Does a large mass change its entropy when its shape alters as it falls in a gravitational field?

Question

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1. What the question is really asking

A cloud of many molecules (a “big mass”) is dropped from far away and falls toward the Earth.
Because the Earth’s gravitational field is not perfectly uniform, different parts of the body feel slightly different accelerations. This tidal effect stretches or squeezes the body as it falls, i.e. its shape changes.

The problem asks:

When the shape of the falling body changes, does its thermodynamic entropy go up, go down, or stay the same?

In other words we must decide whether the internal state of the body becomes more disordered (higher entropy), more ordered (lower entropy), or unchanged as a result of the tidal deformation.

2. Detailed reasoning – step by step

2.1 Free‑fall in General Relativity

Local inertial frame – In GR an object that is in free‑fall experiences no net force in a sufficiently small neighbourhood. In that tiny region the laws of physics are exactly those of special relativity; there is no “gravitational work” acting on the centre of mass.
Tidal forces – The Earth’s field is not perfectly uniform. Over a finite size (L) the differential acceleration is

[ \Delta a \sim \frac{2GM}{R^{3}}\,L , ]

where (M) and (R) are the Earth’s mass and the distance of the centre of mass from the Earth’s centre. This differential acceleration produces a tidal potential

[ \Phi_{\text{tidal}}(\mathbf{x}) \;=\; -\frac{GM}{2R^{3}}\; (x^{2}+y^{2}-2z^{2}) , ]

which tries to stretch the body along the radial direction and compress it in the transverse directions.

Thus the body is not a rigid point mass; it is a continuous medium that feels a small but finite stress field.

2.2 Work done by tidal stresses

The tidal field does mechanical work on the body when it deforms. The infinitesimal work element is

[ \delta W = \sigma_{ij}\,\delta \varepsilon_{ij}\, dV , ]

where

(\sigma_{ij}) = stress tensor generated by the tidal field,
(\varepsilon_{ij}) = strain tensor describing the deformation,
(dV) = volume element.

Integrating over the whole body gives the total mechanical work

[ W = \int_{V} \sigma_{ij}\,\Delta\varepsilon_{ij}\, dV . ]

If the material is perfectly elastic and the deformation is carried out reversibly (no internal friction, no plastic flow), the work is stored as elastic potential energy and can be recovered when the body returns to its original shape. In that ideal case the process is adiabatic and the thermodynamic entropy of the body does not change:

[ \boxed{ \Delta S_{\text{elastic}} = 0 } . ]

2.3 Real materials – irreversibility

Real macroscopic bodies are not perfectly elastic:

Phenomenon	What it does to the energy budget
Visco‑elastic damping (internal friction)	Converts part of the mechanical work into heat.
Plastic deformation / micro‑cracking	Permanent rearrangement of molecules, also generates heat.
Thermal conduction to surroundings	May carry heat away, but the total entropy of the isolated system (body + environment) still increases.

When any of the above irreversible channels are present, a fraction (\eta) (0 < (\eta) ≤ 1) of the tidal work is turned into internal thermal energy (Q). The first law for the body alone reads

[ \Delta U = Q - W_{\text{out}} , ]

but because the body is not doing useful external work ((W_{\text{out}}=0) for a freely falling isolated piece), we have

[ \Delta U = Q = \eta\,W . ]

The resulting temperature rise (\Delta T) (for a body of heat capacity (C)) is

[ \Delta T = \frac{Q}{C} = \frac{\eta\,W}{C}. ]

For a reversible (quasi‑static) heating process the entropy change is

[ \Delta S = \int \frac{\delta Q_{\text{rev}}}{T} \approx \frac{Q}{\bar T}, ]

where (\bar T) is a characteristic temperature during the heating. Hence

[ \boxed{ \Delta S_{\text{real}} \;=\; \frac{\eta\,W}{\bar T} \;>\;0 } . ]

Because (\eta>0) for any realistic material, the entropy of the falling mass increases.

2.4 Order‑of‑magnitude estimate

Take a concrete example: a solid sphere of radius (L=0.1\;\text{m}) (mass (m\sim 10\;\text{kg})) falling from a height where the Earth’s radius is (R\approx 6.4\times10^{6}\,\text{m}).

