C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 of 4. CONCEPT OF WORK W:.... CONCEPT OF POTENTIL ENERGY U:... 3. Force of a Spring... 3 4. Force with several variables.... 4 5. Eamples for Path Dependent Forces.... 4 6. Path Independent Force and Gradient:... 5 7. Conservative Forces, del operator, perfect derivative.... 6 8. Divergence div, gradient grad, curl.... 8 9. Potential energ of the gravitational force....0 0. Curl of n Central Force..... Summar for the Relationship Between Potential Energ and a Conservtive Force. Review and Summar for Phsics:...3 4. Circulation and other advanced calculus stuff (Optional)...7 5. Path Integral to Find the rea of a Closed Curve....7 6. Proof of Stokes Law: Quick and dirt proof:...8 7. Green s Theorem...9 8. Stokes' theorem:... 0. Remarks on determinants: (Optional: For connoisseurs onl!)...3 Work, potential energ, and path integrals, differential vector operators. CONCEPT OF WORK W: The definition of work b a constant force F acting over a distance r is: (.) W F r F F F z Fr cos z ssume that we drag a crate of mass m along the floor, eerting a force F at an angle with the horizontal, without accelerating the crate. The horizontal force component is then equal and opposite to the force of friction. If we drag the crate from =a to =b, the work done b the force F is equal to F d, where d=b-a. The work done b friction is F d. If we have a non-constant force F()=-k in one dimension, like the spring force, and want to calculate the work done b this force while it moves an object of mass m from point to B on the -ais, we proceed as follows:
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 of 4 infinitesimal amount of work: dwi F i i n n W dwi F i i (.) i i If we make the Δ intervals smaller and smaller, while allowing n to go to infinit, we form an infinite sum of tin rectangles which is equal to the integral of the function F() taken between the points and B. Eample: Force of a spring: F ( ) ki (k is the spring constant) If an applied force Fa epands the spring without acceleration, it must be equal and opposite to the spring force itself. Let us calculate the work done b the applied force: (.3) W k d k k The work done b this force can be recovered. If the outside applied force stops, the spring will snap back and release all the work done on it b the applied force. This work is equal to the negative of the work done b the spring. It is also called the potential energ U of the spring U spring k ; (.4)U Wspring U spring k (.5) Work done b the spring = negative work done b the applied force.. CONCEPT OF POTENTIL ENERGY U: If we lift a bo of mass m up b a distance h without acceleration, the work done b the lifting force is mgh, whereas the work done b gravit is mgh. If we drop this bo, it can again do work due to gravit. This possibilit is captured in the word potential energ: U(). We sa that the bo has the potential energ U=mg, and the potential energ-change U of the bo while being lifted from a to b, is (.6) U mgb mga mg mgh The potential energ of the bo increases while the gravitational force does negative work on the bo. In order to talk about a potential energ function U() we need to specif a reference point for U. This reference point is the point where U is equal to 0. Thus, in the case above, =0 is the proper reference point, and we define (.7) U( ) mg Let us shed some more light on the relationship between the force which does the work, and its potential energ function. We stick to the eample in which a mass m is lifted up b a person. The force of gravit does work: (.8) W F r mgj j mg mgh 0 Note that the work done b gravit is equal to the negative change in the potential energ of gravit. h
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 3 of 4 (.9) Wg mgh U Thus, we associate a function U() with this force and call it the potential energ function U() for the force mg. force which has a potential energ function, from which it derives, is called conservative. One of its characteristics is that the work done along an closed loop is alwas 0. The work done b gravit in the above eample is 0 if the object is first moved up b a distance h and then down to its original point. One can see easil that: F mgj du (.0) F mg The force is the negative derivative with respect to d of the potential energ function U()=mg. Saing that the (one dimensional) force is the negative derivative of its potential energ is the same as saing that the work of the conservative force is the negative of its change in potential energ. Memorize this new law: Whenever the work done b a force leads to a full recoverable energ, the work done b this force is equal to the negative change in the potential energ. (When a force does work, it does so at the epense of its potential energ.) We will stud this in great detail in this chapter. (.) W U The force associated with such a potential energ is called conservative for reasons we eplain later. If we have a non-constant force F()=-k in one dimension, like the spring force, and want to calculate the work done b this force while it moves an object of mass m from point to B on the -ais, we proceed as follows: (.) infinitesimal amount of work: dwi F i i n n i i i i i W dw F If we make the Δ intervals smaller and smaller, while allowing n to go to infinit, we form an infinite sum of tin rectangles which is equal to the integral of the function F() taken between the points and B. 3. Force of a Spring Eample: Force of a spring: F( ) ki (k is the spring constant) B k k W F( ) d B U ( U B U ) (.3) B If we set the origin of the potential energ difference at 0, we can define a potential energ function (the potential energ of a spring) as:
