I Gradient of a time-dependent function

  • I
  • Thread starter Thread starter Sturk200
  • Start date Start date
  • Tags Tags
    Function Gradient
AI Thread Summary
The discussion centers on a challenge in the Landau & Lifshitz derivation of fields from the Lienard-Wiechert potential, specifically regarding the relationship between primed and unprimed coordinates. The main issue is finding an expression for the gradient of time t', with participants debating the derivation of the term involving the unit vector in the equation for the gradient of R(t'). One participant suggests that the extra unit vector term arises from the relationship between t' and t, while another argues that this reasoning conflicts with the established equality. The conversation highlights the complexity of transitioning between coordinate systems in this context. Clarifying these relationships is essential for accurately deriving the fields from the potential.
Sturk200
Messages
168
Reaction score
17
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!
 
Physics news on Phys.org
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}##?
 
RUber said:
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:

$$\nabla t' = -\frac{1}{c}\nabla R(t')$$.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top