Vectors & Scalars: Understanding Directional Properties

[StephenPrivitera]

When we discuss vector properties in one dimension, most texts will drop the i, j and k notation and use instead positive and negative signs to denote direction. A scalar value to me is an absolute value. It should have no sign whatsoever. No one can ever have a "negative" speed because that would imply a direction. So why is it that scalar properties like work and potential energy can have negative and positive signs? The concept "negative work" implies to me that work is done in the direction opposite motion (if motion is in the positive direction). Friction does negative work. Since work is associated with a force, and force is a vector quantity, it would almost make sense to me to assign a direction to work. Clearly, "3J" of work is distinct from "-3J" of work - both have different consequences on the motion of the body of interest. If we redefine the axes so that motion is along the negative axis and the force is directed positively, then we still have a negative value for work.
If an object has -3J of gravitational potential energy, this means the object has lost 3J of energy due to its movement in the gravitational field. The problem seems similar to the one with work. Change in PE is associated with a change in position, and position is a vector. This merely denotes that the object's PE is three units to the left of some arbitrary origin. This idea doesn't really denote a direction for the energy (eg. 3J @ NE! - which wouldn't make sense) but it does give the direction of energy transfer (from greater to lesser values of potential energy). So my question is this: how can the negative (or positive) sign be interpreted as giving the direction of a vector quantity sometimes, but not always?

 

[mathman]

The distinction you are making seems to be between a one dimensional vector and an absolute value. What one calls a scalar is a matter of convention, more than anything else. The most common usage is to identify scalar with one dimensional vector, i. e. both positive and negative numbers. As an aside, mathematicians sometimes use complex numbers as scalars in certain situations.

 

[chroot]

This is purely a matter of semantics. A one-dimensional vector IS a scalar.

- Warren

 

[Hurkyl]

A scalar value to me is an absolute value. It should have no sign whatsoever. No one can ever have a "negative" speed because that would imply a direction.

I think the concept you are thinking of is a magnitude, not a scalar. In typical settings, magnitudes are always scalars, but not vice versa.

 

[StephenPrivitera]

Originally posted by Hurkyl
I think the concept you are thinking of is a magnitude, not a scalar. In typical settings, magnitudes are always scalars, but not vice versa.

This makes sense. The fact that speed is not negative is due to the fact that speed is defined as the magnitude of the velocity. I tended to think of speed as the "scalar form" of velocity, distance as the "scalar form" of displacement.
But to help clear things up further, can anyone give me a precise definition of "scalar" and "vector?"

 

[HallsofIvy]

As far as "potential energy" and "work" are concerned, they can be positive or negative because only the change in either has any physical meaning. You can choose the 0 point to be anything you like.

 

[Hurkyl]

Well, part of the problem is that there are two meanings for the term "scalar" (which I had forgotten when I wrote my post)


The first definition is effectively what I gave before. The term "scalar" appears in the definition of a vector space:

A vector space is the tuple (V, F, *) where V is an additive group whose elements we call vectors and F is a field whose elements we call scalars and * is a binary operation from F x V -> V we call scalar multiplication such that... (snip rest of definition)

So if r is a scalar and v is a vector, then we can perform scalar multiplication and compute r * v.


The vector space Rⁿ is an abbreviation for (Rⁿ, R, *); the field of scalars for Rⁿ is simply the real numbers R.



For the second meaning, I'm less sure about the precise definition, so I'll cut / paste from mathworld:

A one-component quantity which is invariant under rotations of the coordinate system.

Mathworld doesn't tend to give completely rigorous definitions, though, so I'm not entirely sure if scalars by this definition have to be scalars by the previous definition... but usually the only scalars we care about are R or C, so the question is generally moot in practice.


Anyways, a magnitude is defined to be the norm of a vector, and norms are defined to be nonnegative real numbers.

 

[pmb]

StephenPrivitera asked

When we discuss vector properties in one dimension, most texts will drop the i, j and k notation and use instead positive and negative signs to denote direction. A scalar value to me is an absolute value. [...] So my question is this: how can the negative (or positive) sign be interpreted as giving the direction of a vector quantity sometimes, but not always?

