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For a perfectly elastic body, Bulk modulus always remains constant and is defined as, $$B=-V_i \frac{\Delta P}{\Delta V} \tag{1}$$ Which means, $$B \left(\frac{V_f -V_i}{V_i}\right)= -(P_f-P_i)$$

But, the definition of bulk modulus in a lot of places is given as, $$B=-V\frac{dP}{dV}\tag{2}$$ In particular, this definition is used to show that for an ideal gas in an isothermal process, we get $B=P$, i.e., the bulk modulus is equal to the pressure for an ideal gas in constant temperature.

Using the definition ($2$) for an elastic body, we get, $$\frac{B}{V}dV = -dP$$ $$\Rightarrow \int_{V_i}^{V_f} \frac{B}{V}dV= -\int_{P_i}^{P_f}dP$$ $$\Rightarrow B\ln\frac{V_f}{V_i}= -(P_f -P_i) $$

Clearly, this doesn't confirm the definition ($1$).

What is a general definition of bulk modulus that we can get ($1$) for an elastic body and ($2$) for an ideal gas? Also, the general definition should not allow us to use definition ($2$) in the case of an elastic body.

If there is no such general definition, why call the quantities in ($1$) and ($2$) with the same name 'Bulk modulus'?

GiorgioP-DoomsdayClockIsAt-90
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Navneet
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2 Answers2

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If $V_f \approx V_i$, then $$\ln\left(\frac{V_f}{V_i}\right) = \ln \left( 1 + \frac{V_f - V_i}{V_i} \right) \approx \frac{V_f - V_i}{V_i}.$$So the two formulas are functionally equivalent so long as the volume does not change significantly during the process.

I suspect that your equation (1) implicitly assumes that $\Delta V = V_f - V_i$ is small compared to $V_i$ (or $V_f$.) The equation (2) is the more general relation.

I should also note that the bulk modulus is not necessarily constant with respect to volume (or pressure), so the integration you have written following (2) is not necessarily correct. For example, for an ideal gas $B = P = NkT/V$, which is very obviously dependent on $V$ or on $P$.

Michael Seifert
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  • For an ideal elastic body, bulk modulus should not change with respect to volume or pressure it should be a constant equal to stress/strain, that is what hooke's law say. Also for an ideal elastic body $\Delta V$ can be as large as possible and equation (1) will hold. – Navneet Mar 01 '24 at 20:21
  • Besides if bulk modulus depends on pressure or volume when Wikipedia mentions bulk modulus of Aluminium is 76 GPa, what is the pressure or volume they are talking about?? – Navneet Mar 01 '24 at 20:25
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    When a single value is reported for the bulk modulus, it's almost certainly at 1 atm and around room temperature, with a specific volume of 1/density, where the density is also measured at those conditions. Real materials deviate from ideality, and their stiffness is pressure and temperature dependent. Eq. (1) is obviously not the general equation; it implies that a pressure increase equal to the bulk modulus would make the material disappear! But it is simpler to calculate than the more general Eq. (2) and is often accurate enough for condensed matter in typical applications. – Chemomechanics Mar 01 '24 at 20:41
  • For an ideal elastic body equation (1) should be the general equation, shouldn't it ? If it is not the general equation then neither are other hooke's laws for different moduli of elasticity in a perfectly elastic body? For example if we compress an ideal spring of length L and spring constant k with force of magnitude Lk from both sides the spring will disappear, it doesn't mean we stop studying about ideal spring or hooke's law for ideal spring. Similarly, when we are talking about ideal elastic body equation (1) should be valid?? – Navneet Mar 01 '24 at 21:10
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    No, Eq. (1) is not the more general equation, and I explain why here. If you load an ideal elastic material from zero pressure ($P_i=0$) to $P_f$ in two successive stages ($0$ to $P_f/2$ and then immediately $P_f/2$ to $P_f$), Eq. (1) predicts a volume change of $-V_iP_f/B$ (the whole load) but also $-V_i(1-P_f/2B)(P_f-P_f/2)/B$ (the two half-loads), a different number. This is nonsensical. Eq. (2) correctly predicts $V_i[\exp(-P_f/B)-1]$ and $V_i[\exp(-P_f/2B)\exp(-P_f/2B)-1]=V_i[\exp(-P_f/B)-1]$, which is consistent. – Chemomechanics Mar 01 '24 at 22:43
  • @Chemomechanics Such contradictions can be avoided if we assume that equation (1) can only be used when initial pressure is some reference pressure, i.e., $P_i=P_0$. But I get the point you want to convey about true stress-strain vs engineering stress-strain. Finally, I wanted to ask if we compare true stress-strain curve and engineering stress-strain curve in proportional limit say for aluminum, where will we find a straight line? That is Hooke's law talks about which stress-strain being proportional in elastic body. – Navneet Mar 02 '24 at 09:04
  • The contradiction exists even if $P_i\neq 0$; it just makes the expressions longer. Try writing them. Hooke's law for real materials is reasonably accurate only for small deformations. I show the deviation between true strain and engineering strain here, – Chemomechanics Mar 02 '24 at 09:12
  • Let us continue this discussion in chat. – Navneet Mar 02 '24 at 13:05
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  1. No doubts,- differential forms are always superior over $\Delta$ variants, simply because differential equation form takes into account instant resulting value. So, $$B=-V_0\frac{dP}{dV}\tag{2}$$ describes instant bulk modulus, which is very handy if for some reason bulk modulus of material is function of time, anisotropic or etc, so to say if $B=B(x_1,x_2,x_3,\dots)$. $\Delta$ variant can only get you to the average bulk modulus, no more, no less. Sure, sometimes average is the only thing you care, then it's ok.

  2. You have made an error in integration. $$\int_{V_0}^{V_f} \frac {B}{V_0} dV = B \frac{V_f - V_0}{V_0},$$

in other words,- it's not logarithm of volume ratios, because $V_0$ is not same volume integration variable, but rather initial volume- wiki mentions it, but alas uses wrong notation for initial volume, which may get confused somebody upon integration. So the result actually matches (1).

Agnius Vasiliauskas
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  • For an ideal gas you have $B=-V\frac{dP}{dV}=P$ if you integrate this you should get $P_i×V_i=P_f×V_f$. The only way we can get this result is by assuming $V$ is same volume integration variable as $dV$. – Navneet Mar 02 '24 at 17:44
  • @Navneet Check wikipedia "offcial" definition of bulk modulus. Citation: ${V}$ is the initial volume of the substance – Agnius Vasiliauskas Mar 02 '24 at 18:40
  • That doesn't work for ideal gas. Imagining $V$ is the initial volume and bulk modulus is pressure $P$ try to prove that $P_i×V_i=P_f×V_f$ by integration. You will realise why V shouldn't be the initial volume. – Navneet Mar 02 '24 at 18:46
  • But on the other hand, technically, if $V$ is not an initial volume, then it follows that $B$ is not a bulk modulus as per standard definition, right? At least if you change definition on the fly, then do not brag that (1) doesn't follows from it. You can't make it follow if you mangle expressions anyway you like and after that expect "usual" behavior of them. – Agnius Vasiliauskas Mar 02 '24 at 19:35
  • The Wikipedia article you are quoting uses the equation (2) to say that bulk modulus is just the pressure of an ideal gas in an isothermal process and on the other hand the very first reference of the Wikipedia article defines bulk modulus as defined in (1). I am not mangling anything, I saw the bulk modulus defined in two ways and asked a question about it. – Navneet Mar 03 '24 at 02:52