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Emilio Pisanty
Emilio Pisanty
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For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

CHARGE DISTRIBUTION A: A charge of +3 at point (0,0,0)

CHARGE DISTRIBUTION B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

  • Charge distribution A: A charge of +3 at point (0,0,0)

  • Charge distribution B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution:

(REAL CHARGE DISTRIBUTION) = (CHARGE DISTRIBUTION A) + (CHARGE DISTRIBUTION B).$$ (\text{real charge distribution}) = (\text{charge distribution A}) + (\text{charge distribution B}). $$

By the superposition principle:

(REAL E-FIELD) = (E-FIELD OF CHARGE DISTRIBUTION A) + (E-FIELD OF CHARGE DISTRIBUTION B) $$ (\text{Real }\mathbf E\text{ field}) = (\mathbf E\text{ field of charge distribution A}) + (\mathbf E\text{ field of charge distribution B}). $$

And, since the multipole expansion also obeys the superposition principle:

(REAL MONOPOLE TERM) = (MONOPOLE TERM OF CHARGE DIST A) + (MONOPOLE TERM OF CHARGE DIST B)

(REAL DIPOLE TERM) = (DIPOLE TERM OF CHARGE DIST A) + (DIPOLE TERM OF CHARGE DIST B)

etc\begin{align} (\text{real monopole term}) & = (\text{monopole term of distribution A}) + (\text{monopole term of distribution B}),\\ (\text{real dipole term}) & = (\text{dipole term of distribution A}) + (\text{dipole term of distribution B}),\\ (\text{real quadrupole term}) & = (\text{quadrupole term of distribution A}) + (\text{quadrupole term of distribution B}), \end{align} and so on.

The field of charge distribution A is a pure monopole field, while the field of charge distribution B has no monopole term, only dipole, quadrupole, etc. Therefore,

REAL MONOPOLE TERM = MONOPOLE TERM OF CHARGE DIST A

REAL DIPOLE TERM = DIPOLE TERM OF CHARGE DIST B

REAL QUADRUPOLE TERM = QUADRUPOLE TERM OF CHARGE DIST B

etc \begin{align} (\text{real monopole term}) & = (\text{monopole term of distribution A}), \\ (\text{real dipole term}) & = (\text{dipole term of distribution B}),\\ (\text{real quadrupole term}) & = (\text{quadrupole term of distribution B}), \end{align} and so on.

Even though it's unintuitive that the real charge distribution has a dipole component, it is not at all surprising that charge distribution B has a dipole component: It is two equal and opposite separated charges! And charge distribution B is exactly what you get after subtracting off the monopole component to look at the subleading terms of the expansion.

[edited for clarity]

For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

CHARGE DISTRIBUTION A: A charge of +3 at point (0,0,0)

CHARGE DISTRIBUTION B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution:

(REAL CHARGE DISTRIBUTION) = (CHARGE DISTRIBUTION A) + (CHARGE DISTRIBUTION B).

By the superposition principle:

(REAL E-FIELD) = (E-FIELD OF CHARGE DISTRIBUTION A) + (E-FIELD OF CHARGE DISTRIBUTION B)

And since the multipole expansion also obeys the superposition principle:

(REAL MONOPOLE TERM) = (MONOPOLE TERM OF CHARGE DIST A) + (MONOPOLE TERM OF CHARGE DIST B)

(REAL DIPOLE TERM) = (DIPOLE TERM OF CHARGE DIST A) + (DIPOLE TERM OF CHARGE DIST B)

etc.

The field of charge distribution A is a pure monopole field, while the field of charge distribution B has no monopole term, only dipole, quadrupole, etc. Therefore,

REAL MONOPOLE TERM = MONOPOLE TERM OF CHARGE DIST A

REAL DIPOLE TERM = DIPOLE TERM OF CHARGE DIST B

REAL QUADRUPOLE TERM = QUADRUPOLE TERM OF CHARGE DIST B

etc.

Even though it's unintuitive that the real charge distribution has a dipole component, it is not at all surprising that charge distribution B has a dipole component: It is two equal and opposite separated charges! And charge distribution B is exactly what you get after subtracting off the monopole component to look at the subleading terms of the expansion.

[edited for clarity]

For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

  • Charge distribution A: A charge of +3 at point (0,0,0)

  • Charge distribution B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution:

$$ (\text{real charge distribution}) = (\text{charge distribution A}) + (\text{charge distribution B}). $$

By the superposition principle: $$ (\text{Real }\mathbf E\text{ field}) = (\mathbf E\text{ field of charge distribution A}) + (\mathbf E\text{ field of charge distribution B}). $$

And, since the multipole expansion also obeys the superposition principle:

\begin{align} (\text{real monopole term}) & = (\text{monopole term of distribution A}) + (\text{monopole term of distribution B}),\\ (\text{real dipole term}) & = (\text{dipole term of distribution A}) + (\text{dipole term of distribution B}),\\ (\text{real quadrupole term}) & = (\text{quadrupole term of distribution A}) + (\text{quadrupole term of distribution B}), \end{align} and so on.

