Errata and amendments: A Most Incomprehensible Thing: Notes Towards a Very Gentle Introduction to the Mathematics of Relativity by Peter Collier

First Edition published July 2012. Second Edition published July 2014. The version number, which can be found on the copyright page, is of the form v*** for ebooks (Kindle and EPUB books) and v***CS or v***IS for the paperback editions. For example, a version number v61CS paperback would contain the errata listed in Section (3) plus all the subsequent Sections. The current published versions are: v112 for the second edition ebooks (Kindle and EPUB) and v74IS for the second edition paperback, so errata for these books are to be found in Section (7) The author apologises for these errors and amendments and for any inconvenience caused. Please contact the author to update this list: incomprehensiblething @ gmail.com.

Contents 1 First Edition: versions 95 (ebook), 55 (paperback) and earlier

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2 First Edition: versions 98 (ebook), 58 (paperback) and earlier

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3 First Edition: versions 101 (ebook), 61 (paperback) and earlier

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4 First Edition: versions 105 (ebook), 66 (paperback) and earlier

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5 First Edition: versions 107 (ebook), 68 (paperback) and earlier

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6 First Edition: versions 108 (ebook), 70 (paperback) and earlier

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7 Second Edition: versions 112 (ebook), 74 (paperback) and earlier

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1 First Edition: versions 95 (ebook), 55 (paperback) and earlier 1. Section 1.14.1 Problem 1.23. Equation dy = sin θ sin φdr + r cos θ cos φdθ + r sin θ sin φdφ should read: dy = sin θ sin φdr + r cos θ sin φdθ + r sin θ cos φdφ. Thanks to Naeem Rashid.

2. Section 5.3.6 Change “where β = 0, 1, 2, 3 = t, x, y, z” to “β = 0, 1, 2, 3 = t, x, y, z, for example”.

1

2 First Edition: versions 98 (ebook), 58 (paperback) and earlier 1. Section 1.7.1 sin 25◦ equals 0.4226 not 0.4426, therefore sin 25◦ = 0.4426 = c=

8 = 18.075. 0.4426

should read: sin 25◦ = 0.4226 = c=

8 c

8 c

8 = 18.930. 0.4226

Thanks to Derek Steel.

sin 140◦ equals 0.643 not 0.623. Thanks to Derek Steel.

2. Section 3.4.3 Equation 3.4.15 vx ct = cγ t − c 0

should read:





vx ct = cγ t − 2 . c 0





Also, the equation after Equation 3.4.15 should read: vx 0 = cγ t − 2 . c 



Thanks to James Bowie.

3. Section 5.1 This section has been significantly rewritten. Rather than give piecemeal amendments, delete from the start of the paragraph starting “Unfortunately, the vectors used in general relativity” to the end of the section. Insert: “Unfortunately, the vectors used in general relativity are not directed line segments stretching from one point to another in space. Instead, each vector is located at a single point in spacetime. In fact, each point in spacetime is itself a vector space and home to an infinite number of vectors. The vector space will be both a tangent space (home to those objects known as contravariant vectors), and a cotangent space (home to those objects known as one-forms). Contravariant vectors and one-forms should be thought of as different representations of the same geometrical object at a point in spacetime. Details to follow, but for a contravariant vector, think of a tangent vector to a parameterised curve; for a one-form, think of a gradient of a scalar field. We’ll see later how the metric tensor is used to convert a vector to its corresponding one-form and vice versa. The reason that the simple vectors in Section 1.13 and the more abstract vectors we are now talking about are both loosely referred to as ‘vectors’ is that they both obey the rules that define a vector space. In brief, a vector space consists of a group of objects (call the group X, for example) that can be added together and multiplied by a scalar, and the result will be another member of the group X. Up until now, our indices have referred to particular coordinate systems: x, y, z for Cartesian, r, θ, φ for spherical, etc. Differential geometry demands a more abstract use of indices, where they can refer to any permissible coordinate system. Similarly, in general relativity, because we are dealing with curved spacetime, there are no preferred coordinate systems and we need to be able to transform from any one coordinate system to any other (the terminology is that we are using general coordinates). So if xµ are the old coordinates and y µ are the new coordinates (µ = 0, 1, 2, 3), then any functions linking xµ and y µ are 2

