Bulk Modulus, How are these formulas equivalent?

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

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

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.

• "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. – Floris Aug 28 '14 at 15:18
• 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? – DWade64 Aug 29 '14 at 3:50
• @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. – David Hammen Aug 29 '14 at 8:54

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

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