# Non geometric vectors in electronic/electrical engineering

#### studiot

Joined Nov 9, 2007
4,998
For those who still think the only sorts of vectors are objects with magnitude and direction please demonstrate the magnitude and direction of the row and column vectors in the matrix analysis of a simple circuit attached. Each has 6 components.

I have solved the circuit by the matrix equation MI = R, where I is a column vector and R is a row vector.

Please note the physical dimensions ( quantities) change half way through the vectors as the first three equations are about a current balance (KCL) and the last three are about a voltage balance (KVL), so all the terms in the vector are not in the same units.

The fact that the maths (arithmetic) still works out shows how careful you must be transferring from the arithmetical world of numbers to the physical one of units.

#### PRS

Joined Aug 24, 2008
989
Studiot, how did you draw that diagram? I'm looking for a way to do the very same thing. Sorry for not answering your question, but I can't.

#### studiot

Joined Nov 9, 2007
4,998
Studiot, how did you draw that diagram? I'm looking for a way to do the very same thing
Pen, paper and scanner, sorry about the quality.

#### PRS

Joined Aug 24, 2008
989
Thanks. I get it. I need a scanner. #### Ratch

Joined Mar 20, 2007
1,070
studiot,

For those who still think the only sorts of vectors are objects with magnitude and direction please demonstrate the magnitude and direction of the row and column vectors in the matrix analysis of a simple circuit attached. Each has 6 components.
It does have magnitute and direction in 6-dimensional space.

Please note the physical dimensions ( quantities) change half way through the vectors as the first three equations are about a current balance (KCL) and the last three are about a voltage balance (KVL), so all the terms in the vector are not in the same units.
The relationships are still in terms of the 6 currents. By the way, a 7th relationship I2 = I6 could also be used.

The fact that the maths (arithmetic) still works out shows how careful you must be transferring from the arithmetical world of numbers to the physical one of units.
Yes, that is true no matter how you represent physical relationships with abstract mathematical equations.

Ratch

#### studiot

Joined Nov 9, 2007
4,998
The relationships are still in terms of the 6 currents
Rubbish

The first three equations - I have written them out for you - all have the dimensions of amps.

The second three equations all have the dimensions of volts. Unless of course you measure battery voltage in amps in your neck of the woods?

By the way, a 7th relationship I2 = I6 could also be used.
Yes I could have chosen other equations in several ways.

But, as I'm sure you know, there are never enough node equations to solve the network on their own.

It does have magnitute and direction in 6-dimensional space.
So what actually is the magnitude of something which has 3 current legs and three voltage legs? What units is that magnitude measured in?

A similar question arises in relation to the 'direction'

#### Ratch

Joined Mar 20, 2007
1,070
studiot,

The second three equations all have the dimensions of volts. Unless of course you measure battery voltage in amps in your neck of the woods?
The independent variables are all in amps. The dependent variables are in amps or voltage or anything else you can relate to the independent variables.

So what actually is the magnitude of something which has 3 current legs and three voltage legs? What units is that magnitude measured in?
I would say six different magnitudes which have a value of the six different dependent variables. Whatever the units of the dependent variables are, three currents and three voltages.

A similar question arises in relation to the 'direction'
Whatever the metric of the six dimensional space is.

Ratch

#### studiot

Joined Nov 9, 2007
4,998
The independent variables are all in amps.
Ratch I just don't believe you have such a remarkably inept viewpoint of maths, physics and elctrical engineering.

Your statement would imply you believe that the (ideal) battery voltage depends upon the other circuit values, not the other way around!

I would say six different magnitudes
So now your definition of vector includes objects that can have (six) different magnitudes. How very confusing, especially as the magnitude is meant to be invariant under co-ordinate transformation.

What is really the case is that there are certain physical quantities that, although they are complete individual entities, we can, and find it convenient to, resolve into components aligned along the spatial co-ordinate axes. We can also represent these entities as little (or big) arrows, positioned in space and develop the whole of (the very useful) vector calculus. Examples would be forces, tangent and normal vectors......

