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Questions, comments and answers.
16 Nov 2004
metric connections etc.
Note in the following i use the symbol ^ to denote superscript indices
and the symbol _ to denote subscript indices. If there is a easier way to add
equations to your feedback set up please let me know. Thanks.
my queation concerns your lecture notes on metric connections (27/10/04). In
this lecture u showed how
del_a *(g^ab * del_b *phi) = g^ab*del_a*del_b*phi
To do this u said that (which is fine)
del_a*(g^ab * del_b *phi)=(del_a * g^ab)*del_b*phi+g^ab*del_a*del_b*phi
The result is obvious if del_a*g^ab = 0....But if so how?
U proceed to then show that (which is fine)
del_a*g_bc*g^cd = (del_a*g_bc)*g^cd + g_bc*del_a*g^cd
Now i get that del_a*g_bc = 0 by the definition of the metric connection and
that
del_a*g_bc*g^cd = del_a*delta^(d)_(b)
so that
del_a*g_bc*g^cd = del_a*delta^(d)_(b)= g_bc*del_a*g^cd
BUT then u claim that del_a*delta^(d)_(b)= 0 ... how?
So in summary i don't understand how
(i) del_a*g^ab = 0
and
(ii) del_a*delta^(d)_(b)= 0
Thanks!!
Reply So, I worried about this when I gave the lecture, and I put it on a problem sheet, this is actually question 8 on the problem sheet one and I should talk about it in the tutorial class, in the meantime:
\nabla_a (g_{bc}g^{cd})=\nabla_a\delta^d_b
Now, by the Leibnitz rule
\nabla_a(g_{bc}g^{cd})=(\nabla_a g_{bc})g^{cd}+g_{bc}(\nabla_a g^{cd})
and, since \nabla_a g_{bc}=0 by definition of the metric connection and assuming that the metric is invertible (couldn't define g^{cd} otherwise) we just need to show:
\nabla_a\delta_b^d=0
Now, assuming the delta function is a tensor (will discuss in the tutorial)
\nabla_a\delta_b^d=\partial_a \delta_b^d+\Gamma_{ae}^d\delta^e_b-\Gamma_{ab}^e\delta_e^d
but \partial_a\delta_b^d=0 since the delta is a constant and contracting over the deltas
\nabla_a\delta_b^d=\Gamma_{ab}^d-\Gamma_{ab}^d=0
as required.
As for adding equation to feedback the usual thing is to use some sort of pseudo latex pidgin as you did in your question and I did in my reply. It is usually easy enough to follow, if not, copy it out by hand. As you see I have the habit of putting backslashes before words for symbols, this is what happens in Latex but here it doesn't much matter if you leave them out. The main thing missing here is to write \frac{a}{b} for the fraction a over b. Note also, \nabla is the vertex down triangle used for covariant derivative and putting the first letter of the word for a greek letter in capitals means the greek letter is capital.
18 Nov 2004
Normal coordinates.
Regarding your lecture on normal coordinates why is
\A^{a')_{c,b} = \Q^{a'}_{bc}
Is the Q here a
connection? Also why does \gamma^{a}_{bc} imply g_{ab,c} ??
Reply No, Q is not a connection, or a tensor or any sort of
covariant object, it is an array of parameters for a specific
coordinate transformation. Notice, for example, that it mixes prime
and unprime indices, so it doesn't relate to an object in a space, but
to a coordinate transformation. What I am doing is setting up a
particular family of coordinate transformations parameterized by Q and
then choosing Q so that for that it gives coordinates with zero
connections. I didn't show that the g_{ab,c} where zero in class, I just
said that it followed. In fact, it follows from writing the connection
coefficients in terms of the metric terms and getting rid of the
overall metric tensor at the front using invertibility of the metric giving:
g_{ab,c}+g_{ac,b}-g_{bc,a}=0
for all a,b and c. This means that
g_{bc,a}+g_{ac,b}-g_{ab,a}=0
also, adding gives the result.
24 Nov 2004
Sinead's baby.
When is Dr. Ryan's baby due? The first years have a pool going and I
want in.
Reply Wouldn't we all like to know. Too much money is at stake for this information to be available.
13 December 2004
metrics.
