求解：爱因斯坦狭义相对论论文中某无穷小量推导步骤

Let's walk through this step slowly—Einstein is using first-order Taylor series approximation here, a standard trick when dealing with infinitesimal quantities like the tiny $x'$ he mentions. Here's the play-by-play breakdown:

1. Label the tiny time increments first #

To simplify the original equation, let's name the time shifts explicitly:

• $\Delta t_1 = \frac{x'}{c - v}$ (the time shift in the right-hand side's argument)
• $\Delta t_2 = \frac{x'}{c + v}$ (the extra time shift in the left-hand side's second term)

Since $x'$ is infinitesimally small, both $\Delta t_1$ and $\Delta t_2$ are also tiny (they scale directly with $x'$). We only need to keep first-order terms in our approximations—higher-order tiny terms (like $(x')^2$) vanish when $x'$ approaches 0.

2. Taylor-expand the left-hand side (LHS) #

The LHS is $\frac{1}{2}[f(0, t) + f(0, t + \Delta t_1 + \Delta t_2)]$. For the second term $f(0, t + \Delta t_1 + \Delta t_2)$, we're only changing the time argument (the spatial argument stays 0), so we use the one-variable Taylor expansion for the time derivative:
$$f(0, t + \Delta t_{\text{total}}) \approx f(0, t) + \frac{\partial f}{\partial t} \cdot \Delta t_{\text{total}}$$
where $\Delta t_{\text{total}} = \Delta t_1 + \Delta t_2 = \frac{x'}{c - v} + \frac{x'}{c + v}$.

Plugging this back into the LHS:
$$\text{LHS} \approx \frac{1}{2}\left[f(0,t) + f(0,t) + \frac{\partial f}{\partial t} \left(\frac{x'}{c - v} + \frac{x'}{c + v}\right)\right]$$
Simplify that to:
$$\text{LHS} = f(0,t) + \frac{1}{2} \left(\frac{1}{c - v} + \frac{1}{c + v}\right) x' \frac{\partial f}{\partial t}$$

3. Taylor-expand the right-hand side (RHS) #

The RHS is $f(x', t + \Delta t_1)$. Here, we're changing both the spatial argument (from 0 to $x'$) and the time argument (from $t$ to $t + \Delta t_1$). We use the two-variable first-order Taylor expansion:
$$f(a + \Delta a, b + \Delta b) \approx f(a,b) + \frac{\partial f}{\partial a} \cdot \Delta a + \frac{\partial f}{\partial b} \cdot \Delta b$$
Applying this to our RHS (where $a=0$, $\Delta a=x'$; $b=t$, $\Delta b=\Delta t_1$):
$$\text{RHS} \approx f(0,t) + \frac{\partial f}{\partial x'} \cdot x' + \frac{\partial f}{\partial t} \cdot \frac{x'}{c - v}$$

4. Set LHS = RHS and simplify #

Substitute our expanded LHS and RHS back into the original equation:
$$f(0,t) + \frac{1}{2} \left(\frac{1}{c - v} + \frac{1}{c + v}\right) x' \frac{\partial f}{\partial t} = f(0,t) + \frac{\partial f}{\partial x'} \cdot x' + \frac{\partial f}{\partial t} \cdot \frac{x'}{c - v}$$

First, subtract $f(0,t)$ from both sides—they cancel out. Then, divide both sides by $x'$ (valid because $x'$ is infinitesimal but non-zero):
$$\frac{1}{2} \left(\frac{1}{c - v} + \frac{1}{c + v}\right) \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x'} + \frac{1}{c - v} \frac{\partial f}{\partial t}$$

That's exactly the equation Einstein arrives at! The core idea is that for tiny changes in the function's arguments, we can approximate the function using its partial derivatives—higher-order terms become negligible as $x'$ shrinks to 0.

内容的提问来源于stack exchange，提问作者Truth-seek