the potential is starting to look like this and once omega exceeds omega C the potential inverts and inverted parabola is unstable at the origin so this is exactly what happens the stiffness is going down as omega approaches omega C it's becoming like a spring with a smaller and smaller spring constant gets more and more flabby and therefore the time period is longer and longer and it diverges at omega equal to omega C beyond that it's not going to sustain oscillations because it's turned around the other way okay that yeah but we haven't really written the phase portrait down just some things near the origin but this is good enough it tells us all that we need to know okay.


Hi and welcome to the session. Let us solve the following question together. It says find the range of each of the following functions. Now the function says 2-3x where x belongs to real number and is greater than 0 and the second function is x square plus 2 where x is a real number and the third function fx is equal to x where x is a real number. In this question we need to find the range for each of these sub parts one by one. So let us solve the question. Now in the first part we are given the function fx as 2-3x where x belongs to real number and is greater than 0 right. So for finding the range we have to find the values of fx for different values of x which belongs to the real number and is greater than 0. So we have fx is equal to 0. So that means 2-3x is equal to 0 that gives us the value of x as 2 by 3 right. Again if x is greater than 2 by 3 fx will always be negative right. So we conclude that fx can take all the real values between minus infinity to 2. Now why 2 is written over here? It's because we have taken the negative aspect of it but if we take the positive aspect of it that will be x is less than equal to 2 by 3 then our fx will always be in positive term and will be less than 2 right. So that is why here we have taken it from minus infinity to 2 and our range will be minus infinity to 2 that is with the open intervals. So this completes the first part of the question. Proceeding on with the second part here we are given the function fx as x square plus 2 where x is a real number. Now as x belongs to real number therefore x square is always greater than equal to 0 isn't it? So we can conclude that our fx is greater than equal to 2. So hence the range will be it will be the closed interval from one end and an open interval from the other end right. And now here if x is a real number fx can take all the real values that is greater than equal to 2 and hence the range of the function is the set of all real numbers that is greater than 2 so it is including 2 in it but then the interval can go on to the infinity and hence it is an open interval from this side. So this becomes the answer of our second part. Proceeding on with the third part we are given the function fx is equal to x and where x belongs to real number. So therefore it is a very simple fx belongs to all the real values and hence our range will be all the real values that it can take. So this completes the question that was given to us. I hope you enjoyed the session and make use of the definition of the range that is known to you. Bye for now.


Okay, so hello everybody, and thank you for the invitation. So I'm Nicolas Leverny, I'm currently in Marseille, and I will talk to you today about the impact of initial perturbations on the first upper passage time of random walks. So basically, I will consider a random walk, pretty general, looking for a target here in red. And I asked a question similar to what Satya talked before. How long will the worker take to find the target? You can rephrase the question, how long will the worker find the target for the first time? So that's why it's a question of first passage time that I will abbreviate FPT in the following slides. So this question of first passage actually appears in many domains, so you can ask the question what's the time for a predator to find a prey, or also what's the time for a transcription factor inside the nucleus to find a specific part of a strain of DNA. But also in a totally different domain, what's the time for a given asset price to reach a threshold. So this question are first passage time problems. A few definitions first. This first passage time is naturally random variable because the walk is itself random, so it has a given distribution that I will call f and a given mean value that is called the mean first passage time. And it is known that in case the walk is non-Markovian, the FPT statistics is ideal and trivial to determine it. And that's the purpose of my talk actually. So generally speaking, you can separate this FPT problem in two big classes. In the first class, the walker is not confined, so you can go everywhere before finding the target. And in this case, typically the density of FPT is algebraic with a decay exponent that is called the persistence exponent in case of Brownian motion, this theta is equal to one-half. And due to this heavy tail, the mean first passage time is infinite. And in this case, in this point, typically one is interested in the value of this exponent, this persistence exponent. On the other side, if your walk is confined, then your walker cannot go very far before touching the target and the FPT density is typically exponential with a finite mean first passage time and one is interested in the computation of this mean first passage time. And actually this second part will mainly be the purpose of the talk of Thomas Guerin tomorrow. And in this talk, I will focus on this first part, unconfined random walk. And in particular, I will focus on this exponent, theta. And if you want a nice review about this exponent and the different way you can use to calculate it, you can read this good review about it, written by Satya in particular. Okay. So the motivation for the walk I will present to you today actually is this paper, published at the end of the 90s, about fluctuating interfaces. So we consider an interface that is fluctuating with classical dynamics with the deterministic part here. So h is the height of a given interface. You have this fractional operator here. For example, if you take h equal 1 divided by 4, you have an Edward Wilkinson membrane or in the context of polymer, it's just a rauss polymer. It's a simple spring polymer. And driven by a Gaussian white nose and we take the temperature as t equal 1 as a reference. Then you can pick one point of one special abscissa and you ask the question, what's the time for this point to cross the level y equal 0? And as you said, this persistence exponent. And what was known actually at the end of the 90s that in case you take initially an interface that is a thermo equilibrium, then this exponent is 1 minus h. It has been proven particularly