physics2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Derivation

Laplacian of 1/r as a Delta Function

D-283 Home PU-305 Threads fields Depends on divergence-theorem
Statement

For the Newtonian potential kernel \( \phi(\mathbf{r}) = 1/r \) with \( r = |\mathbf{r}| \) in three dimensions, the Laplacian vanishes pointwise for \( r>0 \) yet integrates to a nonzero total, and these facts are reconciled only in the distributional sense: \[ \nabla^2\!\left(\frac{1}{r}\right) = -4\pi\,\delta^3(\mathbf{r}). \] The right-hand side is a Dirac delta concentrated at the origin, and the identity is to be read under an integral against a smooth test function.

Why it matters

This one line is the seed of all of electrostatics. It says that \( -\frac{1}{4\pi r} \) is the Green's function of the Laplacian, so the potential of an arbitrary charge distribution is a convolution \( \varphi(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,d^3r' \), and Poisson's equation \( \nabla^2\varphi = -\rho/\varepsilon_0 \) follows by applying \( \nabla^2 \) under the integral.

Beyond electrostatics the same kernel underlies Newtonian gravity, steady diffusion, and the static limit of the wave and Helmholtz equations. The apparent paradox — a function whose Laplacian is zero everywhere except one point, yet which is not harmonic — is the classic gateway to distribution theory in physics, and getting it right is what separates a formal manipulation from a correct one.

