Recommendaton for Clarifying Special Relativity

[Dale]

No one is saying it does.

Phy_man did.

a reference frame is an equivalence class of coordinate systems that all share the same measures of spatial distances, temporal intervals, speeds, angles, etc.

I have never heard that definition of a reference frame. I have always heard a reference frame described as a smooth set of orthonormal vectors at each point, i.e. a tetrad or vierbien. From the tetrad you could find an equivalence class of coordinate systems which all share the same tetrad as their coordinate basis, but the tetrad itself does not pick out any specific coordinate system, so using the tetrad is not the same as using the coordinate systems.

we've been sometimes referring to reference frames informally as coordinate systems

I do that too . That may be the confusion.

Note that those saying we can dispense with frames are also saying that components have no physical meanings - not just that coordinates have no physical meanings.

I am not saying that. Components are themselves vectors, a given vector can always be expressed as a sum of components. As vectors, components "live" in the tangent space.

Coordinates "live" in the manifold itself, not the tangent space. In general, they do not follow the laws of vector addition. They are not themselves vectors.

You can do physics without coordinates (if you are masochistic), but I don't see any way to do physics without components. Coordinates certainly make things easier, and from a coordinate system it is easy to obtain a tetrad and the associated components whenever needed, but they are indeed not "foundational", so you always have a choice to use them or not.

 

[Dale]

I think we all agree your abstract example doesn't give you a unique coordinate system. But this misses the point of the O&R quote.

That example was a response to Phy_man's assertion that a coordinate system and a set of basis vectors are "identically the same thing". They are clearly not the same thing.

 

[DrGreg]

I think we all agree your abstract example doesn't give you a unique coordinate system. But this misses the point of the O&R quote.

It's not just an issue of uniqueness. There are other examples of tetrads that cannot be associated with any coordinate system.

Consider the frame field associated with a Born-rigid rotating cylinder in flat spacetime. Set the timelike basis vector along the worldlines of all the observers at rest relative to the cylinder and three spacelike vectors orthogonal to that, in the radial, tangential and axial directions. If you choose a single observer you can find a local coordinate system associated with that observer's tetrad only, but you can't find an extended single coordinate system compatible with all of the tetrads everywhere.

 

[TrickyDicky]

It's not just an issue of uniqueness. There are other examples of tetrads that cannot be associated with any coordinate system.

Consider the frame field associated with a Born-rigid rotating cylinder in flat spacetime. Set the timelike basis vector along the worldlines of all the observers at rest relative to the cylinder and three spacelike vectors orthogonal to that, in the radial, tangential and axial directions. If you choose a single observer you can find a local coordinate system associated with that observer's tetrad only, but you can't find an extended single coordinate system compatible with all of the tetrads everywhere.

But of course, they are called nonholonomic or noncoordinate basis and it's an abstract notation frequently used.
But we have been talking here about measuring components, that is, giving numerical values as the mathematical representation of physical tensors(that are geometric objects) and for doing that it seems to me one needs coordinate basis and these components will depend on the coordinate system used. So surely depending on the level of abstraction one can dispense with coordinates but when giving specific numerical values one depends on the specific coordinate system used and as have been commented several times if measuring physical quantities in SR one ultimately is always referring to inertial coordinates.

 

[TrickyDicky]

That example was a response to Phy_man's assertion that a coordinate system and a set of basis vectors are "identically the same thing". They are clearly not the same thing.

Clearly. But you were who equated what O&R said about measuring the components of a physical tensor with setting up a basis when it is not the same thing, the latter is a more abstract and broader concept and doesn't depend on coordinate systems while the former depends on the chosen reference frame(ordered coordinate basis).

 

[Dale]

But we have been talking here about measuring components, that is, giving numerical values as the mathematical representation of physical tensors(that are geometric objects) and for doing that it seems to me one needs coordinate basis and these components will depend on the coordinate system used.

One needs a basis, but not necessarily a coordinate basis.