rigorously at the end of the 90s by this mathematician, Molchan. But actually, you are not constrained to take initially an interface that is a thermo equilibrium. For example, you can take an equilibrium for a different temperature. For example, you can take the interface perfectly flat, which is an equilibrium for t equal 0. And in this case, the interface will slowly relax and this slow relaxation can interfere with the FPT statistics. And actually, what happens in this case, so in this paper, the authors have shown that if you take this flat interface, you get a persistence exponent that is different from 1 minus h. And actually, this is pretty general. We have done it for actually many initial temperatures and you see, so this is the survival priority, so just the integral of the first passage time density. And you see that the slope, so the persistence exponent, is different for any initial temperature. This is actually pretty surprising because even though the interface is slowly reaching an equilibrium, you still see an impact of the initial quench on the FPT statistics at arbitrary large times because it changes the exponent. So what we wanted to know is, can we predict the influence of the initial preparation on the FPT statistics? In particular, can we predict the value of this exponent theta? And this is actually an important question because in many experiments, for example, you have to prepare your sample to be able to affect the kinetics. This is because your interface has a zero mode. It becomes arbitrarily slow. No, it's... What do you mean? Oh, okay. I think it's pretty general. We can discuss about it. Okay, so we have addressed this question for a special class of processes, a lot of Gaussian processes. Gaussian is nice because they are entirely determined by two quantities. The first one is the correlator and the second one is the mean value. And we assume that the process is not biased so that the mean value is constant. You have no drift in the motion, so it's equal to the initial position of the random worker. And I also assume that at large time my process becomes scale invariant. So you have this scaling for here. I also assume at the beginning that the motion is not confined. It implies that this exponent should be positive because the mean square displacement should diverge with time so that my process is not smooth so h should be smaller than 1. And I have a last hypothesis. It is that at large time my process should become a process with stationary increments meaning that if I wait a large time t and I start to observe my process starting at t, then I have a process that does not depend on t anymore. And actually you see that with this hypothesis the only possible correlator for a long time process is the one of fractional Brownian motion. So for those of you who are not familiar with it, fractional Brownian motion is a generalization of Brownian motion but non-Marcovian. Actually it's exactly Brownian motion for h equal 1.5, but for smaller h we have anti-persistent motion and a really sub-diffusive motion and for h larger than 1.5 you get a persistent motion which is a strongly super-diffusive. These are examples of process in this class so you have this poem of quenched interfaces that I talked to you in the slide before and also the fractional Brownian motion that is constrained on its path for example you can assume that you have a fractional Brownian motion that is constrained to be constant for t negative and then you let it evolve but as you have this constraint this will affect the posterior probabilities and actually these both examples they are Gaussian, they are scaling variant and they are reaching some stationary increment process. All right, so now I want to compute this persistence exponent and to do so it's not an easy task but we will use another a method that we have actually already used in the case of confined random walk to get the mean first passage time and the key idea was to introduce this quantity y pi, actually it's a stochastic process and it's the position of the worker the FPT. So we have a trajectory that starts at the initial position touch here the target and then in red you have the future trajectory and you can sum over all possible first passage time and you have this process in the future of the FPT and actually in the case of confined random walk we make a direct relation between this mean first passage time and the statistics of this process and we wanted to know if we can do the same for unconfined process to get persistence exponents. So I won't show you all derivations we start with this equation that is actually just like a renewal type equation saying that if my worker is on the target at time t at least once before you touch the target and you make a partition over this first passage time tau so it simply buys formula then I introduce my process in the future of the FPT that is defined in this way so in the position of the worker after the FPT and you sum over all possible FPT I make the difference between the two which I integrate over time and this is actually what you get so this is still not very defined but what you can show is that generally this quantity and you make t large you can see that this scales as t at the power 1 minus theta minus h and this behavior actually constrains the left hand side and you can show that first both terms should scale the same way for large t and also that the difference between the two is to scale as 1 t at the power h plus theta and this is the key observation because it shows that actually theta is related to the speed at which the probability density in the future of the FPT converges toward the initial probability density so actually with this this equation you have a strategy because now to get theta we want to know actually this p pi so how to do so actually we will use an approximation which is that the process in the future of the FPT so this y pi is a Gaussian stochastic process with an unknown covariance function sigma pi so you can see on the right that this approximation is pretty precise and this sigma pi actually you can get a closed equation for it so still it's not easy to solve but it's a closed equation and actually what do matters actually it's only this equation for large time because we want to get this deviation between sigma pi and sigma naught and with this you can do by studying this equation so first we use the scaling variance of the covariance by introducing this function z and we expand everything for large time because it should become negligible compared to the initial coefficient sigma and at the end of the day what you get is an integral equation for this auxiliary function z theta and you get an integral for actually any choice of theta and you can solve this equation numerically and what