Analysis of Single Domain Particles. Kevin Hayden UCSB Winter 03

Transcription

1 Anlyi of Single Domin Prticle Kevin Hyden UCSB Winter 3

2 Prefce Thi pper cme bout becue of my curioity with mgnet. I think every child begin to wonder bout the mgic within two piece of metl tht tick to the refrigertor nd tht lo tick together or repel ech other with jut hlf rottion. Feeling like I know o little bout mgnet, which hve huge impct on life we know it well believing much, much more i to come from them, I ought out Dr. Crlo Grci- Cerver to guide me in tudying them. I hd previouly found tht he know gret del bout computer hrd drive nd therefore mgnetim. The reerch w very intereting nd chllenging nd wht I'm writing bout i unlike nything I've ever ttempted before. Thnk you very much for teching, guiding, nd puhing me through out the coure Crlo. The purpoe of thi pper i to tke my work nd preent it in wy tht i redble for nyone who would like to know ome bic mgnetic propertie. There will be lot of Clculu involved, but hopefully the ide nd reult will be cler even if the reder h little or no experience with the mth involved. Some thing will be technicl, but I will try to explin nd implify them becue I ve found tht mth eqution don t lwy pek for themelve.

3 Single Domin Prticle Wht i ingle domin prticle? It i imply n object with one mgnetic direction throughout tht entire object. So no mtter wht point you pick on, or in, the object, the mgnetic direction, m, i lwy pointing the me wy nd t the me trength. Figure But how do we know which wy the mgnetiztion i pointing? Thi depend on few fctor like the object hpe nd both internl nd externl mgnetic energie. The hpe we re intereted in i the Ellipoid becue it cn be imilr to the hpe of one bit of torge pce on computer' hrd drive. From thi we cn try to undertnd torge pce' mgnetic propertie. Although it i not uncommon for dektop computer to hve -gigbyte hrd drive, which i 9.6x bit or 96 billion bit, one hould find "one bit" hrd drive propertie intereting, even though it i obviouly le complicted. Figure Auming -gigbyte hrd drive conit of billion byte, which in't correct, but they re uully old tht wy.

4 b c Our ellipoid i given by the following formul: Ω = + +. Where b c, b, nd c re the length of the ellipoid' rdiu in the "", "b", or "c" xi direction, repectively. The "ey xi" i the longet xi nd i the direction tht mgnetiztion will nturlly point. Energy Now, let look t the energie of our prticle. There re four type: ) Exchnge Energy, denoted by Ex ) Unixil Energy, denoted by E Ku 3) Externl Field Energy, denoted by E H 4) Sty Field Energy, denoted by E d. The totl energy, tot, i the um of thee energie: unit re in Joule). E Etot Ex EKu EH Ed Exchnge energy i given by: = (ll of their E = A m dv. It i the energy penlty for x chnging the direction of mgnetiztion becue of tomic interction. Here Jr () [ m, m, m3] m = = i the trength nd direction of mgnetiztion of the ellipoid nd A i mteril contnt (different mteril hve bigger penltie thn other). But why would chnging the mgnetic direction require energy? Well in nture, object tend to be in tte tht require the let mount of energy ccording to it urrounding. It' like how op bubble re lwy phere while floting round in the ir. They only pop becue omething, like touching it with you finger, mke it' hpe chnge o tht the new hpe require more energy thn it vilble to hold the inner ir, nd o it pop. Similrly, there i certin direction tht the mgnetiztion nturlly wnt to point in nd it require the input of externl energy to chnge tht direction. Recll tht m i the mgnetiztion vector of the ellipoid Ω (figure ). Alo, note tht Ω i ingle domin prticle nd m i the me for ll point in Ω. Therefore m i contnt. Unixil Energy i given by: EKu = Ku( m + m3 )dv. It i the mgnetic Ω energy of the object nd depend upon the internl orgniztion of the object tom or lttice tructure but not on the overll hpe of Ω. Here, Ku i mteril contnt nd Ω The following energy formul were obtined from: Mgnetic Domin; The Anlyi of Mgnetic Microtructure by Alex Hubert nd Rudolf Schäfer.

