Hour Exam No.1. p 1 v. p = e 0 + v^b. Note that the probe is moving in the direction of the unit vector ^b so the velocity vector is just ~v = v^b and


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1 Hou Exam No. Please attempt all of the following poblems befoe the due date. All poblems count the same even though some ae moe complex than othes. Assume that c units ae used thoughout. Poblem A photon with fequency f in a paticula "laboatoy efeence fame" is obseved fom a space pobe that is moving at % of the speed of light in the laboatoy fame. The photon is moving in the diection of the spacelike unit vecto ^n while the pobe is moving in the diection of the spacelike unit vecto ^b. a. Find the foumomentum components of the photon in the laboatoy fame. Answe a p hf (e + ^n) whee e is the time unit vecto in the lab fame. b. Find the components of the space pobe s fouvelocity in the laboatoy fame. Answe b u e + ~v p v Note that the pobe is moving in the diection of the unit vecto ^b so the velocity vecto is just ~v v^b and u e + v^b p e + v^b p v v At % of the speed of light, v so the numeical answe is u e + ^b q q e + ^b p e + ^b
2 c. Use esults (a) and (b) to nd an expession fo the photon fequency that is obseved fom the space pobe. Answe c In the pobe s efeence fame, the photon foumomentum takes the fom p hf (e + ^n ) The timelike unit vecto used by the pobe is just the pobe s velocity e u p e + : ^b Pick out the tem that we want by using the dot poduct: p e hf f h p u h (hf (e + ^n)) f p f (e + ^n) p f p f e e + f p f + ^n ^b p e + ^b e + ^b ^n ^b cos whee is the angle between the photon diection and the pobe diection as seen fom the lab fame.
3 Poblem Show, using a (+++) signatue spacetime metic, a. that the dot poduct of the foumomenta q; p of any two paticles (massive o massless) must obey the inequality q p : Answe a If one one of the paticles is massive, its foumomentum take the fom p mu whee u is its fouvelocity. But, in the estfame of that paticle, and u e q p q e + q i e i me mq e e mq The time component q of a paticle s momentum is its enegy, which will always be positive. The mass is also positive, so in this case we have q p < : The only othe possible case is whee both paticles ae massless. foumomenta then take the foms p hf (e + ^n) q hf (e + ^n ) Thei whee ^n and ^n ae unit vectos in the diection of motion of each paticle. In that case, q p hf (e + ^n ) hf (e + ^n) h ff (e e + ^n ^n) h ff ( + cos ) whee is the angle between the two unit vectos. This expession is negative o zeo fo all values of so q p
4 b. that a collection of massive paticles cannot decay into a single photon. Answe a Let the paticles have momenta p ; p ; :::; p n and suppose that they decay into a photon with momentum q. Foumomentum consevation then equies Dot both sides of this equation with q. p + p + ::: + p n q q p + q p + ::: + q p n q q Evey one of the tems on the left is negative de nite (not zeo) because they each involve a massive paticle foumomentum. The ighthand side is zeo, so the equiement is impossible.
5 Poblem Conside thee linea mappings of a vecto space V into itself: A : V! V B : V! V C : V! V The vecto space is spanned by a set of n basis vectos, fe i g and its dual space ^V is spanned by the dual basis foms! k. a. Regad these mappings as tensos and wite expessions fo thei tenso components in tems of the basis vectos and foms. Answe a Take the basis vectos to be e i and the basis foms to be! j and associate with A the tenso that assigns the value (A (v)) A (v) to the pai (v; ) A (v; ) (A (v)) The components of this tenso ae then Simlaly, A j i A e j ;! i A (e j )! i B j i C j i B e j ;! i B (e j )! i C e j ;! i C (e j )! i b. Regad the composite mapping C BA as a tenso and nd an expession fo its components in tems of the components of A; B; C. Answe b We need to nd the components K s ((C B A) e s )!! (A (B (C (e s ))))
6 Stat fom the inside and note that the components of the vecto C (e s ) ae C (e s )! i C s i and expand that vecto in tems of the basis C (e s ) C s i e i Now inset that expansion into the expession and pull out the coe cients: K s! A B C i s e i C s i! (A (B (e i ))) Next, expand B (e i ) in tems of its components B (e i ) B i j e j and inset that expansion into the expession and pull out the coe cients: K s C i s! A B j i e j C s i B i j! (A (e j )) C s i B i j A j This expession is, of couse, a matix poduct. aay of components, we have found that Using [] to denote a squae [C B A] [C] [B] [A]
7 Poblem Suppose that an electical cuent I uns along the z axis of a Minkowski coodinate system. a. Expess the esulting magnetic eld components as functions of the Minkowski coodinates, t; x; y; z. It is OK to consult an E&M book hee. Answe a Fom the E&M book, we lean that the magnetic eld at a distance fom the cuent has magnitude. kbk I Assume a ighthanded Catesian coodinate system x; y; z. If we adopt pola coodinates and in the x; y planes, then x cos ; y sin and the eld points tangent to the cicles at constant, t; z; in the diection of inceasing. At this point, you could daw a pictue and use plane geomety and tig. I pefe to nd the unit vecto ^ tangent to these cicles. That unit vecto is elated to the tangent vecto though some nomalization function N chosen so that ^ N ^ ^ N To gue out the dot poduct, use the chain ule fo patial deivatives to get in tems of the Catesian basis vectos since we know thei dot poducts: dx d x + dy d y sin x + cos y y x + x y
