Talk:Scalar (physics)/Archive 1

Archive 1

Moved from talk:scalar

NOTE - The first portion (up to but not including "Restoring deleted material") of the following discussion has been moved from the Talk:Scalar page.

Definition

could someone add a lamens definition to the top paragraph? User:Spencerk

You mean the intro? Don't you mean three defs? --MarSch 11:07, 12 Jun 2005 (UTC)

I tried adding a layman's definition to the intro for scientific scalars. Does it help? Brisvegas 04:28, 12 November 2005 (UTC)

The phrase "A scalar or scalar quantity in physics is a physical quantity that can be described by a single element of a number field such as a real number..." is not in layman's terms.

Try this for layman's terms: "A scalar in physics is a single number that represents a quantity." There ya go. Many here have forgotten how to speak plain English. Your version does nothing mine doesn't, except to obscure the obvious. Scalars are simple, and the adornment is completely unnecessary. WP is not a college textbook and should not read like one. I realize from reading these articles that many of you are, no doubt, "educators" in some form or fashion. But that does not grant you license to regurgitate weasel words throughout this encyclopedia. Please reserve the obfuscation for the classroom, where it can feed your egos in a properly controlled and profitable environment. Many ordinary folks come here looking for straightforward answers, and nothing more. 98.194.39.86 (talk) 05:38, 29 July 2017 (UTC)

Please, note the subtle discrimination between "being" and "being described by". What you consider to be "plain English in laymen terms" is not sufficiently precise to satisfy the targeted scrutiny. Physics is not Math, "temperatures", e.g., are not "numbers". The required discrimination is in no way weasely or obfuscating, but a necessary cognitive effort. I do not want to comment on what you are looking for here, but this encyclopedia seems to look further. Purgy (talk) 06:58, 29 July 2017 (UTC)

This definition of a scalar quantity is mathematically incorrect. A number field is specifaclly a finite extension of the rational numbers. This does not include transcendental numbers like π or e. I am not well-read on the physical topic, but I would suggest changing "a single element from a number field" to "a complex number". Depending on the context, one can also use the notion of a global field athough I doubt that this is appropriate since this also includes function fields of non-zero characteristic. 5 February 2020

Discussion at Wikipedia talk:WikiProject Mathematics/related articles

This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:10, 12 Jun 2005 (UTC)

Recent changes to this article

In the section "In physics" of this article, the following text exists:

Four scalars together don't make a vector and neither are the components of a vector scalars. The reason for this confusion is that physicists are used to working with only a single coordinate system, usually Euclidean coordinates and with respect to such a coordinate system the components of a vector are scalars. In relativity, each observer carries a coordinate system with her and each observer uses her own coordinates to describe the components of vectors.

I don't think this is correct. In physics, a vector is precisely a quantity with three/four scalar components, unlike what this paragraph implies. And if you change coordinates, a vector stays the same. MarSch, in physics a vector is not a differential operator. Oleg Alexandrov 03:34, 15 Jun 2005 (UTC)

Problem here is "scalar" in math and in physics. In physics one is usually referring to transformations under the Lorentz and related transformations, and a scalar (like the rest mass of the neutron) can't change. Components of vectors and higher dimensional objects do change. In math I suppose a component of a vector, since it is just one number, is a scalar, but not in physics. By the way, what about complex numbers as scalars? I do not see an objection, and thinks like admittance are often complex. Pdn 05:10, 15 Jun 2005 (UTC)

I rewrote the physics-scalar section. It's not perfect but we need to make the following points:

• The thing that distinquishes scalars from vectors, and all other tensors, is the behavior of their components under a coordinate transformation. Scalars remain unchanged.
• vector components are not scalars - they change when the coordinate system changes.
• physical vectors may have any number of components (1 to infinity). Many physical vectors are spatial or space-time vectors and happen to have 3 or 4 components, but quantum mechanics has all sorts of vectors in all sorts of spaces, with complex components, etc.
• Physicists are not "used to working with a single coordinate system" and their way of defining vectors, tensors, etc. is not a "fallacy"

Also, can we dispense with the him/her nonsense? A technical article is no place to put one's political correctness or lack thereof on display. Technical articles can be easily written without these pronouns. PAR 06:11, 15 Jun 2005 (UTC)

Great job, PAR. I think you explained the scalar/vector difference very clearly. --MarSch 15:08, 15 Jun 2005 (UTC)

I like the new text too. Oleg Alexandrov 19:08, 15 Jun 2005 (UTC)

Well, good, then. But I don't know about the thing that comes next, that says "electric charge, but not charge density" as an example of a scalar. Charge density is a scalar field. I would expect something like "electric charge but not electric field" or something. PAR 02:41, 16 Jun 2005 (UTC)

I've fixed that and some other nonsense.