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 4 of 4 (.4) U( ) k Following the eample of a constant force with a path in one dimension we define the potential energ in general as the function U(,,z) such that ever component of the force is the negative partial derivative of the potential energ. The reason for the negative can be found in the fact that the potential energ is diminished, when the force is being applied to do work. (From a purel mathematical point of view the negative sign is not necessar.) Thus, we obtain the -component of the spring force in the above eample (.4): U (.5) F ( ) ( k ) k 4. Force with several variables. If the force in question has more than one variable we must use the scalar product between the force F F (,, z), F (,, z), F (,, z) dr (.6) z d, d, dz to give us the infinitesimal amount of work dw: dr and the infinitesimal displacement vector d, d, dz dw F dr In order to avoid the subscripts for the three components of the force vector it ma be convenient F P,, z i Q,, z j R(,, z) k to use the notation Definition of work for a variable force on a variable path: or: F dr W F d F d F dz path path from to B from to B z F dr W P,, z d Q,, z d R(,, z) dz (.7) path from to B path from to B This integral is a path integral and depends in general not onl on the initial and final point but also on the path taken b the force F. s we shall see shortl, such a path integral taken over a closed loop is not necessaril equal to 0 (as is the case for a regular integral). In the previous case of the spring force we had onl one component of the force in the - direction and the force had onl the variable. The other eample was that for the force of gravit on the surface of the earth. The integral in such and similar cases never depends on the path, but onl on the initial and final point. Work, in these cases, is therefore alwas a simple integral. 5. Eamples for Path Dependent Forces. To have an illustration for a case in which the work is indeed path-dependent consider the following force:
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 5 of 4 (.8) F(, ), from (0,0) to (,) F(, ) dr We integrate from (0,0) to (,), first from to B: d d 0d 0d 0 from (0,0) to (,) from (0,0) to (,) D C Then from B to C: from (0,0) to (,) 0 d from (0,0) to (,) d d B Net we integrate from to D to C and get /. Now we integrate from to C on the diagonal, where =. We get: from (0,0) to (,) d d d d d 3 from (0,0) to (,) from (0) to () The answers are not the same, proving that the integral is path-dependent. Furthermore, if we integrate over a whole loop we get 0 if we go from to B to C and then back to on the diagonal. But we get / if we integrate along to B to C to D to. This clearl shows that we are not dealing with the normal kind of integral which alwas ields 0 for (.9) f ( ) d 0 The net eample is a little bit more complicated but shows again that the integral is path dependent. F 3 i 4 j from (0,0) to (,) path ) = (.0) path ) from (0,0) to (,0) and then from (,0) to (,) path 3) = 4 We get the following results: path ) 7, path ) 8, path 3) 7.33 6. Path Independent Force and Gradient: Now, let us use the force
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 6 of 4 (.) F(, ), Calculate the work done b this force along the path = from (0,0) to (,), along =0 from (0,0) to (,0) and then from (,0) to (,). Then along the path = from (0,0) to (,). You will find the same result, namel ½. Now, note the the components of this force can be obtained b taking the partial derivatives of S S the function S(, ) and, Now, we know that the total or perfect derivative of S is equal to: (.) ds d d This, in turn means that it is eas to integrate: (.3) and,, r ds S ds S S S r S is a class of functions called in phsics, potential energ functions, or potential functions. (.4) force function F,, z is said to have a potential function U(,,z) if its components are the partial derivatives of the potential function U. In phsics, the are the negative partial derivatives. F U, F U, F U z ; z It is convenient to define some new quantities: (.5) The del vector operator,, z With this notation we can write the force, deriving from a potential function as: (.6) U U U F gradu U,, z This is the scalar product between the vector operator del and the scalar function U. The result is a vector. 7. Conservative Forces, del operator, perfect derivative. For some forces, called conservative, the work integral is alwas path independent. Such forces can be written as the negative derivative of a scalar function called potential energ function U(,,z).