Assumptions
Three spatial dimensions.The prefactor \( 4\pi \) is the solid angle of the unit sphere and is dimension-specific; in \( d \) dimensions the kernel is \( r^{2-d} \) (or \( \ln r \) in 2D) and the constant changes, so the identity as written is a 3D statement.
The identity is distributional, not pointwise.If read as an ordinary function equation it is false: the left side is \( 0 \) for every \( r>0 \) and undefined at \( r=0 \), while the right side is zero away from the origin but "infinite" there. Only when both sides act on a smooth test function do they agree.
Test functions are smooth and compactly supported.The delta is defined by \( \int \delta^3(\mathbf{r})f(\mathbf{r})\,d^3r = f(\mathbf{0}) \) for \( f\in C_c^\infty \). Dropping smoothness or decay invalidates the integration-by-parts moves that transfer derivatives onto \( f \), and boundary terms at infinity no longer vanish.
Origin-centred singularity.The result localises the source at \( \mathbf{r}=\mathbf{0} \); a shifted kernel \( 1/|\mathbf{r}-\mathbf{r}'| \) simply relocates the delta to \( \mathbf{r}' \), giving \( \nabla^2|\mathbf{r}-\mathbf{r}'|^{-1} = -4\pi\delta^3(\mathbf{r}-\mathbf{r}') \).
Derivation
1
\[ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\,\hat{L}^2 \]
Write the Laplacian in spherical coordinates. Because \( 1/r \) depends only on \( r \), the angular part \( \hat{L}^2 \) annihilates it and only the radial operator survives. A
2
\[ \nabla^2\!\left(\frac{1}{r}\right) = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\cdot\left(-\frac{1}{r^2}\right)\right) = \frac{1}{r^2}\frac{\partial}{\partial r}(-1) = 0,\qquad r>0. \]
Differentiate directly. \( \partial_r(1/r) = -1/r^2 \), so \( r^2\partial_r(1/r) = -1 \) is constant and its derivative is zero. The kernel is harmonic away from the origin. A
3
\[ \int_{V} \nabla^2\!\left(\frac{1}{r}\right) d^3r = \oint_{\partial V} \nabla\!\left(\frac{1}{r}\right)\cdot d\mathbf{A} \]
The pointwise result cannot be the whole story, or the kernel would be harmonic everywhere and hence not \( 1/r \). To probe the origin, integrate over a ball \( V \) of radius \( R \) centred at \( \mathbf{0} \) and apply the divergence theorem to \( \nabla^2 = \nabla\cdot\nabla \). B
4
\[ \nabla\!\left(\frac{1}{r}\right) = -\frac{\hat{\mathbf{r}}}{r^2},\qquad d\mathbf{A} = \hat{\mathbf{r}}\,R^2\sin\theta\,d\theta\,d\varphi \]
Compute the gradient (radial, since the field is central) and the outward area element on the sphere \( r=R \). The gradient is exactly the Coulomb field of a unit point source up to constants. A
5
\[ \oint_{\partial V}\!\left(-\frac{\hat{\mathbf{r}}}{r^2}\right)\!\cdot\hat{\mathbf{r}}\,R^2\,d\Omega = -\frac{1}{R^2}\,R^2\!\oint d\Omega = -4\pi \]
Evaluate the flux. The \( R^{-2} \) from the field cancels the \( R^2 \) from the area, leaving \( -\oint d\Omega = -4\pi \) — independent of \( R \). The flux is the same through every sphere, however small. B
6
\[ \int_V \nabla^2\!\left(\frac{1}{r}\right)d^3r = -4\pi \quad\text{for every ball containing }\mathbf{0}. \]
Combine Steps 3 and 5. Integrating the Laplacian over any region that contains the origin gives \( -4\pi \); over any region that excludes it, Step 2 gives \( 0 \). A "function" that is zero everywhere yet integrates to \( -4\pi \) is precisely a delta. B
7
\[ \int \left[\nabla^2\!\left(\frac{1}{r}\right)\right] f(\mathbf{r})\,d^3r \;\equiv\; \int \frac{1}{r}\,\nabla^2 f(\mathbf{r})\,d^3r \]
To make this rigorous, define the distributional Laplacian by moving both derivatives onto a smooth compactly-supported test function \( f \) via Green's second identity. The two integrations by parts each produce boundary terms; those at infinity vanish because \( f \) has compact support. C
8
\[ \int_{r>\epsilon} \frac{\nabla^2 f}{r}\,d^3r = \oint_{r=\epsilon}\!\left[\frac{1}{r}\nabla f - f\,\nabla\frac{1}{r}\right]\!\cdot(-\hat{\mathbf{r}})\,dA \]
Excise a small ball \( r<\epsilon \) where \( 1/r \) is singular and apply Green's identity on \( r>\epsilon \), where \( \nabla^2(1/r)=0 \) kills the volume term. The inner-sphere normal points toward the origin, hence \( -\hat{\mathbf{r}} \). C
9
\[ \oint_{r=\epsilon}\frac{1}{\epsilon}\,\partial_r f\,dA \sim \frac{1}{\epsilon}\,4\pi\epsilon^2\,\overline{\partial_r f}\to 0, \quad \oint_{r=\epsilon} f\,\frac{1}{\epsilon^2}\,dA \to 4\pi f(\mathbf{0}) \]
Take \( \epsilon\to 0 \). The first surface term scales as \( \epsilon \) and vanishes; the second, using \( \nabla(1/r)\cdot(-\hat{\mathbf{r}}) = 1/\epsilon^2 \) on the inner sphere and \( \frac{1}{4\pi\epsilon^2}\oint f\,dA\to f(\mathbf{0}) \), survives with a sign giving \( -4\pi f(\mathbf{0}) \). C
10
\[ \int \left[\nabla^2\!\left(\frac{1}{r}\right)\right] f\,d^3r = -4\pi f(\mathbf{0}) = \int \big[-4\pi\,\delta^3(\mathbf{r})\big]f\,d^3r \]
Since this holds for every admissible \( f \), the two distributions are equal. This is the defining property of the delta, so the identity is established. B
Result
\[ \nabla^2\!\left(\frac{1}{r}\right) = -4\pi\,\delta^3(\mathbf{r}) \]

Reading. The Newtonian kernel \( 1/r \) is harmonic everywhere except at the origin, where it hides a point source of strength \( -4\pi \). Applied to \( -\frac{1}{4\pi r} \), the Laplacian returns exactly \( \delta^3(\mathbf{r}) \), which is why \( G(\mathbf{r}) = -\frac{1}{4\pi r} \) is the Green's function of \( \nabla^2 \). Physically, the flux of the Coulomb field through any closed surface enclosing the charge is the same — Gauss's law in miniature.

