Chapter 4 Contravariance, Covariance, and Spacetime Diagrams


 Egbert Horn
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1 Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz Trnsformtion, it is necessry to drw coordinte systems tht re skewed to ech other rther thn to use the trditionl orthogonl coordinte systems. It is therefore pproprite to digress for moment nd look into some of the chrcteristics of skewed coordinte system. For simplicity nd clrity we will strt our discussion in two dimensions, lter we will extend the discussion to more thn two dimensions. Consider the skewed coordinte system shown in figure 4.. We will use the stndrd nottion tht is used in reltivity nd use superscripts to lbel the coordintes x l nd x s shown. (x does not men x squred, it is just different mens of lbeling the coordintes, the reson for which, will become cler in moment.) A series of lines everywhere prllel to these coordinte xis estblishes spce grid. The intersection of ny of these two lines estblishes set of coordintes for ny prticulr point considered. Let us now drw the vector r in this coordinte system. Now let us find the components of the vector r in this skewed coordinte system. But how do we find the components of vector in skewed coordinte system? r x x Figure 4. A skewed coordinte system. () Rectngulr Components of Vector in n Orthogonl Coordinte System. First, let us recll how we find the components of vector in n orthogonl coordinte system. To find the xcomponent of the vector we project r onto the xxis by dropping perpendiculr line from the tip of r down to the xxis s shown in figure 4.(). Its intersection with the xxis, we cll the xcomponent of the vector. Note tht the line perpendiculr to the xxis y y y r r j y j i x x () (b) Figure 4. An orthogonl coordinte system. i x x 4
2 is lso prllel to the yxis. The ycomponent is found by projecting r onto the yxis by dropping perpendiculr line from the tip of r to the yxis. Its intersection is clled the ycomponent of the vector r. Also note tht the line perpendiculr to the yxis is prllel to the xxis. In terms of the unit vectors i nd j, nd the x nd ycomponents, the vector r cn be expressed s r ix + jy (4.) The set of vectors i nd j re sometimes clled set of bse vectors. Implied in the representtion of the vector r by eqution 4. is the prllelogrm lw of vector ddition, becuse ix is vector in the xdirection nd jy is vector in the ydirection. Moving these vectors prllel to themselves genertes the prllelogrm, nd the digonl of tht prllelogrm represents the sum of the two vectors ix nd jy s shown in figure 4.(b). Exmple 4. Rectngulr components of vector. A vector r hs mgnitude of 5 units nd mkes n ngle of with the xxis. Find the x nd ycomponents of this vector. Solution The xcomponent of the vector r is found s x r cos θ x 5 cos x 4.33 y y r The ycomponent of the vector r, is found s y r sin θ y 5 sin y.50 θ x x Figure 4.3 The rectngulr components of vector. To go to this Interctive Exmple click on this sentence. (b) Contrvrint components of vector. Now let us return to the sme vector in the skewed coordinte system. We introduce new system of bse vectors l nd s shown in figure 4.4. The bse vector l is in the direction of the x l xis nd is in the direction of the x xis. The bse vectors nd will be clled unitry vectors lthough they don t necessrily hve to be unit vectors. We return to the originl question. How do we find the components of r? For the orthogonl system, the perpendiculr from the tip of r ws perpendiculr to one xis nd prllel to the other. For the skewed coordinte system the prllel of one xis is not perpendiculr to the other. So there ppers to be two wys to find the components of vector in skewed coordinte system. For the first component let us drop line from the tip of r, prllel to the x xis, to the x l xis. This component will be clled the contrvrint component of the vector r nd will be designted s x nd is shown in red in figure 4.4(). Drop nother line, prllel to the x xis, to the x xis. This gives the second contrvrint component x, which is lso shown in red in figure 4.4(). In terms of these contrvrint components the vector r cn be written s 4
