Derivative for a Galilean Tranformation

[doggydan42]

Homework Statement

Using the chain rule, find a, b, c, and d:
$$\frac{\partial}{\partial x'} = a\frac{\partial}{\partial x} + b\frac{\partial}{\partial t}$$
$$\frac{\partial}{\partial t'} = c\frac{\partial}{\partial x} + d\frac{\partial}{\partial t}$$

Homework Equations

Chain rule:
$$\frac{\partial f(x,t)}{\partial x'} = \frac{\partial x}{\partial x'}\frac{\partial f}{\partial x} + \frac{\partial t}{\partial x'}\frac{\partial f}{\partial t}$$
The same form can be used for t'.

Gaililean tranformation
$$x' = x - vt$$
$$t' = t$$

The Attempt at a Solution

For x',
$$a = \frac{\partial x}{\partial x'} = \frac{\partial}{\partial x'} (x' + vt) = 1$$
$$b = \frac{\partial t}{\partial x'} = \frac{\partial}{\partial x'} (\frac{1}{v}(x-x')) = -\frac{1}{v}$$

Similarly, for t',
$$c = \frac{\partial x}{\partial t'} = \frac{\partial}{\partial t'} (x' + vt) = \frac{\partial}{\partial t'} (x' + vt')= v$$
$$d = \frac{\partial t}{\partial t'} = \frac{\partial t'}{\partial t'} = 1$$

 

[eys_physics]

What is your question?

 

[doggydan42]

What is your question?

I do not know what is wrong. When I submitted my solution, it was wrong.

I am also confused. Since if t' = t, then shouldn't the partials be equivalent? That would mean c=0 and d=1.

 

[eys_physics]

You have that $t=t'$, so $t$ doesn't have any dependence on $x'$. Therefore,
$$b=0$$
In your derivation of $b$ you are missing that $x$ depends on $x'$.

 

[doggydan42]

You have that $t=t'$, so $t$ doesn't have any dependence on $x'$. Therefore,
$$b=0$$
In your derivation of $b$ you are missing that $x$ depends on $x'$.

So overall,
a = 1,
b= 0,
c = v,
d = 1?

 

[eys_physics]

Yes, it should be correct.