permissible as long as (a) the functions are differentiable, and (b) each point in spacetime is uniquely labelled by a set of four numbers. We move freely between different coordinate systems by using the transformation properties (involving partial derivatives of the coordinate functions) of contravariant vectors and one-forms. Contravariant and one-form basis vectors are also defined in terms of derivatives of the coordinate functions. We don’t need to go into details, but can just note that contravariant basis vectors are tangent to the coordinate curves (along which only one of the coordinates changes), and one-form basis vectors are gradients of the coordinate surfaces (on which only one of the coordinates remains constant). Bases of this sort (unlike the constant non-coordinate bases of Cartesian coordinates) that change with the coordinates, are known as coordinate bases. These are the bases we’ll be implicitly using from now on. However (thankfully!), the transformation properties of the components of vectors and one-forms (and tensors in general) are basis-independent, meaning we usually don’t need to worry too much about basis vectors and basis one-forms. Crucially, basis-independence means that if a tensor equation is true in one coordinate system it will be true in all coordinate systems. Because we tend to refer only to the components of vectors, one-forms, etc, we’ll continue our habit of loosely referring to a ‘vector V α ’ or a ‘one-form Vα ’ when, strictly speaking, this notation refers to the components of those objects. Though not totally accurate, this terminology is good enough for our purposes. Although, in the context of general relativity, we can’t meaningfully talk about directed line segments stretching from one point to another in space, we can define an infinitesimal displacement vector in spacetime: d~x = dxµ eµ . The power of the mathematics that follows is that it allows us to manipulate d~x and end up with physical measurable quantities (time, distance, velocity, momentum, etc). Any contravariant vector or one-form is the product of its components and a basis of some kind. Contrav~ ) so we can say, using the Einstein ariant four-vectors are often represented by an arrow over the letter (egV summation convention, that ~ = V α eα , V ~ where V α and eα are respectively the components and basis vectors of V. One-forms are often represented by a tilde over the letter (eg V˜ ) so we can say, again using the Einstein summation convention, that V˜ = Vα eα , where Vα and eα are respectively the components and basis one-forms of V˜ . A contravariant vector acts linearly on a one-form (and vice versa) to give a scalar (a real number). This works because the relationship between the basis vectors and basis one-forms is defined (using the Kronecker delta) by the equations eα (eβ ) = δβα . Therefore, for any one-form and vector  

~ P˜ V

= Pα V β eα eβ = Pα V β δβα = Pα V α

= P0 V 0 + P1 V 1 + P2 V 2 + P3 V 3 , which is a scalar. Before looking at the transformation properties of contravariant vectors and one-forms we’ll see what happens to scalar fields when we change from one coordinate system to another.” 4. Section 5.3.3 Delete the final paragraph. 5. Section 5.4 First bullet point should read: “Because the transformation properties of tensor components are basisindependent, if a tensor equation is true in one coordinate system it will be true in all coordinate systems.” 3

6. Section 5.4 After “leading us to the famous mass-energy equation E = mc2 ” insert: “It’s worth making a small detour here in order to see just why the metric tensor is involved when taking the scalar product of two vectors. We first need to note that the components gij of the metric are defined in terms of the coordinate basis vectors as gij = ei · ej or, in spacetime, using Greek indices gµν = eµ · eν . Using this definition, and recalling that an infinitesimal displacement vector in spacetime is given by d~x = dxµ eµ , we can see that the scalar product of this vector with itself will be (d~x)2 = (dxµ eµ ) · (dxν eν ) = (eµ · eν ) dxµ dxν = gµν dxµ dxν . This equation is the spacetime version of (4.2.1) dl2 = gij dxi dxj , the line element for an n-dimensional manifold that we met in the previous chapter. In a similar fashion, the scalar product of two different vectors must then be given by ~ ·W ~ = (V µ eµ ) · (W ν eν ) = gµν V µ W ν . V Hence the need for the metric tensor.” 7. Section 5.4.1 After second equation, delete “We’ll see shortly, when looking at tensor algebra, that”. 8. Section 5.4.2 After equation Tγ = Aαγ B α , insert: “A single tensor can also be contracted by summing over identical upper and lower indices. For example, we can contract a tensor T αα to give another tensor T αα = X  , which is a contravariant vector. Let’s see how this contraction works. The tensor X  = T αα will transform as follows: ∂x0α ∂x0 ∂xη βλ T η ∂xβ ∂xλ ∂x0α ∂xη ∂x0 βλ ∂x0 = T η = δβη λ T βλη . β λ ∂x ∂x ∂x Remember that the x coordinates (represented by the β, λ, η indices) are independent of each other. The Kronecker delta (here used as a tensor with an upper and lower index) in the final term tells us that the derivative of any of these coordinates with respect to any other is zero, and the derivative of a coordinate with respect to itself is 1 (if x = x then dx dx = 1). We can therefore write X 0 = T 0αα =