In all cases the magnitudes is given by the square root of the sum of the squares of the components.

Attempting to apply this process to my circuit/equation yields.

For the column vector

'magnitude' = √(1.6^2 + 0.6^2 + 1^2 + 0.2^2 + 0.4^2 + 0.6^2) = 2.1166

Now since all the 'components' in this column vector are in amps it is reasonable to suppose this 'magnitude' is also in amps.

So what relation does 2.1666 amps have to the circuit, bearing in mind the total draw from the battery is 1.6 amps?

For the row vector

'magnitude' = √ (0 amps^2 + 0 amps^2 + 0 amps^2 + 1 volt^2 + 1 volt^2 + 1 volt^2 = √3 (√amps^ + volts^2)

So what does the number √3 mean in relation to the circuit and what are (√amps^2 + volts^2)?

The whole idea is meaningless.

You stated that these vectors have magnitude and direction.

Prove your words by displaying these quantities and explaining what they mean in relation to the circuit.

#### Ratch

Joined Mar 20, 2007
1,070
studiot,

Your statement would imply you believe that the (ideal) battery voltage depends upon the other circuit values, not the other way around!
Since the battery voltage is already defined, the independent variables are constrained to make the battery voltage what it is.

So now your definition of vector includes objects that can have (six) different magnitudes. How very confusing, especially as the magnitude is meant to be invariant under co-ordinate transformation.
The magnitudes change as you define the voltage and currents at different branches and nodes. What is confusing about that?

What is really the case is that there are certain physical quantities that, although they are complete individual entities, we can, and find it convenient to, resolve into components aligned along the spatial co-ordinate axes. We can also represent these entities as little (or big) arrows, positioned in space and develop the whole of (the very useful) vector calculus. Examples would be forces, tangent and normal vectors......
Yes, I have no conflict with that.

Attempting to apply this process to my circuit/equation yields.

For the column vector

'magnitude' = √(1.6^2 + 0.6^2 + 1^2 + 0.2^2 + 0.4^2 + 0.6^2) = 2.1166

Now since all the 'components' in this column vector are in amps it is reasonable to suppose this 'magnitude' is also in amps.

So what relation does 2.1666 amps have to the circuit, bearing in mind the total draw from the battery is 1.6 amps?

For the row vector

'magnitude' = √ (0 amps^2 + 0 amps^2 + 0 amps^2 + 1 volt^2 + 1 volt^2 + 1 volt^2 = √3 (√amps^ + volts^2)

So what does the number √3 mean in relation to the circuit and what are (√amps^2 + volts^2)?

The whole idea is meaningless.

You stated that these vectors have magnitude and direction.

Prove your words by displaying these quantities and explaining what they mean in relation to the circuit.
First you propose in the title of this thread that they are "nongeometric vectors", then you start doing math to find the magnitude as if they had a orthogonal spatial relationship with each other. Which is it? I said they had a magnitude and direction is six dimensional space. I would not know how to describe the direction in multidimensional space, but the magnitude must be the dependent variable as I said before.

Ratch

#### studiot

Joined Nov 9, 2007
4,998
the independent variables are constrained
Surely if the variables are constrained they are not independant?

What then are the dependant variables?

I am still concerned that you are suggesting that Kirchoff's Voltage Law is not about voltage but about current.

KVL is still applicable in a circuit where there is voltage, but no current.

You have finally admitted that there are at least two types of vector. One with '6 magnitudes' and one with a single magnitude.
You can do vector calculus with the latter but not the former.

The phrases curl I or div R have no meaning,
where I and R are my column and row vectors respectively.

This has been a consistent bone through several threads, where you have maintained there is only one sort of vector.

Finally, for your information, in some branches of Physics, vectors whose entries have different units are represented by a single compound entity upon which you can perform vector calculus.

In relativity,for instance, several 4-vectors are formed using the Lorenz relation

x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$ - c$$^{2}$$t$$^{2}$$ = 0

where x,y and z are position coordinates and t is time.

