Solving Dimensional Analysis Problems in Stress Energy Momentum Tensor

[Dale]

So what function or vector am I differentiating with respect to mu? If you could show an example of the process, I'd also greatly appreciate that. Thanks for your help.

Sorry about that. The notation is often confusing. I will walk through how I think of it. Please don't consider this "official" notation in any way:

Let's start with a simple example, the usual metric on the surface of the unit sphere: $ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu} = d\theta^2 + \sin^2(\theta) d\phi^2$. Now, at each point, $(\theta,\phi)$ on the sphere we can construct vectors in the tangent space:
$\partial_{\theta} = d\theta = (1,0)$
$\partial_{\phi} = d\phi = (0,1)$
this is the coordinate basis.

Now, we want to determine if the coordinate basis is orthonormal. So we take a bunch of inner products:
$g_{\mu\nu}d\theta d\phi = 0$ so the vectors are orthogonal
$g_{\mu\nu}d\theta d\theta = 1$ so $d\theta$ is normalized
$g_{\mu\nu}d\phi d\phi = \sin^2(\theta)$ so $d\phi$ is not normalized.

So the coordinate basis is not orthonormal. It is orthognoal but one of the components is not normalized. However, without changing our coordinates we can construct a new basis:

$e_0 = d\theta = (1,0)$
$e_1 = d\phi/\sin(\theta) = (0,1/\sin(\theta))$

If you take all of the inner products it is fairly easy to show that this basis is orthonormal.

 

[pervect]

Can you give a link? I'm not familiar with the IAU's take on this, but it sounds interesting.

Some papers on the topic:

"Units of Measurement in relativistic context" by Bernard Guinot.

"Time scales in the context of general relativity" by Bernard Guinot
http://rsta.royalsocietypublishing.org/content/369/1953/4131.full.pdf

Some authors consider the relativistic coordinates as dimensionless[14] , others
give a special name to their unit, such as the ‘TCB second’ or a global name such
as ‘graduation unit’. I was myself in favour of the latter name. However, after
long discussions with eminent metrologists, Quinn and de Boer, I agreed that
it was simpler to name ‘second’ the graduation unit. Thus, more generally, all
quantities having the dimension of time have the second (without any qualifier)
as their unit, even if they have different natures,

"Relativistic scaling of astronomical quantities and the system of astronomical units"
http://www.aanda.org/articles/aa/full/2008/06/aa7786-07/aa7786-07.html

Moreover, since a consensus exists, also enforced by the IAU Resolutions (1991), that TT, TCG, TCB, and related quantities are all measured in the SI units (or in ``SI-compatible units of graduation''; see Sect. 2.1), it is quite natural to extend this rule to TDB and TDB-compatible quantities and to make the semantics symmetric and clear. In Sect. 5 below one can find additional arguments against the concept of ``TDB units''.

"Units of relativistic time scales and associated quantities"
http://arxiv-web3.library.cornell.edu/pdf/0907.5100v1.pdf

On review upon re-reading these papers though, while some authors do advocate as I do that relativistic coordinates are dimensionless, it would be a mistake to say there is any actual agreement on this point.

The IAU does say that the SI second should be the unit of time. Using TCB-seconds or other non-SI seconds to express coordinate times in special units would go against this resolution. I will refer to the practice of appending a prefix to the seconds to define a different time unit as "foo-seconds".

So, upon review, I would say that:
1) The IAU, and a growing consensus, is against the use of "foo seconds", and for the use of SI seconds. Similar remarks apply to distance units.
2) Always give a metric (or a reference to a metric) if possible.
3) If you've specified a metric, there isn't any logical necessity to give units to your coordinates, but some might do so anyway. You'll probably get away with either as long as you're self consistent and you have a metric so people can figure out what you meant.

 

[space-time]

Sorry about that. The notation is often confusing. I will walk through how I think of it. Please don't consider this "official" notation in any way:

Let's start with a simple example, the usual metric on the surface of the unit sphere: $ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu} = d\theta^2 + \sin^2(\theta) d\phi^2$. Now, at each point, $(\theta,\phi)$ on the sphere we can construct vectors in the tangent space:
$\partial_{\theta} = d\theta = (1,0)$
$\partial_{\phi} = d\phi = (0,1)$
this is the coordinate basis.

Now, we want to determine if the coordinate basis is orthonormal. So we take a bunch of inner products:
$g_{\mu\nu}d\theta d\phi = 0$ so the vectors are orthogonal
$g_{\mu\nu}d\theta d\theta = 1$ so $d\theta$ is normalized
$g_{\mu\nu}d\phi d\phi = \sin^2(\theta)$ so $d\phi$ is not normalized.

So the coordinate basis is not orthonormal. It is orthognoal but one of the components is not normalized. However, without changing our coordinates we can construct a new basis:

$e_0 = d\theta = (1,0)$
$e_1 = d\phi/\sin(\theta) = (0,1/\sin(\theta))$

If you take all of the inner products it is fairly easy to show that this basis is orthonormal.