The field of charge distribution A is a pure monopole field, while the field of charge distribution B has no monopole term, only dipole, quadrupole, etc. Therefore, \begin{align} (\text{real monopole term}) & = (\text{monopole term of distribution A}), \\ (\text{real dipole term}) & = (\text{dipole term of distribution B}),\\ (\text{real quadrupole term}) & = (\text{quadrupole term of distribution B}), \end{align} and so on.

Even though it's unintuitive that the real charge distribution has a dipole component, it is not at all surprising that charge distribution B has a dipole component: It is two equal and opposite separated charges! And charge distribution B is exactly what you get after subtracting off the monopole component to look at the subleading terms of the expansion.

edited for clarity
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Steve Byrnes
Steve Byrnes
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For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

CHARGE DISTRIBUTION A: A charge of +3 at point (0,0,0)

CHARGE DISTRIBUTION B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution:

(REAL CHARGE DISTRIBUTION) = (CHARGE DISTRIBUTION A) + (CHARGE DISTRIBUTION B). The monopole term in

By the field is actually exactlysuperposition principle:

(REAL E-FIELD) = (E-FIELD OF CHARGE DISTRIBUTION A) + (E-FIELD OF CHARGE DISTRIBUTION B)

And since the multipole expansion also obeys the superposition principle:

(REAL MONOPOLE TERM) = (MONOPOLE TERM OF CHARGE DIST A) + (MONOPOLE TERM OF CHARGE DIST B)

(REAL DIPOLE TERM) = (DIPOLE TERM OF CHARGE DIST A) + (DIPOLE TERM OF CHARGE DIST B)

etc.

The field of charge distribution A. Therefore is a pure monopole field, while the field of charge distribution B has no monopole term, only dipole, quadrupole, etc. terms all together have to add up to the field of charge distributionTherefore,

REAL MONOPOLE TERM = MONOPOLE TERM OF CHARGE DIST A

REAL DIPOLE TERM = DIPOLE TERM OF CHARGE DIST B

REAL QUADRUPOLE TERM = QUADRUPOLE TERM OF CHARGE DIST B

etc.

It's not so suprisingEven though it's unintuitive that the real charge distribution has a dipole term should be involved in describing the field ofcomponent, it is not at all surprising that charge distribution B has a dipole component: It'sIt is two equal and opposite separated charges! And charge distribution B is exactly what you get after subtracting off the monopole component to look at the subleading terms of the expansion.

[edited for clarity]

For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

CHARGE DISTRIBUTION A: A charge of +3 at point (0,0,0)

CHARGE DISTRIBUTION B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution. The monopole term in the field is actually exactly the field of charge distribution A. Therefore, the dipole, quadrupole, etc. terms all together have to add up to the field of charge distribution B.

It's not so suprising that a dipole term should be involved in describing the field of charge distribution B: It's two equal and opposite separated charges!

For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

CHARGE DISTRIBUTION A: A charge of +3 at point (0,0,0)

CHARGE DISTRIBUTION B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution:

(REAL CHARGE DISTRIBUTION) = (CHARGE DISTRIBUTION A) + (CHARGE DISTRIBUTION B).

By the superposition principle:

(REAL E-FIELD) = (E-FIELD OF CHARGE DISTRIBUTION A) + (E-FIELD OF CHARGE DISTRIBUTION B)

And since the multipole expansion also obeys the superposition principle:

(REAL MONOPOLE TERM) = (MONOPOLE TERM OF CHARGE DIST A) + (MONOPOLE TERM OF CHARGE DIST B)

(REAL DIPOLE TERM) = (DIPOLE TERM OF CHARGE DIST A) + (DIPOLE TERM OF CHARGE DIST B)

etc.

The field of charge distribution A is a pure monopole field, while the field of charge distribution B has no monopole term, only dipole, quadrupole, etc. Therefore,

REAL MONOPOLE TERM = MONOPOLE TERM OF CHARGE DIST A

REAL DIPOLE TERM = DIPOLE TERM OF CHARGE DIST B

REAL QUADRUPOLE TERM = QUADRUPOLE TERM OF CHARGE DIST B

etc.

Even though it's unintuitive that the real charge distribution has a dipole component, it is not at all surprising that charge distribution B has a dipole component: It is two equal and opposite separated charges! And charge distribution B is exactly what you get after subtracting off the monopole component to look at the subleading terms of the expansion.

[edited for clarity]

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Steve Byrnes
Steve Byrnes
  • 16.9k
  • 1
  • 49
  • 87

For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

CHARGE DISTRIBUTION A: A charge of +3 at point (0,0,0)

CHARGE DISTRIBUTION B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution. The monopole term in the field is actually exactly the field of charge distribution A. Therefore, the dipole, quadrupole, etc. terms all together have to add up to the field of charge distribution B.

It's not so suprising that a dipole term should be involved in describing the field of charge distribution B: It's two equal and opposite separated charges!