Physics 523, General Relativity Homework 4 Due Wednesday, 25 th October 2006

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1 Physis 523, General Relativity Homework 4 Due Wednesday, 25 th Otober 2006 Jaob Lewis Bourjaily Problem Reall that the worldline of a ontinuously aelerated observer in flat spae relative to some inertial frame an be desribed by tλ, α α sinhλ and xλ, α α oshλ,.a. where λ is an affine parameter of the urve with αλ its proper length i.e. the time as measured by an observer in the aelerated frame. Before, we onsidered α to be onstant and only varied λ. We are now going to onsider the entire non-surjetive urvilinear map from two-dimensional Minkowski to-spae to itself defined by equation.a.. a Consider the differential map from t, x-oordinate harts to λ, α-oordinate harts implied by equation.a. lines of onstant α are in the λ-diretion, and lines of onstant λ are in the α-diretion. We are to show that wherever lines of onstant α meet lines of onstant λ, the two urves are orthogonal. To show that the two urves ross orthogonally, we must demonstrate that their tangent vetors are orthogonal at points of intersetion. This is not partiularly hard. Beause orthogonality is a frame independent notion, we may as well ompute this in t, x-spae. The lines of onstant λ parameterized by α are given by lα α sinh λ, α, osh λ, whih has the assoiated tangent vetor l lα sinhλ, oshλ. α.a.2 Similarly, lines of onstant α parameterized by λ are ϑλ α sinhλ, α oshλ, whih obviously has the assoiated tangent We see at one that ϑ ϑλ λ α oshλ, α sinhλ..a.3 g l, ϑ αosh λsinhλ + α sinhλoshλ 0..a.4 t t x x Figure. The orthogonal urvilinear oordinate harts whih ould be used by a uniformly aelerated observer in Minkowski spaetime. The red urves indiate surfaes of onstant α and the blue urves indiate surfaes of onstant λ. The diagram on the left shows the oordinate path expliitly onstruted in Problem, and the diagram on the right extends this onstrution to the whole of Minkowski spae minus lightone of an observer at the origin.

2 2 JACOB LEWIS BOURJAILY b We are to show that the map speified by equation.a. gives rise to an orthogonal oordinate system that overs half of Minkowski spae in two disjoint pathes. We should also represent this oordinate system diagrammatially. From our work in part a above, we know that the tangent vetors to the lines of onstant λ and α are given by e λ α oshλ e t + α sinhλ e x and e α sinhλ e t + oshλ e x..b. Therefore, the differential map where greek letters are used to indiate λ, α-oordinates is given by Λ µ α osh λ α sinhλ n..b.2 sinhλ oshλ We see immediately that the Jaobian, detλ α 0 whih implies that the λ, α oordinate system is good generially where it is defined. That it is orthogonal is manifest beause e λ e α 0 by part a above. Note that the harts of.a. are not well-defined on or within the past or future lightones of an observer at the origin: the urves of α onstant, the hyperbolas, are all time-like and outside the past and future lightones of an observer at the origin; and the lines of λ onstant are all spaelike and oinident at the origin. It does not take muh to see that these oordinates have no overlap within the past and future lightones of the Minkowski origin. The oordinate system spanned by λ, α is shown in Figure. We are to find the metri tensor and its assoiated Christoffel symbols of the oordinate harts desribed above. Using equation.b., we an diretly ompute the omponents of the metri tensor in λ, α oordinates g µν g e µ, e ν where λ is in the 0 -position α 2 0 g µν... 0 The Christoffel symbols an be omputed by hand rather quikly in this ase; but we will still show some rough steps. Reall that the omponents of the Christoffel symbol Γ µ νρ are given by µ Γ µ νρ e eν µ x ρ e µ. Again making use of equation.b., we see that e α α 0 Γα αα Γλ αα Slightly less trivial, we see e α λ oshλ e t sinhλ e x a e λ Γ λ αλ Γ λ λα α ;..3 and, e λ λ α sinhλ e t + α osh λ e x α e α Γ α λλ α, and Γ λ λλ óπǫρ ǫδǫι πoι ησαι

3 PHYSICS 523: GENERAL RELATIVITY HOMEWORK 4 3 Problem 2 We are to find the Lie derivative of a tensor whose omponents are T ab. Although we are tempted to simply state the result derived in lass and found in numerous textbooks, we will at least feign a derivation. Let us begin by realling that the omponents of the tensor T are given by T ab TEa,E b,e where the E s are basis vetor- and one-form fields. Now, by the Leibniz rule for the Lie derivative we know that X TE a,e b,e X T E a,e b,e X E a,e b,e E a, X E b,e E a,e b, X E. 2.a. Now, the first term on the right hand side of equation 2.a. gives the omponents of X T, whih is exatly what we are looking for. Rearranging equation 2.a. and onverting our notation to omponents, we see X T ab X T ab T αb X E a α T aβ X E b β T ab γ X E γ. 2.a.2 Now, we an either use some identities or just simply reall that X E a α Xa x α and X E γ Xγ x. We now have all the ingredients; putting everything together, we have X T ab Xδ T ab x δ T αb X a x α T aβ X b x β + T ab X γ γ x. 2.a.3 2.a.4 óπǫρ ǫδǫι πoι ησαι Problem 3 Theorem: Ating on any tensor T, the Lie derivative operator obeys U V T V U T [U,V ] T. 3.a. proof: We will proeed by indution. Let us suppose that the theorem holds for all tensors of rank less than or equal to r s for some r, s. We laim that this is suffiient to prove the hypothesis for any tensor of rank r s+ or r+ s. The indution argument is idential for the two ases our argument will depend on whih index is advaning so it is not neessary to expound both ases. Now, all rank r+ s tensors an be written as a sum of tensor produts between r s rank tensors T indexed by i and rank 0 tensors E, again indexed by i. That is, we an express an arbitrary r+ s tensor as a sum of i T i E i where i is just an index label! But this ompliation is entirely unneessary: by the linearity of the Lie derivative, it suffies to show the identity for any one tensor produt in the sum. Making repeated use of the linearity of the Lie derivative and the Leibniz rule, we see U V V U T E U V T E + T V E V U T E + T U E, U V T E + V T U E + U T V E + T U V E V U T E U T V E V T U E T V U E, [U,V ] T E + T [U,V ] E, [U,V ] T E ; where in the seond to last line we used the indution hypothesis appliable beause both T and E are of rank r s or less. The savvy reader knows that an arbitrary r s tensor an not be written as a tensor produt of r ontravariant and s ovariant piees; however every r s tensor an be written as a sum of suh tensor produts: indeed, this is exatly what is done when writing omponents of the tensor.

