# Phasors are Vectors

Discussion in 'Math' started by studiot, Nov 5, 2008.

1. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Since there have been repeated claims that phasors are not vectors I decided to test this against the mathematical definition of a vector that has been in use for more than a century now.

According to my literature, this codification of vectors was first introduced by Peano in 1888 (Calcolo Geometrico) and strengthened by Weyl in 1918 (Space-Time-Matter).

This definition has been in use ever since.

A non-empty set of objects, T,
taken with a second non-empty set of objects F,
and two operations connecting these objects
is said to be a vector space and the objects of T are said to be vectors
if the following conditions (numbered 1 through 10) are satisfied by each and every object in T and F.

The objects of F are called scalars and are often (but not always) real or complex numbers.

One of the two operations is a rule which takes any two vectors u and v from set T and connects them to a third vector w, also from T.
We call this addition and use the + sign to denote it.
Thus w = u + v

The second operation connects vectors in the first set with scalars from the second and is called multiplication by a scalar. It takes any vector v from T and any scalar a from F and outputs the vector av, also in T.

I have used bold capitals for objects of V . the vectors
and normal, lower case for the objects of F .the scalars

The 10 axioms to be satisfied are
(1) if u and v are in T then w = (u+v) is in T
(2) u +v = v + u
(3) u + (v+w) = (u + v) + w
(4) There is an vector in T called the zero vector such that for all v in T
0 + v = v + 0 = 0
(5) For each and every vector, v in T there is a vector v such that
v + (-v) = 0
(6) If k is any scalar in F and v is any vector in T then there is a vector kv in T
(7) K(u + v) = ku = kv
(8) (k+l)v = kv + lv
(9) k(lv) = (kl)v
(10) There exists a scalar 1, in F, such that 1v = v

If I can find (define) a set of object obeying these rules I can call the object vectors.
Of course the rules are framed to include all the normal geometric and other sorts of vectors as subsets.

2. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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To make life easy I have chosen to represent phasors as simple sinusoids, so that

A phasor is {a sin (v+p)} ............Note the whole thing in braces is the phasor or vector object

Where a and p are real constants and v is a variable. When considering two phasors there is no loss of generality to take one as {a sin(v)} and lump all the phase into one constant p.

(1) sin(v+2p) + sin(v)
= 2sin (0.5[2v + 2p])cos(0.5[2p])
=2 cos(p) sin(v)
but 2cos(p) is a constant so {sin(v) + sin(v+2p)} is in T
(2)reversing the order of addition leads to the same expressions, but the cos term is now negative. If we note that cos (-x) = cos (x) we satisfy condition 2

(3) Obvious by substition

(4) Clearly v = 0, p = 0 satisfies 4

(5) sin(v) + sin (-v) = 2 sin (0) cos (0) = 0, satisfying 5

(6) By definition from property of real number.

(7) By definition from property of real number.

(8) By definition from property of real number.

(9) By definition from property of real number.

So I can claim that mathematically phasors conform to the standard definition of a vector.

3. ### Ratch New Member

Mar 20, 2007
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studiot,

Does a phasor have a direction? Does it have a dot product? How about a cross product? And how about a box product or a triple cross product? A phasor has some vector like properties, but not all.

Ratch

4. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Are these serious questions and are you capable of understanding the mathematical answer?

5. ### Ratch New Member

Mar 20, 2007
1,068
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studiot,

Of course they are serious questions. Can you provide a mathematical answer? It would be interesting to see.

Ratch

6. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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No
No
No
No

I did not define a multiplicative structure on T, so why are you asking questions about something that was not in the original post?

7. ### Ratch New Member

Mar 20, 2007
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studiot,

You claim that a phasor is a vector, based on some properties that they both have. I listed some operations that can be done with vectors that don't even exist or make sense with phasors. Therefore my contention is that a phasor is not a vector even though it may share some properties. If it were, you could answer yes to the above questions.

Ratch

8. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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NO NO NO NO NO NO

I have taken the definition of a vector that has been standard for more than 100 years and shown how a particular representation of a phasor complies with all the requirements of this definition.

You have ignored all this and tried to substitute a Ratch definition, which of course it does not satisfy.

So what?

9. ### Ratch New Member

Mar 20, 2007
1,068
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studiot,

I did not define a dot product, cross product, box product, etc. How about a vector field? Ever hear of a phasor field? If it does not walk like a duck, quack like a duck, etc. ... . Well, you know the rest.

Ratch

10. ### steveb Senior Member

Jul 3, 2008
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469
I'd like to make a couple of observations.

1. With phasors, real power is P=|V|*|I|*cos(theta). This looks like a dot product to me.

2. We usually say that phasors have no real direction, but in a rotary motor, the current angle in the phasor representation can have real physical meaning and can be related to flux direction.

3. You can take the cross product of phasors; for example, in an induction motor, the steady state torque is proportional |Ir|*|Is|*sin(theta) where Ir is rotor current and Is is stator current and theta is the angle between them. The direction even comes out correct since torque is also the cross product of force and length of the moment arm.

I think we have to remember that in physics we use vectors to represent physical quantities. Some vector operations have physical meaning and some don't depending on the physics of the situation. In the case of phasors we are using vectors in an abstract space that does not correspond with real physical space; or it may correspond in special cases like rotary motors.

