Gradient of a time-dependent function

In summary, the conversation discusses a problem with finding an expression for the gradient of t' in Landau & Lifshitz's derivation of fields from the Lienard-Wiechert potential. The equation for t' is related to the unprimed coordinates by a given equation, and the conversation focuses on finding the extra unit vector term in the second equality of the expression for the gradient. The possibility of writing ##\nabla t = \frac{-c}{R}\mathbf{R}## is also discussed.
  • #1
Sturk200
168
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!
 
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  • #2
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
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:

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

Related to Gradient of a time-dependent function

What is the gradient of a time-dependent function?

The gradient of a time-dependent function is a vector that represents the rate and direction of change of the function over time. It is calculated by taking the partial derivatives of the function with respect to each independent variable.

How is the gradient of a time-dependent function used in science?

The gradient of a time-dependent function is used in many fields of science, including physics, engineering, and economics. It is used to model and analyze complex systems that change over time, such as heat transfer, fluid flow, and market trends.

Can the gradient of a time-dependent function be negative?

Yes, the gradient of a time-dependent function can be negative. A negative gradient indicates that the function is decreasing in value over time, while a positive gradient indicates that the function is increasing.

How is the gradient of a time-dependent function related to the concept of slope?

The gradient of a time-dependent function is essentially the multi-dimensional version of slope. It represents the steepness or rate of change of the function over multiple independent variables, while slope only considers the change over a single independent variable.

What is the difference between the gradient of a time-dependent function and the gradient of a static function?

The gradient of a time-dependent function changes over time, while the gradient of a static function remains constant. Additionally, the gradient of a time-dependent function is a vector, while the gradient of a static function is a scalar.

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