important fluid stuff read before exam

important fluid stuff read before exam - Lecture 5 Vector...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 5 Vector Operators: Grad, Div and Curl In the first lecture of the second part of this course we move more to consider properties of fields. We introduce three field operators which reveal interesting collective field properties, viz. the gradient of a scalar field, the curl of a vector field. the divergence of a vector field, and There are two points to get over about each: The mechanics of taking the grad, div or curl, for which you will need to brush up your multivariate calculus. The underlying physical meaning — that is, why they are worth bothering about. In Lecture 6 we will look at combining these vector operators. 5.1 The gradient of a scalar field Recall the discussion of temperature distribution throughout a room in the overview, where we wondered how a scalar would vary as we moved off in an arbitrary direction. Here we find out how. If is a scalar field, ie a scalar function of position in 3 dimensions, then its gradient at any point is defined in Cartesian co-ordinates by  ©§¤   ¦ ¨¦  ¦ ¨¦ ¢ ©§¥¤£¡ &$" 2 )4 2 )1 7 8) 6  ¡ ( ( 5¨ ¡ ( ( 30¤ ¡ ( (  ¡ '%#! It is usual to define the vector operator which is called “del” or “nabla” 7  ( ( ) 6 2 ¨ ( ( ) 4 2 ¤ ( ( )1  9 59  ¦¨ §4) 6  (( 2 ) 4 ¨ (( 7  ©§¥¤¢ ¦ ¨¦ 9@ 80% 7   0% # # 7  &%  ) 6  2 ) 4 ¨ 2 )1 ¤ &%  ¡ 9 ¥ 6'  5 £ 2 £ 2 £ $ 0 '   0% ¤ ¤ ( (  ¨ ¤ 2 )1 ¤ ( &%  ) 6 # ( 2 ) 4 ¨# ( 2 )1 ¤# (  ¡ 9 ¥ #  0% ( ¦  ( 0% ( ( 2%   ( &#% 8# (( ¨ ( 3#%  ¨# ( ¤ (( &#%  #¤ (( %' 0()(&#% and similarly for #( 1 ¡ , so exists. As . also, 7 (  $ 0   ¢ But #  , where is a function of alone so  £  2 £ ¨ 2 £ ¥¤¢  ! ! 6) " 2 ) 4 £  2 )1     %" 2 ¨ £  2 ¤  ¢ §  ( ( ) 6 2 ¨ ( ( ) 4 2 ¤ ( ) 1 ( ¦ ¡ ¡ ¡9 ¥   ¡  , where is constant. 7  ©  ) 6  © 2 ) 4 ¨ © 2 )1 ¤ ©   £  2 £ ¨ 2 £ ¤ ¢ § )6  ( ( ) ¨ ( ) ¤ (  ¡ 9 254 21 (¦ ( £ 2£¨ 2£¤ © 7 )1 ¤  £ ¤ § ¨) 6  ( ( 2 5) 4 ¨ ( 3)1 ¤ ( 2 ( (  £ ¥  £  ¡9 ¦ 4. 3. ¡ 2. ¥ ¡ £ ¤¤ 1. Worked examples of gradient evaluation ¢ Note immediately that is a vector field! Without thinking too carefully about it, we can see that the gradient of a scalar field tends to point in the direction of greatest change of the field. Later we will be more precise. ¡9 &" ¡¡ '%$#! 7¡9 Then LECTURE 5. VECTOR OPERATORS: GRAD, DIV AND CURL 60 5.2. THE SIGNIFICANCE OF GRAD 61 ¡ ¤ ¦¢ ¡¨©¡ £ ¡¨ ¦ ©§¡ grad ¤¢ ¥¡ £ PSfrag replacements ¡ ¨ Figure 5.1: The directional derivative 5.2 The significance of grad ¡ If our current position is in some scalar field (Fig. 5.1), and we move an infinitesimal distance , we know that the change in is   ( ' ¡ ( ) 6 2 ¨ ( ' ¡ ( ) 4 2 ¤ ( ' ¡ ( )1 ¢  ¡ 9 ¡  7 % ¡ 9  ¡ ¡   % ) 6 2 ¨ % ) 4 2 ¤ % )1 ¢   % 7  %  ¡ ( ( 2 ¨ % ¨¡ ( ( 2 ¤ % ¤¡ ( (  ¡ and is also given by the scalar product ,  % % )' % % % Now divide both sides by % %  But we know that so that the change in %  % ¡ 9  ¡% 7 %   %  % But remember that , so is a unit vector in the direction of . This result can be paraphrased as: has the property that the rate of change of wrt distance in a particular direction ( ) is the projection of onto that direction (or the component of in that direction).  ¡ &$" ¡ '%#!  &$" ¡ '%#! $" ¡ & '! ) % )' ¡ % The quantity is called a directional derivative, but note that in general it has a different value for each direction, and so has no meaning until you specify the direction. We could also say that LECTURE 5. VECTOR OPERATORS: GRAD, DIV AND CURL ¡ 62 &$" ¡ '%#! At any point P, points in the direction of greatest change of at P, and has magnitude equal to the rate of change of wrt distance in that direction. ¡ 0.1 0.08 0.06 0.04 0.02 0 4 4 2 2 0 0 −2 −2 −4 surface, there is no change in 7 &$" ¡ '%#! ) % '  %    %%  ¡ 9  is a tangent to the surface, so this result shows that is everywhere NORMAL to a surface of constant ¡ © But ¡  %)'  % © %  )' ¡ % ¢ ¨ 7 ¨ ¤ $ ¨ ¦§¤¥£ ¡  ¦©¦§¥¤¢£¡ ¦ ¨¦ ©§¥¤¢ If we move a tiny amount within that iso. So for any in the surface – that is the locus . gradU Surface of constant U These are called Level Surfaces Surface of constant U ¡ Another nice property emerges if we think of a surface of constant for ¡ −4 , so 5.3. THE DIVERGENCE OF A VECTOR FIELD 63 5.3 The divergence of a vector field The divergence computes a scalar quantity from a vector field by differentiation. ) 6 ! ¡ 2 ) 4 £ ¡ )1  ¢ 2 ¡  %¦ ¨ ¦ ¥¤¢  is a vector function of position in 3 dimensions, that is If then its divergence at any point is defined in Cartesian co-ordinates by 9 ( ¨( ¤(  £ ! ¡ ( 2 £ ¡ ( 2  ¡ ( ¦¥¤& We can write this in a simplified notation using a scalar product with the differential operator: , vector ¨ § § (( 9 6) 2 ¨ ( ( ) 4 2 ¤ ( ( )1 ¦  ¦¥¤& £ Notice that the divergence of a vector field is a scalar field. ¢ Examples of divergence evaluation   ) 6  2 ) 4 ¨ 2 )1 ¤ !  ¢  1) ¤  ¢ ('&$ % ©  £   #¢ !   ! £    £  2 £ ¨ 2 £ ¥¤¢     !     !  '  ¢ ¦¥¤&  ¨ ¤    © 7 !"£  £ ' ©  !  ¤ © 7 £    £  2 £ ¨ 2 £ ¥¤¢  ¤ 2 £ !   £  2 £ ¨ 2 £ ¥¤ 7¢ ©  £ !   £  2 £ ¨ 2 £ ¥¤ 7¢ ¤ ¤ ( (    ¤('( £ ! £5 2 £ ¨ 2 @¥¤ 7¢ ¤ ! 6'  £ ¤   )  ' The terms in is ¢ © component of ' , for constant We work through example 3). The © 1) 2) 3) 4) div , and we need to find of it. and are similar, so that 5.4 The significance of Consider a typical vector field, water flow, and denote it by . This vector has magnitude equal to the mass of water crossing a unit area perpendicular to the direction of per unit time. Now take an infinitesimal volume element and figure out the balance of the flow of in and out of . ) 0% ) 0% LECTURE 5. VECTOR OPERATORS: GRAD, DIV AND CURL %¨%¤ % 64 ) 0%  % ¤ % To be specific, consider the volume element in Cartesian co-ordinates, and think first about the face of area perpendicular to the axis and facing outwards in the negative direction. (That is, the one with surface area .) )4  % ¤ %  3¡% ¨  ¨ z dz dS = -dxdz j dS = +dxdz j y dx dy x Figure 5.2: Elemental volume for calculating divergence. ¡ ¢ ) 4  The component of the vector normal to this face is , and is pointing inwards, and so the its contribution to the OUTWARD flux from this surface is ¢ £ ¦ ¤ %  %  ¥¨¢ ¨ ¢ ¡ ¡  ¨¥¢ ¢ ¡ %  ¤ where means that is a function of . (By the way, flux here denotes mass per unit time.) A similar contribution, but of opposite sign, will arise from the opposite face, but we must remember that we have moved along by an amount , so that this OUTWARD amount is ¨ %¤ % § ¨ % ¦¨¢ ¡ ( ( 2 ¡ ¢ ¦ ¦  ¤ %  %  ¨ % 2 ¥¨¢ ¡ ¢ ¥   % ¤ % ¨ % ¦¨¢ ¡ ( ( ) 0% ¡¨ ( ( ¡¤ ( ( ¦ ) 0% § ¥¨ ¡ ( ( ¢ ¦ )% 0§ 9 2 ¢ ¡¨ ( ( 2 § ¦ So we see that % Summing the other faces gives a total outward flux of ¨ ¢ The total outward amount from these two faces is © § ¥ £¡ ¤ ¢¨¦¢ ¤¢ 5.5. THE LAPLACIAN: OF A SCALAR FIELD 65 The divergence of a vector field represents the flux generation per unit volume at each point of the field. (Divergence because it is an efflux not an influx.) Interestingly we also saw that the total efflux from the infinitesimal volume was equal to the flux integrated over the surface of the volume. (NB: The above does not constitute a rigorous proof of the assertion because we have not proved that the quantity calculated is independent of the co-ordinate system used, but it will suffice for our purposes.) $ ¡  ('% $  The Laplacian: !  5.5 of a scalar field &$" ¡ '%#! Recall that of any scalar field is a vector field. Recall also that we can compute the divergence of any vector field. So we can certainly compute , even if we don’t know what it means yet. Here is where the operator starts to be really handy. $" £  ¡ & '! ¢ ¦¥¤& ¡ § §  (( § ¡ §  (( 6) 2 ¨ ( ( 6) 2 ¨ ( ( ( )4 2 ¤ ( ( )4 2 ¤ ( )1   ( ) ¦ §( 6 )1   ( ¦¦ §( § £ ¡ £ (( 2 £ ¡ ¨£ (( ¡ § ££ (( 2 ££¨(( 2 ¨ ( )4 2 ( 6) 2 ¨ ( ( ) 4 ( 2 £¤( ¡£ ( 2£¤( £( ¤ ( )1 (¦ 2 ¤() 9 1 ¦   ¡ 9¢ ¦ 9  ¦ ¦    This last expression occurs frequently in engineering science (you will meet it next in solving Laplace’s Equation in partial differential equations). For this reason, the operator is called the “Laplacian” ¡ § ££ (( 2 ££¨(( 2 ££¤(( ¦ £ "  ¡£ " Laplace’s equation itself is ©  ¡ £ " )6 § ¦¨§ ¡ ( (  ¢ ¤ ¡ ( ( ) ¨ (  §  ( )  (  ¨ ( ¦ 2 4 § ¨ ¡ ( ¦¡ ( ¦ 2 1 § ¢ ¥¡¨ ¡ ( ¢ ¡ ¨ ¡ ¦(§ ¡ ¦ ¦ ¦ ¦ £ £ ¨ ¨  ¨ ¨ &" ©¨" ¢ ¦¦¦ ¨ § ¢ § § § ¦¦¦¦ !  $   ¦ ¦ ) § 6 ) § 4 ) 1§   ¡ 9 ¦ We can follow the pseudo-determinant recipe for vector products, so that ¤  ¢ ¥" ¡9 9 ¢ £ ¡ This gives the curl of a vector field ¡ 9 cross it with a vector field § So far we have seen the operator applied to a scalar field vector field . We are now overwhelmed by an irrestible temptation to ; and dotted with a 9 9 5.6 The curl of a vector field   © § %@¤ 2 ££ ¨ 2 @¥¤¢  2   ¦ ! ©  © £ £ £  ¨ ¤  £ 6' £ ' 2 ©  ¢   ' © ¢  ¤ £ ¨ 2 £ ¥¤ 7¢ 7¤ ' 2 £ !   £  2 £ ¨ 2 ! £ ¥¤¢   £ !   £  2 £ ¨ 2 £ ¤ 7¢ ¤  ¤ ( (  and £    £   2 " Adding up similar terms for £   £  2 £ ¨ 2 £ ¤¢ ¤(( ¤((  £     £  2 £ ¨ 2 £ ¥¤¢ 0 ' © Let’s prove example (3) (which is particularly significant – can you guess why?).  £ ¨ ' 2 ¤ © '© © !  ¤ 5 2 £ ¨ 2 @¤ !   £ ¢ ¨ £ ¤ £ £ ¡£ " 1) 2) 3) ¡ ¡£ " Examples of evaluation LECTURE 5. VECTOR OPERATORS: GRAD, DIV AND CURL ¢ 66 5.7. THE SIGNFICANCE OF CURL 67 ¢ Examples of curl evaluation 9 ¡ )4 £ ¨ ¤ © )6  £ © 1 ¨ @¤ ) )£ ) 4 ¤ 6 £ 2 ¨ )1¤¨ ¤ ©  1) 2) 5.7 The signficance of curl ) 4 ¤ 2 )1 ¨  Perhaps the first example gives a clue. The field is sketched in Figure 5.3(a). (It is the field you would calculate as the velocity field of an object rotating with .) This field has a curl of , which is in the r-h screw sense out of the page. You can also see that a field like this must give a finite