You've misinterpreted the "-" sign in the velocity example you gave. For example: If the velocity vector is V = -3i

The vector is -i. The number 3, in this case '3' is a sca;ar which is the 'magnitude' of the velocity, multiplies that vector which gives the direction.

By definition - In general - A scalar is called a "tensor of rank zero." It's a one component object which remains unchanged upon a change in coordinates.

If it's a "Cartesian" scalar then it's any number which remains unchanged upon a rotation of the Cartesian coordinate system. If it's a Lorentz scalar it's any number which remains unchanged upon a Lorentz transformation.


I explained this to "DavidW" back a while ago when he had serious trouble understanding this. I put it on the web

www.geocities.com/physics_world/scalar.htm

However the definition I've given above is used throughout physics and mathematics. Especially in relativity.

A scalar is also called "an invariant" for obviouss reasons.

Pete

 

[Tyger]

People often use the word scalar

to designate the fourth component of a four vector, which must greatly confuse others who are just learning about physics and relativity. Since boosts and turns are related by being part of Poincare's Group one needs to learn whether this is what is meant when someone says scalar.

Electric charge has the interesting property that it is a "true" scalar, invariant under boosts and turns, and at the same time is the fourth component of a four vector. Energy on the other hand just acts as a fourth component.

 

[Sting]

A scalar is called a "tensor of rank zero." It's a one component object which remains unchanged upon a change in coordinates.

You beat me to it

To embellish this concept a little further, a vector is a tensor of rank one.

 

[StephenPrivitera]

Originally posted by Sting
You beat me to it

To embellish this concept a little further, a vector is a tensor of rank one.

Excuse my ignorance. Does anyone care to explain what a tensor is?

 

[Hurkyl]

A tensor is (to gloss over a lot of details) a thing that does not depend on a coordinate system.

For example, "The first coordinate of the vector v" is not a scalar, because if we changed coordinate systems (say... by rotating them) the first coordinate of v would be something different.


However, "4" is a scalar, because if we change coordinate systems, "4" is still "4".


Similarly, the vector <1, 2, 3> is not a tensor because if we change coordinate systems, <1, 2, 3> will refer to a different vector.

However, "The displacement from point A to point B" is a tensor (in Euclidean geometry), because if we change coordinate systems, the vector in the new coordinates is still "The displacement from point A to point B".


(note that "vector" as used in the context of tensors is more restrictive than "vector" used in the context of vector spaces in general)

 

[quantumdude]

Originally posted by chroot
This is purely a matter of semantics. A one-dimensional vector IS a scalar.

I don't think so. The way I learned this formally was in QM and GR courses, so may not be the way mathematician's prefer, but I was taught to think of vectors as being defined in terms of a set of transformations. For instance, vectors in Euclidean 3-space are defined by their transformation properties under rotation and parity.

Specifically talking about parity, a vector x transforms as:

&Pi;x=-x

whereas a scalar S transforms as

&Pi;S=S.

This would distinguish vectors and scalars even in the 1D case.

 

[pmb]

Originally posted by Tom
I don't think so. The way I learned this formally was in QM and GR courses, so may not be the way mathematician's prefer, but I was taught to think of vectors as being defined in terms of a set of transformations. For instance, vectors in Euclidean 3-space are defined by their transformation properties under rotation and parity.

Specifically talking about parity, a vector x transforms as:

&Pi;x=-x

whereas a scalar S transforms as

&Pi;S=S.

This would distinguish vectors and scalars even in the 1D case.

In classical mechanics, special and general relativity, tensor analysis and differential geometry - "scalar" is a quantity which remains unchanged by a coordinate transformation.

Pmb

 

[DavidW]

Originally posted by pmb

I explained this to "DavidW" back a while ago when he had serious trouble understanding this. I put it on the web

www.geocities.com/physics_world/scalar.htm

Your site is a crank site, not a valid reference. You need a graduate education before you may write about graduate level matters such as general relativity. I wrote the textbook that explains it correctly.
http://www.geocities.com/zcphysicsms/

 

[DavidW]

Originally posted by pmb
In classical mechanics, special and general relativity, tensor analysis and differential geometry - "scalar" is a quantity which remains unchanged by a coordinate transformation.