you get now is this kind of thing so this is a resolution for different theta and you see that this theta, this function z is always diverging for small arguments and we argue that the correct criterion to select the solution was that z should be regular for small arguments and you can show that this is actually necessary in particular for theta minus 2h negative because if you don't have this condition rho which is the difference between sigma pi and sigma will diverge for small arguments ok so this is a good criterion and you can see here that theta should be between these two curves so in between 0.27 and 0.3 and you can also check directly the numeric simulations where we have observed this process in the future of the SPT so these are the simulation you measure the covariance and this is the dose that you get which is that in between these two curves it's consistent alright so at the end what you get for the persistent system are these curves so for point interfaces the first part is the same one as in Kruger and collaborators paper it's an interface that is flat initially with different earth exponent h and you see that our theory the black line is in good argument with the simulation in blue and the second part is a fixed h and it should be divided by 8 for different initial temperature and you see that once again you have a good argument and so these are non perturbative results but you can also have perturbative results around burning motion and you get this formula so it looks a bit complicated but the first part actually 1-h is what you get if you have a stationary process so if you have no aging in the in the process this is much a result if the second part here actually you see that it is consistent but Kruger and collaborators get for t equals 0 which is absolutely not obvious because the methodology is totally different but you can recover this formula and you get also the second order term so these are for the other kind of process so it's exactly the same it works very well so it's a small difference between the two is that now the difference between c time 1-h only appears at second order in h minus 1 half so these we have tried for all possible parameters and you see with the theory and the stimulation and you see that the argument is really good except in this small box here for which actually I've told you before that maybe the criterion to set the execution is not a good one so to conclude first you have seen that transit aging naturally arise when you prepare a non Markovian process and this can affect deeply the first passage term statistics link to the way the process in the future of the FPT converges to the initial process and we have developed a non participative scheme to compute this theta so I want to thank my collaborator so Thomas Guérin with you and his PhD student Tony Mendez and also my previous advisor in Paris Olivier Benichu and Raphael Vatery and I thank you all for your attention thank you very much for this nice talk you most likely already said it but as far as I understand you take an infinite system yeah so what does T equal 1 mean I mean why does this relax in a finite time the big T was the temperature if you look at the relaxation time of the system you say it doesn't matter which initial condition you take it doesn't matter actually but doesn't the relaxation time of this membrane diverge with the length the membrane is infinite so it has an infinite let's assume now I prepare an initial condition where I put my membrane with two gaussians separated by some delta with this initial condition there is no gaussian distribution at any time and you cannot describe a system with a single variance so how does how do this apply as far as I understand you said you can treat general initial conditions so I need to have a Gaussian process at the end so we need to start with a gaussian Gaussian and scaling variance okay so the initial condition must be gaussian okay then and the other thing because this is so understand that if there is no relaxation time then you never forget the initial condition yeah so this is the message yeah okay thanks oh it's fine yeah that's true are there any other questions or online okay a brief question so I see you use this process Y which has no drift right so it's the expected value in the future is the same as the initial value so this is like a martingale process do you are you familiar with this because the fact that Y is equal to consult a Y by the expected value of Y without constraints equals to its initial value this appears a lot in martingale theory I don't know if you know you're aware of this I assumed that the meaning is constant but there is no drift after T so if I understand correctly the martingale would say that if you condition the evolution on T at time T plus tau the value would be T so not at zero because he says it's at zero maybe this is a corollary of the general no I was wondering if you made any connection with martingales in terms of this stopping time theorem no I don't maybe it's not true are there any more questions then let's thank you again


Welcome to another edition of Unfettered Freedom, your weekly GNU slash Linux news video podcast. Packing so much freedom into each episode, it ought to be illegal. If I packed any more freedom into this podcast, I would have a disheveled appearance and I would pick at my feet. On this edition of Unfettered Freedom, Blender 2.9 has been released and it is an impressive release. It makes me want to learn Blender, maybe one day. So Thunderbird, they now have OpenPGP enabled by default. We're also going to discuss a really nice free and open source alternative to something like Google Authenticator. We're going to talk about FreeOTP which is an open source two-factor authentication application. We're also going to discuss the ArchWiki's list of applications page. Now this is really fantastic for new to Linux users. The ArchWiki has a list of applications page where they break down some of the most popular applications available. Also, Hurricane Laura and the several days that I went without electricity and some of the lessons I learned going through what I did with that storm. All this and more on Episode 5 of Unfettered Freedom. I am your host, Derek Taylor, also known as DT or Distrotube on YouTube and on Library. This podcast, as well as all of the video content on the Distrotube channel, it is community sponsored and because of the community support that I received, there are no corporate sponsors or product shilling of any kind in these episodes. And I can think as I please, I can say things that need to be said. In short, I can be independent and if you'd like to support my work, I'd greatly appreciate it. Please subscribe to Distrotube over on Patreon. And our first story is going to be the big release