Units check. Treat \( r \) as a length \( [\mathrm{m}] \). Then \( 1/r \) has units \( \mathrm{m}^{-1} \) and \( \nabla^2 \) carries \( \mathrm{m}^{-2} \), so the left side is \( \mathrm{m}^{-3} \). The delta \( \delta^3(\mathbf{r}) \) is defined by \( \int\delta^3\,d^3r = 1 \) with \( d^3r \) in \( \mathrm{m}^3 \), so \( \delta^3 \) itself is \( \mathrm{m}^{-3} \); the numeric \( 4\pi \) is dimensionless. Both sides are \( \mathrm{m}^{-3} \). Consistent.

Limiting cases
  • Far from the origin (\( r\to\infty \)): the delta is exactly zero and \( \nabla^2(1/r)=0 \) pointwise — the potential is source-free, as it must be in charge-free regions.
  • Shifted source: \( \nabla^2\frac{1}{|\mathbf{r}-\mathbf{r}'|} = -4\pi\,\delta^3(\mathbf{r}-\mathbf{r}') \); superposing over \( \mathbf{r}' \) builds Poisson's equation for arbitrary \( \rho \).
  • Yukawa / screened limit: \( \nabla^2\frac{e^{-\mu r}}{r} = \mu^2\frac{e^{-\mu r}}{r} - 4\pi\delta^3(\mathbf{r}) \); as \( \mu\to 0 \) the bulk term dies and the pure Coulomb result is recovered, showing the delta is a robust short-distance feature.
  • Regularised kernel \( 1/\sqrt{r^2+a^2} \): its Laplacian is the smooth, normalised function \( -3a^2(r^2+a^2)^{-5/2} \), which integrates to \( -4\pi \) for all \( a \) and tends to \( -4\pi\delta^3 \) as \( a\to 0 \) — an explicit nascent delta.
Breaks when
  • Wrong dimension. In two dimensions the fundamental solution is \( \frac{1}{2\pi}\ln r \) with \( \nabla^2(\ln r) = 2\pi\delta^2(\mathbf{r}) \), and in \( d\neq 3 \) the kernel is \( r^{2-d} \) with a different constant. The \( -4\pi \) is specific to 3D and the identity fails verbatim elsewhere.
  • Read pointwise. If you insist the equation holds as an ordinary function equality, it is simply false: away from the origin the left side is \( 0 \), and at the origin \( 1/r \) is not even defined, so no classical value of \( \nabla^2(1/r) \) can equal a delta. The statement lives only in the space of distributions.
  • Non-decaying test functions / no compact support. The integration-by-parts steps assume boundary terms at infinity vanish. Against \( f \) that does not decay (e.g. a plane wave or a constant), those surface integrals survive and the clean \( -4\pi f(\mathbf{0}) \) result is spoiled.
  • Naive differentiation under the integral. Writing \( \nabla^2\int\frac{\rho}{|\mathbf{r}-\mathbf{r}'|}d^3r' = \int\rho\,\nabla^2\frac{1}{|\mathbf{r}-\mathbf{r}'|}d^3r' \) and then setting the integrand to zero (because "\( \nabla^2(1/r)=0 \)") gives \( \nabla^2\varphi=0 \) everywhere — wrong. The delta contribution at \( \mathbf{r}=\mathbf{r}' \) is exactly what is being dropped.
Failure modes
  • The "zero everywhere" trap. Computing \( \nabla^2(1/r)=0 \) for \( r>0 \) and concluding it is zero everywhere — forgetting the single point \( r=0 \) carries all the content.
  • Dropping the \( r^2 \) in the radial Laplacian. Using \( \frac{\partial^2}{\partial r^2}(1/r) = 2/r^3 \neq 0 \) as "the Laplacian" instead of the correct \( \frac{1}{r^2}\partial_r(r^2\partial_r) \). The full operator gives \( 2/r^3 - 2/r^3 = 0 \); the missing term is the classic sign of error.
  • Sign errors on the constant. Writing \( +4\pi\delta^3 \) instead of \( -4\pi \), usually from mishandling the inward normal on the excised inner sphere in Green's identity.
  • Wrong delta normalisation. Treating \( \delta^3(\mathbf{r}) \) as \( \delta(r) \) (a 1D radial delta) — these differ by \( 4\pi r^2 \) and produce factor-of-\( 4\pi \) and dimensional errors downstream.
  • Assuming \( 4\pi \) is universal. Carrying the 3D constant into 2D or \( d \)-dimensional problems where the solid-angle factor \( \Omega_{d-1} \) is different.
Discussion

