Thank you for saying the title so I don't have to My name is Tatiana Bradley I am a student at UC Irvine and this is joint work with my advisor Staszurecki and my colleague Jayu Shu I'm going to be talking about strong asymmetric password authenticated key exchange So I'm going to explain what the problem is which is password authentication over the internet Explain how we can use crypto to do it better and show our protocol which we believe is one of the best ways to use Crypto to do password authentication better. I'll show you some building blocks, and then I'll discuss some open questions that still remain So how is password authentication done in practice now? It's basically just password over TLS so If we have a server Charles, he has a password file, which is a which has a salt and a hashed salted hash of the password and he's going to prove to Ada that he has the right public key using a certificate authority and then they're going to use TLS to establish a shared a session key, which is just going to be a symmetric key And so all of this has nothing to do with pad passwords yet But then now Ada is going to simply encrypt her password under that shared symmetric key Charles decrypts it and checks if it is matches the hashed password. So it's very simple Well, there's at least two problems with this method One is of course that it relies on PKI and we all know there are many attacks on PKI One example is phishing Ada believes she's talking to Charles, but she's actually talking to a malicious server I mean if that's the case, then she basically just sends her password to a malicious server and all of her Security regarding her password is lost. The other problem, which is a little bit more subtle is that at least for some number of milliseconds or whatever Charles holds the password of Ada in the clear So that means if Charles's server is corrupted at any point while he's checking that password The adversary who corrupts the server learns what that password is so we'd like to be able to do better with crypto and Of course we can because we're at crypto So our first attempt is going to be to use password authenticator key exchange, which is a well-established primitive It's been studied in the game-based simulation based models as well as the universally composable model So how does that work? We assume Ada and Charles each have a password they insert it into a paked protocol And then they get the same key if their passwords match. Otherwise, they get an independently random key and You can think of this as just like a secure computation of a password equality function with the addition of getting keys So what are the good things about this? No pki attacks because there's no pki. So we're good for that problem with TLS and The other good thing is that Charles doesn't get Ada's password in the clear during the protocol because it's your secure computation So her password is private But of course in order to be a server Charles needs to store the password on his server that he inserts into the protocol so basically We have an even worse problem with the server with the server storing a password in the clear because it's so it's stored for a long time not just a couple seconds so We hope to do a little bit better than this and the here we have something called asymmetric pake and This has also been studied in the game base and you see models And this is a simple extension of pake in which instead of storing just a password Charles the server stores a hash of that password so you can think of f here as a one-way function We of course have the same pros of pake normal pake and then we get the added pro that if the server is compromised then The adversary has to do an offline dictionary attack in order to learn the password based on that password file but because the Function f is a public fixed function Any computation for that offline dictionary attack can occur before the actual server compromise so basically the adversary who has some time on their hands can do a bunch of computation and then once They get access to the server immediately determine what a lot of the passwords are so we want to do even better than that and This is where we have strong asymmetric pake and this is recently proposed at Eurocrypt by by my advisor and my colleagues and 2018 and I The way this works is probably how you thought it would which is instead of holding a regular a unsolved or unrandomized function we use a randomized function, so we just add a salt a random salt and Now we get the benefit that there's no possibility of offline dictionary attack pre-computation because the salt is per user and It's not known beforehand So finally we are able to use crypto and get all the benefits of a password over TLS, but none of the shortcomings, so this is great But what's our contribution here because this is already existing so just to recap we have password over TLS the current De facto standard. This is nice because it's standard. It uses TLS. It's efficient. It's modular. Basically, it doesn't matter What hash function you use it doesn't matter what encryption scheme you use you just encrypt the password But as I said it has the two problems that I mentioned So opaque the protocol proposed at Eurocrypt Proposes a universally composable strong a-pake that of course is independent of PKI It has three rounds and it's based on an interactive assumption And what our contribution is is that we're proposing a new UC strong a-pake and we're able to shave off one round We have a similar number number of exponentiations, but we don't have to hash onto a group We we have a number of assumptions including ROM We also use GGM, but that is isolated into an offline part of the protocol and we're also proposing some new some new Primitives the UC tight one-way function and the implicit statement conditional seek him and we're showing how we can adapt a existing paradigm for pakes and for UC pakes into Strong a-pake and also shave off one half of that paradigm. So It's a nice little result there So let me get into it explain each one of these things. So let's start with the tight one-way function So this is in the UC model So we have something called f-staff and that's the idealized functionality of a salted tight one-way function and The salted tight one-way function has some passwords some secret password that it knows and it has some notion of Being compromised and it has some notion of having a password attack password guests being Happening against it so in the case that it's compromised is where it's just sends a message. Okay, I'm compromised and in the proof or in any kind of proofs using this we need to When we hear that the server has been compromised We need to sort of generate some kind of password file that looks legitimate to Pretend to have been a real server that has a password file So we have to generate that based on usually just some kind of random Password, but we don't know what the password is so we have to generate that without knowing the password and then the the other functionality of the F-staff is that you can ask it is this the right password? So it's basically like a point function where you're saying is this the password it says yes or no And I learned whether it's the password and then using that functionality as a simulator We need to translate the real attacker who's doing some kind of offline Local computation we need to translate that into a password guess So for example if the one-way function is just a hash function It's saying hash of password X And then we say okay since we're a random oracle we know what that password We know what that password was and then we say is this the right password? Yes or no, and then if it is right we program the password file to have to be Correspond to that correct password guess if it's not we just give it some kind of some kind of junk So as I've been saying I gave an example with the random oracle model It seems like with this model you're going to need some kind of idealized computation Yeah, so this is just a description of what a tight one-way function is Basically you can think of the tight one-way function as being You if you make a password the the best the best strategy for Learning the password from the hash password is simply to guess all the passwords trial hash them and so forth so brute force You can't like shave off a factor an exponential factor something But yeah, so as I was saying it's It seems like we're going to need some kind of idealized model for this so the first example I gave is the hash function and That's what password over TLS basically uses Because the server is storing that hashed password Then the other Example is the hash in the exponent type where you have a group element That's the a random group element That's the salt and you raise it to the hash of the password and that and we're using a variant of this and it's been used before It requires a programmable generic group model in order to show that it's a staff But we're okay with that because this tight one-way function is only surrounding the sort of offline part of the protocol in which the adversary is Compromising the server it's not during the actual protocol So this this does raise some questions such as are there can we prove a tight lower bound for some other arithmetic one-way functions beyond the one? I just described There is a one caveat and this is something technical that you can read about more in our paper But we're actually using a variant of this hash in the exponent Which is similar to the Bon Ivoian signature function and the reason for this is because we don't want the password file to be malleable and The reason for this is that we don't want the adversary who steals the password file to be able to Randomize the password file and then pretend to be the adversary without us being able to tell that they're pretending to the adversary So it's kind of a technical point for the proof But from now on in the talk, I'll just sort of use this app staff as an abstract primitive Alright, so I told you that we are working with a existing paradigm the Encryption plus SPHF paradigm. So let me show you how that works with regular peak If we have if we go back to our regular setting where we have a pass we have a password on both side Unless opposed that Ada wants to prove that she has the right password so Ada sends a commitment to her password using a public key encryption scheme and The public keys in the CRS so she she commits her password using the public key encryption scheme and then Charles creates a Statement that's in a think of like an NP language. He creates a statement that says okay This commitment of a does encrypts the same password as the one that I have so it encrypts my password And of course the statement could be true or it could be false and Then he plugs the statement into this magic SPHF box the SPHF SPHF has two algorithms. It has the hash algorithm and the P hash algorithm and P hash stands for projection hash The hash algorithm takes in a statement and it outputs a value and a hash projection The value is like a secret value the hash projection. You can think of it as kind of like a key so the hash projection You can send that to someone and then they can use the P hash value along with a witness for your statement