Principle of Material Objectivity
So, welcome back to one more lecture on Introduction to Soft Matter. Last time we were discussing issues concerning constitutive modeling and in particular discussing the principle of objectivity.(Refer Slide Time: 0:46)So, the one of the last things, that we were reading out was this line by Professor Suo, where he says that, “We construct variables in variant with respect to rigid body motion, later we will use this variables to construct Rheological models invariant with respect to rigid body motion.”(Refer Slide Time: 1:05)And here we have just stated that equivalent observers, two observers, who are basically related by a general rigid body motion.(Refer Slide Time: 1:14)
So, rigid body motion, you can show that, I am not going to show it here. But it is, it can beshown that rigid body motion, basically involves, a rotation and a translation. So, let us say youhave two frames F is your current frame and we will denote all quantities in another frame by*F and there these two frames are equivalent observers and they are related to each other by arigid body motion.Now, rotation between two frames, rotation between two frames can be represented by a matrix,so three dimensional, if we have 3-dimensional space, then the rotation matrix will also be 3 x 3.So, the rotation between two frames can be represented by a matrix, let us call it Q and I just putdouble bar here to indicate that this is a matrix t ( ()Qt ), where T QQ I = . So, Q is also oftencalled an orthogonal matrix.So, now if you have in the rotated frame you can represent, let us say, the new location for aparticle, let say you have a particle located x and you have another frame, which is rotated andtranslated with respect to the previous one. Then the, what is the new position for that same pointin the rotated frame, in the rotated and translated frame?So, we can actually write, this relationship that * x , which is the position of the same point in therotated frame can be given as some translation we will just called by c(t) just shows that it can bea function of time plus the position of this point with respect to some origin multiplied by Q,where Q is the rotation matrix. And here no loss of generality is incurred and I just make a notehere.* x = c(t) +Q(t)(x − o)So, no loss of generality is incurred by assuming that O is fixed. So, let us say, that you have ameasured field , in the previous, in the reference frame F and that field now becomes * in thenew field in the new frame. So, let a measured field be * in the frame * F . And let this be ascalar quantity for the time being, if it is a scalar quantity, then we say that the scalar fields areinvariant, if they retain the same value and they have changed a frame.So, we will say the scalar fields, such as density, for example are invariant to a change of frame,which implies that, in the rotated frame and the translated frame, if you measure in the *F frameyour measuring a density, which is * , that should be the same as the density that you aremeasuring in a another frame, in an equivalent frame. So, from this we can make our first rulefor scalars and that rule says that hence for invariant scalars * = .So, this is an important point right, because even though, this is a very simple idea, you mustunderstand or we must have a situation or a constitutive relationship, where the change of framecannot affect simple quantities like viscosity. So, whether you are measuring the viscosity ofwater on Planet Earth or Mars or somewhere else, that it is a material property, it cannot changewith just because you have changed your different special location and maybe at a different pointof time, you cannot change that.Again, as I said, this is an axiom. So, this is something that we have believe is true and webelieve, and we build our equations and our theory based on this. So, for invariant scalars, thissimple relationship holds, but this relationship will obviously not remain as simple for otherquantities.(Refer Slide Time: 8:22)So, let us now discuss the case of vectors. So, let us take our old idea of vector, which is that youhave two points and this is the displacement vector between the two points, the point here isgiven by some x these is a coordinates in one frame, reference frame, when you change to adifferent reference frame the coordinates also changed. So, you have now, this coordinates in thenew frame.Similarly, this is another point let us say this is y and in the rotated frame in the other frame,you have a rigid body translated and rotated frame, you have * y as the new value. Now,displacement we have to understand is something, that is truly frame in variant, becauseotherwise how will you even discuss lengths, if displacements are not the same in all for allobservers, how can you even agree upon a length.So, displacement is the, so the position vector of a point may not be independent of the frame,because that is tied down where your measuring it from. But the displacement vector betweentwo points that is truly a frame invariant. So, what we call separation is frame invariant, so wehad just returned above the equation for the change of reference frame.So, let us write down what * x is in terms of x , so we had said that x* = c(t) +Q(t)(x − o) .Similarly, I can write it down for y , the same equation I am going to replace x for y now.* y = c(t) +Q(t)( y − o)So, what is the, so we know that y , this displacement vector is going to be invariant. So, if Isubtract the two, I end up getting * * ( y − x ) = Q(t)(y − x) . So, this is important here now, so this,since we know that the displacement vector is invariant, we know that if we have to calculate theinvariant, if this form is the form of invariance.(Refer Slide Time: 11:48)So, we now what we are going to do is? We are going to say looking at this is that a vector isinvariant when it transforms as. So, let us say you have a vector u , when u transforms asu* = Q(t)u this rule. Now, there is one thing that I just like to point out and that is sometimes anissue between different books and different presentations by different authors, that here what weare trying to do is, this y* etc. this relates to the position to the coordinate values.So, if you have we can think of it as may be a column vector with three points for x, y and zcoordinates. So, these are actually the values itself and it does not, it is not the abstract quantityvector itself. So, I am taking, I am adopting a scheme, that is or I am adopting a formulation thatis taken by Phan Thein and some of the other authors, some other authors like Suo they say thatit is better to write in terms of the components. So, I just wanted to point that out.