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 7 of 4 The generalized relationship between a conservative force F and its potential energ U is given in terms of the differential operator grad,, z U U U (.7) F gradu,, U(,, z) i j k z z The general formula for work as defined b formula (.7) has an integrand which looks eactl like the perfect derivative of a function; let us call that scalar function S(,, z): This function was introduced during our discussion of uncertainties. We defined the total or perfect derivative as: S S S (.8) ds d d dz z For eample, if S (,,z) = 3 3 z -/ we get 3 3 S(,, z) z (.9) 3 3 6 9 3 ds d d dz 3 z z z Note that this S is a scalar function, not a vector. Whenever we are faced with a path-integral containing an integrand like the one above we would immediatel know how to integrate it, namel: 3 3 6 9 3 (.30) d d dz ds S( b) S( a) 3 a to b along z z a to b z path C We need to find a wa to determine whether the factors in front of the d, d, and dz are the partial derivatives of a function S(,, z). We tr to integrate the first term with respect to, the second with respect to and the third with respect to z. If all integrated terms are the same we have succeeded in finding the S, and we can write the integrand simpl as ds. This means we need to investigate whether we can write the path-integral (.7) in the following form: (.3) a to b along C a to b along C a to b along C S S S F,, zdr d d dz z b grads dr ds S( b) S( a) a In order to find an eas intuitive answer to this question we introduce the concept of vector operators. The are an etension of the concept of the derivative operator d. If we appl this d operator to a scalar function S() we are to take the derivative of this function. The basic vector
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 8 of 4 operator is called the del operator and is defined as: as the differential vector operator (deloperator, an upside down Greek Δ) (.3) i j k,,,, z This vector operator must operate on some z z mathematical quantit to its right; it can be a scalar function, a vector function, or another differential operator. In the case of a vector function it can operate through a dot product or a cross-product. In the case that the del operator acts on a scalar function, it is called gradient or simpl grad. U U U gradu z U U U U z (.33) (,, ),,,, z To simplif the writing of partial derivatives one often defines that (.34) ; ; z z Do not mi this notation up with subscripts for vector components, like for eample which is the -component of the vector or vector function,, ; each of the components is a function of,, and z. For eample: z (.35) z (,, z),, z z z the vector component z is a function of,,z is the partial derivative with respect to of the function following the operator; for eample, if U(,) = 3 the operator ields U(,) = 3. 8. Divergence div, gradient grad, curl. Here is a summar of the three major uses of the del operator:
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 (.36) a),, z,, z U U U b) U gradu gradient of U,, ; U is a scalar function U(,,z) z E E Ez c) E dive divergence of the vectorfield E= E,E,E z d) i j k z B curlb z i ( Bz zb ) j zb Bz k B B B B B z 9 of 4 B is a vector field B, B, B ; each component is a function of,, and z. z d)if the vector B is a vector in two dimensions and the curl turns into a vector in the z-direction: i j k B curlb 0 k B B B B 0 --------------------------------------------------------------------------------------------------------- Let us recapitulate the basic idea: If a path integral can be written as the integral of a perfect derivative, we can integrate it easil and the integral depends onl on the initial and final point. This also means that the integral around an closed loop is 0. (.37) a to b along C a to b along C a to b along C S S S F,, zdr d d dz z b grads dr ds S( b) S( a) a If the force involved in the calculation of the work is the gradient of a continuous scalar function, then its curl must be equal to 0. One can convince oneself of this fact easil b realizing that the curl of the gradient of a scalar function is alwas equal to 0:
