What Are the Differences Between These Vector Potential Formulas?

In summary, Equation 1) is approximately the same formula as the one you use to know it is just in another system of units (your original equation is in SI units while 1) is probably in Heaviside units), is . Also the term 1/r has gone outside of the integral, an approximation that is valid for distances far away from the source of current density. Equation 2) is about the time derivative of A and that's the non conservative constituent of the electric field vector, not of the magnetic field vector. And if it was a space derivative of A? Would it have any sense? Do you know something about this form of the Poynting vector? This is the Poynting vector for
  • #1
enerieire
8
0
I'm having some trouble with vector potential formulas. Ihave always used this one :
A=(μ/4pi )*(J/r) dV,
where r is a distance.

I don't understand where this formulas comes from:
1) A=(1/cr)*∫JdV
where c is the speed of light;

2) H=(1/cr) Ȧ
where H is the magnetic field's vector.

Can someone please help me?
Thanks
 
Physics news on Phys.org
  • #2
Equation 1) is approximately(read further on) the same formula as the one you use to know it is just in another system of units (your original equation is in SI units while 1) is probably in Heaviside units) is . Also the term 1/r has gone outside of the integral, an approximation that is valid for distances far away from the source of current density.

I can't help you though with equation 2) because from what I know ##\dot{A}## is the time derivative of A and that's the non conservative constituent of the electric field vector, not of the magnetic field vector.
 
  • #3
And if it was a space derivative of A? Would it have any sense?

Do you know something about this form of the Poynting vector?
S= c/4pi H^2 n
 
  • #4
enerieire said:
Do you know something about this form of the Poynting vector?
S= c/4pi H^2 n
This is the Poynting vector for a linearly polarized wave. In the case of linearly polarized wave you can prove that the electric field has essentially the same equation as the magnetic field but its direction is always perpendicular to that of the magnetic field. So for example if the propagation of the wave is in the z direction, and if the magnetic field vector is on the x direction ##B=B(r,t)\hat{x}## then the electric field vector is in the y-direction and ##E=cB(r,t)\hat{y}##.
 
Last edited:
  • #5
Thank you very much for the answers!
I'm going to post another problem about Poynting's flux in a new question.
 
  • #6
This is the Biot-Savart Law for the vector potential for magnetostatics. The derivation starts from the static Maxwell equations for the magnetic field (here written in Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\frac{\mu}{c} \vec{j}.$$
From the first equation we can write
$$\vec{B}=\vec{\nabla} \times \vec{A},$$
but ##\vec{A}## is defined only up to a gradient field (gauge invariance for the special case of magnetostatics). This can be used to impose one additional condition on ##\vec{A}##. As we shall see in a moment, the following Coulomb-gauge condition is particularly convenient in this case:
$$\vec{\nabla} \cdot \vec{A}=0.$$
Now we use the inhomogeneous equation (Ampere's Law):
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\frac{\mu}{c} \vec{j}.$$
Now using the Coulomb-gauge condition this simplifies finally to
$$\Delta \vec{A}=-\frac{\mu}{c} \vec{j}.$$
This is the same equation as for the electrostatic potential, just for every Cartesian component of ##\vec{A}##. Thus we know the solution via the superposition of Coulomb fields (more formally it's the use of the Green's function of the Laplace operator):
$$\vec{A}(\vec{x})=\frac{\mu}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Note that this is only consistent, if
$$\vec{\nabla} \cdot \vec{j}=0,$$
which is charge conservation for the static case. If this integrability condition (which follows also from the Maxwell equation (Ampere's Law)) is fulfilled, then the found solution also fulfills the Coulomb-gauge condition, as it must be for consistency.
 
  • #7
vanhees71 said:
$$\vec{A}(\vec{x})=\frac{\mu}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$

Thank you,
even if my doubt was about this form of the formula:

A=(1/cr)*∫JdV
 
  • #8
This formula is obviously not fully correct. It's only valid for ##r=|\vec{x}| \gg R## (where ##R## is the radius within which you have a non-vanishing ##\vec{j}##). It's the first term in the magnetostatic multipole expansion.
 
  • #9
vanhees71 said:
This formula is obviously not fully correct. It's only valid for ##r=|\vec{x}| \gg R## (where ##R## is the radius within which you have a non-vanishing ##\vec{j}##). It's the first term in the magnetostatic multipole expansion.
okay, now it's clear.
Could you help me with this other connected problem?
https://www.physicsforums.com/threads/flux-of-the-poyntings-vector.900480/
 

Related to What Are the Differences Between These Vector Potential Formulas?

1. What is a potential vector?

A potential vector is a mathematical vector used to represent a physical quantity, such as force or velocity, that has both magnitude and direction.

2. What is the formula for calculating potential vector?

The formula for calculating potential vector is given by: P = mgh, where P is the potential energy, m is the mass of the object, g is the gravitational acceleration, and h is the height of the object.

3. How is potential vector related to electric fields?

Potential vector is closely related to electric fields, as it represents the potential energy per unit charge at a given point in an electric field. The direction of the potential vector is opposite to the direction of the electric field.

4. Can potential vector be negative?

Yes, potential vector can be negative. This indicates that the potential energy is decreasing as the object moves in the direction of the vector. A positive potential vector indicates an increase in potential energy.

5. How is potential vector used in physics?

Potential vector is used in physics to calculate and analyze the potential energy of objects in various systems, such as gravitational and electric fields. It is also used in determining the direction of forces acting on an object in a given system.

Similar threads

Replies
5
Views
2K
Replies
2
Views
325
Replies
10
Views
3K
Replies
1
Views
857
Replies
4
Views
712
  • Electromagnetism
Replies
1
Views
788
  • Atomic and Condensed Matter
Replies
3
Views
601
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Electromagnetism
Replies
5
Views
1K
Replies
17
Views
3K
Back
Top