For a field $\theta \equiv \theta(t,x,y)$, is $\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\partial\theta}{\partial t}$ for independent $x$,$y$,$t$? Consider a field $\theta$ defined over its independent variables $x$, $y$ and $t$.
$$\theta \equiv \theta(t,x,y) \tag{1}$$
Taking the derivative with respect to $t$,
$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\partial\theta}{\partial t}  +  \frac{\partial\theta}{\partial x} \frac{\mathrm{d}x}{\mathrm{d}t}  +   \frac{\partial\theta}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} \tag{2}$$
My colleague said that $\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\partial\theta}{\partial t}$ because $x$ and $y$ are independent of $t$.
I wasn't too sure about that and gave the example that if we have a temperature field $\theta(t,x,y)$ defined over a two dimensional plane, equation (2) holds good but it doesn't mean anything till we define a trajectory through the plane parametrized by $x \equiv x(t)$ and $y \equiv y(t)$. Then $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $\frac{\mathrm{d}y}{\mathrm{d}t}$ take finite values and you can calculate an analytical expression for $\frac{\mathrm{d}\theta}{\mathrm{d}t}$. My colleague said my intuition was coming from particle mechanics (which he is right about) but for a field $\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\partial\theta}{\partial t}$. However, I still think that to say $\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\partial\theta}{\partial t}$ is like defining a trajectory where you are standing at a fixed point ($x$,$y$) and evaluating $\frac{\mathrm{d}\theta}{\mathrm{d}t}$ there. In this case, I would agree with him. However, if $x$ and $y$ are independent of $t$, $\frac{\mathrm{d}x}{\mathrm{d}t}$ should be undefined and can't be zero because for a small change in $t$, the corresponding change in $x$ could be anything since $x$ and $t$ are independent of each other.
Can someone please shed some light on my issue regarding how do I make physical sense of $\frac{\mathrm{d}\theta}{\mathrm{d}t}$ when $x$ and $y$ are independent of $t$.
 A: The issue is with notation and abuse of language. First of all $\theta$ is a function of 3 variables, so writing $\frac{d\theta}{dt}$ is just wrong. Yes, people write it all the time, but it is wrong. The two $\theta$'s mean completely different things on the two sides of the equation.
The chain rule tells us the following. Suppose we have a (differentiable) parametrized curve $\gamma:\Bbb{R}\to\Bbb{R}^2$. Then, we can construct a NEW function $\Theta:\Bbb{R}\to\Bbb{R}$ as $\Theta(t)=\theta(t,\gamma(t))=\theta(t,\gamma_1(t),\gamma_2(t))$. It may probably be a good idea to write $\Theta_{\gamma}$ to emphasize that this function is constructed from $\theta$ and the curve $\gamma$... but I won't simply for ease of typing. Then,
\begin{align}
\Theta'(t)&=(\partial_1\theta)_{(t,\gamma_1(t),\gamma_2(t))}\cdot 1 +
(\partial_2\theta)_{(t,\gamma_1(t),\gamma_2(t))}\cdot \gamma_1'(t)+
(\partial_3\theta)_{(t,\gamma_1(t),\gamma_2(t))}\cdot \gamma_2'(t)\tag{i}
\end{align}
Or, written in slightly more classical notation,
\begin{align}
\frac{d\Theta}{dt}\bigg|_t&=\frac{\partial \theta}{\partial t}\bigg|_{(t,\gamma_1(t),\gamma_2(t))}+
\frac{\partial \theta}{\partial x}\bigg|_{(t,\gamma_1(t),\gamma_2(t))}\cdot
\frac{d\gamma_1}{dt}\bigg|_t+
\frac{\partial \theta}{\partial y}\bigg|_{(t,\gamma_1(t),\gamma_2(t))}\cdot
\frac{d\gamma_2}{dt}\bigg|_t\tag{ii}
\end{align}
Or if you go one step further and omit the points of evaluation (which are actually important, since this equation expresses what happens only along the curve $\gamma$), we can write
\begin{align}
\frac{d\Theta}{dt}&=\frac{\partial \theta}{\partial t}+
\frac{\partial \theta}{\partial x}\frac{d\gamma_1}{d t}+
\frac{\partial \theta}{\partial y}\frac{d\gamma_2}{d t}\tag{iii}
\end{align}
The sloppiest of all is
\begin{align}
\frac{d\theta}{dt}&=\frac{\partial \theta}{\partial t}+
\frac{\partial \theta}{\partial x}\frac{dx}{d t}+
\frac{\partial \theta}{\partial y}\frac{dy}{d t},\tag{iv}
\end{align}
because the $\theta$ on the LHS really should be $\Theta$ which is a completely different function, and the symbol $x$ has two meanings. One is an arbitrary placeholder for the second entry of the function $\theta$, and also as the first component of the parametrized curve. Similarly, $y$ has two meanings. ALso, the points of evaluation are not indicated at all.
If you're still not convinced of the dangers of such condensed notation/ are unfamiliar with it, perhaps Partial Derivatives of Functions of Functions might help.

