negative distance

[bored_teen]

would negative distance occur in a theoretical situation where the length of a straight line between two points is greater than that of a non-straight line?

[insane_alien]

no. distance is always positive. displacement however can be negative as is also has direction as well as magnitude.

[Sayonara]

A straight line is by definition the shortest possible route between two points, so no route can be shorter. This means that if you have a straight line which is longer than a non-straight line between the same two points, then something has to be "false" about the straight line. For example: it leaves Point A, goes through a wormhole, and approaches point B from the other side of the room. In which case it is not what you would call "the" straight line between the two points, but it is certainly "a" straight line which is longer than any number of non-straight lines between A and B.

 

Whether or not you have negative distance involved depends on how you measure the scenario - it's a matter of differing perspective rather than a difference in what you are actually measuring. You can't, for instance, have an anti-metre; however you can have -1m, which is simply one metre away from where you start measuring, in the opposite direction to what you consider to be "forwards". That would be the displacement, or translation, which is just the difference between points in a co-ordinate system, so the signage (positive/negative) is arbitrary. But whatever system you use has to be consistent.

[ajb]

You can have "negative distance". This kind of thing is found in special and general relativity. Mathematically a metric is how one defines the distance between near by points. Usually, the metric is defined as to always give positive distance. However, you can relax this and this is exactly what is needed in relativity.

[alan2here]

where the length of a straight line between two points is greater than that of a non-straight line?

For example?

[Bignose]

For example?

 

Any space with a null metric line running through it. It may be faster to go to the null line -- travel up the null line (which by definition takes no distance) -- and then travel to the second point than to take the straight line between points.

 

For example, imagine a space with a null metric line running along the x=y, 45 degree line. To travel from (3,3.5) to (0,0) you can either go directly to (0,0) for a total of 4.609 units or you can travel to any point on that line, say (3,3) and then travel the null line to (0,0). I.e. the distance between (3,3) and (3,3.5) is 0.5 In this case, the shortest distance between (3,3.5) and (0,0) is along the line that intersects (3,3.5) and is perpendicular to x=y.

 

Depending on the metric of the space, straight lines aren't always fastest. On the surface of a sphere, the shortest distances are great circles. On the surface of a cylinder, the shortest distances are helices. It all depends on the metric of the space.

[alan2here]

Any space with a null metric line running through it. It may be faster to go to the null line -- travel up the null line (which by definition takes no distance) -- and then travel to the second point than to take the straight line between points.

I don't really understand that but I think you are referring to a shortcut like a teliport.

 

In which case yes. A wall that you can't move though with a bridge over the wall would also achieve the same effect. Only in this example a straight line is still the quickest way of getting between the two points even if it is not possible to take it.

[ajb]

Bignose: How do you define a straight line on a curved surface?

 

But what Bignose says is correct and basically what I said. If you use a pseudo-Riemannian metric then you can have positive, negative and zero distances between distinct points.

[Pete]

Bignose: How do you define a straight line on a curved surface?

 

But what Bignose says is correct and basically what I said. If you use a pseudo-Riemannian metric then you can have positive, negative and zero distances between distinct points.

I disagree. Distance is positive by definition. The magnitude of the fundamental force is correctly called the interval between two points. The term distance only refers to proper distances and even then only certain circmstanes. In relativity a lot of people refer to the square root of the inteval as distance but when this is done it is really a misnomer.

 

Consider special relativity as an example: The first fundamental form is defined as

 

[math]ds^2 = g_{\alpha \beta}d^{\alpha} d^{\beta}[/math]

 

Some physicists call this the metric where g is the metric tesor and [math]g_{\alpha \beta}[/math] are the components of the metric tensor. The term first fundamental form is most often used in differential geometry texts/papers. I usually choose the metric in Minkowski coordinates as

 

g = diag(1, -1, -1, -1)

 

Consider the 4-position 4-vector X = (ct, x, y, z) = (ct, r) where r = (x, y, z). The spatial distance, D, between two events is defined as

 

[math]D = \sqrt{x^2 + y^2 + z^2}[/math]

 

The proper distance is defined only for two events which have a spacelike spacetime seperation. In such case the proper distance, [math]\lambda[/math], is given by

 

[math]\lambda = \sqrt{-s^2}[/math]

 

Pete

[ajb]

Maybe I should have put distance in quotation marks. But the fact remains we have time-like, light-like and space-like separation of points in Minkowski space.

