When Do Imaginary Numbers Occur In Physics?

Thread Starter

Glenn Holland

Joined Dec 26, 2014
686
I'm wondering about which situations in physics do imaginary numbers occur.

In other words what are all situations that result in a negative number appearing inside the square root sign.
 

MrChips

Joined Oct 2, 2009
19,774
I'm wondering about which situations in physics do imaginary numbers occur.

In other words what are all situations that result in a negative number appearing inside the square root sign.
There is a misconception of imaginary numbers.

Imaginary numbers are not imaginary. They are real!

It is like negative numbers. What do negative numbers really mean? If you have -2 apples, what does that mean?

There is such a thing as negative frequency. What would -200Hz mean?

Hence what does -4 + j3 mean?
- is a mathematical operator
+ is a mathematical operator
j is a mathematical operator

These operators allow us to solve real world problems using well established math operations.

"Imaginary" numbers allow us to solve real world problems. They are use extensively in all fields of physics, engineering and electronics, among many other fields.

One of the first places where imaginary numbers are encountered in electronics is in solving for the output of a simple RC filter. It is also used in understanding Power Factor in AC circuits.
 

Thread Starter

Glenn Holland

Joined Dec 26, 2014
686
I was looking for examples where a negative quantity appears in the square root sign.

I believe reactive power is considered and imaginary quantity, however there's nothing imaginary about reactive power. It just isn't available to be used by the consumer.

Personally, I would prefer the word "Virtual" instead of imaginary. Virtual is the opposite if actual so it may be more practical to use that terminology. However, I'm in no position to change a convention that's been around over 100 years.
 

MrChips

Joined Oct 2, 2009
19,774
Ummm...no, they are imaginary. Otherwise, they'd be real.
I am using the word "real" in the sense that they exist.
It is unfortunate that we use the word "imaginary" when describing "complex numbers".

I don't know what would make a better word to descibe 4 + j5. Pictorially, when imagining these as two vectors, they are orthogonal to one another, in other words, they are 90-degrees apart.
 

BR-549

Joined Sep 22, 2013
4,938
Imaginary numbers are used when trying to mathematically describe the effect/change on a ratio.....whose components.....have been separated in time.
 

MrChips

Joined Oct 2, 2009
19,774
Imagine a canoe being paddled at 4 mph and there is a cross current of 3 mph. What is the true velocity of the canoe?
You can say the velocity of the canoe is 4 + i3.
The magnitude of the velocity would be 5 mph.
 

Papabravo

Joined Feb 24, 2006
12,702
They occur whenever a polynomial of third order or higher has fewer intersections with the horizontal axis than the order of the polynomial would allow, e.g. a third order polynomial can have one, two, or three intersections with the horizontal axis. In the first case there is one real root and one complex conjugate pair of roots, in the second case there is a single real root and a second real root with multiplicity two, in the third case there are three distinct real roots.
 
Last edited:

WBahn

Joined Mar 31, 2012
24,980
There is a misconception of imaginary numbers.

Imaginary numbers are not imaginary. They are real!

It is like negative numbers. What do negative numbers really mean? If you have -2 apples, what does that mean?

There is such a thing as negative frequency. What would -200Hz mean?

Hence what does -4 + j3 mean?
- is a mathematical operator
+ is a mathematical operator
j is a mathematical operator

These operators allow us to solve real world problems using well established math operations.

"Imaginary" numbers allow us to solve real world problems. They are use extensively in all fields of physics, engineering and electronics, among many other fields.

One of the first places where imaginary numbers are encountered in electronics is in solving for the output of a simple RC filter. It is also used in understanding Power Factor in AC circuits.
I would say that j is a constant, not an operator, just as the 4 and the 3 are constants.

There is also an implied operator between the j and the 3, namely multiplication.
 

ErnieM

Joined Apr 24, 2011
8,010
"j" may be considered either a constant or an operator depending on your point of view. I typically visualize it as a right angle turn when extending a one dimensional number line into two dimensional Cartesian co-ordinates.

My favorite take on imaginary numbers comes from Isaac Asimov. Here is the best version of the story I too remember:
Isaac Asimov and imaginary numbers said:
Apropos extensions to the number system, I remember a story Isaac Asimov
once told about the square root of minus one. I read it in one of his
books when I was still at school, and found it very striking. But I've
never seen it repeated anywhere else, and I don't remember the title of
the book, or know whether it's still in print, so I thought I'd post the
story here as I remember it, in case it turns out to be new to some of
you. (The story is a bit unlikely, and Asimov comes out of it perhaps
rather too well, but I quite liked that at the time:)

Anyway, apparently Asimov was studying science and mathematics; but he
had a friend who was studying philosophy, and they often had lunch
together immediately following a lecture that formed a part of his
friend's course.