 

[TrickyDicky]

One needs a basis, but not necessarily a coordinate basis.

And how do you decompose the tensor into specific numerical components with a noncoordinate basis?

 

[Dale]

You use the contraction of the tensor with each basis vector, the same as you would with a coordinate basis. You use a coordinate basis and a non-coordinate basis the same way. The only difference is that you don't relate the basis vectors to changes in a coordinate.

Consider the earlier example of the Born-rigid rotating cylinder observers. Two observers at different points on the cylinder can each have a clock and three rods oriented as described above. Using those clocks and rods they can each measure the four-momentum of a nearby object with known mass. They have no need to relate those measurements to a coordinate system that assigns e.g. different $\theta$ to the different observers.

Remember, the numerical values of the components identify a vector in the tangent space. The numerical values of the coordinates identify an event in the manifold. They are different types of objects in different spaces.

 

[TrickyDicky]

You use the contraction of the tensor with each basis vector, the same as you would with a coordinate basis. You use a coordinate basis and a non-coordinate basis the same way. The only difference is that you don't relate the basis vectors to changes in a coordinate.

Consider the earlier example of the Born-rigid rotating cylinder observers. Two observers at different points on the cylinder can each have a clock and three rods oriented as described above. Using those clocks and rods they can each measure the four-momentum of a nearby object with known mass. They have no need to relate those measurements to a coordinate system that assigns e.g. different $\theta$ to the different observers.

I'm not arguing you cannot contract tensors in an abstract coordinate free way, I asked about physical measures with specific numerical components and you have not answered it.

 

[Bill_K]

I'm not arguing you cannot contract tensors in an abstract coordinate free way, I asked about physical measures with specific numerical components and you have not answered it.

There is nothing "abstract" about the tetrad components of a tensor, TrickyDicky. You're not making sense. Numerical means what, you want them in SI units??

 

[Dale]

I'm not arguing you cannot contract tensors in an abstract coordinate free way, I asked about physical measures with specific numerical components and you have not answered it.

Then I don't understand your question. I thought that I answered it clearly and directly.

Please re-phrase your question.

EDIT: are you aware that the contraction of a tensor with a basis vector is a scalar, i.e. a number? What I described above gives you numbers representing the outcome of the measurement.

 

[TrickyDicky]

Then I don't understand your question. I thought that I answered it clearly and directly.

Please re-phrase your question.

EDIT: are you aware that the contraction of a tensor with a basis vector is a scalar, i.e. a number? What I described above gives you numbers representing the outcome of the measurement.

Exactly, and that scalar is obtained from other scalars that are the coefficients of the components of the basis vector and of the tensor in some coordinate system, they are projections of the geometrical object on the coordinate axis. Otherwise you are nor using numbers, you May be using index abstract notation that refers to abstract slots in the tensor and thia of course requires no coordinate system until you want to apply it to some specific physical problem to obtain a certain number in a calculation.

 

[Dale]

they are projections of the geometrical object on the coordinate axis

No, they are projections of the geometrical object on the basis (which may or may not be a coordinate basis). The coordinate axis is not a basis, in fact, some very common coordinates don't even have a coordinate axis (e.g. what is the r axis in spherical coordinates?).

Again, the numerical values of the components identify a vector in the tangent space. The numerical values of the coordinates identify an event in the manifold. They are different types of objects in different spaces.

 

[TrickyDicky]

No, they are projections of the geometrical object on the basis (which may or may not be a coordinate basis). The coordinate axis is not a basis, in fact, some very common coordinates don't even have a coordinate axis (e.g. what is the r axis in spherical coordinates?).

