Exercício 1. Mostre que a equação de onda c^{2} \partial_{x}^{2} f(x, t)=\partial_{t}^{2} f(x, t) é invariante sob as transformações de Lorentz, isto é, via x^{\prime}=\gamma(x-V t) e t^{\prime}=\gamma\left(t-V x / c^{2}\right) obtenha c^{2} \partial_{x^{\prime}}^{2} f^{\prime}\left(x^{\prime}, t^{\prime}\right)= \partial_{t^{\prime}}^{2} f^{\prime}\left(x^{\prime}, t^{\prime}\right) \operatorname{com} f^{\prime}\left(x^{\prime}, t^{\prime}\right)=f\left(x\left(x^{\prime}, t^{\prime}\right), t\left(x^{\prime}, t^{\prime}\right)\right). Para isto, use
\begin{aligned} \frac{\partial}{\partial t} & =\frac{\partial t^{\prime}}{\partial t} \frac{\partial}{\partial t^{\prime}}+\frac{\partial x^{\prime}}{\partial t} \frac{\partial}{\partial x^{\prime}}=\gamma\left(\partial_{t^{\prime}}-V \partial_{x^{\prime}}\right) \\\ \frac{\partial}{\partial x} & =\frac{\partial t^{\prime}}{\partial x} \frac{\partial}{\partial t^{\prime}}+\frac{\partial x^{\prime}}{\partial x} \frac{\partial}{\partial x^{\prime}}=\gamma\left(\partial_{x^{\prime}}-\frac{V}{c^{2}} \partial_{t^{\prime}}\right) \end{aligned}

Passo 1

O enunciado pede para mostrar que a equação de onda
c^{2}\,\partial_{x}^{2}f(x,t)=\partial_{t}^{2}f(x,t)
mantém a mesma forma em coordenadas de Lorentz
x'=\gamma(x-Vt),\quad t'=\gamma\!\bigl(t-\tfrac{Vx}{c^{2}}\bigr),
ou seja, que em variáveis primadas vale
c^{2}\,\partial_{x'}^{2}f'(x',t')=\partial_{t'}^{2}f'(x',t'),
onde f'(x',t')=f\bigl(x(x',t'),\,t(x',t')\bigr).
Para isso, vamos usar as regras de cadeia já dadas:
\frac{\partial}{\partial t} =\gamma\bigl(\partial_{t'}-V\,\partial_{x'}\bigr), \quad \frac{\partial}{\partial x} =\gamma\bigl(\partial_{x'}-\tfrac{V}{c^{2}}\partial_{t'}\bigr).

Passo 2

Expressamos \partial_{t}^{2}f aplicando duas vezes \partial/\partial t=\gamma(\partial_{t'}-V\partial_{x'}).
Primeiro,
\partial_{t}f =\gamma\bigl(\partial_{t'}f'-V\,\partial_{x'}f'\bigr).
Agora aplicamos outra vez \partial_{t}:
\partial_{t}^{2}f =\gamma\bigl(\partial_{t'}-V\partial_{x'}\bigr)\bigl[\gamma(\partial_{t'}f'-V\partial_{x'}f')\bigr] =\gamma^{2}\bigl(\partial_{t'}-V\partial_{x'}\bigr)^{2}f'.
Em forma expandida,
\partial_{t}^{2}f =\gamma^{2}\bigl(\partial_{t'}^{2} -2V\,\partial_{t'}\partial_{x'} +V^{2}\,\partial_{x'}^{2}\bigr)f'.

Passo 3

De modo análogo, para \partial_{x}^{2}f usamos \partial/\partial x=\gamma(\partial_{x'}-V/c^{2}\,\partial_{t'}).
Primeiro,
\partial_{x}f =\gamma\bigl(\partial_{x'}f'-\tfrac{V}{c^{2}}\partial_{t'}f'\bigr),
e então
\partial_{x}^{2}f =\gamma^{2}\bigl(\partial_{x'}-\tfrac{V}{c^{2}}\partial_{t'}\bigr)^{2}f' =\gamma^{2}\Bigl(\partial_{x'}^{2} -2\tfrac{V}{c^{2}}\partial_{x'}\partial_{t'} +\tfrac{V^{2}}{c^{4}}\partial_{t'}^{2}\Bigr)f'.

Passo 4

Construímos c^{2}\partial_{x}^{2}f-\partial_{t}^{2}f em função de derivadas primadas:
\begin{aligned} c^{2}\partial_{x}^{2}f-\partial_{t}^{2}f &=\gamma^{2}\Bigl[c^{2}\bigl(\partial_{x'}^{2} -2\tfrac{V}{c^{2}}\partial_{x'}\partial_{t'} +\tfrac{V^{2}}{c^{4}}\partial_{t'}^{2}\bigr)\\\ &\quad\;-\bigl(\partial_{t'}^{2} -2V\,\partial_{t'}\partial_{x'} +V^{2}\,\partial_{x'}^{2}\bigr)\Bigr]f'\\\ &=\gamma^{2}\Bigl[(c^{2}-V^{2})\partial_{x'}^{2} -(1-\tfrac{V^{2}}{c^{2}})\,c^{2}\partial_{t'}^{2}\Bigr]f'\\\ &=\gamma^{2}(c^{2}-V^{2})\Bigl[\partial_{x'}^{2} -\tfrac{1}{c^{2}}\partial_{t'}^{2}\Bigr]f'. \end{aligned}

Passo 5

Usamos agora \gamma=1/\sqrt{1-V^{2}/c^{2}}, logo
\gamma^{2}(c^{2}-V^{2}) =\frac{c^{2}-V^{2}}{1-\frac{V^{2}}{c^{2}}} =c^{2}.
Portanto
c^{2}\partial_{x}^{2}f-\partial_{t}^{2}f =c^{2}\Bigl[\partial_{x'}^{2} -\tfrac{1}{c^{2}}\partial_{t'}^{2}\Bigr]f' =c^{2}\partial_{x'}^{2}f'-\partial_{t'}^{2}f'.

Resposta

Mostramos que
c^{2}\partial_{x'}^{2}f'(x',t')=\partial_{t'}^{2}f'(x',t'),
ou seja, a equação de onda é invariável sob transformações de Lorentz.

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