5 m nd "c" xi'. m 3 re the trength of mgnetiztion in the direction of the ellipoid' "b" nd Externl Field Energy i given by: E = J d. It i the interction H im V H ex Ω energy between the direction of the mgnetiztion, m, nd the externl mgnetized field vector Hex = [ h, h, h3]. Here, i contnt. Note tht if the externl field i uniform, then thi energy only depend on the verge mgnetiztion nd not on the hpe of the mgnetized object. Stry Field Energy i given by: E d = d H i J dv. It i the energy of the Ω mgnetic field tht Ω crete. The hpe nd ize of Ω i wht decribe the Stry Field J Energy. Here, Hd = N i = [ d, d, d3] nd J = m, where J nd re contnt. N i clled ymmetricl demgnetizing tenor. In thi ce N i 3x3 digonl mtrix, which will be looked t more crefully little lter. For the ke of informtion, when one dd up the Externl Field Energy with the Stry Field Energy, one get the mgnetic field energy. Now, to mke thing eier to work with, let convert every thing to polr co- co in in in co φ ], the previou ordinte nd expnd. Letting m = [ ( θ ) ( φ ), ( θ ) ( φ ), ( ) formul become: x ( [co( θ)in( φ), in( θ)in( φ), co( φ)] ) (ince m i contnt) Ω E = A dv = EKu = Ku(in ( θ)in ( φ) + co ( φ)) dv Ω E = J h co( θ)in( φ) + h in( θ)in( φ) + h co( φ) dv H Ω Ω 3 J Ed = dco()in() θ φ + din()in() θ φ + d3co() φ dv We cn get the totl energy by umming thee energie. But firt, notice tht ll the integrl re over the me domin nd ll integrnd re contnt (becue m,, nd Hex re contnt for our Ω ). Thi men tht we cn pull out ll the integrnd from the integrl, um them, nd end up multiplying ll of them by the volume of Ω. Let the volume of Ω = W (i.e. dv = W ). Then the totl energy involved with our object i: Ω H d

6 + + 3 J d Ku J d Ku J d Etot = W[ co ( θ )in ( φ) + in ( θ)in ( φ) + co ( φ) Jhco( θ )in( φ) Jhin( θ)in( φ) Jhco( φ )] Recll tht the direction of mgnetiztion point in the direction tht require the let mount of energy. Therefore we hould minimize E tot to find which direction re poible for the mgnetiztion to point. 3 Energy Minimiztion Firt, to implify thing omewht, let minimize with no Externl Field Energy. Tht i, let Hex = [h, h, h 3 ] = [,, ]. To check where E tot i minimized, one mut pply the t nd nd prtil derivtive tet. So we hve: W d d d3 Etot = [ co ( θ )in ( φ) + ( Ku+ )in ( θ)in ( φ) + ( Ku+ ) co ( φ)] E tot E And we need to look t: tot E =, tot =, nd E θ Det E φ tot tot E tot φ. E tot The reder could verify: Etot d d = ( Ku + )(co( θ )in( θ)in ( φ)) E J d J d tot = ( Ku + )(co ( θ )in ( φ) in ( θ)in ( φ )) Etot d d d = [ co ( θ ) + ( Ku + )in ( θ) ( Ku + 3 )]co( φ)in( φ ) E J d J d J d = [ co ( θ ) + ( + )in ( θ) ( + )](co ( φ) in ( φ )) tot 3 Ku Ku φ

7 Etot d d = ( Ku + )(co( θ )in( θ)co( φ)in( φ)) φ Thu we find tht: Etot = for π 3π θ =,, π, φ [, π] nd φ =, π θ [, π ) π E tot θ =,, π = for φ [, π ] By the t Etot E prtil derivtive tet, when = nd tot =, with tndrd ngle for θ nd φ, E tot will hve minimum or mximum. To check if it i minimum, we mut pply the nd prtil derivtive tet. The nd prtil derivtive tet Etot E tot φ tte tht E tot i minimized when H( Etot ) = i poitive definite. Firt Etot Etot φ E we need the determinnt of H( E tot ) to be poitive. Next ince tot = for ll the φ vlue tht we mut check, we will only need to find when E tot > nd E tot > π 3π θ = θ = θ = π θ = θ = θ = π,, π φ [, π ) φ [, π ),, π φ = π, nd to φ = φ = π φ = E determine if we hve minimum. If tot E = or tot =, then the determinnt of will be zero nd nd prtil derivtive tet will be inconcluive. H( E tot ) for Firt for θ = or π, we hve φ [, π ) E tot =, o we hve no concluion. θ = E For π, we hve tot E > nd tot d d > when Ku + > nd φ = d3 d Ku + > repectively. Thu minimum for E tot h been found.