Pmb

Scalar is just something with magnitude and no direction. Invariant is the term you are confusing with scalar. As I already explained to you, invariants are scalars, but not all scalars are invariant. Energy is an example of a variant scalar. Charge is an example of an invariant scalar.

 

[quantumdude]

Originally posted by pmb
In classical mechanics, special and general relativity, tensor analysis and differential geometry - "scalar" is a quantity which remains unchanged by a coordinate transformation.

Pmb

Right. And parity is one of those coordinate transformations.

 

[quantumdude]

Originally posted by DavidW
Scalar is just something with magnitude and no direction. Invariant is the term you are confusing with scalar. As I already explained to you, invariants are scalars, but not all scalars are invariant.

No, the "directionless magnitude" definition of a scalar only works in Euclidean space. More generally, a scalar is something that is invariant under a coordinate transformation, like pmb said. Crank site or no, he is right on that one.

Energy is an example of a variant scalar. Charge is an example of an invariant scalar.

Under which set of transformations? In Euclidean space energy is a scalar for sure, but in Minkowski space it is the timelike piece of a 4-vector, and the components of vectors are certainly not scalars.

 

[pmb]

Originally posted by DavidW
Scalar is just something with magnitude and no direction. Invariant is the term you are confusing with scalar. As I already explained to you, invariants are scalars, but not all scalars are invariant. Energy is an example of a variant scalar. Charge is an example of an invariant scalar.

Totally wrong. Please stop posting this misinformation.

What you've written is not how the term scalar is defined. Pick up your copy of "Essential Relativity," by Wolfgang Rindler and turn to page 65. Rindler clearly states

.. scalar invariant (often shortened to just "scalar" or "invariant"), i.e. a real number independent of the coordinate system..

That the example you used above is bogus is example in "Special Relativity," Albert Shadowitz, Dover Pub., page 129

The simplest kind of such law is one that assigns the same number to a quantity measured by all Galilean observers: the quantity is an invariant... Such a quantity is called a scalar, and any equation involving only scalars will be form invariant. A scalar is a number, but not all numbers are scalars: the kinetic energy of a particle is a number, but a number which is different for different observers; so kinetic energy is not a scalar.

Let's get down to the facts:

Fact #1 - The term "scalar" is defined as "tensor of rank zero" - It's a *definition* of the term "scalar" and as such it's impossible for it to be incorrect.

Fact #2 - Every single text that can be found in both special and general relativity, classical mechanics, and electrodynamicsc defines the term as described in Fact #1 above.

Fact #3 - davidw has never been able to find a relativity text or a mechanics text or an electrodynamics text etc. which says differently.

Those are the facts - For referances and the relevant quotes of the definition see

www.geocities.com/physics_world/scalar.htm

Once more - A "scalar" is a tensor. A tensor of rank 0.


Pete

 

[pmb]

Originally posted by Tom
No, the "directionless magnitude" definition of a scalar only works in Euclidean space. More generally, a scalar is something that is invariant under a coordinate transformation, like pmb said. Crank site or no, he is right on that one.

It's not a crank site. One simply has to look at it to find that out. This person is a bit of a trouble maker - I've explained this to him a million times over the last 3-4 years - his own GR text tells him he's wrong - he's either unable or unwilling to learn the meaning of this term.

Pmb

 

[pmb]

Originally posted by Tom
Right. And parity is one of those coordinate transformations.

Sorry Tom. My applogies. I thought that it read -S when you actually posted S. My mistake.

Pete

 

[quantumdude]

Originally posted by pmb
It's not a crank site. One simply has to look at it to find that out.

Sorry, what I meant was, "I don't know if it's a crank site, but regardless pmb is correct."

 

[pmb]

Originally posted by Tom
Sorry, what I meant was, "I don't know if it's a crank site, but regardless pmb is correct."

Thank you. I just wanted to be clear on this point. davidw has been following me from forum to forum doing this same thing for the past 4 years. He's under the impression that because he took a couple more courses than I didn in grad school that automatically makes him right and everyone else wrong.

