Understanding EM Wave Equations

[quietrain]

hi, i have a problem understanding why the wave equations are as such

if wave is moving left, it is
f(z,t) = Acos(kz + wt - d)

if wave moving right ,
f(z,t) = Acos(-kz -wt + d)

finally i don't know what this represent
f(z,t) = Acos(kz - wt + d)

where A is constant ,
k is wave number,
z is the direction of propgation of wave
w is angular frequency
t is time
d is phase constant

in particular, i have trouble visualizing why if wave is moving right, we + d? and vice versa, left means -d? and what's the 3rd equation? general one?

thanks!

 

[tiny-tim]

hi quietrain!

if wave is moving left, it is
f(z,t) = Acos(kz + wt - d)

the direction the wave is moving means the direction a particular part of the wave is moving, eg a particular peak of the wave …

a particular peak (or any part) has a fixed value of cos …

and if you fix cos, then you must fix kz + wt,

ie kz + wt = constant, or z = -(w/k + constant/k)t, which is to the left

finally i don't know what this represent
f(z,t) = Acos(kz - wt + d)

that has kz - wt = constant, or z = (w/k + constant/k)t, which is to the right

f(z,t) = Acos(-kz -wt + d)

cos(-x) = cos(x), so that's the same as f(z,t) = Acos(kz + wt - d)

 

[quietrain]

i can't visualize it :(

but basically, i just make the argument of cos 0? by letting z assume some value to achieve that? (but why do i want to make the argument 0?)

so if my z ends up with -ve, then its to the left? if its +ve then its to the right?

this reminds me of an earlier post i made about the delta dirac function where someone told me i don't have to visualize whether the graph is shifting left or right, all i needed to do was to make the argument of the delta function 0.

 

[quietrain]

ah i found it, its by homology

In both cases the delta function is zero unless x=ax-a is a shift right by a, and a-x=-x+a is a reflection of the x-axis and then a shift left (which means a shift right). But just try to see it this way, the integral is zero unless the argument of the delta function is zero. That happens when (above) x=a. But notice that:

$$\int_{\mathbb{R}}f(x)\delta(x+a)dx=f(-a)$$

I can do that without thinking about how the function's been shifted but rather just seeing where the delta function is equal to zero. Its the same in 3D. When is $$\vec{r}-\vec{r'}=\vec{0}$$ same time that $$\vec{r'}-\vec{r}=\vec{0}$$

 

[yungman]

cos 0 is just a convient point to track, it is the peak. $\omega t - kz$ is the connection between time and displacement. By making $(\omega t - kz)$= constant, you can track the point on the sine wave moving as time change.

For example, if t increase, in order for $\omega t - kz$ = constant, z has to increase, that show the wave is traveling forward ( to the right ). If it is $( \omega t + kz)$, then if t increase, z has to go in the negative direction in order to keep $\omega t + kz$= constant.

All it is that it is easiest to use the peak as the reference point so $\omega t - kz$ = 0 where $cos(\omega t - kz) = 0$.

$cos( -\omega t + kz)$ is really using in math only. We usually use $cos( \omega t -kz) \;\hbox { which is the }\; \Re[e^{j(\omega t - kz)}]$ in phasor form study more commonly used in engineering electromagnetics for transmission lines. But that is a totally different big topic that electrodynamics don't have to deal with ( consider yourself lucky!)

 

[quietrain]

i went to think about it ,

for example we have a line x = 1

if now i do a x-5, then wouldn't it be x = -4 now.

so x-5 shifts the graph to the left?

so why is it when kz - wt, it shifts to the right?

 

[quietrain]

cos 0 is just a convient point to track, it is the peak. $\omega t - kz$ is the connection between time and displacement. But making $(\omega t - kz)$= constant, you can track the point on the sine wave moving as time change.

For example, if t increase, in order for $\omega t - kz$ = constant, z has to increase, that show the wave is traveling forward ( to the right ). If it is $\omega t + kz$, then if t increase, z has to go in the negative direction in order to keep $\omega t + kz$= constant.

All it is that it is easiest to use the peak as the reference point so $\omega t - kz$ = 0 where $cos(\omega t - kz) = 0$.

wow i learn something new again

so basically i just need to see how t and z is related in the equation?

t always increases right? so if z is +ve, wave moves right

if z is -ve, wave moves left?

but now comes the problem , how do i take into account the phase constant?

do i just include it in the z to make the argument of cos = 0 like in tim's post?

but now, what does a +ve or -ve phase constant tell me? my textbook says its a "delay"?