If
\ds^2 = \dr^2 + r^2 d \theta^2 + r^2 \sin^2 \theta d \phi^2
for spherical coordinates, why is
\ds^2 = r^2 d \theta^2 + r^2 \sin^2 \theta d \phi^2
for S^2 the two sphere.
Reply A two-sphere here can be thought of as the sphere of constant radius in {\bf R}^3, so, the two-sphere corresponds to r=const so $dr=0$.
13 December 2004
Isometries of minkowski space
If
\eta_{ \mu^' \nu^'} = \A^{ \mu }_{ \mu^'} \A^{ \nu }_{ \nu^'} \eta_{ \mu \nu}
= L_{ \mu^' \mu } \eta_{ \mu \nu } L_{ \nu^' \nu }
where
L = L_{ \mu^' \mu } = A^{ \mu }_{ \mu^'} ... (matrix)
how does this become
\eta = \L \eta \L^{T}
surely \L^{T} = L_{ \mu \mu^' } with \mu is a dummy variable ... (transpose of
matrix L)
Reply I amn't sure I have understood your question. I am moving from the index form to the matrix form, so \eta L^T implies the second index of the eta is summed with the first index of the L^T, now, L_{\nu'\nu}=L^T_{\nu\nu'} as you point out, so \eta_{\mu\nu}L_{\nu'\nu}=\eta_{\mu\nu}L^T_{\nu\nu'} which is (\eta L^T)_{{\mu\nu'}
13 January 2005
Weyl. In the tutorial this week you
defined the weyl tensor as being W_{abcd} = R_{abcd} + A ( R_{ac}
g_{bd} + R_{bd} g_{ac} - R_{ad} g_{bc} - R_{bc} g_{ad} ) + BR (g_ac}
g_{bd} - g_{bc} g_{ad}) where A and B are constants.
How did you
exactly arrive at this expression? I understand from the question
that the weyl tensor is a combination of the riemann tensor, ricci
tensor and ricci scalar but why is it the particylar combination of
these terms above. For example knowing the Ricci tensor is symmetric
why is the ricci tensors R_{ab} and R_{cd} not contained in the wely
tensor expression. Why are these terms combined with metrics aswell?
also how do you get the ricci scalar terms with the two metrics?
Reply Ok, so I feel bad about this question, it is so
ill-defined; I have improved it since I was given it myself but it is
still kind of an annoying question. The idea is that the Weyl tensor
is traceless and linear in the Riemann tensor and has the symmetries
$W_{abcd}=-W_{bacd}$ and $W_{abcd}=W_{cdab}$. Now, a bit of thought
shows that the above is the most general such expression before you
put the traceless in, for example, you asked about $R_{ab}$ but that's
not going to work because $R_{ab}=R_{ba}$ so nothing you do with this
is going to have the correct symmetry properties, so, choose another
$R_{ac}g_{bd}$, we need the metric bit to make up the full four
indices. Now, this doesn't have any of the symmetris, to get the $ab$
symmetry you need $R_{ac}g_{bd}-R_{bc}g_{ad}$, that is switching $a$
and $b$ should give minus. Now, switching $a$ and $b$ with $c$ and $d$
needs to give plus, add terms to make that happen and you end up with
the thing above. Something similar works for the Ricci scalar terms. $Rg_{ab}g_{cd}$ doesn't work because it is symmetric in $ab$, consider $Rg_{ac}g_{bd}$ and then think about imposing the symmetries.
14 January 2005
exam Can you give an indication of the
format of the exam this year, especially with regards to this worrying
"short questions" concept that seems to be emerging in other Tp style
maths courses. Also, cheeky I know, but can you tell us what the
average mark/distribution of marks was like amongst your students last
time you gave this course? It'll help us know what to expect this time
around. Thanks.
Reply The short and long question format
was introduced on a trial basis last year on the recommendation of the
extern. There was considerable discussion of the new system and a wide
mixture of opinions, however, on balance it was decided not to
continue with the system. A new sentence was added to the exam
guidelines saying that whenever it was practical, exam questions
should have an easier, more straightforward part and a harder,
trickier part. Last time I taught this course the average mark was
very close to the overall average, 2% lower, and the spread was a
slightly higher, as you would expect comparing against an average. In
other words, the marks were very much in line with the other papers.