For the record: the components of a vector depend on the choice of coordinate system; the vector itself does not. A 25-mph wind from the northeast is a 25-mph wind from the northeast regardless of which coordinate system you use to express that fact, even though the scalar components themselves change. Michael Hardy 22:20, 16 Jun 2005 (UTC)

I agree that this article was in a rather poor shape until Paul (PAR) did some work on it. However, I did not see anyware in the article being said that the vector changes as you change the coordinates. I thought that the physics paragraph (written by Paul) which is now deleted was fine. Oleg Alexandrov 23:14, 16 Jun 2005 (UTC)

I really have to revert this. Please read the following points:

• A scalar is NOT something that has a magnitude, but not a direction. A tensor has a magnitude, but does not have a direction, and its not a scalar.
• The state of a quantum system is a complex vector in a Hilbert space. The state vector components are complex numbers. Therefore it has no components in physical space, so it has no "direction" in physical space.
• Please read the paragraph carefully. It says that a vector is an entity that is independent of the coordinate system. The components of a vector depend on the choice of coordinate system. Therefore the components change when the coordinate system changes. Since a scalar is something that does not change with a change in the coordinate system, it follows that the components of a vector are not scalars. The idea that single number is a scalar is the mathematical concept of a scalar, I think. It is the computer idea of a scalar, but not the physics concept of a scalar.PAR 00:31, 17 Jun 2005 (UTC)

The components of a vector obviously depend on the coordinate system. To say that a vector depends on the coordinate system, and even to use that as the definition of the concept of a vector, is nonsense and misses the point. Michael Hardy 23:06, 17 Jun 2005 (UTC)

Ok - I agree. I was more or less addressing the article, not anyone in particular. PAR 04:19, 18 Jun 2005 (UTC)

--Wilmac 19:21, 13 September 2005 (UTC)

In reading the above back-and-forth, I am more concerned than ever that WP has outlived its usefulness. It is simply a case of too many cooks with huge egos, in the same kitchen. Not every article about math or science has to be written at a graduate level. Pseudo-intellectualism is destroying this property. Does the phrase: "Keep it simple." ring a bell with anyone? 98.194.39.86 (talk) 05:44, 29 July 2017 (UTC)

Recent deletion

There are cases when more information does not make for a better article. Case in point, what I just deleted from the article. Here we have a huge paragraph in the article about scalar. The paragraph in question talks about pseudoscalars, which, since this is not the main topic of the article, should be talked only briefly.

However, what happened instead, is a lenghy ranting about pseudoscalars, which later switches to a particular case of pseudoscalars which is the signed volume, then reaching into tensors, their duals, all with parentheses and parentheses inside parentheses. Why on earth does one have to torture the reader like that? Oleg Alexandrov 15:33, 31 August 2005 (UTC)

I don't mean to torture anyone. It's just that pseudoscalars and pseudovectors, and pseudo-whatevers are like some kind of mickey-mouse invention that serve to simplify calculations but obscure the really neat symmetry and unity underlying the whole subject. That paragraph was my evidently unsuccessful attempt to transmit this idea. To treat pseudoscalars as some sort of wierd scalar when they are in fact nothing but simple stand-ins for a perfectly normal 3-tensor needs to be transmitted somehow, I think. Tell me where I lost it, ok? Maybe a separate article on pseudoscalars? PAR 19:19, 31 August 2005 (UTC)

Please try not to inject your POV too much into Wikipedia; standard textbooks (at least, the ones that bother at all with such symmetry considerations) don't seem to regard pseudovectors as "mickey mouse". (To my mind, you're just sidestepping the question. For example, sure, we need not consider the fact that angular momentum L is a pseudo-vector, as long as we only ever talk about r×p instead of L, but disallowing L is not much of a solution.) —Steven G. Johnson 21:17, 13 September 2005 (UTC)

I wasn't injecting "mickey mouse" into an article, just the talk page, and we don't need to be formal and non-POV here. I am not disallowing L, I'm just saying that it is an antisymmetric 2-tensor which transforms like any other tensor (scalar=0-tensor, vector=1-tensor). The dual of L is a pseudo-vector, and it is this pseudo-vector which is used as a stand-in for the angular momentum 2-tensor. It contains all the same information as the 2-tensor, but in condensed form. The problem is that you have to specify "special" rules for its transformation. This is fine, but it should not obscure the fact that the angular momentum is not some kind of wierd vector that has to have special ad hoc rules for its transformation, it is just a 2-tensor that transforms just like any other member of the family.

