- #1
TomServo
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- TL;DR Summary
- I'm confused about how the derivative with respect to time transforms under a Galilean transformation.
I'm studying how derivatives and partial derivatives transform under a Galilean transformation.
On this page:
http://www.physics.princeton.edu/~mcdonald/examples/wave_velocity.pdf
Equation (16) relies on ##\frac{\partial t'}{\partial x}=0## but ##\frac{\partial x'}{\partial t}=-v##
But this seems like a contradiction to me. If you swap primed/unprimed you get ##\frac{\partial t}{\partial x'}=0## but ##\frac{\partial x}{\partial t'}=v##, in which case you have ##x=vt+x_0## and ##t=t'=\frac{x-x_0}{v}##. Thus ##\frac{dt'}{dx}=\frac{\partial t'}{\partial x}=\frac{1}{v}##, in violation of Eq. (16).
So where have I gone wrong? Thanks.
On this page:
http://www.physics.princeton.edu/~mcdonald/examples/wave_velocity.pdf
Equation (16) relies on ##\frac{\partial t'}{\partial x}=0## but ##\frac{\partial x'}{\partial t}=-v##
But this seems like a contradiction to me. If you swap primed/unprimed you get ##\frac{\partial t}{\partial x'}=0## but ##\frac{\partial x}{\partial t'}=v##, in which case you have ##x=vt+x_0## and ##t=t'=\frac{x-x_0}{v}##. Thus ##\frac{dt'}{dx}=\frac{\partial t'}{\partial x}=\frac{1}{v}##, in violation of Eq. (16).
So where have I gone wrong? Thanks.