[Pete]

Maybe I should have put distance in quotation marks.

That would be a wise thing to do.

 

I posted that message because I was merely trying to clarify the terminology a bit. The idea of calling the quantity [math]\Delta s[/math] in

 

[math]\Delta s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2[/math]

 

distance has always be a pet peeve of mine. When it is referred to as such it becomes difficult to explain to the layman that it really isn't the distance between the two events, hence the quotations. I believe that some authors of SR texts ractually do refer to it as "distance." It leads people to believe that an object at rest "moves" a finite distance even though it is at rest. It gives rise to phrases like "moving through spacetime" and "the speed of a tardyon through spacetime is always c" which only serves to confuse people in my humble opinion.

 

Such a choice of words Proper distance between two events can't always be defined whereas the actual distance can always be defined. The fact that it is sometimes impossible to define "distance" this way only serves to confuse the layman.

But the fact remains we have time-like, light-like and space-like separation of points in Minkowski space.

While the physicist understands this very well it becomes hard for the layman to understand this idea.

 

I have to think about this some more so please make sure that it hasn't changed during the time you respond to it. Thanks.

 

Best wishes

 

Pete

[D H]

I disagree. Distance is positive by definition.

I agree with Pete here. Calling the result of some operator a "distance" when the operator doesn't satisfy the definitions of a distance operator is abuse of terminology. For some operator [math]d(x,y),\,x,y\in X[/math] to qualify as a "distance" operator (aka a metric), the operator must be

1. positive-definite,
2. symmetric, and
3. satisfy the triangle inequality [math]d(x,z)\le d(x,y)+d(y,z)\,\forall x,y,z\in X[/math]

 

Some physicists and mathematicians use the term "Cartesian tensor" to describe things that live in a Cartesian space and look a bit like tensors. Some use the term "Cartesian-tensor" to describe these things because Cartesian tensors are not tensors. A Cartesian-tensor has some, but not all, of the characteristics of a tensor. Similarly, a metric tensor is not a metric. It has some, but not all, of the characteristics of a metric. In particular, a metric tensor satisfies the condition for a metric only for points that are infinitesimally close to one another. Integrating a metric tensor yields a metric.

 

The Minkowski metric is not even a metric tensor. It is a pseudo-Reimannian metric tensor. A pseudo-Reimannian metric tensor has some properties that make it look kind of like a metric tensor -- except it isn't positive definite and doesn't satisfy the triangle inequality. It might still have some useful mathematical properties, but a pseudo-Reimannian metric tensor isn't a metric tensor, and a metric tensor is not a metric. The operator called the Minkowski metric would be better named the Minkowski pseudo-Reimannian metric tensor. Since that is too much to say, we say Minkowksi metric for short. It is, however, always good to remember that this is abuse of terminology.

[Pete]

The spatial distance, D, between two events is defined as

 

[math]D = \sqrt{x^2 + y^2 + z^2}[/math]

I must say that its a bit irritating not being able to go back and correct a post. In the present case the error here is that D is the distance betwen an event located at the origin of the coordinate system and an event located at the point (x, y, z).

 

Pete

[ajb]

I am quite happy to refer to such things as a distance, but the I know what is meant by that. The truth being that in geometry (relativity) the (pseudo-)Riemannian metric is used to classify elements in the tangent bundle over a point on the space-time. Locally, this gives you the notion of "separation distance" between two near by points. In geometry, the "distance" between near by points does not (canonically) come for free. You have to specify how this is done, i.e. you need to define a metric.