The 2-tensor L is the wedge product of r and p. The cross product is the dual of the wedge product, so the pseudo-vector L is the cross product of r and p. The cross product is just a stand-in for the wedge product. If you go to 4 dimensions, you can go nuts trying to remember whether you are dealing with a cross product or a tensor-vector product, etc, etc, and then it becomes obvious what to do. Forget about the duals and the pseudos and all their special rules, and just deal with the straight objects themselves. Its a lot cleaner.PAR 22:05, 13 September 2005 (UTC)

Hi Paul. I definitely overreacted in my ranting. Sorry. Yes, I think an article about pseudoscalars could do the trick. On the scalar article one could just briefly mention the similarities/differences between scalar and pseudoscalar, and refer to the other article for details. And in the new article, instead of a long paragraph one could have several paragraphs explaining this in detail, and maybe one section about signed volume, which I think deserves a bit more attention. What do you think? Oleg Alexandrov 20:53, 31 August 2005 (UTC)

Hi Oleg - I was not deeply offended, because upon rereading it, you were basically right. We have both complained about this sort of thing in the past. Besides, the "dual" link in the discussion just added to the confusion. I won't have time for a while to cook up a new pseudoscalar article, but I could put up a stub, unless you have an idea for an article. PAR 01:47, 1 September 2005 (UTC)

Hi Paul. Again, I appologize. And you are very right that actually writing things is better than ranting. I will work on pseudoscalar tomorrow (Thursday) morning. Oleg Alexandrov 05:23, 1 September 2005 (UTC)

I copied some of the contents I deleted to pseudoscalar. I copied little, because I do not understand most of it. For example, it says that If we think of volume as a 3-tensor.... I know little about tensors, but isn't a tensor linear in its components? That is, if one takes the tensor of three vectors, and then one multiplies the first vector by negative one, shouldn't the tensor change sign? But then, how can the tensor be the volume? I definitely got something wrong, but now you relise that I am not competent to copy that text. :) Oleg Alexandrov 02:12, 2 September 2005 (UTC)

The ONLY difference between pseudotensors and tensors is that pseudotensors change sign when orientation is reversed and tensors do not. Thus this distinction only exists on orientable spaces. Since this is the only difference I don't think pseudoscalar should be split off. --MarSch 12:15, 2 September 2005 (UTC)

The pseudoscalar page existed even before. Also, the concept has a different but related meaning in abstract algebra. So, saying some words about pseudoscalar here, and then referrring to pseudoscalar for more details, is the right thing I think. Oleg Alexandrov 15:48, 2 September 2005 (UTC)

Pseudotensors

I know this sounds complicated, but its not, its excellent.

Take for example, a 2-dimensional Euclidean space. There are three kinds of objects:

• 0-tensors (scalars) for example S
• 1-tensors (vectors) for example V_i where i=1,2
• 2-tensors for example T_ij where i and j = 1,2

There are transformation matrices, R_ij which transform these objects from one coordinate system to another.

• $S'=S\,$
• $V'_{i}=R_{ij}V_{j}\,$
• $T'_{ij}=R_{ai}R_{bj}T_{ab}\,$

where repeated indices are summed. Look at an antisymmetric 2-tensor

$X={\begin{bmatrix}0&A\\-A&0\end{bmatrix}}$

subject it to an inverting transformation R

$R={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}$

and you have:

$X'={\begin{bmatrix}0&-A\\A&0\end{bmatrix}}$

Notice its the same except the sign of A is changed. There is a one-to-one relationship between every pseudo-scalar and every antisymmetric 2-tensor. The pseudoscalar associated with the above 2-tensor X is just A. In other words the pseudoscalar A is the dual of the 2-tensor X, and vice versa. When you subject a perfectly normal antisymmetric 2-tensor (X) to an inverting transformation, its the same as changing the sign of its associated pseudoscalar!!! A pseudoscalar is just a shorthand device for an antisymmetric 2-tensor!!!. Instead of having a 2-tensor and saying it transforms perfectly normally, we use its associated pseudoscalar and have special transformation properties. The 2-tensor is "real" thing, the pseudoscalar is a device.

When we go to 3 dimensions there are three objects: scalars, vectors, 2-tensors and 3-tensors. Now a pseudoscalar is the dual of an antisymmetric 3-tensor. A pseudovector is the dual of an antisymmetric 2-tensor. The classical magnetic field is not a pseudovector, it is an antisymmetric 2-tensor, which transforms perfectly normally, and the pseudovector magnetic field is a shorthand device for representing it, along with its special transformation rules. PAR 20:46, 2 September 2005 (UTC)

Thanks Paul. But my question remains, is it true that one can think of volume as of a tensor? Oleg Alexandrov 21:53, 2 September 2005 (UTC)

Hi Oleg - yes, in three space, the volume is an antisymmetric 3-tensor, and the pseudovolume is a pseudoscalar. Just like an area, the volume has two "sides". I guess if you lived in 4 dimensions, you could see that.