I looked at both your post and aleazk's post and used your method to try and derive the orthonormal basis for the metric I am using. Here is the metric tensor:

g₀₀= -1
g₁₁= 1
g₂₂= b² + l²
g₃₃= (b² + l²)sin²(Θ)

where
x⁰= ct (notice that I have changed it from just t to ct.)
x¹= l
x²= Θ
x³= Φ

Now, I started by setting up some vectors in a tangent space:

dx⁰= < -1, 0, 0, 0 > (I did negative 1 because of the whole sign signature thing)
dl= < 0, 1, 0, 0 >
dΘ= < 0, 0, 1, 0 >
dΦ= < 0, 0, 0, 1 >

I checked to see if all these vectors were orthogonal with each other and normalized by doing the metric tensor element times the corresponding vectors. All of the vectors were orthogonal to each other, but...

g_ΘΘdΘdΘ= b² + l²

and

g_ΦΦdΦdΦ= (b² + l²)sin²(Θ)

Since those were not normalized, I set:

dΘ= < 0, 0, 1/squrt(b² + l²) , 0 >

and

dΦ= < 0, 0, 0, 1/squrt((b² + l²)sin²(Θ)) >

These new vectors for dΘ and dΦ are normalized and they are still orthogonal to the others.

Therefore, my orthonormal basis is:

e₀ = < -1, 0, 0, 0 >
e₁ = < 0, 1, 0, 0 >
e₂= < 0, 0, 1/squrt(b² + l²) , 0 >
e₃= < 0, 0, 0, 1/squrt((b² + l²)sin²(Θ)) >

Is this correct? I think it is, but aleazk seemed to get something slightly different (or I am just interpreting his post incorrectly). Also, if it is correct, where would I go from here and how would this help me derive my Einstein tensor and not have to worry about units?

 

[aleazk]

I looked at both your post and aleazk's post and used your method to try and derive the orthonormal basis for the metric I am using. Here is the metric tensor:

g₀₀= -1
g₁₁= 1
g₂₂= b² + l²
g₃₃= (b² + l²)sin²(Θ)

where
x⁰= ct (notice that I have changed it from just t to ct.)
x¹= l
x²= Θ
x³= Φ

Now, I started by setting up some vectors in a tangent space:

dx⁰= < -1, 0, 0, 0 > (I did negative 1 because of the whole sign signature thing)
dl= < 0, 1, 0, 0 >
dΘ= < 0, 0, 1, 0 >
dΦ= < 0, 0, 0, 1 >

I checked to see if all these vectors were orthogonal with each other and normalized by doing the metric tensor element times the corresponding vectors. All of the vectors were orthogonal to each other, but...

g_ΘΘdΘdΘ= b² + l²

and

g_ΦΦdΦdΦ= (b² + l²)sin²(Θ)

Since those were not normalized, I set:

dΘ= < 0, 0, 1/squrt(b² + l²) , 0 >

and

dΦ= < 0, 0, 0, 1/squrt((b² + l²)sin²(Θ)) >

These new vectors for dΘ and dΦ are normalized and they are still orthogonal to the others.

Therefore, my orthonormal basis is:

e₀ = < -1, 0, 0, 0 >
e₁ = < 0, 1, 0, 0 >
e₂= < 0, 0, 1/squrt(b² + l²) , 0 >
e₃= < 0, 0, 0, 1/squrt((b² + l²)sin²(Θ)) >

Is this correct? I think it is, but aleazk seemed to get something slightly different (or I am just interpreting his post incorrectly). Also, if it is correct, where would I go from here and how would this help me derive my Einstein tensor and not have to worry about units?

Yes, it's correct, and it's equivalent to mine (though you forgot to take the square root of sin²(Θ)). In metrics with these signatures, the definition of 'orthonormal' is with a -1 for the inner product of the e₀ vector with itself (this is because this vector is timelike). Also, you don't need that -1 in the components, it's just a +1 (the vector is future directed). Second, be careful with the notation: the metric is a purely dual tensor, i.e., you contract it with tangent vectors, not dual vectors like dΘ. If you want to use the dual basis given by the dΘs, etc., you need to contract with the inverse metric then. I actually used the dual basis, and that's why the result looks different. But the dual basis to the tangent vector basis you got, is the basis (of the dual space) I got previously.

If you want to use this to calculate curvature, you will have to study the method (the so called 'tetrad method'). You can find it in Wald's or Carroll's textbooks.

As for the units, I have not followed the discussion here. In my case, I use the normal practice of taking G= 1 and c=1. After I finish the calculation, I insert the adecuate factors in order to get the right units. You can find the details of this in appendix F of Wald.