 

In relativity, it appears necessary to loosen the definition of a metric to a pseudo-metric, that is you lose the positive definite property. From a geometric point of view (at least locally) this is not such a big jump. I don't see any real difficulty here. (Though, not all manifolds can be equipped with a Lorentzian signature pseudo-metric).

 

If we just restrict ourselves to special relativity, we are defining "distance" to be the geometric invariant associated to the Lorentz group. Therefore, I think it is being pedantic to insist on "distance" being strictly positive. Maybe you want to call it pseudo-distance or something?

 

I have never come across the term "Cartesian tensor" before. How are they defined?

[Pete]

Some physicists and mathematicians use the term "Cartesian tensor" to describe things that live in a Cartesian space and look a bit like tensors. Some use the term "Cartesian-tensor" to describe these things because Cartesian tensors are not tensors. A Cartesian-tensor has some, but not all, of the characteristics of a tensor.

I disagree. Why do you believe that Cartesian tensors are not tensors?

 

Cartesian tensors are most definitely tensors because they fit the definition of a tensor in the most rigorous sense, i.e. it has all the characteristics of a tensor. What characteristic do you believe it doesn't have?

 

A Cartesian vector can be defined as a geometric quantity whose components transorm in a certain way (i.e. tensorially) under an orthogonal transformation.

 

A Cartsian tensor can be defined in two different, but equivalent, ways. One way a Cartesian tensor can be defined is as a multilinear map from 1-forms and Cartesian vectors into real numbers (i.e. scalars). They can also be defined by how their components transform under an orthogonal transformation.

 

A Lorentz 4-vector is defined as a geometric quantity whose components transorm in a certain way (i.e. tensorially) under a Lorentz transformation.

 

A Lorentz tensor can be defined as a multilinear map from 1-forms and Lorentz 4-vectors into real numbers (i.e. scalars). They can also be defined by how their components transform under a Lorentz transformation.

If we just restrict ourselves to special relativity, we are defining "distance" to be the geometric invariant associated to the Lorentz group. Therefore, I think it is being pedantic to insist on "distance" being strictly positive. Maybe you want to call it pseudo-distance or something?

If one defines "distance" in that way then I wholeheartedly agree that there is absolutely no reason to insist on it being positive. I sure hope I didn't give you the impression that I thought otherwise?

I have never come across the term "Cartesian tensor" before. How are they defined?

You've probably seen/used them but they may not have been called that. A Cartesian tensor is a geometric quantity whose components transform tensorially under an orthogonal transformation. If you have Jackson EM text then he defines them there. I believe that he calls them rotational tensors though. They are also defined in Goldstein. I gave a definition of them in one of my web pages, i.e. in

 

http://www.geocities.com/physics_world/gr_ma/tensors_via_analytic.htm

 

The following are Cartesian tensors

 

1) Stress tensor

2) Moment of Inertia tensor --

http://www.geocities.com/physics_world/mech/inertia_tensor.htm

3) Tidal force tensor -- Moment of inertial tensor

http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm

4) Metric tensor

5) Maxwell's stress tensor --

http://scienceworld.wolfram.com/physics/MaxwellStressTensor.html

6) Stress tensor

 

Pete

Edited July 26, 2008 by Pete
multiple post merged

[ajb]

So they are only "tensors" under the orthogonal group [math]O(3)[/math] which is a subgroup of [math]Diff(E)[/math], where [math]E[/math] is the 3 dimensional Euclidean space?

 

So in particular, all tensors are Cartesian tensors, but not vice versa. So, being a Cartesian tensor is much more restrictive than being a tensor.

 

In exactly the same way you could define Lorentz tensors as tensor objects under the Lorentz transformations. So again, all tensors are Lorentz tensors, but not all Lorentz tensors are tensors.

 

It makes me wonder if "most natural examples" of Cartesian and Lorentz tensors are actually tensors. Have we got any examples to hand?