That to get that same value so What's happening here is that? Charles is showing That so if only if Ada has the witness for her commitment Can she recover that? That value and also she can't recover it if she encrypted the wrong password because the statement will not be in the language So that shows that that proves to Charles that Ada has the right password But then Charles has to show Ada the same thing. So you just do the same thing symmetrically on the other side And that's and that's how the SPHF plus encryption paradigm works What about with a pic so with a pic we just do a simple transformation we instead of having Charles have the password he has the hash of the password and he his commitment his statement is that Ada's commitment encrypts the pre-image of The pre-image of his password file and he instead In he encrypts the password file instead of the password and so forth. It's just kind of a simple transformation So that all that's all fine. And what about for essay pic we can do the same thing but with the salts but What we notice here is that because we basically need to use ROM because of the staff as I said we need to use ROM for the staff so If we use ROM We can shave off a whole half of this protocol Oh and just to just for some context this kind of thing can be done without the random oracle model in general But if we say okay, it's fine to use ROM then we can shave off half of this protocol. So how do we do that? Well, we do we do need to change some things so in particular We now have Ada now has no guarantee that Charles has the right password file Because she proved that she has the password, but he didn't prove anything So we need to embed inside of this this Hash projection some more information. So we need CCA security. We need it to be non malleable and we also need it to be able to carry a payload an encrypted payload so Because he because Charles needs to be able to send his password file over to Ada So she can check that he does have the right one and has to be non malleable so that he can't Pretend he had the password file when he didn't really have it And then this sort of changed SPHF is going to be what we call an implicit statement seek him All right, so the implicit statements seek him is works pretty similar to the SPHF We have a send algorithm. That's like the hash algorithm and a receive algorithm. That's like the p-hash algorithm The send algorithm Transforms a statement into a key and a ciphertext the receive algorithm Transforms a witness and that ciphertext into a statement and a key so here It's called implicit statement because the receiver does not need to know what the statement is In order to generate the value and in fact she retrieves the statement in the in the course of doing the receiver algorithm and So that's great because in our setting the receiver who's going to be the client Ada Doesn't know what the password file is. She needs to grab it and check it and And then this is a trap door conditional seek him and so that means we have a trap door algorithm that can use a trapdoor and the CRS to compute the same values that Ada can compute and They look indistinguishable from Ada's values. This is just used in the proof and We also have simulation stoutness which you can think of like CCA and that's that the key appears to be rant the key K that the Sender generates appears to be random if the statement he Sends is not in fact in the language even given the presence of this trapdoor receiver So you can think of that like a decryption oracle and Finally we have statement privacy which says that the statement is hidden without the witness and this is The thing I just described is fairly similar to implicit zero zero knowledge Proposed by Benhamuda at all But the new thing here is this statement privacy notion and that's that the statement is hidden Given the cybertext you cannot determine what the statement is unless you have the witness and that's important because if the statement is revealed then the statement reveals the password file and Then it's as though the adversary Compromise the server without even having to do any work. So basically just by listening on the network They would get the password file and we don't want that to be the case All right, so how do we make the implicit statement trapdoor seek him? I'm not going to go through all the details But you can think of it like if you're familiar with the Fujisaki Okamoto transform from CPA to CCA public key encryption You can think of it like that basically what we're doing is we are We're Encrypting we're using that That hash value that we got that we know is the same if those two passwords if the passwords match We're using that to encrypt the randomness that we use to make the hash and also the statement itself And also just macking everything so that nothing can be changed and then Then Ada now knows that if Charles sent her a valid cipher text That he does have a valid. He does have a valid password file and nothing was changed So basically if we tie that all together what that gives us is this is like the picture before except now We're using the trapdoor seek him and the send and receive function instead and Just to recap we know that Ada authenticates herself using Using the basic seek him security, which is that if she's able to decrypt the Charles is Charles a cipher text then she must know The the witness and the statement must be correct. So she must have Encrypted the correct password and then Charles also authenticates to Ada because he's including his encrypted password file and the payload so we We basically just combine The the work that is done by the SPHF, which is the Authenticating Ada with the authenticating Charles into the same cipher text. So that's how we shave it down to two rounds And this is our protocol. Oh and just to just as a note We instantiate the PKE with a lifted Algamol, which is just a slight variant of Algamol and with our instantiation of the Seek him based on SPHF. We're able to get a very efficient protocol What I want to highlight for my summary is that we are in our instantiation is is quite efficient Not only does it have two rounds, but it also in terms of variable variable base Exponentiation which are much more expensive than fixed base exponentiation We only have one or one to two per party And so if you think of regular Diffie-Hellman key exchange, you would have one variable base per party So we're actually getting pretty close to that plain unauthenticated Diffie-Hellman and That leads into One thing I'm particularly interested in which is integrating this work with TLS. That's our last open question But some other Cool open questions are whether there are other tight one-way function constructions Like with different assumptions and maybe we could even get rid of GGM and ROM It's a possible to shave off the rounds from SAP even further Would we be able to make a seek him faster by using DHIS type transforms that of FO? Finally, could we extend it to two-factor authentication or TPEG? So yeah, that's all I have. Thank you so much for your time All right, we can have some time for questions. Please come to the mic if you have one So I would have a quick one regarding But I guess the last bullet point integration So my understanding is that OPEC is on its way to potentially being standardized at least it's an working group item of the CFRG Do you have any similar plans with that protocol? We don't have any specific plans regarding that and I am aware that OPEC is On it's on the road too, but I I would like to I don't have specific plans But I would like to get in on that if possible. Yeah, cool. Okay. All right. There are no other questions. Thanks. Thanks


Welcome to lecture series on advanced geotechnical engineering and we are in module 4 which is on stress strain relationship and shear strength of soils. Primarily we have actually discussed about the in the previous lecture about the Mohr circles but in this module 4 lecture 3 we will be concentrating and introducing ourselves to you know what is stress path and how these stress paths can be plotted and introduction to PQ space. The introduction to principal stress space and stress paths in PQ space so in this particular lecture of the module 4 we will be concentrating on principal stress space and stress paths in PQ space. So in the previous lecture we have discussed about the total stress circles and effective stress circles and we said that both in case of total stress circles total stress and effective stresses the Mohr circle diameter remains same. And when the pore water pressure is positive when u is positive the total stress circle effective stress circle is on the left hand side of the total stress circle but when the pore water pressure is say negative then the effective stress circle is on the right hand side of total circle. So it can be seen here tau versus sigma and sigma dash and where in the effective stress circle is actually obtained by using this half sigma 1 dash plus sigma 3 dash is equal to half sigma 1 sigma 3 minus u and half sigma 1 minus sigma 3 dash is equal to half sigma 1 minus sigma 3. So if you look into this here the total stress and effective stress more circles have the same radius or same diameter but are separated along the sigma axis by amount equal to the pore water pressure either it is positive or negative. In case of negative the effective stress circle on tau sigma space will be on the right hand side of the total stress circle and in case if the pore water pressure is positive then the effective stress circle is on the left hand side of the total stress circle. So the total stress and effective stress more circles have the same radius but are separated along the sigma axis by an amount equal to the pore water pressure. And inability of the pore water to resist shear stress so that the shear stresses are resisted entirely by the contact pore stress between the soil. So this explanation is given the shear stresses are not affected by the pore pressure. You can see that the shear stresses are not affected by the pore water pressure they are actually same both in case of total stress and effective stresses and this can be physically explained by the fact that the inability of the pore water pressure to resist shear stress so that the shear stresses are resisted entirely by the contact pores between the soil grains only. So you know we can see from this graph and then for the discussion we had in the previous lecture we can understand that the shear stresses are not affected by the pore water pressure. So this can be explained physically by the fact that inability of the pore water pressure to resist shear stress and so that the shear stresses are resisted entirely by the contact forces between soil grains. So now let us introduce ourselves to principal stress space. So the principal stresses sigma 1, sigma 2, sigma 3 experienced by a point in our soil continuum can be used as Cartesian coordinate to define D in a three dimensional space and which is actually called as principal stress space. The principal stresses sigma 1, sigma 2, sigma 3 experienced by a point in our soil continuum can be used as Cartesian coordinates to define a point D in a three dimensional space and which is actually called as principal