One day, Asimov happened to arrive early, decided to drop in on the
philosophy lecture, and slipped into the room, to find himself faced
with two lists on the blackboard. One was headed "Realists" and the
other "Mystics", but Asimov was astounded to see the word
"Mathematicians" under the heading "Mystics". So much so, that he
walked forward, put up his hand and asked for an explanation from the
philosophy professor.

"Ah, so we have a mathematical guest who thinks he's a realist!" said
the professor, with a condescending smile. "Mathematicians, my young
friend, are mystics, because they believe in the existence of unreal
objects; for example they believe in the existence of such a thing as
the square root of minus one."

"I know we *call* the square root of minus one imaginary, but it's just
as real as other numbers," said Asimov indignantly. "And we're
certainly not mystics!"

"Very well," said the professor, smirking a little, "if you say that
the square root of minus one is as real as other numbers, come down here
and show us all. Give me the square root of minus one pieces of chalk!"
And all the philosophy students started to laugh.

For a moment Asimov didn't know what to say, and went a bit red, while
the students' laughter grew louder, and the professor's smirk grew
bigger. But then he suddenly shouted out, "OK! I'll do it!" and there
was an immediate silence.

"I'll do it," Asimov went on, "on one condition. And the condition is
that you give me a half piece of chalk to do it with."

"All right," said the professor, a little puzzled. What was going on?
He broke a fresh stick of chalk in two. "Here's your half piece of
chalk. Now what are you going to do?"

"Ah, but wait a minute," said Asimov. "You haven't kept your side of
the bargain yet. This isn't a half piece of chalk. This is *one* piece
of chalk!"

And the students gradually began to laugh again.

"It's a half piece of chalk!" said the professor, getting a bit
flustered. "A new piece of chalk has a regulation length, and this is
half that regulation length, so this is a half piece of chalk."

"Well, now you're springing a *very* arbitrary definition on me," said
Asimov, beginning to enjoy himself. "The *regulation length of a piece
of chalk* enters philosophical debate about reality. Really??? But
even if I were to accept your definition, how can you be sure that this
piece of chalk is *exactly* half the regulation length? You just broke
it casually in two. You'd have to go to a very great deal of trouble to
divide it in half - in fact, you couldn't possibly do what you said
you'd do *exactly*, so I say you don't know what "one half" really
means, in the real world. But either way, if you think you can use a
number like one half to count pieces of chalk, you're wrong."

The professor was speechless.

"But I'm sure you'll agree that doesn't mean that one half isn't a real
number, just because you can't use it to count pieces of chalk with,"
said Asimov. "I'll tell you what: when you have a better idea of what
you mean by one half, or even what you mean by reality for that matter,
we can discuss the square root of minus one."

And he left, and waited for his friend outside in future.
 

WBahn

Joined Mar 31, 2012
24,980
"j" may be considered either a constant or an operator depending on your point of view. I typically visualize it as a right angle turn when extending a one dimensional number line into two dimensional Cartesian co-ordinates.
So, considering it as an operator, explain the notion

j = √-1

An operator needs an operand. What is the operand?

Is √-1 then also an operator? If so, then isn't √4 also an operator?

If an operator can be equal to a constant, then what are the operators '+' and '/' equal to?

An operator maps from a domain to a range. What is the domain of the 'j' operator? What is it's range?
 

WBahn

Joined Mar 31, 2012
24,980
Or here's perhaps a bigger problem with considering 'j' as an operator.

When you fully evaluate an expression, you execute all operators on their respective operands. The end result is a quantity without any operators left.

If you multiply 3 A by 5 V you get 15 W.

But if j is an operator, then 4j is an expression and can't be a result of fully evaluating an expression. That is why j is probably best seen as a unit -- namely the imaginary unit.

Admittedly, this gets into ambiguous territory. How do you "fully evaluate" x = 3 + 4j, for instance? How do you get rid of the '+' operator? One way to dance around this is to say that, like the negative sign, the sign here is not actually an operator, but merely a delimiter within the representation of a complex number. We could have written x as <3,4> (and, in fact, in many instances do exactly that). From that perspective, you would then say that "3 + 4j" is either a single, complex value, or that it is the sum (making the + sign an operator) of a real number and an imaginary number. But then you could also say that the imaginary number 4j is either just an imaginary number (i.e., a quantity with a magnitude of 4 and a unit of j) or you could say that it, itself, is an expression, namely the multiplication of the real number 4 with the imaginary number j. We are then free to interpret it according to which ever makes sense for our purposes at the moment and that the use of the addition and (implied) multiplication operators within the representation of a complex number is coincidental and arbitrary, but very natural since it allows this intuitive and trivially simple mapping back and forth an expression that evaluates to a complex number and the complex number it evaluates to.

