0
$\begingroup$

Bulk Modulus is defined as

$$ B = \frac{VdP}{-dV}$$

Where $V$ is volume and $P$ is pressure. It is also defined as,

$$ B = \frac{\rho dP}{d\rho}$$

Where $\rho$ is density. Question: How are these equivalent?

If $V = \frac{m}{\rho}$, $dV = \frac{m}{d\rho}$

Substituting into the first equation of the bulk modulus, I get $$ B = \frac{\frac{m}{\rho}dP}{\frac{m}{d\rho}}$$

or

$$ B = \frac{d\rho dP}{\rho} $$

$\endgroup$
3
$\begingroup$

If $V = \frac{m}{\rho}$, $dV = \frac{m}{d\rho}$

This is the source of your error. You can re-write the above as $\rho V = m$, and this yields $\rho dV + Vd\rho = 0$, or $V \frac d {dV} = -\rho\frac d {d\rho}\,$ as a differential operator. This leads directly to the alternate form for the bulk modulus $B = \rho \frac{dP}{d\rho}$.

Being a bit more formal, from $B=-V\frac{dP}{dV}$, the chain rule dictates that $B=-V\frac{d\rho}{dV}\frac{dP}{d\rho}$. From $\rho V = m$, $V\frac{d\rho}{dV} + \rho = 0$, or $V\frac{d\rho}{dV} = -\rho$, once again yielding $B = \rho \frac{dP}{d\rho}$.


You need to be very careful when you use "physics math." It can get you in trouble.

$\endgroup$
  • $\begingroup$ "physics math". First time I heard that term, but I understand what you mean. Reminds me of "HR math" (which is definitely dubious - you would know it when you saw it). But yes, you nailed it. $\endgroup$ – Floris Aug 28 '14 at 15:18
  • $\begingroup$ I have a follow up question regarding your second line. $\rho$V = m. Then you did some type of product rule to get $\rho dV + Vd\rho = 0$. Usually the product rule is done on functions. So what is volume a function of? Because $dV$ without a denominator is not a derivative. I know $dV$ and $d\rho$ are differentials, so is there some sort of product rule for differentials? Why was mass, m, considered a constant and set to zero? $\endgroup$ – DWade64 Aug 29 '14 at 3:50
  • $\begingroup$ @DWade64 - The natural impulse in measuring bulk modulus would be to think of pressure as the independent variable. After all, that is what the experimenter is varying; the experiment inevitably measures how volume changes as applied pressure changes. What one will get is a nice smooth monotonic curve. That's an invertible function, and that in turn means one can (at least mathematically) treat volume as the independent variable of the experiment. $\endgroup$ – David Hammen Aug 29 '14 at 8:54
3
$\begingroup$

You have to use the differentials properly: If $V = \frac{m}{\rho}$, then

$$ \mathrm{d}V = \frac{\partial V}{\partial \rho}\mathrm{d}\rho = -\frac{m}{\rho^2}\mathrm{d}\rho$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.