8 θ = π E For π, we hve tot E > nd tot d d > when Ku + > nd φ = d d3 > Ku + repectively. Another minimum for E tot h been found. For π 3π θ = or, we hve π φ = d3 d E tot > nd E tot > > repectively. A 3 rd minimum for E tot h been found. d d when > Ku + nd E All the previou inequlitie ocited with tot E > nd tot > re dependnt on the vlue of contnt, nd minly on H d = [ d, d, d3] of the Stry Field Energy. Recll tht Hd depend on the hpe nd ize of Ω. According to thi, E tot i then minimized in either the "" or "b" xi direction. Since we hve lredy defined the -xi the "ey xi", d will be mll enough ( we will ee lter) for the inequlitie to hold nd hve the direction of mgnetiztion point in one of the direction of the "ey xi". Now, let minimize E tot when n externl field i pplied in the direction of the "ey xi". Tht i, let Hex = [h, h, h 3 ] = [h,, ]. Then: E tot W J d J d = [ co ( θ )in ( φ) + ( Ku+ )in ( θ)in ( φ ) d3 + ( Ku + ) co ( φ) h co( θ)in( φ)] E So let' find when tot E =, tot = We hve:, nd when H( E tot ) i poitive definite. Etot d d = [( Ku + )co( θ )in( φ) + J h ]in( θ)in( φ )

9 E J d J d tot = ( Ku + )(co ( θ ) in ( θ))in ( φ) + J h co( θ)in( φ ) E J d J d J d [ co ( ) ( )in ( ) ( )]in( ) co( ) co( φ ) tot = θ + Ku + θ Ku + 3 φ h θ E J d J d J d = [ co ( θ ) + ( Ku + )in ( θ) ( Ku + )](co ( φ) in ( φ)) + J h co( θ)in( φ) tot 3 φ Etot d d = ( Ku + )co( θ )in( θ)co( φ)in( φ) + J h in( θ)co( φ ) φ It directly follow tht: Etot = for θ =, π, φ [, π ] nd φ =, π θ [, π ) E tot = for π 3π θ =, φ =, π π φ = nd θ [, π ) Recll tht to check if E tot i t minimum, one mut pply the nd prtil derivtive tet. Etot E tot θ θ φ So we mut find when Det E >, tot E >, nd tot >. Etot E tot φ θ = or π For, we hve Det J φ = d d3 or π = Ku (it' undeterminble). π 3π ± Jh θ = or For, we hve Det J ( 3 ) d d J h Jh = < (no minimum). φ = or π ± d d θ = Ku + + J h For π, we hve Det which φ = d3 d Ku + + J h d d d d3 men E tot i minimized when Ku + + J h > nd Ku + + h >.

10 θ = π For π φ = men E tot d d Ku + J h one get Det which d d 3 Ku + J h d d d3 d i minimized when Ku + > + h nd Ku + > + J h. π 3π d d θ = or Ku Finlly, for, we hve Det π J d J d φ = d d d3 d men E tot i minimized when > Ku + nd >. 3 which Combining thee reult of no externl energy nd n externl field long the "ey xi", we find tht the direction of mgnetiztion, m, cn only be in the direction of θ = θ = π the "ey xi". The "ey xi" correpond to the ngle π nd π. But φ = φ = wht hppen when we chnge the only thing tht in't contnt? Tht i, wht hppen when the pplied field i increed, decreed, or revered? The Hyterei Loop From the previou minimiztion inequlitie, we found tht m will be pointing in θ = the direction π, which i the poitive "ey xi" direction, whenever the externl φ = Ku Ku field i trong enough o h > mx[( d d 3 ),( d d ) ] J J i tified. θ = π Converely, the direction of m will be π, the negtive "ey xi" direction, φ = Ku Ku whenever h < min[( d 3 d),( d d) ] + J + J i tified.

