Show Derivative of Function Takes Same Form as

[Saladsamurai]

Homework Statement

Homework Equations

Chain Rule

The Attempt at a Solution

So we have that $f = f(x,t)$ as well as the transformations $x = x' + Vt'$ and $t = t'$

By the chain rule:

$$\frac{\partial{f}}{\partial{t'}} = <br /> \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t'}} +<br /> \frac{\partial{f}}{\partial{t}}\frac{\partial{t}}{\partial{t'}}$$

$$\Rightarrow<br /> \frac{\partial{f}}{\partial{t'}} = <br /> \frac{\partial{f}}{\partial{x}} * <br /> \left [ \frac{\partial{x}}{\partial{t'}}+V+t'\frac{\partial{V}}{\partial{t'}}\right ]<br /> + \frac{\partial{f}}{\partial{t}}\frac{\partial{t'}}{\partial{t'}}$$

$$\Rightarrow<br /> \frac{\partial{f}}{\partial{t'}} = <br /> \frac{\partial{f}}{\partial{x}} * <br /> \left [ \frac{\partial{x}}{\partial{t'}}+V+t'\frac{\partial{V}}{\partial{t'}}\right ]<br /> + \frac{\partial{f}}{\partial{t}}$$

Not really sure what the next move is? Is the above equation in its simplest form? Or can I do something more with it?

Also, I don't really see how I go about transforming $\rho'$ and $v'$ into the x-t coordinates?

 

[CompuChip]

Note that there are only partial derivatives.
So then isn't
$$\frac{\partial x}{\partial t'} = V$$?

 

[hunt_mat]

You made a mistake:
$$\frac{\partial f}{\partial t'}=V\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}$$
as:
$$\frac{\partial x}{\partial t'}=V\quad\frac{\partial t}{\partial t'}=1$$

 

[Char. Limit]

You made a mistake:
$$\frac{\partial f}{\partial t'}=V\frac{\partial f}{x}+\frac{\partial f}{\partial t}$$
as:
$$\frac{\partial x}{\partial t'}=V\quad\frac{\partial t}{\partial t'}=1$$

Shouldn't that be $$\frac{\partial f}{\partial x}$$?

 

[tiny-tim]

Hi Saladsamurai!

(have a rho: ρ and a curly d: ∂ )

Not really sure what the next move is? Is the above equation in its simplest form? Or can I do something more with it?

Now you put in ∂x'/∂t etc, which you can read off the Galilean transformation.

Also, I don't really see how I go about transforming $\rho'$ and $v'$ into the x-t coordinates?

But the question tells you that … ρ' = ρ, v' = v - V.