In Special Relativity and Classical Field Theory by Susskind, he says that we can imagine a function of $(x+ct)$, then he says that we can consider its derivatives and easily see that $$\frac{\partial F(x+ct)}{\partial t} = c\frac{\partial F(x+ct)}{\partial x}$$ If I think of examples, I can understand why this relation is true, however, I am unsure how to prove it. Could someone provide a hint/some insight to this?
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asked Sep 4, 2022 at 0:53
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If $f(x,t) = f(x + ct)$, then the chain rule gives \begin{align} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial (x+ct)}\frac{\partial (x+ct)}{\partial x} = \frac{\partial f}{\partial (x+ct)}\\ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial (x+ct)}\frac{\partial (x+ct)}{\partial t} = c\frac{\partial f}{\partial (x+ct)} \end{align} so \begin{align} \frac{1}{c}\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \end{align}
