Larmor Formula: Shaking Photons from Charges

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  • เผยแพร่เมื่อ 18 พ.ย. 2024

ความคิดเห็น • 14

  • @jesselajeunesse4857
    @jesselajeunesse4857 6 ปีที่แล้ว +2

    I was given this formula at the end of Astrophysics lecture today. It was so helpful to see the derivation and explanation (I'm an Advanced Chemistry major, not physics). Thank you!

  • @AaronRobertParsons
    @AaronRobertParsons  9 ปีที่แล้ว +2

    HitAndMissLab, you are right that this formula is expressedly non-relativistic. In the relativistic form, there is a "beaming" effect. Because of relativistic length contraction, radiation is preferentially emitted parallel to the axis of motion,
    Rather than perpendicular. Moreover, the Doppler shifting of photons makes emission in the forward direction more energetic than in the reverse direction.

    • @jokerhhhify
      @jokerhhhify 9 ปีที่แล้ว +1

      +Aaron Parsons thank you sir

  • @anwarshiekh7142
    @anwarshiekh7142 8 ปีที่แล้ว +1

    Feynman felt it was wrong (as did Pauli)
    From the 'Feynman lectures on Gravitation'
    Lecture 9, P123
    Larmor equation (9.1.1)
    This is usually derived from calculating the flow from Poynting's theorem far away, and it is only valid for cyclic motions, or at least motions which do not grow forever in time (as a constant acceleration does). It does not suffice to tell us "when" the energy is radiated. This can only be determined by finding the force of radiation resistance, ...
    Feynman's version had P ~ da/dt v, rather than a^2; which avoids the paradox 'does a charge on a table radiate, as it is equivalent to an accelerating charge'

  • @georgeveropoulos2489
    @georgeveropoulos2489 5 หลายเดือนก่อน

    hi very nice presentation In r direction there not radiation?

  • @jainecruz7470
    @jainecruz7470 2 ปีที่แล้ว

    Hellou! awesome video and perfect explanation! I'm trying to find/understand a way of determining the frequency of the EM wave generate by the accelerated charge, considering that one knows the aspects of that acceleration. Have you any idea of how to do so? Maybe the lenght of that ring (ct) have that information?

  • @solidsnake19902342
    @solidsnake19902342 9 ปีที่แล้ว +2

    Larmor derive it in 1897 i can't imagine !!!

  • @HitAndMissLab
    @HitAndMissLab 9 ปีที่แล้ว

    But the outer and inner shell are not strictly concentric. If delta-v is large, say, 0.5*c, than there will be an offset +/- (delta-v * delta-t). Since we are talking particles here, that is realistic assumption.
    How would that change the outcome?

  • @Snoopies622
    @Snoopies622 2 ปีที่แล้ว

    Problem with this formula: A charge oscillating in a sinusoidal way, at the location farthest from the equilibrium position its acceleration is maximum, yet P=Fv must be zero since v=0. How can it be radiating energy at this moment when it's not absorbing energy to radiate? Please resolve.

  • @souravchakravarty2748
    @souravchakravarty2748 8 ปีที่แล้ว

    How will the electric field act in case of a charge moving with a constant velocity?

    • @SEAPanel
      @SEAPanel 4 ปีที่แล้ว

      If the charge moves, then the qv component is nonzero, and the resultant force changes.

  • @Jkauppa
    @Jkauppa 4 ปีที่แล้ว

    you cannot determine if you are in an elevated ground within the "ether", if the ether is itself elevated

  • @veronicanoordzee6440
    @veronicanoordzee6440 2 ปีที่แล้ว

    @ 17:51 By pulling the plates of a capacitor apart you transfer energy to the plates, NOT to the field. In W = F . dx, dx is the displacement of the plates. So where is the field-energy? P.S. "In the end nothing is moving" (18:08). But you just said you moved the plates!