1
$\begingroup$

This maybe because I am missing a whole concept itself, but how can vectors have a phase? I have been studying forced oscillations and I read that when multiplying a vector with i (sqrt(-1)) , the vector is rotated by 90 degrees clockwise. Why is that true? How does rotating vectors change their phase? ( I read that the driving force of a forced oscillation always has a phase difference with the oscillation itself).

$\endgroup$
1
  • $\begingroup$ Vector is a very vague term... In which vector space are the vectors you're talking about? Multiplying a vector by $i$ isn't defined for every vector, it has to be a vector living in a complex space. $\endgroup$
    – Miyase
    Aug 14, 2022 at 8:19

3 Answers 3

4
$\begingroup$

I've chosen to concentrate on your most basic question: "[...] how can vectors have a phase?" I'd recommend that you draw diagrams to help you follow my argument.

In your study of oscillations you will have met solutions of the form $$y=A \sin (\omega t +\epsilon)\ \ \ \ \ \ \text {and} \ \ \ \ \ \ x=A \cos (\omega t +\epsilon).$$ What do we mean by $\sin\theta$ and $\cos\theta$ ? They are the projections on to the $y$ and $x$ axes of a 'radial line' of unit length at angle $\theta$ anticlockwise from the $x$ axis. [This is an extension of the elementary 'opp/hyp' and 'adj/hyp' definitions; an extension that gives meaning to sine and cosine for angles of any size, both positive and negative, and gives us our familiar wiggly graphs of $\sin\theta$ or $\cos\theta$ against $\theta$.]

We can, if we wish, regard the radial line as a vector, and its projections as its $y$ and $x$ components.

Now in the case of oscillations,$$\theta=\omega t+\epsilon.$$ Our radial line – and let's give it a length $A$ and symbol $\mathbf A$ – is now a vector $rotating$ with angular velocity $\omega$, because its orientation, $\theta$, changes linearly with time. Its orientation at time $t=0$ is at angle $\epsilon$ anticlockwise from the positive $x$ axis. We call its angle to the positive $x$ axis its phase. So at time $t$ its phase is $\omega t+\epsilon$. The projections of this rotating vector or $phasor$ on to the $y$ and $x$ axes are none other than $$y=A \sin (\omega t +\epsilon)\ \ \ \ \ \ \text {and} \ \ \ \ \ \ x=A \cos (\omega t +\epsilon).$$ [One advantage of this vector viewpoint is that it gives us a simple method of adding together oscillations of the same frequency, even if their amplitudes and phases are different. We simply add the phasors as vectors:$$\mathbf A_{sum}= \mathbf A_1+\mathbf A_2.$$ in which $\mathbf A_1$ and $\mathbf A_2$ are the phasors whose $y$ (or $x$) components are $y_1=A_1 \sin (\omega t +\epsilon_1)$, $y_2=A_2 \sin (\omega t +\epsilon_2)$ (or $x_1=A_1 \cos (\omega t +\epsilon_1)$, $x_2=A_2 \cos (\omega t +\epsilon_2))$. The value of the resultant oscillation at time $t$ is the $y$ (or $x$) component of $\mathbf A_{sum}$.]

A closely related method of dealing with oscillations is to consider our phasor (rotating vector) to be rotating in the complex plane. In that case its tip is the complex number $$z(t)=A\left[\cos(\omega t+\epsilon)+i\sin(\omega t+\epsilon)\right].$$ But we have Euler's relationship $$\cos\theta+i\sin\theta=e^{i\theta}.$$ So turning the vector by $\frac{\pi}2$ anticlockwise gives us $$A\left[\cos\left(\theta+\tfrac{\pi}2\right)+i\sin\left(\theta+\tfrac{\pi}2\right)\right]=Ae^{i\left(\theta+\tfrac{\pi}2\right)}=Ae^{i\theta}\times e^{i\tfrac{\pi}2}.$$ But $$e^{i\tfrac{\pi}2}=\cos \tfrac{\pi}2 +i\sin\tfrac{\pi}2 = i$$ So to turn the vector by $\frac{\pi}2$ anticlockwise on the complex plane we multiply it by $i$.

$\endgroup$
3
  • $\begingroup$ As I'm typing in latex equations, each lower case "i" is automatically changed to upper case "I", and usually needs changing back more than once. This is, I believe, new. Any idea why it's happening? $\endgroup$ Aug 14, 2022 at 11:52
  • $\begingroup$ That's odd. But I think it's your browser (or maybe your OS), not the site software. Do you have any spelling auto-correction stuff enabled? $\endgroup$
    – PM 2Ring
    Aug 14, 2022 at 11:59
  • $\begingroup$ @PM 2Ring Many thanks. I'll look into auto-correction, as it might be enabled by some annoying default! $\endgroup$ Aug 14, 2022 at 12:03
1
$\begingroup$

Consider:

$\vec{r}(t) = \vec{A}_{0}\cos(\omega t)$

The vector $\vec{A}_{0}$ will osscillate periodically with frequency $\omega$

Now, you say that multiplying this by i, will change the phase of this vector, by rotating it.

This is half true, we rotate this vector in the complex plane, when it is in complex form. The real component of this new vector will have the same oscillation frequency, but different phase.

In complex form:

$$\vec{r}(t) = \vec{A}_{0}e^{i\omega t}$$

Using eulers formula see this is just:

$\vec{r}(t) = \vec{A}_{0}cos(\omega t) +\vec{A}_{0}\sin(\omega t) i $

The real of this qauntity is just our original function.

The complex number i can be written in polar form

$$i = e^{i\frac{\pi}{2}}$$

This can also be shown using eulers formula.

Multiplying these two:

$$[e^{i\frac{\pi}{2}}] [\vec{A}_{0}e^{i\omega t}]$$

$$\vec{A}_{0}e^{i(\omega t+ \frac{\pi}{2})}$$

If your comfortable with eulers formula you can immediately see that multiplying by i has rotated this in the complex plane by $\frac{\pi}{2}$

Taking the real part of this vector:

$$\vec{r}(t) = \vec{A}_{0}\cos(\omega t+ \frac{\pi}{2})$$

Part 2:

Forced osscillations almost always uses complex functions to solve the equations, the real part of this answer then is our solution. That is because you can when using complex numbers, the driver force is changed to a complex force you can then show that the real of the solution satisfies the original equation with the real driving force

$\endgroup$
1
$\begingroup$

In math, 2D vectors and complex numbers are pretty much the same (isomorphic space). $$ \overrightarrow{V} = \begin{cases}a \\b \end{cases} \Longleftrightarrow z=a+ib$$ When rotations are involved, complex numbers offer a clear advantage if they are expressed in exponential form . $$ z=a+ib= \rho \big(cos( \theta )+i.sin( \theta )\big) = \rho e^{i. \theta } $$ If you want to rotate a vector by an angle $ \varphi $, you go complex and multiply the complex form by $ e^{i. \varphi } $ $$ \overrightarrow{V}= \rho e^{i. \theta } \longrightarrow \Re ( \overrightarrow{V}, \varphi )=\rho e^{i. \theta }. e^{i. \varphi }=\rho e^{i. (\theta+\varphi ) }$$ In your question, $\varphi= \frac{ \pi }{2} $, so: $e^{i. \varphi }=e^{i. \frac{ \pi }{2}}=i$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.