What is the Four-Vector Potential of a Moving Charge?

In summary, the four-vector potential of a moving charge is given by A = (q/4πε0)((q/4πε0c^2√(z^2+a^2))(c,vs) and the z-component of the electric field is E_z = -(qz/4πε0(z^2+a^2)^(3/2))k. To find the electric potential to first order of x, the four-potential is used to calculate R · U = γ(-cr + aωΔxsin(ωt)) and the electric potential is given by φ = (-qc/4πε0)(1/[c√(a^2+z^2)+
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Homework Statement



(a) Find the four-vector potential of a moving charge
(b) Find source time and z-component of electric field
(c) Find electric potential to first order of x and hence electric field[/B]

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Homework Equations

The Attempt at a Solution



Part(a)
[/B]
[tex]\phi = \frac{q}{4\pi \epsilon_0 r} [/tex]
[tex] \vec A = 0 [/tex]

Consider world line of aritrarily moving charge, where the vector potential is only dependent on what charge is doing at source event. The relevant distance is source-field distance ##r_{sf}##. We represent this in 4-vector displacement ## R = (ct, \vec r)## where ##t = \frac{r_{sf}}{c}##. Consider ##R \cdot U = \gamma(-rc + \vec r \cdot \vec v)##. It gives the right form in rest frame. Thus:

[tex]A = \frac{q}{4\pi \epsilon_0} \frac{ \frac{U}{c} }{-R \cdot U} [/tex]Part (b)

[tex] t_s = \frac{r_{sf}}{c} = \frac{\sqrt{z^2+a^2}}{c} [/tex]
[tex]R \cdot U = \gamma (-ct + \vec r \cdot \vec v) = -\gamma r c = -\gamma c \sqrt{z^2 + a^2} [/tex]

Thus the four-potential is
[tex]A = \frac{q}{4\pi \epsilon_0 c^2 \sqrt{z^2 + a^2}} (c, \vec v_s) = (\frac{\phi}{c}, \vec A)[/tex]
To find electric field,
[tex]E_z = -\frac{\partial \phi}{\partial z} - \frac{\partial A_z}{\partial t}[/tex]
z-component of ##\vec A## is zero, so
[tex]E_z = -\frac{\partial \phi}{\partial z} = -\frac{qz}{4\pi \epsilon_0 \left( z^2 + a^2 \right)^{\frac{3}{2}}} \hat k[/tex]Part (c)

[tex]r = \sqrt{ z^2 + \left[ a cos(\omega t) - \Delta x \right]^2 + a^2 sin^2(\omega t) } [/tex]
[tex]r = \approx \sqrt{ z^2 + a^2 - 2a \Delta x cos(\omega t) } \approx \sqrt{z^2 + a^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{z^2 + a^2} \right][/tex]

For ##R \cdot U##, we have

[tex]R \cdot U = \gamma(-rc + \vec r \cdot \vec v) = \gamma \left[ -cr + a\omega \Delta x sin (\omega t) \right] [/tex]

[tex] \phi = -\frac{qc}{4\pi \epsilon_0} \frac{1}{ c\sqrt{a^2 + z^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{a^2 + z^2} \right] + a\omega \Delta x sin(\omega t) }[/tex]

[tex]\phi \approx \frac{-qc}{4\pi \epsilon_0} \frac{1}{c\sqrt{a^2+z^2} + \Delta x \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2}} \right] } [/tex]

[tex] \phi \approx \frac{-qc}{4\pi \epsilon_0} \left[ 1 - \frac{a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2+z^2}} }{c\sqrt{a^2+z^2}} \Delta x \right] [/tex]

[tex]E_x = -\frac{\partial \phi}{\partial x} = \frac{qc}{4\pi \epsilon_0 c^2 (a^2 + z^2) \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2 }} \right] } [/tex]

Can't seem to match the final expression - what worries me is the lack of ##\omega^2## in the answer.
 
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  • #2
What happened to the x-component of the vector potential?
 

FAQ: What is the Four-Vector Potential of a Moving Charge?

1. What is the four-potential of a moving charge?

The four-potential of a moving charge is a mathematical concept that describes the electromagnetic field surrounding a charged particle as it moves through space. It combines the electric potential and magnetic vector potential into a single four-dimensional vector.

2. How is the four-potential of a moving charge calculated?

The four-potential of a moving charge can be calculated using the Lorenz gauge condition and the equations of motion for the charged particle. It can also be calculated using the Maxwell's equations for electromagnetism.

3. What is the significance of the four-potential in electromagnetism?

The four-potential is significant in electromagnetism because it allows us to accurately describe and predict the behavior of charged particles in motion. It also provides a framework for understanding the relationship between electric and magnetic fields.

4. How does the four-potential change in different reference frames?

The four-potential is a relativistic quantity, meaning it changes in different reference frames. In particular, it undergoes a Lorentz transformation when viewed from different frames of reference, which takes into account the effects of time dilation and length contraction.

5. What are some real-world applications of the four-potential?

The four-potential has a wide range of applications in areas such as particle physics, electrodynamics, and relativity. It is used in the development of technologies such as particle accelerators and electromagnetic radiation detectors. It also plays a crucial role in understanding the behavior of charged particles in space, such as the interaction between the solar wind and Earth's magnetic field.

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