stress space. So we said that sigma 1 which is the major principal stress and the sigma 3 is the minor principal stress and sigma 2 is intermediate principal stress. So this point in the principal stress space only displays the magnitude of the principal stresses and cannot fully represent the stress tensor because the three data establishing the directions of the principal stresses are not included. So the point in the principal stress space display the magnitudes of the principal stresses and cannot fully represent the stress transfer because the three data establishing the directions of the principal stresses are not included. Now let us consider for example a case of the point D where it has actually has coordinates like along sigma 12 and along sigma 2 axis 6 units and along sigma 3 axis let us say it has got 3 units. Then this forms the as can be seen from this figure here and D is the point in the principal stress space where sigma 1 is in this direction and sigma 2 and sigma 3. So this D is represented in the principal stress space with 12 units along sigma 1 axis and 6 units along sigma 2 axis, 3 units along the sigma 3 axis. Now this principal stress space the division of principal stress tensor into spherical and deviator parts can be done like this within the matrix 12 6 3 which is equal to matrix of sigma 1 sigma 2 sigma 3 and which is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 into matrix of 1 1 1 plus sigma 2 minus sigma 3 by 3 where matrix with 0 1 minus 1 plus sigma 3 minus sigma 1 by 3 with matrix of minus 1 0 1 plus sigma 1 minus sigma 2 by 3 1 minus 1 0. So by putting the respective units like for sigma 1 12 the units which we have defined here 12 6 3 in the example here then we actually get sigma 1 plus sigma 2 plus sigma 3 and once we get simplified by that this 7 will come then sigma 2 minus sigma 3 and sigma 3 minus sigma 1 and sigma 1 minus sigma 2 after simplification this represents the matrix like 7 7 7 plus matrix of 0 1 minus 1 plus 3 0 minus 3 plus 2 minus 2 0. So this represents in the principal stress space OA AB BC CD so which is nothing but OA is from here to here which is you know has 7 units here 7 units here 7 units here that point A and AB which is having I can say that towards the sigma 2 axis 1 unit here and towards the sigma 3 axis minus 1 unit here so this is B here and then when we have BC which is similarly 3 0 minus 3 so that is represented here similarly CD which is nothing but which is represented here then which is equivalent to OD so OD is nothing but what we have actually represented in the principal stress space. So the principal stress space is particularly favored for the representation of the theories of the yield strength particularly perfectly plastic materials so the principal stress space is particularly favored for representation of the theories of the yield strength of perfectly plastic materials and for a perfectly plastic material the principal stress axis can be converted into Cartesian axis as x is equal to 1 by root 3 sigma 1 plus sigma 2 plus sigma 3 is equal to root 3 p and similarly y is equal to 1 by root 2 sigma 2 minus sigma 1 and z is equal to 1 by root 6 2 sigma 3 minus sigma 2 minus sigma 1 so the principal stress space is particularly favored for representation of the theories of the yield strength of perfectly plastic materials. Now let us look the introduction to the stress paths in PQ space so before that let us consider two examples we have on the left hand side in the figure two marbles or which can be represented like a two spherical grains which are actually in figure on the left hand side here which is actually on the one grain is sitting on the other and assume that there is a vertical force is acting in the direction of the vertical direction and in this case the grain which is bottom grain is has the same position but upper grain is actually being pushed with a force applied at an angle theta the force applied at an angle theta. So in order to represent this one let us say that we keep on increasing this from 0 to certain F then the path of this load what we say that the direction the load path or force path traverses from 0 to A in this direction that means that in the vertical direction. In this case the force is actually applied at an angle which is theta so which actually traverses in this direction so this is for the path of the experienced by this marble where it can undergo tilting it can undergo riding and depending upon the frictional interaction between the two grains. So line O A is called the load path or force path and line O B represents a load path for the example which is actually shown here with a force applied at an angle so it is important to note that the response stability and failure of the system depends on the force path. So for the point B where which actually has got the y intercept of phi n phi f sin theta x intercept of f cos theta that is actually represented here. So in this particular example when the for the two grains which are either grains spherical grains or marbles when it is actually applied force in the vertical direction the load path is actually shown here and similarly the load path O B is actually shown with a force applied at an angle theta and this is for the force applied at an angle. Now we know that soils of course are not marbles but the underlying principle is same. So when you extend this principle the soil fabric can be thought of as a space frame with the soil particles representing the members of the frame. So soil particles representing the members of the frame and the particle contacts representing the joints. So here what we are having is that the soil fabric can be thought of as a space frame with the soil particles representing the members of the frame and the particle contacts representing joints. So the response stability and failure of the soil fabric or the space frame depend upon the stress path to which the soil specimen or soil mass is subjected. So what we have done is that from the example whatever we have taken for the spherical grains or marble grains we are connecting to from the our soil where the soil fabric is assumed as the space frame, the soil fabric means the particular arrangement is assumed as a space frame with the soil particles representing the members of the frame and particle contact points are represented as the joints. So the frame is actually connected with the joints and these joints are actually represented in the soil as the contact points at the grain to grain contacts and members are actually represented by soil particles. So the response and stability of a failure of the soil fabric or the space frame depend upon the stress path. So the stress parts are presented in a plot showing the relationship between the stress parameters and provide a convenient way to allow geotechnical engineer to study the changes in stresses in soil caused by loading conditions. So the stress parts are represented in a plot showing the relationship between stress parameters and provide a convenient way to allow geotechnical engineer to study the changes in stresses in soil caused by the different types of loading conditions and different combinations can be considered. So let us consider an isotropic compression loading condition in one, this loading condition is defined as one where in the isotropic loading condition where delta sigma 1 is equal to that means that vertical stress and horizontal stress sigma 1 delta sigma 1 and delta sigma 3 that is in the horizontal direction they both are same. So here in this particular chart where Q which is represented as P is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 and Q is equal to sigma 1 minus sigma 3. So here consider the isotropic loading condition one, so by using incremental form of the stress variants so we can say that the stress invariant of, invariance of isotropic compression are so delta P1, delta P1 is nothing but the incremental increase in P1 that in the P space is equal to delta sigma 1 plus 2 delta sigma 3 because delta sigma 2 is equal to delta sigma 3 we have taken it as delta sigma 1 plus 2 delta sigma 3 by 3 is equal to where because delta sigma 3 is equal to delta sigma 1 then we are putting in terms of delta sigma 1. So because of that delta sigma 1 plus 2 delta sigma 1 by 3 which simplifies to delta sigma 1 and delta Q1 is equal to delta sigma 1 minus delta sigma 3 which is equal to delta sigma 1 and delta sigma 1 so it is 0 and the stress variants at the end of the loading one are so it increments with P0 so P0 plus delta P1 so where P0 is equal to 0 initial conditions are 0 so delta sigma 1 that delta P1 is equal to delta sigma 1 and Q1 is equal to Q0 plus delta Q1 both in initial and at the end of the loading they both are same so in that case what will happen is that Q1 is 0. So the stress path for this loading condition where isotropic compression is involved it traverses from O to A so it traverses from O to A where the slope of this line is equal to 0 because delta sigma 1 that is delta sigma 1 delta Q1 by delta P1 so delta Q1 is equal to 0 and delta P1 is delta sigma 1 so 0 by delta sigma 1 it is 0 so the slope of this line is 0 and this is OA is equal to stress path for the loading one that is loading one is isotropic compression which is indicated on the where we have got the stresses which are identical in all directions then that is indicated as path OA. So if you are having earth pressure at rest condition and with K0 is equal to 1 let us say and then in that case vertical stress is equal to horizontal stress in that case with isotropic kind of compression conditions are simulated then you know the stress path actually is along this P line along the P axis with delta Q is equal to 0 so this is for you know for this OA which is for isotropic compression. Let us considering the loading condition 2 where sigma 3 is constant sigma 3 is constant that means that there is a sigma 3 which is applied but there is no change in delta sigma 3 the sigma 3 is constant means rate of change of delta sigma 3 is constant that so delta sigma 3 is equal to 0 which is the loading condition is indicated within 2 here and delta sigma 1 is greater than 0 that means that once we held with certain sigma 3 with out at constant sigma 3 then you know the we increase the pressure. So in that case this loading condition is 2 where sigma 3 is equal to constant and with the delta sigma 3 is equal to 0 but continue to increase sigma 1 in that case the increase in the stress variance for loading 2 R delta P 2 is equal to delta sigma 1 plus 2 into delta sigma 3 that is said delta sigma 3 is equal to 0 so what we have is the delta sigma 1 by 3 and delta q 2 is equal to delta sigma 1 minus delta sigma 3 delta sigma 3 is equal to 0 so what we have delta sigma 2 is equal to delta sigma 1 and the stress variant at the end of the loading 2 R P 2 is equal to P 1 plus delta P 2 where P 1 is obtained as delta sigma 1 that is here in this case in the previous condition delta sigma 1 and delta sigma 1 by 3 so this simplifies to 1 plus 4 by 3 into delta sigma 1 so with simplifies to 4 by 3 into delta sigma 1 and q 2 is equal to q 1 plus delta q 2 which is nothing but q 1 initially we have seen in the previous case q 1 is equal to 0 so what we can say that q 1 is