Density log and photoelectric factor (Pe)

Quick refresher

The source is caesium-137, and its photons have an energy of 662 keV. Photons lose energy in Compton scattering on electrons (they bounce off electrons). Photons that have grown weak are absorbed in the photoelectric effect (an atom takes the whole photon).
The number of scattering events is proportional to the number of electrons per unit volume. So the tool measures electron density. After calibration, its reading for most minerals and for water is practically equal to the bulk density. For salt, gypsum and gas the difference is noticeable.
The near and far detectors respond differently to the mudcake and to the irregularities of the wall. By comparing them, the tool applies the correction $\Delta\rho$ . A large correction and a washout on the caliper log (a record of the hole diameter) are a signal not to trust the density.
Density porosity $\phi_D$ is total porosity. The fluid in the calculation is the mud filtrate with the residual hydrocarbons of the flushed zone (the zone next to the wall, where the filtrate has pushed the original fluids away).
$P_e$ rises steeply with atomic number: quartz 1.8, dolomite 3.1, calcite 5.1 barn/electron (a barn is a very small unit of area used in nuclear physics). It depends little on porosity and fluid, so it points to the mineral. Barite in the mud makes $P_e$ unusable.
Gas lowers the density and makes $\phi_D$ too high. Together with the neutron log, this gives a characteristic separation of the curves (Neutron log).

Key relationships

\rho_e = \rho_b\,\dfrac{2\sum Z}{M}

electron density index of a substance

\rho_a = 1.0704\,\rho_e - 0.1883

apparent density of a tool calibrated on limestone saturated with fresh water

\phi_D = \dfrac{\rho_{ma} - \rho_b}{\rho_{ma} - \rho_f}

porosity from the density log

P_e \approx (Z/10)^{3.6},\qquad U = P_e\,\rho_e

photoelectric factor of an element and the volumetric cross section, which mixes by volume

Common pitfalls

Using the density of formation water or oil for the fluid, instead of the density of the filtrate in the flushed zone.
Calculating porosity in washouts and wherever the correction $\Delta\rho$ is large.
Using $P_e$ when the mud is weighted with barite.
Mixing $P_e$ in proportion to the volumes. What mixes by volume is $U = P_e\rho_e$ , not $P_e$ itself.

After this lesson you can

explain how gamma rays are scattered and absorbed in rock and why the detector count falls as density increases
distinguish electron density, apparent density and bulk density, and explain why rock salt looks “lighter” on the log than it really is
describe how the compensated two-detector tool is built and what the correction $\Delta\rho$ means
calculate porosity from density, taking the filtrate density into account, and judge the reliability of the readings from the correction and the caliper log
calculate the photoelectric factor of a mineral and of a rock and use $P_e$ to identify lithology

Gamma rays in rock: scattering and absorption

The source of the density tool, caesium-137, emits gamma rays of a single energy, 662 keV. In the rock, the photons meet the electrons of atoms. In this energy range, one of two events happens. A third known process, the production of an electron–positron pair, is possible only at energies above 1.02 MeV. It does not matter for this source.

Compton scattering: A photon collides with an electron, gives part of its energy to it and flies on in a different direction. The chance of such a collision is almost the same for the electrons of any atoms. So the number of scattering events per unit path is proportional to the number of electrons per unit volume, which is the electron density. This is the main process at energies from hundreds of keV to a few MeV.
Photoelectric absorption: An atom absorbs a photon of low energy, usually below 100 keV, completely, and ejects an electron. The chance of absorption rises very rapidly with the atomic number $Z$ of the element (the number of protons in its nucleus). Heavy atoms (calcium, iron, barium) absorb photons far more strongly than light ones.

The path of a photon from the source to the detector is a chain of Compton scattering events. After each one, the photon has less energy. Photons that have weakened to tens of keV are absorbed. A dense rock contains more electrons. So photons scatter more often, lose energy faster and are more often absorbed before they reach the detector. At the source-to-detector distances chosen in the tool, the denser the rock, the lower the count.

Detector count and electron density (calibration relationship)

\ln N = a - b\,\rho_e

where:

$N$ count rate of the detector in the energy window of Compton scattering, counts per second
$\rho_e$ electron density of the rock in the volume the detector sees (see the next section), g/cm³
$a,\ b$ constants of the detector, found by calibration; $b$ grows with the distance from the source to the detector, —; cm³/g

Units: $N$ in counts/s (natural logarithm), $\rho_e$ in g/cm³

Holds when

An approximate relationship found by experiment, for a fixed tool design, energy window and pad contact.
The distance to the detector is greater than a few photon mean free paths (the average distance a photon travels between collisions). Right next to the source the relationship is reversed: a denser rock scatters more photons to the detector.
The rock is uniform in the volume of investigation, with no mudcake or standoff.

A change in density by $\Delta\rho_e$ changes the count by a factor of $\e^{-b\,\Delta\rho_e}$ . For the far detector $b$ is larger. It is more sensitive to density and sees deeper, but fewer photons reach it, so the statistical scatter is higher. The near detector sees less deep and therefore reacts more strongly to mudcake. The tool uses this difference for compensation.