Differential forms exist on any diff manifold, pseudoscalars only on orientable diff. manifolds. --MarSch 18:17, 25 September 2005 (UTC)

Is speed a scalar?

Some moron on one of my physics duscussion groups for the last few days has been trying to tell me that static gas pressure is a scalar force. He left in a huff because I tried to explain through common sense how he was wrong. So I started thinking maybe others were just as confused, and now I know on the internet that they'll let just about any half-baked loon who thinks he knows physics to post.

Speed is a vector. Vectors have direction in a certain distance and speed is distance / time. Electric charges have a vector, or how else could they interact with electromagnetic fields? Think about it. (post by Wilmac 19:21, 13 September 2005 (UTC))

Um, no. (Would you also say that "mass must have a vector, or how else could it interact with gravitational fields?") One needn't descend into a long metaphysical discussion here; there is a precise mathematical definition of what a scalar is, and anyone who has a long argument about whether some X is a scalar doesn't understand the concept. Charge is a scalar (under the rotation group, although not under the Lorentz group). —Steven G. Johnson 21:17, 13 September 2005 (UTC)

Anyway, speed is a scalar; its vector equiavalent is velocity. An interesting thought about the electromagnetic fields though. Brisvegas 04:31, 12 November 2005 (UTC)

This is why I've always thought physics requires a license like medicine or psychology. If you take a physics class, you would know that the electric charge is a Coloumb, which does have a force that effects other particles. Mass and charge are different, that's why they have different ubits. Like charges repel, and unlike charges attract. The attraction is a finite force and the direction is the movement toward or away.

There is no such thing as a "scalar force." If you said such a thing in a physics conference, they would take you away for disturbing the real science. This is not to discourage you from your perpetual motion or anti-gravity, or whatever it is anyone here thinks will work. :-) --Wilmac 19:39, 22 September 2005 (UTC)wilmac@aapt.net.au

Stevenj is correct. Classical electric charge does not change with a change in coordinate system, therefore it is not a vector. Velocity is a vector. The absolute value of the velocity is the speed and speed is not a vector because, again, it does not change its value with a change in coordinate system. PAR 22:05, 13 September 2005 (UTC)

I am sorry, but you are wrong. Coloumb is a force, and all forces have vectors. Just rub your head with a balloon and stick it to a wall to see for yourself.

Your analogy is the kind of over-simplification that authors use in science books for the general public. There is no such thing as a scalar force. Scalars are components of a force and nothing more. Your knowledge can get you through High School, but it would never cut it in college.--Wilmac 15:44, 14 September 2005 (UTC)

The only way to understand physics is from the ground up, and that requires hard work. Science is not called a "discipline" for nothing.. I have little patience with people who don't even understand the basics. A full understanding requires a certain amount of linear algebra and general coordinate geometry. That is where vectors come into their own.

Don't just post what you plagiarize from a book. Understand physics so that your misunderstandings don't confuse other layman. Scalars are only for units which you can count. --Wilmac 19:21, 13 September 2005 (UTC)wilmac@aapt.net.au

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The above discussion was moved from the original scalar article

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Sometimes a single number is just a single number for the vast majority of cases that concerns the vast majority of people. Keep it simple, folks. 98.194.39.86 (talk) 05:51, 29 July 2017 (UTC)

Restoring important material

I have restored deleted material because it is important. PLEASE NOTE: In physics a vector is not simply "something with a magnitude and a direction". This has been gone over time and again on this page. The essence of a physical vector is its behavior under a coordinate transformation. A vector need not exist in physical space. There are statistical mechanical vectors in phase space. There are quantum mechanical vectors in isospin space, etc., etc. It follows that the essence of a scalar is something that is invariant under a coordinate transformation, and is intimately tied to the vector space of which it is a part.

The part concerning dimensions is also important. I did not write it, and I am not satisfied with it but until it can be improved, it should not be deleted.

Please do not remove these sections without discussion, they are there for a good reason. PAR 23:14, 1 January 2006 (UTC)

I undestand the point about coordinate system invariance, but the text is too verbose. Over-explaining does not make the concept any clearer, on the contrary. As for the dimensions/units stuff, it is not relevant to this article because the concept of physical quantity, an hence of scalar quantity, are independent of the choice of units, and possibly even of the concept of measurement. After all, a body has a definite mass independently of whether anyone measures it or not. (Indeed, one flaw of the current text, and of the 180 miles/h example, is that it makes scalarity seem an attribute of the way you measure the quantity.) The discussion about units should be moved to some other article, perhaps measurement or physical quantity. Jorge Stolfi 13:36, 2 January 2006 (UTC)

I have added the talk page from the old scalar article. You can see by reading it that there has been a lot of discussion about how the article should be written. The text is not too verbose, it does not over-explain, it explains. There are a large number of people who think a physical scalar is just a number and that a vector is a number along with a direction in physical space. I once thought that this misconception could be explained in the main text, but it became clear than it was a pervasive misconception, and needed to be cleared up or at least questioned immediately.

