# Positivity of $dt$?

## Background

I had an idea to write time $$dt$$ in the first law of thermodynamics. So let us proceed with the $$1$$'st law in differential form:

$$dw = T dS - d U$$

where $$w$$ is the work, $$T$$ is temperature, $$S$$ is the entropy and $$U$$ is the internal energy.

Now, let the work be in terms of pressure and volume:

$$P dV = T dS - d U$$

Dividing by pressure :

$$dV = \frac{T}{P} dS - \frac{dU}{P}$$ Taking the hodge dual:

$$dt = \star ( \frac{T}{P} dS - \frac{dU}{P} )$$

The below is an attempt. We continue with:

$$dt = \star ( \frac{T}{P} dS - \frac{dU}{P} )$$

Using,

$$dS = \frac{\partial S}{\partial x} dx + \frac{\partial S}{\partial y} dy + \frac{\partial S}{\partial z} dz + \frac{\partial S}{\partial t} dt$$

And similarly:

$$dU = \frac{\partial U}{\partial x} dx + \frac{\partial U}{\partial y} dy + \frac{\partial U}{\partial z} dz + \frac{\partial U}{\partial t} dt$$

We have:

$$dt = \sum_{i} ( \frac{T}{P} \frac{\partial S}{\partial x} - \frac{1}{P} \frac{\partial U}{\partial x} ) \star dx_i$$

where $$dx_0 = cdt$$, $$dx_1 = dx$$, $$dx_2 = dy$$ and $$dx_3 = dz$$

## Question

Can anyone prove or disprove that if $$dS > 0$$ then $$dt > 0$$? Using:

$$dt = \sum_{i} ( \frac{T}{P} \frac{\partial S}{\partial x} - \frac{1}{P} \frac{\partial U}{\partial x} ) \star dx_i$$

• Apologies for the massive edit but I wasn't sure if I should answer my own question and then ask a new one or just do a massive edit. – More Anonymous Feb 14 at 12:46
• What does $dt>0$ mean from the viewpoint of differential geometry? – NDewolf Feb 14 at 12:46
• Yeah, that's not true. $\text{d} V$ is not the volume form on some manifold that you would integrate over to get the volume of your region/system/whatever. It's a one-form in your space of thermodynamic variables -- just as $\text{d} S$ or $\text{d} U$ are one-forms. – Toffomat Feb 14 at 12:53
• This paper (arxiv.org/abs/0712.0868) might shed some light on the issue. The problem is that the volume $V$ should not be viewed as some spatial structure, but instead as a thermodynamic variable. The manifold on which you are working is accordingly not the spatial (4D) spacetime and as such there is not an immediate relation between $V$ and $t$. – NDewolf Feb 14 at 12:59
• Integrating $dV$ should still give you the volume (change), however it is not related to the volume form on the ordinary spacetime manifold. It is just (the differential of) some coordinate on the "thermodynamic" phase space. – NDewolf Feb 14 at 13:42