Questions on tensors in GR and the Newtonian limit

[dyn]

Hi. I am self-studying GR and have many questions. Here are a few. If anyone can help me with any of them I would be grateful.
1 - What is the difference between T^u _v and T_v^u ?

2 - I have read that the order of indices matters in tensors but when transforming tensors from one coordinate system to another it involves products of partial derivatives but the order of these doesn't matter so why does the position of indices matter ?

3 - When looking at the derivation of the Newtonian limit using g_ab = n_ab+h_ab could anyone explain how to get from
Γ^a ₀₀ = -1/2 g^ad ∂g₀₀/ ∂x^d to the following equation
Γ^a ₀₀ = -1/2 n^ad ∂h₀₀ / ∂x^d
I don't understand what has happened to the h^ad term

4 - I have been looking at the model solution to a question involving the Schwarzschild solution and timelike geodesics for radial motion only which states that the coordinate r does not correspond to physical distance from the gravitational source. What does it correspond to then ?

5 - In SR almost everything seems to boil down to comparisons between 2 inertial frames ie. the unprimed and primed frames. In the GR I have done so far I see no mention of any other frames. There are no primed frames anywhere. Is basic GR only concerned with one frame ?

 

[Geofleur]

I will try to answer question 1. When there is a metric available (and usually there is), you can use it to "raise" and "lower" indices of a vector, or more generally, a tensor. Consider a vector $v = v^i e_i$. I have used the summation convention here, where an index is summed over if it appears exactly once in an upper position and exactly once in a lower position. The $e_i$ may not be orthogonal to one another in general, but we can still define a "cobasis" $e^j$ so that $e^j \cdot e_i = \delta^j_i$. Then, the same vector can be written in terms of the cobasis as $v = v_i e^i$. Using the metric $g$, one can show that $v_i = g_{ij} v^j$. There is a single object, the vector $v$, and it can be represented by either $v^i$ (when using the original basis $e_i$) or by $v_i$ (when using the cobasis $e^i$). A tensor is the same way! There is one object, $T$, and we can write it as $T_{ij} e^i \otimes e^j = T^i{_ j} e_i \otimes e^j = T^{ij} e_i \otimes e_j = T{_i}^j e^i \otimes e_j$. We need tensor products of basis and cobasis vectors now because the tensor is rank two (as opposed to the vector, which has rank one). All of these different ways of writing the tensor components refer to a single object, $T$, but expressed in terms of different bases.

 

[dyn]

For a rank 2 tensor ie. a matrix how does T^u_v differ from T_v^u in terms of rows and columns of the matrix ?

 

[PeterDonis]

1 - What is the difference between T^u_v and T_v^u ?

Do you understand the difference between upper and lower indexes? If you do, the difference between the two tensors you've written should be obvious, since the index order is reversed.

I have read that the order of indices matters in tensors but when transforming tensors from one coordinate system to another it involves products of partial derivatives but the order of these doesn't matter so why does the position of indices matter ?

The order of indexes matters because each index corresponds to a "slot" in the tensor that can be contracted with either a vector or a covector (depending on whether the index is lower or upper). Reordering the indexes means reordering the slots, which means reordering the set of vectors and covectors that the tensor is contracted with. That changes the physical meaning of the contraction, and since all actual observables are expressed as contractions with no free indexes (i.e., scalars), changing the order of indexes means changing which actual observable you are talking about.

Note that none of this has anything to do with how the tensor transforms between coordinate systems.

When looking at the derivation of the Newtonian limit using g_ab = n_ab+h_ab could anyone explain how to get from
Γ^a₀₀ = -1/2 g^ad ∂g₀₀/ ∂x^d to the following equation
Γ^a₀₀ = -1/2 n^ad ∂h₀₀ / ∂x^d
I don't understand what has happened to the h^ad term

It got dropped because it's second order (because it has a factor of $h^{ad}$ and a factor of $h_{00}$), and the Newtonian limit only includes first order terms.

