Wednesday, 28 November 2018

Commentary 2.6 Expanding Universe

In section 2.6 we met a special case of the Robertson-Walker metric for an expanding Universe. Exciting! The section was a 'playground' for our previous work. From the metric we got to
0 < q < 1. Even I could integrate that equation, but Carroll produced
by changing the variables around. As you really want how x changes as t progresses, the obvious equation
is much more informative and useful. x0 is a constant of integration, in fact where our test photon starts.

I could then get some graphs and understand what he was talking about. t must be positive. (One can't take a fractional power of a negative number.)
Two light cones
x0  is the point on the X axis where the light cone starts. The red x(xo) lines are the solutions for x0 = -7 the one on the left (paler) being the negative part. The x(x1) lines are for x0 = 7. The light cones always and only intersect at some t > 0 midway between their start points. There are no solutions for t < 0.

If we set q = 0 then we get our flat Minkowski space and we can have t < 0. The diagram becomes
Light cones in Minkowski spacetime 

The light cones intersect in the past (bottom open triangle) as well as in the future (top open triangle).

In the spirit of the playground, it is amusing to show the first diagram again with (impossible) t < 0. Excel does a great job!
Big Bang

Read in more detail at Commentary 2.6 Expanding Universe.pdf. Includes link to spreadsheet.

Monday, 26 November 2018

Commentary 2.5 Metric, Rocket and 4-velocity (four-velocity)

The observer, abandoned by the rocket.
Before equation (2.51) near the end of section 2.5, Carroll writes: "Take a very simple example, featuring an observer with four-velocity U and a rocket flying past with 4-velocity V. What does the observer measure as the ordinary three-velocity, v, of the rocket? In special relativity the answer is straightforward. Work in inertial coordinates (globally not locally) such that the observer is in the rest frame and the rocket is moving along the x-axis. Then the four-velocity of the observer is
Uμ  =  (1 , 0 , 0 , 0) 
and the four-velocity of the rocket is
Vμ =(γ , γv , 0 , 0)
where v is the three-velocity and
γ = 1 / (√(1 - v2 ) ) 
so that
= √(1 - γ-2)
"

It is easy enough to understand the first and last equations but the two middle ones were not so easy.

His (2.51) itself starts
γ =  -ημν UμVν
"since η00 = -1"

that seems to come out of nowhere and had me thinking.

It all ends well with the gloriously tensorial equation

v = √(1 - (Uν Vν)2 )

which allows me to proceed to section 2.6 on An Expanding Universe!

The answers, and suggestions for improvement (!), are in Commentary App A Mapping S2 and R3.pdf.

Saturday, 24 November 2018

Commentary 2.4 Tensors again - transformation law for general tensors

In section 2.4 the we meet the powerful but intimidating transformation law for general tensors (his 2.30)

that is followed by a simple example with a rank (0,2) tensor in a 2 dimensional coordinate system with a transformation to another 2 dimensional system. He derived the the transformed tensor without using the above law but encouraged us to do so and see if the answer was the same. I did and it was. The tensors and the transformation become much simpler and comprehensible.

I also explained more about his equations on the way, including where the missing ⨂'s were, and tried to simplify the transformation law. Putting the primes higher up in the chain helps a bit:


but that was as far as it went.

Read it all 3 pages here at Commentary 2.4 Tensors again.pdf

Thursday, 22 November 2018

Rules for tensors, matrices and indices

This all started because I was confused about the difference between co- and contra-variant vectors and how to turn tensors into matrices. There seemed to be conflicting definitions, most of which I have now resolved. The exploration took me almost two months, when I intended a few days. I have learnt a lot.

During the two months I created some word macros to help write 211 equations.I also heavily revised Exercise 1.07 Tensors and VectorsCommentary on Appendix A: Mapping S2 and R3 and had a lively discussion on Physics Forums: Question on co- / contra-variant coordinates.

I have now found about 25 useful rules for tensors, vectors, indices and matrices. They come first, then many notes and examples finally a contents at the end.