Electron density, apparent density and bulk density

The tool senses electrons, but we need mass. The link between them is the ratio of the number of electrons to the mass of the substance. For most elements this ratio is nearly the same. For oxygen, silicon, calcium and carbon, the number of electrons (the atomic number $Z$ ) is about half the atomic mass. Hydrogen stands out sharply: it has twice as many electrons per unit mass. It is convenient to use the electron density index. This is the electron density recalculated so that it equals the ordinary density when $2Z/M = 1$ .

Electron density index

\rho_e = \rho_b\,\frac{2\sum Z}{M}

where:

$\rho_e$ electron density index, g/cm³
$\rho_b$ bulk density of the substance, g/cm³
$\sum Z$ number of electrons in the molecule: the sum of the atomic numbers of its atoms, —
$M$ molar mass of the molecule, g/mol

Units: densities in g/cm³; the ratio $2\sum Z/M$ is dimensionless when $M$ is in g/mol

Holds when

For a mixture of minerals and fluids the indices add in proportion to the volume fractions, as the densities themselves do.

For quartz, calcite and dolomite, the ratio $2\sum Z/M$ differs from one by less than 0.3%. For rock salt it is about 0.96, for water about 1.11 and for methane about 1.25. So, counted in electrons, water and hydrocarbons are “denser” than they are by mass, and salt is “lighter”. The calibration of the tool (below) corrects this for water and limestone, but not for all substances.

We calibrate the tool so that in the most common situation, limestone saturated with fresh water, it reads the true density. For this, the electron density index is converted into the apparent density $\rho_a$ , the one that is recorded on the log.

Apparent density, which the tool reads

\rho_a = 1.0704\,\rho_e - 0.1883

where:

$\rho_a$ apparent density (the one the tool reads), g/cm³
$\rho_e$ electron density index of the rock, g/cm³

Units: g/cm³

Holds when

Calibration on limestone saturated with fresh water: for calcite and for water $\rho_a$ coincides with the true density.
The constants 1.0704 and 0.1883 come from requiring $\rho_a = \rho_b$ for these two substances. So for other substances the equality is approximate.

For ordinary reservoirs (sandstones, limestones and dolomites with water or oil in the pores), the apparent density differs from the bulk density by no more than 0.01 g/cm³, so we take it directly as $\rho_b$ . We apply a correction for salt, gypsum, coal and gas-bearing formations.

Worked exampleWhy rock salt looks lighter on the log than it really is

Given (assumed data)

Quartz $\mathrm{SiO_2}$: $\rho_b = 2.65$ g/cm³; $\sum Z = 14 + 2\cdot 8 = 30$
Rock salt $\mathrm{NaCl}$: $\rho_b = 2.165$ g/cm³; $\sum Z = 11 + 17 = 28$
Fresh water $\mathrm{H_2O}$: $\rho_b = 1.00$ g/cm³; $\sum Z = 2\cdot 1 + 8 = 10$
Atomic masses, g/mol: H 1.008; O 16.00; Na 22.99; Si 28.09; Cl 35.45

Solution

Molar masses: quartz 60.09, salt 58.44, water 18.016 g/mol.
Ratios $2\sum Z/M$ : quartz $60/60.09 = 0.9985$ , salt $56/58.44 = 0.9582$ , water $20/18.016 = 1.1101$ .
Electron density indices $\rho_e = \rho_b\cdot 2\sum Z/M$ : quartz 2.646, salt 2.075, water 1.110 g/cm³.
Apparent density $\rho_a = 1.0704\,\rho_e - 0.1883$ : quartz 2.644, salt 2.032, water 1.000 g/cm³.

Answer: The tool reads 1.000 g/cm³ for water, 2.644 for quartz (0.006 g/cm³ below the true value) and 2.032 g/cm³ for rock salt instead of 2.165.

What it means. Sodium and chlorine have fewer electrons per unit mass than the elements used to calibrate the tool, so salt looks lighter by 0.13 g/cm³. This is not a tool error. It follows from the fact that the tool counts electrons. That is why reference tables give an apparent density of 2.03 g/cm³ for salt, and why charts and interpretation software work with the apparent densities of minerals. For quartz the difference is a few thousandths of a g/cm³, and we ignore it. The data in this example are assumed for illustration, and the atomic masses are rounded.

True and apparent density of substances

Substance: Calcite
$2\sum Z/M$: 0.9991
$\rho_b$ , g/cm³: 2.71
$\rho_a$ , g/cm³: 2.710

Substance: Dolomite
$2\sum Z/M$: 0.9978
$\rho_b$ , g/cm³: 2.87
$\rho_a$ , g/cm³: 2.877

Substance: Anhydrite
$2\sum Z/M$: 0.9990
$\rho_b$ , g/cm³: 2.96
$\rho_a$ , g/cm³: 2.977

Substance: Gypsum
$2\sum Z/M$: 1.0222
$\rho_b$ , g/cm³: 2.32
$\rho_a$ , g/cm³: 2.350

Substance: Oil $\mathrm{CH_2}$
$2\sum Z/M$: 1.1407
$\rho_b$ , g/cm³: 0.85
$\rho_a$ , g/cm³: 0.850

Substance: Methane
$2\sum Z/M$: 1.2467
$\rho_b$ , g/cm³: 0.20
$\rho_a$ , g/cm³: 0.079

Substance	$2\sum Z/M$	$\rho_b$ , g/cm³	$\rho_a$ , g/cm³
Calcite	0.9991	2.71	2.710
Dolomite	0.9978	2.87	2.877
Anhydrite	0.9990	2.96	2.977
Gypsum	1.0222	2.32	2.350
Oil $\mathrm{CH_2}$	1.1407	0.85	0.850
Methane	1.2467	0.20	0.079

Calculated with the two formulas above. Gypsum contains water of crystallisation (water built into its crystals), so it looks heavier than it is. For methane the difference is huge. This is why gas in the flushed zone lowers the tool reading so much (more in the advanced section).

