Vectors & Scalars: Understanding Directional Properties

[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.

 

[DavidW]

Originally posted by ahrkron
DavidW,

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

Tell that to pmb. He is the one first called me names. If he can do it, I can do it.

 

[DavidW]

Originally posted by pmb
davidw wrote


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.

You are lying. You are not educated enough to teach me any physics. You only have a bachelor of arts degree. I was the one who proved that a general scalar is NOT a tensor of rank zero. You are STILL confusing "Lorentz scalar" ie an invariant with the more general term.

 

[pmb]

davidw wrote [flames]

The moderator has requested that you to stick to the physics - Please do so.

 

[quantumdude]

Originally posted by DavidW
I am making sense. You just aren't paying attention to what it is I am saying.

No, I get your drift. I just think you're wrong.

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.

That's fine. It still doesn't take away from the fact that a scalar is a rank 0 tensor by definition in so many textbooks, whether or not we are talking about "Lorentz scalars".

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.

I am not saying that a scalar is the same as a "Lorentz scalar". A scalar is a rank 0 tensor in whatever vector space you are talking about. Once again, if you are talking about Euclidean 3-space then a "scalar" (aka a "Euclidean scalar") is anything that is invariant under rotations and parity. If you are talking about Minkowski 4-space, then a "scalar" (aka a "Lorentz scalar") is anything that is invariant under the full Lorentz group. It is not a simple matter of laziness, as many authors who equate "scalar" with "invariant" give special attention to the issue.

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.

Actually, I don't like repitition, and I wish you would open your mind so that I could stop repeating myself. For the ump-teenth time: When I say "scalar", I do not necessarily mean "Lorentz scalar", but I do mean "invariant".

Lorentz scalar = rank 0 tensor

Now you're being the lazy one. This is wrong, as written. It should be:

Lorentz scalar=rank 0 tensor under the full Lorentz group

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

As I've explained repeatedly, this definition is not used above the high school level because it is not useful. When you study some more advanced physics, you will see that.

Tell that to pmb. He is the one first called me names. If he can do it, I can do it.

PMB has not called you a name in this thread, not even once. If he did it at another message board, then that is not our concern. But you have repeatedly insulted him in this thread, and that is our concern. Keep it up, and I'll see your membership at PF canceled by the end of the week.

 

[pmb]

Originally posted by Tom
PMB has not called you a name in this thread, not even once. If he did it at another message board, then that is not our concern. But you have repeatedly insulted him in this thread, and that is our concern. Keep it up, and I'll see your membership at PF canceled by the end of the week. [/B]

Actually I encountered this person several years ago. Someone at another forum asked what a tensor was. I told then using a scalar as an example in that it was a tensor of rank zero. I.e. I wrote

"A scalar is a tensor. It is a number which is independant of the coordinate system. [...] A scalar is said to be a tensor of rank zero."

The funny part is that his own text states
"of "Essential Relativity," Wolfgang Rindler. From page 65

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

Pete

 

[DavidW]

Originally posted by Tom
Keep it up, and I'll see your membership at PF canceled by the end of the week.

Threats do not constitute valid arguements, and your threat backfired. If you tell me the link to cancel my registration, I will leave. You will loose a physicist and kept a crank. But that's ok because you are not open to hear the whole story behind the physics or the truth about pmb yet and that is not my fault. Eventually pmb will flame everyone who dissagrees with his kooky physics viewpoints at this site just as he does everywhere. If you don't believe me then either ask the others in any of the other physics boards at which he posts or read the posts he makes there for yourself. Here I leave you with a few google physics groups links to the kinds of "conversations" you will look forward to him eventually having with people here as well.

http://groups.google.com/groups?hl=en&lr=&ie=ISO-8859-1&q=+pmb++jerk&meta=


http://groups.google.com/groups?hl=...q=+pmb++ass&meta=group=sci.physics.relativity

http://groups.google.com/groups?hl=en&lr=&ie=ISO-8859-1&q=+pmb++sparky&btnG=Google+Search&meta=

 

[pmb]

Originally posted by DavidW
Eventually pmb will flame everyone who dissagrees with his kooky physics viewpoints at this site just as he does everywhere.

Please do not start trouble here. I've never flamed anyone who hasn't spent a great deal of energy flaming me to the breaking point. It's just never happened. And here people don't flame others when they disagree - I've never started anything anywhere and I never will. Your obssesive nature is not welcome here. You want to discuss physics? Then that's fine. But do so without flaming and lying.

 

[quantumdude]

David W has elected not to receive Private Messages, so regrettably I have to address this in the open forum.