∂x0 βλ ∂x0 βλ ∂x0 λ T = T = X , η β ∂xλ ∂xλ ∂xλ meaning the contracted tensor transforms, as it should, as a contravariant vector.” X 0 = T 0αα = δβη

9. Section 5.4.2 Problem 5.3. Delete “representing two arbitrary coordinate systems” and replace with “representing an arbitrary coordinate system”. Also, after “true in any coordinate system” insert: “A similar argument shows that any tensor equation if true in one coordinate system will be true in all coordinate systems.” 4

3 First Edition: versions 101 (ebook), 61 (paperback) and earlier 1. Section 1.5 Last but one bullet point: “γ will only be a tiny bit less than 1.” should read: “γ will only be a tiny bit more than 1.” Thanks to Don Berry.

Also, final bullet point: “γ cannot equal zero because v < c, and therefore v/c can never equal 1.” should read: “γ cannot equal zero because the numerator (the top bit of the fraction) is a non-zero constant, ie equals 1.” Thanks to Jeremy Roach.

2. Section 1.5.1 Figure 1.1. The origin should be at the lowest point of the surface. Thanks to Don Berry.

3. Section 1.6 Problem 1.3. The solution should read: q

(3x + 8)2 √ x2 3x + 8 . ±7 = x Taking the left-hand side as +7, multiply both sides by x √

49 =

7x = 3x + 8 4x = 8 8 x = = 2. 4 If we take

√

49 = −7, we obtain another solution, ie x = −4/5.

Thanks to Bob Bazely.

4. Section 1.10.1.4 Figure 1.15 The origin should be at the lowest point of the surface. Thanks to Don Berry.

5. Section 3.2.7 Equation 3.2.2 v =w+u should read: u = w + v. Thanks to Jeremy Roach.

6. Section 3.4.9 Replace two occurrences of z1 0 = z1 with ∆z1 0 = ∆z1 . Thanks to James Bowie.

5

7. Section 3.4.10 Problem 3.10. “The frame”

(∆x2 +∆y2 +∆z 2 ) ∆t2

term is actually your velocity measured by an observer in the Earth

should read: (∆x2 +∆y2 +∆z 2 ) “The term is actually the square of your velocity as measured by an observer in the Earth ∆t2 frame”. Also, equation v = 1 − 1.25 × 1013 should read:





v = 1 − 1.25 × 10−13 . Thanks to James Bowie.

8. Section 3.6.5 A missing minus sign means that equation ˆ

c2 m 2 should read:

c2 m − 2

ˆ

1 √ ds s 1 √ ds, s

meaning √ c2 m s + c should read:

√ −c2 m s + c,

and q

c2 m 1 − (v/c)2 + c should read:

q

−c2 m 1 − (v/c)2 + c. Thanks to James Bowie.

9. Section 5.4.1 Equation 5.4.1 m T 0µ1 µα21...µ α2 ...αn =

∂x0µm ∂xβ1 ∂xβ2 ∂xβm ∂x0µ1 ∂x0µ2 m . . . × . . . × T ν1 ν2β...ν ν ν ν 0α 0α 1 β2 ...βn ∂x 1 ∂x 2 ∂x m ∂x 1 ∂x 2 ∂x0αm

should read (a slight mix up with the m0 s and n0 s): m T 0µ1 µα21...µ α2 ...αn =

∂x0µ1 ∂x0µ2 ∂x0µm ∂xβ1 ∂xβ2 ∂xβn m . . . × . . . × T ν1 ν2β...ν . 1 β2 ...βn ∂xν1 ∂xν2 ∂xνm ∂x0α1 ∂x0α2 ∂x0αn

Thanks to James Bowie.

10. Section 5.4.2 Problem 5.2. Equation 

Aa = 1, r2 0 should read:







Aa = 1, r3 , 0 . Thanks to James Bowie.

6

11. Section 5.4.2 Problem 5.4. Equation ∂Vα Vβ = g αβ ∂xλ ∂xλ should read (there’s a missing ∂): ∂Vα ∂V β = g . αβ ∂xλ ∂xλ Thanks to James Bowie.

