HOMEWORK 2 SOLUTIONS

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HOMEWORK 2 SOLUTIONS PHIL SAAD 1. Carroll 1.4 1.1. A qasar, a istance D from an observer on Earth, emits a jet of gas at a spee v an an angle θ from the line of sight of the observer.

1 HOMEWORK 2 SOLUTIONS PHIL SAAD 1. Carroll A qasar, a istance D from an observer on Earth, emits a jet of gas at a spee v an an angle θ from the line of sight of the observer. The apparent spee of this jet is the perceive velocity in the irection perpeniclar to the line of sight, v app x / t. At a time T after the jet is emitte, it has reache a perpeniclar istance of x vt sinθ. Now to fin t. If the qasar was emitte at time t 0, the observer sees it being emitte at time t 0 + D/c. At time t 0 + T, the jet has reache the perpeniclar istance vt sinθ an it takes L/c time for light from the qasar at that point to reach the observer. So the time ifference between the observer seeing the qasar emit the jet an the observer seeing the jet reach the perpeniclar istance of vt sinθ is We have t t 0 + T + L/c t 0 + D/c T + L/c D/c L D vt cosθ 2 + vt sinθ 2 D 1 2 vt D cosθ + vt D 2 Date: October

2 HOMEWORK 2 SOLUTIONS 2 Qasars are generally pretty far from the Earth, so the approximation D >> vt is a goo one. We can rop the thir term in the sqare root, an o the binomial expansion to first orer, getting This then gives s An L D vt cosθ t T + L/c D/c T v c T cosθ v app x t vt sinθ T v c T cosθ v sinθ 1 v c cosθ Let s fin the angle that maximizes this for any v. 0 θ v appθ, v v cosθ max θmax 1 v c cosθ max v sinθ max v 1 v c cosθ max 2 c sinθ max cosθ max 1 v c cosθ max v c sin2 θ max cosθ max v c An sing sincos 1 x 1 x 2, we have v app θ max, v This can efinitely be greater than c for v < c. 1 γv 1 v2 c 2 2. Scalar Gravity 2.1. We want a relativistic generalization of newtons gravitation law for a scalar fiel φx µ. Newton s law is simply 2 t 2 x i i φx i. This gives an obvios generalization, 2 τ 2 xµ µ φx µ This is inconsistent thogh. Let s mltiply both sies by m an rewrite this as τ P µ m µ φ. If we ot both sies with P µ, we have P µ τ P µ mp µ µ φ 1 2 τ P µp µ mp µ µ φ 1 2 τ m2 mp µ µ φ 0 mp µ µ φ This can t be tre in general. We can fix this eqation by aing a term to the right sie that cancells this. We arrive at τ P µ m µ φ + xµ x ν τ τ νφ Or eqivalently, τ P µ m µ φ + P µ P ν m 2 νφ

3 HOMEWORK 2 SOLUTIONS 3 Let s see if this has the correct nonrelativistic limit. For small v, we have t τ. We also have x µ /t being small, since we are in natral nits. This means we can ignore the terms qaratic in the the velocity an so we have t P i m i φ t P 0 m 0 φ + xν t νφ m 0 φ + φ m 0 φ + 0 φ + v i i φ mv i i φ t The first term is the Newton force law. In the secon term, we se the fact that 0 0, ẋ φ φ, an the chain rle φ/t φ/ t + φ/ x i x i / t. Ths gives s the corresponing law for the change in energy. Integrating over time an sing the chain rle on the right sie, we have As we expect. P 0 m φ 2.2. We want to show that this eqation of motion gives no bening of light. The first thing we mst o is se a ifferent parameter than τ, as we cannot parametrize a nll path by the proper time. Let s se coorinate time. Mltiplying both sies of or eqation of motion by τ/t, we have τ t τ P µ m τ µ φ + P µ P ν t m 2 νφ t P µ m µ φ + P µ P ν γ m 2 νφ We then can say P 0 mγ, so we have t P µ m2 µ P 0 φ + P µ P ν m 2 νφ m2 P 0 µ φ P µ P ν P 0 µ φ We can now take the correct limit. We have m 2 0, bt P 0 is hel fixe. This gives s t P µ P µ P ν P 0 ν φ We see that the change in the for momentm over time is proportional to the for momentm, which means it scales changes energy bt oesn t change irection! 2.3. Here s another way to approach the problem that is a bit more sophisticate. It makes it mch clearer that light oesn t ben. First, I ll talk abot the action for a point particle. This action is x S m τ m λ η µ x ν λ λ This action is annoying for two reasons. First, sqare roots can be kin of a bother. More importantly, this oesn t make sense in the massless limit. Let s consier a ifferent action S 1 1 λ 2 e η x µ x ν λ λ em2 where e is another egree of freeom that epens on λ. This action makes sense in the massless limit. The eqations of motion are then obtaine by varying with respect to e an the x µ. First we vary e δs 1 λ 1 2 e 2 η x µ x ν λ λ m2 δe Which gives s e 1 x η µ x ν m λ λ Plgging this back into the action gives s the normal point particle action.