 

[pervect]

I looked at both your post and aleazk's post and used your method to try and derive the orthonormal basis for the metric I am using. Here is the metric tensor:

Now, I started by setting up some vectors in a tangent space:

dx⁰= < -1, 0, 0, 0 > (I did negative 1 because of the whole sign signature thing)

No, keep your basis vector positive, or you will flip the direction of time, which will be unnecessary and confusing. The minus sign is handled by the set of inner products of the basis vectors.

Given basis vectors $(e_0), (e_1), (e_2), (e_3)$, where the subscripts inside the parenthesis serve to identify the basis vectors and do not represent subscripts or superscripts as in usual tensor notation. Then we have.

$(e_0) \cdot (e_0)$ = -1, while $(e_1) \cdot (e_1) = (e_2) \cdot (e_2) = (e_3) \cdot (e_3) = 1$

because $(e_0)$, which we would write in abstract index tensor notation as $(e_0)_a$ or $(e_0)^a$ depending on whether it was covariant or contravariant, is timelike.

dl= < 0, 1, 0, 0 >
dΘ= < 0, 0, 1, 0 >
dΦ= < 0, 0, 0, 1 >

I checked to see if all these vectors were orthogonal with each other and normalized by doing the metric tensor element times the corresponding vectors. All of the vectors were orthogonal to each other, but...

g_ΘΘdΘdΘ= b² + l²

and

g_ΦΦdΦdΦ= (b² + l²)sin²(Θ)

Since those were not normalized, I set:

dΘ= < 0, 0, 1/squrt(b² + l²) , 0 >

good

and

dΦ= < 0, 0, 0, 1/squrt((b² + l²)sin²(Θ)) >

A probable typo here, you should have:
dΦ= < 0, 0, 0, 1/squrt((b² + l²)sin(Θ)) >

as the square root of sin^2 is sin

 

[space-time]

Yes, it's correct, and it's equivalent to mine (though you forgot to take the square root of sin²(Θ)). In metrics with these signatures, the definition of 'orthonormal' is with a -1 for the inner product of the e₀ vector with itself (this is because this vector is timelike). Also, you don't need that -1 in the components, it's just a +1 (the vector is future directed). Second, be careful with the notation: the metric is a purely dual tensor, i.e., you contract it with tangent vectors, not dual vectors like dΘ. If you want to use the dual basis given by the dΘs, etc., you need to contract with the inverse metric then. I actually used the dual basis, and that's why the result looks different. But the dual basis to the tangent vector basis you got, is the basis (of the dual space) I got previously.

If you want to use this to calculate curvature, you will have to study the method (the so called 'tetrad method'). You can find it in Wald's or Carroll's textbooks.

As for the units, I have not followed the discussion here. In my case, I use the normal practice of taking G= 1 and c=1. After I finish the calculation, I insert the adecuate factors in order to get the right units. You can find the details of this in appendix F of Wald.

Thank you. Do you know where I can find Wald's textbook or have a link that you can post by any chance?

 

[PeterDonis]

keep your basis vector positive, or you will flip the direction of time

If you're using a (-+++) signature, you can't do this with both the vector and the covector. See below.

$(e_0) \cdot (e_0) = -1$ ...

...because $(e_0)$, which we would write in abstract index tensor notation as $(e_0)_a$ or $(e_0)^a$ depending on whether it was covariant or contravariant, is timelike.

If $(e_0)^a (e_0)_a = -1$, then both of them can't possibly be positive; one has to be negative, which means either the vector or the covector has to have a negative 0 component. I've always seen it be the covector that is negative; i.e., if we stipulate that $(e_0)^a > 0$ means time is moving forward, then we must have $(e_0)_a < 0$ for time moving forward if we are using a (-+++) signature.

 

[pervect]

If you're using a (-+++) signature, you can't do this with both the vector and the covector. See below.
If $(e_0)^a (e_0)_a = -1$, then both of them can't possibly be positive; one has to be negative, which means either the vector or the covector has to have a negative 0 component. I've always seen it be the covector that is negative; i.e., if we stipulate that $(e_0)^a > 0$ means time is moving forward, then we must have $(e_0)_a < 0$ for time moving forward if we are using a (-+++) signature.

Good point, I stand corrected.

add.

Though I saw a recent example in Wald that did make the convector positive (pg 121, eq 6.1.6a sets $(e_0)_a = f^{1/2} (dt)_a$. I think your way has a lot to recommend it though.

 

[space-time]

If you're using a (-+++) signature, you can't do this with both the vector and the covector. See below.
If $(e_0)^a (e_0)_a = -1$, then both of them can't possibly be positive; one has to be negative, which means either the vector or the covector has to have a negative 0 component. I've always seen it be the covector that is negative; i.e., if we stipulate that $(e_0)^a > 0$ means time is moving forward, then we must have $(e_0)_a < 0$ for time moving forward if we are using a (-+++) signature.

Hey Peterdonis, do you know where I can find Wald's textbook or where I can find the method to use orthonormal bases to help me derive the Einstein tensor without having to worry about units?