Force on a dipole in a non-uniform electric field

In summary, it seems that the two formulae for the force are equivalent if we assume the dipole moment is constant, but if the dipole moment varies then the F=∇(p.E) approach implies we must differentiate the dipole moment components, whilst the F=(p.∇)E approach does not.
  • #1
fayled
177
0
I'm going to explain my understanding about a bit of a contradiction I can't resolve, and I was hoping somebody could help me understand it. Sorry about the bombardment of vector components!

There appears to be two methods we can use to calculate the force (effectively the same in fact, but one is a bit of a shortcut):

1. Find the potential energy of the dipole using
U=-p.E
=-pxEx-pyEy-pzEz.
Then simply use F=-U to get the force. This comes out as
F=(px∂Ex/∂x+py∂Ey/∂x+pz∂Ez/∂x)i+(px∂Ex/∂y+py∂Ey/∂y+pz∂Ez/∂y)j+(px∂Ex/∂z+py∂Ey/∂z+pz∂Ez/∂z)k
Now this method has given me correct answers when I have started with a specific vector function E, found its corresponding U and then got F

2. The 'shortcut'. We can use a vector calculus identity so that
F=-U=-(-p.E)=(p.E)=(p.)E. All this means is we can find F without the intermediate stage of needing U. All we need is E. However this in component form gives
F=(px∂Ex/∂x+py∂Ex/∂y+pz∂Ex/∂z)i+(px∂Ey/∂x+py∂Ey/∂y+pz∂Ey/∂z)j+(px∂Ez/∂x+py∂Ez/∂y+pz∂Ez/∂z)k

So the same vector F comes out in two different ways. Now, I can't find any reason that makes them equal, but I recall using the first correctly for a specific field, and the second method I'm assuming would have given me the correct answer too, so why aren't they agreeing with one another for a general dipole in a general field in component form? Thanks for any help!
 
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  • #2
Anybody? :)
 
  • #3
Your electrostatic field has zero curl, so the following holds:
[tex]
\frac{\partial E_x}{\partial z} = \frac{\partial E_z}{\partial x}, \quad \frac{\partial E_y}{\partial x} = \frac{\partial E_x}{\partial y}, \quad \frac{\partial E_z}{\partial y} = \frac{\partial E_y}{\partial z}
[/tex]
It comes directly from the definition of curl:
http://en.wikipedia.org/wiki/Curl_(mathematics)#Usage

Edit:
Just a reminder that the identity you're using in (2) is:
[tex]
\nabla(\mathbf{u}\cdot\mathbf{v}) = (\mathbf{u}\cdot\nabla)\mathbf{v} + (\mathbf{v}\cdot\nabla)\mathbf{u} + \mathbf{u}\times(\nabla\times\mathbf{v}) + \mathbf{v}\times(\nabla\times\mathbf{u})
[/tex]
If you're not working with an electrostatic field, then things get a bit more complicated.
 
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  • #4
fayled said:
2. The 'shortcut'. We can use a vector calculus identity so that
F=-U=-(-p.E)=(p.E)=(p.)E.

∇(p.E) = (p.∇)E + (E.∇)p + p x (∇ x E) + E x (∇ x p)

so in you last step you seem to have omitted no less than 3 extra terms!
 
  • #5
rude man said:
∇(p.E) = (p.∇)E + (E.∇)p + p x (∇ x E) + E x (∇ x p)

so in you last step you seem to have omitted no less than 3 extra terms!
That's fine, though, for a permanent dipole moment and electrostatic field.
 
  • #6
milesyoung said:
Your electrostatic field has zero curl, so the following holds:
[tex]
\frac{\partial E_x}{\partial z} = \frac{\partial E_z}{\partial x}, \quad \frac{\partial E_y}{\partial x} = \frac{\partial E_x}{\partial y}, \quad \frac{\partial E_z}{\partial y} = \frac{\partial E_y}{\partial z}
[/tex]
It comes directly from the definition of curl:
http://en.wikipedia.org/wiki/Curl_(mathematics)#Usage

Edit:
Just a reminder that the identity you're using in (2) is:
[tex]
\nabla(\mathbf{u}\cdot\mathbf{v}) = (\mathbf{u}\cdot\nabla)\mathbf{v} + (\mathbf{v}\cdot\nabla)\mathbf{u} + \mathbf{u}\times(\nabla\times\mathbf{v}) + \mathbf{v}\times(\nabla\times\mathbf{u})
[/tex]
If you're not working with an electrostatic field, then things get a bit more complicated.

Ah so simple! I've been trying to get my head around this for ages - thanks :)

Edit: A further thought - I'm not sure if I am going to be very clear here so if it doesn't make sense let me know.

Suppose we have a dielectric medium in which the dipole moment varies with position. Then although we have shown the two formulae for the force to be equivalent, in this case the F=(p.E) approach implies we must differentiate the dipole moment components, whilst F=(p.)E approach does not. Is this not a contradiction (my book seems to suggest it is)?

The only way I can think of resolving this is: in deriving both of these expressions we assumed the dipole moment was constant - i.e I took the dipole moment components from out of the derivatives when applying the gradient to p.E and when using the vector calculus identity for the second approach, we assumed the divergence and curl of the dipole moment were zero. Therefore we should in fact not be relying on any of these expressions for this situation. My book however states that the second approach would be valid here - it seems somewhere that may have 'got lucky' possibly. This is a bit sketchy but just all I can think of.
 
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  • #7
What book are you reading?