 

[PeterDonis]

For a rank 2 tensor ie. a matrix how does T^u_v differ from T_v^u in terms of rows and columns of the matrix ?

The second tensor would be the transpose of the first if both were viewed as matrices. But a matrix representation doesn't really capture the distinction between upper and lower indexes.

 

[dyn]

In terms of a matrix I still don't understand the difference between T^a_b and T_b^a and T_ab.
as regards the Newtonian limit the h^ad term is multiplied by the derivative of h₀₀. Is this classed as a 2nd order term. Just because h^ad is small doesn't necessarily mean that its derivative is small ?

 

[PeterDonis]

In terms of a matrix I still don't understand the difference between T^a_b and T_b^a and T_ab.

There isn't really; that's why I said the matrix representation doesn't really capture the difference between upper and lower indexes. To see the difference, it helps to look at the simplest example: a vector $v^a$ as compared to a covector $v_a$.

A vector $v^a$ is a set of numbers that transform in a particular way when you change coordinates; if we change from coordinates $x^u$ to new coordinates $x'^u$, the vector $v^a$ transforms like this:

$$
v'^b = \frac{\partial x'^b}{\partial x^a} v^a
$$

A covector $v_a$ (also called a "1-form") is a set of linear maps from vectors to numbers. The transformation law for covectors is this:

$$
v'_b = \frac{\partial x^a}{\partial x'^b} v_a
$$

Notice that the coefficients here are the inverses of the coefficients in the vector transformation law above.

The following Wikipedia page has a decent discussion:

https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors

Just because h^adis small doesn't necessarily mean that its derivative is small ?

Yes, it does; if the derivative of $h^{ad}$ isn't small, then $h^{ad}$ itself, even if it starts out small, won't stay small for long.

In any case, that's not the comparison being made. The comparison being made is between the product of $n^{ad}$ and the derivative of $h_{00}$, and the product of $h^{ad}$ and the derivative of $h_{00}$. The latter is much smaller than the former.

 

[Orodruin]

In terms of a matrix I still don't understand the difference between T^a_b and T_b^a and T_ab.

These are all components of different tensors (if you have a metric they are related). In the first one, the first index has been raised and in the second the second index has been raised. In matrix notation it corresponds to multiplying with the metric from the left and right, respectively, although I would suggest not always trying to use matrix representation because it only works for rank 2 tensors and gives you the impression that tensors and matrices are one and the same..

I would also make sure that I was fluent in how tensors are used in SR before going on to GR.

as regards the Newtonian limit the h^ad term is multiplied by the derivative of h₀₀. Is this classed as a 2nd order term. Just because h^ad is small doesn't necessarily mean that its derivative is small ?

Everything involving h is considered small and may be expanded in, h itself as well as its derivatives.

 

[Orodruin]

Oh yes, as it has not been pointed out: If T is symmetric, i.e., if $T_{ab} = T_{ba}$, then $T^a_{\phantom a b} = T_b^{\phantom b a}$.

 

[bcrowell]

4 - I have been looking at the model solution to a question involving the Schwarzschild solution and timelike geodesics for radial motion only which states that the coordinate r does not correspond to physical distance from the gravitational source. What does it correspond to then ?

It can be interpreted as the circumference divided by 2 pi. It wouldn't make sense to define it as a physical distance from the center because there is no measurement process that could measure such a distance.

5 - In SR almost everything seems to boil down to comparisons between 2 inertial frames ie. the unprimed and primed frames. In the GR I have done so far I see no mention of any other frames. There are no primed frames anywhere. Is basic GR only concerned with one frame ?

Frames are not fundamental in relativity and never necessary. They are sometimes a convenience in SR. In GR global frames don't exist. Einstein thought that frames needed to play a central role in relativity. This was one of the things that history proved Einstein wrong about.

 

[PeterDonis]

I am self-studying GR

What resources are you using?

In the GR I have done so far I see no mention of any other frames.

