# Different formulas, same physics?

1. Jul 25, 2012

### nonequilibrium

Hello,

Simple but perhaps odd question.

$e^{i(kx+\omega t)}$ and $e^{i(-kx-\omega t)}$

The first formula describes a wave (with wavevector k and angular frequency omega) travelling to the left. And the second formula? I thought that it physically describes exactly the same situation (negative wavevector travelling to the right, being physically the same (what else does it mean?) to the positive wavevector travelling to the left), however come to think of it these two formulas are mathematically very different, and it is not the case that they differ only be constant phase or something like that. So what's up?

2. Jul 26, 2012

### Simon Bridge

These are the phasor descriptions. One is rotating clockwise and the other anti-clockwise, so for the same k and ω you'd get a different relative phase.

z1(x,t) = exp(i(kx+wt)) = cos(kx+wt) + i.sin(kx+wt)

z2(x,t) = exp(-i(kx+wt)) = cos(kx+wt) - i.sin(kx+wt)

... the real parts are the same is what you are thinking right?

Normally a sinusoidal wave travelling left to right would be Acos(k(x-vt)) = Acos(kx-wt) becauyse v=fλ - notice the minus sign in there? Now if you want to go the other way you have Acos(k(x+vt))=Acos(kx+wt) ... so now, why would you have a minus sign in front of the wave number: what would be the physical meaning of that?

3. Jul 26, 2012

### PhilDSP

From Euler's identity

$e^{i\pi} + 1 = 0$

The basic relationships Simon Bridge mentions can be derived (or rather directly from Euler's formula):

$e^{ix} \ \ \ = cos \ \ x \ + \ i \ sin \ \ x$
$e^{-ix} \ = cos \ \ x \ - \ i\ sin \ \ x$

Last edited: Jul 26, 2012
4. Jul 26, 2012

### nonequilibrium

Simon, interesting. Indeed, when one uses exp(i(kx-wt)) to describe a physical wave, one implicitly means the real part is the physical part, and indeed both waves in my OP have the same real part so describe the same physics, however since they're mathematically non-trivially different expressions, the result I get depends on which complex version I choose for the calculations (as I experienced first hand yesterday)! This is kind of a surprise to me; implies the complex method is trickier than it seems... Thank you.

5. Jul 26, 2012

### mikeph

Sorry to butt in, but I don't really understand the difference- if they represent physically identical waves, why do your results differ? Specifically why would two different analyses of the same problem yield two different results? Surely one is wrong and the other one represents the truth?

Or is that a simplification?

6. Jul 26, 2012

### Simon Bridge

@mr. vodka: Perhaps it would help to show the situation where you got different results depending on the representation :)

You do have to be careful to make sure that your math describes something physical ... what does it mean, physically, when you put a minus sign in front of the wave number?

Have a look at what happens in each representation if you start with an initial phase angle of π/2 (resulting in a sine rather than a cosine wave)?

How does the phase of each wave vary (at a particular x) with time?

7. Jul 26, 2012

### Studiot

The general solution for the wave equation

$$\psi = {C_1}\exp \left[ {i\left( {\omega t - kz} \right)} \right] + {C_2}\exp \left[ { - i\left( {\omega t - kz} \right)} \right]$$

in complex exponential format has two arbitrary complex constants C1 & C2
This means there are four constants in all ReC1; Im C1; ReC2; Im C2
In normal circumstances we require a wave with a real amplitude. This leads to boundary conditions that can only provide two constants.
We can do this by restricting C2 to the complex conjugate of C1

There are then 4 ways to write the wave, each with two arbitrary constants.

$$\begin{array}{l} \psi = C\exp \left[ {i\left( {\omega t - kz} \right)} \right] + {C^*}\exp \left[ { - i\left( {\omega t - kz} \right)} \right] \\ \psi = A\cos \left( {\omega t - kz + \phi } \right) \\ \psi = {B_1}\cos \left( {\omega t - kz} \right) + {B_2}\sin \left( {\omega t - kz} \right) \\ \psi = {\mathop{\rm Re}\nolimits} \{ D\exp [i\left( {\omega t - kz} \right)]\} \\ {\mathop{\rm Re}\nolimits} \{ C\} = \frac{1}{2}{B_1} = \frac{1}{2}A\cos \phi \\ {\mathop{\rm Im}\nolimits} \{ C\} = - \frac{1}{2}{B_2} = \frac{1}{2}A\sin \phi \\ {\mathop{\rm Re}\nolimits} \{ D\} = A\cos \phi \\ {\mathop{\rm Im}\nolimits} \{ D\} = A\sin \varphi \\ \end{array}$$

Unless you get the complex conjugate in your second expression you will get a mismatch.

Edit
Note it is conventional to write wt-kz not wt+kz. This really does not matter since k, w can be positive or negative so it will amount to the same thing and the sign will be taken care of in the calculation.

Last edited: Jul 26, 2012
8. Jul 26, 2012

### Simon Bridge

Students are often more comfortable with kz-wt ... from the derivation from the space distribution.

An arbitrary distribution in space, at t=0, may be: $\psi(z,0)=f(z)$ ... if it travels in the +z direction with speed v then $\psi(z,t)=f(z-vt)$.

For a sine wave, we write $f(z)=A\cos(kz):k=2\pi/\lambda$
So, if we want this to travel in the +z direction it becomes $$\psi(z,t)=A\cos(k(z-vt))$$... and we realize that $kv=\omega$ because $v=\nu \lambda$ and $\omega = 2\pi \nu$.

Of course, from the symmetry of the cosine: $\psi(z,t)=A\cos(kz-\omega t) = A\cos(\omega t - kz)$ so it's all six of one and a half-dozen of the other :)

But I was thinking that, although there are several ways that the two phasor representations can be considered different, you only see the difference mattering in things like quantum mechanics where the complex part can be responsible for the interference terms. So I am curious to find out what the actual problem was that meant the two representations did not give the same answer. Classically, something that relies on the phase angle would do this.

It may even be something as simple as making a mistake... nothing deep from the math at all.

9. Jul 26, 2012