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This is an extension of the question asked here, with the same setup.

Question 1

In 3D compressibility is: $$\beta = -\frac{1}{V} \left (\frac{\partial V}{\partial P} \right )_T$$

In 1D compressibility is: $$\beta = -\frac{1}{L}\frac{\partial L}{\partial F}=\frac{1}{\kappa x_{eq}}$$

My question is the units seem to be different in the 3D case vs the 1D case. In 3D, we have: $$\frac{1}{V} \Longrightarrow \frac{1}{m^3}$$ $$\frac{\Delta V}{\Delta P} = \frac{\Delta V}{\frac{\Delta F}{A}} \Longrightarrow \frac{m^3}{\frac{N}{m^2}}$$ So in total we get: $$\frac{1}{m^3} \times \frac{m^3}{\frac{N}{m^2}} = \frac{m^2}{N}$$

In 1D we have: $$\frac{1}{L} \Longrightarrow \frac{1}{m}$$ $$\frac{\partial L}{\partial F} \Longrightarrow \frac{m}{N}$$ So in total we get: $$\frac{1}{m} \times \frac{m}{N} = \frac{1}{N}$$

Which seems to suggest the units of compressibility are different, which confuses me because shouldn't they have the same units?

Question 2

Why does $P \to F$ from 3D to 1D? This was asked in the link attached but there was no answer given there.

Question 3

Why is $$F = - \frac{d V}{d x}?$$ I know that force is the derivative of potential energy, but here $V$ is potential, not potential energy. And from here, potential energy and potential have different units, so if force is the derivative of potential energy, it seems to me that it cannot be the derivative of potential too.

What am I missing here?

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    $\begingroup$ Please limit your posts to one question. The first two questions are related, so they're fine, but the third should really be posted as its own question. That said, in that formula, $V$ is the potential energy not the electric potential. Sometimes people are a little lazy with their language and use the word "potential" when they mean "potential energy", and that can be confusing, but one gets used to it. In the context of electrostatics, the electric field is the derivative of the electric potential, and if we multiply both sides by a test charge $q$, we get that equation above back. $\endgroup$
    – march
    Commented Nov 4, 2022 at 22:54
  • $\begingroup$ Thanks for the feedback! I will be sure to separate the questions in the future. $\endgroup$ Commented Nov 4, 2022 at 23:13
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    $\begingroup$ @fertilizerspike - solids are measurably compressible. With enough force one can compress them quite nicely. $\endgroup$
    – Jon Custer
    Commented Nov 5, 2022 at 0:56
  • $\begingroup$ Disagree but okay $\endgroup$ Commented Nov 5, 2022 at 3:32
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    $\begingroup$ @fertilizerspike The measurability of isothermal compressibility (or its inverse, the bulk modulus) of solids is a fact, not a matter of opinion (see, for instance values at en.wikipedia.org/wiki/Bulk_modulus ) $\endgroup$ Commented Nov 5, 2022 at 11:17

1 Answer 1

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It's unremarkable for properties of systems of different dimensionality to have different units. For 1D, 2D, and 3D idealized systems exposed to equiaxial, equibiaxial, and equitriaxial stress* (meaning the same normal stress applied in every available dimension), the compressibility has units of $\frac{1}{\mathrm{N}}$, $\frac{\mathrm{m}}{\mathrm{N}}$, and $\frac{\mathrm{m}^2}{\mathrm{N}}$, respectively. No conflict arises because these units are never used simultaneously; a system can't be simultaneously modeled as 2D and 3D, for example.

*$\sigma_{xx}$, $\sigma_{xx}=\sigma_{yy}$, and $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-P$, respectively, where the pressure $P$ is taken as negative (tensile stresses as positive). In 1D, if a certain cross-sectional area $A$ is assumed, then $F=-\sigma_{xx}A$ might be used instead for an axial compressive force.

Does that answer your first two questions? I don't understand your third question, which reuses $V$ (volume?) to probably mean a potential energy, sometimes called a potential.

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