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Disclaimer: I'm an engineer, not a mathematician

I recently had a fierce discussion (lots of blood) on electronics.stackexchange about phase shifts.

The impedance of a resistor is real, that of a coil imaginary ($j\omega L$), so that adding 3V across a resistor and 4V across a coil in series results in a 5V overall.

Now John (let's call him John) explained the same by using a sine and a cosine function, claiming they're at 90°. I can see where he gets this, and I tried the following to explain that the sine of a number and the cosine of a number are scalars, not vectors, so they can't have a phase difference:

$$\sin(\omega t) = \dfrac{e^{j \omega t} - e^{- j \omega t}}{2j}$$

This is easy for me to visualize: $e^{\omega t}$ and $e^{- \omega t}$ are phasors rotating in opposite directions in the complex plane. Their difference is a vector on the imaginary axis. Dividing by $j$ rotates that vector by $\pi /2$ clockwise, so that it moves from the imaginary axis to the real axis. And then it's a sine, a scalar. Sitting nicely next to the cosine, no phase difference.

My problem is that half a second ago it was still a vector. How does it become a scalar? Or do "being a vector on the real axis" and "being a scalar" mean the same thing?

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    They are orthogonal, in the sense of the $L^2$ inner product. (This is why we can decompose functions into Fourier series!)2012-05-24
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    A vector is (by definition) an element of a vector space. $R$ forms a vector space over itself, so the distinction that something is either scalar or vector is not exactly valid.2012-05-24
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    The sine of a number is a number and can't have a phase difference with another number. The sine as a function over the real line is another matter. The sine function and the cosine function obviously have a $90^\circ$ phase difference, or I don't know what phase difference is.2012-05-24
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    I think the answer is yes: being a vector on the real axis and being a scalar is the same, this is because $\mathbb C$ is not just a vector space over $\mathbb R$ but a field extension. So that $1$ is a very special vector in $\mathbb C$ (as a real plane), it is its identity as a field. The subspace spanned by $1$ over $\mathbb R$ is then distiguished: it is both the "real axis" and the "scalars" -if you just think of $\mathbb C$ as a real plane.2012-05-24
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    @Rahul - Yes, that's what John also used as argument: sin(x + $\pi$/2) = cos(x). But that only means that the sin and cos then are equal. Sin(37°) = 0.6 and cos(37°) = 0.8. How are 0.6 and 0.8 90° apart??2012-05-24
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    A link to the original comment would be very useful, so that we can see what John was saying, and pick up on any details that you missed.2012-05-24
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    I think the comments show that there is some ambiguity in the meaning of "at right angles" - but also that orthogonality is a rich mathematical concept with many applications. The naïve thought that - in a right-angled triangle with hypotenuse 1, sin and cos are literally at right angles is related, as also is the fact that the projections of points on the unit circle onto orthogonal axes track sin and cos at right angles.2012-05-24
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    @Hurkyl - The question is [here](electronics.stackexchange.com/questions/32269/). John is Telaclavo, I am stevenvh (I must have my existing account here merged with this new one). Most of the discussion has been deleted, however, in the name of peace :-)2012-05-24
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    @user32112, you have to distinguish between the sine *of a particular number* and the sine function itself. Your argument is kind of like saying, it's not true that the President of the United States is elected every four years, because Barack Obama is the President of the United States, and Barack Obama is not elected every four years.2012-05-24
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    @Rahul - I think I can see that. I see the curve of both functions, and they're indeed $\pi$/2 apart. That $\pi$/2 is in the *argument* of the function, in the function's value there's nothing of that left. I understand that the sine value 0.6 is Obama, but this guy claims that Obama and Bush are orthogonal. Two scalars!2012-05-24
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    @user32112: thanks for the link. It helped clear up what he was saying. (and it let me re-derive a neat fact I had long forgotten too! See the edit to my answer)2012-05-24

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