The tidal acceleration across the sphere is

[ \Delta a \sim \frac{2GM}{R^{3}}L \approx \frac{2\times 6.67\times10^{-11}\times 5.97\times10^{24}}{(6.4\times10^{6})^{3}}\times 0.1 \approx 3\times10^{-6}\;\text{m/s}^{2}. ]

The associated stress (roughly (\rho L \Delta a) with (\rho\approx 8000\;\text{kg/m}^{3})) is

[ \sigma \sim \rho L\,\Delta a \approx 8000 \times 0.1 \times 3\times10^{-6} \approx 2.4\times10^{-3}\;\text{Pa}, ]

tiny compared with typical material strengths (MPa).
The work done over a fall of (\Delta h\sim 10^{3}\,\text{m}) is roughly

[ W \sim \sigma \times (\text{strain}) \times V \sim \sigma \times \frac{\Delta a}{g}\times V \sim 2.4\times10^{-3}\times \frac{3\times10^{-6}}{9.8}\times 4/3\pi (0.1)^{3} \approx 10^{-12}\,\text{J}. ]

Even if all of this ((\eta =1)) became heat, the temperature rise of a 10 kg steel piece ((C\approx 500\;\text{J K}^{-1})) would be

[ \Delta T \approx \frac{10^{-12}}{500}\; \text{K} \sim 2\times10^{-15}\,\text{K}. ]

The corresponding entropy increase is minuscule ((\Delta S \sim 10^{-18}\,\text{J/K})), but non‑zero in principle. The calculation shows why the effect is usually ignored, yet it does not contradict the thermodynamic conclusion.

2.5 Entropy of the whole system

If we consider the combined system “body + Earth + gravitational field”, the total entropy always increases (or stays the same) because the second law applies to isolated systems. The tiny entropy increase of the body is compensated (or exceeded) by the (essentially zero) change in the Earth’s entropy, so the net change is still non‑negative.

3. Final answer

In an ideal, perfectly elastic, loss‑free material the shape change caused by tidal forces is reversible; the entropy of the falling mass remains unchanged ((\Delta S = 0)).
In any realistic material the tidal deformation produces irreversible heating (through internal friction, plastic flow, etc.). This converts a small amount of the tidal work into thermal energy, raising the body’s temperature and increasing its thermodynamic entropy ((\Delta S > 0)).

Hence the entropy of a large mass that deforms while free‑falling will either stay the same (if the deformation is perfectly reversible) or increase (if there is any dissipation). Practically, because all real substances have some dissipation, the entropy increases, although the amount is astronomically tiny for ordinary sizes and heights.

4. Common mistakes

Mistake	Why it is wrong	How to avoid it
Assuming gravity does work on the whole body like a point mass.	A freely‑falling centre of mass feels no net force; only tidal gradients can do work.	Remember that in GR free‑fall is inertial; only spatial variations of the field produce stresses.
Equating “change of shape” with “increase of disorder”.	Shape change can be completely reversible (elastic stretching) and does not automatically raise entropy.	Distinguish reversible (elastic) from irreversible (viscous, plastic) deformations.
Neglecting the need for a heat flow term in the entropy change.	Entropy change (dS = \delta Q_{\text{rev}}/T) requires a heat exchange; mechanical work alone does not change entropy unless it is dissipated.	Explicitly check whether the mechanical work is converted into heat (i.e. (\eta>0)).
*Thinking the second law forces the entropy to always* increase locally.**	The second law applies to isolated systems; a subsystem (the falling body) can keep its entropy constant if the process is reversible.	Clarify the system boundaries and whether the process is reversible or irreversible.
Using Newtonian “gravity does work = mgh” for an extended body.	That expression assumes a uniform field acting on a point mass; it ignores tidal stresses.	Use the tidal potential (\Phi_{\text{tidal}} \propto (GM/R^{3})\,x^{2}) to compute stresses and work.

By keeping these points in mind, you can correctly assess entropy changes for deforming bodies in gravitational fields.

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