equal to 0 plus delta q 2 is equal to you know what we have done here increase in the stress variant for the type of loading condition which we have considered it is delta sigma 1 so q 2 is equal to delta sigma 1 so AB is the stress path for the loading 2 AB is the stress path for the loading 2 AB is the stress path for the loading 2 and the slope of the stress path AB is delta q 2 by delta P 2 delta q 2 is delta sigma 1 and delta P 2 is delta sigma 1 by 3 so you know by simplifying this what we get is the 3 so that means that the slope which is you know which actually has got 3 vertical one horizontal 3 vertical one horizontal is the slope for this line stress path from A to B for the condition where the loading where delta sigma 3 is equal to 0 and sigma 1 is continue to increase. So this is for a type of loading condition as can be seen this is initial condition isotropic compression and then there is an increase like this this is the stress path and for the next to this thing what we do loading condition 3 where delta sigma 1 is equal to 0 but we continue to increase let us say delta sigma 3 that means that we are date of change of delta sigma 1 is not there delta sigma 1 is maintained constant that is sigma 1 is maintained constant delta sigma 1 is equal to 0 in that case here what we are doing is that we are actually trying to see some sort of you know expansion where delta sigma 3 is actually some it also present something like a squeezing. So in this case for the loading 3 what will happen is that sigma 1 is constant that means that delta sigma 1 is equal to 0 but continue to increase delta sigma 3 and that means that delta sigma 3 is greater than 0 so increase in the stress variance for loading 3 r delta p 3 is equal to 0 plus that is delta sigma 1 is equal to 0 plus 2 delta 3 divided 2 sigma delta sigma 3 divided by 3 which is simplifies to 2 by 3rd of delta sigma 3 and delta q 3 is equal to 0 minus delta q sigma 3 which is nothing but 0 minus delta sigma 3. So there is an increase delta sigma 3 is being increased so you know delta q 3 is equal to minus delta sigma 3 and the stress variance at the end of loading 3 r p 3 is equal to p 2 plus delta p 3 is equal to 4 by 3 delta sigma 1 p 2 was obtained has here in this case p 2 is obtained has 4 by 3 delta sigma 1 4 by 3 delta sigma 1 plus 2 by 3 delta sigma 3 which is actually delta p 3 here. So q 3 is equal to q 2 plus delta q 3 delta q 3 q 2 was obtained as delta sigma 1 so by substituting here delta sigma 1 and delta sigma 3 you know with delta sigma 1 is equal to 0 what we will have is minus delta sigma 3. So this b c the stress path for the loading 3 which is shown here the stress path for the loading 3 is in this direction. So you can see that this path now takes a downward turn here because the loading condition here is that there is no change in sigma 1 that is delta sigma 1 change rate of change of then sigma 1 is 0 delta sigma 1 is 0 but delta sigma 3 is being increased that is greater than 0. So with that you can see that how the slope of the stress path b c is nothing but delta q 3 by delta p 3 which is nothing but minus delta sigma 3 by 2 third of delta sigma 3 with that what we get is the minus 3 by 2. So minus 3 by 2 the slope is actually indicated here 3 vertical 2 horizontal that is the slope which actually runs down here for stress path b c for the loading condition where delta sigma 1 is 0 sigma 1 is not sigma 1 is constant and delta sigma 1 is 0 and delta sigma 3 is continue to that is greater than 0. So that is for loading condition 3 but till now this is an example which we have taken and then we try to plot the stress paths but when we connect it to the soil when it is under drained condition or undrained condition it can be under during consolidation stage if the drainage is allowed then the consolidation takes place and during shearing if the drainage is not allowed then the excess pore water pressure develops and this is also depends upon the type of the soil if you are having normally consolidated soil then you have positive pore water pressure and when we have over consolidated soil there can be possibility that you know the native pore water pressures they do develop. So if the soil pore water is allowed to drain from the soil sample the increase in the stress carried by the pore water is called as excess pore water pressure delta u will continuously decrease to 0 and the soil solids will have to support all of the increase in the applied stresses. So if the soil pore water pressure is allowed to drain from the soil sample and the increase in the stress carried out by the pore water is called the excess pore water pressure and will continue this excess pore water pressure will continue to decrease g to 0 and the soil solids will have to support to the all the pressure which is increased by the increase in the applied stresses. Now let us assume that loading condition one represents the isotropic consolidation and since the excess pore water pressure delta u1 dissipates as the pore water pressure drains from the soil the mean effective stress at the end of each increment of loading one is equal to the mean total stress because delta p1 dash is equal to delta p1 minus delta u1 and at the end of consolidation because when the pore water pressure excess pore water pressure which is generated is dissipates to delta u1 when it is tending to 0 then we can say that delta p1 dash is equal to delta p1 which is nothing but which is both the effective stress path ESP and total stress path both are identical for the isotropic consolidation and which is represented by OA that is for during isotropic consolidation both OA represents for the effective stress path as well as for the total stress path. So in case of isotropic consolidation the slope of the stress path which is 0 and which is along the p dash axis and which actually the effective stress path and total stress path you know stress paths are identical and they follow in one line and this is for the total stress path and effective stress path for the isotropic consolidation. Now assume that during loadings 2 and 3 the sample is undrained now during loading during loadings in 2 and 3 the sample is undrained then we can see how things will change. Now let us assume that if the during loading so this is you know AB represents the stress path for total stress path and BC what we have deduced that is for the total stress path and then for effective stress path with sample remains undrained you know the effective stress path traverses from A to B dash and the effective stress path for this condition where delta sigma 1 is equal to 0 and delta sigma 3 greater than 0 the effective stress path traverses vertically down with slope almost 90 degrees you can see that BC dash and AB dash for the loading condition where delta sigma 3 is equal to 0 and where delta sigma 3 is greater than 0 you can see that how this path traverses downwards like this and this traverses upwards like this. But however this we will discuss while you know after having introduced the triaxial test for the stress paths in PQ space stress variable S and T are the two dimensional variables that do not capture the effect of sigma 2 and the whereas the mechanical response of soil is mainly expressed in terms of P and Q which are defined as the following which are defined as follows in terms of principal stresses. So this P and Q mechanical response of soil is mainly expressed by taking into sigma 1, sigma 2, sigma 3 and the stress variables S and T are basically they are two dimensional variables and basically they do not capture the effect of sigma 2 and the thus the P and Q are normal and shear stresses that are representative of the three dimensional state stress state at a point. So the P and Q are the normal and shear stresses and that are representative of the three dimensional stress state at a point where P is given by 1 by 3 sigma 1 plus sigma 2 plus sigma 3 Q is equal to 1 by root 2 root over sigma 1 minus sigma 3 whole square plus sigma 1 minus sigma 2 whole square plus sigma 2 minus sigma 3 whole square. So in the two dimensional when you wanted to convert let us say that so this is in terms of Q which is 1 by root 2 into root over sigma 1 minus sigma 3 whole square plus sigma 1 minus sigma 2 whole square plus sigma 2 minus sigma 3 whole square. Now let us say that we have a Mohr circle with sigma 1 as the major principal stress and sigma 3 as the minor principal stress and sigma 1 minus sigma 3 is the diameter of the Mohr circle and the center of the Mohr circle on the sigma axis at a distance sigma 1 plus sigma 3 by 2 and let us assume that this is the stress point S T and the point here we can see that this point A which is nothing but when you join the pole here this is the major shear stress plane, this is called the major shear stress plane. And sigma this coordinate of this point is along the sigma axis it is sigma 1 plus sigma 3 by 2 along the tau axis it is sigma 1 minus sigma 3 by 2. So when you have let us say with the constant sigma 3 let us say that when sigma 1 is increasing then in that case what will happen is that with the constant sigma 3 you know the Mohr circle actually travels us like this. So you will actually get sigma 1 1 1 sigma 1 1 2 sigma 1 1 3 like this. So whenever we actually join these points which are actually of maximum shear stress then that indicates something like you know when you join this line and this is actually inclined at 45 degrees so a stress path is a plot in tau sigma or tau sigma space of the progression of S T points that means that when different states of soil stress states of the soil and the final stress state can be at failure, final stress state can be at failure and representing the loading process of a on a sample. So you know this is one stress point which is actually shown here but when we have got number of series of Mohr circles and they represent each and every time the Mohr circle is actually having new diameter and new stress points and but when you join these two and that actually represents the so called the stress path for this condition which is actually being defined. So a stress path basically is a plot which is drawn in the tau sigma space which is connecting a progression of S and T points where S is equal to S and T which we have defined from the two dimensional point of view representing the loading process on a sample. So where you have the S is equal to half sigma 1 plus sigma 3 by 2 and T is equal to half sigma 1 minus sigma 3 by 2.  