Maths of GPS: Understanding Wave Packet Group Velocity

[guillefix]

I am doing my extended essay on the maths involved in GPS, and I am trying to understand the concept of the group velocity of a wave packet, http://www.mathpages.com/home/kmath210/kmath210.htm" they explain it more or less, but I still don't get how do they differenciate dw/dk to get c/n-ck/n^2(dn/dk), and in other book I read it says it is v+k(dv/dk). And what is exactly group velocity, the velocity of the envelope of the wave resulting from adding two waves or what?

Thank you for your help.

 

[johng23]

Well, you should be able to show that those two statements are equivalent just by writing the second in terms of the index of refraction.

Strangely I always have trouble with this derivation. I think it's because I'm not careful from the beginning about distinguishing between free space wavelength and \lambda(\omega) in the various equations that relate w,k,n etc. If you start from the relation w/k = c/n you should be able to arrive at the first equation you posted easily.

The one thing I don't understand, which I'm sure is some basic calculus that I'm forgetting, is this: why is dn/dk not \lambda_{0} /2\pi? I understand that n(\lambda) is some complicated function for a given material which depends on electronic resonances, etc. But it's a true statement that k(\omega)=2\pin(\omega)/\lambda_{0}, and I can't remember the exact mathematical reason why I can't do the differentiation.

 

[guillefix]

I am confused because when you integrate dw/dk i think you should get ck/n + ck/n, and using the formula you mentioned w=ck/n. And why is it that when you derivate ck/n + ck/n, you consider the firs n as a constant ant the second n as a function of k?

Thank you

 

[johng23]

I guess you are trying to work backwards from the answer, but I wouldn't recommend that because you need to do integration by parts, for the same reason you need to do chain rule to solve the problem initially. Make sure you can get that expression when you differentiate w=ck/n, then you can try to work backwards if you want.

 

[nasu]

For both forms , you can start from the definition of the group velocity and write omega in terms of the phase velocity. (using $ v_{ph}=\omega/k)

$ v_g=\frac{d\omega}{dk} = \frac{d (kv_{ph})}{dk}$
In order to get the group velocity in terms of phase velocity (the second form in OP), you just need to apply the chain rule:
$ v_g=v_{ph}+k\frac{dv_{ph}}{dk}$

To express the group veolocity in terms of index of refraction (first form in OP), you can write $ v_{ph}=c/n
and again apply chain rule:
$ v_g=\frac{d (kv_{ph})}{dk}=\frac{d}{dk}\frac{kc}{n}=\frac{c}{n}-\frac{kc}{n^2}\frac{dn}{dk}$

 

[nasu]

The one thing I don't understand, which I'm sure is some basic calculus that I'm forgetting, is this: why is dn/dk not \lambda_{0} /2\pi? I understand that n(\lambda) is some complicated function for a given material which depends on electronic resonances, etc. But it's a true statement that k(\omega)=2\pin(\omega)/\lambda_{0}, and I can't remember the exact mathematical reason why I can't do the differentiation.

The reason is that \lambda_0 is a function of k and not a constant.

 

[johng23]

The reason is that \lambda_0 is a function of k and not a constant.

No because in my terminology \lambda_0 is the free space wavelength 2\pic/\omega.

 

[nasu]

No because in my terminology \lambda_0 is the free space wavelength 2\pic/\omega.

And still a function of omega or k, depending on your choice of variable. It's not a constant.
Even in free space, the components of the wave packet have different wavelengths so different values for lambda_0 .
What do you think is
d$ \lambda_0 $/d$ \omega $, according to your definition ?
Is it zero? Not at all.
Then what is
d$ \lambda_0 $/dk?

 

[guillefix]

For both forms , you can start from the definition of the group velocity and write omega in terms of the phase velocity. (using $ v_{ph}=\omega/k)

$ v_g=\frac{d\omega}{dk} = \frac{d (kv_{ph})}{dk}$
In order to get the group velocity in terms of phase velocity (the second form in OP), you just need to apply the chain rule:
$ v_g=v_{ph}+k\frac{dv_{ph}}{dk}$

To express the group veolocity in terms of index of refraction (first form in OP), you can write $ v_{ph}=c/n
and again apply chain rule:
$ v_g=\frac{d (kv_{ph})}{dk}=\frac{d}{dk}\frac{kc}{n}=\frac{c}{n}-\frac{kc}{n^2}\frac{dn}{dk}$

Thank you, I think that this is what I needed to understand it, but isn't what you are using the product rule?

 

[nasu]

d(fg)/dx=(df/dx)*g+f*(dg/dx)

I suppose "product rule" is the more appropriate name. Chain rule refers to the case of compound functions.