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 0 of 4 curl( grad S) S(,, z) 0 i j k S S S S S S i j k z z z z z (.38) S S S z s ou learn in calculus, if S is a continous function in, and z, then the mied second S S derivatives are the same: etc. z z Pa particular attention to the third term, which is the term of the curl of a function in the variables and onl! It implies that a test for the path-independence of a force function in two variables consists in checking whether F F 0 or whether F F (.39) a to b along C F(, ) F, i F, j If F, F, a to b along C F(, ) dr F, d F, d then the integral is path independent. More generall, if the curl of the vector function under the path integral is 0, then the integral is path independent and the force can be written as the perfect derivative of a scalar function. In phsics this scalar function S(,,z) is called the potential energ function U(,,z) of the force associated with it. The relationship is further defined as the negative gradient of U. Such a force is then called conservative: (.40) F gradu U ---------------------------------------------------------------------------------- 9. Potential energ of the gravitational force. Here are some problems to get ou acquainted with the cross product and vector operators To practice the concepts of the vector operators a little bit more, consider for eample, the gravitational force, which is a vector field (vector function with several variables). In this eample the origin of the force is located at the center of the mass M. The force F is the force eerted b the mass M on the small mass m.
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 (.4) (.4) r Fg i j zk mmg 3 r u r r r mmg z i j zk z The -component of the vector field F is mmg mmg F ; the -component is F z z u ; 3 3 of 4 Let us calculate the partial derivative of this function with respect to : (This leads to the calculation of the curl of F.) It is, using product rule and chain rule, which hold for partial derivatives just as for regular derivatives: (.43) F 3/ mmg z 3 / z 5 / Similarl for F,, z; Fz,, z and so on. z For eample: mmg mmg 3 (.44) F 3 5 z z Double-check that the curl of the force of gravit equals 0: F curlf 0 0. Curl of n Central Force. g One can show that an force that is directed along the radial connection between the objects involved, and onl depends on the distance r from a central point, is a conservative force. Such forces are called central forces. g f() r Central force: F f ( r) ur g( r),, z ; g( r) ; r z r Prove that the curl of such a force is equal to zero!
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 of 4. Summar for the Relationship Between Potential Energ and a Conservtive Force We know that the relationship for ever potential energ function U and the associated conservative force F is F gradu (.45) vector function: F F, F, F each component is a function of,, and z z scalar function: U=U(,,z) U U U F gradu,, U, U, zu U z ssume that we know that the potential energ U of,, and z for the gravitational force is given b: (We will prove this later.) mmg mmg (.46) U( r) ; with the reference at ; U( r ) 0 r z One can check that according to (.45) F =- U The component of the force of gravit F is given b: (.47) mmg mmg mmg F U () r 3 3 z z z which is indeed the correct component of F g To repeat the general definition of work for a variable force on a variable path in several dimensions: (.48) F dr W F d F d F dz path path from to B from to B z The line integral above is equal to the work W done b the force F along a path, which must be defined with the path-integral. The result of this integral is in general path-dependent. If this integral is path-independent we call the associated force F a conservative force. There are several was in which one can find out whether a force is path independent or not. If one can find a function U(,,z), through inspection and/or guessing, for eample, such that U,, z U,, z U,, z (.49) F,, z ; F,, z ; Fz,, z z then the force is conservative and does not depend on the path taken.