So the representative of the successive states of the stress as sigma 1 increases with sigma 3 constant. So when you have the sigma 3 constant and this is for Mohr circle A, Mohr circle B, Mohr circle C, Mohr circle D, Mohr circle E. So series of Mohr circles, so the representation of the successive states of the stress as sigma 1 increases with sigma 3 constant points A, B, C, D, E they represent the same stress conditions in both the diagrams. So this points which are actually because it is not possible to indicate the number of circles when we do the conditions with different sigma 3 values. So when you will be having number of Mohr circles on the tau sigma space so that is difficult and can lead to confusion. So for that what it is actually simplified is that the stress path only simplifies a path where in point joining the points where the maximum shear stress on the Mohr circle is generated. So that points actually are picked up here and the A, B, C, D, E is actually known as a stress path for the condition where the sigma 3 constant and sigma 1 increases. So this is ST diagram or QP diagram where Q is equal to sigma 1, sigma 3 by 2 and Q is equal to sigma 1 minus sigma 3 by 2 where Q is equal to also T is equal to sigma 1 minus sigma 3 by 2. So the duplication of successive states of stress that exist in a specimen is represented by a series of Mohr circle that is what we have discussed and plot with the series of stress points and when we connect these points is called as a stress path. The locus of the stress points this is also called as the locus of the stress point. So the plot this stress path is nothing but a plot with a series of stress points and connecting these point when you connect these points and that is called as a stress path which is here indicated again as A, B, C, D, E which is actually inclined at 45 degrees because this is nothing but the plane of tau max and 45 degrees to the principal plane. So the principal plane, major principal plane is here and minor principal plane here. So it is actually 45 degrees to the principal plane. So in this case if you look into it and this is the major principal plane and this is the minor principal plane. So this stress path for this condition is inclined at 45 degrees to the principal plane. So the plane of tau max, so this is nothing but the plane of tau max and 45 degrees to the principal plane. So this stress path represents a state of stress and successive states of stress and the stress path need not be a straight line depending upon how we actually vary the incremental stresses. Sometimes we can actually vary non-linearly also. In that case the stress path actually is also formed as a curve. So stress path need not be a linear version, it can be also need not be a straight line, it can be non-linear also. So here when we have that ds which is nothing but small change in sigma 1 plus sigma 3 by 2 that is half d sigma 1 by ds is indicated as d sigma 1 plus d sigma 3 by 2 and dt is indicated as half d sigma 1 minus d sigma 3. When sigma 1 changes d sigma 3 is equal to 0, when sigma 1 changes that means that you are actually having sigma 3 but d sigma 3 is 0 then in that case ds is equal to half d sigma 1 and dt is equal to half d sigma 1. So you can see that ds and dt both are same, ds and dt are both are same, ds and dt are both are same. The change in ds and change in dt both are same. When sigma 3 changes and d sigma 1 is equal to 0 then ds is equal to d sigma 3 by 2 and dt is equal to minus 1 by 2 into d sigma 3. So dt is equal to, so if you look into here, in this direction sigma 3 is unchanged here and sigma 1 increase then the path goes like this, sigma 1 increase the path goes like this and sigma 3 unchanged, sigma 1 decreased then the path comes down like this. And when you have sigma 1 unchanged that is d sigma 1 is equal to 0 and sigma 3 decreases that means that you are actually releasing the confining pressure then this comes down like this. When you have sigma 1 unchanged there is certain sigma 1 and sigma 3 is continuous then the path actually goes down 0 here. So the stress path actually changes from this direction either this side or this side depending upon how the sigma 1 sigma 3 is the stress conditions which are actually changing on the sample. So in this particular slide what we have tried to explain is that ds is equal to of d sigma 1 plus d sigma 3 by 2 and dt is equal to of d sigma 1 minus d sigma 3 by 2. So here when sigma 3 changes and only d sigma 3 is equal to 0 then ds is equal to d sigma 1 by 2 and dt is equal to d sigma 1 by 2. When only sigma 3 changes d sigma 1 is equal to 0 then ds is equal to d sigma 3 by 2 only sigma 3 changes and d sigma 1 that no change sigma 1 remains unchanged then dt is equal to shear stress actually decreases so you can say that this is you can see that the path is actually coming down see here is coming down. So all possible stress paths when only one of the sigma 1 or sigma 3 changes are straight lines at 45 degrees with the horizontal with dt by ds is equal to 1 when sigma 1 alone changes when dt by ds is equal to minus 1 when sigma 3 alone changes. So here important point to be noted here is that all possible stress paths when only only one of the sigma 1 or sigma 3 changes are straight lines at 45 degrees and with horizontal and with dt by ds is equal to 1 the dt by ds is equal to 1 dt by ds is equal to 1 when sigma 1 alone changes and dt by ds is equal to 1 when sigma 3 alone changes. So this observation need to be noted. Now let us take an example problem where we have got initial condition which is hydrostatic condition sigma v is equal to sigma h you have got some cylindrical sample where you have applied vertical stress and horizontal stress that means that the sample is confined with in hydrostatic pressure which is actually having that means that the sample is actually with a pressurized system and then it will be subjected to sigma v in vertical direction sigma h in other direction but when it is with the hydrostatic condition the sigma v is equal to sigma h and during loading or unloading it can be like when you have the initial conditions and delta sigma v can increase or you can actually have delta sigma h can increase or decrease that means that here when you have the conditions are that during loading and unloading the sample will be experienced in something like increasing in delta sigma 1 decreasing delta sigma 1 and or when this is maintained constant there is increase in the delta sigma h or decrease in the delta sigma h. So the stress parts for this case with the different stress points parts for initially hydrostatic condition. So as we have said that initially hydrostatic condition means you know that path actually starts along the p axis or this is s is equal to sigma 1 plus sigma 3 by 2 q is equal to t is equal to sigma 1 minus sigma 3 by 2. So you can see that we have got different paths now a that is the if this is let us say a point t ta, tb, tc, td, te and tf are the different stress parts and out of this you can see that path a which is with condition delta sigma h is equal to delta sigma v and path b what we have done is that we have maintained delta sigma h is equal to half delta sigma v and in stress parts c what we have done is that delta sigma h is equal to 0 and delta sigma v is increasing that is what we have discussed in the previous example. And path d where delta sigma h is equal to minus delta sigma v here we are actually maintaining delta sigma h as minus delta sigma v and e where delta sigma h decreases and delta sigma v is equal to 0 that means that there is a certain sigma v on the sample and the delta sigma h is being decreased and in path f where delta sigma h increases and delta sigma v decreases. So this example you know this is the solution for this example problem when you actually have got you know variable loading conditions and this can be worked out like this the initial conditions of the for all the soil all the stress parts are p0 is equal to sigma v plus sigma h by 2 and with that sigma v is equal to sigma h.  So when we have you know this when we substitute for this conditions what we have for the stress path a in order to obtain the stress path a what we do we what we have been given the loading condition is that you know initially sigma v is equal to sigma h and delta sigma v is equal to delta sigma h that means that q f is equal to sigma v plus delta sigma v minus sigma v because sigma h is equal to delta sigma v minus delta sigma v by 2.   So we can see that the stress path the stress path a moves out of the p axis by an amount equal to delta sigma v is equal to delta sigma h that means that the stress path which is you know if you take q by delta q by delta pf it is 0 0 by delta sigma v so that means that the stress path actually travels us along this and it moves by a distance delta sigma h is equal to delta sigma v and this is nothing but the q f and pf are nothing but the coordinates where q f is equal to 0 that means that is on point is on the p axis or s axis which is nothing but s is equal to sigma 1 plus sigma 3 by 2. Similarly we have you know the stress path b the initial conditions are same but what we have done is here is that delta sigma h is equal to delta sigma v by 2 that means that delta sigma h is equal to delta sigma by 2 the stress path b is actually traversing at an inclination of we can see that 18.4 degrees.   So with that what we got is that 1 by 4 delta sigma v and similarly pf is equal to when you simplify this we get sigma v plus 3 by 4 delta sigma v. So these points are nothing but the coordinates of the at the end of the stress path q f and pf are the points coordinates of the at the end of the stress path b and q and p both increase by delta sigma v by 4 and 3 delta sigma v by 4. So both are actually increasing you can see that this is increased by delta sigma v by 4 and this is increased by delta 3 by 4 of delta sigma v 4 this implies that this stress path as a slope of you can see that this by this what q by p we get the slope as 1 by 3 or 18.4 degrees. So hence we can actually draw now this stress path b if t b as the you know with an incline and with the coordinates of what we discuss is 1 by 4 delta sigma v and then 3 by 4 delta sigma v it is increased by that much amount and then that is the stress path for the loading condition where delta sigma h is equal to delta sigma v by 2. The third loading condition here for stress path c is delta sigma h is equal to 0 that is no change in the sigma h but sigma v continue to increase.  