4 4 JACOB LEWIS BOURJAILY It now suffies to show that the identity holds for all 0 forms and all 0 tensors 2. We will atually begin one-step lower and note that equation 3.a. follows trivially from the Leibniz rule for 0-forms. Indeed, we see that for any 0-form f, U V f U V f + V U f, [U,V ] f + V U f, U V V U f [U,V ] f. 3.a.2 Now, to finish our proof, we laim that the identity holds for any 0 and 0 tensors, say X and Y, respetively. Reall that a one form Y is a funtion mapping vetor fields into salars i.e. YX is a 0-form. For our own onveniene, we will write YX X,Y. From our work immediately above, we know the identity holds for X,Y : U V V U X,Y [U,V ] X,Y. 3.a.3 Beause the Leibniz rule obeys ontration, we an expand out the equation above similar to as before. Indeed, almost opying the equations above verbatim we find U V V U X,Y U V X,Y + X, V Y V U X,Y + X, U Y, U V X,Y + V X, U Y + U X, V Y + X, U V Y V U X,Y U X, V Y V X, U Y X, V U Y, U V V U X,Y + X, U V V U Y, [U,V ] X,Y + X, [U,V ] Y. In the last line, we referred to equation 3.a.3 and expanded it using the Leibniz rule. Almost a footnote-omment: the last two lines do not preisely prove our required theorem as they stand: to identify the two piees of eah sum we need one small trik replae one of X or Y with the basis one-forms or vetor fields, and the result for the other beomes manifest. Therefore, beause equation 3.a. holds for all one-forms and vetor fields, our indution work proves that it must be true for all tensor fields of arbitrary rank. 2 You should probably suspet this is overkill: the indution step seemed to make no obvious use of the fat that r, s. And, as shown below, the identity is almost trivially true for the ase of salars. Nevertheless, it is better to be over-preise than inorret. In the famous words of Blaise Pasal to a mathematiian friend: I have made this letter longer beause I have not had the time to make it shorter.

5 PHYSICS 523: GENERAL RELATIVITY HOMEWORK 4 5 Problem 4 The torsion and urvature tensors are defined respetively, TX, Y X Y Y X X Y and RX, Y X Y Y X [X,Y ]. 4.a. We are to prove a TfX, gy fgtx, Y, b RfX, gy hz fghrx, Y Z, for arbitrary funtions f, g and h, and vetor fields X, Y and Z. Theorem a: TfX, gy fgtx, Y. proof: In both of the required proofs, we will make repeated uses of the defining properties of the ovariant derivative and of the Lie derivative. In partiular, we will need the following properties of the onnetion: X Y is a tensor in the argument X. This means that as an operator, fx+gy f X + g Y. 2 X Y obeys the Leibniz rule in Y. Speifially, this means X fy XfY + f X Y. This implies that X Y is linear in Y whih follows when f is a onstant. We are almost ready to prove the identity by brute fore in a ouple of lines. Let s just prepare one more trik up our sleeve: we will need [fx, gy ] fx gy g fx Y + fx g Y, Let us begin: TfX, gy fx gy gy fx fx gy, f X gy g Y fx fx gy, g Y fx + fxgy, gfy X gxy f + fxgy. fxgy + fg X Y gy fx fg Y X + gfy X + gxy f fxgy, fg X Y fg Y X gf X Y, TfX, gy fgtx, Y. 4.a.2 Theorem b: RfX, gy hz fghrx, Y Z. proof: We have already olleted all of the properties and identities neessary to straightforwardly prove the theorem. Therefore, we may proeed diretly. RfX, gy hz { fx gy gy fx [fx,gy ] hz, {f X g Y g Y f X fg [X,Y ] f Y Xg + g XY f hz, {fxg Y + fg X Y gy f X fg Y X fg [X,Y ] fxg Y + gy f X hz, {fg X Y fg Y X fg [X,Y ] hz, fgrx, Y hz, fg { X Y hz + h Y Z Y XhZ + h X Z fg { X Y hz + Xh Y Z + h X Y Z Y XhZ [X, Y ] hz h [X,Y ] Z, Y h X Z h Y X Z X Y hz + Y X hz h [X,Y ] Z { fg hrx, Y Z + X Y hz Y XhZ X Y h Z + Y XhZ, RfX, gy hz fghrx, Y Z., 4.b.