I was using the common example with the standard basis, didn't mean that coordinates must necessarily be fixed axis. I would have thought you know enough about this to understand what I meant. I like how this is explained in the wikipedia entry for Euclidean vectors decomposition:
"As explained above a vector is often described by a set of vector components that add up to form the given vector. Typically, these components are the projections of the vector on a set of mutually perpendicular reference axes (basis vectors). The vector is said to be decomposed or resolved with respect to that set.
Illustration of tangential and normal components of a vector to a surface.
However, the decomposition of a vector into components is not unique, because it depends on the choice of the axes on which the vector is projected.
Moreover, the use of Cartesian unit vectors such as (x, y, z) as a basis in which to represent a vector is not mandated. Vectors can also be expressed in terms of the unit vectors of a cylindrical coordinate system (rho, phi, z) or spherical coordinate system (r, theta, phi). The latter two choices are more convenient for solving problems which possesses cylindrical or spherical symmetry respectively.

The choice of a coordinate system doesn't affect the properties of a vector or its behaviour under transformations.

A vector can be also decomposed with respect to "non-fixed" axes which change their orientation as a function of time or space. For example, a vector in three-dimensional space can be decomposed with respect to two axes, respectively normal, and tangent to a surface (see figure). Moreover, the radial and tangential components of a vector relate to the radius of rotation of an object. The former is parallel to the radius and the latter is orthogonal to it.[9]

In these cases, each of the components may be in turn decomposed with respect to a fixed coordinate system or basis set (e.g., a global coordinate system, or inertial reference frame)."

Again, the numerical values of the components identify a vector in the tangent space. The numerical values of the coordinates identify an event in the manifold. They are different types of objects in different spaces.

Components of vectors and tensors at points of the manifold (events) is what we have been referring to all the time haven't we? And anyway for Minkowski space the distinction you are drawing is superfluous, both spaces are equivalent(isomorphic).
If V is an n-dimensional vector space over a field F. A choice of an ordered basis for V(since you said you were always referring to ordered basis or frames) is equivalent to a choice of a linear isomorphism φ from the coordinate space Fn to V.

 

[Dale]

I would have thought you know enough about this to understand what I meant.

Clearly I am not very skilled at knowing what you meant when it differs from what you actually wrote. I am not even very good at knowing what you meant when it is what you wrote.

Components of vectors and tensors at points of the manifold (events) is what we have been referring to all the time haven't we? And anyway for Minkowski space the distinction you are drawing is superfluous, both spaces are equivalent(isomorphic).

Even for Minkowski space the distinction is not superfluous. The manifold itself is not a vector space. E.g. (neglecting gravity) what is the vector sum of the events where my wristwatch read 9:00 am last Tuesday in Germany and where my wristwatch read 9:00 am today in the USA?

Furthermore, a coordinate chart, even on Minkowski space, maps an open subset of the manifold onto an open subset of R4. While R4 can easily be made into a vector space, an open subset cannot in general since the usual R4 vector addition can result in an element of R4 which is outside the open subset.

In contrast, the tangent space is a legitimate vector space. Vector addition, contraction, etc. are all legitimate operations on members of the tangent space. E.g. the vector sum of the four-momentum of an electron traveling at .9c in the direction of Andromeda and a proton traveling at .6c in the direction of Sagittarius is well-defined.

By specifying a basis of vectors in the tangent space you can map the tangent space to R4, note this is not an open subset of R4, but the whole R4. The usual vector operations in R4 correspond directly with the usual vector operations in the tangent space.

So there is a clear distinction between the components of a vector in the tangent space and the coordinates of an event in the manifold. They have different mathematical structures, and you don't need to set up coordinates in the manifold to set up a basis in the tangent space.

If V is an n-dimensional vector space over a field F. A choice of an ordered basis for V(since you said you were always referring to ordered basis or frames) is equivalent to a choice of a linear isomorphism φ from the coordinate space Fn to V.

True, but irrelevant. In GR the vector space V is the tangent space and the coordinates map from an open subset of the manifold M to an open subset of R4, not from V to R4.

 

[WannabeNewton]

You never need coordinates to express a single tensor in terms of components relative to some basis for the linear space the tensor comes out of. That's just plain old linear algebra. Now, if you have a frame field on space-time, the individual basis vector fields in the frame field are geometric objects which are independent of any choice of coordinates as usual and the values of the components of tensor fields expressed in terms of the frame field will be independent of the choice of coordinates.