So this also increases by amount you can see that both they are q f and p f they are the coordinates of stress path c and both increase by magnitudes which is equivalent to delta sigma v by 2 and delta sigma v by 2. So this implies that the slope is 1 so q by p which is 1 which implies that the slope actually the stress path actually has got a slope of 45 degrees. So now this Tc which is the condition where this delta sigma h no change in the cell pressure and that is the lateral horizontal pressure and delta sigma v increases and that is nothing but this stress path Tc where the coordinates of this which increases by the point which is what we have shown as delta sigma v by 2 and delta sigma v by 2. Now we have stress point D so in this stress path D the condition what has been maintained is that delta sigma h is nothing but minus delta sigma v, the stress path D where the condition here is maintained is that delta sigma h is minus delta sigma v. So by using the same principles substituting in this and writing here with sigma h is equal to sigma v and you know when we put this what is happened is that you actually have increased by delta sigma v so this actually has increased by delta sigma v and this point is nothing but sigma v plus delta sigma v plus we have sigma v minus delta sigma v because the delta sigma h is equal to minus delta sigma v. So with that what we get is that you know these two will get cancelled then what we have is that 2 sigma v by 2 which is nothing but sigma v. So the values of PQ are the coordinates of the end of the stress path D and Qp both increase by delta sigma v and sigma v and this implies that the stress path D has a slope of 90 degrees. So that is the reason why you can see that the stress path is actually traversing vertically up with a coordinate here what you can see that delta sigma h is equal to minus delta sigma v it is actually traversing up here with a 90 degrees slope. So for E let us consider for stress path E where delta sigma v is equal to 0 that is the change in vertical stress is 0 but in this self pressure delta sigma h is actually decreased by some amount. So by using the same principles where Qf is equal to sigma v plus delta sigma v minus sigma h plus delta sigma h by 2 by substituting here what we have what we will get is that sigma v plus 0 that is delta sigma v is equal to 0 minus sigma v minus delta sigma h by 2. So wherein we get Qf is equal to delta sigma h by 2 and Pf is equal to sigma v plus 0 plus sigma v minus delta sigma h by 2 wherein we have got sigma v plus delta sigma h by 2. So we can see that the values of Q and P are increased both by delta sigma h by 2. So this implies that the slope of E has a slope of minus 1 and 45 degrees so this traverses for this load condition which is what we have discussed is that like this. So when delta sigma h decreases and with delta sigma h decreases sigma h decreases with delta sigma v is equal to 0 you can see that this traverses in this direction towards the Q axis and which is with inclined at 45 degrees and increased by the same amount both in delta Q and delta P and when you have delta sigma h is equal to delta sigma v that is when you have delta sigma h is equal to 0 and delta sigma v increases it traverses in this direction when you have in this direction that means that here what is actually happening is that with a constant sigma 1 when you have and when delta sigma h is actually the when your sample is actually losing the confinement that means that there is sort of the sample is subjected to a tension. So you can see that is actually happening here whereas in this case where delta sigma h is actually 0 there is no change in the cell pressure or the horizontal pressure but when delta sigma is increasing this is actually towards the compression and this is towards the tension. So you can look into this the way what we have done is that we try to you know see you know these plot these test paths by actually following the fundamental principles wherein in this particular example wherein it is subjected to initially hydrostatic conditions are considered but suppose if you are not having hydrostatic conditions initially when say delta sigma v is not equal to delta sigma h or sigma v is not equal to sigma h then that means that you know the origination of this test path location itself somewhere here and from there again it actually follows the you know different paths can be drawn. So for the in this particular example what we have considered is that the initial condition is hydrostatic compression where we have actually sigma v is equal to sigma h and when we have this when we have got different variations of delta sigma v and delta sigma h then we have actually drawn and then we have actually also seen how this procedure can be adopted for plotting this stress paths. So you can see that here the stress paths are actually plotted and then they are nothing but the locus of the you know points on the Mohr circle and instead of this actually avoids the series of Mohr circles when you actually have different stress conditions. So the stress paths are the you know the straight lines joining the locus of the stress states on the Mohr circle and they are actually need not be straight line and depending upon how we exercise for example when you have delta sigma h is equal to half let us say delta sigma v square by half delta sigma v square then in that case it can actually undertake as a curve also. So you actually get a non-linear variation also so not necessarily the straight lines but when we have these conditions of these you know delta sigma h is equal to delta sigma by 2 and when we have delta sigma h 0 and delta sigma v increases and when we have delta sigma h decreases and delta sigma v 0 then it actually follows and it is actually nothing but the plane of maximum shear stress wherein it actually has got and it actually inclined at a 45 degrees to the principal plane. So in this particular lecture what we have done is that we discussed very briefly about the principal stress space and then we have tried to introduce ourselves to stress paths and seen some examples where how these stress paths can be plotted in pq space or st in the two dimensional space where s is equal to q is equal to sigma 1 minus sigma 3 by 2 and p is equal to s is equal to sigma 1 plus sigma 3 by 2. So further we will actually discuss on this and then when we introduce ourselves to the traction compression test then we will try to see when we have got a different drainage conditions undrained and drained how we can actually draw effective stress paths and total stress paths and with some examples and that makes very clear.


Welcome to the lecture number 14 of the course, Quantum Mechanics and Molecular Spectroscopy. As usual, we will have a quick recap of contents of lecture number 13 before we proceed with the lecture number 14. We started with the time depend perturbation and we showed that this was equal to i h bar q by m a del, where del is nothing but d by dr, where r is general coordinate, ok. It could be x, y, z or whatever, ok r, ok. Now, this we could transform to q e naught by omega m cos omega t e dot p, ok, because you should realize i h bar del minus i h bar del is equal to operator p, ok. So, now what you are looking at is the dot product between the electric field vector and the momentum vector. This is nothing but dot product and momentum vector, ok. That means, for this to h prime t to be non-zero, the electric field and momentum should be in the same direction or should have enough projection on to each other, ok. Now, we know that r h naught, h naught commutator, ok, which I derived in the last class is equal to i h bar by m. So, one can replace this momentum operator by r h naught, ok. Having all done, we come across this equation f h prime of t i.  Minus i e naught cos omega t divided by h sigma over n q n epsilon n r n i minus e i f sigma over n q n epsilon r n i, ok. These are the integrals. Now, you will see that nuth sigma over n q n r n, this is nothing but dipole moment of the molecule mu or the atom, ok. So, this is just tells you the, how the charge is distributed, ok. So, when we do that and do little bit of rearrangement, what you get is f h prime t i equals to e naught cos omega t omega i f pi omega integral f, where h bar omega i f is equal to delta e is equal to e i minus e f, ok. There is a e f minus e i, but there is a negative sign. So, I kind of cancel that, ok. And there is an i here. So, there is a ratio of omega i f to omega and there is this integral, ok.  Now, e naught into e vector is going to give you electric field e. So, f.  Mu also is a vector by the way the dipole moment is a vector, ok. So, essentially you know if you leave out this, ok, it depends on this integral. So, this integral has been transformed with some prefactor into this integral, ok. So, you can think of it like this f h prime of t i is proportional to f e dot mu, ok. And this integral is called transition moment integral. So, f e dot mu i, ok is one of the most important integrals in spectroscopy. It is a transition moment integral and that will determine whether the transition will take place or not. If transition moment integral goes to 0, ok, the selection rule is decided by this.  Let us suppose that, ok, there is an electric, there is a electromagnetic radiation which is going like that, ok. So, this is the direction of propagation r, ok. And your electric field vector e is along this, ok. Of course, when it goes down it becomes minus e. So, the electric field vector is going up and down, ok, in this direction e and me. So, it changes sign. Now, if you take a diatomic molecule, let us consider a single diatomic molecule AB. And since the diatomic molecule it is linear, the dipole moment is in this direction. So, if you consider a as a positive and b as negative, the negative direction is it. So, it is, ok, positive, negative, ok. So, a. So, now, you can see that if the propagation direction is r and the dipole moment is along the r. So, now, think of it like this. So, if your light is propagating like that and your molecule AB is like that, ok. Now, what is the e? e is perpendicular. So, your epsilon dot. So, this is your mu and this is the director. So, and your epsilon vector is like this. Epsilon dot mu is equal to 0 because the dipole moment of the molecule AB and your vector epsilon are perpendicular to each other. That means, under the circumstances, the molecule will not absorb light or there will be no transition. On the other hand, if AB is like this, if you have AB like this and your propagation direction is like this and your electric field vector is like this. Now, epsilon and mu are parallel to each other. So, the dot product will be maximum. So, this will be epsilon dot mu will be not equal to 0 or any direction that is has a projection. That means, so this will have some projection on to epsilon. Even that case epsilon dot mu will not be equal to 0, ok. That means, the transition moment integral that is F, transition moment integral is equal to E naught dot mu. This will become nonzero. So, which means, if the dipole moment and epsilon, the electric filter should have projection on to each other, only then the transition moment integral will not to 0. So, that tells you the direction in which the electric field vector and the dipole moment have to be aligned with respect to each other, ok.  