My suggestion is to stick with ##\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}##, since it's the more general form. IIRC, it's strictly only valid for dipoles with very short displacement vectors (since its derivation uses a first-order Taylor expansion of the electric field), but it works for induced (nonconservative) electric fields as well.
 
  • #8
milesyoung said:
What book are you reading?

My suggestion is to stick with ##\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}##, since it's the more general form. IIRC, it's strictly only valid for dipoles with very short displacement vectors (since its derivation uses a first-order Taylor expansion of the electric field), but it works for induced (nonconservative) electric fields as well.

It's Griffiths, in a footnote on page 165.

I will quote it:
'In the present context Eq. 4.5 could be written more conveniently as F=(p.E). However it is safer to stick with (p.)E because we will be applying the formula to materials in which the dipole moment (per unit volume) is itself a function of position and this second expression would imply (incorrectly) that p too is to be differentiated.

Eq. 4.5 is F=(p.)E.

I don't really understand why there seems to be an inconsistency between the two formulae given as we have shown them to be the same (assuming constant p) but it is quite evident there is for non-constant p.
 
  • #9
fayled said:
It's Griffiths, in a footnote on page 165.

I will quote it:
'In the present context Eq. 4.5 could be written more conveniently as F=(p.E). However it is safer to stick with (p.)E because we will be applying the formula to materials in which the dipole moment (per unit volume) is itself a function of position and this second expression would imply (incorrectly) that p too is to be differentiated.

Eq. 4.5 is F=(p.)E.

I don't really understand why there seems to be an inconsistency between the two formulae given as we have shown them to be the same (assuming constant p) but it is quite evident there is for non-constant p.
To be honest, I've never really given the form ##\mathbf{F} = \nabla(\mathbf{p}\cdot\mathbf{E})## much thought. You'd have to start out with a potential field, so that precludes induced electrical fields, which is why I think most tend to avoid it. In the third edition of Griffiths, he also takes the approach of using a first-order Taylor expansion of ##\mathbf{E}## to derive ##\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}## on page 164.

Say you had a neutral atom in an electrical field such that it's polarized and you get an induced dipole with a moment proportional to the electrical field ##\mathbf{p} = \alpha \mathbf{E}##, as shown in Griffiths on page 161. That makes it seem like ##\mathbf{p}## itself is a vector field, but as far as I can tell, it's understood that ##\mathbf{p} = \alpha \mathbf{E}(\mathbf{r}_0)##, where ##\mathbf{r}_0## is the position vector of the dipole at some instant in time. As such, you shouldn't include it as a vector field in the coordinates ##(x,y,z)## when determining ##\nabla(\mathbf{p}\cdot\mathbf{E})##, i.e. ##\mathbf{p}## is assigned a value depending on its current position in the electrical field.

I think that's what Griffiths means in the footnote - at least, that's all that makes sense to me.
 
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  • #10
milesyoung said:
To be honest, I've never really given the form ##\mathbf{F} = \nabla(\mathbf{p}\cdot\mathbf{E})## much thought. You'd have to start out with a potential field, so that precludes induced electrical fields, which is why I think most tend to avoid it. In the third edition of Griffiths, he also takes the approach of using a first-order Taylor expansion of ##\mathbf{E}## to derive ##\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}## on page 164.

Say you had a neutral atom in an electrical field such that it's polarized and you get an induced dipole with a moment proportional to the electrical field ##\mathbf{p} = \alpha \mathbf{E}##, as shown in Griffiths on page 161. That makes it seem like ##\mathbf{p}## itself is a vector field, but as far as I can tell, it's understood that ##\mathbf{p} = \alpha \mathbf{E}(\mathbf{r}_0)##, where ##\mathbf{r}_0## is the position vector of the dipole at some instant in time. As such, you shouldn't include it as a vector field in the coordinates ##(x,y,z)## when determining ##\nabla(\mathbf{p}\cdot\mathbf{E})##, i.e. ##\mathbf{p}## is assigned a value depending on its current position in the electrical field.

I think that's what Griffiths means in the footnote - at least, that's all that makes sense to me.

Ah I see, it's not that there's an inconsistency then, it's just that one of them can trick you into wanting to differentiate p. Thankyou :)
 

Related to Force on a dipole in a non-uniform electric field

What is a dipole?

A dipole is a pair of equal and opposite charges separated by a small distance. This creates a neutral system with a positive and negative end, also known as poles.

How is force calculated on a dipole in a non-uniform electric field?

The force on a dipole in a non-uniform electric field is calculated by taking the product of the electric field and the dipole moment. This force can then be broken down into its components, one parallel to the electric field and one perpendicular to it.

What is the direction of the force on a dipole in a non-uniform electric field?

The direction of the force on a dipole in a non-uniform electric field depends on the orientation of the dipole with respect to the electric field. If the dipole is aligned with the electric field, the force will be in the same direction as the electric field. If the dipole is perpendicular to the electric field, the force will be in a direction perpendicular to both the dipole and the electric field.

How does the strength of the electric field affect the force on a dipole?

The strength of the electric field directly affects the magnitude of the force on a dipole. A stronger electric field will result in a larger force on the dipole, while a weaker electric field will result in a smaller force.

Can a dipole experience a net force of zero in a non-uniform electric field?

Yes, it is possible for a dipole to experience a net force of zero in a non-uniform electric field. This can occur when the dipole is oriented in a specific way, such as being perpendicular to the electric field, or when the electric field is uniform.

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