3 x x x x cos α α x r α r α x θ α α x x cos α x x α O x () Contrvrint components (b) Prllelogrm lw Figure 4.4. Contrvrint components of vector. x r x + x (4.) We observe from figure 4.4(b) tht the vectors x nd x dd up to the vector r by the prllelogrm lw of vector ddition. So tht eqution 4. is vlid representtion of vector in the skewed coordinte system. Exmple 4. Contrvrint components of vector. A vector r hs mgnitude of 5.00 units nd mkes n ngle of with the xxis. If the skewed coordinte system, figure 4.5, mkes n ngle α , () find the contrvrint components of this vector, nd (b) express the vector in terms of its contrvrint components. Solution. The contrvrint components of the vector r re found from the geometry of figure 4.5. The contrvrint component x is found by observing from tringle I sin α r sin (α θ) (4.3) x Upon solving for the contrvrint component x we get r sin ( α  θ) x (4.4) x 5.00 sin( ) 5.00 (0.643) 5(0.684) sin x 3.4 x (α θ) α θ α α x x x x x Figure 4.5 Contrvrint components. r I x r sin (α θ) x II r sin θ The contrvrint component x is found by observing from tringle II tht 43
4 sin α r sin θ x (4.5) nd upon solving for the contrvrint component x we get r sin θ x (4.6) x 5.00 sin (0.500) sin x.66 5(0.53) b. The vector r cn now be written in terms of its contrvrint components from eqution 4. s r As check tht these re the correct contrvrint components of the vector r, let us determine the mgnitude r from this result. We cn no longer use the Pythgoren Theorem to determine r, becuse we no longer hve right tringle s we do in the cse of rectngulr components. We cn however pply the lw of cosines to the tringle of figure 4.4(b) to obtin r (x ) + (x ) + x x cos α (3.4) + (.66) + (3.4)(.66) cos( ) (.7) + (7.08) + (8.) (0.34) r 5.00 We see tht we do get the correct result for the mgnitude of the vector r. To go to this Interctive Exmple click on this sentence. (c) Covrint components of vector. For second representtion of the components of vector in skewed coordinte system, we now drop perpendiculr from the tip of r to the x xis intersecting it t the point tht we will now designte s x, nd is shown in yellow in figure 4.6(). We will cll x covrint component of the vector r. Now drop perpendiculr from the tip of r to the x xis, obtining the second covrint component x, lso shown s yellow line. We now hve the two vector components x nd x. But these vector components do not stisfy the prllelogrm lw of vector ddition when we try to dd them together hed to til, s is obvious from figure 4.6(b). Tht is, by dding the vectors from hed to til, you cn see tht x + x will not dd up to the vector r. In fct you cn see tht the sum would be ctully greter thn the mgnitude of r, nd would not be in the correct direction. Therefore in terms of these components r x + x At first glnce it therefore seems tht the only wy we cn find the components of vector tht is consistent with the prllelogrm lw of vector ddition is to drop lines from the tip of r tht re prllel to the coordinte xis, thereby obtining the contrvrint components of vector. 44
5 x x r α θ () x x (b) Figure 4.6 Covrint components of vector. However there is still nother wy to determine the components of the vector r nd tht is to estblish new coordinte system with unit vectors e nd e where e is perpendiculr to nd e is perpendiculr to. This new bse system is shown in figure 4.7(), long with the old bse system. The bse vector e defines the direction of new xis x, while the bse vector e defines L L L L () (b) Figure 4.7 Introduction of some new bse vectors. 45
6 new xis x. We now drop perpendiculr from the tip of r to the x xis, but insted of terminting the perpendiculr t the x xis, we continue it down until it intersects the new x xis. (Note tht the perpendiculr line is perpendiculr to the x xis but not the x xis.) We cll the projection on the new xis, L. Now drop perpendiculr from the tip of r to x, then extrpolte it until it crosses the new x xis t the point L. Then we cn see from figure 4.7, tht the vectors Le nd Le will dd up by the prllelogrm lw of vector ddition to r Le + Le (4.7) Cn we express the lengths L nd L, in terms of the covrint components x nd x? Referring bck to figure 4.4(), we first note tht the ngle α is the mesure ngle of the mount of skewness of the coordintes. The covrint component x cn be seen to be composed of two lengths, i.e. x x + x cosα (4.8) while the covrint component x is composed of the two lengths From the bottom tringle in figure 4.7() we observe tht x x + x cosα (4.9) nd hence L x x L (4.0) And from the upper tringle in figure 4.7() we hve Therefore L x x L (4.) Replcing equtions 4.0 nd 4. into eqution 4.7 gives r x e + x e (4.) or upon slightly rerrnging terms, we cn write this s r x e + x e If we now define the two new bse vectors nd e sin α e sin α (4.3) (4.4) 46