The heart of the matter is that \( 1/r \) is locally integrable in three dimensions — \( \int_0^R \frac{1}{r}\,4\pi r^2\,dr = 2\pi R^2 \) is finite — so it defines a perfectly good distribution even though it blows up at the origin. Distributions are differentiated by moving derivatives onto test functions, and when you do so for \( \nabla^2 \), the near-origin behaviour that classical calculus cannot see is captured exactly by the delta. The pointwise Laplacian tells you the field is source-free in the bulk; the flux integral tells you a source is hidden at the point; the distributional identity unifies both into a single equation.

The number \( 4\pi \) is not arbitrary decoration — it is the surface area of the unit sphere, i.e. the total solid angle. Its appearance here is the same \( 4\pi \) that shows up in Gauss's law \( \oint\mathbf{E}\cdot d\mathbf{A} = q/\varepsilon_0 \) and in the Gaussian-unit Coulomb law. Choosing to define the Green's function as \( -\frac{1}{4\pi r} \) is precisely what absorbs this geometric factor so that \( \nabla^2 G = \delta^3 \) comes out clean; the SI/Gaussian split in electromagnetism is largely a bookkeeping choice about where this \( 4\pi \) is allowed to live.

The practical payoff is convolution. Because \( \nabla^2\left(-\frac{1}{4\pi r}\right)=\delta^3 \) and the Laplacian is linear and translation-invariant, the solution of \( \nabla^2\varphi = -\rho/\varepsilon_0 \) with decaying boundary conditions is \( \varphi = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d^3r' \). Every superposition-of-point-charges argument in electrostatics is this identity in disguise, and the same structure recurs for the Helmholtz Green's function \( \frac{e^{ikr}}{r} \) and the retarded Green's function of the wave equation.

At the deepest level this is a statement about the analytic structure of harmonic functions and the codimension of the singularity. In \( d \) dimensions the fundamental solution scales as \( r^{2-d} \) so that \( \nabla^2 r^{2-d}\propto\delta^d \), with the constant set by \( \Omega_{d-1} \); the logarithm in 2D is the borderline case \( d=2 \) where \( 2-d=0 \). One can also derive the identity by Fourier transform: \( \widehat{(1/r)}(\mathbf{k}) = 4\pi/k^2 \), so \( \widehat{\nabla^2(1/r)} = -k^2\cdot 4\pi/k^2 = -4\pi \), a constant, whose inverse transform is exactly \( -4\pi\delta^3(\mathbf{r}) \) — the same result seen from momentum space, where it reflects the fact that the Coulomb propagator behaves as \( 1/k^2 \).

Common misconceptions. The delta is not "\( \nabla^2(1/r) \) being infinite at a point in the ordinary sense" — it is a linear functional, meaningful only under an integral. Nor does \( \nabla^2(1/r)=0 \) for \( r>0 \) contradict the identity: pointwise vanishing away from a measure-zero set is entirely consistent with a distribution supported on that set. And the delta is genuinely necessary — it is not an optional regularisation you may discard when convenient.