Distinguishing different frames by putting a prime on quantities in one of them is only one convention. In GR it's more usual to just use different letters to distinguish coordinates in different charts.

 

[dyn]

Thanks for all your replies. For resources I am working through lecture notes from universities and using books such as Schutz , Hobson and Carroll.
If T^u_v is the transpose of T_v^u what would T_uv be then ?
I found the following in some notes - "the order of free indices is not important so S^a_b = T^ac_cb + W_b^a
Is this correct ? It seems wrong to me as order of indices is supposed to matter ?

 

[PeterDonis]

If T^u_v is the transpose of T_v^u what would T_uv be then ?

As I have said several times now, the matrix interpretation of tensors does not really capture the difference between upper and lower indexes. If you want to understand the difference between $T^u{}_v$ and $T_{uv}$, trying to interpret them as matrices will not help you.

You need to understand the difference between vectors and covectors. Please read post #7 and the Wikipedia page I linked to there. (I don't have Schutz or Hobson, but I have read Carroll's online lecture notes; I don't think he spends a lot of time on the difference between vectors and covectors. Misner, Thorne, and Wheeler go into it in detail.)

I found the following in some notes

Reference please?

It seems wrong to me as order of indices is supposed to matter ?

The expression as written is weird, but I don't think it's wrong as it stands, because it's a relationship between different tensors. What would be wrong would be to say that, for example, $S^a{}_b$ and $S_b{}^a$ were the same tensor in the general case. This will only be true in the special case of $S$ being a symmetric tensor.

 

[stevendaryl]

I will try to answer question 1. When there is a metric available (and usually there is), you can use it to "raise" and "lower" indices of a vector, or more generally, a tensor. Consider a vector $v = v^i e_i$. I have used the summation convention here, where an index is summed over if it appears exactly once in an upper position and exactly once in a lower position. The $e_i$ may not be orthogonal to one another in general, but we can still define a "cobasis" $e^j$ so that $e^j \cdot e_i = \delta^j_i$.

This is a little off-track from the original post, but I'm a little confused about your explanation. The way you are describing it, $e^j$ is a vector, just like $e_j$. But isn't it a covector? I understand that when there is a metric, vectors and covectors can be interconverted, but they aren't the same thing.

 

[Geofleur]

This is a little off-track from the original post, but I'm a little confused about your explanation. The way you are describing it, eje^j is a vector, just like eje_j. But isn't it a covector? I understand that when there is a metric, vectors and covectors can be interconverted, but they aren't the same thing.

The $e^i$ are in fact vectors. They are like the reciprocal lattice vectors in crystal diffraction. When there is a metric present, there is a one-to-one correspondence between one-forms and vectors and one can identify them. We look for the vector w such that $dx^i(v) = \langle w, v \rangle$ for any vector v. w turns out to be exactly $e^i$. Instead of writing $dx^i(v) = v^i$ we can then write $e^i \cdot v = v^i$. We can even write things like $e^j = g^{ij} e_i$, even though the e's are vectors and not components.

I was trying to avoid the use of one-forms because I thought that might be more confusing. This perspective of opting for reciprocal basis vectors instead of one-forms I got from Don Koks' book, Explorations in Mathematical Physics.

 

[stevendaryl]

The $e^i$ are in fact vectors

Well then I have been misreading equations involving $e^i$. I thought that it was supposed to be a basis of one-forms. I guess as long as there is a metric, that confusion doesn't matter much.

 

[Geofleur]

Right. When there is not a metric, vectors and one-forms are certainly different kinds of objects.

 

[Geofleur]

I should probably point out that, even when there is a metric present, sometimes there are good reasons to use forms $dx^i$ instead of the $e^i$. Just because one can identify them doesn't mean one always should. When integrating, forms are much more natural. In curvilinear coordinates, for example, it's nicer to deal with the exterior derivative of a one-form than it is to deal with the curl of a vector.