7 then the vector r cn be written in terms of the covrint components s r x + x (4.5) Since eqution 4.5 is just eqution 4.7, but in different nottion, it lso stisfies the prllelogrm lw of vector ddition. Exmple 4.3 Covrint components of vector. A vector r hs mgnitude of 5.00 units nd mkes n ngle of with the xxis. If the skewed coordinte system mkes n ngle α , () find the covrint components of this vector, (b) express the vector in terms of its covrint components, nd (c) find the vlues of the bse vectors. Solution. To express the vector in terms of its covrint components we use eqution 4.5 r x + x The covrint component x is found from figure 4.7() s while the covrint component x is found from figure 4.7() s x r cos θ (4.6) x 5.00 cos ( ) 5.00 (0.866) x 4.33 x r cos (α θ) (4.7) x 5.00 cos ( ) 5.00 cos ( ) x 3.83 As check let us find the covrint component x from eqution 4.8 in terms of the contrvrint components s x x + x cosα (4.8) The vlues of x nd x were determined in Exmple 4. to be nd x r sin (α θ) (4.4) sin α x 5.00 sin( ) 3.4 sin 70 0 x r sin θ sin α x 5.00 sin sin 70 0 (4.6) Replcing these vlues into eqution 4.8 gives for the covrint component x x x + x cosα (4.8) x cos
8 x 4.33 The covrint component x is found from eqution 4.9 in terms of the contrvrint components s x x + x cosα (4.9) x cos x 3.83 Notice tht the components re the sme for either procedure. b. The vector r in terms of the covrint components is obtined from eqution 4.5 s Hence, r x + x r As check tht these re the correct covrint components of the vector r, let us determine the mgnitude r from this result. We cn no longer use the Pythgoren Theorem to determine r, becuse we no longer hve right tringle s we do in the cse of rectngulr components. We cn however pply the lw of cosines to the tringle of figure 4.4(b) to obtin r (x) + (x) xx cos α r (4.33) + (3.83) (4.33)(3.83) cos 70 r 4.70 units But something is wrong here. We know the mgnitude of r should be 5 nd it is not. The trouble is tht the unitry vectors re not unit vectors but unitry vectors. They re not equl to one. We hve to tke these vectors into ccount. They re tken into ccount by using the bse vectors e nd e which re unit vectors. c. The bse vector is found from eqution 4.3 s e e e sin e (4.3) The bse vector is found from eqution 4.4 s e e e sin e (4.4) Notice tht the bse vectors nd re not unit vectors, nd their vlue will vry depending upon the ngle α tht the coordintes re skewed. If we wish we could lso write the vector in terms of the unit vectors e nd e by using equtions 4.3 nd 4.4. r x + x r r (4.33)(.06)e + (3.83)(.06)e 48
9 r (4.6)e + (4.08)e Recll from eqution 4.7 nd figure 4.7b tht r Le + Le (4.7) where L x times the mgnitude of the unitry vector,. Tht is nd L x L (4.33)(.06) 4.6 L (3.83)(.06) 4.08 As check tht these re the correct covrint components of the vector r, let us determine the mgnitude r from this result. As we showed erlier for the contrvrint vector, we cn no longer use the Pythgoren Theorem to determine r, becuse we no longer hve right tringle s we do in the cse of rectngulr components. We cn however pply the lw of cosines to the tringle of figure 4.7(b) to obtin r (L) + (L) LL cos α r (4.6) + (4.08) (4.6)(4.08) cos 70 r 5.0 r 5.00 units Notice tht we get the sme mgnitude of 5 units s we did in exmples 4. nd 4.. We could lso use eqution 4.7 by first determining the vlues of L nd L from the equtions nd L x / sin α L (4.33) / sin (70) L (4.6) L x / sin α L (3.83) / sin (70) L (4.08) nd plcing these into the lw of cosines we get r (L) + (L) LL cos α r (4.6) + (4.08) (4.6)(4.08) cos 70 r 5.0 r 5.00 units Agin notice tht we get the sme correct result for the mgnitude of the vector r. To go to this Interctive Exmple click on this sentence. To summrize our results, eqution 4. is the representtion of the vector r in terms of its contrvrint components, r x + x (4.) while eqution 4.5 is the representtion of the vector r in terms of its covrint components, i.e. 49