Pete

 

[StephenPrivitera]

So tensor=invariant and scalar=invariant but tensor[x=]scalar

Originally posted by Hurkyl
However, "The displacement from point A to point B" is a tensor (in Euclidean geometry), because if we change coordinate systems, the vector in the new coordinates is still "The displacement from point A to point B".

Why? Wouldn't this be the same idea as velocity? A rotation of the axes would change the coordinates of the displacement.

QUOTE:
A scalar is a number, but not all numbers are scalars: the kinetic energy of a particle is a number, but a number which is different for different observers; so kinetic energy is not a scalar.
=======
So as far as numbers go, you have scalars and nonscalars, where nonscalars are variant (and aren't vectors) and scalars aren't? Kinetic energy is a nonscalar number?


__________________
BTW, you won't be seeing any more posts from me. I'm going to have to buy a new computer after I throw this one off the roof! I've had to retype this post three times.

 

[quantumdude]

Originally posted by StephenPrivitera
So tensor=invariant

No.

and scalar=invariant

Yes.

but tensor[x=]scalar

Yes.

What you've missed in this discussion is that a scalar is a tensor of rank 0. Rank 0 tensors are invariant, but higher rank tensors are not. For instance, a rank 1 tensor is a vector, which certainly is not invariant under rotations or parity.

Why? Wouldn't this be the same idea as velocity? A rotation of the axes would change the coordinates of the displacement.

Yes, a rotation will change the components of the velocity. But no one claimed that velocity (a vector, aka rank 1 tensor) should be invariant.

So as far as numbers go, you have scalars and nonscalars, where nonscalars are variant (and aren't vectors) and scalars aren't? Kinetic energy is a nonscalar number?

You need to be careful about which space and set of transformations you are talking about when you say that something "is a scalar".

KE is a scalar in Euclidean space under rotations and parity. That is because it is formed by taking the inner product of two vectors, like so:

KE=(1/2)mv^.v

KE is not a scalar in Minkowski 4-space under the full Lorentz group, because we include boosts in that set of transformations.

__________________
BTW, you won't be seeing any more posts from me. I'm going to have to buy a new computer after I throw this one off the roof! I've had to retype this post three times.

Are you sure it's your computer? You know, your login times out after 900 seconds. So, if you spend longer than that typing your post, the site counts that as inactivity and logs you out, and you lose your post. But that's the website, not your machine.

I type all my long posts on MS Word, then cut and paste.

 

[Hurkyl]

Originally posted by Hurkyl
However, "The displacement from point A to point B" is a tensor (in Euclidean geometry), because if we change coordinate systems, the vector in the new coordinates is still "The displacement from point A to point B".

Why? Wouldn't this be the same idea as velocity? A rotation of the axes would change the coordinates of the displacement.

Right... but the trick is that A and B also change their coordinates, and the changed coordinates of the displacement correctly gives the displacement from A to B in the new coordinates.

 

[pmb]

Tom wrote

What you've missed in this discussion is that a scalar is a tensor of rank 0. Rank 0 tensors are invariant, but higher rank tensors are not. For instance, a rank 1 tensor is a vector, which certainly is not invariant under rotations or parity.

Unfortunatey the term "invariant" is an overloaded term. I.e. it has different meanings according to who's using it.

Some people look at tensors are geometrical objects which map vectors and one-forms to the real numbers in a linear way. In this view some will then refer to vectors as "invariant" quantities.

For example: From "Introduction to Tensor Calculus for General
Relativity," Edmund Bertschinger, MIT

See - arcturus.mit.edu/8.962/notes/gr1.pdf

Note that we have introduced vectors without mentioning coordinates or coordinate transformations. Scalars and vectors are invariant under coordinate transformations; vector components are not. The whole point of writing the laws of physics (e.g., F = ma) using scalars and vectors is that these laws do not depend on the coordinate system imposed by the physicist.

Same thing with the term "covariant" - yet another overloaded term. It's useful to know these differences so one can understand the authors point of view - in this case Bertschinger's point of view


re - "You need to be careful about which space and set of transformations you are talking about when you say that something "is a scalar"."