15 January 2005
Schwarzchild solution etc. In your
lecture (last term) on the schwarzchild solution you showed that
R_{\mu \nu} = 0 . This result you got from the Einstein equation for
T_{\mu \nu} = 0 (vacuum). Does this mean that R_{\mu \nu} = 0 for \mu
\not= \nu only . The reason I ask is because you then go on and
calculate R_{\mu \nu} fo \mu = \nu. (i.e. R_{00} ,R_{rr}, R_{\theta
\theta}, R_{\phi \phi}.
Reply Ok, so exactly, I work out
these things and then set them to zero to give the equations. The only
complication was, that instead of considering $R_{00}=0$ and so on, it
turned out to be better to look at $R_{rr}/A+R_{00}/B=0$, but only because this equation was easier to solve; really the main point was that this was zero because $R_{rr}$ was zero and so was $R_{00}$.
17 January 2005
Weyl tensor You stated in the tutorial
last wek that the weyl tensor has the following symmetries W_{abcd} =
-W_{abdc} W_{abcd} = W_{cdab}.does it also have the symmetry W_{abcd}
= -W_{bacd}?
Reply Okay so the third symmetry follows from the other two:
W_{abcd}=W_{cdab}=-W_{cdba}=W_{bacd}
hence W_{abcd}= -W_{bacd}!
27 January 2005
Klein-Gordon Equation In your lecture on
the Klein-Gordon equation you calculated the Lagrangian density of the
Klein-Gordon field as L = \frac{1}{2} \mu l^{2} ( \frac{\partial
\phi}{\partial t} )^2 - \mu g l (1- \cos \phi) - \frac{1}{2} k (
\frac{\partial \phi}{\partial x} )^2 which becomes (for small $\phi$)
L = \frac{1}{2} \mu l^{2} ( \frac{\partial \phi}{\partial t} )^2 - \mu g l
\phi^{2} - \frac{1}{2} k ( \frac{\partial \phi}{\partial x} )^2
= \mu l^{2} [ \frac{1}{2} ( \frac{\partial \phi}{\partial t} )^2 -
\frac{1}{2}c^{2} ( \frac{\partial \phi}{\partial x} )^2 - \frac{1}{2}
m^{2} \phi^{2} ] ... eqn(1)
where $m = \sqrt{ \frac{2 g}{l} }$ ;
$c = \frac{1}{l} \sqrt{ \frac{k}{\mu} }$ ( I think ) You then said the
general expression of this Lagrangian density was L = \frac{1}{2}
\eta^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi + \frac{1}{2}
m^{2} \phi^{2} .... eqn(2)
Q: My question is how did you get the
general form eqn(2) from eqn(1) and in eqn(2) why is it $+ m^{2}
\phi^{2}$ when its $- m^{2} \phi^{2}$ in eqn(1).
With your general
KG Lagrangian density you obtain the KG eqn
\partial_{\mu}
\partial^{\mu} \phi - m^{2} \phi = 0 ... with the above question about
the sign should this be \partial_{\mu} \partial^{\mu} \phi + m^{2}
\phi = 0. I looked it up on the web and it had a plus sign in the
general eqn(howreliable the web source was I can't say)
Thanks.
Reply Ok, so cos\approx 1-\half phi^2 means you are right and I
was wrong. The problem with web sources is that they might use a
different metric, but in this case the web source seems to have
correct. The best way to say this is that the potential energy should
increase with phi and the Lagrangian is minus the potential. I always
get this wrong. I think I changed to the correct form the next day,
but, again, this was a mistake.
28 January 2005
Einstein Hilbert action In class you defined the EH action as S = \int R \sqrt{g} d^{4} x should this have a \frac{1}{16 \pi} (G=1) term before the integral.
Reply I put the 1/16\pi in as a seperate factor in the total action, this is one convention, some people include it in the definition of the EH action.
29 January 2005
tutorials Could you please do each
tutorial question out step by step. The notation is hard to grasp so
putting "plus other terms that cancel" as a step does not help when I
go back to try and do it myself and find i don't know how these other
terms are got. Please put up on your webpage solutions to this years
tutorials in an easy to follow, step by step format with each change
of index explained. Please don't fly through your tutorials as that
defies the purpose
Reply Thanks for your email, I will bear these points in mind. I do think though that the main problem is that few people attempt the problem sheet in advance and noone asks questions in the tutorial, which is the main point. I don't know how to address this, but, if I skip a step and you feel it is unclear, ask there and then, otherwise I will assume it isn't a problem. I know I am behind posting solutions, however, if you follow the link to the previous time the course was taught, most of the solutions are available there.