12. Section 6.3 The section starting “Finding the covariant derivative of a contravariant vector or tensor field is equivalent to” should read: “Using an identical upper and lower index (ie a dummy index), the covariant derivative of a contravariant vector or tensor field is equivalent to the field’s divergence. Then, in flat space using Cartesian coordinate, the covariant derivative (6.3.6) ∂V α ∇α V α = + V γ Γαγα ∂xα reduces to ∂V α ∇α V α = , ∂xα which is the same ... ” Thanks to Steve Burk.

13. Section 6.5 Problem 6.6. Equation y= should read:

c x− a

c y = x+ a





cb + ad a



ad − bc . a 

Thanks to James Bowie.

14. Section 7.5.3 Change “We mentioned earlier that the covariant derivative of a tensor field is equivalent to the divergence of that field.” to: “We mentioned earlier that, using a dummy index, the covariant derivative of a tensor field is equivalent to the divergence of that field.” Also, the section starting “In the flat spacetime of special relativity, the fundamental laws of conservation of energy” should read: “In the flat spacetime of special relativity, the fundamental laws of conservation of energy and momentum can indeed be expressed by saying ∂T µν =0 ∂xµ or, using comma notation, µν T,µ = 0. These equations describe conservation of energy when ν = 0, and conservation of the ith component of momentum when ∂T µi = 0. ∂xµ In other words ... ” Also, equation T;νµν = 0 should read: µν T;µ = 0.

7

4 First Edition: versions 105 (ebook), 66 (paperback) and earlier 1. Section 1.13.2 In the first paragraph delete “(the Greek letter nabla)” and insert at the end of the paragraph: “(Nabla is from the Greek word for a Phoenician harp, which nineteenth century mathematicians thought the inverted Delta symbol ∇ resembled.)” Thanks to Stuart Collins.

2. Section 1.15 In the rules of index notation, the first sentence of Rule 4 should read: “Dummy indices appear twice in a term, either as subscripts in Cartesian vectors/tensors, or once as superscript and once as subscript in the more general vectors/tensors we tend to use in this book.” Thanks to Michael Guggenheimer.

3. Section 3.4.2 After “Strictly speaking, the brackets in (3.4.13) mean we are referring to matrices” insert: “(From now on we won’t bother to use this clunky bracket notation for single index vectors.)” 4. Section 3.6.5 To avoid confusion with the speed of light, the constant of integration c should be C. Thanks to Frits van der Laan.

5. Section 3.6.6 At the end of the first paragraph (after “which is the particle’s mass energy Eo ”) insert: “It can be shown theoretically and has been verified experimentally that, providing no external forces act, total relativistic energy is conserved in all inertial frames, irrespective of whether mass or kinetic energy are conserved. In high-speed particle collisions, for example, mass, kinetic energy, even the total number of particles may not be conserved, but the total relativistic energy of the system will be.” Suggested by Frits van der Laan.

6. Section 5.4 Delete everything between (but not including) “A type (0, 0) tensor is therefore just a number, a scalar, 106 for example” and “Tensors of rank 3 can be thought of as ... ” and insert: “A type (1, 0) tensor (aka a contravariant vector) is conventionally written in column form, eg    

Vα =

a b c d

   . 

A type (0, 1) tensor (aka a one-form) is conventionally written in row form, eg Xβ = (e, f, g, h) . These two tensors can be multiplied together (known as taking the tensor product) to form a rank 2 tensor, which can be written as a 4 × 4 matrix:  h

i

  

T αβ = [V α Xβ ] = 

ae af be bf ce cf de df

ag bg cg dg

ah bh ch dh

   . 

So, for example, T 30 = V 3 X0 = de. In a similar fashion, two contravariant vectors or two one-forms may also be multiplied together to give a rank 2 tensor.” 7. Section 5.4.2 After “where we have again used the Kronecker delta as a tensor with an upper and lower index” insert: 8

“Double contraction to zero In order to understand this somewhat counter-intuitive result we need to introduce the tensor properties of symmetry/antisymmetry. A rank 2 tensor is symmetric if Sµν = Sνµ . Such a tensor can be represented by a symmetric matrix     

w a b c

a x d e

b c d e y f f z

   . 