The tool: pad, two detectors and the correction Δρ

The source and detectors sit in the pad (also called a skid), a heavy shielded block that an arm presses against the borehole wall. The shields let photons through only via the windows that face the rock. This stops the radiation from travelling directly along the tool and through the mud. The pad has a wedge shape and is pressed hard, so it cuts through soft mudcake. The opening of the arm also gives the hole diameter. But a thin layer of mudcake, or mud in the irregularities of the wall, almost always remains. It falls into the volume the tool sees, and the tool counts it just like the rock.

SchematicA schematic drawing of a two-detector density tool (not to scale). The pad with the source and two detectors is pressed against the wall through the mudcake, and the arm braces against the opposite wall. The broken lines are photon paths: scattering in the formation (dots) and absorption (cross).

A single detector cannot tell mudcake from formation. That is why the compensated tool has two. The near detector sees less deep and reacts more strongly to mudcake and standoff (the gap between the pad and the wall). The far detector sees deeper and gives the main measurement. If there is no mudcake, the readings of both match after calibration. If there is mudcake, they move apart. They move further apart when the mudcake is thicker and when its density differs more from that of the formation. Plot the far-detector count against the near-detector count. The points without mudcake lie on one line, the “spine”. Mudcake pushes the points sideways along the “ribs”. Each rib corresponds to one formation density, almost whatever mudcake has distorted it. So we find the corrected density from the pair of counts, without knowing either the thickness or the density of the mudcake.

The tool records the corrected bulk density $\rho_b$ (the RHOB curve). It also records the correction $\Delta\rho$ itself (the DRHO curve), which was added to the far-detector reading. Light mudcake lowers the reading, so the correction is positive. Mudcake with barite is heavier than the formation, so the correction is negative. The $\Delta\rho$ curve, together with the caliper log, is the main indicator of the quality of the density.

The litho-density tool sorts photons by energy. In the high-energy window, photons are lost only through scattering, so the count depends only on the electron density. In the low-energy window, photoelectric absorption adds to the scattering, so the count also depends on the atomic number. The ratio of the counts in the two windows gives $P_e$ (the PEF curve). The vertical resolution (the thinnest bed a curve shows clearly) is about 15–45 cm for density, depending on processing, and 5–10 cm for $P_e$ . More photons reach the detectors of the litho-density tool than in the first two-detector tools, so its curves are less noisy. Tools with three detectors add one more, very close to the source. It improves the compensation in rough holes.

From density to porosity

The bulk density of a clean rock is a mixture of the matrix density and the fluid density, each weighted by its volume: $\rho_b = (1 - \phi)\,\rho_{ma} + \phi\,\rho_f$ . Solving this equality for porosity gives the main formula of the method, the same as in the topic Porosity.

Porosity from the density log

\phi_D = \frac{\rho_{ma} - \rho_b}{\rho_{ma} - \rho_f}

where:

$\phi_D$ porosity from the density log (total: it includes isolated pores too), fraction
$\rho_b$ bulk density from the log after the correction $\Delta\rho$ , g/cm³
$\rho_{ma}$ grain density of the matrix: quartz 2.65, calcite 2.71, dolomite 2.87, g/cm³
$\rho_f$ density of the fluid in the flushed zone: filtrate with residual hydrocarbons, g/cm³

Units: g/cm³ or kg/m³: any density units, the same for all

Holds when

The matrix is known, and there is no shale or heavy minerals (otherwise a correction is needed).
The pores in the volume of investigation (10–15 cm from the wall) are filled with a fluid of known density.
The reading is corrected for mudcake and hole, and the apparent density is practically equal to the bulk density, which is not so for salt, gypsum and gas.

The lower the density for the same mineralogy, the more pores there are. The denominator $\rho_{ma} - \rho_f$ is about 1.6–1.9 g/cm³, so a density error of 0.01 g/cm³ gives a porosity error of about 0.6 percentage points.

Which fluid should we put into the formula? The tool sees the flushed zone (Borehole environment: invaded zone and mudcake). The pores there are filled with mud filtrate that contains some residual oil or gas (oil or gas left behind). The filtrate of a water-based mud has a density from 1.0 g/cm³ (fresh) to 1.1–1.2 g/cm³ (salty). For a sodium chloride solution at surface conditions, $\rho_{mf} \approx 1 + 0.73P$ , where $P$ is the mass fraction of salt. The filtrate of an oil-based mud is lighter, about 0.8 g/cm³. Residual oil hardly changes the density of the fluid. Residual gas lowers it strongly.

