Partial Derivatives multivariable

[PFuser1232]

I am quite new to the topic of multivariable calculus. I came across the concept of "gradient" (∇), and although the treatment was somewhat slapdash, I think I got the hang of it. Consider the following case:

$z = f(x,y,t)$

$∇z = \frac{∂z}{∂t} + \frac{∂z}{∂y} + \frac{∂z}{∂x}$

If we're dealing with a physical scenario, wherein the variables x, y and t have different units, how is it possible to compute ∇z?

 

[jtbell]

z=f(x,y,t)z = f(x,y,t)

∇z=∂z∂t+∂z∂y+∂z∂x∇z = \frac{∂z}{∂t} + \frac{∂z}{∂y} + \frac{∂z}{∂x}

No, the gradient is a vector. Each partial derivative is one component of the vector, and is multiplied by the unit vector in that direction. Also, I've never seen anyone use t (time) as part of a gradient (at least not in physics!) so I'll use u = f(x,y,z) instead: $$\vec \nabla u = \frac{\partial u}{\partial x} \hat x + \frac{\partial u}{\partial y} \hat y + \frac{\partial u}{\partial z} \hat z$$ Compare this to a generic vector $$\vec A = A_x \hat x + A_y \hat y + A_z \hat z$$ (Maybe your textbook uses $\hat i$, $\hat j$, $\hat k$ for the unit vectors instead of $\hat x$, $\hat y$, $\hat z$)

If we're dealing with a physical scenario, wherein the variables x, y and t have different units, how is it possible to compute ∇z?

In physics, the gradient is always in terms of position coordinates: x,y,z in rectangular (Cartesian) coordinates, r,θ,φ in spherical coordinates, etc. For non-Cartesian coordinates, the formula for the gradient is different than above, partly to ensure that each component has the same units.

 

[DrClaude]

Usually in physics the gradient implies a derivative of spatial coordinates only. Remember also that the gradient is a vector operator, such that
$$
\nabla f(x,y,t) = \frac{\partial f(x,y,t)}{\partial x} \mathbf{i} + \frac{\partial f(x,y,t)}{\partial y} \mathbf{j}
$$
where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors along the x and y axes respectively.

 

[PFuser1232]

No, the gradient is a vector. Each partial derivative is one component of the vector, and is multiplied by the unit vector in that direction. Also, I've never seen anyone use t (time) as part of a gradient (at least not in physics!) so I'll use u = f(x,y,z) instead: $$\vec \nabla u = \frac{\partial u}{\partial x} \hat x + \frac{\partial u}{\partial y} \hat y + \frac{\partial u}{\partial z} \hat z$$ Compare this to a generic vector $$\vec A = A_x \hat x + A_y \hat y + A_z \hat z$$ (Maybe your textbook uses $\hat i$, $\hat j$, $\hat k$ for the unit vectors instead of $\hat x$, $\hat y$, $\hat z$)
In physics, the gradient is always in terms of position coordinates: x,y,z in rectangular (Cartesian) coordinates, r,θ,φ in spherical coordinates, etc. For non-Cartesian coordinates, the formula for the gradient is different than above, partly to ensure that each component has the same units.

This clears it up, thanks!

 

[jtbell]

I most often deal with the gradient in situations where there's no time dependence, e.g. electrostatics ($\vec E = -\vec \nabla V$), but DrClaude's interpretation of your original example is spot on. Just apply the derivatives to the spatial variables only, and carry the t's along as if they were constants.

 

[DrClaude]

I most often deal with the gradient in situations where there's no time dependence, e.g. electrostatics ($\vec E = -\vec \nabla V$), but DrClaude's interpretation of your original example is spot on. Just apply the derivatives to the spatial variables only, and carry the t's along as if they were constants.

To give some context, one field where this is the case is quantum mechanics. Take for instance the Schrödinger equation for a free particle:
$$
\begin{align}
i \hbar \frac{\partial}{\partial t} \psi(x,y,z,t) &= -\frac{\hbar^2}{2m} \nabla^2 \psi(x,y,z,t) \\
&= -\frac{\hbar^2}{2m} \left[ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right] \psi(x,y,z,t)
\end{align}
$$
where the time derivative is treated separately from the spatial derivatives. (Note that $\nabla^2 = \nabla \cdot \nabla$, so it results in a scalar.)

Mohammed, you can also look up the D'Alambertian,[/PLAIN] which is used in special relativity when all 4 coordinates must be treated on the same footing. You will see that the speed of light is introduced to make it such that all the derivatives have the same units.

 

[PFuser1232]

To give some context, one field where this is the case is quantum mechanics. Take for instance the Schrödinger equation for a free particle:
$$
\begin{align}
i \hbar \frac{\partial}{\partial t} \psi(x,y,z,t) &= -\frac{\hbar^2}{2m} \nabla^2 \psi(x,y,z,t) \\
&= -\frac{\hbar^2}{2m} \left[ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right] \psi(x,y,z,t)
\end{align}
$$
where the time derivative is treated separately from the spatial derivatives. (Note that $\nabla^2 = \nabla \cdot \nabla$, so it results in a scalar.)

Mohammed, you can also look up the D'Alambertian,[/PLAIN] which is used in special relativity when all 4 coordinates must be treated on the same footing. You will see that the speed of light is introduced to make it such that all the derivatives have the same units.

So, in the case of the Schrödinger equation, do we consider t to be constant (considering variations of spatial coordinates only at a particular time t)?

 

[Matterwave]

So, in the case of the Schrödinger equation, do we consider t to be constant (considering variations of spatial coordinates only at a particular time t)?

No, we can't consider time a constant in the Schrödinger equation because otherwise the partial derivative with respect to time, on the left hand side, would just be 0. When we take the gradient, we are taking a partial derivative with respect to spatial coordinates. Recall from elementary multivariable calculus that to take the partial derivative with respect to one variable, we take the derivative as if the other variables are constant. But of course, the other variables are not actually constant, we just treat them as such when we are taking the partial derivative.

 

[quantumtimeleap]

So, in the case of the Schrödinger equation, do we consider t to be constant (considering variations of spatial coordinates only at a particular time t)?

if you're talking about calculating spatial derivatives, yes you keep t constant when differentiating. but in general t cannot be a constant. it is one of the independent variables, the four coordinates, (x,y,z,t)