Understanding for group velocity and phase velocity?

[lanbeiming]

understanding for group velocity and phase velocity??

Hi, everyone. I learned physics for 3 years, but up to now I still cannot understand clearly the difference between the group & phase velocity. What is exatly "phase"? and what is exactly "group". From book I read the Vph corresponds to the transmisssion of phase and the Vg corresponds to the transmission of node or based on the velocity of signal modulation by some approximation. But still feel abstract about these words.

Also, if we use a light source and a detector locate at two position with distance Zo, and put a material between them. and in some cases Vph could <Co. However the result by using this material is just the delayed phase of the signal. What does this mean? By what method could we observe this less velocity? Does this mean that the time duration from the source to the signal would smaller than Zo/Co?

Many thanks!

 

[jtbell]

This page has an applet which shows a wave packet whose group velocity is greater than the phase velocity. Can you see the two different speeds?

http://gregegan.customer.netspace.net.au/APPLETS/20/20.html

 

[Rohitasch]

Hi, everyone. I learned physics for 3 years, but up to now I still cannot understand clearly the difference between the group & phase velocity. What is exatly "phase"? and what is exactly "group".

v_g and v_p explanations suck in textbooks, especially the long detour they take. Its pretty simple v_g/c = c/v_p means that the dimensions used can be written in unit terms, ie, light-seconds covered per second for v_g (the particle's velocity) and seconds spent per light-second for v_p (by the particle with velocity v_g, measured from its own frame of reference).
So, v_g/v_p = (v_g^2)/(c^2). Also, v_g/c = (Compton wavelength of particle)/(deBroglie wavelength of particle) and the reciprocal of this is equal to v_p/c.
In other words, while the particle travels {lambda_Compton} distance, the information of its presence travels {lambda_deBroglie} distance.
Note that nothing is traveling faster than c here, inspite of what bull a lot of textbooks have.

 

[Antiphon]

This is not rigorous but gets the idea across.

Imagine water waves heading slowly toward a shore. Single out a peak The group velocity is the speed at which the peak travels.

Now imagine the the waves come into shore at an angle of 45 degrees. Locate the point where the crest of the wave you picked intersects the shoreline. This point moves along the shore as the wave comes in. The phase velocity is like the speed of this intersection as it moves laterally along the shore.

If the wave is nearly perpendicular to the shore, the phase velocity can approach infinity. The phase velocity is not the movement of energy which is more closely associated with the group velocity.

 

[lanbeiming]

This page has an applet which shows a wave packet whose group velocity is greater than the phase velocity. Can you see the two different speeds?

http://gregegan.customer.netspace.net.au/APPLETS/20/20.html

Nice applet!

 

[lanbeiming]

This page has an applet which shows a wave packet whose group velocity is greater than the phase velocity. Can you see the two different speeds?

http://gregegan.customer.netspace.net.au/APPLETS/20/20.html

jsut one more question, can you show me the meaning of shutter-down？i'm not sure so see that the pulses can't be used to send messages faster than light in vacuum.

thanks!

 

[lanbeiming]

v_g and v_p explanations suck in textbooks, especially the long detour they take. Its pretty simple v_g/c = c/v_p means that the dimensions used can be written in unit terms, ie, light-seconds covered per second for v_g (the particle's velocity) and seconds spent per light-second for v_p (by the particle with velocity v_g, measured from its own frame of reference).
So, v_g/v_p = (v_g^2)/(c^2). Also, v_g/c = (Compton wavelength of particle)/(deBroglie wavelength of particle) and the reciprocal of this is equal to v_p/c.
In other words, while the particle travels {lambda_Compton} distance, the information of its presence travels {lambda_deBroglie} distance.
Note that nothing is traveling faster than c here, inspite of what bull a lot of textbooks have.

thanks for reply, but i don't understand this:
v_g/c = (Compton wavelength of particle)/(deBroglie wavelength of particle) and the reciprocal of this is equal to v_p/c.

how could the left-hand-side ratio and right-hand-side ratio(p/mc) be linked? could you show me the duduction or some references?

 

[lanbeiming]

This is not rigorous but gets the idea across.

Imagine water waves heading slowly toward a shore. Single out a peak The group velocity is the speed at which the peak travels.

Now imagine the the waves come into shore at an angle of 45 degrees. Locate the point where the crest of the wave you picked intersects the shoreline. This point moves along the shore as the wave comes in. The phase velocity is like the speed of this intersection as it moves laterally along the shore.

If the wave is nearly perpendicular to the shore, the phase velocity can approach infinity. The phase velocity is not the movement of energy which is more closely associated with the group velocity.

thanks, but if you don't mind, could you tell me how the velocity of intersection points are related to the phase velocity?

 

[lanbeiming]

i still have a question, if we put a generater and detecter on both sides of a, say cubic-like, material, with size say,a*a*a, then the time we measured from the generator to the detecter would really be different from a/Co ?

 

[torquil]

jsut one more question, can you show me the meaning of shutter-down？i'm not sure so see that the pulses can't be used to send messages faster than light in vacuum.

thanks!

I think it is because it is not possible to block the large pulses using the shutter? In order to create a message, e.g. sending a stream of bits, you need to be able to block some of the pulses to be able to generate both 0s and 1s. But when I try to block the pulse in that applet, it is recreated on the other side of the shutter... So I'll just be sending a steady stream of 1s...

Am I right?

 

[Rohitasch]

thanks for reply, but i don't understand this:
v_g/c = (Compton wavelength of particle)/(deBroglie wavelength of particle) and the reciprocal of this is equal to v_p/c.

how could the left-hand-side ratio and right-hand-side ratio(p/mc) be linked? could you show me the duduction or some references?

Well, we break up h. So,
h = m c $$\lambda_{C}$$ and,
h = m v $$\lambda_{B}$$.
Here, v is the velocity of the particle with mass m and lambda_{C} and lambda_{B} are its Compton and D'Broglie wavelengths respectively. Also, $$v_{g}$$ = v.
Cancelling off m and juggling the rhs and lhs terms gives the ratio.