Worked examplePorosity of limestone with a salty filtrate, and a point that must not be trusted

Given (assumed data)

Rock: limestone, $\rho_{ma} = 2.71$ g/cm³
Filtrate: water-based mud, 120 g of NaCl per 1 kg of solution ( $P = 0.12$ )
Bit diameter: 216 mm (8½ in)
Point A: $\rho_b = 2.47$ g/cm³, $\Delta\rho = +0.02$ , diameter 216 mm
Point B: $\rho_b = 2.38$ g/cm³, $\Delta\rho = +0.05$ , diameter 211 mm
Point C: $\rho_b = 2.19$ g/cm³, $\Delta\rho = +0.16$ , diameter 254 mm

Solution

Filtrate density: $\rho_{mf} \approx 1 + 0.73\cdot 0.12 = 1.088$ g/cm³.
Point A: $\phi_D = \dfrac{2.71 - 2.47}{2.71 - 1.088} = 0.148$ ; with fresh water (1.00) it would have been 0.140.
Point B: $\phi_D = \dfrac{2.71 - 2.38}{2.71 - 1.088} = 0.203$ (with fresh water, 0.193). The diameter is 5 mm smaller than the bit. This is mudcake about 2–3 mm thick, and a correction of +0.05 is usual.
Point C: formally $\phi_D = 0.32$ . But the hole is enlarged to 254 mm, and the correction of +0.16 g/cm³ is large. The pad did not touch the wall, and the tool saw the mud in the washout.

Answer: The porosity of the limestone is about 14.8% at point A and 20.3% at point B. The value of 32% at point C is an artefact of the washout (a false result), and we do not use it.

What it means. If we used fresh water instead of the salty filtrate, the porosity would come out lower by 0.8–1.0 percentage points. This is about 5% of its value, and the error would be the same all along the section. The error in the matrix density is usually larger (Porosity). For point C, we take porosity from methods that are less sensitive to washouts, or we exclude the interval. The formula for the filtrate density is approximate and applies to surface conditions: at formation temperature the filtrate is slightly lighter. The data in this example are assumed for illustration.

Shale, heavy minerals and gas shift the result in predictable directions. The density of shale (2.2–2.7 g/cm³) differs from the matrix density, so in shaly beds the density porosity is distorted. Pyrite (an iron sulfide, 5.0 g/cm³) and siderite (an iron carbonate, 3.9 g/cm³), even in small amounts, raise the density and lower the porosity. Residual gas in the flushed zone lowers the density and makes $\phi_D$ too high. We see this when we compare with the neutron log (Neutron log). The topics Porosity from the density log and Gas, shale, carbonate and pressure effects on porosity logs discuss in detail the choice of parameters and the corrections.

Photoelectric factor and lithology

The photoelectric factor $P_e$ is the cross section for photoelectric absorption of gamma rays per electron of the substance. We give it in barns per electron (1 barn = $10^{-24}$ cm²). The cross section measures the chance of an interaction: the larger it is, the more often a photon is absorbed. The factor depends strongly on the atomic number and does not depend on density, so it shows which elements the rock is made of. For an element, a power law describes it well.

Photoelectric factor of an element

P_e \approx \left(\frac{Z}{10}\right)^{3.6}

where:

$P_e$ photoelectric factor of the element, barn/electron
$Z$ atomic number of the element, —

Units: barn/electron

Holds when

An approximate relationship for the photon energies at which the tool measures photoelectric absorption.

Oxygen ( $Z = 8$ ) gives about 0.45, silicon (14) 3.4, calcium (20) 12.1, barium (56) about 490. Doubling the atomic number multiplies the factor by $2^{3.6} \approx 12$ .

Photoelectric factor of a molecule (mineral)

P_e = \frac{\sum_i A_i Z_i P_{e,i}}{\sum_i A_i Z_i}

where:

$A_i$ number of atoms of the $i$ -th element in the molecule, —
$Z_i$ atomic number of the $i$ -th element; $A_i Z_i$ is the number of its electrons in the molecule, —
$P_{e,i}$ photoelectric factor of the $i$ -th element, barn/electron

Units: barn/electron

Holds when

The factor of a mineral is the average of the factors of its elements, weighted by the number of electrons.

Quartz $\mathrm{SiO_2}$ gets a factor of 1.8, and calcite $\mathrm{CaCO_3}$ gets 5.1. The difference comes from a single heavy calcium atom in place of silicon.

For a rock made of a matrix and a fluid, we cannot mix $P_e$ by volume. It refers to a single electron, and a mineral and water have different numbers of electrons per unit volume. What mixes by volume is the volumetric photoelectric cross section $U$ , the product of $P_e$ and the electron density.

Volumetric photoelectric cross section and its mixing

U = P_e\,\rho_e, \qquad U = \phi\,U_f + (1 - \phi)\,U_{ma}

where:

$U,\ U_f,\ U_{ma}$ volumetric photoelectric cross section of the rock, the fluid and the matrix, barn/cm³
$P_e$ photoelectric factor, barn/electron
$\rho_e$ electron density index, g/cm³ (in the role of the number of electrons per unit volume)
$\phi$ porosity, fraction

Units: barn/cm³; for several minerals $U = \sum_i V_i U_i$ by volume fractions

Holds when

Pores and minerals are distributed uniformly in the volume of investigation; there is no barite in the mud or the mudcake.

