# About Tensor Calculus

Dear Friends

I have two questions to do about Tensor Calculus:

1) Is there any program to calculate Christoffel Symbols, Riemann and Ricci Tensors and everything about Tensor Calculus (Free or paid)?

2) When in an exercise anyone asks to use the Euclidean metric or Riemann metric, what is the difference? Is it only the presence of € (indicator)?

Thank you for your help.

Good day.

## Answers and Replies

dx
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1. Yes. http://grtensor.phy.queensu.ca/ [/URL]

2. I don't know what € is, but a Riemannian metric is any metric with signature (n,0), and a Euclidean metric is a special Riemannian metric of the form ds² = dx[SUB]1[/SUB]² + dx[SUB]2[/SUB]² + ... + dx[SUB]n[/SUB]².

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a Euclidean metric is a special Riemannian metric of the form ds² = dx1² + dx2² + ... + dxn².

If I've undrestood this right, a Euclidean metric takes that form in an orthonormal coordinate system, but other forms in non-orthonormal coordinate systems, e.g. $ds^2 = dx^2 + dy^2$ becomes $ds^2 = dr^2 + r^2 \; d \theta^2$ in plane polar coordinates. Is there an intrinsic definition of Euclidean metric that doesn't refer to a special coordinate system, or is the full definition of a Euclidean metric: a metric that has the above form in an orthonormal coordinate system? Similarly for a Euclidean metric tensor. Is it defined as a metric tensor whose matrix representation, in orthonormal coordinates, is a diagonal matrix with diagonal entries all 1, or is there a definition that doesn't refer to coordinates?

dx
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I'm not sure if 'Euclidean metric' has a conventionally accepted frame invariant meaning. I think its just a term used to describe things which look like the Pythagorean formula in some coordinate system. Of course any such coordinate system will be orthonormal since

g(∂i, ∂j) = (dx1² + dx2² + ... + dxn²)(∂i, ∂j) = δij.

Okay, thanks. So if a metric has Pythagorean form, it's Euclidean, and the form guarrantees that an orthonormal coordinate system is meant.

Since the word tensor is used in this context to mean a tensor (or tensor field) with respect to the tangent space (or if tensor field, the tangent bundle) of a manifold, presumably a Euclidean metric tensor is a frame invariant object that can have coordinate representations besides the diagonal matrix $I$ with entries $\delta_{ij}$ that is its coordinate representation in an orthonormal system. And since the metric can be worked out from the metric tensor, I'm guessing a Euclidean metric can also have non-Pythagorean representations, and that it's the existence of a Pythagorean representation that makes a metric Euclidean, rather than the metric happening to be in Pythagorean form. Is that right?

atyy
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For Euclidean, how about a Riemannian manifold with signature X and all components of the curvature tensor zero erverywhere?

Thanks for your help.

Till the next. Cheers.

g(∂i, ∂j) = (dx1² + dx2² + ... + dxn²)(∂i, ∂j) = δij.

I think I may like this notation more than the notation I'm familiar with. Can you reference a text that utilizes it?

dx
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You mean using ∂i for the tangent vector ∂/∂xi?

I'm not sure if 'Euclidean metric' has a conventionally accepted frame invariant meaning. I think its just a term used to describe things which look like the Pythagorean formula in some coordinate system. Of course any such coordinate system will be orthonormal since

g(∂i, ∂j) = (dx1² + dx2² + ... + dxn²)(∂i, ∂j) = δij.

Should I be thinking then of a metric generally as a coordinate-dependent thing: orthonormal metric (i.e. "Euclidean metric"), polar metric etc., rather than as a feature of the underlying set of the manifold? Is the following correct? Iff the underlying set of a Riemannian manifold is the point set of Euclidean space En, it can be given a global coordinate system in which the metric takes this form, the "Euclidean metric", everywhere. But in any Riemannian manifold, even one whose underlying set is not the point set of Euclidean space, it's possible, for any given point, to pick a coordinate system in which the metric is the "Euclidean metric" at that point, but not necessarily at all other points. And a Euclidean space En can be referred to coordinate systems in which the metric is not the "Euclidean metric" everywhere, and even to coordinate systems in which the metric is not the "Euclidean metric" anywhere. (And if so, would "orthonormal metric" be a suitably unambiguous alternative name?)

dx
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Should I be thinking then of a metric generally as a coordinate-dependent thing: orthonormal metric (i.e. "Euclidean metric"), polar metric etc., rather than as a feature of the underlying set of the manifold?

