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physics 122 Lecture 02 October 1st 2013_final

physics 122 Lecture 02 October 1st 2013_final - Lecture 2...

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Lecture’2’ October’1,’2013’
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COLLECTIVE’BEHAVIOR’ By’collec;ve’behavior’we’mean’mo;on’that’ depends’not’only’on’local’condi;ons’but’on’ the’state’of’the’plasma’in’remote’regions’as’ well’ Local’concentra;ons’of’+ ve’and- ve’charges’give’rise’to’E’Felds.’ Mo;on’of’charges’give’currents’and’hence’magne;c’Felds.’These’Felds’ a±ect’the’mo;on’of’the’plasma’far’away’ Compare’with’molecules’of’air’no’EM’ gravita;onal’forces’are’negligible’just’ collisions’
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Concept’of’temperature’ Ο Π Ξ Μ Ν Λ = kT mu A u f 2 2 1 exp ) ( Figure I.4 where f is the number of particles per cm 3 with velocity between u and u + du . 2 2 1 mu is the kinetic energy k is the Boltzmann’s constant k = 1.38 × 10 -23 J/ o K The number n , or number of particles per cm 3 , is given by = . ) ( du u f n The constant A is related to the density n by 2 / 1 2 Ο Π Ξ Μ Ν Λ π = m n A The width of the distribution is characterized by the constant T which we call the temperature. To see the exact meaning of T , we can compute the average kinetic energy of particles in this distribution.
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= du u f du u f mu E av ) ( ) ( 2 1 2 Defining 2 / 1 ) / 2 ( m kT v th = and , / th v u y = we can express f ( u ) and E av as = = dy y Av y y mAv E v u A u f th th av th ) exp( ) exp( 2 1 ) / exp( ) ( 2 2 2 3 2 2 The integral in the numerator is integrable by parts [ ] [ ] + + + = = dy y dy y y y dy y y y ) exp( 2 1 ) exp( 2 1 ) exp( 2 1 ) exp( 2 2 2 2 Canceling the integrals we have kT mv Av mAv E th th th av 2 1 4 1 4 1 2 3 = = Ο Π Ξ Μ Ν Λ =
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Temperature’ ’(in’3-D)’ E av =3KT/2’ Since’ T ’and’ E av ’are’so’closely’related,’it’is’customary’in’plasma’ physics’to’give’temperatures’in’units’of’energy.’ To’avoid’confusion’on’the’number’of’dimensions’involved’it’is’not’ E av ’but’ rather’the’energy’corresponding’to’ kT ’that’is’used’to’denote’the’ temperature.’ For’ kT ’=’1’ eV’=’1.6’×’10 -19 ’J,’we’have’ Thus’the’conversion’factor’is’1’ eV’=’11,600’ o K’ By’a’2-eV’plasma’we’mean’that’ kT ’=’2’ eV,’or’ E av ’=’3’ eV’in’(3D)’ 600 , 11 10 38 . 1 10 6 . 1 23 19 = × × = T
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Debye’Length’ DEBYE’SHIELDING’
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Debye Length ε 0 2 φ = 0 2 x 2 = e n i n e ( ) This comes from: ∇⋅ E = σ 0 E = −∇ E = ∇⋅ −∇ ( ) = −∇ 2
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f ( u ) = A exp 1 2 mu 2 + q φ " # $ % & ' kT e ( ) * + , - ( ) e e kT e n n exp = ε 0 d 2 dx 2 = en exp e kT e " # $ % & ' " # $ % & '− 1 ) * + + , - .
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physics 122 Lecture 02 October 1st 2013_final - Lecture 2...

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