Yes! Exactly. In special relativity one might qualify the scalar by calling it a Lorentz scalar. Another good example would be the Coulomb potential. It's a Cartesian scalar (invariant under orthogonal transformations) but not a Lorentz scalar (in this case it's a component of the 4-potential)

Pete

 

[DavidW]

Originally posted by Tom
No, the "directionless magnitude" definition of a scalar only works in Euclidean space. More generally, a scalar is something that is invariant under a coordinate transformation, like pmb said. Crank site or no, he is right on that one.

No he isn't. The definition "works" just fine no matter whether space is Euclidean or not. That is irrelevent.

Under which set of transformations? In Euclidean space energy is a scalar for sure, but in Minkowski space it is the timelike piece of a 4-vector, and the components of vectors are certainly not scalars.

Transformation are irrelevant. I didn't say that energy was an invariant. I said that it was a variant scalar.

 

[DavidW]

Originally posted by pmb
Tom wrote

Unfortunatey the term "invariant" is an overloaded term. I.e. it has different meanings according to who's using it.

Some people look at tensors are geometrical objects which map vectors and one-forms to the real numbers in a linear way. In this view some will then refer to vectors as "invariant" quantities.

For example: From "Introduction to Tensor Calculus for General
Relativity," Edmund Bertschinger, MIT

See - arcturus.mit.edu/8.962/notes/gr1.pdf

Same thing with the term "covariant" - yet another overloaded term. It's useful to know these differences so one can understand the authors point of view - in this case Bertschinger's point of view


re - "You need to be careful about which space and set of transformations you are talking about when you say that something "is a scalar"."

Yes! Exactly. In special relativity one might qualify the scalar by calling it a Lorentz scalar. Another good example would be the Coulomb potential. It's a Cartesian scalar (invariant under orthogonal transformations) but not a Lorentz scalar (in this case it's a component of the 4-potential)

Pete

You really should stop with your crank info. I see that you have already confused enough people here.

 

[pmb]

Originally posted by DavidW
No he isn't. The definition "works" just fine no matter whether space is Euclidean or not. That is irrelevent.




Transformation are irrelevant. I didn't say that energy was an invariant. I said that it was a variant scalar.

Which is a contradiction in terms. Pick up your copy of "Essential Relativity," by Wolfgang Rindler and turn to page 65.

Are you at least aware of the fact that you're not using a definition that is not used in relativity/tensor analysis etc?

 

[pmb]

Originally posted by DavidW
You really should stop with your crank info. I see that you have already confused enough people here.

More nonsense and flames! Nobody has to take my word for it - all they have to do is see the list of referances I've provided for times like this when you post such enormous BS


There you have it folks - How accurate do you think this davidw character is in everything else if he can't comprehend a simple definition?


waite - Do you comprehend what Rindler means by

.. scalar invariant (often shortened to just "scalar" or "invariant"), i.e. a real number independent of the coordinate system..

Or don't you even read your own text?

Tell you what - Open MTW and turn to pg 480 and tell us how the authors define "scalar"

I'd also love to know where you got the idea that "scalar" is defined as you claim it is?

Give one relativity text that agrees with you! Or a mechanics text or a mathematical physics text or a differential geometry text etc.


We're waiting.

 

[quantumdude]

Originally posted by DavidW
No he isn't. The definition "works" just fine no matter whether space is Euclidean or not. That is irrelevent.

No. Once again, energy is a scalar in Euclidean space under rotations and parity. It is not a scalar in Minkowski 4-space under the full Lorentz group. Clearly, what you call a scalar depends on the spacetime you are talking about.

Transformation are irrelevant. I didn't say that energy was an invariant.

You need to go back and hit the books. A scalar is a rank 0 tensor, and that label only makes sense if you specify a set of transformations under which the object is a rank 0 tensor.

I said that it was a variant scalar.

If you mean that it varies under coordinate transformations and are still calling it a scalar, then you are talking nonsense.

 

[quantumdude]

Originally posted by DavidW
Transformation are irrelevant. I didn't say that energy was an invariant. I said that it was a variant scalar.

As has been stated repeatedly and with emphasis, this is not compatible with any accepted definition of "scalar" (beyond Halliday and Resnick level, that is). But what is really perplexing me is this:

Why the hell are you making such a big deal over a definition?