So, for example, S02 = S20 = b, S23 = S32 = f , etc. Symmetric tensors we meet in this book include the metric tensor, the energy-momentum tensor, the Ricci tensor and the Einstein tensor. Conversely, a rank 2 tensor is antisymmetric if the tensor is of the matrix form  0  −a    −b −c

Aµν = −Aνµ . If you think about it, this is only possible if a b 0 d −d 0 −e −f

c e f 0

   , 

where the diagonal terms are all zero. The Riemann curvature tensor, written as Rhijk , is antisymmetric on the first and second pair of indices, meaning Rhijk = −Rhikj = −Rihjk . The double contraction of a symmetric tensor S µν and an antisymmetric tensor Aµν is zero, ie Aµν S µν = 0. We can show this as follows: Aµν S µν = (−Aνµ ) (S νµ ) = − (Aνµ S νµ ) . As µ and ν are dummy indices, we can swap them to give Aµν S µν = − (Aµν S µν ) , which can only be true if Aµν S µν = − (Aµν S µν ) = 0. This may be clearer if we write out the double contraction in component form. We’ll use two simple tensors, a symmetrical one   x a b h i   S ij =  a y c  , b c z and an antisymmetrical one 

0 d  [Aij ] =  −d 0 −e −f



e  f , 0

giving Aij S ij = (0 × x) + da + eb − da + (0 × y) + f c − eb − f c + (0 × z) = 0.” 8. Section 5.4.2 Problem 5.2. Equation g

φφ



Bφ =

should read: g

φφ



Bφ =

  1 2 2 sin θ cos θ r2

  1 2 cos θ . 2 r2 sin θ

Thanks to Frits van der Laan.

9. Section 5.5

9

This section has been rewritten. It should now read: “We’ve now looked at scalars, vectors, one-forms and tensors, and we know that these objects live on the manifold. This is all quite abstract. Can we picture these things in the real world? Visualising four-dimensional spacetime is beyond the imaginations of most of us, so we’ll simplify the situation and model a manifold using a rough and ready mental picture of a real two-dimensional surface. I’m going to visualise a pebble, but any smooth surface will do – an apple, a bottle, the bodywork of a car etc. The surface has to be smooth because it needs to be differentiable, so use an imaginary file to remove any imaginary sharp edges from your imaginary surface. Picture the surface in your mind (or put a real pebble on the table in front of you) and using our imagination we’ll construct our simple model of a manifold. The surface of the pebble is our manifold, or curved space. It’s two-dimensional because if we choose the right coordinate system we only need two coordinates to describe any point on the surface of the pebble. Next, imagine we stretch a red elastic net over the pebble to represent an arbitrary x, y coordinate system, followed by a blue elastic net to represent an arbitrary x0 , y 0 coordinate system. Now we can describe any point on the surface of the pebble using both x, y and x0 , y 0 coordinates. Let’s consider a point P on the surface of the pebble. First, because our manifold is a Riemannian manifold, we can assign a symmetric metric tensor gµν = gνµ to P and to every other point on the surface of the pebble. In the jargon, we say we have defined a metric tensor field over the surface of our pebble (a tensor field assigns a tensor to each point of a manifold, just as a vector field assigns a vector to each point). The metric defines the separation of infinitesimally adjacent points on the manifold. Because our manifold is two-dimensional, the metric will be a 2 × 2 matrix. We don’t know what the metric actually is, but we do know its coefficients will not be constant (otherwise our pebble would be flat!), and that the metric for the red coordinate system will be different to the one for the blue. Next, we can use our coordinates to construct a scalar field on the surface of the pebble. I’m going to make one up, call it S, and say S = x2 + 3y. Now we can assign a scalar value S to the point P. For example if x = 7, y = 2, then S = 55. By taking the gradient of S we can define a one-form (5.3.4) at P 

∇S =

∂S ∂S , ∂x ∂y



.

But we can define an infinite number of scalar fields, and therefore an infinite number of one-forms at P. We therefore have a vector space at P, which is called the cotangent or dual-space. Now draw a number of imaginary curves on the surface of the pebble, each going through P. These are our parameterised curves. If we were clever enough, each curve could be expressed in terms of a parameter (t for example), so x = f (t) and y = f (t) . Now take a piece of paper, draw a point on it and carefully draw a number of infinitely short (just do your best) straight lines through that point. Rest the paper on the pebble so that the point on the paper is touching point P on the pebble. We can see that the straight lines drawn on the paper are tangent to the curves drawn on the pebble at point P. The direction of each tangent is given by (5.3.3) and defines a contravariant vector. There are an infinite number of parameterised curves we could draw on the pebble, and therefore an infinite number of tangent vectors passing through P. We therefore have a vector space at P, which is called the tangent space. At any point on the surface of the pebble we can now, if we wish, take a contravariant vector or one-form and use the transformation equations to flip between the red and blue coordinate systems. But, as stressed earlier, in practice we don’t need to worry about physically calculating transformation equations. The prime advantage of defining the geometry of the pebble’s surface in terms of general coordinates is that we can then proceed to define tensors that are true in any coordinate system, tensors that describe both the pebble’s curvature (the Riemann curvature tensor) and physics (the field equations, for example). Then, if we can somehow ascertain the metric, we can use these tensors to calculate measurable physical quantities.” 10. Section 6.2 10