Getting directly to the case you're asking about, let us fix a point $(a,b)\in\Bbb{R}^2$, and suppose we consider the constant curve $\gamma:\Bbb{R}\to\Bbb{R}^2$, defined as $\gamma(t)=(a,b)$. Now, if you plug this into equations (i) or (ii), you'll see that the function $\Theta$ is defined as $\Theta(t):=\theta(t,\gamma(t))=\theta(t,a,b)$, and that its derivative is given by
\begin{align}
\frac{d\Theta}{dt}\bigg|_t&=\frac{\partial \theta}{\partial t}\bigg|_{(t,a,b)}+0+0.
\end{align}
Well, this should be obvious even from the very definition of a partial derivative. The meaning of the partial derivative symbol $\frac{\partial \theta}{\partial t}\bigg|_{(t,a,b)}$ or $(\partial_1\theta)_{(t,a,b)}$ means we fix all the other entries, so that we're left with a single-variable function  $\Theta(\cdot)=\theta(\cdot, a,b)$, and then calculate the derivative of this function at $t$. This is literally definition. There's no need to talk about fields/particle motion or anything else.
Note however, that you should not fall into the trap of saying $\frac{d\Theta}{dt}=\frac{\partial \theta}{\partial t}$. On the LHS, $\Theta:\Bbb{R}\to\Bbb{R}$ is a function of one variable, so its derivative $\Theta'=\frac{d\Theta}{dt}:\Bbb{R}\to\Bbb{R}$ is also a function of one variable. On the RHS, $\theta:\Bbb{R}^3\to\Bbb{R}$ is a function of 3 variables, hence its partial derivative $\partial_1\theta=\frac{\partial \theta}{\partial t}:\Bbb{R}^3\to\Bbb{R}$
is also a function of 3 variables. A function of one-variable can never be equal to a function of three variables.
So, you adivsor's claim that $\frac{d\theta}{dt}=\frac{\partial \theta}{\partial t}$ (or even with my slightly more precise notation, the "equation" $\frac{d\Theta}{dt}=\frac{\partial \theta}{\partial t}$) are strictly speaking wrong. They are only equal if you know what you're talking about.
A: You should forget about the case when $x$ and $y$ depend on $t$ (which is a more complicated case) and just come back to the definition of partial derivatives: $\frac{\partial \theta}{\partial t}$ means exactly that you fix $x$ and $y$ and take the derivative of $\theta$ with respect to $t$.
Your confusion comes from the fact that in physics, notations are ambiguous in that you don't clearly distinguish between the case where $x$ and $y$ are variables, and when they are functions (of t in this case). In the setting you are asking about, $x$ and $y$ are variables, and the expressions $\frac{\partial x}{\partial t}$ and $\frac{\partial y}{\partial t}$ are not meaningful.
A: Other answers are correct. The symbol $d\theta/dt$ only has definite sense whenever we have chosen a class of curves passing through each point of the plane $x, y$. Then, that derivative is understood as the derivative of $\theta$ along the curves (and is calculated by the chain rule).
However, in field theory, the class of curves is usually implicitly inderstood to be the motion of particles subject to the field $\theta$. This means that, related to the field $\theta$, there's a system of differential equations of motion.
The formula $d\theta/dt= \partial\theta/\partial t$ can then be deduced from those differential equations.
For example, if $\theta$ was the energy function and $x, y$ were canonical coordinates of phase space, then the equations would be:
$$dx/dt= \partial \theta/\partial y$$
$$dy/dt= -\partial \theta/\partial x$$
And then the chain rule would imply that $d\theta/dt= \partial \theta/\partial t$.
But, since your advisor says that $x$ and $y$ are constant with $t$, then it must be what other answer says, and those curves are constant.
A: In my "early" time I could solve the dilemma (which in Thermodynamics become even more intricated)
by keeping firm the pivot of the total differential
$$
d\theta  = \frac{{\partial \theta }}{{\partial x}}dx + \frac{{\partial \theta }}{{\partial y}}dy +
 \frac{{\partial \theta }}{{\partial z}}dz +  \cdots \quad \left| {\,x,y,z\;{\rm independent}} \right.
$$
Then it follows that:

*

*if $\theta$ does not depend on $z$, or if $z$ is a constant (held constant) and not a variable
$$
d\theta \left( {x,y} \right) = d\theta \left( {x,y,z = c} \right) = \left. {d\theta \left( {x,y,z} \right)} \right|_{\,z = c}
  = \frac{{\partial \theta }}{{\partial x}}dx + \frac{{\partial \theta }}{{\partial y}}dy + 0dz
$$


*if $z$ actually depends on $x,y$
$$
\begin{array}{l}
 d\theta \left( {x,y,z(x,y)} \right) = \frac{{\partial \theta }}{{\partial x}}dx + \frac{{\partial \theta }}{{\partial y}}dy +
 \frac{{\partial \theta }}{{\partial z}}\left( {dz} \right) =  \\ 
  = \frac{{\partial \theta }}{{\partial x}}dx + \frac{{\partial \theta }}{{\partial y}}dy + \frac{{\partial \theta }}
{{\partial z}}\left( {\frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy} \right) =  \\ 
  = \left( {\frac{{\partial \theta }}{{\partial x}} + \frac{{\partial \theta }}
{{\partial z}}\frac{{\partial z}}{{\partial x}}} \right)dx + \left( {\frac{{\partial \theta }}{{\partial y}} +
 \frac{{\partial \theta }}{{\partial z}}\frac{{\partial z}}{{\partial y}}} \right)dy \\ 
 \end{array}
$$


*if $x,y, \ldots$ actually depend on a further variable $t$
$$
\begin{array}{l}
 d\theta \left( {x(t),y(t), \cdots } \right)
 = \frac{{\partial \theta }}{{\partial x}}\frac{{dx}}{{dt}}dt + \frac{{\partial \theta }}
{{\partial y}}\frac{{dy}}{{dt}}dt +  \cdots  =  \\ 
  = \left( {\frac{{\partial \theta }}{{\partial x}}\frac{{dx}}{{dt}}
 + \frac{{\partial \theta }}{{\partial y}}\frac{{dy}}{{dt}} +  \cdots } \right)dt
 = \frac{{d\theta }}{{dt}}dt \\ 
 \end{array}
$$


*if $\cdots$