Originally posted by DavidW
Threats

No, it's called a "warning". Perhaps you are unaware, but at PF moderators are called "Mentors". That's what I am, and so is ahrkron. Between the 2 of us, you have been given 2 warnings in this thread alone.

do not constitute valid arguements,

I know that, and I didn't intend for it to be an argument. I intended for it to be corrective action against your unacceptable conduct.

and your threat backfired.

So does that mean you intend to keep insulting pmb in spite of my warning?

If you tell me the link to cancel my registration, I will leave.

Don't worry. We'll cancel it for you if the situation does not improve.

You will loose a physicist and kept a crank.

You are obviously not a physicist. You don't even understand the simple explanations I've given you as to why you are mistaken.

But that's ok because you are not open to hear the whole story behind the physics or the truth about pmb yet and that is not my fault.

Who do you think you are, the Internet Messiah? You think we all need you to tell us about physics and about PMB? First of all, I learned what I've been telling you from grad school, not this website or any of its members. Second, I don't give a rat's ass about hearing your version of the truth about PMB. Frankly, I think you need to get a life.

Eventually pmb will flame everyone who dissagrees with his kooky physics viewpoints at this site just as he does everywhere. If you don't believe me then either ask the others in any of the other physics boards at which he posts or read the posts he makes there for yourself. Here I leave you with a few google physics groups links to the kinds of "conversations" you will look forward to him eventually having with people here as well.

http://groups.google.com/groups?hl=en&lr=&ie=ISO-8859-1&q=+pmb++jerk&meta=


http://groups.google.com/groups?hl=...q=+pmb++ass&meta=group=sci.physics.relativity

http://groups.google.com/groups?hl=en&lr=&ie=ISO-8859-1&q=+pmb++sparky&btnG=Google+Search&meta=

I already told you that anything he has said at other message boards is no concern of mine. I am a moderator at Physics Forums, not anywhere else. If PMB or anyone else becomes a problem here, we will deal with it, just as I am dealing with you.

This is your third and final warning: Cut the crap, or we will cut it for you. If you come back with so much as one more sentence fragment of your nonsense, my recommendation goes into the Administrator for your dismissal.

 

[DavidW]

Too late, I just reported you recomending your dismissal for threatening me and asked for a link for where I could cancel my own memebership due to your cranky responces in the face of physics facts and the truth. You are not a physicist. I am, but you lost me. Guess your stuck with pmb and yourself (prabably you are pmb under a second name) who merely speculates what actual physics is about.

 

[chroot]

To whom did you report the moderator? The moderator? :D

- Warren

 

[quantumdude]

Originally posted by DavidW
Too late, I just reported you recomending your dismissal for threatening me

David, I am a moderator here. One of my duties is to issue warnings. I did not threaten you, I just did my job, and the Administrator has told me he will back me up on it.

Seriously, what did you hope to accomplish with this? I got this position because the Administrator recognized me as one of the more knowledgeable members. You, on the other hand, are just some raving looney who walked in off the street. Surely you knew this was a losing battle?

and asked for a link for where I could cancel my own memebership

Like I said, we'll handle it for you ASAP. You'll be able to tell your changed status by your new title, "Cracker".

due to your cranky responces in the face of physics facts and the truth.

You simply do not know what you are talking about.

As I pointed out, we were talking about matters of definition here. There is no inherent "truth" or "factuality" in definitions. I only tried to explain to you what the standard convention is and that it differs from what you hold to. You need to study more physics, simple as that.

Seriously David, you blew it at this message board over nothing.

You are not a physicist.

Not yet, but I am working on my thesis.

I am, but you lost me.

You flatter yourself too much. Your profile says that you have an MS in physics and teach at the college level. I hate to burst your bubble, but an MS does not a physicist make. Hell, I am more advanced in my education than you, and I won't call myself a physicist until I finish my PhD.

Guess your stuck with pmb and yourself (prabably you are pmb under a second name)

Hear that ladies? He's got all that, and psychosis too. What a catch! LOL

who merely speculates what actual physics is about.

Some speculation is OK, but I prefer the "study and research" approach. Give it a shot sometime, it will really help you out.

 

[On Radioactive Waves]

LMAO! Its always funny to see physics arguments get s personal. Davidw, you shouldn't hold a grudge like that. It just looks like you have issues with PMB, not like you are trying to discuss the issue here. And oh yeah, that quite curtious of you to cancel your own membership to a place you don't like, instead of just leaving. Ha ha ha ha ha.