Dispersion: Phase Velocity Versus Group Velocity, PHYS 372

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  • Опубліковано 23 січ 2025

КОМЕНТАРІ • 33

  • @aieousavren
    @aieousavren Місяць тому +1

    EXTREMELY clear explanation. I always had vague ideas of phase velocity and group velocity, but this is the first time I've ever actually followed the derivation for e.g. the phase velocity. THANK YOU!! ❤ Super grateful!

  • @adityabaghel1270
    @adityabaghel1270 11 місяців тому +5

    SO DAMN CLEAR EXPLANATION I CAN'T BELIEVE, THANKS HELL A LOT

  • @victorwallace8974
    @victorwallace8974 3 роки тому +12

    So clearly explained. I get it now. 20 minutes well spent.

  • @nguyensontung5923
    @nguyensontung5923 8 місяців тому +3

    20 minutes and it solves my 2-day problem. Big thanks!

  • @Marzart_Marseille
    @Marzart_Marseille Місяць тому +2

    Very nice explained! All your vedios!

  • @Mushicus
    @Mushicus 2 роки тому +3

    I've been trying to design, essentially, a microwave cavity resonator and this has helped me conceptualize the wave pattern so much! Thank you!

  • @augustineokekeoma1750
    @augustineokekeoma1750 2 роки тому +4

    Thank you so much for your hardwork. I hv finally understood this.

  • @chevestong
    @chevestong 8 місяців тому +2

    16:09: Dr. Remillard says "the group velocity is LESS than the phase velocity", which I believe was a mistake, since it's written that the group velocity is GREATER than the phase velocity, which is true for d v_p / d omega > 0.

    • @stephenremillard1
      @stephenremillard1  8 місяців тому +1

      You are right. What is written is correct. Thanks for pointing that out.

  • @M_0892
    @M_0892 3 роки тому +3

    Great! Lots of visual examples. Thx a lot!

  • @tasminkhan6165
    @tasminkhan6165 10 днів тому

    At 11:02, why did we take the phase of the cosine to be directly equal to zero instead of the differential of it (with respect to t) to be equal to zero like how we did for phase velocity.

  • @밤고구마-z3i
    @밤고구마-z3i 3 роки тому +8

    What an amazing lecture! Thanks a lot!

  • @myasterr
    @myasterr 5 місяців тому +1

    Fantastic explanation. Many thanks!

  • @andrealiu8650
    @andrealiu8650 2 роки тому +2

    Great video and truly helpful, thank your!

  • @MRF77
    @MRF77 2 роки тому +2

    I wish you had all your QM lecture organized in your QM playlist. But thanks for amazing lecture.

  • @michaellovejoy8751
    @michaellovejoy8751 3 роки тому +3

    Thank you for this video! Very helpful!

  • @valor36az
    @valor36az 3 роки тому +3

    Excellent

  • @thomasolson7447
    @thomasolson7447 6 місяців тому

    It's a little bit different from what I taught myself using quadratics. That one in the middle (9:46) is a Second Kind type, or Fibonacci-like discrete homogenous sequence, even though it has that plus sign. That would make the magnitude equal to one, but that can be manipulated to r^((t-1)/2). I don't know how that changes, given the outside term. Is that a cubic? Are they triangle waves in 3d? The 'e' on the outside is vector angle addition. The magnitude is 1. That one is easier, r^t. There should be another function that pairs with this. Ψ(n+1)+Ψ(n-1)+f(n)=0 (I'm too lazy to do notation correctly). I might be wrong though, given that it is cubic.
    Anyway, that's wrong. You can't do the 2cos(dwt-dkx) thing. ChatGPT always simplifies that function, but it's wrong, I checked. It will work if time and displacement is an integer. It becomes a complex number when they are rational (fractional). 2cos(dwt-dkx) doesn't appear to become a complex number. Standup Maths: "Complex Fibonacci Numbers" kind of addresses it.
    Ψ(n+1)/Ψ(n) where n = -2 -i*2.. 2+i*2 should be a magnetic field. Three poles, I'm guessing, project it on a sphere. You might need to customize the tool you use to graph it because it's cubic.