10 r x + x (4.5) So in skewed coordinte system there re two types of components contrvrint nd covrint. The contrvrint components re found by prllel projections onto the coordinte xes while the covrint components re found by perpendiculr projections. Contrvrint components re designted by superscripts, x i, while covrint components re designted by subscripts, xi. The bse vectors nd re not unit vectors even if nd re. The distinction between contrvrint nd covrint components disppers in orthogonl coordintes, becuse the xes re orthogonl. Tht is, in orthogonl coordintes, projection which is prllel to one xis, is lso perpendiculr to the other. Let us now return to the spcetime digrms we discussed in chpter nd see how these concepts of covrince nd contrvrince re pplied to these spcetime digrms. 4. Different Forms of The Spcetime Digrms Figure.9 showed the reltion between the S nd S frmes of reference in spcetime. The S frme ws the sttionry frme nd S ws frme moving to the right with the velocity v. The ngle θ, of figure.9 ws given by eqution.3 s θ tn But we lredy sid tht there is no frme of reference tht is bsolutely t rest, nd yet our digrm shows the preferred sttionry frme, S, s n orthogonl coordinte system while the moving frme, S, is n cute skewed coordinte system. So it seems s if the rest frme is specil frme compred to the moving frme. However, the principle of reltivity sys tht ll frmes re equivlent. Tht is, there should be no distinction between frme of reference tht is t rest or one tht is moving t constnt velocity v. Figure 3.9 should be modified to show tht there is no preferentil frme of reference. We showed in Chpter, tht if body is t rest nd body moves to the right with velocity v, tht this is equivlent to body being t rest nd body moving to the left with the velocity v. Another equivlence is to hve n rbitrry observer t rest between nd nd body cn move to the right with velocity v/ with respect to the frme t rest nd body cn move to the left with velocity v/. We cn incorporte these generlities by redrwing figure 3.9 for S with the τ xis now mking n ngle θ/ with the originl τxis, nd by showing second observer, S, moving to the left of the sttionry observer with the velocity v/. This is shown in the spcetime digrm of figure 4.8 s the τ xis mking n ngle θ/ with the τxis. The ngle θ/ is computed in the sme wy s the computtion for the τ xis, tht is, v c θ/ tn v/ c θ/ tn v/ c S frme moving to the right with velocity v/ with respect to S frme S frme moving to the left with velocity v/ with respect to S frme In this wy the S frme will be moving t the velocity v with respect to the S frme. Similrly n x xis cn be drwn t n ngle θ/ from the xxis. Notice tht the x nd τ re found in the sme wy tht we found x nd τ, except tht x nd τ hve negtive slopes, indictive of the motion to the left. These new x  nd τ xes generte new cute skewed coordinte system, S, locted in the fourth qudrnt, s seen in figure 4.8() nd 4.8(b). Note tht the S coordinte system is shown in blue while the S coordinte system is shown in red. Also notice tht becuse of the symmetry, the scles re the sme in the S frme s they re in the S frme, which is of course different to the scle in the S frme s we showed before. Also note tht in figure.9, θ ws the ngle between τ nd τ becuse S ws moving t the speed v with respect to the S frme. Now 40
11 notice tht θ/ is the ngle between τ nd τ becuse S is now moving t the speed v/ with respect to the S frme of reference. Also note tht θ is now the ngle between τ nd τ becuse S is now moving t the speed v with respect to S. θ/ θ/ θ/ θ/ θ θ/ θ/ θ θ/ θ/ () (b) Figure 4.8 Reltion of S nd S frme of references. Also note from figure 4.8 tht the x xis is orthogonl to the τ xis since but α 90 0 θ. Hence, nd τ 0x α + θ τ 0x 90 0 θ + θ τ 0x 90 0 Similrly, the τ xis is perpendiculr to the x xis, since but φ 90 0 θ/. Hence, nd x 0τ φ + θ/ x 0τ 90 0 θ/ + θ/ x 0τ 90 0 The fct tht the x xis is orthogonl to the τ xis, nd the τ xis is orthogonl to the x xis, should remind us of how the x xis ws perpendiculr to the x xis nd the x xis ws perpendiculr to the x xis in figure 4.7 in our study of some of the chrcteristics of covrint nd contrvrint components. We will return to this similrity shortly. Figure 4.8 shows tht the S nd S frmes re symmetricl with respect to the S frme of reference, but not with respect to ech other. Both frmes should lso mesure the sme velocity of light c, which is ssured if the light line OL were to bisect both sets of coordinte xes. Also note tht becuse of the symmetry of both S nd S frmes, they would both intersect the scle hyperbols t the sme vlues. Hence, the scle on the S frme is the sme s the scle on the S frme. We cn modify figure 4.8 by reflecting the x xis in the fourth qudrnt, through the origin of the coordintes, to mke n x xis in the second qudrnt, s shown in figure 4.9. Note tht now 4