You can use vectors to represent finite rotations of a rigid body about a fixed point. However, here even the simple operation of vector addition does not have physical meaning since rotations don't add vectorially according to Newton.

So, lack of physical meaning of vector operations, does not prevent us from using vectors to represent physical states.

11. ### Ratch New Member

Mar 20, 2007
1,068
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steveb,

Yes, and Pr=|V|*|I|sin(theta). The reactive power is the absolute value of the cross product. I don't know if this proves anything other than phasors have a conformal similarity to certain electrical quantities. You made some good points, however. Have you ever heard whether a phasor field exists as does a vector field? I suppose you could claim that phasors are special set of converted vectors that don't have all the properties of full vectors. This is a slippery subject that involves arguments over definitions and properties. I am going on vacation starting Nov 7 and should be back on Nov 17, so we have one more day to argue about this.

Ratch

12. ### steveb Senior Member

Jul 3, 2008
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469
Actually, this is not a good example of cross product because reactive power does not have direction perpendicular to voltage and current. However, I agree that this doesn't prove anything. The point of my post was just that a phasor meets the criteria to be represented as a vector, although there is not always a physical meaning to the vector operations. I used the example of finite rotations which can be represented by vectors even though there is no vector operation (that I know of) which has physical meaning.

Honestly, I never really thought about whether phasor fields exist, but why is that relevant? A vector field is a field of vectors. Some physical problems can be described by a vector field. A phasor can be represented by a vector, it doesn't need a field of vectors. What do you consider to be "all the properties of full vectors"? I'm not even sure what a "full vector" is. I've seen the mathematical definition of a vector, and that for a vector field. The only question at hand is whether a phasor can be represented by a vector, which it can, although not necessarily in a space that correlates with physical space.

Yes, but why should we argue over the definition of a vector? People smarter than we have already defined it.

Last edited: Nov 6, 2008
13. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Steve

Please do not allow yourself to be diverted from the point of the original post.

Mathematicians have defined an object with a particular set of properties and called it a vector.
These properties are chosen for very good reasons and do not include operations that only apply to certain types of vector.
In particular they do not include any sort of cross or vector product. Some vector spaces have them, some do not.
Again the dot product is more formally known as an 'inner product' and again is not universal.
Note also that my published list of properties does not guarantee an inverse or identity vector, only an identity scalar.

Ratch

Do have a good holiday.
While you are on holiday ponder this.

Your (geometric) vectors exist in a multidimensional space of real numbers called R$^{n}$.
A dot product can be defined for all values of n
However there is no such thing as a cross product or vector triple product etc for R$^{2}$ - two dimensional vectors for instance.

The concept does not work in all dimensions, which is why mathematicians have excluded it.

Once again have a good holiday.

14. ### steveb Senior Member

Jul 3, 2008
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469
My apologies. I thought that I was reasonably within the bounds of the original post. Even now I'm not sure why my comments are not relevant. My basic point was that phasors are represented by vectors and that this is not related to whether or not there is physical meaning to certain vector operations. But anyway, I'll say no more about it.

Yes, I know. I never meant to imply that I disagreed with anything you said in this thread.

15. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Please don't take offence, Steve. That was not my objective.

The important point of the post is that, just as the whole of Euclidian geometry can be deduced from just 5 postulates,

A good deal of engineering mathematics flows from the 10 postulates (some authors condense them to 5). For instance allowing the objects (vectors ) of set T to be continuous functions, not merely sticks in 3d space, leads to the derivation of spectral analysis (Fourier, Legendre, etc series) Fourier and Laplace transforms and much much more.

I'm sure you will agree that this is a prize worth pursuing. This is why the definitions were chosen as they were and I am happy to develop this train of thought.

I would also agree with anyone who said that the choice of the word 'vector' to name objects with these properties was unfortunate. Unlearning schoolboy vector ideas cost me much angst when doing my applied maths degree.

Incidentally Exeter (UK) is just next door to me.

16. ### jonjkj New Member

Feb 15, 2008
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Not going to qualify this as all knowing but as far as the electrical field is concerned phasors and vectors are used interchangable. As for the for questions that were answered no well the answers are yes. You can check "Alternating Current Fundamentals" and "Handbook for Electricity Metering" for your self as I doubt you will take my word for it.

17. ### Ratch New Member

Mar 20, 2007
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jonjkj,

Can they be used interchangeably? Is the magnitute and direction of a quantity like force (vector) the same as the magnitude and phase difference of two voltages (phasor). Can a voltage by itself, which is the energy density of a charge, have a magnitude and direction? I believe that phasors have SOME vector like properties, but they are not quite vectors. Ratch

Last edited: Jan 23, 2009
18. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
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If one objects to the evolution of language, perhaps one should write a polite letter to the publishers of the cited texts.

19. ### Ratch New Member

Mar 20, 2007
1,068
4
Thingmaker3,

Uh huh, and what do the publishers know about it? They did not write the text. The authors did. So cut out the middleman. But what good would it do anyway? It's not practical. Are they going to publish a correction? I doubt it. But why did the word "phasor" come into being in the first place. How could it do so if it meant the same as "vector". I think we should decide amongst ourselves if phasor and vector are the same. You know why I don't think so. How about some other thoughts on this subject?

Ratch

20. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Since we are talking about subleties of language, writing to some is definitely a practical activity as opposed to a theoretical one.

The word you required in your context is 'practicable'