GREPhysics.NET: GR8677 #59

GR8677 #59

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Wave Phenomena $\Rightarrow$ }Group Velocity

Group velocity is $v_g=d\omega/dk$ . So, take the derivative of the quantity $\omega = \sqrt{c^2 k^2 + m^2}$ to get

$v_g=d\omega/dk = \frac{c^2k}{\sqrt{c^2 k^2 + m^2}}. <br />$

Use the above equation to test the 5 choices:

(A) As $k \rightarrow 0 \Rightarrow \omega = 0$ , not infinity. The first condition doesn't work, no need to test the second (don't have to remember L'Hopital's rule).

(B) Wrong for the same reason as (A).

(C) Wrong because $v_g$ approaches $0$ , not $c$ , as $k \rightarrow 0$ .

(D) As $k \rightarrow \infty$ , $v_g \approx \frac{kc^2}{\sqrt{c^2 k^2}}=c$ , since $m \ll (ck)^2$ . So, $v_g$ doesn't tend towards $\infty$ . This choice is wrong.

(E) This is it. The conditions work (and it's the only choice left).

Alternate Solutions

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Comments

chrisfizzix
2008-10-03 13:16:24

I solved this by some quick common sense. The group velocity is just the velocity of the particle associated with the wave motion. For any physical system, the group velocity can never be larger than c. Thus, any answer that has $v \rightarrow \infty$ is out - goodbye A, B, D. The propagation vector k is absolutely related to group velocity, and if $k \rightarrow 0$ then the wave better not be going anywhere. Thus, C is out, leaving E.

neon37
2008-11-05 00:07:59

Hi I read this article in wiki http://en.wikipedia.org/wiki/Group_velocity that seems to suggest group velocity can be greater than speed of light. It says quote:
Since the 1980s, various experiments have verified that it is possible for the group velocity of laser light pulses sent through specially prepared materials to significantly exceed the speed of light in vacuum. However, superluminal communication is not possible in this case, since the signal velocity remains less than the speed of light.

well i can sorta visualize it but if someone is expert in this subject could you please explain it so that I can confirm.

Thanks

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