value to the line integral around the complete loop  )6 © 7 y ax (y+dy) ¢ £¡ y+dy dy ay (x) x y ay (x+dx) %  © ¦ © ¦ ©  y dx x+dx x ax (y) (a) (b) §  ¦ ¨¦¤  §© ¥ Figure 5.3: (a) A rough sketch of the vector field . (b) An element in which to calculate curl. In fact curl is closely related to the line integral around a loop. % 4  ¢¡ The circulation of a vector round any closed curve is defined to be and the curl of the vector field represents the vorticity, or circulation per unit area, of the field.   LECTURE 5. VECTOR OPERATORS: GRAD, DIV AND CURL ¨ % ¤ 68 % Our proof uses the small rectangular element by shown in Figure 5.3(b). Consider the circulation round the perimeter of a rectangular element. The fields in the direction at the bottom and top are ¡ § ¨   ¨ % 2 ¢¨ ¡ § is a function of , and the fields in the ¨ ¦ ¨ % § ¨ ¡ ( ( 2  ¥¨¢ ¤ direction at the left ¤ % ¥¤¢ ¡ ( ( 2  ¥¤¢ ¢ ¡   ¤ % 2 ¥¤¢ ¢ ¡ § &¤$ &¤$ ¡  ¢¨  ¢¨ ¡ § ¡ where denotes and right are §  ¤¢ ¢ ¡ Starting at the bottom and working round in the anticlockwise sense, the four contributions to the circulation are therefore as follows, where the minus signs take account of the path being opposed to the field:  ¢  ¨ %  ¥¤¢ ¦¡   % % ) 6 ¨ % ¤  ¤% ¡%   ¡ 9¢ ¨ % ¤ % § ¦¨§ ¡ ( (  ¢ ¤ ¡ ( ( ¦ % % ¨( ¢ 2 § ¡  ¢ % % ¢ ¤ ( 2 ¦¡ 2 % § ¡  2 ¢¤¤ § ¨ § ¡ (  ¥¨¢ ¦ ¡ £¨ § ¤ ¦¡ (  ¢¤ ¢ 5¡  ¤  ¢¨ ¦ ¢  ¨ %  ¥¤¢ ¦¡    ¤ %  ¨ % 2 ¢¨ § ¡   ¨ %  ¤ % 2 ¥¤¢ ¦¡  2  ¤ %  ¢¨ § ¡  2      %  where . NB: Again, this is not a completely rigorous proof as we have not shown that the result is independent of the co-ordinate system used. 5.8 Some definitions involving div, curl and grad A vector field with zero divergence is said to be solenoidal. A vector field with zero curl is said to be irrotational. A scalar field with zero gradient is said to be, er, constant. Revised Sep 2005 Lecture 6 Vector Operator Identities In this lecture we look at more complicated identities involving vector operators. The main thing to appreciate it that the operators behave both as vectors and as differential operators, so that the usual rules of taking the derivative of, say, a product must be observed. There could be a cottage industry inventing vector identities. HLT contains a lot of them. So why not leave it at that? , and describe key aspects of vectors fields, they arise often First, since in practice, and so the identities can save you a lot of time and hacking of partial derivatives, as we will see when we consider Maxwell’s equation as an example later. Secondly, they help to identify other practically important vector operators. So, although this material is a bit dry, the relevance of the identities should become clear later in other Engineering courses. " ¤  ¡   ¢ )6 2  ¢ ) 4 2 § ¨ (  (  ¡ £' ( ' ¦ ¦  ( '¡ ( ¨ ( '¡ ( ¦  ¨ ¦ )( 4 ( )( 6 ( ¦ ¦ ¥ ¤& £ ¢ £ &$" '%#! 6.1 Identity 1: curl grad ¦  ¦  ( ¨ ( )1  ¡£ ( ¦ ¤(' ¡(¦ ¦ ¤ ' ¦  ¡ 8¡ 9 9 )( 1 ( ¦¦ ¢ ¨ ( ('£ (   ( ¨ ('£ ( as . Note that the output is a null vector. 