For water $U \approx 0.4$ barn/cm³, for oil about 0.1, and for ordinary rock-forming minerals roughly 5 to 15. The fluid adds almost nothing, so the $P_e = U/\rho_e$ of a porous rock stays close to the $P_e$ of its matrix.

Photoelectric factor and density of minerals and fluids

Substance: Quartz
$P_e$ , barn/electron: 1.81
$\rho_b$ , g/cm³: 2.65
$U$ , barn/cm³: 4.8

Substance: Dolomite
$P_e$ , barn/electron: 3.14
$\rho_b$ , g/cm³: 2.87
$U$ , barn/cm³: 9.0

Substance: Calcite
$P_e$ , barn/electron: 5.08
$\rho_b$ , g/cm³: 2.71
$U$ , barn/cm³: 13.8

Substance: Anhydrite
$P_e$ , barn/electron: 5.05
$\rho_b$ , g/cm³: 2.96
$U$ , barn/cm³: 14.9

Substance: Rock salt
$P_e$ , barn/electron: 4.65
$\rho_b$ , g/cm³: 2.165
$U$ , barn/cm³: 9.7

Substance: Pyrite
$P_e$ , barn/electron: 17.0
$\rho_b$ , g/cm³: 5.00
$U$ , barn/cm³: 82

Substance: Barite
$P_e$ , barn/electron: 267
$\rho_b$ , g/cm³: 4.5
$U$ , barn/cm³: about 1070

Substance: Fresh water
$P_e$ , barn/electron: 0.36
$\rho_b$ , g/cm³: 1.00
$U$ , barn/cm³: 0.40

Substance: Oil
$P_e$ , barn/electron: 0.12
$\rho_b$ , g/cm³: 0.85
$U$ , barn/cm³: 0.12

Substance	$P_e$ , barn/electron	$\rho_b$ , g/cm³	$U$ , barn/cm³
Quartz	1.81	2.65	4.8
Dolomite	3.14	2.87	9.0
Calcite	5.08	2.71	13.8
Anhydrite	5.05	2.96	14.9
Rock salt	4.65	2.165	9.7
Pyrite	17.0	5.00	82
Barite	267	4.5	about 1070
Fresh water	0.36	1.00	0.40
Oil	0.12	0.85	0.12

The factors of the minerals, calculated with the two formulas above, agree with the reference values to within hundredths. Clay minerals give from 1.8 (kaolinite) to 6 and more (iron-rich chlorite). So shales usually have a $P_e$ of 2–4, with a large scatter.

Worked exampleSandstone or limestone at the same density

Given (assumed data)

Density reading: $\rho_b = 2.36$ g/cm³, pores filled with fresh filtrate
Hypothesis 1: quartz sandstone, $\rho_{ma} = 2.65$ g/cm³, $\rho_e = 2.646$ g/cm³
Hypothesis 2: limestone, $\rho_{ma} = 2.71$ g/cm³, $\rho_e = 2.708$ g/cm³
Water: $\rho_e = 1.110$ g/cm³

Solution

Factors of the elements $(Z/10)^{3.6}$ : silicon 3.36, calcium 12.1, oxygen 0.448, carbon 0.16; hydrogen has about 0.0003 barn/electron.
Quartz: $P_e = \dfrac{14\cdot 3.36 + 16\cdot 0.448}{30} = 1.81$ ; calcite 5.08, water 0.36 barn/electron.
Volumetric cross sections $U = P_e\rho_e$ : quartz 4.78, calcite 13.8, water 0.40 barn/cm³.
Density porosity: sandstone $(2.65 - 2.36)/1.65 = 0.176$ , limestone $(2.71 - 2.36)/1.71 = 0.205$ .
Factor of the rock $P_e = U/\rho_e$ with $U$ and $\rho_e$ mixed by volume: sandstone 1.69, limestone 4.63 barn/electron.

Answer: At the same density of 2.36 g/cm³, the sandstone shows

P_e \approx 1.69

and the limestone 4.63 barn/electron. Their porosities are 17.6 and 20.5%.

What it means. A porosity of about 20% lowered the $P_e$ of the limestone by only 9% compared with dense calcite, whereas the two minerals differ by almost a factor of three. That is why $P_e$ gives the lithology and density gives the porosity, and each curve complements the other. For mixtures of quartz, calcite and dolomite, and for shaly rocks, we use the $P_e$ – $\rho_b$ crossplot (Lithology from the photoelectric factor and spectral gamma ray). The data in this example are assumed for illustration.

CalculatedCalculated photoelectric factor of a porous rock against porosity for a matrix of calcite, dolomite and quartz with the pores filled with fresh water (mixing by

U

). The points are the two hypotheses from the example above.

Because its resolution is high, $P_e$ picks out thin beds and fractures. It also reveals heavy minerals: pyrite and siderite raise $P_e$ far above the values of ordinary minerals.

When density cannot be trusted

The tool sees a thin layer of rock behind the wall and is pressed against the wall by a pad. So its readings depend on the condition of the hole more than those of other tools. Before calculating anything from density, look at the caliper log, the $\Delta\rho$ curve and the repeat section (a second pass over the same interval).