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 3 of 4 Eample : (.50) F a, b ; W ad bd B a b B from, to B, B Each term can easil be integrated separatel, and our potential function in this case is clearl equal to: (.5) a b U (, ) a b + an arbitrar constant Note that it does not matter for the previous statement if we add a constant to this function. We sa that: The potential energ function is determined up to a constant. There are general methods to find the original function. You learn this in the class on differential equations. We limit ourselves here to the simple cases used in phsics.. Review and Summar for Phsics: For an conservative force: (.5) W F dr gradu dr U dr du U path path path from to B from to B from to B B B W U where and B are the initial and final points of the integration path. In the case of a If the curl of a function F is 0, then its path integral is path independent. It derives from a scalar function U(,,z) which depends onl on the coordinates. Such a force is called a conservative force (non-dissipative) and is equal to gradu. conservative force the path does not matter, and the work depends onl on the initial and final points. We have seen that there is a mathematical test to see if the function F is a conservative vector function. If the function F(, ) has onl two components, the partial derivatives must obe the following: (.53) F, F, F, d F, d is path independent if: In the general case (force in three directions), we must have that the curl of F is 0. (.54)
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 4 of 4 3. Kinetic energ and the law of conservation of mechanical energ: Whenever the sum of the eterior forces is not equal to zero the mass on which the forces are acting will be accelerated according to Newton s second law: i F ma m i 0 dv dt We obtain the work (.55) b appling the path integral to both sides of this equation: dv dr W= F dr m dr m dv m v dv i 0 0 0 i dt dt path B path B path B path vvb (.56) fter the variable transformation from r to v, using dr=vdt, W= m v dv= m v dv v dv v dv 0 0 z z path v v path v v B B this is a path integral but we can readil integrate it which means that it is path independent. The reason for this can be seen easil: Just like the function in the line integral d+d derives from,so does the path (.57) v v function v, v derive from and depends onl on the initial and final velocit. Therefore we get the final result that the work done b all eterior forces, including frictional, non-conservative forces is equal to the change in kinetic energ of our moving mass between the initial and final point. (.58) (.59) W m0 (vb v ) K change in kinetic energ K. K= m0v Work-energ theorem: The work done b all forces eterior to a mass m along a specific path is equal to the change of kinetic energ of the mass between the initial and final point of the path. F d i r W= K KB K path to B i
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 5 of 4 Let us now assume that one of the forces in the path integral is a conservative force whereas the other is not: (.60) F gradu and Fdr gradu dr du U c (.6) or P c path path from P P to P W= F dr ( F F ) dr U W K nc nc i c nc path B i path B W U K We again use the fact that the total derivative of a scalar field U(,,z) is equal to du U U U (.6) du gradu dr d d dz z If all forces are conservative, the left side is 0 and we have the law of mechanical energ conservation: K U 0 K K U U 0 U K U K of the particle with mass m. for an two points and during the motion (.63) nother wa of saing the same thing is that the total mechanical energ K+U of a particle is the same at all times. When the particle loses some kinetic energ it gains the same amount in potential energ, and vice versa. When we consider macroscopic effects like friction, the work done b such dissipative forces is alwas negative and we can sa that, see (.6): Wnc f r K U (.64) The work done b friction forces is equal to the change in kinetic plus potential energ of the sstem. It is a negative value. rock of mass 500g is dropped b a distance of 00 m, and buries itself in the ground b a distance of 5.0 cm, calculate the average force of resistance over this distance of 5.0 cm. (3 significant figures) Solution: f 0.5 0 0 0 mgh 0.59.800.5 f 37N This work is usuall dissipated as heat, which cannot be recovered to do some kind of work.
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 6 of 4 For eample, when a bullet of mass 5 grams fling with a velocit of 550m/s, strikes a wall, most of the kinetic energ of the bullet is turned into heat energ Q (also sound and deformation energ). If all forces are conservative we arrive at the important theorem for conservation of energ: (.65) If all forces are conservative the total change of potential energ of all of these forces plus the total change in kinetic energ of the object on which these forces are acting is equal to 0. Or, the total energ of the sstem is conserved. (Hence the name conservative force.) K U 0 K K U U 0 K U K U End of regular phsics stuff for 30. The remaining pages are strictl optional.