69 LECTURE 6. VECTOR OPERATOR IDENTITIES 70 ¡ ('( ¨('( ¤('( ('( ¨('( ¤('( ¦ ¦ ¦ ¦ ¦ £ (¡ ¨ ¡( £  ( (  ¤¦¡( £  ( ( ¢ ¦ £¡( ¢ ¦ ¨ 2  § (¡ ¨ ( 2 ¤ ¨ ¡ ( ¨ ( £(  ¢ ¢   ¦(¡ ¤ ( ¢ ¦ ¦ ¦ ¦  ¨ ¨ (¡ ¤ ( ¡ £ ( ¦¦  § ©  ¡ 9 9   6.3 Identity 3: div and curl of § 6.2 Identity 2: div curl  ¢ ¤ ¡  ¢ ¡ Suppose that is a scalar field and that is a vector field and we are interested in the product . This is a vector field, so we can compute its divergence and curl. of a fluid is a scalar field, and the instantaneous velocity For example the density of the fluid is a vector field, and we are probably interested in mass flow rates for which we will be interested in . is given by: The divergence (a scalar) of the product  ¢ £ ¡ 9  7 ¡  ¡ 9¢ 2 , and the result should be a ¦¥ ¡ 9 ¡   ¡¢ ¡ 9 6.4 Identity 4: div of §  " §   ¡ ¡ '9& %$¢ #! 2 ¢  2 § ¦¥¤¢9£ & ¡¡   ¡ ¢ ¡  ¢ ¤ ¢ £ In a similar way, we can take the curl of the vector field vector field: Life quickly gets trickier when vector or scalar products are involved: For example, it is not that obvious that ¢ ¢ ¤ ¤ £ § ¥"    §  ¥" £¡  ¨¡ ¢ ¦¥¤& § To show this, use the determinant: §¢ ( (    §  ¢ 7 7 7 ¦ ¨ §¢ & ¢ " ¨ ¨ ¨ ¢ ¢ ¡  ¢  ¦¡   ( 2  ¨  ¦¡  §  ¥¡  ¨ ( 2  ¢  ¥¡  § § ¨ ¨ curl ¦ ( ¨ ¢ ¡ ¤ (   ¢ ¤ § ¦  ¥"    $ 777 ¦ ¦ ¦ ¦ ¦ ¨ § ¢    ¨ ¥¡ § ¦¡ ¢¡  ¤ ( ' ( ¤ ( ' ( ¤ ( ' (  © ¦ ¦ ¦ ¦ ¦ ¦ ©§¥£ ¡ ¤ ¨¦¢ ¤§ ¢ 71 ¦ ¦ ¦ ¦ ¦ ' )( 6 ( ¨' )( 4 (   ¤' )( 1 ( ¦ § ¦ § § ¨  ¡  ¢  ¦¡ ¨  ¥¡  §  ¨ ¡ ¢  ¥¡  ¨  ¢ ¡ §¢ ¦ ¦ ¦ ¦ ¡ ¢ ¥¤"  ¢ )1 so the component is ¦ ¦¥ 6.5 Identity 5:    6.5. IDENTITY 5: § §¨  ¥¡ ¢  ( (   §  ¢ ¡  ¢  ¦¡ ¢ ¨ ( ( §  ¨  ¦¡ which can be written as the sum of four terms:  ¤ ( ' § ¡ ( ¢ §   ¤ ( ' § (¢ § ¡ § ¢  §  ( ( ¨ ¡ 2 ¨ ( ( ¦¡ ¦  § ¡ §  ( ( ¨  2 ¨ ( ( ¢  ¦ 2 § ¨  ¡ ( ( 2 ¦¨¢ ¡ ( ( ¦ §   § ¨   ( ( 2 ¢ ¨  ( ( ¦ § ¡ Adding same with this): §  9     9  §  2 §  § 9 ¢   §  9 ¢   ¨¡ ¢ § can be regarded as new, and very useful, scalar differential operator.  9   6.6 Definition of the operator !     9 ¡ where to the first of these, and subtracting it from the last, and doing the to the other two terms, we find that (you should of course check This is a scalar operator, but it can obviously can be applied to a scalar field, resulting in a scalar field, or to a vector field resulting in a vector field: 7 ¢ ( ( ( 2 ¨( ¡ ¨ ¥ for you to derive    " ¢ ¡    ( 2 ¤(   ¡ ¡  9   § ¦ 6.7 Identity 6: The following important identity is stated, and left as an exercise: £ "  ¦¥¤& '%$#! £&" )6 ¨ ¡£ " ¢ 2 ) 4 ¦¡ £ "   ¥"  ¢ ¥" ¢ ¤ ¤ 2 )1 § ¡£ "   £ ¢  where " LECTURE 6. VECTOR OPERATOR IDENTITIES 72 Example of Identity 6: electromagnetic waves Q: James Clerk Maxwell established a set of four vector equations which are fundamental to working out how eletromagnetic waves propagate. The entire telecommunications industry is built on these. ( ( ¦ §( ¦ ¢ §( 2  © ¥¦¤& £ ¦¥¤& £ ¢£ ¡ £  ¤¤ ¥¥"   ¨¤ ©¥"  ¢ £ ¢  In addition, we can assume the following, which should all be familiar to you: , , , where all the scalars are constants. , and with zero Now show that in a material with zero free charge density, conductivity, , the electric field must be a solution of the wave equation  ¤ ""! © 3£  ¤   ¤ 3    ¨  £ §(          ¤ £ " ¥ ¦ ¤£ ( § §¤¦ ( (     ¦ %(¦ (       ¢ ¦   ¤  &¨" £ ¢ %( ( !   ¤ £ "  ©  ¡¥ ¤& ¤£ ¢ ¢ ¢ ¤ ¤ £ &$" ¤ ¤   %( ' ¨ ¢(      ¢ ¥" ¡ ¤ £ "  £¦¥¤& '%#!  ©¤¥"  ¥"  ¦ £ ¢ ¢  ¦¥£¤& &'%$"#!  ¤" £ ¤" £ ¤  %¢  ¦%( ' ¤ ( ' ¢      2     § ( ' ' ( 2    ¥¨¥" ££ ¢¢ ¦  ¢ ¦ ©   ¦§( ¨ (  ¢     ¨ §(   ¢ ¢ ¦¥( ¤&  ©¤£¤¢ ¦¥" ¤&  £ £  ¢ ©  ¢ £ © ¡¥¦¤& ¥ © ¨$¦¥£¤& ¤£ ¤ £ £  ¡ ¢ 7  £¦¥¤& ¥  £  £¦¥£¤&      ¤     ¢ ¦¥¤&  ¡¥ ¤& '   £¤ ( ¢    ¡   # ¤ £ " © ¢ 7  £ §( ¦ A: First, a bit of respect. Imagine you are the first to do this — this is a tingle moment. ¢ But we know (or rather you worked out in Identity 6) that , and using (c) " so interchanging the order of partial differentation, and using (a) This equation is actually three equations, one for each component: ' £( £ §( ¦ ¨ ¢ ' '        # §  and . § £ ' ( " and so on for : ¢ IN CURVILINEAR CO-ORDINATE SYSTEMS 6.8 Grad, div, curl and ¤ ¥£ ¡ ¢ 6.8. GRAD, DIV, CURL AND 73 in curvilinear co-ordinate systems It is possible to obtain general expressions for grad, div and curl in any orthogonal curvilinear co-ordinate system by making use of the factors which were introduced in Lecture 4. We recall that the unit vector in the direction of increasing , with and being kept constant, is ¦ © ¨ § (¦   ¦ ¦ ¦ ¦ ¦ (¦  § ¦   (( ©    where is the position vector, and ¦ § ¦ is the metric coefficient. Similar expressions apply for the other co-ordinate directions. Then © ¨ § ¦ 7 %   ¦ 2 %  ¤ ¦ 2 %      % 6.9 Grad in curvilinear coordinates © % © ( ¨ ¨ §§ 2 % ( 2 % (  ¡ %   %  ¡ 9 ¡( ¡( ¡(  ©¦ ¨¦ §¢ ¡  ¡  ¢ ¡  ¡ Noting that and a scalar field obtained previously , and using the properties of the gradient of It follows that © % © ( 2 % ¨ ( 2 % § (  %   ¦ 2 % ¤  ¦ 2 %  ¦ ¢  ¡ 9 ¨ § © ¨ §   ¡( ¡( ¡(   ©¨§ % % % The only way this can be satisfied for independent , , is when ) © ¦ ¨ ¦  § ¦ ( 2 )¤ ( © 2 ) ( ©  ¡ 9 ¡( © ¡( ¡( 6.10 Divergence in curvilinear coordinates Expressions can be obtained for the divergence of a vector field in orthogonal curvilinear co-ordinates by making use of the flux property. We consider an element of volume . If the curvilinear coordinates are orthogonal then the little volume is a cuboid (to first order in small quantities) and ) 0% ©% ¨ § ¦ ¦ 7 % %  ¦  ! ) 0% LECTURE 6. VECTOR OPERATOR IDENTITIES 74 w h (v) dw w h (v+dv) dw w y h (v) du h (v+dv) du u u h v dv u The scale params are functions of u,v,w Figure 6.1: Elemental volume for calculating divergence in orthogonal curvilinear coordinates However, it is not quite a cuboid: the area of two opposite faces will differ as the scale parameters are functions of , and in general. So the net efflux from the two faces in the direction shown in Figure 6.1 is ©§ % %  ¦  ¦  ( )¤ ©%¤§% % ¨ ¦ ( 2 ¦ ¨  ¡ ¢ © ¡ ¢ ¨§ ( ¡ ©¨§ % % % ¦ ¨ %  ¨ ¦ ( 2  ¦ %  ¨ ¡ ( ¨ ¢ ( ¡ ¨   ¦( ¡ ¢(  2 ¡  which is easily shown by multiplying the first line out and dropping second order terms (i.e. ). By definition div is the net efflux per unit volume, so summing up the other faces: ¨ 3£ % ¢ ©% %¤%  ¨§ §  ¦ ©¨§ % %¤%  §  ¦ ¢(  ¢(  © ¨ 2 2  ¦ ( ¡ ¦(   ¦  ¨ ¢ ¡ ¢ (  ¦ © ¢( 2  ¦  ¢ ¡ 2 ¦  ¦ ( ¡ ¦(  ¢( §  ¦ (  ¡ § ¢(  ¦ (  ¡ ¦ ¦ © ¨%§ % ¤% )£ 0% ¦¥¤&   ¦¦¦ ¦¥¤& ¥ £ So, finally, §¦ ©  ¦ ( ¡¢ ( ¨ § 2  ( 2  (  ¦  ¦ ¡ ¢ (  ¦ ¦  ¡ ¢ ( ¦ ©  ¦  ¦ ¦ £  ¥ ¤& 6.11. CURL IN CURVILINEAR COORDINATES 75 6.11 Curl in curvilinear coordinates  Recall from Lecture 5 that we computed the unit area from component of curl as the circulation per ¨% ¤ % § ¥¨§ ¡ ( (  ¢ ¤ ¡ ( ( ¦ %   By analogy with our