(Refer Slide Time: 13:44)So, good, displacement is invariant, but what about velocity, is it also and variant? So, let us lookat, let us ask also simple question, the question is by the way this Q is question. So, I may bewrite down question, that is we have already another Q. So, the question is how does velocitytransform? So, we know that, we have previously written that, expression, we let us just rewritethis.And this is also a function of time, we have to discuss the velocity. So, velocity, your transformvelocity, is nothing but the derivative of this, the** dxvdt= and here this is remember that this Qcan also be a function of time and you have x also have a function of times. So, you have toapply chain rule.So, first time maybe we will just take the derivative of ()Qt so we have ()Qt•, and then thesecond time we take the derivative of ( ) x t . So, we have ()xt•and o is fixed as we had just saidpreviously, there is no loss of generality in assuming that is just fixed and you have ()ct•.** ( )[ ( ) ] ( )[ ( )] ( )dxv Q t x t o Q t x t c tdt• • •= = − + +So, this implies, if you just clean out this expression with this quantity is the velocity, in theother frame. So, you have transpose, this is a velocity plus so I am just rearranging the terms hereand you will probably can understand why I am doing that.And we had just said that a vector is invariant if it transforms as this equation previously. So, thiswas our important equation. But we see that when we apply to velocity, you now have this anextra set of contribution here. So, velocity is not, is not frame invariant that is something youshould expect also, if you think about it velocity is a very important frame dependent, framesensitive quantity.(Refer Slide Time: 16:52)So, what about acceleration? So, we just may be do one more question. What about acceleration?So, we know that, we have already written that expression * v Q(t)v Q(t)[x(t) o] c(t)• •= + − + . So,*a , which is now going to be the derivative of this where we are going to have to take thederivative again. So, let us see what we get?** ( ) ( ) ( )[ ( ) ] ( ) ( ) ( )dva Q t v Q t a Q t x t o Q t v t c tdt• •• • ••= = + + − + + .So, this time, we differentiate the other term, we get acceleration, and differentiating the firstone, and in the second case, second term I am differentiating the first one applying chain rule,plus sorry, by the way I missed this. So, we can see that contrary to our expectation had this beeninvariant, that only this particular term would have been there, but we have so many other terms.So, acceleration is also not frame in different.So, we will make a note. So, acceleration is also so not frame indifferent, actually framesensitive. So, we see that despite the fact that displacement is frame invariant, its time derivativeis not. So, that also we should make us question as to are there other type of time derivatives,which might be, we will look at that later.(Refer Slide Time: 20:10)So, we have, we discussed scalars and we discussed vectors, how vectors should transform. So,the obvious question is how should tensors transform? So, we ask now, how should a frame indifferent, a frame invariant tensor transform? By the way you might have noticed that I haveused a way a word frame indifferent here and somewhere I am using the word frame invariant.Now, most authors from what I have seen they tend to use the two interchangeably at differenttimes. Although sometimes the principle of frame invariance or indifference is stated as adifferent principle. So, how should of frame invariant tensor transform? So, what is a tensor bythe way? We know what vectors are, but what are tensors? A tensor is something that cantransform one vector into another vector.So, let us say you have a vector x , y , the tensor T is a transformation of this vector into anothervector. And we want to understand how frame invariant tensors should transform, a what shouldbe the rule that should be governing that.(Refer Slide Time: 21:54)So, let us say in the other frame, this is *x , this is * y , and you have another tensor now atdifferent tensor and this will transform it into. So, that transform tensor should do the same job,this should take these displacement vectors and transform it to some other vector.(Refer Slide Time: 22:27)So, we wrote down before, let us write down the relationships that we have written before also,we let us say * y = c(t) +Q(t)( y − o) . And then * x = c(t) +Q(t)(x − o) . So, we are already justwriting down the expressions that we had agreed upon earlier. So, when we take this, then weknow that you have this is the same as the expression that we had previously written, which isgoing through that exercise one more time.So, similarly, your * * (b − a ) = Q(t)(b − a) . And this represents now the transformed. So, this isthe T should act on this quantity and convert it to **()ba− . So, * * * T ( y − x ) should give me thisquantity now, but this is T ( y − x) . So, and now I am going to skip a step here and I want you tosort of complete this, Q was the orthogonal matrix, in fact I can use that property and then I canassert that T * Q T Q = T , that is one step I have skipped and I have deliberately skipped. So, thatyou can fill in the gaps here.(Refer Slide Time: 25:19)And this implies that your * T T = QTQ and we are going to this as the rule to say. So, this we aregoing to do that attend, so we are going to say, so we are going to use this to make our rule. Atensor T is invariant when it transforms under a change of frame according to this relationship.Okay.So, what we have been able to do is today we came up with 3 different relationships one forscalars, one for vectors and one for tensors. And we said that if they need to be frame invariant,then there is a certain transformation law that should govern them and we found out what that is.We also discovered that the velocity, that while displacement of separation is frame invariantvelocity is not, neither is acceleration, they are frame sensitive terms.So, now that we have established the rule for tensors we will look at some quantity, some tensors, and we try to establish whether they are in frame invariant. So, we will do that in the next class. So, we will stop here today. Thank you.
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