Signs of unreliable density

Sign: Caliper log clearly larger than the bit
What is happening: the pad moves away from the wall and the tool sees mud: $\rho_b$ is too low, $\Delta\rho$ is large
What to do: exclude the interval or take porosity from other methods

Sign: Rough wall, frequent small washouts
What is happening: jagged $\rho_b$ and $P_e$ , jumps in $\Delta\rho$
What to do: compare with the neutron and sonic logs, do not take the extremes

Sign: Thick mudcake without barite
What is happening: $\Delta\rho$ is positive
What to do: within the range of compensation the correction can be trusted; check the mudcake thickness against the caliper log

Sign: Barite in the mudcake or mud
What is happening: $\Delta\rho$ is negative, $P_e$ is much too high
What to do: do not use $P_e$ , check the density

Sign: Gas in the flushed zone
What is happening: the density is too low; this is a property of the formation, not an error
What to do: interpret together with the neutron log

Sign: Pyrite, siderite
What is happening: the density and $P_e$ are higher than for the main matrix, so porosity is underestimated
What to do: take the mineral into account in the matrix density

Sign	What is happening	What to do
Caliper log clearly larger than the bit	the pad moves away from the wall and the tool sees mud: $\rho_b$ is too low, $\Delta\rho$ is large	exclude the interval or take porosity from other methods
Rough wall, frequent small washouts	jagged $\rho_b$ and $P_e$ , jumps in $\Delta\rho$	compare with the neutron and sonic logs, do not take the extremes
Thick mudcake without barite	$\Delta\rho$ is positive	within the range of compensation the correction can be trusted; check the mudcake thickness against the caliper log
Barite in the mudcake or mud	$\Delta\rho$ is negative, $P_e$ is much too high	do not use $P_e$ , check the density
Gas in the flushed zone	the density is too low; this is a property of the formation, not an error	interpret together with the neutron log
Pyrite, siderite	the density and $P_e$ are higher than for the main matrix, so porosity is underestimated	take the mineral into account in the matrix density

The remaining checks are common to all nuclear methods. The tool is calibrated on reference blocks of known density (for example, aluminium and magnesium). It is checked on field standards before and after the job. A short interval is recorded again as a repeat section. The count of photons is random, so at too high a logging speed the curve becomes noisy (Gamma ray log and natural gamma ray spectrometry). Finally, density is compared with other methods. In a clean water-bearing formation of known lithology, the porosities from the density and neutron logs should be close.

Where else density is needed

Porosity is the main application (Porosity from the density log).
Lithology, from $P_e$ and together with the neutron and sonic logs (Lithology and porosity from crossplots, Lithology from the photoelectric factor and spectral gamma ray).
Gas, from the characteristic separation from the neutron porosity (Neutron log).
Evaporites (rocks left behind when salty water dries up): rock salt (apparent density 2.03 g/cm³), anhydrite (2.98) and gypsum (2.35) can be recognised by density almost without doubt.
Overburden pressure: the density of the rocks, averaged from the surface down to the required depth, gives the weight of the layers above (Pore pressure and overburden pressure).
Seismic data: we need the product of density and elastic-wave velocity, the acoustic impedance, for synthetic seismograms (seismic traces computed from logs) and for tying reflections to the well (Seismic waves, acoustic impedance and reflections).
Mechanical properties of rocks: together with the compressional and shear wave velocities, density gives the elastic moduli (Rock mechanics for hydraulic fracturing).
Coals: a very low density (1.2–1.5 g/cm³) picks out coal seams.

Terms and symbols

Bulk density $\rho_b$ (RHOB): The mass of a unit volume of rock together with the fluid in its pores. It is the main result of the method.
Electron density index $\rho_e$: The number of electrons per unit volume, expressed in g/cm³: $\rho_b\cdot 2\sum Z/M$ .
Apparent density $\rho_a$: The reading of a tool calibrated on limestone with fresh water. For ordinary reservoirs it is practically equal to $\rho_b$ .
Correction $\Delta\rho$ (DRHO): The correction for mudcake and wall irregularities, found from the two detectors. It is the main indicator of the quality of the density.
Photoelectric factor $P_e$ (PEF): The photoelectric absorption cross section per electron, in barn/electron. It is an indicator of mineral composition.
Volumetric cross section $U$: The product $P_e\rho_e$ , in barn/cm³. Unlike $P_e$ , it mixes in proportion to the volumes.

Common mistakes and misconceptions

Wrong

“The fluid density in the porosity calculation is the density of formation water or oil.”

In fact

The tool sees the flushed zone, where the pores are filled with filtrate that contains residual hydrocarbons. Use the density of the filtrate, taking its salinity into account. When gas is present, also apply a correction for residual gas.

Wrong

“The $P_e$ of a rock is the average of the $P_e$ of the matrix and the fluid, weighted by volume.”

In fact

What we weight is the volumetric cross section $U = P_e\rho_e$ . We then get the $P_e$ of the rock as $U/\rho_e$ . The $U$ of the fluid is negligibly small, so the $P_e$ of a porous rock is close to the $P_e$ of its matrix.

Wrong

“The tool has applied the correction $\Delta\rho$ , so the density is correct.”

