I have the following relationship between variables: $$\varepsilon = \Psi -\frac{v^2}{2}\tag{1}$$
A while back, I had an integral with respect to $v$ and I wanted to convert it to an integral with respect to $\varepsilon$. I used the Jacobian method of converting differential elements.
- $$\mathrm{d}v = \det\left[\frac{\partial v}{\partial \varepsilon}\right]\mathrm{d}\varepsilon$$ $$\frac{\partial \varepsilon}{\partial v} = -v \implies \frac{\partial v}{\partial \varepsilon} = -\frac{1}{v} \implies \mathrm{d}v = -\frac{1}{v}\mathrm{d}\varepsilon$$ $$\implies \mathrm{d}\varepsilon = -v\mathrm{d}v\tag{2}$$
I was satisfied with equation (2) and used it to prove an expression in a research paper. Now, I wonder what would have happened if I need to change $\mathrm{d}\varepsilon$ to $\mathrm{d}\Psi$. In such a case,
- $$\mathrm{d}\varepsilon = \det\left[\frac{\partial \varepsilon}{\partial \Psi}\right]\mathrm{d}\Psi$$ $$\frac{\partial \varepsilon}{\partial \Psi} = 1$$ $$\implies \mathrm{d}\varepsilon = \mathrm{d}\Psi\tag{3}$$
Then, the prediction for the relationship between $\mathrm{d}v$ and $\mathrm{d}\Psi$ from eqn (2) and eqn (3) would be: $$-v\mathrm{d}v = \mathrm{d}\varepsilon = \mathrm{d}\Psi$$ $$\text{i.e.,}\ \ \ \ \ \ -v\mathrm{d}v = \mathrm{d}\Psi\tag{4}$$
However, if I directly use the Jacobian method to find out the relation between $\mathrm{d}v$ and $\mathrm{d}\Psi$,
- $$\mathrm{d}\Psi = \det\left[\frac{\partial \Psi}{\partial v}\right]\mathrm{d}v$$ $$\frac{\partial \Psi}{\partial v} = v$$ $$\implies \mathrm{d}\Psi = v\mathrm{d}v\tag{5}$$
Clearly, eqn (4) and eqn (5) don't match. So, there is some inconsistency here in the way I am relating the differentials. Could someone shed some light on what I'm doing wrong, because I was able to prove the expression in the research paper using eqn (2).
Assume that all quantities depend on each other. This amounts to not making the assumption that any partial derivatives are $0$. Since you seem interested in the differential $\mathrm{d}\Psi$, first solve for $\Psi$: $$ \Psi = \varepsilon + \frac{v^2}{2} $$ Now, use the (multivariable) chain rule: $$ \mathrm{d}\Psi = \frac{\partial\Psi}{\partial\varepsilon}\, \mathrm{d}\varepsilon + \frac{\partial\Psi}{\partial v}\, \mathrm{d}v = 1\, \mathrm{d}\varepsilon + v\, \mathrm{d}v $$ Your various formulas arise from assuming certain partial derivatives vanish. For $(2)$, (perhaps without realizing it) you assume that $\mathrm{d}\Psi = 0$, so $\mathrm{d}\varepsilon = -v\, \mathrm{d}v$. For $(3)$, you assume that $\mathrm{d}v = 0$, so $\mathrm{d}\Psi = \mathrm{d}\varepsilon$, etc.
Once you assume that more than one of these differentials vanish, then they all vanish, so your equations amount to the tautology $0 = 0$, which is of course equivalent to $0 = -0$.