Gradient of a time-dependent function

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  • Thread starter Sturk200
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  • #1
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Hi!

I am struggling with what I think is probably a fairly simple step in Landau & Lifshitz derivation of the fields from the Lienard-Wiechert potential. We have the potential in terms of a primed set of coordinates but the fields are defined in terms of derivatives with respect to unprimed coordinates; the two are related by

$$t'+{R(t')\over{c}} = t$$

where R is the distance from the point charge to the field point at time t' and time t is the moment of observation.

The step that I am having trouble with is in finding an expression for the gradient of t'. Landau has:

$$\nabla t' = -\frac{1}{c}\nabla R(t') = -\frac{1}{c} \bigg(\frac{\partial R}{\partial t'} \nabla t' + \frac{\textbf{R}}{R}\bigg)$$

The first equality obviously follows from the equation above. The second equality is where I am stumped. I would think that it should be simply

$$\nabla R(t') = \frac{\partial R}{\partial t'} \nabla t'$$.

Does anyone know where that extra unit vector term comes from?

Thanks!
 

Answers and Replies

  • #2
RUber
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It looks to me like that came from the unprimed t in your original equation.
## t' = -\frac{R(t')}{c} +t##
so
##\nabla t' = -\nabla \frac{R(t')}{c} +\nabla t##
Is it reasonable to write ##\nabla t = \frac{-c}{R}\mathbf{R}##?
 
  • #3
168
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It looks to me like that came from the unprimed t in your original equation.
## t' = -\frac{R(t')}{c} +t##
so
##\nabla t' = -\nabla \frac{R(t')}{c} +\nabla t##
Is it reasonable to write ##\nabla t = \frac{-c}{R}\mathbf{R}##?

I don't think that makes sense, since it is not consistent with the first equality:

$$\nabla t' = -\frac{1}{c}\nabla R(t')$$.
 

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