11 Thi implie tht whichever wy m i pointing, the direction will not chnge if i pointing in the me direction m or if equl. But if one pplie trong enough h in the oppoite direction of m, h will force m to witch direction. Thi i becue Ω would no longer be in tte of minimum energy. The end reult i tht if the externl energy,, i mde lrge enough, it will dominte the inequlitie nd force the h direction of mgnetiztion, m, to be in the me direction decribing i when wek force of the inequlitie of minimiztion, wek o m will not chnge direction. h h h. A finl itution worth i pplied in the oppoite direction of m. Bed on h, will not ffect the energy minimiztion nd For the following figure, we dopt the convention of tting tht if m i pointing in the poitive direction, then it direction i, nd if it' pointing in the negtive direction. Figure 3 i grph of the phyicl property of mgnet tht i clled the Hyterei Loop. h To explin the Hyterei Loop, let y m' direction i nd poitive. If we were to grdully tke we trt to mke where h h Figure 3 i lrgely to zero, then m' direction will till be. Now negtive, m' direction will remin until one reche the vlue J Ku J Ku h < min[( d d ) +,( d d ) + ] J. Then the direction of m witche 3 to (long the dhed line on the left) becue the force of h would force E to no longer be in tte of minimum energy. A we keep increing direction, the direction of m will till be -. Then we bring get to where h h in the negtive bck to zero nd trt to i poitive gin, m till point in the - direction. But oon we get Ku Ku to vlue for h where the inequlity h > mx[( d d3),( d d) ] J hold, the direction of m witche bck to (long the dhed line on the right). So, m h h tot

12 doen't follow the me pth when it chnge direction. Thi i how the "loop" of the Hyterei Loop i creted nd why mgnet hve "memory". Tht i to y: once mgnet i mgnetized in certin direction, it will point in tht direction until n externl force mke it point in nother direction. Thi fct form the bi of how computer hrd drive work ince it i wht give computer it "memory". The writer of hrd drive crete n externl mgnetic field, h, nd tell the direction, m, of the bit which wy to point. One direction i red "" nd the other i red "". The direction tht hrd drive' bit re pointing i how computer interpret thee ' or ' it red file. Shpe nd Size It i now time to explore the propertie of N (from Stry Field Energy). Specificlly will exmine why d mut be mll when the -xi i the "ey xi". Recll tht the Stry Field Energy i E d = d H ijdv where Ω N J Hd = Ni = [ d, d, d3]. Becue of the dot product, N = i wht Nb N c ign how much mgnitude of J goe into ech entry of [ d, d, d3]. Ech digonl entry of N, i dependent on the hpe of the Ellipoid. N' digonl entrie re given by the following formul: bc N = ( η) ( η)( b η)( c η) d η bc Nb = ( b η) ( η)( b η)( c η) d η bc Nc = ( c η) ( η)( b η)( c η) d η Let' briefly tlk bout ize. If we et = b = c(mking the ellipoid phere), then we would find tht N = Nb = Nc =. Since we did not ign n ctul length to 3 the length of the phere' rdiu, it ugget tht hpe, not ize, i wht ffect the vlue of N. To better tudy the hpe of n object, we cn tudy the rtio of length between, nd c. To do thi efficiently, we mut pply chnge of vrible. Let, η = x b b c dη = dx nd let λ = nd =. Thu:

13 bc η 3 bc = ( + x) ( + x)( b + x)( c + x) dx N = ( η) ( η)( b η)( c η) d 3 bc 6 b c = ( x) ( x)( x)( x) d x 3 bc 5 = ( + x) ( + x)( λ + x)( + x) dx 3 bc = ( x) ( x)( λ x)( x) dx bc = ( x) ( x)( λ x)( x) dx b c = ( x) ( x)( λ x)( x) dx λ = ( x) ( x)( λ x)( x) dx A imilr proce cn be performed for Nb nd λ N ( ) ( ) ( ) = + x λ + x + x 3 λ ( ) ( ) ( N ) b = + x λ + x + x λ ( ) ( ) ( N ) c = + x λ + x + x 3 3 Nc dx dx dx with the following reult: Now we re redy to ee the effect of hving different ize rtio. There re two min propertie of N, N, N, which re follow: ) N + N + N = b c ) < N, N, N < b c b c Property ) implie tht ech entry of N i dependent on the other entrie. The following figure demontrte thi by fixing nd letting λ vry. b c (Note: In the following grph, b/ i ctully λ = nd c/ i ctully =.)

14 Figure 4

15 Figure 5

16 Figure 6

17 Since it my hve been difficult to ee propertie ) nd ) from the previou digrm, here i digrm of N, Nb, nd Nc on the me pge. Alo, to mke it eier to ee, the following digrm conider only one vlue of nd h hd the logrithmic function pplied to the vlue of λ, giving the grph gentler curve. To ee property ), look t log ( b/) 3 Figure 7 = on ech curve. Notice tht i little bit more.6, Nb i lmot, nd tht Nc i little le thn.4. Summing thee vlue give the vlue. To ee property ), imply note tht ech curve i between nd. Now let ee why d, of the Stry Field Energy, i mll when the -xi i the J J "ey xi". Recll tht Ni = [ d, d, d3], where i contnt, N N = Nb, nd tht the "ey xi" i the longet xi. Thu, ince the -xi N c i the "ey xi", b/ nd c/ re both le thn nd becue of the dot product d i entirely dependnt on N. So reclling Figure 4, we cn ee tht, nd therefore d, N N