 

 

Actually, this is not a million miles away from some of my research work. But kind of in reverse. I am thinking about tensors on a manifold [math]M[/math] not only via there transformations under [math]Diff(M)[/math], but under a larger group that contains [math]Diff(M)[/math].

 

On a small pedantic issue, there is no well established definition of a "geometric object". They are defined by there action under [math]Diff(M)[/math] or a subgroup of it. Tensors carry a linear representation, other objects may have more complicated representations.

Edited July 26, 2008 by ajb

[Pete]

So they are only "tensors" under the orthogonal group [math]O(3)[/math] which is a subgroup of [math]Diff(E)[/math], where [math]E[/math] is the 3 dimensional Euclidean space?

Yes. Please correct me if I'm wrong but that is true of all tensors, i.e. that they pertain to a certain class of transformations.

So in particular, all tensors are Cartesian tensors, but not vice versa.

Huh? I don't understand!? I'd say that all Cartesian tensors are tensors but not all tensors are Cartesian tensors.

So, being a Cartesian tensor is much more restrictive than being a tensor.

Exactly.

In exactly the same way you could define Lorentz tensors as tensor objects under the Lorentz transformations. So again, all tensors are Lorentz tensors, but not all Lorentz tensors are tensors.

Again, me being confused ->

 

It seems to me that all Lorentz tensors are general tensors but not all general tensors are Lorentz tensors. I've always been confused as to what types of transformations are allowed in general tensors in general relativity though. Do you have an answer to that?

It makes me wonder if "most natural examples" of Cartesian and Lorentz tensors are actually tensors. Have we got any examples to hand?

Do you mean besides the examples I gave above??

 

Pete

[ajb]

A tensor on a manifold [math]M[/math] is defined as a "geometric object" which carries a (passive) linear representation of [math]Diff(M)[/math]. (This is the best way I can think of defining them). That is the standard "rules" for changing coordinates.

 

So, all tensors transform as a "tensor" under any subgroup of [math]Diff(M)[/math] Try for example [math]0(N)[/math]. Consider infinitesimal transformations that are (strictly) linear. You will see that all tensors transform as "tensors".

 

If we define a "sub-tensor" to be an object that carries a (passive) linear representation of a subgroup of [math]Diff(M)[/math] then is it true that these sub-tensors are tensors? I don't think so

 

So all tensors are Cartesian, but all Cartesian tensors are not tensors. The space of tensors sits inside the space of Cartesian tensors. So, what I should have said is that being a tensor is more restrictive than being a Cartesian tensor. My mistake.

 

I am just wondering if the examples you gave are in fact simply tensors.

[Pete]

A tensor on a manifold [math]M[/math] is defined as a "geometric object" which carries a (passive) linear representation of [math]Diff(M)[/math]. (This is the best way I can think of defining them). That is the standard "rules" for changing coordinates.

Sorry but you lost me.

 

Do you happen to know of a text which actually gives that as a definition ver batim?

 

You're using terminology that I'm not familiar with (either I haven't learned it yet or a Iearned it and forgot it). What is "[math]Diff(M)[/math]"? What is a (passive) linear representation of [math]Diff(M)[/math]?

 

If you could dirtect me to a text which explains this then I can take it from there. Does Wald use this terminology that you've used here?

 

Regarding the following

So all tensors are Cartesian, but all Cartesian tensors are not tensors.

Can you please give me an example of a Cartesian tensor which is not a tensor? That would be strange indeed because the Cartesian tensor a tensor by definition.

 

Pete

[ajb]

[math]Diff(M)[/math] is the diffeomorphism group of the manifold [math]M[/math]. It consists of all smooth invertible maps [math]f: M \rightarrow M [/math].

 

In local coordinates you can express this as [math] f: x^{\mu} \rightarrow y^{\alpha}[/math]. More explicitly that means [math]f^{*}(y^{\alpha}) = y^{\alpha}(x)[/math].