#### studiot

Joined Nov 9, 2007
4,998
Another point, which underlines the difference between Maths and Physics and the reason why you should not apply mathematical routines blindly in Physics is the following.

If I were to add another equation about a ficticious seventh current I7

eg I2-I6+I7 = 3

there would be no mathematical inconsistency and all I1 thorugh I6 would remain the same, but I7 would appear to be 3.

This is obviously a nonsense in physical terms.

It also underlines the fact that we need to be careful we neither gain something nor loose something when we equate an expression to zero.

#### Ratch

Joined Mar 20, 2007
1,070
studiot,

Surely if the variables are constrained they are not independant?

What then are the dependant variables?
Any equation constrains it variables. That is how they are solved.

Whatever you define them to be. Some definitions are naturally more convenient. x^2+y^2 = y would seem to be more convenient for x and y to be the best choice for independent variables.

I am still concerned that you are suggesting that Kirchoff's Voltage Law is not about voltage but about current.

KVL is still applicable in a circuit where there is voltage, but no current.
As long as you compute the loop voltages with a constant current, in this case zero, KVL is not violated.

You have finally admitted that there are at least two types of vector. One with '6 magnitudes' and one with a single magnitude.
You can do vector calculus with the latter but not the former.
I don't think I said that, or at least I did not want to give you that impression. I consider the matrix you presented as a set of 6 different vectors defining 6 different magnitudes.

The phrases curl I or div R have no meaning,
where I and R are my column and row vectors respectively.
So?

This has been a consistent bone through several threads, where you have maintained there is only one sort of vector.
Vectors are defined to be what they are. No need to repeat that definition. No matter what kind or sort of vector you are talking about, it should follow the definition.

In relativity,for instance, several 4-vectors are formed using the Lorenz relation

x + y + z - ct = 0

where x,y and z are position coordinates and t is time.
And so?

If I were to add another equation about a ficticious seventh current I7

eg I2-I6+I7 = 3

there would be no mathematical inconsistency and all I1 thorugh I6 would remain the same, but I7 would appear to be 3.

This is obviously a nonsense in physical terms.

It also underlines the fact that we need to be careful we neither gain something nor loose something when we equate an expression to zero.
Well, you have another variable to solve, and another equation to help solve it. Looks like a no gainer. Ratch

#### studiot

Joined Nov 9, 2007
4,998
So, 'the ignorant mock that which they don't understand.'

I don't think I said that, or at least I did not want to give you that impression. I consider the matrix you presented as a set of 6 different vectors defining 6 different magnitudes.
No I specifically referred to my row vector and column vector when I presented questions about vectors.

Not the matrix.

Further you also did this when you stated (correctly) that the coulmn vector components are all currents.

The matrix of course is a set of coefficients, none of which are currents.

You should be concentrating on the last equation as presented in the first post.

MI = R

Where M is a matrix, I and R are vectors.

The rest was explanation as to where they came from.

#### Ratch

Joined Mar 20, 2007
1,070
studiot,

So, 'the ignorant mock that which they don't understand.'
Or don't think is relevant.

No I specifically referred to my row vector and column vector when I presented questions about vectors.

Not the matrix.
OK, I stand corrected.

Further you also did this when you stated (correctly) that the coulmn vector components are all currents.

The matrix of course is a set of coefficients, none of which are currents.
Yes, they are coefficients for the currents.

You should be concentrating on the last equation as presented in the first post.

MI = R
I did. It is a typical way to solve a set of simultaneous linear equations.

The rest was explanation as to where they came from.
I know where they came from. I checked the math and found it to be correct.

Ratch

#### studiot

Joined Nov 9, 2007
4,998
I did. It is a typical way to solve a set of simultaneous linear equations.
It also contains two objects we are both agreed are vectors.

I further claim these vectors are not amenable to vector calculus, unlike geometric vectors.

#### Ratch

Joined Mar 20, 2007
1,070
studiot,

I further claim these vectors are not amenable to vector calculus, unlike geometric vectors.
I'll agree to that.

Ratch