18 i mll when b/< nd c/<. The phyicl interprettion of thi i tht the mller the vlue of N implie mll mount energy, d, i ocited with "ey xi" when it i the me direction the mgnetiztion. Summry of Bic We hve now een tht ingle domin prticle hve uniform mgnetic direction, they cn be very mll, nd tht there re everl different form of energy tht determine the propertie of it mgnetiztion. We lo found tht the direction of mgnetim will nturlly point long the "ey xi", the longet line through the prticle, when there i no externl force. After tht, we looked t the propertie of the mgnetic direction when n externl force i pplied long the "ey xi" nd dicovered of the Hyterei Loop property. Thi loop i the property tht llow mgnet to retin it direction depite n oppoing externl force. Finlly, we w how ize h no effect on mgnetic direction nd tht it i ctully dependnt on the prticle hpe, or it reltive xi length decribed by the formul N, Nb, nd N c. So the direction of mgnetiztion minly depend upon the mgnet compoition, hpe, nd urrounding externl force. Anlyi A jut tted in the Summry of Bic, ingle domin prticle cn be mll. We re ctully very intereted in wht hppen when the "ey xi" i much longer thn the other. Tht i wht hppen when λ nd. We cre bout wht hppen when λ or becue thi correpond to procedure like decreing the ize of bit on computer hrd drive to llow for more memory in the me mount of pce. However, thi led to problem with the formul: 3 λ ( ) ( ) ( N ) x λ x = x dx 3 λ ( ) ( ) ( N ) b x λ x = x dx λ ( ) ( ) ( N ) c x λ x 3 = x dx Notice tht lower vlue of the integrl i zero nd tht when we tke lim or lim of N, λ Nb, nd Nc, the integrnd i. So we mut mke ure tht the integrl exit for ech formul λ or. Recll tht from Figure 4-6, we hd N, N, nd Nc when λ. Therefore, we will be looking for propertie of the integrl tht correpond with the preceding entence nd ech formul' leding coefficient. b

19 λ, fix, nd ume ( ) Conider N with out the coefficient N λ, = Cλ α, for ome C, then we cn olve for α uing logrithmic propertie: α N ( ) N ( λ, ) = Cλ α λ, Cλ α α N ( λ, ) = C ( λ ) = (, ) log N λ = α log ( ) N ( λ, ) ( ) ( N( λ )) N ( λ ) log ( ) log, log, Thi cn be done in the me mnner for Nb nd Nc. Furthermore, we only need to tudy the integrl uing λ becue λ nd re ymmetric between the formul. (Notice tht λ nd re ymmetric in while λ nd re ymmetric between N c nd ). N Numericlly we cn olve for α, which will ct guide to wht to expect for the ymptotic behvior of the integrl in N, Nb, nd Nc λ or. The next figure how the behvior of α λ with =. (Note: In the following b c grph, b/ i ctully λ = nd c/ i ctully =.) = α N b Figure 8

20 From the top grph, which correpond to N ( λ,) = Cλ α, we cn ee tht α λ. Thi men tht we hould expect the integrnd of to converge to ome contnt C nd it integrl will exit. We hould expect the me reult for well ince α λ. However, for N we get tht α. Thi men tht C Nb ( λ, ) =, which men the integrnd of Nb goe to infinity λ nd therefore λ the integrl hould not exit. It doe not look good for N b t thi point, but mybe we cn find more deirble nwer with ome ymptotic nlyi. Firt, let tudy b 3 λ ( ) ( ) ( N ) x λ x = x dx rigorouly λ to confirm the numericl reult. So fix >, nd let f x = + x + x. Since f ( x ) i independent of λ, we need to only to tudy N( λ ) λ w replced with λ. We wnt to how tht we cn repreent ( ) N Nc 3/ ( ) ( ) ( ) / expnion of the form: N( λ ) = N( ) + λ + bλ+ O( λ), where ( ) term tht converge to zero fter thn λ λ. tht Notice tht ( ) f ( x) x for ll x >, which men tht f ( ) f ( x) f ( x) hve the following inequlity ( ) ( x) f = dx, where λ + x N λ with Tylor O λ include ll ( ) / f x h infinitely mny continuou derivtive, f =, nd λ + x f x i bounded for < x <, we hve for x nd dx = 5/ x 3 f ( x) N( ) = dx= x + x + x for ll x f ( x) dx < x x i bounded for ll x. We lo x >. Since dx x = nd. Furthermore, ince f ( x) x, we finlly get the inequlity ( ) dx ( ) 3/ pproximted once > i fixed. x f x dx x <. Therefore, exit nd cn be numericlly For the next term in the ymptotic expnion of N ( λ ), we need to plit the f ( x) f ( x) integrl in the following mnner: N( λ ) = dx+ dx. The integrl on λ+ x λ+ x the right h no problem ince the lower limit i nd doe not crete ingulrity in the f x h nice integrnd. So let turn our ttention to the integrl on the left. Since ( ) propertie, we cn get good ide of wht' hppening by looking t the integrl without