 

There are two representation of [math]Diff(M)[/math] that need to be considered. Passive, which is just changing the (local) coordinates at a point and active which is really mapping the point to another point.

 

The standard way to define a tensor is to define it (the components really) as carrying a linear representation of the passive diffeomorphisms. This is just the change of coordinate rule I assume you know. For example for vectors the components change as [math]\overline{X}^{\alpha} = X^{\mu}\frac{\partial y^{\alpha}}{\partial x^{\mu} }[/math]. And so on for higher degree tensors.

 

 

A Cartesian tensor is not a genuine tensor. It need only transform linearly with respect to [math]O(n)[/math] and not [math]Diff(M)[/math]. So all tensors are Cartesian tensors, but all Cartesian tensors are not tensors.

 

Wald may not use the exact language I have used, but he will have defined tensors by the change of (local) coordinates.

 

One salient point here is that this definition is implemented point by point on a manifold. I guess this makes it not so clear how we can say that two objects are "tensors" (or more generally geometric objects) by comparing how they transform at two different points. (Maybe this suggests something more like a groupoid should be used to define tensors? But never mind.)

 

As for an example, I think we could construct something. But I am not sure what "natural" examples there are. Let me think about it.

[D H]

So they are only "tensors" under the orthogonal group [math]O(3)[/math] which is a subgroup of [math]Diff(E)[/math], where [math]E[/math] is the 3 dimensional Euclidean space? So in particular, all tensors are Cartesian tensors, but not vice versa.

That's exactly right.

 

So, being a Cartesian tensor is much more restrictive than being a tensor.

But that's exactly wrong. Think of it this way: *2=3+ is a string of mathematical symbols. It is not, however, a well-formed formula in the elementary algebra. All well-formed formulae can be expressed as a string of symbols, but not all strings of symbols are well-formed formulae. The concept of a well-formed formula is much more restrictive than the concept of a string of symbols.

 

Similarly, the concept of a Cartesian tensor is much less restrictive than the concept of a tensor. For example, an absolute position vector is a Cartesian tensor but it is not a tensor.

[ajb]

That's exactly right.

 

 

But that's exactly wrong. Think of it this way: *2=3+ is a string of mathematical symbols. It is not, however, a well-formed formula in the elementary algebra. All well-formed formulae can be expressed as a string of symbols, but not all strings of symbols are well-formed formulae. The concept of a well-formed formula is much more restrictive than the concept of a string of symbols.

 

Similarly, the concept of a Cartesian tensor is much less restrictive than the concept of a tensor. For example, an absolute position vector is a Cartesian tensor but it is not a tensor.

 

Yes, I realised my mistake. See my post after that one.

 

I have got a simple example.

 

[math]\frac{\partial X^{\mu}}{\partial x^{\nu}}[/math] where [math]X^{\nu}[/math] is the component of a vector.

 

This is a Lorentz (or O(N)) tensor but is not a tensor under general coordinate changes.

Edited July 26, 2008 by ajb

[Pete]

Before I go on let me state that, at least for the moment (until I am convinced otherwise) I firmly hold to the idea that a Cartesian tensor is a tensor in the most rigorous sense. If I'm being slow here its because I'm not used to the terminology that you are using and because you seem to be defining the term tensor in a manner which is different than every source of definition that I'm aware of. I therefore have to sort out this terminology and notation so I can determine if what you have defined here is identical to the definition of a tensor of rank (j, k) as a multilinear map of j dual vectors and k vectors to real numbers. This is the definition of the term tensor as it is normally found in the relativity and differential geometry/tensor analysis literature.

 

For this reason I ask for your continued patience.

[math]Diff(M)[/math] is the diffeomorphism group of the manifold [math]M[/math]. It consists of all smooth invertible maps [math]f: M \rightarrow M [/math].

This is where I have a problem with the definition that you are describing. You are adding in the extra criteria that the relavent coordinate transformations are all the diffeomorphisms in [math]Diff(M)[/math]. This is not part of the definition of tensor as found in texts that I'm aware of.