Worked examples
1
Example A — Poisson's equation for a point charge. Show that \( \varphi = \dfrac{q}{4\pi\varepsilon_0 r} \) satisfies \( \nabla^2\varphi = -\rho/\varepsilon_0 \) with \( \rho = q\,\delta^3(\mathbf{r}) \).
Set up: apply the identity to the explicit Coulomb potential. A
2
\[ \nabla^2\varphi = \frac{q}{4\pi\varepsilon_0}\,\nabla^2\!\left(\frac{1}{r}\right) = \frac{q}{4\pi\varepsilon_0}\big(-4\pi\,\delta^3(\mathbf{r})\big) = -\frac{q}{\varepsilon_0}\,\delta^3(\mathbf{r}). \]
Pull constants out, insert the identity, and cancel \( 4\pi \). A
3
\[ -\frac{q\,\delta^3(\mathbf{r})}{\varepsilon_0} = -\frac{\rho}{\varepsilon_0},\qquad \rho = q\,\delta^3(\mathbf{r}). \]
Numbers: take \( q = 1.60\times10^{-19}\,\mathrm{C} \), \( \varepsilon_0 = 8.85\times10^{-12}\,\mathrm{F\,m^{-1}} \). Then the source density is \( \rho/\varepsilon_0 = (1.60\times10^{-19}/8.85\times10^{-12})\,\delta^3 = 1.81\times10^{-8}\,\delta^3(\mathbf{r})\ \mathrm{V\,m} \). B
\[ \nabla^2\varphi = -\frac{q}{\varepsilon_0}\,\delta^3(\mathbf{r}),\qquad q = 1.60\times10^{-19}\,\mathrm{C} \]

Reading. The Coulomb potential is the exact Green's-function solution of Poisson's equation with a point source at the origin; the identity is what converts the smooth-looking \( 1/r \) into a localised charge.

1
Example B — Flux through a sphere from a nascent delta. Verify \( \displaystyle\int_{V}\nabla^2\!\left(\frac{1}{r}\right)d^3r = -4\pi \) numerically using the regularised kernel \( \dfrac{1}{\sqrt{r^2+a^2}} \) with \( a = 0.10\ \mathrm{nm} \), integrated over all space.
Set up: compute the Laplacian of the softened kernel and integrate; the result must be \( a \)-independent. B
2
\[ \nabla^2\frac{1}{\sqrt{r^2+a^2}} = -\frac{3a^2}{(r^2+a^2)^{5/2}} \equiv g_a(r). \]
Apply \( \frac{1}{r^2}\partial_r(r^2\partial_r) \) to \( (r^2+a^2)^{-1/2} \); the \( r^{-2} \)-type singularity is removed by \( a \), giving a smooth, everywhere-defined function. B
3
\[ \int g_a\,d^3r = -\int_0^\infty \frac{3a^2}{(r^2+a^2)^{5/2}}\,4\pi r^2\,dr = -12\pi a^2\!\int_0^\infty\frac{r^2\,dr}{(r^2+a^2)^{5/2}}. \]
Insert the spherical volume element \( 4\pi r^2\,dr \); the integral is standard, \( \int_0^\infty r^2(r^2+a^2)^{-5/2}dr = \frac{1}{3a^2} \). B
4
\[ \int g_a\,d^3r = -12\pi a^2\cdot\frac{1}{3a^2} = -4\pi \approx -12.566. \]
Numbers: the \( a^2 \) cancels, so for \( a = 0.10\ \mathrm{nm} \) — or any \( a \) — the total is exactly \( -4\pi = -12.566 \), confirming the delta normalisation as \( a\to 0 \). A
\[ \int_{\mathbb{R}^3}\nabla^2\frac{1}{\sqrt{r^2+a^2}}\,d^3r = -4\pi \quad(\text{all }a) \]

Reading. The regularised kernel is a smooth stand-in whose Laplacian is an ordinary bell-shaped function of total weight \( -4\pi \); shrinking \( a \) concentrates that weight onto the origin, realising \( -4\pi\delta^3(\mathbf{r}) \) as an explicit limit of smooth functions.