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 7 of 4 4. Circulation and other advanced calculus stuff (Optional) dvanced stuff (just for fun; omit it out if ou have not et had vector calculus): n interesting path integral is the following which is equal to the area inside a closed curve, which is given parametricall: (.66) d d For eample an ellipse has the parametric representation =a cos(t) and = b sin(t) where a and b are the major and minor ais, and t varies from 0 to π: d asin t dt; d bcost dt a cost bcost dt bsint a sin t dt cos ab t absin t dt ab dt ab ab 0 The proof of the formula requires the knowledge of Stokes law: 5. Path Integral to Find the rea of a Closed Curve. Calculate the work done for the following force along the path =(0, 0) to B=(, 0) along the - ais, then from B to C=(,) along the -ais, then back to along the hpotenuse. F, ; W W d d; 0 and d 0 to B to B: W 0 B to C: = d=0
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 8 of 4 d 0 C to on the hpotenuse =/; d=d/ 0 d 0 d The total integral equals the area inside of the described loop. The formula = d d can be used to calculate the area inside an closed loop. Obviousl, if we interpret this as the work done b a force <-/, />, the force would be non-conservative. 6. Proof of Stokes Law: Quick and dirt proof: Let us do this directl and calculate the circulation (line integral) of a vector field E P, i Q, j around a rectangle with sides and. The lower left corner point of the square has the coordinates (a,b). We must calculate P, d along the horizontal lower and upper segments of the rectangle, and Q, d along the vertical segments on the right and left of the rectangle. To illustrate: If P(,)= and Q(,)=, we integrate from to B to C to D to, starting with the lower left corner at =(a,b). To choose the path along the -ais we set =b. Thus the first term of the integral is given b P, P, b b. When we move from (a+, b) to (a+, b+ ) we set have a fied value, namel a+. Onl the Q component contributes: Qa, a. Net we follow the path from (a+, b+ ) back to (a, b+ ) and the integral becomes: P(, b ) ( b ). Last, we integrate from (a, b+ ) down to (a,b): Q( a, ) a.,,, E ds P d Q d The total circulation is equal to: (.67) E ds P, b Q a, P(, b ) Q( a, ) Let us assemble the -and -components P and Q:, (, ) (, ), (.68) P b P b P b P b Q a, Q( a, ) Q a, Q( a, ) -ais
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 9 of 4 -ais We multipl the first epression b /, and the second b / : P(, b ) P, b (.69) Q a, Q( a, ) In the limit both of these terms can be written as the respective partial derivatives: Q a, Q( a, ) Q a, Q, dd P(, b ) P, b P, b P, dd Thus, the whole circulation becomes: Q, P, E ds d d On the right hand side we have now the approimation of a double integral: Circulation or line integral of the vector function E =P i Qj around the closed loop.,,,, Q P Q P E(, ) ds Pd Qd d d d d 7. Green s Theorem This is Green's theorem, which is nothing but the third component of Stokes' theorem in the plane: The circulation of an vector-field E around a closed loop is Q, P, equal to the surface integral of.
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 Note that, P, is the third component of Q curle E, E P,, z i Q,, z j R,, z k Regular Stokes theorem in three dimensions: The line integral of a vector field along a closed loop is equal to the surface integral of its curl : E ds E d. The surface is an surface which has the closed loop as its boundar. It the vector field is a funtion of and onl, the curl is reduced to the z-component. The theorem is then called Green's theorem. Q, P, E ds d d E z 0 of 4 Eample: Calculate the circulation of the vector field B 5, 7, 0 around a rectangle with sides a and b: nswer: curlb 0, 0, k The third component of this vector is parallel to the surface vector on the rectangle, and it is a constant. Therefore, the surface integral is the simple product k dd k - a b. 5d 7d 50d 7ab 5ba ab; (0,0) to ( a,0) to ( a, b) to (0, b) to (0,0). The Green theorem (Stokes theorem in the plane) is usuall stated in terms of a vector field with components P=E and Q=E ; P(,) is the -component, Q(,) is the -component: Q P E ds P, Q ds Pd Qd d d If we choose Q= and P=-, the rhs becomes a surface integral: Q P d d d d With this choice we arrive at a wa to calculate the surface of an area b means of the circulation around this area:
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 of 4 This circulation is equal to the surface integral of curlb. But we know that curlb j.the surface integral of j is 0 0j d 0I. Therefore, we have 0 0I rb 0I B r circle d d d d For eample, an ellipse can be written in parametric form as: acos ; bsin Show that the area of an ellipse equals =abπ Eample: Now, assume that we have a current densit entering into the page. We know from earlier calculation that a magnetic field surrounds the current densit. This means that the direction of the magnetic field around an loop, circling the current densit is parallel to the tangential unit vector: B Bu The circulation of this vector is given b: Bds Brd rb (.70) End of intermission. circle circle 8. Stokes' theorem: The flu of the curl of a vector field through an (simple) open surface is equal to the circulation of the vector-field around a closed boundar of the surface. (In short: Surface integral of the curl of the vector-function equals line-integral of the function= circulation.) Think of the surface and the rim in analog to a butterfl net and the rim that bounds it. The butterfl net can take on an shape, as long as it refers to the same rim. If the vector field (, ) B d curlb d B ds the flu of curlb through an simple surface with the closed boundar the circulation of B around the closed boundar of an surface B is a function of and onl, flu integral turns into a surface integral, with Bd B B k dd k B B dd curlb B B B k and the which is equal to the
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 of 4 circulation of the vector B around this area in the - plane. This law is then called Green's theorem, see below. Q P dd Pd Qd or B B dd B d B d This mathematical law of vector calculus is correct for an vector field (with the correct mathematical assumptions). The epression "circulation" refers to a line integral for a closed loop of an kind. We call the closed surface and its boundar. (.7) Stokes' theorem: The surface integral of the curl of E is equal to the line-integral along the line forming the boundar of the surface of the vector E. If ou let E=(P, Q, R) be a vector function with an and component, namel P(,) and Q(,) the Stokes theorem turns into Green's formula d ddz, ddz, dd E d Edl i j k i j k Q P E curle 0 0 k E E 0 P Q 0 Green's formula: which for Q= and P=- becomes: Q P Q P E d k ddk dd Q P dd Pd Qd dd d d area bounded b a curve : (.7) rea dd d d 9. The Curl of Tangential Velocit One can derive an interesting relationship between angular velocit, velocit in circular motion, and the curl as follows. Calculate the curl of velocit in circular motions: (Don t ask if ou must know this. What is there to know?). In pure circular motion the velocit vector is alwas tangential. Usuall we write this velocit as v= ru
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 3 of 4 When we are dealing with a three-dimensional problem, like the rotation of a sphere, then the motion of a point along an circle on the sphere can better be described in terms of a crossproduct, namel d v= r r dt The angle is the angle between the radius vector r with its origin on the ais of rotation, (.73) d and the point moving in circular fashion. The vector is perpendicular to dt the plane of motion along the ais of rotation. (.74) v=? with v= r,, z ; r=,, z ;,, ; z ω (.75) v= r z, z,,, z z z i j k v= z z z z z To summarize: (.76) v r The curl of velocit in circular motion equals : v= or v curl 0. Remarks on determinants: (Optional: For connoisseurs onl!) Let me emphasize some important facts about the cross product which we use here for the definition of the curl: In the definition of the curl through the determinant we must bear in mind that the multiplication between a derivative operator and a function is not commutative, for eample: (.77) Bz Bz Some of the properties of a determinant in the definition of the cross product are therefore not automaticall applicable to the definition of the curl as a determinant.
C:\phsics\30 lecture-giancoli\ch 07&8a work pathint vect op.doc 4/3/0 4 of 4 i j z z B z C, C, Cz i j k B Bz B B Bz B B B B k (.78) i Bz z B j Bz z B k B B z C B B ; C B B ; C B B z z z z z If we can factor out a certain scalar value from the vectors in the determinant, we can pull them to the front of the determinant. This is not correct if the vector is a vector operator. (.79) ssume that,, and B=5,, z z i j i j B 5 z 5 z C, C, C z z k k z End of intermezzo on determinants. ------------------------------------------------------------------------------------------ End of advanced stuff.