 

Recall your comment above

This is a Lorentz (or O(N)) tensor but is not a tensor under general coordinate changes.

I disagree with this view. I believe the correct phrase to be used here is as follows -The Lorentz tensor is not a general tensor, but it is still a tensor, i.e. it is a Lorentz tensor. The name given to a tensor is related to the type of coordinate transformations which apply. This is a very important difference.

 

As for the reason why the not all diffeomorphisms can be used in the set of coordinate transformations. The physical problems for which tensors were created/defined to be useful in have a restriction on the kinds of transformations which can be used. For instance, if you made the attempt to apply any sort of coordinate transformation to the components of the Minkowski metric you will find that there will be coordinate transformations which do not leave the spacetime interval invariant. I believe that this is an important restriction on what qualifies as a valid coordinate transformation. As a simple example recall that you can't use the Galilean transformation to transform the components of the Minkowski metric since this will not leave th spacetime interval invariant.

 

It is for reasons like this that the term allowable coordinate transformation is used. In the example above the Galilean transformation is not an allowable transformation.

The standard way to define a tensor is to define it (the components really) as carrying a linear representation of the passive diffeomorphisms.

I disagree. I know of no source which allows all diffeomorphisms to be used in coordinate transformations.

So they are only "tensors" under the orthogonal group [math]O(3)[/math] which is a subgroup of [math]Diff(E)[/math], where [math]E[/math] is the 3 dimensional Euclidean space?

I disagree with this too for the reasons stated above. You have been thinking of tensors as 'being tensors only under a given group of transformation' which is not how this works. A geometrical object is defined according to whether the components transform tensorially or not. The type of tensor is defined by the type of allowable coordinate transformations used. Thus a geometric object is not a tensor under this group of transformations etc. The correct phrase is that these geometrical objects are say, Lorentz tensors while these other geometrical objects are, say,Cartesian tensors.

On a small pedantic issue' date=' there is no well established definition of a "geometric object".

[/quote']

I believe that this is one of those things where the term is taken as undefined and only certain properties attributed to it. E.g. the term point is undefined but we assign it the property that it has no spatial extension.

 

I forgot to mention that the generalization of the Cartesian tensor is called the Affine tensor. These are geometric objects whose components transform under a generalized (arbitrary dimensional space) orthogonal transformation. A Cartesian tensor is therefore a specfic example of an affine tensor.

 

 

Please keep in mind that the notion of allowable coordinate system is somewhat ambigous in the literature. The question on what kinds of coordinate transformations hold does not seem to have a unique answer. Please keep this in mind when you consider my comments above regarding allowable coordinate systems. Thanks.

 

Best wishes (and thanks for your patience)

 

Pete

Edited July 26, 2008 by Pete

[ajb]

Before I go on let me state that, at least for the moment (until I am convinced otherwise) I firmly hold to the idea that a Cartesian tensor is a tensor in the most rigorous sense. If I'm being slow here its because I'm not used to the terminology that you are using and because you seem to be defining the term tensor in a manner which is different than every source of definition that I'm aware of. I therefore have to sort out this terminology and notation so I can determine if what you have defined here is identical to the definition of a tensor of rank (j, k) as a multilinear map of j dual vectors and k vectors to real numbers. This is the definition of the term tensor as it is normally found in the relativity and differential geometry/tensor analysis literature.

 

Ok, so this is the definition of a tensor at a point. I am using sloppy terminology here and mean explicitly by a tensor a tensor field. Either way, in geometry one needs to worry about changing the coordinates. Equivalent to the definition you have given, is the definition as an "object" that changes in a particular way under changing the coordinates. (More precisely, tensors are invariant, but the components and basis change in a particular way).

 

To me this definition is the most "geometric" and is the easiest to generalise to things like supermanifolds etc.

 

For this reason I ask for your continued patience.