30 January 2005
Q6 2nd tutorial You never did Q6 from the second tutorial sheet
Reply Did so.
4 February 2005
Maximum Do you know any example in classical mechanics where the action of a system is extremised?
Reply I seem to have missed this when it was sent to me, sorry, this is a good question, I will think about it and add a reply; I may to consult my collegues.
9 February 2005
Problem sheetsThe links to the last few assignments, such as
http://www.maths.tcd.ie/~houghton/TEACHING/442/PS-04-05/prob4.pdf
don't work.
Reply Thanks for that, I forgot to set the permissions, fixed it now.
17 Febuary 2005
Weak field limit Just wondering if given that $g_{ab} = \eta_{ab} + h_{ab}$ does this mean that $g^{ab} = \eta^{ab} +h^{ab}$ and if so could you explain why? Thanks!
Reply Actually g^{ab}=\eta^{ab}-h^{ab} where h^{ab}=\eta^{ac}\eta^{bd}h_{cd} because we should have g^{ab}g_{bc}=\delta^a_c and indeed
(\eta^{ab}-h^{ab})(\eta_{bc}+h_{bc})=\delta^a_c-h^{ab}\eta_bc+\eta^{ab}h_{bc}
to leading order in h and expanding out what you mean by h^{ab} you can
see the two h terms cancel as required.
19 Febuary 2005
Q2 set4 Just having some trouble with part of the calculation.
We have $\Gamma^{c}_{ab} = \frac{1}{2} \eta^{dc} ( \partial_{a} h^{c}_{b} +
\partial_{b} h^{c}_{a} - \partial^{c} h_{ab} )$
and
\R_{ac} \approx \partial_{d} \Gamma^{d}_{ac} - \partial_{a} \Gamma^{d}_{dc}
which is fine, you then have
\Rightarrow 2R_{ac} = \partial_{a} \partial_{d} h^{d}_{c} + \partial_{c}
\partial_{d} h^{d}_{a} - \partial^{2} h_{ac} - \partial_{a} \partial_{c} h
but when I do it out myself explicitly I get as far as
2R_{ac} = \partial_{a} \partial_{d} h^{d}_{c} + \partial_{c} \partial_{d}
h^{d}_{a} - \partial^{2} h_{ac} -\partial_{a} \partial_{d} h^{d}_{c} -
\partial_{a} \partial_{c} h - \partial_{a} \partial^{d} h_{dc}
= \partial_{c} \partial_{d} h^{d}_{a} - \partial_{a} \partial^{d} h_{dc} -
\partial^{2} h_{ac} - \partial_{a} \partial_{c} h
just wondering what the steps are with playing around with the indices to get
what you got above. Thanks.
Reply Ok, so you have a sign error in the sixth term of your expression for R_{ab}, correct that and use the fact that
\partial_{a} \partial^{d} h_{dc}=\partial_{a} \partial_{d} h_{c}^d
19 Febuary 2005
Q3 set4 If T_{ab} = \mu \delta(x) \delta(y) diag(1,0,0,-1) how is T = \eta^{ab} T_{ab} = -2 \mu \delta(x) \delta(y) ?
Reply Well [\eta^{ab}]=diag(-1,1,1,1) and diag(1,0,0,-1)diag(-1,1,1,1)=diag(-1,0,0,-1) which has trace -2, ie \eta^{ab}T_{ab} is the trace of the multiple of the two matrices.
19 Febuary 2005
Linearized Einstein equations What exactly is the linearized einstein equation??
Reply It is defined by the question on the problem sheet, it is the expansion of the Einstein field equation to linear order in $h_{\mu\nu}$ where the metric is $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$. In other words it is the linearization for a small deformation away from Minkowski space: it is the leading order equation satisfied by the deformation. Of course, you could also have considered different linearizations by replacing $\eta_{\mu\nu}$ with some other solution to the Einstein equation.
22 Febuary 2005
Q3 sheet 4 In Q3 tutorial sheet 3 how did
you get the matrix diag[0,1,1,0] in the linearized einstein equation?