In fact

The correction can be trusted only within the range of compensation. A large correction together with a washout on the caliper log means that the density in the interval is unreliable.

Wrong

“Low density in shales is a sign of high porosity.”

In fact

More often it is a washout. Shales wash out, the pad moves away from the wall, and the tool sees mud. Check the caliper log and $\Delta\rho$ first.

Wrong

“Rock salt shows 2.03 g/cm³: the tool is lying.”

In fact

The tool measures electron density and is calibrated on limestone with water. Salt has fewer electrons per unit mass, so its apparent density is 2.03 against a true 2.165. This is a known feature of the method, and interpretation allows for it.

In depth: advanced detail

How two detectors cancel the mudcake: a simple model

Let us simplify the tool. Suppose that each detector would read the formation density $\rho_b$ if there were no mudcake. The mudcake has a density $\rho_{mc}$ . It adds its own share of the signal $J$ to the reading of each detector. $J$ is the geometric factor of the mudcake for that detector. For thin mudcake, $J$ grows in proportion to the thickness, and it grows faster for the near detector. Then the readings of the detectors are $\rho_{near} = \rho_b + J_n(\rho_{mc} - \rho_b)$ and $\rho_{far} = \rho_b + J_f(\rho_{mc} - \rho_b)$ . We remove the unknown factor $(\rho_{mc} - \rho_b)$ and get:

Corrected density from two detectors (simplified model)

\rho_b = \rho_{far} + \frac{\rho_{far} - \rho_{near}}{k - 1}, \qquad k = \frac{J_n}{J_f}

where:

$\rho_{near},\ \rho_{far}$ densities from the near and far detectors, each on its own calibration without mudcake, g/cm³
$J_n,\ J_f$ shares of the signal that the mudcake gives to the near and the far detector, —
$k$ how many times more sensitive to mudcake the near detector is than the far one, —

Units: g/cm³

Holds when

Thin mudcake: the shares of the signal are small, and their ratio $k$ hardly depends on the thickness and density of the mudcake.
The readings mix linearly. In real tools the relationship is found by experiment, by measurements in formation models with mudcakes of different thickness and density (these are the “ribs”).

The correction $\Delta\rho = (\rho_{far} - \rho_{near})/(k - 1)$ is proportional to the difference between the two detectors. It does not need the thickness or the density of the mudcake. Its sign shows whether the mudcake is lighter or heavier than the formation.

Worked exampleCorrection for light mudcake and for barite mudcake

Given (assumed data)

Ratio of sensitivities $k$: 3 (assumed value)
Light mudcake: $\rho_{far} = 2.38$ , $\rho_{near} = 2.26$ g/cm³
Mudcake with barite: $\rho_{far} = 2.47$ , $\rho_{near} = 2.59$ g/cm³

Solution

Light mudcake: $\Delta\rho = (2.38 - 2.26)/2 = 0.06$ g/cm³, $\rho_b = 2.38 + 0.06 = 2.44$ g/cm³.
Mudcake with barite: $\Delta\rho = (2.47 - 2.59)/2 = -0.06$ g/cm³, $\rho_b = 2.41$ g/cm³.

Answer: The corrected density is 2.44 g/cm³ with light mudcake (correction 0.06) and 2.41 g/cm³ with barite mudcake (correction −0.06).

What it means. The near detector is “pulled” towards the density of the mudcake more strongly than the far one. From the size and sign of the difference, the tool restores the formation density. While the mudcake is thin, the model works. With thick mudcake or a standoff, both shares grow in a nonlinear way, the ratio $k$ changes, and the correction can no longer be trusted. This is the reason for the rules of thumb for $\Delta\rho$ in the main part of the lesson. The data in this example are assumed for illustration.

Apparent density of gas

For gas the difference between the apparent and the true density is large. For methane, $2\sum Z/M = 20/16.043 \approx 1.2467$ . We put this into the formulas for the electron density index and for the calibration. This gives the apparent density of methane in the pores:

Apparent density of methane for the density tool

\rho_{a,g} \approx 1.334\,\rho_g - 0.188

where:

$\rho_{a,g}$ apparent density of gas: the quantity we put into the porosity formula in place of the fluid density, g/cm³
$\rho_g$ true density of gas at reservoir pressure and temperature, g/cm³

Units: g/cm³

Holds when

The gas is methane ( $1.334 \approx 1.0704\cdot 1.2467$ ). For a heavier gas the coefficients are slightly different.
The calibration formula is linear, and the fractions of the components add up to one. So the apparent densities of the components mix by volume in the same way as the true ones. The apparent density of gas is a calculated quantity for such mixing, not a physical density. It can even turn out negative.

The tool sees methane with a density of 0.20 g/cm³ as a substance with a density of about 0.08 g/cm³. At 0.15 g/cm³ and below, the apparent density of the gas is close to zero or even formally negative. So even a small residual gas saturation of the flushed zone (the share of the pore space still filled with gas) noticeably lowers the tool reading.

Density while drilling

In logging-while-drilling (LWD) tools, the source and detectors are built into a drill collar (a heavy section of the drill string) or into a stabiliser blade (a fin that keeps the string centred). They rotate with the string. There is no pad, and the standoff between the tool and the wall changes with every turn. So we divide the measurements into sectors around the circumference of the hole and apply a separate standoff correction to each sector. In a deviated or horizontal well, the best contact is usually in the lower sector, where the tool lies on the wall.