of Blender 2.90. So Blender is a really neat 3D modeling application. It's for creating animations and really cool stuff like that. I've never really used Blender myself. I've always wanted to learn how to use Blender and every time they have another release and I keep seeing this thing improve and I see some of the amazing things people can do with Blender, the more I want to learn about it. And the other thing we should also mention about Blender is it does have the ability to edit video. You can actually use Blender as a video editor. I don't know how well it does video editing because that's not really the main purpose of Blender, but it does have that capability. So one of the things that immediately impress me is those of you watching the video version of this podcast, when you go and look at the release announcement for Blender 2.90 from Blender's website over at Blender.org, it's not just plain text on a mailing list. It's not just the same old release announcement that you typically get with open source software releases.  I mean this page just is, it's amazing. It's a gorgeous website as you would expect the kind of people that use Blender no doubt would be very good at web development as well. It's just a very well put together release announcement page. Some of the new features and improvements with Blender 2.90 include things like now they have this physically based texture for simulating the colors of the sky. This feature is called Nashita. They also have improved motion blurring, improved shading to avoid that blotchy appearance that you can sometimes get. They have NVLink support now for CUDA and Optics, so of course that's for your NVIDIA cards. You really need powerful graphics cards really to do some of the 3D modeling and applications like Blender. That's one of the things that sometimes people really don't take into account with things like video editing and 3D modeling and animations and things like that is because they're so graphically intensive you actually do need a pretty decent GPU to get the best performance out of these applications. Some other notable changes in Blender 2.90 include a new interactive search menu, drag and drop reordering of modifiers and other stacks, scene statistics available as an overlay in the 3D viewport. There is a new lens distortion model and various other features. One other one that just jumps out at me is that there is now initial Wayland support. Probably not great Wayland support. Most people that do things with graphics and especially video, if you're recording video especially, Wayland is not ready at all. It's nowhere near ready, you kind of have to still use X11. We're going to be using X11 probably for another decade or two on GNU slash Linux unfortunately until Wayland has to come along a lot further. Right now I don't even think you can capture your screen, video capture your screen using Wayland. Until we can get some of that basic stuff working, X11 is going to be around forever. Blender 2.90 is free and open source software. It is available on Linux of course because Linux free and open source software of course is a first class citizen. But Blender is also available on proprietary operating systems like Windows and Mac OS. You can grab the latest versions of Blender from the Blender.org website for those of us on Linux. If you run a distribution that has system D, Blender 2.90 is available right now as a snap from the snap store. And I mentioned that Blender is one of these pieces of software that I have wanted to try out forever. I have wanted to actually learn Blender for a long time and for whatever reason I've just been putting it off and you know I really want to get into maybe doing some animations and things maybe you know for purposes of my videos I just think it would be cool what if I created you know an intro to my videos or outro to my videos some kind of animation you know using Blender I think that would be a neat little project to work on because I think me learning Blender would be a cool thing to document on the channel you know as I learn Blender I can show you guys that are also interested in learning Blender maybe how to use Blender. That might be something I get into maybe here in the near future. And our next story is the new release of Thunderbird now has OpenPGP support and that's enabled by default OpenPGP of course is encryption for your email. Now I have been a Thunderbird user for years almost since the beginning of Thunderbird. So I've been using Thunderbird for a long long time but I've never used encryption with Thunderbird I've actually never used encryption with any email because I've never found the need because I don't typically email anything that has any kind of sensitive information in it because email by its very nature is not secure because the people that invented email. I mean email was invented so long ago it was it really predates modern computing it predates the web by many many many years. So they never had security or privacy in mind when they invented email. So just know that email I don't think will ever be secure but if you want to encrypt your emails OpenPGP is a way to do it and Thunderbird had OpenPGP in several releases prior to this but it wasn't enabled by default and the big deal now is starting with version 78.2.1 OpenPGP is enabled by default. For those of you that are able to run Thunderbird 78.2.1 or later the way you get to the OpenPGP key manager is you have to go to the tools menu and you get to that by clicking on the little hamburger menu on the right side of the Thunderbird window. Now most of your Linux distributions are not going to have 78.2.1 or later in their repos just yet most distributions of course are static release distributions they typically are not shipping the latest and greatest software. On rolling release distributions you may have a hard time finding Thunderbird 78.2.1. You can always go to the Mozilla website go to the official Thunderbird website and grab the latest 32-bit or 64-bit binaries. On Arch Linux I noticed that the Thunderbird package available in the standard Arch repos is still on 68. whatever is the 68 series and the reason is they have been asked to not package anything until the 78.2 release so Arch has been stuck on 68. whatever for a while. Those of you that want the latest releases though on Arch you do have a couple of other options. There is a Thunderbird beta package in the AUR it's actually called Thunderbird dash beta dash bin that's a binary and that really gives you the very latest cutting edge features but pretty good stability it's not the nightly builds. Those of you that want the Thunderbird nightly builds though that is also available in the AUR that is located as Thunderbird dash nightly dash bin and again that's a little bit more bleeding edge and