Regarding the units problem, I have not thought as much about, but it is clear that any physical quantity is more than simply a field element (e.g. a real number) unless it is dimensionless. The mass of an object has no meaning without measurement, and when measured, it is done with respect to a particular set of physical units - the equivalent of a coordinate system. The mass of an object may be 1000 grams or 1 kilogram, and the fact that these two different mathematical field elements represent the scalar mass of one object cannot be dealt with unless the concept of physical units is introduced. The concept of units is ABSOLUTELY VITAL to the concept of a physical scalar, and does not belong elsewhere. We need to improve this section, not discard it. PAR 16:55, 2 January 2006 (UTC)

Well, I stand by my comments above. In particular, the head parag is too verbose; and when one says that 1000 g or 1 kg are the "same mass", one is admitting that the object's mass is a well-defined thing, independently of measurement. Perhaps there is confusion of the concept of a physical quantity with the number that represents it in physical equations and calculations? (That is where the concept of units is all-important.) But I will gladly leave that for physicists to sort out. Jorge Stolfi 18:15, 2 January 2006 (UTC)

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I have asked some other contributors to take a quick look at this discussion (I am not always right), and we can let the consensus rule.

Regarding the units - I am taking the positivist view that mass has no meaning unless it is measureable. The results of that measurement do not give an absolute number, but rather a ratio with respect to a reference mass. This is the ONLY handle that we have on the mass of the object. Similarly, a physical vector has no meaning unless it is measureable. The results of that measurement give a set of ratios with respect to a set of reference (or base) vectors. These vector components are the ONLY handle that we have on that vector. Any physical concept or theory must be independent of the particular reference elements (i.e. coordinate system) used in the measurement of the physical quantites involved. This fact is fundamental, and it is at the core of the meaning of a physical scalar (e.g. mass) or a physical vector. PAR 18:57, 2 January 2006 (UTC)

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I would avoid getting too metaphysical in this page. Readers aren't interested in your opinion, or mine, on such questions. (If you can find a reputable source commenting on the question that's another matter). In the introduction, I would simply say something like:

Informally, a scalar is a quantity characterized by a magnitude only, as opposed to a vector which is characterized by a magnitude and a direction. More formally, a scalar is a quantity that is invariant under rotations of the coordinate system, or a rank-0 tensor. Examples of scalars include mass, temperature, electric charge, and the magnitude of any vector. Further examples and elaborations on this concept are discussed below, including an extension to special relativity in which invariance under both rotations and changes of velocity is considered.

And then go into further detail from there, as necessary. The section on units is confusing and superfluous; concepts of units apply to all physical quantities, and there is no special rule for scalars. The section on relativity is confusing, because it gives no examples of scalars in relativity. The distinction made between "classical physics" and "relativity" is confused, because the distinction is between scalars under the rotation group and scalars under the Lorentz group, not between classical and relativistic physics per se. In short, I think this article needs a lot of work. —Steven G. Johnson 19:46, 2 January 2006 (UTC)

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• I think it would be good if there were more examples than just the speed/velocity, but I think the current article spends too much time on noreast vs northwest, etc, that should be trimmed.
• I don't like the part about units, dimensions, etc.
• I agree that one should not say much about pseudo-scalars, rather than mentioning that there is such a thing and referring to the appropriate article for details. Oleg Alexandrov (talk) 20:02, 2 January 2006 (UTC)

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Steven and Oleg - thanks for your input -

I am absolutely certain of the following, and I only wish I had the verbal skills to make it obvious:

• The components of a physical vector have magnitude only, yet they are most certainly not physical scalars. Therefore the statement "a (physical) scalar is a quantity characterized by a magnitude only" is wrong and misleading.

  Yes and no. The dot product of two vectors is certainly a scalar, and a component of a vector is simply its dot product with the unit vector in the corresponding direction. It's only when you take that direction to be your coordinate axis which varies under rotations, as opposed to being a fixed direction, that it becomes "not" a scalar. In any case, one has to expect that any handwaving definition is going to be imprecise...that's why you follow it with a formal definition. The same is true in any mathematical article — first some intuitive handwaving to convey a general idea to a layperson, followed by a more precise statement. The most you can hope for is that the intuitive notion conveyed is more-or-less correct, for people who don't have the training or inclination to read the precise definition. —Steven G. Johnson 01:51, 3 January 2006 (UTC)

• Physical units are intimately tied to the definition and behavior of a physically measureable quantity, scalar, vector, tensor, whatever. The article will not be complete until this is made clear.