 

[bcrowell]

The $e^i$ are in fact vectors. They are like the reciprocal lattice vectors in crystal diffraction. When there is a metric present, there is a one-to-one correspondence between one-forms and vectors and one can identify them. We look for the vector w such that $dx^i(v) = \langle w, v \rangle$ for any vector v. w turns out to be exactly $e^i$. Instead of writing $dx^i(v) = v^i$ we can then write $e^i \cdot v = v^i$. We can even write things like $e^j = g^{ij} e_i$, even though the e's are vectors and not components.

I was trying to avoid the use of one-forms because I thought that might be more confusing. This perspective of opting for reciprocal basis vectors instead of one-forms I got from Don Koks' book, Explorations in Mathematical Physics.

Either I'm misunderstanding you or you're misunderstanding Koks, or I'm misunderstanding Koks. Amazon let me look at what seems to be the relevant portion of the book (p. 284, first page of section 8.4). Koks refers to the $e_\alpha$ as the "basis vectors," and the $e^\alpha$ as the "cobasis vectors," which makes it seem pretty clear to me that he is following the usual custom of using upper-index notation for components of vectors and the basis for the space of covectors, and lower indices for components of covectors and the basis for the space of vectors.

 

[Geofleur]

He calls them cobasis vectors, but he is very clear that he is talking about vectors of the "arrow" type. I've read the whole book many times, and I'm pretty certain. He even devotes several paragraphs to how one-forms are not needed. In the beginning of the book he introduces the reciprocal lattice vectors, and then he says later on that the cobasis vectors are the same thing.

 

[stevendaryl]

Koks refers to the $e_\alpha$ as the "basis vectors," and the $e^\alpha$ as the "cobasis vectors,"

But maybe "cobasis vectors" is different from "basis covectors". The order of the "co" might be significant.

 

[Geofleur]

Cobasis vectors and basis covectors refer to the same thing. But in Koks' book both refer to honest-to-goodness vectors, while in most other books they refer to one-forms. I started with Koks' approach years ago and then moved on to Frankel's book, The Geometry of Physics. It took me some time to figure out how Koks' one-form-free approach relates to the approach that uses one-forms. It works like I described it above. A one-form eats a vector and spits out a number. When there is a metric, there is a vector $e^i$ corresponding to each basis covector $dx^i$. Then dotting $e^i$ into a vector $v$ will eat that vector and spit out the same number that $dx^i(v)$ does. If you view a tensor as a multilinear map and construct its basis from the $e^i$ and the $e_j$, instead of $dx^i$ and $e_j$, everything works out consistently.

 

[Geofleur]

I just noticed, it even says on the back of Koks' book:

"The book takes a fresh approach to tensor analysis built solely on the metric and vectors, with no need for one-forms."

 

[dyn]

If r is interpreted as the circumference divided by 2pi then where is the origin r = 0 ? Also do the 2 spherical polar angles have their usual meaning ?

 

[bcrowell]

If r is interpreted as the circumference divided by 2pi then where is the origin r = 0 ?

There is no such point. If there were such a point, it would be at the singularity, but the singularity is not considered a point in spacetime. The nonexistence of such a central point is one of the reasons that we can't define r as the distance from the center.

Also do the 2 spherical polar angles have their usual meaning ?

Depends on what you consider to be their usual meaning.

 

[dyn]

Their usual meaning as in the 2 angles in spherical polar coordinates eg. used to describe a unit sphere.

 

[bcrowell]

Their usual meaning as in the 2 angles in spherical polar coordinates eg. used to describe a unit sphere.

That isn't a definition. If want to use a definition that talks about two lines that cross, then that won't work here, because there is no point where the radial lines cross. If you define the radian measure of an angle as arc length divided by r, then that works, but since r is defined here in terms of circumference, then you'd really be defining angles in terms of a proportionality with respect to a full circle.

 

[dyn]

If anyone could tell me what the 2 angles correspond to that would be appreciated. Thanks