Reply Well, from the linearized equations, we are interested T_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}T and we have decided that T=-2\mu\delta(x)\delta(y), see discussion above so
T_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}T=\mu\delta(x)\delta(y)[diag(1,0,0,-1)+diag(-1,1,1,1)]=\mu\delta(x)\delta(y)diag(0,1,1,0)
22 Febuary 2005
Q3 sheet 4 In Q3 again when you say you
substitute into the linearized einstein eqn how did you get your
result? the -16pi etc
Reply Well the -16\pi is in the linearized equation which is given in the question above this one and is
\partial^2h_{\mu\nu}=-16\pi\left(T_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}T\right)
and the right of this equation is zero except when for h_{11} and h_{22}. The idea is that given some sort of boundary conditions, the Laplace equation uniqueness theorems say that the other h_{\mu\nu} are zero since there laplacian is; this arguement is vague because you can't help but get nervous about the boundary conditions given the linearize equations only hold near but not too near the string.
23 Febuary 2005
Lecture link Could you update the lectures
link on the 442 homepage. Its only done as far as lecture 21 and I
find them extremely helpful.
Reply Thanks for this, I completely forgot! I will start work on it as soon as I have cuaght up with the problem sheet solutions.
1 March 2005
2003 paper I was just looking at the 2003 past paper; in the hint for Q1 are the
initial conditions for U supposed to be for V^1 and V^2?
also in question 5(b) i keep getting a factor of t^1/3 rather then t. As the
scale factor a(t)^1/2 is proportional to t^1/3?? Thanks for your help.
Reply Re q1, rhere is something screwy here, that question 1 isn't the version that was on the exam. I think you must be right about the correction, I'll look tomorrow or as soon as I get a chance. I should put solutions to this paper on the web, the only thing that's holding me back is that some of the questions are on problem sheets.
As for question 5, there is an extra factor of $a_0$ you might be missing. So the arguement goes, for light $dt=-ads_{III}$, lets leave out the $III$ for now, so, here with $a=Ct^{2/3}$, $s=3(t_0^{1/3}-t_e^{1/3})/C$ and then use the expression for the red-shift to substitute for $t_e$ giving as it sounds like you got:
s=3t_0^{1/3}\left(1-1/sqrt{z+1}\right)/C
however, $s$ isn't the distance to the star, the metric has the extra $a^2$ factor multiplying ds^2, so the actual distance is $a_0ds$ and this cancels the $C$ and gives the extra $t_0^{2/3}$!
6 March 2005
Sample Exam
Hey,
I seem to remember you said that you'd be giving us a sample exam paper.
I was just wondering if there was any chance we could get it before the end of term?
Reply Ok, its done now.
6 March 2005
Lecture list
would it be possible for you to have updated the lecture material section of your website before or during
the easter break as i find it really helpfull and i plan on catching up with the course over the break. thanks.
Reply Ok, its done now.
10 March 2005
Chain
What's on the end of that chain you always wear?
Reply Keys
11 March 2005
Eqns with cosmological constant
How did you get the matrix for $\Lambda g_{\mu \nu}$ ?
Reply The g_{\mu \nu} is the usual ansatz matrix for the Robinson Walker metrics, it is the standard three-d metric embedded with a scale factor. I amn't sure if this is what you are asking!
16 March 2005
Inflation You mentioned you had a handout
on inflation.. any chance this could be put up on the web over the
break. Thanks. Have a good Easter!!
Reply Yes, I am doing it now.
1 May 2005
problem sheets the answers to Q2,3 in tutorial 6; Q4 in tutorial 7; Q2 in tutorial 8
and all of tutorial nine was not done in class. Could you please post them on
the web asap
Reply Now I am confused, all of tutorial sheet 6, 7 and questions 2 and 3 of 8 are on the web already. Tutorial sheet 9 is in progress.
25 Apr 2005
Exam Papers
Hi, Is there any past papers for this course? I can't seem to get them
off the tcd website-either the links aren't working and/or it wasn't taught
certain years!
Reply It was taught 2002/3 only, before that Petros taught it and the course was quite different. There should be one past paper and one sample paper.
6 May 2005
Sample paper
In regard to problem 1 part a in your sample paper , could you show
explicitedly how Rabcd = Rdcab. Also will the torsion free metric connection
(PS 1b) be examined , hypothetically speaking of course?
