Question

Consider a particle moving along a trajectory in a vector potential field A(x, t) that is a function of the radius-vector x and time t. Show that the time derivative of the field seen by the particle at its current location (the so-called material derivative) is $\frac{dA}{dt} = \frac{\partial A}{\partial t} + (v \cdot \nabla)A$, (1)

          Consider a particle moving along a trajectory in a vector potential field A(x, t) that is
a function of the radius-vector x and time t. Show that the time derivative of the field
seen by the particle at its current location (the so-called material derivative) is
$\frac{dA}{dt} = \frac{\partial A}{\partial t} + (v \cdot \nabla)A$,
(1)

Consider a particle moving along a trajectory in a vector potential field A(x, t) that is
a function of the radius-vector x and time t. Show that the time derivative of the field
seen by the particle at its current location (the so-called material derivative) is
(dA)/(dt) = (∂ A)/(∂ t) + (v ·∇)A,
(1)

Added by Victoria A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Timothy James

01/23/2023

Step 1

In this case, we want to find the material derivative of the vector potential field A(x,t) at the current location of the particle. Show more…

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Don't copy. I want a complete solution. here, v is velocity. Consider a particle moving along a trajectory in a vector potential field A(x,t) that is a function of the radius-vector x and time t. Show that the time derivative of the field seen by the particle at its current location (the so-called material derivative) is dA dt DA (v.V)A Ot (1)