  As you say, units are important for any measurable quantity, not just scalars. Which is precisely why a vague and general discussion of the concept of units does not belong in this article. —Steven G. Johnson 01:51, 3 January 2006 (UTC)

I agree with Steven that the relativity part is not to the point, and I agree with Oleg that the examples might be expanded and that a detailed discussion of pseudoscalars belongs elsewhere. Since I cannot make people understand the importance and meaning of units and measurement in the definition of a physical scalar, I will try to find references and sharpen my dialog on the subject. Until then, I won't argue with anyone who wishes to eliminate that section.

With regard to the rest of the article, I will incrementally start fixing things, unless somebody else does it first :) PAR 20:47, 2 January 2006 (UTC)

Inertial frames/scalars independent of coordinate systems

There seems to be some inconsistency between this article, Coordinate system, and Inertial frame of reference. If any inertial frame is a 'coordinate system' - which is what Inertial frame of reference appears to be saying - then the claim here that KE is independent of coordinate systems is untrue, since we can change an object's apparent speed by changing frames.

AFAICT, this article and 'coordinate system' take it as read that different coordinate systems will still have fixed origins in one preferred frame of reference, while inertial frame of reference does not. (I've mangled that badly, hopefully somebody can understand what I'm saying.) I think I understand the discrepancy in my own head, but not well enough to harmonise the articles; perhaps somebody with a better grasp of the terminology could do this? --Calair 23:17, 29 January 2006 (UTC)

I also wondered about that, Talk:Kinetic_energy#Kinetic energy increase caused by a push depending on frame of reference? says that kinetic energy is not a scalar. The same would apply for speed.--Patrick (talk) 10:29, 14 September 2008 (UTC)

Scalar Particles

I have a question about "scalar particles" which are (as far as I can tell) are particles with 0 spin. (Other particles have a spin vector.) Where should they be put for a definitive definition? Many of the articles seem to refer back to the scalar page, which is where this scalar physics page grew up, and which is not really helpful. This page really isn't either - should a new page be created? ... WhiteHatLurker 00:06, 13 June 2006 (UTC)

Yes, scalar particles are always spin-0 particles. They are quanta of scalar field (physics). linas 23:39, 6 July 2006 (UTC)

They may, however, be isospin vectors, or belong to some non-trivial representation of a Lie group. However, they would still be spin-0 bosons, i.e. Lorentz scalars. linas 01:41, 7 July 2006 (UTC)

Scalars in physics vs. scalars in mathematics

Are there any physical scalars that are represented mathematically as vectors? --Smack (talk) 22:56, 14 June 2006 (UTC)

No. linas 23:35, 6 July 2006 (UTC)

I think you could consider the total charge in some region of space as a mathematical vector, even though it is a scalar in physics. It would be like the position vector, where you add vector displacements to get new positions; here you add changes in total charge to get new totals in charge. Zeroparallax 00:24, 10 August 2006 (UTC)

That's just scalar addition, though. The distinction between scalars and vectors is that vectors have a meaningful direction. For instance, it's meaningful to say that two forces are acting at right angles to one another, but what would it mean to say that two charges are at right angles to one another? --Calair 01:54, 10 August 2006 (UTC)

There is great ambiguity in language. To be more precise, we should be very specific when we speak of vectors and scalars since there are several different types that appear in different contexts. Many people are introduced to scalars as "things with magnitude and no direction" and vectors and "things with both magnitude and direction". This may have been how scalars and vectors were originally defined, and it is appropriate for much of physics, but these ideas have been generalized in several different ways.

In mathematics, when you say a type of object or quantity is a vector, you only mean that instances of that quantity are elements of a vector space and that they follow the eight rules for vector spaces. Total charge, together with changes in charge, follow these rules. Or, perhaps to be more precise, I should just say that changes in charge alone can be considered vectors. Anyway, there is no need for a mathematical vector to have a "direction". (Although, for this example you could consider negative changes to be in one "direction" and positive changes to be in the "opposite direction", in an abstract sense.) As for scalars, I have seen them defined in Byron and Fuller's Mathematics of Classical and Quantum Physics (page 85) as elements of fields, which are defined in abstract algebra. This appears to be consistent with the use of the word scalar in beginning physics.

Also, in classical physics both the concept of a "scalar" and "vector" are generalized from their very visceral and geometric origins. As far as I have read (in books such as Sakurai's Modern Quantum Mechanics, page 232) a vector is a quantity with three components (V_1, V_2, V_3) that transforms by definition like V_i -> sum_ij R_ij V_j, where R_ij is an element of a proper rotation matrix. Conceivably, this could be a collection of three quantities that have no discernable "direction", although I have no example for that right now. Scalars in classical physics are defined to be invariants under proper rotations. (See Jackson's Classical Electrodynamics, Third Edition page 268.)