12 the light line OL does indeed bisect the x,τ xes nd the x,τ xes gurnteeing tht the speed of light is sme in both coordinte systems. The S coordinte system is now n obtuse skewed coordinte system insted of the cute one it ws in the fourth qudrnt. Figure 4.9 should now be used to describe events in the S nd S coordinte systems, insted of figure 3.9 which described events in the S nd S frmes of reference. θ/ θ/ θ θ/ θ/ θ/ θ θ/ θ/ θ θ/ () (b) Figure 4.9 New S nd S frme of references. 4.3 Reciprocl Systems of Vectors We hve discussed the spcetime digrms in two dimensions. We would like to extend tht discussion first into three dimensions nd then into four or more dimensions. In order to extend this discussion we must first discuss the concept of reciprocl systems of vectors. Consider the three dimensionl oblique coordinte system shown in figure 4.3. The three xes re described by the constnt unitry vectors,, nd 3 s shown in the figure. We now define the set of reciprocl unitry vectors s 3 3 (4.6) Figure 4.3 The reciprocl unitry vectors (4.7) (4.8) 4
13 By nture of the cross product of two vectors, nd s cn be seen in figure 4.3, is perpendiculr to the plne generted by nd 3; is perpendiculr to the plne generted by 3 nd ; nd 3 is perpendiculr to the plne generted by nd. Hence,,, nd 3 re clled unitry vectors, while the vectors,, nd 3 re clled reciprocl unitry vectors. Let us now consider combintions of products of these unitry vectors nd their reciprocl unitry vectors. First, let us consider the product nd 3 3 ( 3) ( 3) 3 (4.9) (4.0) But s you recll from vector nlysis, by the vector triple product of three vectors, cyclic interchnge of letters is permissible, tht is, Applying this to our unitry vectors we get Using eqution 4. in eqution 4.0 gives ( b c) b ( c ) c ( b ) (4.) ( 3) 3 3 (4.) Agin using eqution 4.8 we find for the third product ( 3) 3 3 (4.3) The results of equtions 4.9, 4., nd 4.3 shows tht the product of unitry vector nd its reciprocl unitry vector is equl to one, tht is, (4.4) 3 3 When we consider the mixed products of these unitry vectors nd their reciprocl unitry vectors we get 3 ( 3 ) ( 3) ( 3) But s cn be seen in figure 4.3, 3 is perpendiculr to the plne generted by nd 3 nd hence is perpendiculr to the vector nd hence its dot product with is equl to zero. Tht is, Hence (3 ) 3 cos
14 Similrly the mixed product 3 ( 3) 3 But s cn lso be seen in figure 4.3, is perpendiculr to the plne generted by nd nd hence is perpendiculr to the vector nd hence its dot product with is equl to zero. Tht is, ( ) cos Hence the dot product of ( ) is equl to zero, therefore 3 0 In similr wy, ll the mixed products of the unitry vectors nd the reciprocl unitry vectors re equl to zero. Tht is, (4.5) In summry, the reciprocl unitry vectors re defined by equtions 4.6, 4.7, nd 4.8, nd the product of these unitry vectors nd the reciprocl unitry vectors re summrized in equtions 4.4 nd 4.5. Just s equtions expressed the reciprocl unitry vectors in terms of the unitry vectors, the unitry vectors cn be expressed in terms of the reciprocl unitry vectors by the sme reciprocl reltions. Tht is, 3 (4.6) (4.7) (4.8) Since the order of dot product is not significnt, tht is, b b, the combintions of ll the products in eqution 4.9 through 4.5 re the sme. Tht is, nd get (4.9) (4.30) If we pply the sme resoning process to the orthogonl i, j, k, system of unit vectors we j k i i i i ( j k) i i k i j j j i ( j k) i i i j k k k i j k i i 44
15 Therefore, the reciprocl vectors of i, j, k, re the vectors i, j, k, themselves. In fct for ny orthogonl set of unit vectors, whether rectngulr, sphericl, cylindricl etc., the reciprocl unit vectors will be the unit vectors themselves. All orthogonl sets of vectors re selfreciprocl. The only time we will hve reciprocl sets of vectors is when we hve oblique coordinte systems, s we do in our spcetime digrms. In section 4. we nlyzed skewed coordinte system in twodimensions nd showed tht we could represent vector in tht twodimensionl system by using either contrvrint or covrint components of vector. Tht is, we found the vector r could be written in terms of the contrvrint components x nd x s r x + x (4.) nd in terms of the covrint components x nd x s r x + x (4.5) Remember tht the contrvrint components were found by dropping lines tht were prllel to the pproprite xes, while the covrint components were found by dropping lines tht were perpendiculr to the pproprite xes. Now tht we hve estblished three dimensionl skewed coordinte system, we cn now write the vector r, in figure 4.3, in terms of the three dimensionl contrvrint components x, x, nd x 3, nd the unitry vectors,, nd 3 s r x + x + x 3 3 (4.3) r Figure 4.3 Three dimensionl skewed coordinte system. The vector r cn lso be expressed in terms of the covrint components x, x, nd x3 of the vector nd the reciprocl system of unitry vectors,, nd 3 s r x + x + x3 3 (4.3) where the reciprocl unit vectors,, nd 3 re given by equtions 4.6, 4.7, nd 4.8. Exmple 4.4 In section 4. we showed tht we could estblish new coordinte system with unit vectors e nd e where e is perpendiculr to nd e is perpendiculr to. The bse vector e defined 45
16 the direction of new xis x, while the bse vector e defined new xis x. In this new set of coordinte we showed tht vector r could be written in terms of the covrint components s if we defined the two new bse vectors nd r x + x (4.5) e e (4.3) (4.4) Show tht eqution 4.3 is equivlent to eqution (4.6) The reciprocl unitry vector is given by Solution 3 3 But the ngle between the unitry vectors nd 3 is the skew ngle α of the coordintes. Hence, 3 3 sin α sin α Where 3 since they re unit vectors. Also since the ngle between the vectors nd ( 3) is (90 0 α), then ( 3) ( 3) cos (90 0 α) 3 sin α cos (90 0 α) But 3 since they re unit vectors. Therefore However, Therefore ( 3) sin α cos (90 0 α) cos (90 0 α) cos 90 0 cos α + sin 90 0 sin α sin α ( 3) sin α cos (90 0 α) sin α Replcing these vlues in eqution 4.6 gives nd hence sin α 3 3 e Hence the cse shown in section 4. for finding the covrint components of vector is specil cse of the reciprocl system of vectors. 46
17 4.4 Exmple of The Use of Covrint nd Contrvrint Vectors As we hve seen, ny vector cn be written in two wys. One in terms of the contrvrint components nd the other in terms of the covrint components. A vector written in terms of its contrvrint components is clled contrvrint vector. A vector written in terms of its covrint components is clled covrint vector. As n exmple, force vector cn be written s or F F + F Contrvrint Vector ( 4.33 ) F F + F Covrint Vector (4.34) Notice tht the contrvrint vector is represented in terms of the contrvrint components nd the bse vectors nd, while the covrint vector is represented in terms of the covrint components nd the bse vectors nd. Either bse system or both my be used in connection with the vectoril tretment of given problem. As n exmple, the work done in moving n object through displcement r by force F cn be expressed three wys: () in terms of contrvrint vectors (b) in terms of covrint vectors (c) in terms of mixture of contrvrint nd covrint vectors. () Work done using contrvrint vectors. The work done in terms of the contrvrint vectors is Now W F r (F + F ) (x + x ) (4.35) W F x + F x + F x + F x cos0 0 (4.36) cos0 0 (4.37) If the ngle between the two xis is α then Therefore the work done in terms of the contrvrint components is which is not prticulrly simple nd is dependent upon the ngle α. (b) The work done using covrint vectors. The work done in terms of the covrint vectors is Now we showed in equtions 4.3 nd 4.4 tht cos α cosα (4.38) cos α cosα (4.39) W F x + F x + (F x + F x ) cosα (4.40) W F r (F + F ) (x + x ) (4.4) W Fx + Fx + Fx + Fx (4.4) 47
18 Therefore, nd e nd e sin α 0 cos0 (4.43) sin α 0 cos0 (4.44) Now the ngle between nd is (80 0 α) s cn be seen in figure 4.7. Therefore but Therefore 0 cos(80 α ) cos(80 0 α) cos(80 0 ) cos α sin (80 0 ) sin( α) cos α cos α sin α cos α sin α (4.45) (4.46) Substituting equtions 4.43 through 4.45 into eqution 4.4 gives for the work done W Fx + Fx Fx cos α Fx cos α sin α sin α sin α sin α (4.47) which is rther complicted form for the work done (c) The work done using mixture of contrvrint nd covrint components. The work done cn be expressed s the product of the contrvrint force vector nd the covrint displcement vector. Tht is, W F r (F + F ) (x + x ) (4.48) W F x + Fx + F x + F x (4.49) But s cn be seen in figure 4.7, nd shown in eqution 4.30 nd by eqution becuse 0 becuse Replcing these vlues into eqution 4.49 gives W F x + F x (4.50) Eqution 4.50 gives the work done expressed in terms of contrvrint nd covrint components. If we hd expressed the force s covrint vector nd the displcement s the contrvrint vector we would hve obtined W F r (F + F ) (x + x ) 48