Problems
  1. Compute \( \dfrac{\partial^2}{\partial r^2}\!\left(\dfrac{1}{r}\right) \) and separately the full \( \nabla^2\!\left(\dfrac{1}{r}\right) \) for \( r>0 \). Explain in one sentence why they differ.
    Solution \( \partial_r(1/r) = -1/r^2 \), \( \partial_r^2(1/r) = 2/r^3 \). The full Laplacian is \( \frac{1}{r^2}\partial_r(r^2\partial_r(1/r)) = \frac{1}{r^2}\partial_r(-1) = 0 \). They differ by the term \( \frac{2}{r}\partial_r(1/r) = -2/r^3 \): the radial Laplacian is \( \partial_r^2 + \frac{2}{r}\partial_r \), and \( 2/r^3 + (-2/r^3) = 0 \). The plain second derivative omits the geometric \( \frac{2}{r}\partial_r \) piece.
  2. Using the divergence theorem, evaluate the flux of \( \nabla(1/r) \) through a cube of side \( 2L \) centred at the origin. Does the answer depend on \( L \) or on the shape?
    Solution The flux equals \( \int_V\nabla^2(1/r)\,d^3r = -4\pi \) for any closed surface enclosing the origin, cube included, by the identity. It is independent of both \( L \) and shape — this is exactly Gauss's law: the flux depends only on whether the source is enclosed, giving \( -4\pi \) here.
  3. In two dimensions the fundamental solution is \( G_2 = \frac{1}{2\pi}\ln r \). Show that \( \nabla^2 G_2 = \delta^2(\mathbf{r}) \) by integrating over a disk of radius \( R \).
    Solution For \( r>0 \), \( \nabla^2\ln r = \frac{1}{r}\partial_r(r\,\partial_r\ln r) = \frac{1}{r}\partial_r(r\cdot\frac{1}{r}) = \frac{1}{r}\partial_r(1) = 0 \). Flux of \( \nabla\ln r = \hat{\mathbf{r}}/r \) through the circle \( r=R \): \( \oint \frac{1}{R}\,R\,d\theta = \int_0^{2\pi}d\theta = 2\pi \). So \( \int\nabla^2\ln r\,d^2r = 2\pi \), giving \( \nabla^2(\frac{1}{2\pi}\ln r) = \delta^2(\mathbf{r}) \). The \( 4\pi \) of 3D is replaced by \( 2\pi \), the circumference of the unit circle.
  4. Verify the Fourier-space route: given \( \widehat{(1/r)}(\mathbf{k}) = 4\pi/k^2 \), find the transform of \( \nabla^2(1/r) \) and invert it.
    Solution Under Fourier transform \( \nabla^2 \to -k^2 \), so \( \widehat{\nabla^2(1/r)} = -k^2\cdot\frac{4\pi}{k^2} = -4\pi \), a constant. The inverse transform of the constant \( -4\pi \) is \( -4\pi\,\delta^3(\mathbf{r}) \), since \( \int\frac{d^3k}{(2\pi)^3}e^{i\mathbf{k}\cdot\mathbf{r}} = \delta^3(\mathbf{r}) \). This reproduces the identity without excising the origin.
  5. A charge \( q = 2.0\ \mathrm{nC} \) sits at the origin in vacuum. Write \( \rho(\mathbf{r}) \), then confirm \( \int\rho\,d^3r = q \) and compute \( \nabla^2\varphi \) at a field point \( r = 5.0\ \mathrm{cm} \) away.
    Solution \( \rho(\mathbf{r}) = q\,\delta^3(\mathbf{r}) = 2.0\times10^{-9}\,\delta^3(\mathbf{r})\ \mathrm{C} \). Total charge: \( \int\rho\,d^3r = q\int\delta^3\,d^3r = q = 2.0\ \mathrm{nC} \), as required. At \( r = 5.0\ \mathrm{cm} = 0.050\ \mathrm{m} \) the point is away from the origin, so \( \delta^3(\mathbf{r}) = 0 \) there and \( \nabla^2\varphi = -\rho/\varepsilon_0 = 0 \) — the field region is source-free. The full statement is \( \nabla^2\varphi = -\frac{q}{\varepsilon_0}\delta^3(\mathbf{r}) = -\frac{2.0\times10^{-9}}{8.85\times10^{-12}}\delta^3 = -226\,\delta^3(\mathbf{r})\ \mathrm{V\,m} \), nonzero only at the origin.