So it seems that there are three basic types of vectors: physical or geometric vectors ("things with magnitude and direction"), classical physics vectors, and mathematical vectors. (People may even argue that there are other very specific uses of the word vector, such as 3-tuples of numbers.) Now, I would argue that the definition for scalars as "things with 'magnitude' but no direction" is not a good definition, unless the people who make this definition admit that they are using the word "magnitude" in an abstract manner. I say this because the people that use this definition usually turn right around and use positive and negative numbers, which could be said to have both magnitude and sign (and they sometimes even imply direction). Zeroparallax 05:04, 11 August 2006 (UTC)

field density is one number

The article says: "On the other hand, the electric field at a point is not a scalar in this sense, since to specify it one must give three real numbers that depend on the coordinate system chosen." I'm thinking that might be worded better. Field density is one number. A point is three numbers. Together they are four numbers. -- Another Stickler (talk) 02:07, 20 December 2008 (UTC)

Given a point, say P, and an instant, say t, to specify the field E(P, t), you need three numbers: its x component E_x(P, t), its y component, and its z component. (Alternatively, you can give its magnitude, its "elevation" and its "azimuth", etc., but you need three numbers anyway.) -- Army1987 – Deeds, not words. 02:15, 20 December 2008 (UTC)

Yeah, but the example doesn't say it's describing the whole field. That's why it's ambiguous. It could as easily be interpreted as saying it's describing the field "strength" at only one given point. Since the point is given, it could be assumed that the three numbers necessary for locating that point are separately known, leaving only a scalar necessary to hold the field strength value measured at that single point, such as plus 1 volt. Time isn't even mentioned. Since the purpose of the example is to provide a non-scalar, a simpler example would do, else the ambiguity should be tightened up. -- Another Stickler (talk) 03:28, 20 December 2008 (UTC)

I think you're confusing the electric field with the electric potential. The electric field at any given point is a vector. So, given three numbers individating a point (and a fourth number individuating a time), you need three more numbers to give the components of the electric field at that point. So, the example is correct, and not ambiguous. -- Army1987 – Deeds, not words. 13:06, 20 December 2008 (UTC)

I was. Thanks. So, we're really talking about seven numbers. One for time, three to locate the point, and three to describe the force on a hypothetical stationary charge at that point (those last three could be arranged as two for direction and one for magnitude, in which case magnitude would be a scalar). The sentence is correct, but it's not very explanatory. People who know that an electric field is a force field and that forces are vectors will already know the difference between scalar and vector and not need the example. People who don't know the difference cannot be taught the difference by an example which requires one to know the difference already. Now I know how to improve the article. Thanks. -- Another Stickler (talk) 21:46, 20 December 2008 (UTC)

Confusing Sentence

"A physical quantity is not quite expressed as the product of a numerical value and a physical unit, not just a number." This sentence is very confusing to me, and I think it should be reworded. Does the sentence mean "A physical quantity is expressed as the product of a numerical value and physical unit, not merely as a number." or does it mean that it is expressed as such but not quite? I think the first rewording and interpretation that I am suggesting is correct. If you do not think this is the intent of the message, please make the appropriate adjustments to my alterations. Freebytes (talk) 18:44, 4 February 2010 (UTC)

can a scalar quanity be made into a vector quanity by adding a direction to its magnitude: Explain why or why not and give an example —Preceding unsigned comment added by 74.176.147.191 (talk) 20:01, 28 February 2010 (UTC)

Please answer

Could anyone kindly tell me if a scalar quantity can be negative? Please give me the answer on my talk page.--Pritam Laskar (talk) 11:05, 31 October 2011 (UTC)

Definition of Vector

I believe that the definition is incorrect. A vector remains invariant under transformation, however its components transform as do the coordinates under a coordinate transformation. Cf Bernard Schutz (2009), A first Course in General Relativity, ch 2. "Vector analysis in special relativity"., A scalar -- as used by engineers -- represents the magnitude of a vector multiplied by a particular scale. Therefore, it "pins" the quantity to a particular coordinate system. Temperature is not a scalar, but the magnitude of temperature attached to the Celsius scale, is -- at least in the engineering sense. Similarly, speed is the scalar form of velocity. When we attach this magnitude to a scale like "mph" we pin it to the English system. In mathematics, a scalar is simply a real number. This distinction should be made clear.

The idea that vectors transform leads to serious misunderstandings, like the(incorrect) notion that a traveler moving away from earth at a constant velocity near the speed of light would age more slowly than his Earthly twin. The time vector does not change, but its basis "stretches" as a result of a "doppler effect", therefore the traveler appears to age more slowly. For the same reason, the Earthly twin would appear to age more slowly when viewed by the traveler. — Preceding unsigned comment added by 50.42.140.89 (talk) 23:05, 26 February 2012 (UTC)

I think the article now been changed to avoid the first issue you raise (i.e. it doesn't give the impression that vectors, rather than their components, transform under a coordinate rotation - it looks to me like the aim here was to contrast scalars with vectors, rather than to define what a vector is, but in any case the wording in the article at the time was unhelpful).