19 W F x + F x (4.5) In generl the product of contrvrint vector with covrint vector will yield n invrint (sclr) which will be independent of the coordinte system used. Hence, when using skewed coordinte systems, it is n dvntge to hve two reciprocl bse systems. In most of the nlysis done in generl reltivity by tensor nlysis, there will usully be mix of covrint nd contrvrint vectors. Also note tht the unitry vectors nd cn hve ny mgnitude. As n exmple if nd l then the spce grid would look s in figure 4.33(). If on the other hnd nd l, the spce grid would pper s in figure 4.33(b). We see tht this mounts to hving different scle on ech xis. Tht is, unit length on the x xis is twice s lrge s the unit length on the x xis. Although l nd cn hve ny mgnitude in generl, we will lmost lwys let them hve unit mgnitude () l (b) ; Equl Spced Grid Different scle on x nd x xis. Figure 4.3 The unitry vectors nd cn hve ny mgnitude. Note tht the contrvrint vector in eqution 4.33 hs the contrvrint components of the vector, F nd F, nd the unitry bse vectors nd ; while the covrint vector in eqution 4.34 is represented in terms of the covrint components of the vector F nd F nd the reciprocl unitry bse vectors nd. F F + F Contrvrint Vector (4.33 ) nd F F + F Covrint Vector (4.34) Notice tht the unitry bse vectors re described with subscripts, while the reciprocl unitry bse vectors re described with superscripts. Notice tht the product of ech term is product of contrvrint superscript nd covrint subscript. Hence, the vector cn be thought of s consisting of contrvrint nd covrint terms; superscript times subscript for contrvrint vector nd subscript times superscript for covrint vector. Since the product of contrvrint vector nd covrint vector gives us n invrint quntity or constnt, this nottion will give us n invrint quntity for the mgnitude of ny vector. We will see much more of this lter. In generl, the product of contrvrint vector with covrint vector will yield n invrint (sclr) which will be independent of the coordinte system used. Hence when using skewed coordinte systems, it is n dvntge to hve two reciprocl bse systems. We will see tht in most of the nlysis done in generl reltivity, there will usully be mix of covrint nd contrvrint vectors. Summry of Bsic Concepts Skewed Coordinte Systems. In order to show inertil motion tht is consistent with the Lorentz Trnsformtion, it is necessry to drw coordinte systems tht re skewed to 49
20 ech other rther thn to use the trditionl orthogonl coordinte systems. Components of vector in rectngulr coordinte system. The xcomponent of vector is found by dropping perpendiculr line from the tip of r down to the xxis. Note tht the line perpendiculr to the xxis is lso prllel to the yxis. The ycomponent is found by dropping perpendiculr line from the tip of r to the yxis. Also note tht the line perpendiculr to the yxis is prllel to the xxis. Contrvrint components of vector in skewed coordinte system. For the skewed coordinte system the prllel of one xis is not perpendiculr to the other. For the first component we drop line from the tip of r, prllel to the x xis, to the x l xis. This component is clled the contrvrint component of the vector r nd is designted s x. Drop nother line, prllel to the x xis, to the x xis. This gives the second contrvrint component x. The bse vector l is in the direction of the x l xis nd is in the direction of the x xis. The bse vectors nd re clled unitry vectors lthough they don t necessrily hve to be unit vectors. The vector r cn be written in terms of the contrvrint components s r x + x Covrint components of vector in skewed coordinte system. For second representtion of the components of vector in skewed coordinte system, we drop line from the tip of r, perpendiculr to the x xis intersecting it t the point tht we will now designte s x, We will cll x covrint component of the vector r. We now drop line from the tip of r perpendiculr to the x xis, obtining the second covrint component x. We now hve the two vector components x nd x. However, these vector components do not stisfy the prllelogrm lw of vector ddition when we try to dd them together hed to