The sector measurements add up to an azimuthal image of density and $P_e$ around the hole (a picture that shows how the values change around the hole). It shows where the hole crosses the top or the bottom of a bed, from above or from below. This helps to keep a horizontal well inside the bed. The measurement is made soon after the formation is opened, while the invasion of filtrate is shallow. So gas in the formation shows up more strongly on the LWD density than on a wireline record made a few days later.

Connections to other topics

Overview of logging tools — the place of the density log among other methods, depth of investigation and resolution.
Porosity — porosity and its calculation from density; the influence of the chosen lithology.
Neutron log — the paired method: together with density it identifies gas, shale and lithology.
Borehole environment: invaded zone and mudcake — the flushed zone, mudcake and washouts: what a pad tool sees.
Porosity from the density log — choice of the matrix and fluid densities and corrections to density porosity.
Lithology from the photoelectric factor and spectral gamma ray — mineral composition from the photoelectric factor and spectrometry.
Pore pressure and overburden pressure — overburden pressure from the density of rocks.

Check yourself

1
Why does the count of the far detector fall as the density of the rock increases?
1. A.A dense rock itself emits fewer gamma rays
2. B.A dense rock reflects photons back to the source
3. C.The detector works worse at high density
4. D.A dense rock contains more electrons: photons scatter more often, lose energy and are absorbed before they reach the detector
Show the answer
Correct answer: D. A dense rock contains more electrons: photons scatter more often, lose energy and are absorbed before they reach the detector
The number of Compton scattering events per unit path is proportional to the electron density. In a dense rock, photons lose energy faster and are absorbed in the photoelectric effect. So fewer of them reach a detector tens of centimetres away. Photons from the tool's own source make up the count. The natural radiation of the rock is only a weak background, and the processing allows for it.
2
The drilling mud is weighted with barite. Which curve of the litho-density tool deserves the least trust?
1. A.Bulk density $\rho_b$
2. B.All curves are equally reliable
3. C.Photoelectric factor $P_e$
4. D.Caliper log
Show the answer
Correct answer: C. Photoelectric factor $P_e$
Barite has $P_e \approx 267$ barn/electron and a volumetric cross section of about 1070 barn/cm³. Even a thin barite mudcake, or mud in the irregularities of the wall, makes $P_e$ much too high, so we cannot find the lithology from it. The density, meanwhile, is corrected by the (negative) correction $\Delta\rho$ , although we should check that too.
3
Dolomite: $\rho_b = 2.50$ g/cm³, filtrate density 1.05 g/cm³. What is the density porosity?
Show the answer
The numerator is $2.87 - 2.50 = 0.37$ g/cm³, the denominator $2.87 - 1.05 = 1.82$ g/cm³, so $\phi_D = 0.37/1.82 \approx 0.203$ , that is, about 20%. With fresh water (1.00 g/cm³) it would be 0.198, a difference of half a percentage point. With a limestone matrix instead of dolomite it would be only 0.13, one and a half times less.
4
In an interval of shales the density is 2.15 g/cm³, the correction is $\Delta\rho = +0.18$ g/cm³, and the caliper log is 40 mm larger than the bit. What should we do with this density?
1. A.Regard it as unreliable: the pad was not in contact with the wall; use other methods or exclude the interval
2. B.Calculate porosity from it as usual
3. C.Add the correction once more
4. D.Regard the shales as gas-bearing
Show the answer
Correct answer: A. Regard it as unreliable: the pad was not in contact with the wall; use other methods or exclude the interval
A washout in shales together with a large correction is a classic sign of lost contact. There is mud between the pad and the rock, and the tool gives a density that is too low. We do not use such values in quantitative calculations.
5
Why does rock salt show 2.03 g/cm³ on the log when its density is 2.165?
Show the answer
The tool measures electron density and is calibrated on limestone with fresh water. For sodium and chlorine the ratio $2Z/M$ is about 0.96, that is, there are fewer electrons per unit mass than in calcite, quartz or water. The electron density index of salt is about 2.07 g/cm³, and after the calibration conversion $1.0704\,\rho_e - 0.1883$ it comes to 2.03.
6
What will be the $P_e$ of a dolomite with a porosity of 20% whose pores are filled with fresh water? (Dolomite: $P_e = 3.14$ , $\rho_e = 2.86$ ; water: $U = 0.40$ , $\rho_e = 1.11$ .)
1. A.2.5
2. B.2.9
3. C.1.8
4. D.3.1
Show the answer
Correct answer: B. 2.9
The volumetric cross section of the matrix is $3.14\cdot 2.86 \approx 8.98$ barn/cm³. For the rock: $U \approx 0.2\cdot 0.40 + 0.8\cdot 8.98$ , that is, about 7.26 barn/cm³; $\rho_e \approx 0.22 + 2.29 = 2.51$ g/cm³; $P_e = U/\rho_e \approx 2.9$ . The factor has fallen slightly compared with the matrix, but it is still far from both calcite (5.1) and quartz (1.8).

Updated: 19 October 2026