I mean, there is no inherent truth or falsity to definitions. They are just a matter of convention, and what you are saying is going against accepted convention. It's OK to define terms the way you want (I guess), but it's not OK to insist that that is not what you are doing. Just look at any number of books on the subject. If you look at anything in SR, GR, field theory, particle physics, or tensor analysis, you simply will not find anyone applying the term "scalar" to anything other than a "rank 0 tensor", which is an invariant under a specified set of coordinate transformations.

Stop being so dense, man.

 

[pmb]

DavidW claims

Scalar is just something with magnitude and no direction.

Prove it.

 

[DavidW]

Originally posted by Tom
No. Once again, energy is a scalar in Euclidean space under rotations and parity. It is not a scalar in Minkowski 4-space under the full Lorentz group. Clearly, what you call a scalar depends on the spacetime you are talking about.



You need to go back and hit the books. A scalar is a rank 0 tensor, and that label only makes sense if you specify a set of transformations under which the object is a rank 0 tensor.



If you mean that it varies under coordinate transformations and are still calling it a scalar, then you are talking nonsense.

No, you are. Repeating yourself doesn't change anything. I never said that energy was a Lorentz scalar. I said that it was a scalar. Invariance is the difference. Energy is not invairant and is not a Lorentz scalar and I never said that it was. Energy is a variant scalar. Just because a few physicists get lazy and shorted Lorentz scalar to just scalar doesn't mean everyone should. I don't and you can't make me.

 

[DavidW]

Originally posted by pmb
DavidW claims


Prove it.

I already have several times in several places. But fine, once again for the sake of your memory loss

"Recall that Lorentz vectors must be transformed but Lorentz scalars
are automatically invariant under transformations."
http://heppc16.ucsd.edu/ph130a/130a_notes/node33.html

"A scalar
is defined as a quantity that only describes a magnitude (no
direction)."
http://physics.damien.edu:16080/~eyama****a/lab3A.html

 

[quantumdude]

Originally posted by DavidW
I never said that energy was a Lorentz scalar.

I know that. It is a "Euclidean scalar", if I may coin a term.

I said that it was a scalar.

You also said that it does not matter in what spacetime it is a scalar, which is most certainly false. That is, it is false if one accepts that a scalar is a rank 0 tensor, as everybody does. If you don't want to, then fine, but you are using a nonstandard definition.

Invariance is the difference. Energy is not invairant and is not a Lorentz scalar and I never said that it was. Energy is a variant scalar. Just because a few physicists get lazy and shorted Lorentz scalar to just scalar doesn't mean everyone should. I don't and you can't make me.

You aren't making any sense here, and I am wondering if you are even reading the same thread as the rest of us. I never accused you of saying that energy is a Lorentz scalar. I said that a scalar is a rank 0 tensor, and that invariance under *some* set of transformations is implied in that. You disagree, and in doing so you are going against the accepted definition of the term "scalar" as it is used in SR, GR, QM, QFT, particle physics, and tensor analysis. But hey, other than those few obscure fields, I'm sure that everyone uses your definition.

 

[quantumdude]

Obviously, the demand to "prove it" is unreasonable because we are talking about a matter of definition, not a theorem, but even so...

Originally posted by DavidW
"Recall that Lorentz vectors must be transformed but Lorentz scalars
are automatically invariant under transformations."
http://heppc16.ucsd.edu/ph130a/130a_notes/node33.html

All this does is reinforce the point that one has to specify the set of transformations under which something is considered "scalar".

"A scalar
is defined as a quantity that only describes a magnitude (no
direction)."
http://physics.damien.edu:16080/~eyama****a/lab3A.html

The link doesn't work, but all the quote does is present an alternate definition that is not used in most books above the level of Halliday and Resnick. The reason most people don't use that definition is that it is useless. Definitions are usually formulated for the sake of convenience. Where is the convenience of that to relativity, QM, particle physics, etc?

On the other hand, the utility of identifying a "scalar" with a "rank 0 tensor" is immediately seen in all of those fields.

 

[pmb]

davidw claims

I already have several times in several places. But fine, once again for the sake of your memory loss...

Wrong - No soup for you!

You've never proved anything anywhere ... other than repeating that you're right and everyone esle is wrong.