til. Tht is, r x + x, However, if we define two new bse vectors e e nd sin α sin α then the vector r cn be written in terms of the covrint components s r x + x In generl the product of contrvrint vector with covrint vector will yield n invrint (sclr) which will be independent of the coordinte system used. Hence, when using skewed coordinte systems, it is n dvntge to hve two reciprocl bse systems. We will see tht in most of the nlysis done in generl reltivity, there will usully be mix of covrint nd contrvrint vectors. In Summry, in skewed coordinte system there re two types of components contrvrint nd covrint. The contrvrint components re found by prllel projections onto the coordinte xes while the covrint components re found by perpendiculr projections. Contrvrint components re designted by superscripts, x i, while covrint components re designted by subscripts, xi. The bse vectors nd re not unit vectors even if nd re. The distinction between contrvrint nd covrint components disppers in orthogonl coordintes, becuse the xes re orthogonl. Tht is, in orthogonl coordintes, projection which is prllel to one xis, is lso perpendiculr to the other. 40
21 Summry of Importnt Equtions The vector r written in terms of contrvrint components r x + x (4.) Contrvrint component x r sin ( α  θ) x Contrvrint component x r sin θ x The bse vectors e sin α e sin α (4.4) (4.6) (4.3) (4.4) The vector r written in terms of covrint components r x + x (4.5) Covrint component x x r cos θ (4.6) Covrint component x x r cos (α θ) (4.7) Lorentz trnsformtion for spce coordintes x '' x ' + vt' (4.3) v / c Lorentz trnsformtion for the time v t' + x' coordintes. t'' c v / c (4.35) Inverse Lorentz Trnsformtion for spce coordintes x ' x '' vt'' (4.36) v / c Inverse Lorentz Trnsformtion for time v t'' x'' coordintes t' c (4.37) v / c Length contrction formul L L0 v / c (4.46) Time diltion formul t'' t' t ' 0 v / c t '' 0 v / c (4.5) (4.56) The invrint intervl of Spcetime ( s ) ( x ) c ( t ) (4.94) nd ( s ) ( x ) c ( t ) (4.95) The invrint intervl of Spcetime in terms of differentil quntities (ds ) (dx ) c (dt ) (4.96) nd (ds ) (dx ) c (dt ) (4.97) The invrint intervl in fourdimensionl spcetime (ds) c (dt) (dx) (dy) (dz) (4.98) Reciprocl unitry vectors expressed in terms of the unitry vectors 3 (4.6) ( 3) 3 (4.7) ( 3) 3 (4.8) 3 Products of the unitry vectors nd the reciprocl unitry vectors 3 3 (4.4) (4.5) The unitry vectors expressed in terms of the reciprocl unitry vectors, 3 (4.6) (4.7) 4
22 3 3 (4.8) The combintions of ll the products (4.9) nd (4.9) For three dimensionl skewed coordinte system, the vector r is written in terms of the three dimensionl contrvrint components x, x, nd x 3, nd the unitry vectors,, nd 3 s r x + x + x 3 3 (4.3) the vector nd the reciprocl system of unitry vectors,, nd 3 s r x + x + x3 3 (4.3) Work done using contrvrint vectors. W F r (F + F ) (x + x )(4.35) W F x + F x + (F x + F x ) cosα (4.40) Work done using covrint vectors. W F r (F + F ) (x + x ) (4.4) W Fx + Fx Fx cos α Fx cos α(4.47) sin α sin α sin α sin α The work done using mixture of contrvrint nd covrint components. W F r (F + F ) (x + x )(4.48) W F x + F x (4.50) W F x + F x (4.5) The vector r cn lso be expressed in terms of the covrint components x, x, nd x3 of. Why cn t we just use orthogonl systems in our nlysis of reltivity?. Wht is contrvrint vector? 3. Wht is covrint vector? 4. Is unitry vector the sme s unit vector? Questions for Chpter 4 Problems for Chpter 4 5. When using product of two vectors, is it better to hve two covrint vectors, two contrvrint vectors, or one of ech? 4. The Components of Vector in Skewed Coordintes. A vector r hs mgnitude of 5.0 units nd mkes n ngle of with the x xis. Find the rectngulr components of this vector. A vector r hs mgnitude of 5.0 units nd mkes n ngle of with the x xis. If the skewed coordinte system, mkes n ngle α , () find the contrvrint components of this vector, nd (b) express the vector in terms of its contrvrint components. 3. A vector r hs mgnitude of 5.0 units nd mkes n ngle of with the x xis. If the skewed coordinte system mkes n ngle α , () find the covrint components of this vector, (b) express the vector in terms of its covrint components, nd (c) find the vlues of the bse vectors. To go to nother chpter, return to the tble of contents by clicking on this sentence. 4
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