# Misconception in partial derivatives of Lorentz transformation

Let us consider a Lorentz transformation of four vectors from frame S to S' where S' is moving with relative velocity $$\textbf{v}$$ with respect to S. The boost is given by $$t'=\gamma(t-vx), \quad x'=\gamma(x-vt), \quad y'=y, \quad z'=z.$$ The inverse transformation is given by $$t=\gamma(t'+vx'), \quad x=\gamma(x'+vt'), \quad y=y', \quad z=z'.$$ Now here comes the crucial part. Notice that $$\frac{\partial x}{\partial x'} = \gamma, \quad \frac{\partial x'}{\partial x} = \gamma. \tag{1}$$ I have thought about this for a while, but more thoughts always lead me to the same conclusion that this is true.

However, then we have a problem: $$\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} = \gamma \frac{\partial}{\partial x} \tag{2}.$$ This seems fine. Continue: $$\frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} = \gamma \frac{\partial}{\partial x'} \tag{3}.$$ This also seems fine. Continue: $$\frac{\partial}{\partial x'}=\gamma \frac{\partial}{\partial x} = \gamma \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} = \gamma^2 \frac{\partial}{\partial x'}$$

where we have reached a contradiction since $$\gamma^2 \neq 1$$.

Where's have I messed up in (1), (2), (3)?

• how did you get (1)?
– user65081
Nov 30, 2019 at 0:52
• You neglected the time derivatives. Nov 30, 2019 at 0:52
• Can you explain how you got (2)? Nov 30, 2019 at 0:53
• @WillO, no, I mean you obtain (2) from (1)
– user65081
Nov 30, 2019 at 0:57
• (1) is correct. (2) and (3) are incorrect. Nov 30, 2019 at 0:57

You forgot that $$x'$$ is not only function of $$x$$, but also of $$t$$: $$x' = x'(x, t)$$. Similarly, $$x=x(x', t')$$. Hence: $$\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} + \frac{\partial t}{\partial x'} \frac{\partial}{\partial t}$$ and $$\frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} + \frac{\partial t'}{\partial x} \frac{\partial}{\partial t'}$$