"Recall that Lorentz vectors must be transformed but Lorentz scalars
are automatically invariant under transformations."
http://heppc16.ucsd.edu/ph130a/130a_notes/node33.html

So? All that does is prove that I'm right - i.e. I've always told you that a scalar remains invariant under a transformantion. If you really understood that comment then you sure wouldn't be posting that as any kind of proof. Energy is *not* a Lorentz scalar - For example: it depends on the 'direction' of the 4-momentum.

"A scalar is defined as a quantity that only describes a magnitude (nodirection)." http://physics.damien.edu:16080/~eyama****a/lab3A.html [/B]

And how is that different from what I said? This proves that the components of a vector aren't scalars - different directions mean different values of the components - hence the components aren't scalars - contrary to your bad example of energy as being a scalar.

 

[pmb]

Tom wrote

Obviously, the demand to "prove it" is unreasonable because we are talking.

Sorry = That wasn't my intention

davidw has, for the last 4-years, claimed that the definition he keeps stating is the only correct one. I wanted him to prove that this is the case.

That is to say - I asked from him to prove that the definition that he keeps posting is actually a definition that is used in relativity.

According to davidw - Whoever defines a scalar as a tensor of rank zero is a layman.


Pete

 

[quantumdude]

Originally posted by pmb
davidw has, for the last 4-years, claimed that the definition he keeps stating is the only correct one.

Well, that is patently false.

According to davidw - Whoever defines a scalar as a tensor of rank zero is a layman.

LMFAO!

David, I'm going to say it again: The definition you are using is at the level of Halliday and Resnick. When one is ready to stop being a "layman" and graduate to Goldstein, Sakurai, Jackson, Halzen and Martin, Bjorken and Drell, Wald, etc, then one uses the adult definition of scalar=rank 0 tensor. That is the only definition of which I am aware that has any utility in advanced physics.

 

[pmb]

Originally posted by Tom
Obviously, the demand to "prove it" is unreasonable because we are talking about a matter of definition, not a theorem, but even so...



All this does is reinforce the point that one has to specify the set of transformations under which something is considered "scalar".



The link doesn't work, but all the quote does is present an alternate definition that is not used in most books above the level of Halliday and Resnick. The reason most people don't use that definition is that it is useless. Definitions are usually formulated for the sake of convenience. Where is the convenience of that to relativity, QM, particle physics, etc?

On the other hand, the utility of identifying a "scalar" with a "rank 0 tensor" is immediately seen in all of those fields.

If you look at the source of this link this will start to make sense - it's a web page from a high school - I've told davidw many many many times that "scalar" has a different meaning than he learned in high school - I've given him many examples from the physics literature. Seems everyone at sci.physics knows this fact except for him.

Pete

 

[pmb]

Tom wrote

Originally posted by Tom
Well, that is patently false.



LMFAO!

David, I'm going to say it again: The definition you are using is at the level of Halliday and Resnick. When one is ready to stop being a "layman" and graduate to Goldstein, Sakurai, Jackson, Halzen and Martin, Bjorken and Drell, Wald, etc, then one uses the adult definition of scalar=rank 0 tensor. That is the only definition of which I am aware that has any utility in advanced physics.

I'm sorry. Seems I made a mistake. He didn't call me a layman. He called me a "lawman."

See this link (he was posting using the name joe - one of many handles he uses)

www.psyclops.com/hawking/forum/printmsg.cgi?msg=25482

Pete

 

[DavidW]

Originally posted by Tom
I know that. It is a "Euclidean scalar", if I may coin a term.



You also said that it does not matter in what spacetime it is a scalar, which is most certainly false. That is, it is false if one accepts that a scalar is a rank 0 tensor, as everybody does. If you don't want to, then fine, but you are using a nonstandard definition.



You aren't making any sense here, and I am wondering if you are even reading the same thread as the rest of us. I never accused you of saying that energy is a Lorentz scalar. I said that a scalar is a rank 0 tensor, and that invariance under *some* set of transformations is implied in that. You disagree, and in doing so you are going against the accepted definition of the term "scalar" as it is used in SR, GR, QM, QFT, particle physics, and tensor analysis. But hey, other than those few obscure fields, I'm sure that everyone uses your definition.