Reply Sorry for the delay in replying I have been away. So, you work with the metric in normal coordinates, I have outlined it here: sample.NCsoln.pdf. The torsion free metric connection could come up, yes.
7 May 2005>
Bending of light
In regard to problem 1 part a
Regarding the Bending of light calculation is it necessary to:
1. Show how you get the Schwartzchild metric
2. Show how the constants E and J arise from the E-L eqts of the geodesic.
In your first lecture of Hilary term you do the calculation without showing the
above, It would be a very long exam question if we had to show 1 and 2 ?????
Reply No, these would be probably be given in any such question, if they werene't then other large bits would be given to you.
9 May 2005
Exam question
Hi, Will there be a question on the exam on gravitational waves or
Kaluza-Klein theory?
Reply Yes to gravitational waves, and to Kaluza-Klein theory, but any KK question would be very vague: "explain generally the idea behind", that kind of thing, I haven't been able to find good calculational questions on this.
9 May 2005
Urgent! What will be Q9 of the paper? If its
Kaluza Klein, please post up a solution. Is it possible that it will
be gracity waves? You said it would be Kaluza Klein.
Reply
As explained at the time, it will be gravitational waves or Kaluza
Klein, but, if it is Kaluza-Klein it will be an essay style question without any derivations.
16 May 2005
Sample Paper
I assume that in q1 of the sample paper you are asking to define a
normal coord not a geodesic coord. If you are asking for a geodesic coord, then
what is it and how do i prove the riemann identity with it?
Reply Geodesic coordinate is another word for normal coordinate.
18 May 2005
Sample paper solutions Would you have solutions to the sample paper or 2003 past paper that I could have a look at please? Or is there any!
Reply But all the questions in the sample paper are off the problem sheets aren't they? If there is any question you want an answer to, email me and I will try and put up a solution to it
Further questions Em some of them I have
But questions 5,7 and 8 Im having difficulty with. Solutions would be
brilliant!
Do you have solutions to the 2003 paper at all?
And some simple things like whats the difference between Define A Geodesic and
Define Geodesic Coordinates?
Reply Geodesic coordinates are the same as normal coordinates. So question 5 was on problem sheet 5. Questions 7 and 8 are
on problem sheet 6 and 8 respectively. Which problems from 2003 do you
want me to do?
18 May 2005
Q4 PS5
Hi, just working through the tutorials...in Sheet 5 Q.4 calculating the Energy-Momentum tensor, I dont understand the final step that
Tuv = 2(delta)S / (delta)h^(lambda)(mu)
Just seems to come from nowhere! And I dont follow how the result is then got
from it! Thanks!
Reply Are you asking where
T_{\lambda\mu} = 2(\delta)S / (\delta)h^{\lambda\mu}
comes from? That's the formula for the energy momentum tensor, that first
line is being quoted, the second line then follows from the calulation
above by definition of the functional derivative, the derivative is the
coefficient of the variation in the integrand. I think your problem is
just that you have missed a definition.
May 2005
learning and 2004
in Q1 2004 the connection coefficients with a 1 superscript are not
equal. the one with subsrcipt 22 incurs a minus sign as it's the last component
of the connection that contributes. this mistake was then carried through the
whole question.
also, in Q2 can yiu go into more detail of how you got from equation 3 to 4.
theecond term in 3 has the indices unchanged. also could you do out the steps
from 4 to 5 fully
finally can you post a list of things that won,t be given on the exam that
we'll be expected to know eg the connection coefficient formula, the formula
for the riemann tensor, the cosmological equations etc
Reply Thanks, I have corrected the mistake in q1, I had both superscript one connection coefficients wrong. I will check this against my solutions when I am back at my office and check then that it is right now. In q2, you use the definition of the Riemann tensor:
\nabla_a\nabla_c l_b-\nabla_c\nabla_a l_b=R_{acbd}l^d
I can't give a complete list of everything you need to know, that is given by the lectures themselves, you need to know what was lectured, there is a formula sheet given in the notes and I will give a hint for any very difficult integrals.
2 June 2005
Harmonic Gauge Question.
In your question involving the harmonic gauge condition (Set 4 Q2 or
sample paper Q4), could you write out the full 6 terms for the expansion of the
Ricci Tensor? Thanks.
Reply Well the two terms missing from equation (11) are plus and minus
\partial_\mu\partial_\rho h_\lambda^\rho
and they cancel.