derivation of divergence, you will realize that for an orthogonal curvilinear coordinate system we can write the area as . But the opposite sides are no longer quite of the same length. The lower of the pair in Figure 6.2 is length , but the upper is of length ©§ % % ¦  ¦ §¨ %  % 2 ¨ ¢  ¦ § %  ¨ ¢  ¦ y au (v+dv) v+dv h u(v+dv) du dv hu (v) du y u+du u au (v) Figure 6.2: Elemental loop for calculating curl in orthogonal curvilinear coordinates Summing this pair gives a contribution to the circulation   %  % 2 ¨ ¢  ¦  % 2 ¨ ¢  §¨ ¨ ¡  %  ¨ ¢  ¦  ¨ ¢  ¡ § § ¨%§ ¤% §   ¡  (¦ ¢ ( ¨ 2  ¡  (¦  ¢(  ¦   % So the circulation per unit area is §% % ¨ ( ¨   ¡ ¦¢ ( and together with the other pair: ¨ §  ¦  ¦ §   ¡  ( ¦ ¢ (    ¡  (¦ ¢ ( ¦ © ¨§  % %  ¦  ¦ % )6 , and 7  §  ¡ ¡ ( (   § ¡ ( ¢ ( ¦ © )6  2 ) 4 ¢ § ¡ ¤£ ¦  © (  ¦ ¡ (  ¦ ¦ ¦ 2 )1 ¡ 2) © ¦¢ ¥£  © (2 (  § ¨ ¡ ((  ¨ ( ¦ ¡ ( ¦  ¦ § ¦  £ ¢ ¢ ¨ £ " ¨ ¤ $ " ¨  ( 2 ) § §  ¡ (  ¡ (  ¡ (¦ (  ¨ ¡ ( © ¦ ( ¡( 2( ¨¡ 2 ( § § ¡ (   ¡  ¢ ( ¦ © 6)  ¡ ( ( 2 ) © ¡ ¡ ( ( © 2 ) ¡ ( (   ¦ ¨ (2 ( § ¤£ § ¦ ( ¡ ( ¦¦  %§¢¦  ¢ ¦ ¡ ¡£ " ¥¤" £ ¢  ¥ ¤& £  ¡ '%#! &$" © ¦¢ ¦    ¡ £ ¥£ £  2 ¡ £ ¤ £ ¦ £  ¢ ¥£ ¦¢ ¦ ©   ¡ £ ¦¤ 2 ¡ £ ¥£ ¢  . The position vector is in cylindrical polars ¦¥¤& £ $ ¥  ¨¦  ¨¦ §  ¦ ¥  '   ¦ ((  ©¦ ¨¦ §¢ , etc. Here 6.13 Grad Div, Curl, ¦ § ( ¦¦¦  ¡£ ( " © ¡ Substitution of the components of into the expression for immediately (!*?) gives the following expression for the Laplacian in general orthogonal co-ordinates: $" ¡ & '! 6.12 The Laplacian in curvilinear coordinates ¡  ¡  ¦  ¡  ¦  §¦ ¦  ¦  ¦  ¦   © ¦ ¨ ¦ § ¢ ¥" ¤ ¦ § ¦ ¦ ¦ § § ¦ © ¦ )  §  ¦ ) ¤ §¦ )  §  ¦  ¦ ¦ ¦ ¦ ¦ ¢ £ You should check that this can be written as Curl in curvilinear coords: ¨  ¦  ¦ § ) § ¡  (  (   ¨¦ ¢ (   ¡  ¦ ¢ ( ¦ ¦ ©  ¦  ) ¤ § § (  ¡ © (   ¡  ¦ © ¢ (   ¨¦ ¨ ¢ ( ¦ © ¦ ¦  §   ¡ ¦( ¢ (    ¡  ( ¦ ¢ ( ¦ © 2   © ¦ ¨ ¦ § ¢ ¥" ¤ 2 ) ¢  and hence curl is LECTURE 6. VECTOR OPERATOR IDENTITIES 76 F G9 R 53 @6W( 1 1 A9 BS! E¡56W! 1 3 ! 7 53 @6#! 1 R5 C S63 D1 ( &$ " )'B! E5 C ¡63 D1 7 53 ¡64( 1 ( &$ " )'%#! H F9 ( &$ " A ¥9 HF IG9 A 9 ( &$ B@)8%"   IG@T'¥!  P 1 V0 U 1 1 Q0 1 1 20 7 ©  §  ) 6  2 ) 4 ¨ 2 )1 ¥¤¢   . So ¡ 7 )6 ¨ 2 ) 4  ¦ and ¡ ©  ¡ ¥£ ¦¢ ) ¡¡ ¥£ ¦¢ ¦¢ ¥£ ¤ £ ¦  ¤  ¢ ¥ ¦ § ¤¦  ¤ ££ ¦ ££  ¦  ¦¦  ' ¤ ¨ ' ¤ £ ¤' ¦  ( ( ¨ () ( ¤ ( ) ( )6 4 1    ¦    © ¨ ¡ ¥  " ¤ ¦ ¦ ¦ ¦ ¦ ¦ ¢ £ ¦ in (i) Cartesians and (ii) Spherical polars when  ) 6  2 ) 4 ¨ 2 )1 ¥¤¢ ¤  Hence as (ii) In spherical polars, ¥¤"  ¢ A1 (i) In Cartesians Q1 Find . Examples ¦  ¢ ¢ ¨£  " ¨  ¤ $ £ " ¨ § ¡¢ (   ¡ ( (  ¢ ( ¦ )© ¤ ¦   ¤¦   § # ¤ £ ¦  § ¡ ¢ ( (    ¡ ¢ ¡ ( ( ¦ ) £ 2 §  ¡ ¢ ¡ ( (   ¤ £ ¦ § ¡ ¢ ( ( ¦ ) £ £¤ ¦  2 ( ¤ £ ¦  2  ( £  ¡( §¡   ¤£ ¦ ¡¢¤(¦  ©   ¡ £  ¢ (  © (© ) ¡ ( £ 2 ) ( 2 ) ( © ¡( © ¡(© ¡( ¤ £ ¦   ¡ £ ¥£ 2 ¡ £ ¤ £ ¦ ¢ £ ¤ £ ¦ £  ¢  ¦¢   # £ ¤ £ ¦ £  2  ¡ £ ¤ £ ¦ 2 ¡ £ ¥£ ¢ £ ¥£ £  ¢ ¦¢ ¦¢ ¦¢ ¦¢ £ ¥£ 2  ¡ £ ¤ £ ¦ 2 ¡ £ ¥£ ¢ £ ¤ £ ¦ ¢ 2 ¤ ¤£ ¦ ¤£ ¦  ¡ 2 )1 ¡ ¢ ¥ ¢ " ¥ § ¨ ¦    ¡ ¦ ¡¦  ¢ ¥¤" £ ¢   ¥ ¤& £  ¡ '%#! &$" ¤ ¦ ¢ £       ¦¢ ¥£ ¤ £ ¦   ¡£ ¢ ¥ ¢ ¥ © 2 )4 . The position vector is ¤ ¥£ ¦ )6  ©¦ ¨¦ §¢ ¥ ¦¢ ¥£  . Here 6.14 Grad Div, Curl, ¡ ¢ 6.14. GRAD DIV, CURL, in spherical polars IN SPHERICAL POLARS 77  ¡   ) 777 2 777 2   7  © § £     £  2 £ ¨ 2 £ ¥¤ 7¢ ¤ § £    £  2 £ ¨ 2 £ ¥¤¢ ¤ ( (   ¦¥¤& £  ) ¦¦ ¥¢ ¥¢   ¢  ) 4 ¡ ¥£ ¦¢ A2 (i) Using Cartesian coords: Q2 Find the divergence of the vector field where is a constant vector (i) using Cartesian coordinates and (ii) using Spherical Polar coordinates. which is exactly the result in Cartesians. 