However the individual basis vector fields in the frame field must vary smoothly from point to point and for that you need to pick some coordinate chart because the smoothness criterion for vector fields on a smooth manifold is as follows: let $M$ be a smooth manifold and $X:M\rightarrow TM$ a vector field (more precisely a rough vector field), where $TM$ is the tangent bundle. If $(U,(x^{i}))$ is any smooth coordinate chart on $M$ then $X$ is smooth on $U$ if and only if its component functions with respect to this chart are smooth. Of course the smoothness is independent of the choice of coordinate chart since smooth manifolds have maximal smooth atlases (this is not by definition but rather can be proven to be true based on the definition of a smooth atlas, surprisingly without the use of Zorn's lemma).

 

[TrickyDicky]

Even for Minkowski space the distinction is not superfluous. The manifold itself is not a vector space.

I'm not 100% sure you are not trying to be humorous with this statement. Anyway the definitions of Minkowski spacetime I know like Wikipedia's define it as a four-dimensional real vector space (equipped with a Lorentzian metric).

True, but irrelevant. In GR the vector space V is the tangent space and the coordinates map from an open subset of the manifold M to an open subset of R4, not from V to R4.

This thread is about SR, but even in GR there is no global coordinate system for the manifold so the case is very different to the situation we are discussing.

 

[TrickyDicky]

You never need coordinates to express a single tensor in terms of components relative to some basis for the linear space the tensor comes out of. That's just plain old linear algebra. Now, if you have a frame field on space-time, the individual basis vector fields in the frame field are geometric objects which are independent of any choice of coordinates as usual and the values of the components of tensor fields expressed in terms of the frame field will be independent of the choice of coordinates.

Sure, and nobody so far has claimed that they have to be expressed in a unique or canonical coordinates. That is not the point. I'm stressing frame (ordered coordinates)-independence, but as you very well explain below you need to pick some coordinate chart, it seems to me some people conflates frame-independence with being coordinate-free.

However the individual basis vector fields in the frame field must vary smoothly from point to point and for that you need to pick some coordinate chart because the smoothness criterion for vector fields on a smooth manifold is as follows: let $M$ be a smooth manifold and $X:M\rightarrow TM$ a vector field (more precisely a rough vector field), where $TM$ is the tangent bundle. If $(U,(x^{i}))$ is any smooth coordinate chart on $M$ then $X$ is smooth on $U$ if and only if its component functions with respect to this chart are smooth. Of course the smoothness is independent of the choice of coordinate chart since smooth manifolds have maximal smooth atlases (this is not by definition but rather can be proven to be true based on the definition of a smooth atlas, surprisingly without the use of Zorn's lemma).

 

[Dale]

I'm not 100% sure you are not trying to be humorous with this statement. Anyway the definitions of Minkowski spacetime I know like Wikipedia's define it as a four-dimensional real vector space (equipped with a Lorentzian metric).

If Minkowski spacetime is a vector space then what is the vector sum of the events where my wristwatch read 9:00 am last Tuesday in Germany and where my wristwatch read 9:00 am today in the USA?

 

[robphy]

Euclidean geometry, Minkowski spacetime, and Galilean spacetime are really "affine geometries" (often described as "a vector space that has forgotten its origin"). Indeed, there is no distinguished element in any of these spaces. Thus, one cannot add elements or scalar-multiply... although one can subtract two elements (and get a vector).

As mentioned earlier, the tangent space at each point-event is a vector space.

 

[TrickyDicky]

Euclidean geometry, Minkowski spacetime, and Galilean spacetime are really "affine geometries" (often described as "a vector space that has forgotten its origin"). Indeed, there is no distinguished element in any of these spaces. Thus, one cannot add elements or scalar-multiply... although one can subtract two elements (and get a vector).