The rules

  1. A tensor of type (or rank) (k , l) has k upper indices and l lower indices.
  2. If we just say a tensor has rank z, we are being a bit vague: z = k + l.
  3. A scalar is a type (0,0) tensor. Scalars are invariant under coordinate changes.
  4. A vector (or contravariant vector) is a type (1,0) tensor, e.g. i . Each i is a component of the vector V.
  5. A dual vector (or one-form or covariant vector or covector) is a type (0,1) tensor, e.g. ωi
  6. Upper indices are co↑travariant, lower indices are co↓ariant.
  7. Both kinds of vectors can be written as column or row matrices.
  8. When we say a vector is covariant or contravariant we really mean its components are covariant or contravariant. A vector is invariant under coordinate changes. Its components are not. Therefore length and velocity are not real vectors in Minkowski spacetime.
  9. A rank 2 tensor can be written as a two dimensional matrix. Components must be written so that the first index indicates row components and the second index column components. Whether the indices are up or down is irrelevant for this rule.
  10. By 'contracting' two tensors Ti l , R k j we mean multiplying and adding components, e.g. l, k. We would write this Ti l , R l j which would produce Xi j. This is most easily done by multiplying the matrices T, R or  Ti l , R k j . Subject to these rules: 
    • When converting tensors to matrices for multiplication the summed indices must be last and first (or adjacent): Ti Rk j  not Ti k  R j k . Ti k  R j k   can be calculated without matrices but it's horrid. 
    • If you know the matrix T for a rank 2 tensor Ti k, then the matrix of the tensor with indices reversed k i is the transpose of T or TT .
    • You can only contract upper and lower indices. Contracting two upper or two lower indices would give something that is not a proper tensor.
    • The 3 bullets above even apply to the partial derivative matrix jai .. 
    • The metric ημν  is a type (0,2) tensor and it lowers an index.
      The inverse metric ημν raises an index.
      η μν ηνκ  = δ μκ  , the Kronecker delta (identity matrix).
      Replace η by g when in curved spacetime. 
    • The order of the tensors does not matter (unlike matrices). Ti k R j  = R k j Ti k always. TR = RT rarely. The same applies for tensors of any rank including vectors. 
  11. Co-/contra- variant vector components combined give correct magnitude (length) and dot products.
    • Norm (=length²): |v= ημν  Vν Vμ  = Vμ Vμ . 
    • Dot product: v∙w = ημν  Vν  Wμ  = Vμ  Wμ . 
    • The 2 bullets above only work on flat manifolds (all ημν are constant). 
  12. More tensor rules 
    • Raising or lowering an index does not change its left-right order. Free indices must be the same on both sides of the equation. Dummy indices only appear on one side of the equation - they are being summed over.
      Correct: αβμ δ = η μγ T αβ γδ .
      Wrong: T αβδ μ  = η μγ T αβ γδ . Indices changed order.
      Wrong: αβμθ  η μγ αβ γδ . Free indices not the same.
      Wrong: αβμγ  η μγ αβ γδ. Dummy index on both sides. 
    • You may raise and lower dummy indices simultaneously: λ Bλ   = Aσ B σ 
    • There are yet more rules. 
For these rules and over 34 pages of examples and notes see
Commentary 1.1 Tensors matrices and indexes.pdf.

Monday, 12 November 2018

Question on co- / contra-variant coordinates

I am studying co- and contra- variant vectors and I found the video at https://www.youtube.com/watch?v=8vBfTyBPu-4 very useful. It discusses the slanted coordinate system where the X, Y axes are at an angle of α. One can get the components of v either by dropping perpendiculars to the axes (vi) or by dropping a line parallel to the other axis (vi). These give correct results for the norm vi vi and the dot product vi wi. (I have not shown w). So the vi are called contravariant and the vi are called covariant. According to the video Dirac thought this was a great example.

But both vi and vi  contra-vary with a change of scale of the basis vectors. This contradicts some definitions of contravariant and covariant components, e.g. this one on Wikipedia. These definitions say that covariant components co-vary with a change of scale.

Is there a simple resolution to this apparent contradiction?

Submitted as comment on video and
on Physics Forums at https://www.physicsforums.com/threads/covariant-coordinates-dont-co-vary.959888/

Orodruin answered on PF seven hours later with "The covariant and contravariant components belong to different basis vectors." Cryo came up with something a bit more complicated and I followed Orodruin. So here's how it goes:

To get v with the covariant coordinates we must have
v = vxex + vyey      (1)
I have drawn them in on the diagram and ex is clearly smaller than ex, which I have conveniently drawn so that vx = 1. We must also have
 vxex = vxex       (2)
and similarly
vyey = vyey       (3)
which give us
v = vxex + vyey       (4)
which is as it should be.
(3) gives us
vy = ey(ey)-1vy       (5)
where (ey)-1 is the inverse of ey: or ey(ey)-1 = 1.

(5) shows us that vy does indeed vary with ey with a strange constant of proportionality (ey)-1vy.

Back to the drawing board

Orodruin did not like that at his #5 and then gave me a tip at #7 that the dual (or covariant) bases are given by êa = ∇xa. I will use ê for basis vectors in future, following Carroll. From the formula I could calculate the covariant basis vectors in terms of the Cartesian basis vectors because I already had the inverse metric of the covariant system. This was calculated from the transformations of the systems to and from Cartesian.

The answer came:

and draw them properly (x = 1, y = 2)
From (NF12) it is clear that  êx is π/2 - α below the Cartesian X axis. Therefore the angle between êy and êx is a right angle, and the projected line from vy  is parallel to êx as we should have guessed and Orodruin intimated. Likewise êy is parallel to the projected line from vx and always on the Cartesian Y axis. In addition |êi | > |êi |.

Getting back to the original question, what happens to vx if we double the contravariant basis êx to ê'x? We could leave  vx, êx, vy, êy alone and they would still give v. We could double êx and halve vx. Neither option would co-vary. Both would give incorrect values for |v|. We need to find the metric of the primed system so we can just calculate the primed covariant coordinates and bases. We also cannot use the previous technique (calculating from the transformations of the systems to and from Cartesian), because we don't know how to transform primed covariant coordinates to Cartesian - we don't know what the primed coordinates are.

Therefore we use the pullback operator which we met two posts ago in Commentary on Appendix A: Mapping S2 and R3.

If we have our contravariant primed bases, with scale factors a,b so
ê'x = aêx  ,  ê'y = bêy 
which give
v'x  = vx  ⁄ a   ,  v'y =vy  ⁄ b 

Turning all the handles we end up with

v'x = a2v'x  + abv'ycos⁡α
v'y =b2v'y + abv'xcos⁡α

Since v' i  decrease as a, b and v'i  increase as a2  ,b2 ,ab, we can safely say that v'i  increase as êi . They co-vary. It also looks as if they will 'compensate' for v' i , giving correct |v| and v.w. We would just have to follow the trail.

The full details of all the above are contained in sections 5b and Note F of  Commentary 1.1 Tensors matrices and indexes.pdf