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Derivative for a Galilean Tranformation

  • #1
166
18

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

Answers and Replies

  • #2
264
70
What is your question?
 
  • #3
166
18
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.
 
Last edited:
  • #4
264
70
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'##.
 
  • #5
166
18
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?
 
  • #6
264
70
Yes, it should be correct.
 

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