  • @jarlhamm
    @jarlhamm Рік тому +1

    This is fantastic, thank you so much.

  • @sobhisaeed3095
    @sobhisaeed3095 Рік тому

    Amazing lecture! Very efficient, thanks a lot!

  • @alvarodemontes3818
    @alvarodemontes3818 Рік тому

    Thank you, very interesting.
    Where could i find info on the "extreme normal dispersion" ?

  • @official-ikechukvvu
    @official-ikechukvvu 3 роки тому +3

    Thank you for this!

  • @robertcoughlin7604
    @robertcoughlin7604 2 роки тому +1

    14:22 not the product rule, it's an inverse application of the quotient rule

  • @rosarionapoli9765
    @rosarionapoli9765 3 роки тому

    On the Group velocity slide i read "This can exceed the speed of light". I think it's the fase velocity that can exceed the speed of light, and never the Group velocity, because it's the envelope that brings energy/information. (In the case of a particle it's also physically the probability of finding it somewhere, so in some sense the position, that moves at the Group Velocity... Sure a particle can't move faster than light, right?)

    • @stephenremillard1
      @stephenremillard1  3 роки тому +5

      Good question. I'll try my best here. The group velocity of an electromagnetic (EM) wave can exceed the speed of light in vacuum. But energy does not travel with it. There are a few ways to visualize this. It might help to think about a shadow being cast by an object moving near the speed of light. The shadow on the ground can exceed the speed of light in vacuum, c. Things such as shadows and wave group profiles can move faster than c as long as matter and energy don't move with them. Now imagine the front of an EM wave. All waves have a beginning, and that wave front propagates at the speed of light (phase velocity) in the medium. That's the speed of energy/information. The shape, or modulation, of the wave is the result of interference between frequency components, which can have different phase velocities in dispersive media. The destructive interference nodes that define the group, just like a shadow, might be moving faster than the energy - maybe even faster than c. But they don't arrive at the destination earlier than the wave front. Each component carries spectral energy which travels at the speed of light in the medium. It isn't the energy that can travel faster than c. Rather the interference between components of the wave is what can travel faster than c.

    • @m_tahseen
      @m_tahseen Рік тому

      ​​Well explained ... But if the interference between waves travels > c , then doesn't it imply that information has travelled > c ... And that's again an impossibility @@stephenremillard1

    • @stephenremillard1
      @stephenremillard1  Рік тому +2

      Although the interference is the information, it is carried by the energy, which cannot exceed c. You will notice that as a wave pulse travels, the phase fronts might be moving faster than the pulse itself, but they die out at the edge of the pulse. The interference may be moving around faster than c within the pulse, but it will not get there faster than the pulse can get there.

  • @shishaykidane6836
    @shishaykidane6836 Рік тому

    But what is the concept of K(function of lamda)?

    • @stephenremillard1
      @stephenremillard1  Рік тому +1

      k=2*pi/Lambda is the "wave number". Inside of a sinusoid in x, sin(kx), it's a spatial frequency. When used in certain topics, such as Fourier optics and band theory, 2*pi/Lambda might be better referred to as the "spatial angular frequency of the wave".

  • @poecilia1329
    @poecilia1329 2 роки тому

    Great video. But I have a question.
    I wonder why you mentioned E=p**2/2mv. This is classic, not relativistic formula.

    • @stephenremillard1
      @stephenremillard1  2 роки тому +4

      True. But the point being made at 11:20 is that a free particle moves at its group velocity. I prefer not to complicate that matter with a protracted aside about the relativistic dispersion equation, which is a topic in itself. So, sure, you're right, this discussion can only conclude that a nonrelativistic particle's wave function travels at the group velocity.