 

This is where I have a problem with the definition that you are describing. You are adding in the extra criteria that the relavent coordinate transformations are all the diffeomorphisms in [math]Diff(M)[/math]. This is not part of the definition of tensor as found in texts that I'm aware of.

 

Recall your comment above

 

If we work in the category of smooth manifolds, then this is part of the definition. In (classical) physics, it is usually reasonable to assume this. By differential geometry you really do imply you are working in this category. Usually the word manifold is usually synonymous with differential (smooth) manifold unless otherwise stated.

 

If one works outside this then we could have coordinate changes that are not smooth. I agree, but then calculus would become difficult and as physics is mostly calculus...

 

I disagree with this view. I believe the correct phrase to be used here is as follows -The Lorentz tensor is not a general tensor, but it is still a tensor, i.e. it is a Lorentz tensor. The name given to a tensor is related to the type of coordinate transformations which apply. This is a very important difference.

 

Tensor with no other label attached to it formally means a tensor under smooth changes of coordinates, i.e. passive representation of [math]Diff(M)[/math].(Remember we are working on smooth manifolds). But ok, it is quite all right to call something a Lorentz tensor. We all know what that means.

 

For example saying [math]x^{\mu}[/math] is a vector makes me cringe. It is not a vector, but calling it a Lorentz vector I will accept.

 

As for the reason why the not all diffeomorphisms can be used in the set of coordinate transformations. The physical problems for which tensors were created/defined to be useful in have a restriction on the kinds of transformations which can be used. For instance, if you made the attempt to apply any sort of coordinate transformation to the components of the Minkowski metric you will find that there will be coordinate transformations which do not leave the spacetime interval invariant. I believe that this is an important restriction on what qualifies as a valid coordinate transformation. As a simple example recall that you can't use the Galilean transformation to transform the components of the Minkowski metric since this will not leave th spacetime interval invariant.

 

I think we might have to be careful about passive and active.

 

The metric

 

[math]\eta = \eta_{\mu\nu}\delta x^{\nu}\otimes\delta x^{\mu}[/math]

 

is invariant under arbitrary changes of coordinates, that is passive transformations.

 

What Lorentz invariance is really is the isometries of Minkowski space. That is you are looking for diffeomorphism [math]f[/math] such that [math]f^{*}\eta = \eta[/math] under an active transformations. Isometries preserve the length of a vector, that is important. You should also note that the set of all isometries forms a group.

 

So for example, consider Euclidean space for simplicity. The isometries are of the form [math] f: x \mapsto Ax + a[/math] where [math]A\in SO(n)[/math] and [math]a[/math] is an arbitrary vector.

 

Passing to Minkowski space, we get the Poincare group. This should be understood as the isometry group of Minkowski space.

 

Now armed with this, we can now use Poincare/Lorentz transformations as either passive or active.

 

It is for reasons like this that the term allowable coordinate transformation is used. In the example above the Galilean transformation is not an allowable transformation.

 

I disagree. I know of no source which allows all diffeomorphisms to be used in coordinate transformations.

 

See above and the distinction between a mere coordinate change and a "real" diffeomorphism.

 

I disagree with this too for the reasons stated above. You have been thinking of tensors as 'being tensors only under a given group of transformation' which is not how this works. A geometrical object is defined according to whether the components transform tensorially or not. The type of tensor is defined by the type of allowable coordinate transformations used. Thus a geometric object is not a tensor under this group of transformations etc. The correct phrase is that these geometrical objects are say, Lorentz tensors while these other geometrical objects are, say,Cartesian tensors.

 

I believe that this is one of those things where the term is taken as undefined and only certain properties attributed to it. E.g. the term point is undefined but we assign it the property that it has no spatial extension.

 

Best wishes (and thanks for your patience)

 

Pete

 

Same as above.

[Pete]

I forgot another quantity that is referred to as a tensor and that's the susceptibility tensor. Perhaps you've heard of it? It applies to crystals and relates the electric field vector to the polarization vector. I'm not sure if this is a true tensor or not though.

 

Pete