No, it is better and more correct to think of a metric as a coordinate indepent object.

Since in special relativity, we say that the metric is locally Minkowski, I think it would be consistent to say that the metric on a Riemannian manifold is locally Euclidean. But still, as I said, I don't know if there is a conventional used definition.

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Ah, okay. Sounds like I was closer to the mark in #3 and #5 than #10.

Is this more or less what you mean by "define it in this way"? A metric field is a function-valued field (in the sense scalar, vector, tensor, etc. field), or perhaps a field whose value at each point is defined as some kind of coordinate-independent equivalence class of functions... Its value at some point, in some coordinate system might be given by, for example, $ds^2 = dr^2 + r^2 \; d \theta^2$. The metric (i.e. distance function which is the value of the metric field) is said to be Euclidean at this point iff there exists a coordinate system (namely an orthonormal one) in which the coordinates of the metric tensor give the metric the form of Pythagoras's formula. Iff for some n-manifold, there exists a single, global coordinate system according to which the metric everywhere has Pythagorean form, then the manifold is the Euclidean space En. Iff for each point in the underlying set of the manifold, there exists a coordinate system (not necessarily global, i.e. not necessarily the same coordinate system as for all other points) according to which the metric is Euclidean at that point, then the manifold is Riemannian.

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bcrowell
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For software, I use maxima and ctensor. It does everything I need, and it's free and open source.

You mean using ∂i for the tangent vector ∂/∂xi?

Sean Carroll uses the ∂i notation for taking derivatives, but not for coordinate basis vectors. Instead, he moves quickly to using

$$\hat{e}_{(\mu)} = \partial_{\mu}$$

and

$$\hat{\theta}^{(\mu)} = d^{\mu} \ ,$$

as bases in a manner I found too hard to follow. I would be interested in a presentation where the bases dxu are presented as true infinitesimals, and --- I'm not sure what the ∂u would be.

dx
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I would be interested in a presentation where the bases dxu are presented as true infinitesimals, and --- I'm not sure what the ∂u would be.

I wasn't using notation in which the dxu were infinetesimals. I was using operator notation, where dx1² + dx2² + ... + dxn² is short for the operator

$$dx_1 \otimes dx_1 + dx_2 \otimes dx_2 + ... + dx_n \otimes dx_n$$

and the dxu are coordinate 1-forms. So (dx1² + dx2² + ... + dxn²)(∂i, ∂j) would be

$$(dx_1 \otimes dx_1 + dx_2 \otimes dx_2 + ... + dx_n \otimes dx_n) (\partial_i, \partial_j) = \sum_{\alpha} dx_{\alpha}(\partial_i)dx_{\alpha}(\partial_j) = \sum_{\alpha} \delta_{\alpha i} \delta_{\alpha j} = \delta_{ij}$$

If you're interested in GR books that don't use the modern notation of 1-forms and multilinear operators, the only book I can recommend is Landau and Lifgarbagez's "Classical Theory of Fields". That's probably the only book written in 'old tensor' that is still worth reading.

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If you're interested in GR books that don't use the modern notation of 1-forms and multilinear operators, the only book I can recommend is Landau and Lifgarbagez's "Classical Theory of Fields". That's probably the only book written in 'old tensor' that is still worth reading.

Sorry. I wasn't clear. I'm looking for something modern using both differential forms and infinitesimal calculus, although the infinitesimal calculus is secondary. It would be good if exterior derivatives and wedge products were given some notice, as well.

Perhaps this is not such a small wish. You certainly have more modern training than I've had, so I was hoping you might point me in the right direction.

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dx
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Sorry. I guess I wasn't clear. I'm looking for something modern using both differential forms and infinitesimal calculus, although the infintiesimal calculus is secondary. It would be good if exterior derivatives and wedge products were given some notice, as well.

Perhaps this is not such a small wish after all. You certainly have more modern training than I've had, so I hoping you might point me in the right direction.

Some books (that use modern 'calculus on manifolds') that I like are:

"Spacetime, Geometry, Cosmology" by William Burke
"Gravitation" by Misner, Thorne and Wheeler
"Gauge Fields, Knots and Gravity" by John Baez and Javier Muniain

Baez is good for the algebraically oriented, while the other two are good for geometrically oriented readers.