that's for people that want to look at the latest and greatest features being added and they don't mind Thunderbird breaking on them occasionally. And our next story is about this really cool free and open source two-factor authenticator application because really you should be using two-factor authentication wherever you can. Any service that asks you to enable two-factor authentication you really should do it because passwords can be cracked with enough time anybody can crack a password. So with stuff especially services that have very sensitive information about you you really should enable two-factor whenever possible. So the one downside to two-factor authentication for those of us that want to use free and open source software is the application that most people have to use to enable two-factor authentication is the Google authenticator that is like the standard application and obviously it's a Google application most of us don't want to be tied to Google anymore than we already are and it's not free and open source software and you know all of your mainstream big online services are going to ask you to enable two-factor authentication all the various Google services Facebook Twitter any big social media platform they're all going to ask for you to enable that two-factor and they typically just tell you to go get the Google authenticator and that's the way you do it well there is a free and open source alternative to the Google authenticator it's called free OTP it is free as in freedom it is licensed under the Apache 2.0 license and it is developed by Red Hat now I know some people have a problem with Red Hat a lot of people in the free and open source software communities for whatever reason don't trust Red Hat because it's a billion dollar corporation and you know it's got to be evil because they make money well I trust Red Hat a lot more than I trust Google so I will use Red Hat's authenticator program before I use Google's and I've always used Google's because I didn't know there was an alternative until I came across this blog post the other day I didn't actually know this application existed and I am just thrilled by this it is available on both Android and iOS devices and to use the application it's just like any other authenticator app all you do is scan a QR code and that activates your two-factor authentication on whatever website you were trying to activate that on and then it starts to automatically generate security codes every 30 seconds for you by the way I'm an Android user and I did notice that it is also available in the fdroid store so that's great because if you're an Android user and you're looking for free OTP you because you want to get away from the Google authenticator you're probably you're also not using the Google Play Store you're probably using the fdroid store and I love the fact that free OTP is in the fdroid store for those wanting to check out the source code for free OTP the source code is available and published over on GitHub and really the bottom line is you need to use two-factor authentication we really have to these days because again just using a password they're asking for trouble most large websites most large online services are eventually going to have a data breach really to be honest things like Facebook and Twitter and all of that crap you really shouldn't be on those services because they get hacked or if they haven't been hacked they're going to get hacked and if you put sensitive information on those social media sites which you really shouldn't be doing you know you're just asking for trouble at that point but if you have to use sites like that at least enable two-factor authentication it won't save you from when the online service as a whole is hacked but it will at least prevent you you know the individual from being so easily hacked because I see way too many people you know get their Google accounts hacked Twitter accounts hacked and they wonder why you know how did I get my account hacked but you know what was your password you know it's something like one two three four or the word hello it's like really I wonder how they guess that the next story I want to talk about is really just a page straight out of the ArchWiki the ArchWiki has this really cool page called the list of applications and it is a lengthy page it has a very long table of contents but it's basically a list of hundreds of various applications that are available in Arch Linux but probably in pretty much every other Linux distribution as well and it just breaks it down by category and if you're new to Linux you know you just coming to Linux for the first time you just came from Windows or Mac and you're wondering what is available even if you don't run Arch Linux go read this page because it's fascinating and I mean let's just go ahead and pick a category to check out I'm going to check out web browsers because I have no idea what kind of web browsers people use on Linux because I come from Windows and I'd been using Microsoft Edge for the last few years and Internet Explorer before that none those browsers aren't available what do I have well it looks like I have about seven or eight different terminal based web browsers if I wanted to view the web on the command line there's also various graphical based web browsers as well so you have this broken down by what engine they use so if it's based on Gecko you have Firefox and Seamunkey and various Firefox spinoffs like GNU iSCAT and LibreWolf the Tor browser waterfox you have the various blink based web browsers of course we're talking about Chromium and all the Chromium spinoffs like Brave ungoold Chromium Opera Vivaldi you know all that stuff I think one of the most common questions I get people ask about desktop environments and window managers and if I go down to chapter 7.5 desktop environments actually let's go to 7.5.1 window managers and what window managers are available well they actually have so many window managers available they didn't list it on this already very lengthy page they have a link to a list of window managers page and you can see that there are oh at least 50 60 I don't know as there's a lot of window managers available for you in the arch repos so this is a really great page here the list of applications pages so if you're wondering about audio players video players terminal emulators even networking stuff system admin tools whatever it is you do with Linux this list of applications page probably has a heading for that type of application for those of you that watch the video version of this podcast on YouTube or on library I will link to this list of applications wiki page again if you run a good news slash Linux distribution it doesn't necessarily have to be arch I think you probably want to take a look at this page so the final topic that