21 f ( x ) in it: dx = + λ λ. The term + λ h infinitely mny λ + x continuou derivtive o it doen't cue ny problem. However, the firt derivtive of λ i ingulr when λ = nd i reponible for the ingulritie of the integrnd. So we re ble to ue the Tylor' expnion N( λ ) N( ) λ O( ) N ( λ ). Now we jut need to how tht exit. = + + λ to repreent Notice tht ( λ ) N ( ) = lim N nd cn be computed: λ λ ( λ ) ( ) N N x λ + x = f ( x) dx f ( x) = λ λ λ+ x x λ λ+ x x λ = f ( x) dx λ λ+ x x x + λ+ x dx ( ) Next, fter doing the chnge of vrible x = λ dx = λd, one get: N ( λ ) N( ) = λ f d ( λ ) ( ) λ λ λ+ λ λ λ + λ+ λ = f ( λ) d ( ) Uing thi, we cn tke the limit nd olve for : N( λ ) N( ) = lim = f ( ) λ λ ( ) d Thi integrl exit nd i finite ince when we re cloe to the lower limit of zero the integrnd diverge like, which i integrble nd we pproch the upper limit of the integrnd converge to zero like 3. Replcing λ bck with λ, we get N( λ ) N( ) λ O( λ ) = + +. So we hve λ λ N = N( λ) = N( ) + λ+ o( λ ) nd lim N, ince λ λ λ which i jut ome contnt. Therefore ( ) N λ, nd hence lim N λ jut we expected from the numericl computtion. lim N( λ ) N( )

22 λ Next, let tudy N ( ) ( ) ( ) c x λ x 3 = x dx rigorouly λ. For ome fixed vlue >, let ( ) ( ) / ( g x x x) 3/ ( ) = + +. Since g x i g x independent of λ, we need to only to tudy N( λ ) = dx, where λ w λ + x replced with λ. Notice tht thi etup i the me wht we hd for. Then g x nd f ( x ) hve the me propertie, nd the ( ) N λ of nd reult. Therefore lim i the me lim. N N c λ N c ( ) N ( ) will hve the exct me nlyi N λ Finlly, let ee wht i hppening for. Recll tht we re lredy expecting it integrnd to diverge to, nd we could eily be tuck with thi if we were to ttempt the me nlyi tht w done for. However, we cn get ide of it ymptotic behvior N N b λ ince we know the behvior of N nd N λ nd by uing the property N + N + N =. b c We know N + N + N = λ b c But, we cn ee from Figure 6, ( N N ) let N ( λ ) repreent >. Thi men tht ( N N N ) λ c lim + + = too. λ b c lim + = implying tht lim =. But if we c N b λ N with out the leding λ coefficient, we'll hve lim λn ( λ) =. Thi implie tht we mut hve lim N ( λ ) diverge like b when λ λ. λ λ λ = lim. So we hve tht the integrl of λ N b Summry of Anlyi Now we hve n undertnding of the ymptotic behvior of N, Nb, nd Nc for when λ or. Thi i ueful for, nd correpond to, chnging the Single Domin Prticle from 3-D ellipoid to ner -D ellipe poible. Therefore it will be nerly impoible to hve the mgnetic direction point in the 3 rd dimenion, but till cn chnge direction bout the "ey xi" ince it cn trvel in the other two dimenion. Thi i importnt becue computer hrd drive mnufcturer wnt to hve bit tht re mnipulted in the mot retricted environment poible to void ccidentl mgnetic direction chnge of ech individul bit. So n undertnding of the ymptotic behvior of mgnetiztion on ever hrinking bit ize i crucil to undertnd for efficient writing nd efficient corruption protection.