Equation ∂V = xˆ ex + 2ˆ ey ∂y should read:

∂V = xˆ ex + 2ˆ ey + 2ˆ ez . ∂y

Also, in the sentence immediately before Equation 6.2.2, change “in terms of a new set of basis vectors eγ ” to: “in terms of basis vectors eγ ”. Thanks to Fred Mandelkorn.

11. Section 6.5 Problem 6.7. Equation d2 θ d2 φ dφ = = =0 dλ dλ dλ should read:

d2 θ d2 φ dφ = = = 0. 2 2 dλ dλ dλ

Thanks to James Bowie.

12. Section 7.5 Change the three sentences in the paragraph after Equation 7.5.2 from “These world-lines cross two surfaces: AB of constant x and BC of constant time ct. The world-lines crossing AB represent the flux across constant x, (ie, nv from (7.5.2)). The same world-lines also cross BC, a surface of constant time.” to: “These world-lines cross two surfaces: BC of constant x and AB of constant time ct. The world-lines crossing BC represent the flux across constant x, (ie, nv from (7.5.2)). The same world-lines also cross AB, a surface of constant time.” Thanks to James Bowie.

13. Section 10.3 Problem 10.1 (b). Equation τ=

2 × 5.12 × 106 = 0.121 ms 8.47 × 1010

τ=

2 × 5.20 × 106 = 0.123 ms. 8.47 × 1010

should read:

Problem 10.1 (c). Equation 0.121 − 0.103 = 0.018 should read: 0.123 − 0.103 = 0.020. Thanks to James Bowie.

14. Section 10.5 Equation 10.5.4 RS ds = 1 − r 2





RS RS c dt − 2 cdt0 dr − r2 1 + r r 2



02





dr2 − r2 dθ2 + sin2 θdφ2



should read (there’s a superfluous r2 term): RS ds = 1 − r 2





RS RS c dt − 2 cdt0 dr − 1 + r r 2

02



Thanks to James Bowie.

11







dr2 − r2 dθ2 + sin2 θdφ2 .

15. Section 11.3.2 Equation just after Equation 11.3.9

should read:

σ = sin χ if k = +1 R σ = sin−1 χ if k = +1. R

Thanks to James Bowie.

16. Section 11.4.5 The final equation R0 = √ should read: R0 = √

c c = 1/2 4πGρ Λ

c 1 = 1/2 . 4πGρ Λ

Thanks to James Bowie.

5 First Edition: versions 107 (ebook), 68 (paperback) and earlier 1. Section 6.2 Problem 6.1. Delete “We can find the transformation rule for basis vectors using (5.3.2)” and insert: “For the rules of tensor algebra to work, the basis vectors eβ must transform in the same way as the one-form Vβ . So we can find the transformation rule for basis vectors using (5.3.2)” Thanks to Giancarlo Bernacchi.

2. Section 6.3 Problem 6.5 should now read: “Show that the covariant derivative of the Euclidean metric tensor gij for any coordinate system is zero. Substituting the Euclidean metric tensor for Cartesian coordinates (1.14.6) 



1 0 0   [gij ] =  0 1 0  0 0 1 into (6.3.9) ∇β Tµν =

∂Tµν − Tαν Γαµβ − Tµα Γανβ ∂xβ

gives ∂gij − gαj Γαiβ − giα Γαjβ . ∂xβ In Cartesian coordinates the right-hand side equals zero (because the Euclidean metric is constant, and the connection coefficients are zero), therefore ∇β gij = 0. ∇β gij =