I want to discuss is the last week so those of you that have been following me on the YouTube channel know that one week ago when I published unfettered freedom episode for when I published that right before I did I also posted over on my YouTube community tab that my area was about to be hit by a hurricane and I was just warning everybody hey I don't know what's gonna happen I may be without a power for a little while I don't want you guys to worry though I'm not really near the Gulf Coast so I'm no storm surge or massive flooding or anything I shouldn't have to worry about that but with these storms typically there's massive power outages and I was letting everybody know that I might be away for a few days or even a few weeks so if you go to my YouTube community tab you know I posted a picture of the satellite image of Hurricane Lord just before it came ashore near Lake Charles Louisiana Cameron Louisiana down in the southwestern portion of the state now I live about three and a half hours from that area inland I'm actually in the northeast section of the state but this storm really initially it didn't look like it was gonna amount to much they were predicting it might be a category one or category two you know earlier in the week and then this thing blew up almost to a category five in strength right before it made landfall you know 150 mile per hour sustained winds which actually tied it for being the most powerful hurricane to ever strike the coast of Louisiana and we are not immune to hurricanes right we get a lot of these storms but Laura was really quite something special as far as the strength and typically when these storms move inland you know they lose strength rather fast and I often get these storms you know when they hit the coast as cat three cat four storms you know those really strong ones by the time they get here they're typically tropical storms and you know the winds are not as much of a concern really for us it's the rain because the area I live in is very flood prone because I have a lot of rivers nearby by use all throughout the city and if one of these tropical storms move through and then they sit there for a while and just dump rain on us for a couple of days much of the city will be underwater people will lose their houses in that case so it's really the rain we're typically worried about here in this area we typically don't have to worry about strong wind events but this thing Laura got here in such a hurry she still had quite the punch when she got here she was still a category one hurricane I think at the time may be closer to a category two hurricane there was some very powerful winds I was standing outside trying to save some of my stuff before it blew away you know it was blew a door off the carport and it ruined a wood fence we had outside it just blew it away just shattered it it was crazy the wind thousands and thousands of downed trees I've never seen anything like the damage that this did and this I mean we I've been through tornadoes I've seen what they can do but you know tornadoes impact small areas you know this thing you know I can drive hours in each direction and just see tree after tree that's laying down you know thousands and thousands of power poles laid down power lines everywhere and I spent nearly a week without power after this because my city you know I live in the northeastern section of the state the city I live in is a population about 60 70 000 people it's not a small town it's the largest city in this part of the state and the entire city was without power there right after the storm it was crazy I mean no traffic lights essential services there's no supermarkets and gas stations and then when a few came back online there was such a rush to go get essential items like gas and ice and you know things like that batteries you can find extension cords people trying to run generators and you know I got used to just sitting at home in the heat it got very hot here so in this part of the world you know especially in August it's the hottest month of the year and typically temperatures can hover around 100 degrees Fahrenheit with the heat index typically they can get up to 115 degrees with the heat index now thankfully it didn't quite get that hot the hottest days were like around 95 96 Fahrenheit outside which brought the temperature inside the house up to like 89 or 90 degrees which is still extremely hot it's like a sauna in the house very dangerous and you really had to fear for the elderly because when the entire city is without power you know how many people really know their neighbors most people don't know their neighbors and that's what worries me is most people know their next door neighbors or the person right across the street from you because you see them right you just come across them you wave to them maybe you speak to them every now and then but by three four houses down do you know that person no and what if there's you know a lady in her 80s been living by herself and she's just sitting there with no central air you know generator and nothing her house is 90 degrees and she's just laying there you know you fear for those people because in that kind of situation those folks can just go to sleep in their house in that extreme heat and never wake up you know is anybody checking on those folks unfortunately after this hurricane you know when we start evaluating some of the deaths associated with lora many of them are not going to be deaths from the storm because we're going to have people that died from the heat we're going to have a lot of people that died from carbon monoxide poisoning because of all the generators many people don't understand that you can actually run a generator indoors it's a gasoline engine just like your car engine you know car engine you know puts out carbon monoxide same thing with a generator you cannot run that in your house but we had people that did that families you know somebody comes and plugs up a generator in the house and then the entire family goes to sleep and that's it the carbon monoxide builds up and none of them ever wake up many people thankfully didn't die but when they did this it made them very sick and some of those folks are hospitalized but thankfully somebody got to them before they actually passed away from the carbon monoxide poisoning from the storm itself there were people that were killed because of trees you know trees falling on houses or vehicles and then after the storm there was just thousands and thousands of trees you know many of them in the road and we had vehicles striking trees you know you're going down a highway at you know 60 miles an hour and you come up on a tree that's laid across the road you didn't see it and you hit that that tree you know going that fast you can imagine what that would do to a vehicle