2 )1 ¡ ¤£ ¦ ¢ ¡  ) 6 ¨ ¦ 2¢ 4  ) £¤ ¦ ¤ £ ¦  ) ¥£   ¤ £ ¦   ¢  ) 6 ¤ £ ¦  ) 4 ¡6 ¡ ¤ £ ¦ ¥¢ 2 )1 2¡ ¥4¦ £ ¦ ¢ ¢ ¦ ¢ 2   ¥¤"  ¢  Don’t be shocked to see a rotation matrix : we are after all rotating one righthanded orthogonal coord system into another. So the result in spherical polars is  ¤ ¥ )  )6 )1 4  ) )6 )1 4 ¤ ¥ ¢£ ¡ ¢£ ¨ 7 ) 6  2 ) 4 ¨ 2 )1 ¤   ¥" ¨ ¡   )1 ) %! © ¦¢ ¤¦ ¤ ¦  ¤ ¡ ¦ ¦ ¢ ¥¢ £ ¦¡ ¦£ ¢ ¡ ¡ ¤ ££ ¦ ¤ £ ¦ ¡¡ ¥£ ¦¢ ¢£ ¡ ( ¦ ) © © ¦ %£  (  ¦¥¢ ¤£ ¦ )©  ) ¡ ) ¤ ¢£ ¡  0%  ¦ Grinding through we find © ¦ ) ( ¦ ¤  ) © Checking: these two results should be the same, but to check we need expressions for in terms of etc. Remember that we can work out the unit vectors and so on in terms of etc using )1  ¡ ¦ ¢  ¦ ¢   ¢ ) © 2  ¡ ¤ £ ¦ ¦¢ ¦¢ !  ¡ ¥£ ¥£ £   ) 2  ¡ ¤ £ ¦ ¤ £ ¦ £   ¢   © ¦¢ ¦¢ §  ¡ ¥£ ¤ £ ¦ £  ¢ ( (  ¦ ) © 2 §  ¡ ¥£ ¤ £ ¦ £  ¢ ¡ ( ( ¦ ) © ¦ )¤  ¤ £ ¦ ¢ ) ) ¤£ ¦   ¤  ¤ ¥¤"  ¢  )¤ LECTURE 6. VECTOR OPERATOR IDENTITIES 78  ¢ ¦ @¢ ¡ ¥£ 2 § ¦ @¢ ¡ ¤ £ ¦¦ ¢  ¢   ¦¢   ¨ ) ¤ ¦  ¢ ¤ ¦ ¥£ §  ¥£ ¥£   ¥£¦ £ 2 ¢  @¡ ¡ ¤£ £ ¦ ¤ £ ¦ 2 2 § @¡ ¡ ¥£ ¤ £ ¦ ¢¢  ¢  @¦¢ ¨ §  , etc, where is a constant, so now we can ¨ ¢£ ¡ ¡ ¢¦ §  § ¡   ¤ ¥ ¡ ¨¥ ¢¦ § ¨ ¡ ¡ ¢ ¡ ¡ ¢£  § ¡ ¢ ¥ ¡    ¡ ¡ §   ¡ ¡ ¢ # ¡  ¢£ § ¡ ¡  ¢£ ¡ §  ¡ ¢ ¥ ¥ ¡ ¥ ¡ )  )6 )1 4  § ¡  ¡ ¤ ¥ ¡ ¢£ ¡ )©  ¡ ¡ ¥ ¡ ¡ ¢ ¤ ¡ ¡ ¢¦ § ¢£ ¡ ¢¦ § ¢£ ¡ ¡ ¡  ¡ § ¦ ¡ ¨ ¢£ ¡ ¨ ¡ ¡   § ¡ ¡  ¤ ¡ ¢£ ¡ ¢£ ) )6 )1 4 ) ) 46 )1 ¢£ For our particular problem, write down ¡ ) )¤ ¡ ¢£ ¡  ¢£ ¡ § ¡ ¢ ¥ ¡ ¡  ¢£ ¡ ¢ ¥ ¡ ¡ ¡ ¡ ¡ and using the inner product ¢ ¡ ) 6 ¨ ¡ 2 ) 4 ¦¡ 2 )1 § ¢ ) © ¡ § ¢ 2 ) ¤ ¥¡ 2 )  ¡ Now the point is the same point in space whatever the coordinate system, so )  )6 )1 4  ¤ ¢£ ¡ ¢£ ) )6 )1 4 ¡ ¢£ © ¦¢ ¥£ ¤¦ ¤ ¦  ¤ ¡ ¦ ¥£ ¥£ £ ¦ ¢ ¦¡ ¢ ¦£ ¢ ¥£ ¡¡ ¤ ££ ¦ ¤ £ ¦ ¡¡ ¥£ ¦¢  ¦¢ ¤£ ¦ ¡ )©  ) ¡ )¤ ¢£ ¡ and our first task is to find and so on. We can’t do this by inspection, and finding their values requires more work than you might think! Recall  )© ¡ § ¢ 2 ) ¤ #¡ 2 )  ¡  (ii) Using Spherical polars ¡ ¢ 6.14. GRAD DIV, CURL, IN SPHERICAL POLARS 79 LECTURE 6. VECTOR OPERATOR IDENTITIES 80 Now all we need to do is to bash out 2 ¤¦ ¡¡ ( ( ©£ ¦ ! ¨ ) ¥¢ ¤ £ ¦ ©   ¢  ¡ ¤£ ¦ 2 2  ¨ )  ¨ 2  # ( ¤£ ¦ ( ¡¢ ¢  ¢ @¡ ¤ £ ¦   § ¥£  @¡ ¦ ¢ ¢ # £ ¤ £ ¦ ¦ ¥¢ 2 ¢ @¡ ¤ £ ¦ © ¤£ ¦   £ ¡ £ (  ¢ ( £ ©  ¦¥¤& ¤¦ ¦ § ¡ ¥¢  ¢ © £ ¦¢  £ ¥£ ¤ £ ¦ ¤ £ ¦ 2 § @ ¡ ¥£©¢ ¤ £ ¦ ¢  ¦  ¦¥¤& £ ¦¢  ¥£ 2 ¢  ¡ ¤ £ ¦ ¤ £ ¦  2 A bit more bashing and you’ll find § In glorious detail this is 2 §   )  ¡ ¦ ¢ ¤£ ¦   ¦¥¤& £ This is EXACTLY what you worked out before of course. Take home messages from these examples: Just as physical vectors are independent of their coordinate systems, so are differential operators. Don’t forget about the vector geometry you did in the 1st year. Rotation matrices are useful! Spherical polars were NOT a good coordinate system in which to think about this problem. Let the symmetry guide you. Revised Sep 2005 ...
View Full Document

Ask a homework question - tutors are online