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:

<br /> <br /> \int_{\mathbb{R}}f(x)\delta(x+a)dx=f(-a)<br /> <br />

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&#039;}=\vec{0} same time that \vec{r&#039;}-\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!