and the person inside the vehicle one of the interesting things from me going all those days without power is that I got used to it after a while you know those first couple of days kind of actually every day sucked it was probably the worst week of my entire life as far as just discomfort just being in the heat but I had gotten used to not having any electrical device working there for a while and some of you guys were asking me how was I taking a picture on my phone and posting it to youtube because I would post every morning hey this is day three without power day four without power day five and I had people asking well how are you even using your phone how are you charging your phone if you don't have power I did have a car right you you have a car charger in your car right I hope most of you do if you don't you know make sure you have a phone charger in your car and it was great because where I was at in this house I didn't have any kind of phone signal I couldn't call I couldn't get on the internet couldn't do anything on my phone my phone was useless there for five plus days so what I would do is when I got up early every morning I would snap a selfie picture you know to send you guys but I couldn't actually send it from here what I would do is I'd go get in my car and I would drive for like 30 minutes in the air conditioning in the car because I just wanted air but mainly to charge my phone and then once I got away from this kind of dead area because I guess the cell towers around here were all dead too but once I drove away a few miles I could actually use my phone I could get on the internet on my phone and that's when I would post those pictures to YouTube so that's how I did that I was surprised so many people were concerned about how to charge the phone there's actually multiple ways to charge a phone you also have those charging banks that some people keep and those things last forever but I actually am kind of happy that my phone didn't work all those days you know at least while I was in the house because I would have just been sitting there playing on my phone because that's all I could do but instead I was out in the yard you know cleaning up the mess out in the yard I even mowed one day I actually push mowed my yard and the heat knowing I couldn't come inside and relax in any kind of air conditioner but it was like I don't know how long this is going to last is it gonna last another day am I gonna be without power another week another two weeks you know the yard's got to get mowed so I just worked around the yard and cleaned up the house a little bit things that maybe I would have done anyway but many of them I wouldn't have done because I had other things to do right when the world is normal well I can go play on the internet I can go do my YouTube stuff I can go to the gym and work out I couldn't do any of that during this episode and when the power finally came back on and I could be normal again I could do my normal routine it was weird because it's like I had gotten so used to not having to do YouTube every day or go to the gym every day or do this every day you know it was almost like I was a different person when the power came back on and that was why when the power came back on you know the I spent a few hours cleaning things out you know I had a deep freeze that I lost all our food I had throw everything out and that first day I decided I needed to make a video just to make sure I got back in the swing of things and there was only one video I thought I needed to make because it made the most sense I mean what's the one thing that kind of defines me arch Linux I made a arch installation video on the first day back just because I thought maybe if I did that things would start seeming normal again and you know what it did help one interesting part of this week though was you know after five days without power I was really getting tired of the heat sweating and just constantly pouring sweat you know the heat down here is unbearable mainly because of the humidity and after five days somebody had told me that hey Home Depot just got in a shipment of generators a large shipment of generators because there's so many people without power and they've got enough generators for anybody that wants one if you want to go buy one and generators are not cheap you know typically they're going to cost you at least eight hundred dollars you know more than a thousand for some of the better ones so I went down to Home Depot I was like okay I'm tired of sitting in the heat I do have a window unit for an air conditioner you know I can't run central air and heat off of a generator because it draws too much power but I did have a window unit and that's perfect because you know I should just sit in the room with the window unit and cool off a little bit and I thought it would be worth it if I went and spent a thousand dollars on a generator it would be worth it just to have that window unit up and running so I went to Home Depot bought a generator and I was coming back home and I had the porch lights on outside I'd flip them on that way I would know if the lights were back on when I came back and the porch lights were still out so I knew power was not back on so I was like okay well I'm glad I purchased the generator so I back up so I could unload this generator and as I was coming in the door of the house I hear the noise the noise of the power coming back on so luckily that thousand dollar generator I bought I didn't even need to unload it I just left it in the car I returned it the next day you know still new in the package and I thought about keeping it because of the future storms that may or may not happen but it's such an expensive piece of equipment and I just didn't think that it was worth it for me to keep it if I had small children or something like that or I lived in an area where massive power outages for days or weeks at a time were common I might have but this was a very unusual event and this was really like a once in a lifetime kind of storm one thing I did do though in preparation for the future is the National Guard was passing out water and MREs to everybody because you know the entire city's out of power you're talking about like in this parish a hundred thousand people you know everybody was out of power so the National Guard comes in they're giving out food and water or whatever you need you just wait in line line of cars you get there you tell them how many people are in your family and they give you whatever they gave me like five cases of water and four cases of MREs because I asked them hey could I have some for my neighbors you know it's like while I'm here can I pick up some for next door neighbor guy across the street whatever and the National Guard doesn't care they give you as much as you want and sure we'll