So, as I told you this is nothing but your transition moment integral, ok. And I also told you that the dipole moment and the electric field should have projection on to each other. They must be aligned. If they are perpendicular to each other, there will be no the transition moment integral will go to 0, ok. Now, there is another thing that you can do is that if you now this is just a transition, but your transition probability P F of t is given by, ok, 1 by H bar square integral of 0 to t prime dt e to the power of minus i omega i F t integral F H prime of t i whole square. But now we have evaluated this integral as this, ok. So, this will be equal to E naught square E naught square by H bar square omega i F by omega whole square 0 to t prime dt e to the power of minus i omega i F t cos omega t integral F epsilon dot mu i whole square. Now, this is a integral that does not involve time and this is a integral. So, I can separate it out as two integrals. So, this is nothing but E naught square by H bar square omega i F by omega square integral 0 to t prime dt e to the power of minus i omega i F t cos omega t square to F epsilon dot mu i whole square, ok. So, we will know how to evaluate this integral if you know E dot mu operator and if you know the states f and i, ok. And these are just the pre factors. So, now it comes to evaluating this integral, now let us look at that integral. So, the integral 0 to t dt e to the power of minus i omega i F t cos omega t is something that I need to evaluate, ok. Now, this integral this is equal to 0 to t prime dt e to the power of minus i omega i F t cos omega t cos omega t is can always written as e to the power of i omega t minus e to sorry plus e to the power of minus i omega t by 2, is not it cos theta definition is that, e to the power of i theta plus e to power of minus i theta by 2. Now, if I slightly rearrange this equation will become 0 to t prime dt e to the power of if I take i common then I get omega minus omega i F t plus e to the power of minus e to the power of minus i if I take minus i common then I will get omega plus omega i F and there is a half outside I will get 2 integrals. Now, I am going to slightly rearrange this. So, this is equal to half integral 0 to t prime. So, omega i F will become. So, this will become e to the power of i omega F i plus omega t 2 plus e to the power of my, ok. Now, this omega F i if it becomes minus omega F i minus of omega F i and I can take minus of here from outside. So, it will become i into omega F i minus of ok. So, I have these 2 equations now. So, which means I have this is equal to you will get 2 integrals, integral 0 to 2 prime e to the power of i omega F i plus omega t dt plus half integral 0 to t prime e to the power of i omega F i minus omega t, ok. So, this is our next step half of integral 0 to t prime e to the power of i omega F i plus omega t dt plus half integral 0 to t prime e to the power of i omega F i minus omega t dt. Now, what I will do is I will multiply pi and divide by pi for both of them. So, this is equal to pi into 1 by 2 pi integral 0 to t prime e to the power of i omega F i plus omega t dt plus over 2 pi 0 to t prime e to the power of i omega F i minus omega t. Now, I will make a small mathematical trick. Let us suppose you have a some atom or a molecule some system, ok. So, let us say some atom or a molecule and a time t is equal to 0 you switch on the light, ok. So, and a time t is equal to t is equal to t prime you switch off the light. So, what you do is that you start switching on the light and then a time t is equal to t prime equal. So, this is 0 and this is t prime. So, you switch on the perturbation at time t is equal to 0 and switch off the perturbation at time t is equal to 0. And you will see now this will be minus infinity and this will be plus infinity, ok. Once the perturbation is switched off or not switched on, ok, time does not the perturbation does not really add that means, ok. Instead of integrating over from 0 to t because then what happens here it is all zeros no perturbation, ok. So, which means if you have no perturbation it does not add to the integral. So, which means instead of 0 to t prime as you are integral definite integral ends, ok. You can write it as minus infinity to plus infinity. Even though it is from minus infinity to plus infinity the actual region in which the light adds or the light acts on the molecule is only 0 to t prime rest of the time you are just adding zeros, ok. So, it does not really matter, ok. So, I want to make this transformation it will not change any physical principle. However, it will deduce these equation to some standard integrals. So, when I have done that then I can write it as pi into 1 over 2 pi minus infinity to plus infinity e to the power of i omega f i minus omega plus omega into t dt plus 1 over 2 pi integral minus infinity to plus infinity e to the power of i omega f i minus omega t, ok. So, what I have done is I have just changed the limits of your integral instead of going from 0 to t prime I have taken from minus infinity to plus infinity, minus infinity to 0 there is no light acting on it, ok. So, that integral essentially the contribution to that integral is essentially 0, but after t prime to infinity again there is no light. So, the contribution to the integral from t prime to infinity will also be 0. So, essentially what you are doing you are adding lot more zeros and the number is not going to change by adding zeros, but this transformation or this simple mathematical trick allows you to get to a standard integral, ok. So, this is our integral that we need to evaluate, ok. So, I will stop here and continue in the next lecture. Thank you.


What is leukemia? Leukemia is a cancer of the blood-forming organ called the bone marrow. And also it's a cancer sometimes of the lymphoid tissue, which also is part of the immune system. It's estimated that in the U.S. we're going to have 60,000 cases of cancer leukemia this year. And it's one of the leading cause of mortality, being the sixth cause of mortality related to cancer in the U.S. Leukemia is a cancer of the blood-forming organ system called the bone marrow. And it also involves this immune system called the lymphoid tissue. It is suspected that we're going to have 60,000 cases of leukemia diagnosed this year. And it's expected to be the sixth leading cause of mortality in the U.S. Can you explain the different types of leukemias? We have classified the leukemias according to the time and according to the cell of origin. In terms of the time we have acute leukemias and we have chronic leukemias. And in terms of the cell of origin we have the myeloid and the lymphoid. Therefore, we have four types of leukemia in the broad spectrum. Acute myeloleukemia, chronic myeloleukemia, acute lymphocytic leukemia, and chronic lymphocytic leukemia. What are the symptoms of leukemia? The symptoms of leukemia vary according to the type of leukemia we have. For acute leukemia, the symptoms seem to be very fast. For chronic leukemias, the symptoms develop over years or many months. In general, the symptoms for leukemia are related to the side effects in the immune system of the rest of the Balmar reserve. For the immune system, you can have fevers, chills, night sweats, or even multiple infections. For the Balmar compromise, you can have anemia, low platelets, bleeding. There's also some mention of out-immune diseases related to leukemia. There can be enlarged liver, enlarged spleen, an invasion of the leukemic cells into the other organs. How is leukemia diagnosed? The diagnosis of leukemia depends again on what type of leukemia you have. In general, we looked at the physical examination. We looked for enlarged lymph nodes, enlarged liver, over spleen, any signs of infiltration of the leukemia into any part of your body. We will look at your bone marrow, and we'll also look at your floating cells, looking for mutations and determine if the leukemia is present or not, and what type of leukemia there is. What are the risk factors of leukemia? That's a great question and is asked very often in my consultations. The risk factors for leukemia are really vary according to the type of leukemia. In general, we know that we need a genetic predisposition and an environmental factor to produce a leukemia. Some of the risk factors that are already known, for example, are genetic predispositions like Down syndrome, dyskeratosis congenital, these kind of symptoms that change your immune system or make you have more mutations. Other things like exposure to chemicals like benzines or like oil derivatives. Other risk factors include exposure to radiation, prior chemotherapy, and in general, you also will need an environmental factor added to this. How is leukemia treated? Treatment for leukemia depends on how the disease is manifested, either acute leukemia or a chronic leukemia. In general terms for acute leukemia, we tend to use chemotherapy plus, minus other targeted therapies or therapies that are directed to the cancer cells per se. For chronic leukemias, there are some ones that you don't really need to treat unless they become symptomatic or you have complications like chronic lymphocytic leukemia. And for chronic myeloid leukemia, we have special targeted therapies that are taking away the directed chemotherapy per se. The other part of the treatment for leukemias is the coming up with the immune system modulators and the immune system regulators helping us balance in the immune system to help us fight the disease. And the last part of this is the bone marrow transplantation, which we have the autologous bone marrow transplantation and the allogeneic bone marrow transplantation. What can I do to prevent from getting leukemia? The prevention therapies and prevention strategies for leukemia has not been very well developed. We don't really know how to 100% prevent leukemia from happening because, as we said, it's environmental factors plus genetic predisposition. But maybe we can change the environmental factors, exposure to chemicals, radiation therapy before patients that have had also family history of leukemia. These patients should be evaluated by their primary care physicians every year looking for changes in their blood counts or their physical exam that might prompt an evaluation by a hematologist and oncologist. In terms of the general health, it's always encouraged to have a good balanced diet and exercise. At least, it might help modulating the immune system and help us might be preventive for some of these cancers. What is the role of stem cell transplantation in leukemia? Stem cell transplantation is a process in which we give you a new bone marrow to treat the diseased bone marrow. This bone marrow will have the stem cells that will help you produce new cells. This process is usually given to patients with acute high-risk leukemias like acute lymphocytic leukemia or acute myelo leukemia. In this process, we hope to maintain the leukemia at bay by the effect of the graft versus the leukemia effect and also achieving control of the leukemia by the initial chemotherapy that is given to the patient. In general terms, these stem cell transplantations are given to patients with acute leukemias. Can you describe the hematologic malignancies program at NYU Winthrop? We have developed a comprehensive hematological malignancy program that has all of the resources that we need to take care of our patients with the standard of care and above the standard of care. We have a nurse navigator dedicated to help our patients go through all of the layers of care. We have social worker that help us in our patient needs in terms from economical needs to social support. We also have a nurse that is dedicated to us in terms of research and research coordinators and we have developed this team that it's able to provide for our patients in the most comprehensive manner and with the latest scientific advances to take care of our patients the best we can. What is your sense about the future treatment of leukemia? I believe that the leukemia treatment is evolving and every year we get new therapies being approved. The time for personalized treatment of leukemia is here. That is something that we do right now at NYU Winthrop. We look into specific mutations and look for targets that we can use to treat your leukemia. I believe that the leukemia treatment will go to be more specialized, more into targeted therapies, more into immune modulator therapies with the hope of cure. For more information about the NYU Winthrop Malignant Hematology Program call us at 866 Winthrop or visit us at NYUWinthrop.org.