I disagree with the comment "A scalar represents the magnitude of a vector multiplied by a particular scale" - temperature is a scalar, there's no underlying vector quantity of which temperature is the magnitude. Djr32 (talk) 13:14, 27 December 2012 (UTC)

Describing temperature as a scalar (in kinetic theory) is done under the assumption that the kinetic energy of the particles of gas is equipartitioned in all 3 DOF, and thus the component vectors cancel, leaving no net particle vector { that's the reason for T = (2 KE) / (3 k_B) in the equation... a more accurate equation would be T = (2 KE) / (DOF k_B) }... but if the kinetic energy is not equipartitioned in all 3 DOF, temperature cannot definitionally be a scalar quantity. This occurs in nanostructures, for example. For instance, in a quantum well state, electrons will have a zero axial temperature Tz and non-zero transverse temperature Txy. In fact, there are separate states in the quantum well, each with its own electron temperature. Conventionally, the different temperature components are just treated separately and not formed into vector mathematics, but there is no reason why they cannot be. If a tensor has a magnitude and one vector field (i.e., rank 1 tensor), then it is called a vector and has 3 components in 3-D space and 4 components in 4-D space-time... and that's exactly what occurs for temperature when kinetic energy is not equipartitioned in all 3 DOF. — Preceding unsigned comment added by 71.135.33.210 (talk) 01:58, 14 February 2020 (UTC)

Elephant in the room

FWIW, it isn't at all clear from the article how physicists determine whether a scalar is "classical" or "relativistic". For example, temperature seems to be used a lot to illustrate the idea of a scalar quantity. But is it a classical or relativistic scalar? It certainly appears as a scalar in classical theories, but why does/doesn't it make the cut for relativistic theories? 75.139.254.117 (talk) 06:31, 10 March 2017 (UTC)

Errors on this page regarding definition of Scalar

There are many problems with lead section, so I will try to explain them here:

1. "...can be described by a single element of a number field such as a real number, often accompanied by units of measurement."

This statement is OK.

2. "...is usually said to be a physical quantity that only has magnitude and no other characteristics"

This statement clearly contradicts statement 1. For example, real numbers include both positive and negative numbers. Since a negative number can be thought of having the opposite direction as a positive number, clearly negative numbers and positive numbers both have a direction. If we change statement 1 to say "complex number" instead of "real number", then we have another dimension of possible directions for the number, for example the number (1+2i) clearly does not have the same direction as a positive or negative real number. Statement 1 also says "accompanied by units of measurement", which contradicts "only has magnitude and no other characteristics".

3. "...Formally, a scalar is unchanged by coordinate system transformations."

This statement is completely wrong. Imagine a scalar such as the number 5. This number 5 does not specify a coordinate system, so let's assume it means 5 in the positive y direction from traditional 2 dimensional cartesian coordinate system. I'm going to use a very simple example so we don't get confused. Let's transform this coordinate system from "positive y direction" to "positive 2y direction". Now the scalar 5 which was defined in the positive y direction, is the scalar 5/2 defined in the positive 2y direction. Here's a more complicated example but still pretty easy to understand: lets instead assume we take our scalar 5 in the positive y direction and transform coordinate system to positive x direction. Now our scalar 5 in the positive y direction becomes the scalar 0 in the positive x direction. Brian Everlasting (talk) 16:17, 19 April 2018 (UTC)

1. So let's -at least for the moment- agree that this os OK, seting aside constructs, combining some scalars to some vector ... Note however that it does not say that every element of a number field behaves as a scalar(physics). See (3).
2. The "usually" excludes any fundamentalisms as are appropriate in mathy articles, and -in fact- egregious examples of scalars(physics) like e.g., temperature and speed are represented -in suitable units- by positive reals, so fitting to the (vague) statement. Considering a number field as a 1-dim vector space is mathematically correct, but a categorical misconception in this context. Identifying a 2-dim real vector space with a 1-dim complex vector space, and the latter with its scalar number field takes the same dangerous route of mixing categories in physics as above.
3. I do not assume that in physics a single coordinate of the set of coordinates making up a position vector is considered a scalar(physics). The whole set of coordinates transforms according to "vectors".

I do not see this article at the top of its possibilities, and I fully agree to your estimation that edititing it might become too complicated and controversial. See also the history in the TP.

As belongs to the DAB, I removed the double entry of pseudoscalar, but I think that even an angler fish belongs there, if a suitable search term is reasonably close. Purgy (talk) 09:53, 20 April 2018 (UTC)