But (and this is the clever thing about tensors) if this equation is true in Cartesian coordinates, it must be true for the Euclidean metric in all coordinate systems. Actually, it is straightforward to show, by substituting (6.2.6) 1 ∂gρν ∂gµρ ∂gµν Γσµν = g σρ + − µ 2 ∂x ∂xν ∂xρ 



into (6.3.9) ∇β Tµν =

∂Tµν − Tαν Γαµβ − Tµα Γανβ , ∂xβ 12

(and, of course, substituting gµν for Tµν ) that the covariant derivative of the general metric tensor is zero for any coordinate system, ie ∇β gµν = 0. Intuitively, we can also show this by knowing that the covariant derivative of the metric tensor must vanish in the flat spacetime of special relativity (where the metric is constant). Being a tensor equation, ∇β gµν = 0 must also then be true in any coordinate system.” Thanks to Giancarlo Bernacchi.

3. Section 6.4 The equation after “So, using the chain rule, we can write” should read: β DV α dxβ dV α γ α dx + V Γ = = Uβ γβ dλ dλ dxβ dλ



dV α + V γ Γαγβ . dxβ 

Thanks to Giancarlo Bernacchi.

4. Section 6.6 Problem 6.9. The final sentence of the second paragraph should read: “We’ll take one of these, Rθφθφ for example, and calculate it.” Thanks to Giancarlo Bernacchi.

5. Section 7.3 The third paragraph should start “In free fall, of course, the spacetime ... ”. 6. Section 7.5.2 At the end of this section delete “Because of the equivalence principle, any physical law that can be expressed as a tensor equation in special relativity has exactly the same form in a local inertial frame in the curved spacetime of general relativity. We can therefore rewrite” and insert: “The principle of general covariance allows us to replace the Minkowski metric and rewrite”. 7. Section 8.2.2 Delete “The ‘comma goes to semi-colon’ rule (where we use covariant derivatives instead of partial derivatives)” and insert: “The principle of general covariance”.

6 First Edition: versions 108 (ebook), 70 (paperback) and earlier 1. Section 1.7.3 Change the opening sentence from “An identity is a relationship between two equations that is always true.” to “An identity is a relationship between two expressions that is always true.”. Thanks to Walter Vanhimbeeck

2. Section 1.8.2 Property 4. Equation logx (ay ) = y log a should read (there’s a missing subscript): logx (ay ) = y logx a. Thanks to Don Golfen

3. Section 1.9.1 13

Delete the sentence “ie P’s perpendicular distance from the x, y and z axes.”. Thanks to Walter Vanhimbeeck

4. Section 1.10.2.1 ´ ´ Problem 1.11 (c) 3tx dx should read (missing brackets): (3t)x dx. And the solution (yet more missing brackets) ˆ 3tx 3tx dx = +C ln 3t ˆ

should read:

(3t)x + C. ln (3t)

(3t) x dx = Thanks to Thomas Laming

5. Section 1.13 In the final paragraph of this section delete “eg F = Fx + Fy + Fz ” and insert: “eg (Fx , Fy , Fz ) for F”. Thanks to Thomas Laming

6. Section 1.13.3 Equation 1.13.9 (paperback only, not ebooks) ∇2 φ = ∇ · ∇φ =

∂2φ ∂2φ ∂2φ + 2 + 2 ∂x2 ∂y ∂z

2

should read (there’s a superfluous superscript 2): ∇2 φ = ∇ · ∇φ =

∂2φ ∂2φ ∂2φ + 2 + 2. ∂x2 ∂y ∂z

Thanks to Yong Tang

7. Section 1.15. In the Kronecker delta section delete “because, from the definition of the Kronecker delta, when ν = ν (which it always does of course) then δνν = 1 and there are four possible values of ν.” Thanks to Thomas Laming

8. Section 2.5.7 In the sentence immediately before Equation 2.5.18, change “we do work on the object equal to” to: “gravity does work on the object equal to”. Thanks to Thomas Laming

9. Section 2.5.7.3 Poisson’s equation in Cartesian coordinates is given as 2

∇ φ= but should read:

∂2φ ∂2φ ∂2φ , , ∂x2 ∂y 2 ∂z 2

!

= 4πGρ,

∂2φ ∂2φ ∂2φ + 2 + 2 ∂x2 ∂y ∂z

2

∇ φ=

!

= 4πGρ.

Similarly, in Problem 2.5, 

2

∇ φ = GM

∂ ∂ ∂ , , ∂x ∂y ∂z



x, y, z

!

