Vectors and scalars

  • #1
StephenPrivitera
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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?
 

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  • #2
mathman
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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.
 
  • #3
chroot
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This is purely a matter of semantics. A one-dimensional vector IS a scalar.

- Warren
 
  • #4
Hurkyl
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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.
 
  • #5
StephenPrivitera
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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?"
 
  • #6
HallsofIvy
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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.
 
  • #7
Hurkyl
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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.
 
  • #8
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 wich 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
 
  • #9
Tyger
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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.
 
  • #10
Sting
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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 :wink:

To embellish this concept a little further, a vector is a tensor of rank one.
 
  • #11
StephenPrivitera
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Originally posted by Sting
You beat me to it :wink:

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?
 
  • #12
Hurkyl
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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)
 
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  • #13
quantumdude
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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.
 
  • #14
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
 
  • #15


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 text book that explains it correctly.
http://www.geocities.com/zcphysicsms/
 
  • #16
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.
 
  • #17
quantumdude
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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.
 
  • #18
quantumdude
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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.
 
  • #19
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
 
  • #20
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
 
  • #21
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
 
  • #22
quantumdude
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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."
 
  • #23
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
 
  • #24
StephenPrivitera
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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.
 
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  • #25
quantumdude
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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.
 
  • #26
Hurkyl
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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.
 
  • #27
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
 
  • #28
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.
 
  • #29
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.
 
  • #30
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?
 
  • #31
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.
 
  • #32
quantumdude
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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.
 
  • #33
quantumdude
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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.
 
  • #34
DavidW claims
Scalar is just something with magnitude and no direction.

Prove it.
 
  • #35
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.
 

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