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Some books (that use modern 'calculus on manifolds') that I like are:

"Spacetime, Geometry, Cosmology" by William Burke
"Gravitation" by Misner, Thorne and Wheeler
"Gauge Fields, Knots and Gravity" by John Baez and Javier Muniain

Baez is good for the algebraically oriented, while the other two are good for geometrically oriented readers.

William Burke I have heard of. Which one have you studied from, formally?

dx
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William Burke I have heard of. Which one have you studied from, formally?

I've studied Burke thoroughly, doing all the exercises etc., and I've studied part II and part III of MTW ("Physics in Flat Spacetime" and "Mathematics of Curved Spacetime").

To get a gut feeling for 1-forms and tangent vectors, I know of no better book than Burke. He doesn't go too deeply into curvatre and Einstein's equations, instead focusing on studying given spacetimes and cosmological models. There are some unconventional ideas in Burke that I think are worth learning about, like the 'wave diagram' and 'Huygen's construction'.

Burke can serve as excellent preparation for a more detailed discussion of differential geometry and Einstein field equations given in MTW. The first part of Baez is a good presentation of differential forms from an algebraic point of view, while MTW emphasizes the geometry.

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I've studied Burke thoroughly, doing all the exercises etc., and I've studied part II and part III of MTW ("Physics in Flat Spacetime" and "Mathematics of Curved Spacetime").

To get a gut feeling for 1-forms and tangent vectors, I know of no better book than Burke. He doesn't go too deeply into curvatre and Einstein's equations, instead focusing on studying given spacetimes and cosmological models. There are some unconventional ideas in Burke that I think are worth learning about, like the 'wave diagram' and 'Huygen's construction'.

Burke can serve as excellent preparation for a more detailed discussion of differential geometry and Einstein field equations given in MTW. The first part of Baez is a good presentation of differential forms from an algebraic point of view, while MTW emphasizes the geometry.

"curvatre"? Are you British? It's 6 am for me.

OK. I'll buy Burke. Thanks so much for the help. I've have MTW sitting on the self since before I understood special relativity and I'm still too frightened to crack it open. Baez never made any sense to me. I guess I'm one of those geometrical thinkers.

[...] and the dxu are coordinate 1-forms. So (dx1² + dx2² + ... + dxn²)(∂i, ∂j) would be

$$(dx_1 \otimes dx_1 + dx_2 \otimes dx_2 + ... + dx_n \otimes dx_n) (\partial_i, \partial_j)$$

$$= \sum_{\alpha} dx_{\alpha}(\partial_i)dx_{\alpha}(\partial_j) = \sum_{\alpha} \delta_{\alpha i} \delta_{\alpha j} = \delta_{ij}$$

Could we, in principle, choose some other ordered set of basis vectors for the cotangent space, a basis for which this equation isn't true? Would you still call the cotangent vectors of such a basis "coordinate 1-forms"? Would it be necessary to use a different notation then; in other words, does the notation imply this equality?

If you're interested in GR books that don't use the modern notation of 1-forms and multilinear operators, the only book I can recommend is Landau and Lifgarbagez's "Classical Theory of Fields". That's probably the only book written in 'old tensor' that is still worth reading.

If this is old-fashioned notation, how would it be written in modern notation? Is there an ambiguity in the notation, in that $\partial_i$ might stand for some particular coordinate basis tangent vector, conceived of as a derivative operator, e.g. $\partial_1$ or $\partial_2$, but in abstract (slot-naming) index notation, $\partial_i$ could stand for the set $\left \{ \partial_1, \partial_2, ..., \partial_n \right \}$, which some people call the gradient?

dx
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Could we, in principle, choose some other ordered set of basis vectors for the cotangent space, a basis for which this equation isn't true? Would you still call the cotangent vectors of such a basis "coordinate 1-forms"? Would it be necessary to use a different notation then; in other words, does the notation imply this equality?

For any coordinate system {xμ}, the coordinate 1-forms dxμ and the coordinate basis vectors ∂ν always satisfy dxμ⋅∂ν = δμν

If this is old-fashioned notation, how would it be written in modern notation?

No, everything I wrote was in modern notation.