4 HOMEWORK 2 SOLUTIONS 4 Now consier the simplest way to a a scalar fiel interaction to the point particle action. We col se the moifie action S m fφx µ ττ m fτ If f has a taylor expansion like f 1 + aφ +..., this makes a lot of sense, as it incles the normal kinetic term pls interaction terms. I ll leave it general for now thogh. What are the eqations of motion we get from this action? Varying with respect to the x µ we have τ P µ m fp µ m f τ φ µ φ f φ f µ φ + xµ τ x ν τ νφ By looking at the first term on the right sie, we see that in orer for this to have the correct nonrelativistic limit, we want f φ /f 1. This gives f eφ an or eqation of motion is µ φ + xµ τ m τ x ν τ νφ The same as we got the other way. In orer to nerstan the absence of light bening a little better, we shol try an nerstan this a little more physically. With f now chosen as e φ, we go back to the above action S m e φ τ m e φ x η µ x ν λ m e λ λ 2φ x η µ x ν λ λ λ This is the action of a point particle moving in a space with a conformally flat metric, g e 2φ η. This helps explain why light oesn t ben- conformal transformations preserve nll geoesics. To see this in terms of the eqations of motion, we go back to the action that has a well efine massless limit. Replacing η with or new g, an setting m 0, we have S 1 2 x µ λ 1 λ x µ e e2φ η λ Varying e gives s tells s that xµ λ 0 for a massless particle. Varying the xµ gives s δs 1 2 λ x 2 e e2φ µ δx µ λ λ 2 x ν x ν e λ λ µφδx µ Integrating by parts on the first term gives s δs 1 λ 2 λ an so 2 x e e2φ µ λ x ν λ 2 x ν x ν e λ λ µφ δx µ 2 x λ e e2φ µ 2 x ν x ν λ e λ λ µφ This secon part is zero by the eqation from e, an so we have 2 x λ e e2φ µ 0 λ Which says that the irection of the light s tangent oesn t change, it only scales it, as we saw before.

5 HOMEWORK 2 SOLUTIONS 5 3. Diffeomorphism Between The Real Line an Interval 3.1. We want to show that the sbset 1, 1 of R is iffeomorphic to R. In other wors, we nee φ : 1, 1 R Sch that φ is C an bijective, an its inverse is also C. The tangent fnction shol probably come in hany. Let s fin ot if these gys are C. tan πx 2 : 1, 1 R, 2 π tan 1 x : R 1, 1 x tanπx 2 π 2 sec2 πx 2 π 1 2 cos 2 πx 2 Which is continos on the interval 1, 1. Frther erivatives will be sms of terms with either one or sine on top an powers of cosine on the bottom, which has no zeroes in the region of interest. Ths or fnction is smooth in the region of interest. Now lets look at tan 1. 2 x π tan 1 x 2 1 π x Which is continos over the entire real line. We can see from the proct an power rles that any frther erivatives will be of the form n x n 2 π tan 1 x P nx x n Where P n x is a polynomial in x. Ths tan 1 is C, an tangent establishes a iffeomorphism between a finite open interval an the entire real line. Yo can also se φx tanh 1 x 4. Carroll We want to fin a single chart that covers the entire infinite cyliner, R S 1. The first step I will take is mapping the infinite cyliner to the semi-infinite cyliner, 0, S 1. With the obvios choice of having the circle in the x-y plane an having the z coorinate be the hight on the cyliner, x y x y e z

6 HOMEWORK 2 SOLUTIONS 6 Now all we nee to o is fin a single chart that covers the semi-infinite cyliner. This is most easily one by sterographic projection. I take the pnctre plane, R {0} with coorinates an v an pt it throgh the cyliner at a hight L above the bottom. I then take a line from the center of the circle at the bottom which is of rais R of the cyliner an exten it throgh the cyliner. Each line intersects the cyliner an the -v plane exactly once each, establishing a continos bijection between the two given by With the inverse relation x y v R Rv RL x L y L We can see that this maps any point on the cyliner to any point on the pnctre -v plane. These maps are also infinitely ifferentiable on the regions where z is greater than zero, or eqivalently 2 + v 2 is greater than zero, which is evient from the fact that the erivatives of these maps are given jst by power an proct rles an are proportional to the inverses of z or 2 + v 2, so are only singlar when these are zero. Now we want to trn this into the infinite cyliner, which is simply one by taking the log of z an making that the new z. Or maps from the -v plane to the infinite cyliner are given by x y z log R Rv RL which maps R 2 {0} to R S 1, {x, y, z R 3 x 2 + y 2 R 2 }, an v xl e z yl e z

7 HOMEWORK 2 SOLUTIONS 7 which maps the set R S 1, {x, y, z R 3 x 2 + y 2 R 2 } to R 2 {0}. Which are, sing the argments from before, clearly infinitely ifferentiable on the pnctre -v plane an cyliner. 5. Charts on the Sphere 5.1. We consier the charts on S 2 that are jst projections against the x-y or x-z or whatever planes. There are six sch charts neee to cover the spere. In particlar, we focs on the charts that cover the region z > 0 an x > 0 an look at their intersection. v x y, v z y To show that these charts are compatible, we mst fin the fnctions for an v in terms of an v on the region of overlap, an show that they are C., v z, v zx, y r 2 x 2 y 2 r 2 2 v 2 v, v y, v v Both these fnctions are C on the region of interest, since 2 + v 2 is always between zero an the rais of the circle. Differentiating this wol give polynomials over powers of the sqare root, an since the root has no zeroes on the region of interest, the transition fnctions are smooth. They also map open regions to open regions, since the region of interest is the sbset of the sphere with x an z greater than zero. They are also injective an srjective; which is evient in the existence of smooth inverses. Let s look at the inverse fnctions, v x, v xz, y r 2 z 2 y 2 r 2 2 v 2 v, v y, v yz, y y Again, both these fnctions are C on the region of interest an map the open sets of interest to open sets in a one to one an onto way. We know this becase the charts cover open sets, so their overlap is the intersection of those sets, which mst be open. Along with the transition fnctions being smooth, we have that these charts are compatible. The other charts mae from projection are obviosly compatible by symmetry.

Logarithmic, Exponential and Other Transcendental Functions

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