 

[yungman]

i went to think about it ,

for example we have a line x = 1

if now i do a x-5, then wouldn't it be x = -4 now.

so x-5 shifts the graph to the left?

so why is it when kz - wt, it shifts to the right?

THat's what I was trying to explain. You normally set up +ve z towards the right.

 

[yungman]

wow i learn something new again

so basically i just need to see how t and z is related in the equation?

t always increases right? so if z is +ve, wave moves right

if z is -ve, wave moves left?

but now comes the problem , how do i take into account the phase constant?

do i just include it in the z to make the argument of cos = 0 like in tim's post?
yes.
but now, what does a +ve or -ve phase constant tell me? my textbook says its a "delay"?
Just move forward or backward a constant amount. It is just an offset.

Phase constant is considered as just an offset. The idea is the same. So you look at $(\omega t -kz + \phi)$ = constant and you see the phase constant just kind of move the point back or forward by a constant amount.

Takes me a while to figure this out, and it is just that simple!

 

[quietrain]

Phase constant is considered as just an offset. The idea is the same. So you look at $(\omega t -kz + \phi)$ = constant and you see the phase constant just kind of move the point back or forward by a constant amount.

Takes me a while to figure this out, and it is just that simple!

oh so in this case, since t increase, z has to increase for constant, so it is moving right right?

so if phase constant is +ve, means shift right ? -ve means shift left?

so if question says out of phase by 90degrees, means i plus 90 or minus 90?

 

[yungman]

oh so in this case, since t increase, z has to increase for constant, so it is moving right right?

so if phase constant is +ve, means shift right ? -ve means shift left?
I believe so. It is confusing also. +ve means phase lead. It will take less time to reach a point on the positive z direction...like you have a head start with a lead.
so if question says out of phase by 90degrees, means i plus 90 or minus 90?

Just say out of phase in not enough, because it can be lead or lag 90 deg. +90 is leading by 90 degree, -90 is lagging by 90 deg.

This is my understanding, someone might want to double check.

 

[quietrain]

Just say out of phase in not enough, because it can be lead or lag 90 deg. +90 is leading by 90 degree, -90 is lagging by 90 deg.

This is my understanding, someone might want to double check.

common sense tells me that +90 is leading too and vice versa...

but after how i see the wave equations is like, common sense doesn't seem right :(

 

[yungman]

common sense tells me that +90 is leading too and vice versa...

but after how i see the wave equations is like, common sense doesn't seem right :(

+90 is lead is quoted straight from the book, not my interpretation.

 

[yungman]

Think of it this way, let t=0 so the peak at $(-kz+\phi)=0$. So if \phi is +ve, the peak is at +ve z location.

 

[quietrain]

ah i see .. i guess common sense was right after all thanks!

 

[yungman]

ah i see .. i guess common sense was right after all thanks!

I think common sense still work, just takes a lot of effort to develope the "common sense" in this field. I am no expert, I am struggling myself. I responed in more detail in the other post. It just take time to acquire the "common sense". Buy more used books, they all present the material a little different. I have 8 books on this subject! Whenever I don't understand one thing, right away I hit the second and the third book and see whether I get a better explanation.

 

[quietrain]

I think common sense still work, just takes a lot of effort to develope the "common sense" in this field. I am no expert, I am struggling myself. I responed in more detail in the other post. It just take time to acquire the "common sense". Buy more used books, they all present the material a little different. I have 8 books on this subject! Whenever I don't understand one thing, right away I hit the second and the third book and see whether I get a better explanation.

wow... 8 bookS?

thats amazing...

 

[yungman]

I am a self studier, I need all the help I can get. If you go to school, you have your professor to go to, I don't! Right now I am stuck on an equation for two days and can't move on, maybe you can help.

 

[quietrain]

I am a self studier, I need all the help I can get. If you go to school, you have your professor to go to, I don't! Right now I am stuck on an equation for two days and can't move on, maybe you can help.

oh the ones with the many equations one? i died when i saw the string of equations :(

anyway, i may be in sch , but i realize not all profs are good teachers :(

 

[yungman]

oh the ones with the many equations one? i died when i saw the string of equations :(

anyway, i may be in sch , but i realize not all profs are good teachers :(

Yes, that's the one!