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dx
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Is there an ambiguity in the notation, in that $\partial_i$ might stand for some particular coordinate basis tangent vector, conceived of as a derivative operator, e.g. $\partial_1$ or $\partial_2$, but in abstract (slot-naming) index notation, $\partial_i$ could stand for the set $\left \{ \partial_1, \partial_2, ..., \partial_n \right \}$, which some people call the gradient?

If we want to express a directional derivative along some vector v in terms of ∂i, we use vii, where summation is implied. Is that what you meant?

According to Mathworld, an euclidean metric is one that can be represented as a unit matrix.

http://mathworld.wolfram.com/EuclideanMetric.html

This seems to treat the Euclidean metric as coordinate-dependent, contra to dx's post #11. Since $ds^2 = dr^2 + r^2 \; d \theta^2$ isn't the Pythagorean formula (and its inputs aren't even all real numbers, being rather one real variable and two differentials), the author of the article would presumably not call it the Euclidean metric, even though with a change of coordinates, such a metric might take Pythagorean form after all. On the other hand, it says the Euclidean metric is "given by" the identity matrix in Euclidean 3-space, making me wonder if it means that the Euclidean metric would be given by some other matrix in some other space, or whether it's suggesting it's only defined in Euclidean space, and what these definitions would look like in a treatment that distinguished between Euclidean space and its coordinate representations. The article conflates Euclidean n-space with the set of real n-tuples, whereas I gather more precise treatments distinguish between the set of points (the point set) of Euclidean space, its algebraic structure (conceived either as an affine space with an inner product, or as a manifold), and the various ways of representing its points as coordinate n-tuples in Rn.

If we want to express a directional derivative along some vector v in terms of ∂i, we use vii, where summation is implied. Is that what you meant?

Yes. When we express the directional derivartive of a scalar field f in terms of ∂i f, the symbol seems to denote the set of components ∂i f, i = 1,2,...,n (with basis cotangent vectors understood but not written explicitly), since we need to sum over all of them in vii f, so the directional derivative is expressed in terms of all of them.

(Or from another point of view, "abstract index notation" or "slot-naming index notation", ∂i f is an entity which has n components in any coordinate system, although ∂i f doesn't refer specifically to one particular set of components.)

But then might we not also want sometimes to refer to one particular arbitrary component ∂i f, where i is 1 or 2, or whatever, without specifying which component but also without implying the whole set of components, or is that never a necessary distinction to make?

In his Lecture Notes on General Relativity, Sean Carroll writes

The fact that the gradient is a dual vector leads to the following shorthand notation for partial derivatives:

$$\frac{\partial \phi}{\partial x^\mu} \equiv \partial_\mu \phi \equiv \phi_{,\mu}$$

Similarly, in their Introduction to Vector Analysis, Davis and Snider write

The ith component of the gradient of $\phi$ is $\partial_i \phi$.

In Calculus, Berkey and Blanchard use a similar notation, $\textup{D}_{\textbf{u}} f$, for the directional derivative of f along u. I think the partial symbol is used in this way too: $\partial_{\textbf{u}} f$, isn't it?

Footnote: it seems that "gradient", depending on the author, can mean one of various things: (1) a cotangent vector field which acts on a tangent vector to give the directional derivative of a particular scalar field in the direction indicated by the tangent vector; I think this is Carroll's sense; (2) when there's an inner product: the tangent vector isomorphic to gradient1, namely $g(\nabla f,\enspace)$; this is the definition given by Fredrik in #9 and wofsy in #12 https://www.physicsforums.com/showthread.php?t=356954"; Davis and Snider don't introduce the concept of dual spaces and so don't distinguish between gradient1 and gradient2; (3) in Penrose's The Road to reality, gradient seems to be synonymous with what the other books I've seen call a directional derivative, and he uses the term "full gradient" for gradient2.

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da. Which book by William Burke do you recommend first:
Spacetime, Geometry, Cosmology or
Applied Differential Geometry?

dx
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da. Which book by William Burke do you recommend first:
Spacetime, Geometry, Cosmology or
Applied Differential Geometry?

I recommend "Spacetime, Geometry, Cosmology" first.

I recommend "Spacetime, Geometry, Cosmology" first.

Thanks, dx. I ordered it. In fact, I order both, just in case Spacetime is too remedial.

dx
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Thanks, dx. I ordered it. In fact, I order both, just in case Spacetime is too remedial.

Both are excellent. You won't be disappointed.