Joys are gone before they can be possessed or even grasped. That's an interesting line from Jesus. Joys are gone before they can be possessed or even grasped. You know, if you have a glimmer of joy, or you just feel happy for no reason and everything, the ego will try to turn it into some kind of possession. You know, it will try to put the cause outside of your mind onto the world for where, oh, I have joy with this person, or I have joy in this particular place or spot. It will try to immediately tie it to form. And then, but Jesus is saying they're gone before they can be possessed or even grasped. He's just saying how fleeting and ephemeral is those little spots of joy are. I find that true joy, the only way it comes into consistent awareness is through purpose. And Jesus is the way-shower for that. You know, instead of being psychotic, he was one with the Father, I and the Father are one of that. He said the Kingdom of Heaven is within. He demonstrated a sense of light-heartedness, happiness, kindness. He was very uncompromising. He demonstrated the will of the Father on earth, so to speak. And the only way you can be consistently joyful is to hear one voice.  And I have had people recently tell me that they're coming to a more state of consistent peace but they still don't feel like an exuberant joy and it's like a real full joy. And I think that just comes when you become really tuned in to the Holy Spirit and consistently in alignment with that. So, but it's great to just rejoice and celebrate even when you get these miraculous glimmers. And you can just praise God and give all the glory to God and everything, which is the safest thing to do. And then the ego, don't follow it when it tries to say, oh, you have joy because you're with this person or because it's this situation or something because you've got it right because it's not going to become consistent unless you're in the passenger seat and you're not trying to steer the car, you know, when you're in such a place of following where you put the Holy Spirit of Jesus as the driver and just be a happy passenger. It's almost like the Gospel of Thomas, be passers-by with this world. You know, don't attribute the joy to anything in this world because it's so ephemeral. It's just you aren't getting the joy from the images. It's coming from inside.


Mostly Tom Brady said yeah, I am retiring I mean yeah, I am retired But who knows what six months is gonna say I just make the best decision for me in the moment Well, what's gonna happen in a moment when the San Francisco 49ers say we just trade Jimmy G out Why don't you come back home when the Super Bowl that's the team's ready to go Trey Lance can learn from it. Is that gonna happen? That's kind of what it sounds like Well, why are you put him in San Fran Insta Lee though? Well, I think that's just a natural thing because we all seen the photo of him when the San Francisco 49ers gear He's from NorCal and allegedly at one point he wanted to go over there in Shanahan and Lynch had to have the conversation whether or not we want to go with Tom Is this a lease? Is this a rental or do we want to go with Jimmy G and we all thought that Jimmy G was that motherf**ker from the barbershop show Actually turns out it was Fitz magic in Miami, right because we're about Trey Lance. What about the guy stinks


This is stock number 9223X. We are here at Summit Automotive in Fond du Lac, Wisconsin. We are checking out this super cool, super clean 2014 Jeep Wrangler Unlimited Polar Edition. This vehicle has the 3.6 liter Pentastar V6 motor. From this HD video you'll be able to tell that this Jeep is extremely clean all the way around. We got factory fog lights, the black gloss center bumper piece. Front bumper is in excellent condition. We take these HD videos in 1080p so if you have HD capabilities turn them on right now because it's like you're right here next to the vehicle with me checking it out. It has pro comp alloy painted alloy rims, Goodyear Wrangler Duratrac LT31570R17 tires and these tires look like they have close to 70 to 80% of the tread left. It has a Rubicon Express shocks on it and I would guess that's the type of spring that they did as well. So it's got a nice little lift kit on it. You can see the Polar badge right here, 78 degrees south, 106 degrees east badge, little mountains on it. You also get these decals on the hood along with the power dome hood. This hood is in excellent condition. I didn't see any dents or dings on it. Hydro blue is the color. Passengers side rim is in excellent shape, no scuffs or scrapes on that and you can see just how clean that frame and underbody is on this vehicle. Down this side of the Jeep you can see just how clean and reflective the paint is, how clean the body is, the door hinges are in really nice condition. I didn't see any corrosion on them. It has the rock rails which are also part of the Polar package. It has the color match hardtop along with the color match fenders. We take these HD videos so if you are far away or even if you're close by and just cannot make the trip down, you can still see the vehicle, hear the vehicle and have confidence in what you're looking at before even get here. Back tires have just as much tread as the front tires and you can see that the frame and underbody on this Jeep is in really nice shape and it's like that all the way underneath. Those rock rails are in pretty nice condition as well. Back here you can see the hydro blue color just a little bit more with that. It's got a lot of pearl flake in it. This one has a towing package which includes the receiver hitch and wiring. This back rim is in excellent condition as well and this tire has just as much tread as the rest of them which tells me that they rotated their tires pretty regularly on this Jeep. Back here we have the factory subwoofer. This one also comes with the dual top group so there's the soft top for the vehicle and then you have more pieces and parts for that soft top in that box right there. Underneath here we have the little storage area. There's more pieces for that soft top but you have your door hinge bolt and you have your roof bolts in here and actually they have some of the roof bolts in here. Looks like a couple extra of them but very clean back here. You do get the Jeep floor mat which turns into an all-weather floor mat if you flip it upside down and very clean back here. Those seats do fold down for extra storage so a full-size spare tire with a matching rim. Rear bumper is in pretty nice condition. I didn't see any major dents or dings on it and as we go down this side the Jeep just as clean as the passenger side that color match hard top is an excellent condition and once again down this side very very clean. You can see just how glass light and reflective that paint is. Door hinges on this side look like they're in pretty nice condition as well.  You get the little mountain stitching in the backrest on this one. There are no rips or tears on the driver's seat height adjuster. Factory all-weather floor mats. Also part of the Polar package you get this instrument cluster in here with the little mountains on it as well and it's kind of like a ice blue background. You get the white trimmed steering wheel. This is a leather wrap steering wheel. Bluetooth audio controls. You got compass temperature and all your instrument cluster controls right there. It looks like it has auto headlamps on it as well but that steering wheel is in really nice condition.  So all that's part of the Polar package. AM FM series satellite radio capabilities. You can see it has dual heated seats, stability control, power mirrors. You have your four by floor shifter on the floor, automatic transmission, five speed and passenger seat is very clean as well. No rips or tears. I don't think this Jeep's ever been smoked in. Smells very clean inside. The roof is in really nice condition as well. We'll take a quick look at the back seats. Very clean inside. Very nice Jeep outside. Very good looking. Back seats are very clean as well. No rips, no tears back here. You have the latch child safety system and all-weather floor mats. These seats do fold down almost completely flat which is a really nice feature. Both sides do that and I have the Wrangler myself. Love it. Probably the funnest vehicle I've ever had. Let me put it this way. You won't be sorry that you have a Wrangler. Always something to do in them. Engine bay is very clean. Engine runs very smooth. This is a 3.6 liter Pentastar V6 motor bent on Ward's top 10 engine list four different times. This particular Jeep has been fully safety and inspected by our service shop has a fresh oil and filter change. All the fluids have been checked and topped off and this Jeep is 100% ready to go. Really sharp looking. Probably one of the best Wranglers we've had on our lot as far as just a nice stance, a nice lift, nice rims and tires and the color of course. And to see more pictures of this Jeep or one of our other 400 new and used cars, trucks, SUVs, minivans, Wranglers, you name it we got it go to our website www.summitauto.com. Full pictures and descriptions of every single vehicle on our lot. Videos of every single used vehicle that we have all at summitauto.com. Thank you so much for checking out my video. If you like it, want to make this Jeep years give us a call right now<PHONE_NUMBER> ask for one of our sales associates to make this Jeep years today. Once again that number is<PHONE_NUMBER>. If you like the video and want to check out more Wrangler videos go to our YouTube channel which is youtube.com slash summit auto. Remember to like subscribe and share and in fact in a second you will see a likes subscribe button on your left. A link to more Wrangler videos like this one on your right and if you haven't been to our website a link to this Jeep on the bottom to this Jeep on our website click those check us out see what we're all about we look forward to helping you with this super clean 2014 Jeep Wrangler unlimited Polar Edition. Thanks again.