As mentioned earlier, the tangent space at each point-event is a vector space.

Very true. Taken as abstract spaces they are affine. Note however that both Euclidean and Minkowskian geometries being homogeneous allow one to freely choose a point as origin and that is what one does in physics when measuring, and in that sense it is accepted to consider them vector spaces and indeed anybody can check Minkowski space is defined as such for instance in Wikipedia and other sources.

 

[WannabeNewton]

Minkowski space-time is just $\mathbb{R}^{4}$ equipped with the bilinear form $\eta_{ab}$. $\eta_{ab}$ doesn't alter the natural vector space structure of $\mathbb{R}^{4}$; it simply adds a pseudo-inner product structure on top of the vector space structure. As a side note, every finite dimensional real vector space is a smooth manifold.

 

[Dale]

both Euclidean and Minkowskian geometries being homogeneous allow one to freely choose a point as origin

"Allow", yes. "Require", no.

If flat spacetime by itself were a vector space then two events would have a well defined vector sum. They don't. The elements of spacetime are events, and events aren't vectors, therefore spacetime is not a vector space.

Furthermore, coordinate charts do not generally form vector spaces either. Consider spherical coordinates. The r coordinate is strictly positive, so multiplying a valid coordinate by a negative number gives a point in R4 which is outside the open subset covered by the chart.

In contrast, the fields we would want to measure in physics are vectors (and tensors and scalars). As such, their components are also vectors, scalar multiples of some basis vectors. So not only can they be described without coordinates, they can only be described with coordinates if you use the coordinates to generate a basis (and obviously that isn't the only way to generate a basis).

 

[robphy]

Very true. Taken as abstract spaces they are affine. Note however that both Euclidean and Minkowskian geometries being homogeneous allow one to freely choose a point as origin and that is what one does in physics when measuring, and in that sense it is accepted to consider them vector spaces and indeed anybody can check Minkowski space is defined as such for instance in Wikipedia and other sources.

So, I found

http://en.wikipedia.org/wiki/Minkowski_space][/PLAIN] [/PLAIN]
Structure

Formally, Minkowski space is a four-dimensional real vector space equipped with a nondegenerate, symmetric bilinear form with signature (−,+,+,+) ...

and

http://en.wikipedia.org/wiki/Minkowski_space][/PLAIN] [/PLAIN]
Alternative definition

The section above defines Minkowski space as a vector space. There is an alternative definition of Minkowski space as an affine space which views Minkowski space as a homogeneous space of the Poincaré group with the Lorentz group as the stabilizer. See Erlangen program.

Note also that the term "Minkowski space" is also used for analogues in any dimension: if n≥2, n-dimensional Minkowski space is a vector space or affine space of real dimension n on which there is an inner product or pseudo-Riemannian metric of signature (n−1,1), i.e., in the above terminology, n−1 "pluses" and one "minus".

Yes, once you pick a point-event to be the origin, you now have a vector space (the set of displacement vectors from your choice of origin)... without singling out a point-event, you [still] have the affine structure.

Since the universe doesn't distinguish any particular event in Minkowski spacetime [e.g. the game-winning goal at the recent Stanley Cup finals], I prefer not to impose that structure in the mathematical model... until necessary.

(One lesson I learned when studying physics is that it's good to know the MINIMAL structure needed to obtain something. Once you toss everything in [e.g. symmetries, choice of dimensionality, signature]... it's hard to see WHERE a particular feature comes from. This is important (e.g.) in quantum gravity where one tries to deconstruct what we observe and try to find the right generalizations to extend the classical theory to a quantum one.tangentially-related anecdote... My favorite math professor sternly corrected a student who said "a square is a parallelogram with four right-angles" by saying "a square is a parallelogram with ONE right-angle". This was an aha-moment for me.)

 

[Vanadium 50]

Pretty much every possible point has been made (often more than once) and we're just covering the same ground again and again. Also this thread has become a magnet for sockpuppets of banned members and other crackpots, so this thread is closed.