But I also realize that this certainly opens the door to make a similar case for considering 'j' to be either an operator or a constant, depending on which makes the most sense in a given situation. I guess I have to be okay with that. But that still leaves some issues. For instance, since j has to be a unary operator, does it precede to succeed its operand (i.e., is it j4 or 4j). Treating it as a constant with implied multiplication neatly sidesteps this issue since multiplication is a binary operation that is commutative.

To me, it just seems a LOT cleaner to leave j as a constant (the imaginary unit) and then expand the definitions of the existing operators to encompass the larger domain.
 

MrChips

Joined Oct 2, 2009
19,774
I would say that j is a constant, not an operator, just as the 4 and the 3 are constants.

There is also an implied operator between the j and the 3, namely multiplication.
I don't have a problem with that. But consider this. Is minus an operator or is it a multiplication with a constant -1? One could imagine multiplication with -1 turns a vector in the opposite direction, i.e. 180 degrees.

I like to think of multiplication by j turns the vector by 90 degrees.
Multiplying by j again turns it another 90 degrees, same as -1,
i.e. j x j = -1.
 

WBahn

Joined Mar 31, 2012
24,980
I don't have a problem with that. But consider this. Is minus an operator or is it a multiplication with a constant -1? One could imagine multiplication with -1 turns a vector in the opposite direction, i.e. 180 degrees.

I like to think of multiplication by j turns the vector by 90 degrees.
Multiplying by j again turns it another 90 degrees, same as -1,
i.e. j x j = -1.
Coming from a programming language implementation perspective, the unary minus sign can be treated either as part of the literal representation (so -1 is a single, literal value) or as a unary operator, in which case -1 is the operator - and the operand is 1. While this has the same effect as multiplying the operand by -1, it is usually not defined that way. It is a pure operator and it evaluates to the additive inverse of its operand. Each language makes the decision of how they will interpret it and then write the lexer, parser, and code generator accordingly. From a pure math perspective we almost always use the former interpretation -- we have numbers, such as +4 and -6, and the signs in front of the digits are not operators at all, but merely part of the representation, just as a period or a comma is part of the representation in 1,234.56. In most compilers, the notion that -15 is a single, literal constant that happens to be less than zero is the most common interpretation, but there are languages that only allow literal values to be positive integers (these languages are often educational languages or languages associated with MCU's and usually don't include the concept of non-integer values, at least directly).

But when we describe what -6 is, we can do it a couple of ways which, probably unwittingly, imply one interpretation over the other. If we say that -6 is the additive inverse of 6, then some could argue that we are describing an operator. But we could also say that -6 is the integer whose additive inverse is the natural number 6. There we are explicitly defining the concept of the single value represented as "-6". In algebra, we need the unary operator interpretation since we can't incorporate the notion of the negative sign being part of the representation when we have something like -x.
 

BR-549

Joined Sep 22, 2013
4,938
Can you use a vector to describe a marble rolling down a hill?

What can a vector describe?

Can you describe an angular motion with a vector?

Why do we break a fundamental mathematical law to do so? Is it proper to use an mathematical impossibility to solve an equation? (they do it all the time.....and then profess how true math is)

We need a new model to replace a vector.....let's call it an impetus.

It would have linear direction and linear velocity like a vector........but it would have area and/or density for magnitude (mass)......it would have a + (R) and - (L) for perpendicular rotation (spin)........and it would have length for density and duration (time). Duration can have different contexts. Such as period. After all, frequency is just a spin to length ratio.

This impetus would describe and define any field or mass entity. An impetus would be fractal. It can describe the planet Neptune's orbit in our system.......the ashtray on my desk......or the particles the make the ashtray on my desk.
 

WBahn

Joined Mar 31, 2012
24,980
It is right here:

j = j * 1 = √-1
..........^
So the operator has both another implied operator and an implied operand?

And what are the operands of the multiplication operator? An operator and a constant?

Sure seems a LOT cleaner to classify it as a constant and be done with it. Is there anything that is not perfectly well defined with that classification?
 

ErnieM

Joined Apr 24, 2011
8,010
So the operator has both another implied operator and an implied operand?

And what are the operands of the multiplication operator? An operator and a constant?

Sure seems a LOT cleaner to classify it as a constant and be done with it. Is there anything that is not perfectly well defined with that classification?
Yes.

Yes.

As you wish.
 
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