8 x y x + y y x x y y x Now nd the nomalization function N N and the desied unit vecto in the diection of the magnetic eld ^ x y y x The magnetic eld, in tems of x; y; z; t is then B I x y y x I x y y x Take the Minkowski coodinate basis vectos to be e t ; and nd the magnetic eld components to be e x ; e y ; e z B B B Iy (x + y ) B Ix (x + y ) b. Find the equation of motion of a positively chaged paticle that is moving though this magnetic eld in the x z plane. Answe b 8
9 Fist, gue out what the Maxwell tenso components would be. Since thee is no electic eld, F F F F F F The emaining components ae linked to the magnetic components by the elations F F B F F B Iy (x + y ) F F B Ix (x + y ) Because these ae spacelike components in an othonomal fame, we can aise o lowe the indexes without changing anything. Next, wite out the space components of the Loentz foce law in the fom that we had in the notes dp m kmf m sv s whee we found that km must equal the chage on the paticle and v s dxs with t x. Fo a paticle of chage q, we get dp m qf m dx s s To nish the job of poducing equations of motion, we need to expess the space momentum components p m in tems of the deivatives dxs. p m p vm v dx dy Now wite out the equations one by one fo m ; ; : d B dx dy dz dx m dz dx C A qf dy + qf dz q Ix dz (x + y ) 9
10 d B dx dy dz dy C A qf dx + qf dz q Iy dz (x + y ) d B dx dy dz dz C A qf dx + qf dy q Ix dx (x + y ) + q Iy dy (x + y )
11 Poblem In spheical coodinates, the metic tenso on Minkowski spacetime has the fom g + d d + d d + sin d' d' a. Wite the components of this tenso and the invese metic g in the holonomic basis that coesponds to the coodinates x t x x x ' Answe a Thee ae no cosstems, so the tenso is diagonal. the fom Compae the tenso in g g dx dx + g dx dx + g dx dx + g dx dx to the one above with the names of the coodinates changed: g dx dx + dx dx + dx dx + sin dx dx and identify g g g g sin o g g sin sin
12 o g ; g ; g ; g sin b. Nomalize the basis vectos t ; ; ; ' and foms ; d; d' to obtain an othonomal (i.e. Minkowski) fame at each point. Wite out the components of the tansfomation matices that expess the othonomal basis vectos and foms in tems of the holonomic basis objects. Answe b The vectos ae aleady nomal to one anothe and two ae aleady unit vectos, so we just need nomalizing factos fo the othe two: Fom the metic, so and e t ; e e K e J ' e e K K g K e e e J ' ' J g J sin so e sin ' In tems of a tansfomation matix, In index notation, e e e e so that the components ae [U] sin e U U U U U U U U U U U U U U U U U t ' sin
13 The othonomal basis foms ae elated to the holonomic ones though the invese matix elation! T T!! d d! d' sin o In index fom,!!!! T d d d' T! dx U so we can ead o the components fom U U U U U U U U U U U U U U U U sin sin c. The notes discuss the splitting of the Maxwell Field Tenso components into Electic and Magnetic eld components in an othonomal (Minkowski) fame. Show how this splitting woks in a holonomic spheical coodinate fame. In othe wods, expess f s in tems of E k and B j when all of these components ae taken using the holonomic bases. Answe b Fo the othonomal components, f B f B f B f f E f f E f f E These components ae elated to the holonomic components accoding to f s U a U b s f ab ; f U a U f a U a f a ; B j U k j B k ; E j E k j U k
14 The tansfomation matices ae diagonal and we can ead the components fom the ealie esult. The sums collapse to f U U f U f f f U U f U f sin f f U U f sin f f U f f f U f f f U f sin f B U B B B U B B B U B sin B E E U E E E U E E E U E sin and the elationships fo the magnetic components become f sin B sin f B sin f B o f sin B ; f sin B ; f sin B Fo the electic components, the elationships become f E f E f E f E f E sin f E sin f sin E
15 Poblem Fo onefoms ; ; and a vecto v, conside the tenso K v ( + + ) a. Expess the components of TK in tems of the components of ; ; ; v. Answe a Fo any fom and vectos u; v; w this tenso assigns the numbe: K (; u; v; w) v () [ (u) (v) (w) + (u) (v) (w) + (u) (v) (w)] The tace is de ned to be and has components TK (v; w) K (! a ; e a ; v; w) (TK) s TK (e ; e s ) K (! a ; e a ; e ; e s ) v (! a ) [ (e a ) (e ) (e s ) + (e a ) (e ) (e s ) + (e a ) (e ) (e s )] Notice that we just eplaced by! a ; u by e a, v by e and w by e s in the expession fo K (; u; v; w). But v (! a ) v a ; (e a ) a and so on, ae just the components that we want, so (TK) s v a [ a s + a s + a s ] b. Expess the tenso TK as a combination of basisindependent objects such as dot poducts and tenso poducts of the objects ; ; ; v. Answe b The answe can be seen petty diectly fom the index expession above, o you can get it by witing out the de nition of TK (v; w) T K (v; w) v (! a ) [ (e a ) (v) (w) + (e a ) (v) (w) + (e a ) (v) (w)] v a a (v) (w) + v a a (v) (w) + v a a (v) (w) (v ) (v) (w) + (v ) (v) (w) + (v ) (v) (w) T K (v ) + (v ) + (v )
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