I am making sense. You just aren't paying attention to what it is I am saying. I am saying that I disagree with using the general term scalar to specifically mean Lorentz scalar without further qualification. I disagree with doing such a silly thing no matter what theory is the context. For example. Let's say I come up with a theory we'll call "beta theory" which hypothetically yields all forces of nature as byproducts of some odd convoluted electromagnetic field. As such the theory's general equations are ultimately written in terms of only electromagnetic field tensors. Outside the context of the theory the term field may refer to the cause of any of the forces of nature and so field unqualified has a more general meaning than "electromagnetic field". Now in beta theory the term electromagnetic field shows up so often as it is the ultimate source of all forces that later authors get lazy and just start calling electromagnetic field by just field. This doesn't mean that field doesn't really have a more general meaning, it just means that they got lazy. This is the case for Lorentz scalar in terms of relativity just being called scalar. Not every author is so lazy as to let just scalar mean invariant as you claim. Several still write out the full name and call it a Lorentz scalar. Those who don't are lazy no matter what field is being referred to.

 

[DavidW]

Originally posted by Tom
Well, that is patently false.



LMFAO!

David, I'm going to say it again: The definition you are using is at the level of Halliday and Resnick. When one is ready to stop being a "layman" and graduate to Goldstein, Sakurai, Jackson, Halzen and Martin, Bjorken and Drell, Wald, etc, then one uses the adult definition of scalar=rank 0 tensor. That is the only definition of which I am aware that has any utility in advanced physics.

pmb is lying, and I'll say it once again since you like repetition so much that, saying scalar to mean just Lorentz scalar is lazy and you can't make me do it no matter who else has done so.

Lorentz scalar = rank 0 tensor
Scalar in general = an single element, has magnitude, but not direction

 

[DavidW]

Originally posted by pmb
Tom wrote


I'm sorry. Seems I made a mistake. He didn't call me a layman. He called me a "lawman."

See this link (he was posting using the name joe - one of many handles he uses)

www.psyclops.com/hawking/forum/printmsg.cgi?msg=25482

Pete

You are an undereducated cranky liar. Post only one responce and then only to the correct thread. Don't be posting multiple replies here and sending some to google groups hoping I'll miss some of your accusations against me. Such deviousness immoral.

 

[pmb]

Originally posted by DavidW
You are an undereducated cranky liar. Post only one responce and then only to the correct thread. Don't be posting multiple replies here and sending some to google groups hoping I'll miss some of your accusations against me. Such deviousness immoral.

Do not come here and start harrassing me - this is a moderated forum and the games you play everywhere else don't fly here.

 

[ahrkron]

Originally posted by DavidW
You are an undereducated cranky liar. Post only one responce and then only to the correct thread. Don't be posting multiple replies here and sending some to google groups hoping I'll miss some of your accusations against me. Such deviousness immoral.

DavidW,

Stick to the physics, please. Name calling contributes nothing to the issue. If anything, it hurts your credibility.

 

[pmb]

davidw wrote

..., saying scalar to mean just Lorentz scalar is lazy and you can't make me do it no matter who else has done so.

Lorentz scalar = rank 0 tensor
Scalar in general = an single element, has magnitude, but not direction [/B]

This was not your point of view when you called me a layman in response to me explaining what a tensor of rank zero was - you're claim then was that I was confused. You never even hinted that you understood the scalar = tensor of rank zero until I proved it to you. And even after that you keep trying to correct me.

 

[Hurkyl]

Scalar in general = an single element, has magnitude, but not direction

How would you interpret the complex number i as a magnitude? After all, if I'm studying the vector space Cⁿ, the field of scalars of my vector space is C. So how can I consider i a magnitude?

Or to get more abstract, what if I'm living in GF(3)ⁿ? How would you consider an element of GF(3) a magnitude? GF(3)ⁿ isn't even a normed vector space! But, elements of GF(3) are scalars, by the definition of a vector space.


Things change in different contexts! I hope my examples made this clear.


However, if you really wanted to, you could identify the field of scalars (by your definition of scalar, DavidW) with the set of constant functions of coordinate charts, and then I think both your definition and the tensor analysis definition mean the same thing.