load you up however much you can carry so I've got like five cases of water four cases of MREs gave it all away except for one case of MREs and I never opened it because I didn't need to I didn't feel like eating an MRE I've eaten some of them before they're not great but I am going to put back that one case of MREs that I still have just in case just for the next storm because these MREs of course last year so I will store that away just in case I need it in the future and that is it for this edition of Unfettered Freedom this was episode five I try to release Unfettered Freedom on a weekly basis I try to release every Thursday again I am your host Derek Taylor also known as DT over on YouTube library and Patreon if you guys want to help support my work please consider doing so you'll find DistroTube over on Patreon before I go I do need to thank the producers of this episode I need to thank Michael, Gabe, Corbinian, Mitchell, Devon, Fran, Akami Channel, Chuck, Claudio, Donnie, Dylan, George, Caleb, Devils, Lewis, Paul, Scott, and Willie these guys they are my highest tiered patrons over on Patreon without these guys episode five of Unfettered Freedom wouldn't have been possible those of you watching the video version of the podcast all these names you're seeing on the screen right now these are all my supporters over on Patreon because again this channel is supported by you guys the community again look for DistroTube over on Patreon all right guys peace


How's everybody doing welcome back to my channel this week. I'm gonna be tattooing the Aztec guard Which also belongs to the Mayans and all mechas gets alcohol The placement is gonna be the right side of the chest Which I'm gonna be extending it all the way up to the shoulders. So with that being said, let's get this day started Let's go Yelpegue lo traigo como goma si no entendiste Ay te va en otro idioma Cipitufo que perepes a pesepe copo sapas ma palapas Copa mi picofuepe ni picapa ma fam Sopelo folla perepe pa para pa que petepe depe lo poche perepe perepe o cipitufo Chavá! Amoros! Yes sir Let's do it! I'm gonna truth Ya estuvo You ready? That's the way to speak. I'll never read I guess This is it. This is officially your first tattoo Because that first one didn't count No, that first one. So I have two actually Oh two? This one over here. Let me see Oh! Like you can feel the lines and everything Oh yeah I got it from the same guy Damn it!  Like at the house Hanging out Wow Yeah it was there And like everybody like my sisters My cousins You know everybody was there Just trying to get Tattooed by By whoever Because And at the time he was like the only guy tattooing in the neighborhood So it was like Was it in Mexico or here?  They older than me So they were like You know 18, 19 at the time They were a little older But wow Which is didn't We didn't really know where to go Or where to look for tattooed artists Yeah So we're just like Okay he's He's got a machine lens It's like And then the only artist that was willing to tattoo A 14 year old Yeah That's dangerous zone right there Yeah Alright let's go to the middle of the shop We're gonna put on the stencil And don't Were you able to sleep last night?  Is it time yet? Am I late? That is why I'm gonna be covering up I don't know I have no idea how I'm gonna cover it up I'm kind of nervous My goal is for the bottom Of Quetzalcoatl to land On the nipple area Because this is the darkest part of this tattoo So I need this to land here So he can have a So I can black it out You're thinking about getting your chest done I highly recommend for you to black out the nipple It looks weird when you work around it When you leave it open When you leave it negative It looks really weird because it just Looks like a little pepperoni Perfect You land it right The darkest part of this piece Land it right on the nipple That way there's not It's gonna be really easy to black it out What do you want me to start?  I'm sane Well, let's do it You ready? Let's do it I'm gonna have you lift your arm up Like that Do you like that pillow? You want me to take it out? No, it's fine A few times when I felt pain Right around there I was like, okay, he's on the nipple now I'm about to get there I didn't understand anything I can say Anyone can say It's not gonna hurt It's not gonna hurt It's like it tickles in you You're just gonna be mad that You don't have more than two nipples on you Because to experience that more Perfect song I'm gonna start on the What is it? That uriola? Uriola Uriola Uriola I'm gonna start on the uriola Work my way up Or towards the middle Where the nipple is actually located How's that?  That's a good one Whatever reason I think I like Don Amant better Don Amant makes you the rhythm John Sebastian Or Jose Jose Oh, that's a tough one I was gonna say Vicente Fernandez But that's an obvious answer Yeah, no Jose Jose or John Sebastian Because I grew up a little more with John Sebastian And I know, you know, like the poet that he was I mean, oh my God Yeah So I think I would go with With John Sebastian But I do love Jose Jose Yeah, Jose Jose has got roots That's crazy Paquita del barrio Paquita, genzano Paquita, no, pero juires No matter who you put it against I like listening to Rock too I like rock But El Tri was everything Did you ever listen to rock? Yeah, I actually went to see El Tri a couple times No way Yeah, when they came here That's it I went to their latest one Which was maybe four years ago Here at El Capri They came here and I was like I gotta go see them because if I don't go I was gonna Yeah, because they're in the point where they're about to retire I was like, it's now or never Yeah Bro, you're a pimp, bro What the... Oh my God That's it I think that is over Oh my God Yo I'm not gonna lie I think you're the first client I've ever had That handles the nipple like that Yeah We just put some in a conversation You were not gonna make it at some point During the whole session Honestly, yes You were like, I don't think I could make it What am I doing here? I'm about to tap out No way For reals? It was bad What made you not want to say stop?  Barely Barely My Barely made it Thank you for coming back And watching this brand new video I really appreciate you Let me know if you've had Let me know if you've had Let me know if you've had If you've If you've If you've If you've If you've If you've If you've If you've If you've If you've Okay, like saying Jude Let me know if you'd Like family feud Let me know if you'd Let me know if you'd Let me know if you'd Maybe if I say it fast Maybe you'll sound more natural Let me know if you've got Your nipple tattooed Anyway, whatever Let me know if you'd had Let me know if you'd Let me know if you'd had Your nipple tattooed And if you have Let me know your experience Drop a comment Like and subscribe