(x2 + y 2 + z 2 )3/2

should read: 2

∇ φ = GM



∂ ∂ ∂ + + ∂x ∂y ∂z

Thanks to Richard Oppenheim

14



x, y, z (x2 + y 2 + z 2 )3/2

!

.

10. Section 3.3.1 In the fifth and seventh paragraphs, the time t (3 × 10−8 s) should be time t ((1/3) × 10−8 s). Thanks to Walter Vanhimbeeck

11. Section 3.3.6 In the second bullet point under Time Dilation, the sentence “However, observer O0 measures the same event occurring at time ct = 1 on his ct0 axis.” should read (there’s a missing prime): “However, observer O0 measures the same event occurring at time ct0 = 1 on his ct0 axis.” Thanks to Mark Mitchell

12. Section 3.3.6 In the paragraph starting “It’s essential to realise that both time dilation ... ”, delete the sentence starting “An astronaut in a rocket ... ” and insert: “An astronaut in a rocket flying past the Earth near the speed of light would measure the planet as being squashed along her direction of travel (see Figure 3.18). However, measurement is not the same as visual appearance. What an observer of a rapidly moving object actually sees or photographs depends not only on the mathematics of length contraction, but also on the time lag of photons from different part of the object arriving at her eyes or camera, a phenomenon known as the Terrell or Penrose–Terrell effect.” Thanks to Kees Kuijlaars

13. Section 3.4.1 Just before the end of this section, equation x0 = (a1 )2 x0 − v 2 a1 x0 /c2 should read: x0 = (a1 )2 x0 − v 2 (a1 )2 x0 /c2 . Thanks to Walter Vanhimbeeck

14. Section 5.3.4.1 In the final paragraph, “where r, θ, φ are all functions of x, y, z” should read: “where x, y, z are all functions of r, θ, φ”. Thanks to Walter Vanhimbeeck

15. Section 5.4.2 In the Contraction section, change “we can contract a tensor T αα ” to “we can contract a tensor T βα ”. Thanks to Walter Vanhimbeeck

16. Section 6.7.1 After Equation 6.7.2 insert: “The Ricci tensor can also be formed from other contractions of the Riemann curvature tensor (many authors use one and three). However, because of the symmetries of the Riemann tensor, they all give either zero or ±Rµν .” Also, in Problem 6.10, change “and obtain” to “(for example) and obtain”. Thanks to Steve Burk

17. Section 7.5.1 In the final line of this section, “U 0 = 1” should read: “U 0 = c”. Thanks to Walter Vanhimbeeck

18. Section 9.3.2 15

In the final paragraph, “g 11 = 1/ − e2ν = −e−2ν ” should read: “g 11 = 1/ − e2λ = −e−2λ ”. Thanks to Walter Vanhimbeeck

19. Section 9.4.6 Problem 9.5. Equation 9.4.29 should read: d dτ



dr dτ

2

=

d2 r dr d2 r dr d2 r dr + = 2 . dτ 2 dτ dτ 2 dτ dτ 2 dτ

Also, delete equation dx 2GM =− 2 , dr r and delete “which we can rearrange and divide by dτ to give” and insert: “Using the chain rule, we can then say”. And change Equation 9.4.30 to dx dx dr 2GM dr = × =− 2 . dτ dr dτ r dτ Thanks to Bill Parsons

20. Section 11.2.2 In the final paragraph delete “An isotropic universe is necessarily homogeneous, but the reverse isn’t true” and insert: “A universe that is isotropic for all observers is necessarily homogeneous, but the reverse isn’t true”. Thanks to Bill Parsons

7 Second Edition: versions 112 (ebook), 74 (paperback) and earlier 1. Section 1.4 Change the sentence immediately before the equation q =3+t

√ u −5 s v

from “Here’s a function containing five variables” to: “Here’s a function containing five variables (where we are only considering non-negative values of Thanks to John Hudson

2. Section 7.4.2 The geodesic deviation equation β γ D2 ξ µ µ α dx dx + R ξ =0 αβγ Dλ2 dλ dλ

should read

β γ D2 ξ µ µ α dx dx + R ξ = 0. βαγ Dλ2 dλ dλ

Thanks to Frank Cook

3. Section 9.3.4 The final equation 

R33 = sin2 θ e2λ 1 + r ν 0 − λ0





−1



should read (there’s a missing minus sign before 2λ) 

R33 = sin2 θ e−2λ 1 + r ν 0 − λ0 Thanks to Dave Proffitt

16







−1 .

√

s)”.