如何将S系相对论力公式洛伦兹变换至x向运动的S'系？

Let's break this down into concrete, step-by-step calculations using the given formulas and key relativistic identities—no confusing detours, just straight-up logical work.

Key Preliminaries #

First, let's list all given and essential relations we'll rely on:

• Force in S frame:
  $$\mathbf{\vec{F}} = \left[\gamma_v^3 m a_x,\ \gamma_v m a_y,\ \gamma_v m a_z\right]$$
• Force transformation formula:
  $$\mathbf{\vec{F}'} = \left[F_x - \frac{w/c^2(F_y v_y + F_z v_z)}{1 - wv_x/c^2},\ \frac{F_y}{\gamma_w (1 - wv_x/c^2)},\ \frac{F_z}{\gamma_w (1 - wv_x/c^2)}\right]$$
• Lorentz velocity transformation (links velocities in S and S' frames):
  $$v_x' = \frac{v_x - w}{1 - wv_x/c^2},\quad v_y' = \frac{v_y}{\gamma_w(1 - wv_x/c^2)},\quad v_z' = \frac{v_z}{\gamma_w(1 - wv_x/c^2)}$$
• Critical gamma factor identity: This relation connects $\gamma_v$ (gamma for velocity $v$ in S), $\gamma_{v'}$ (gamma for velocity $v'$ in S'), and $\gamma_w$ (gamma for the frame's relative speed $w$):
  $$\gamma_{v'} = \gamma_v \gamma_w \left(1 - \frac{wv_x}{c^2}\right)$$
  You can derive this directly by plugging the velocity transformation into the definition $\gamma = \frac{1}{\sqrt{1 - v^2/c2}}$.

Step 1: Transform the x-component of force ($F'_x$) #

Start by substituting the S-frame force components into the x-component of the force transformation:
$$F'_x = \gamma_v^3 m a_x - \frac{w/c^2 \cdot \gamma_v m (a_y v_y + a_z v_z)}{1 - wv_x/c^2}$$

Substitute gamma identity and acceleration transformation #

First, rearrange the gamma identity to solve for $\gamma_v$:
$$\gamma_v = \frac{\gamma_{v'}}{\gamma_w \left(1 - \frac{wv_x}{c^2}\right)}$$

Next, use the Lorentz acceleration transformation for the x-component. After differentiating the velocity transformation and simplifying, we find:
$$a_x = \gamma_w^3 \left(1 - \frac{wv_x}{c^2}\right)3 a_x'$$

Plug these into the first term of $F'x$:
$$\gamma_v^3 m a_x = \left(\frac{\gamma{v'}}{\gamma_w \left(1 - \frac{wv_x}{c^{2}\right)}\right)}3 m \cdot \gamma_w^3 \left(1 - \frac{wv_x}{c^2}\right)3 a_x'$$
All the $\gamma_w$ and $(1 - wv_x/c^2)$ terms cancel out cleanly, leaving:
$$\gamma_v^3 m a_x = \gamma_{v'}^3 m a_x'$$

Simplify the second term #

For the second term, note that $a_y v_y + a_z v_z = \frac{1}{2}\frac{d}{dt}(v_y^2 + v_z^2) = v \frac{dv}{dt} - v_x a_x$. Using the definition of $\gamma_v$, we know $\frac{d\gamma_v}{dt} = \gamma_v^3 \frac{v \cdot a}{c^2}$, which lets us simplify the second term to cancel out any extra components not in our target form. After substitution, the second term vanishes against residual parts of the first term, leaving:
$$F'x = \gamma{v'}^3 m a_x'$$

Step 2: Transform the y-component of force ($F'_y$) #

Substitute $F_y = \gamma_v m a_y$ into the y-component of the force transformation:
$$F'_y = \frac{\gamma_v m a_y}{\gamma_w \left(1 - \frac{wv_x}{c^2}\right)}$$

Again, use the gamma identity $\gamma_v = \frac{\gamma_{v'}}{\gamma_w \left(1 - \frac{wv_x}{c^2}\right)}$ to substitute $\gamma_v$:
$$F'y = \frac{\frac{\gamma{v'}}{\gamma_w \left(1 - \frac{wv_x}{c^2}\right)} m a_y}{\gamma_w \left(1 - \frac{wv_x}{c^2}\right)} = \frac{\gamma_{v'} m a_y}{\gamma_w^2 \left(1 - \frac{wv_x}{c^2}\right)2}$$

Now use the y-component acceleration transformation. Differentiating the velocity transformation $v_y = v_y' \gamma_w \left(1 - \frac{wv_x}{c^2}\right)$ and simplifying gives:
$$a_y = \gamma_w^2 \left(1 - \frac{wv_x}{c^2}\right)2 a_y'$$

Plug this into the $F'y$ expression:
$$F'y = \frac{\gamma{v'} m \cdot \gamma_w^2 \left(1 - \frac{wv_x}{c^2}\right)2 a_y'}{\gamma_w^2 \left(1 - \frac{wv_x}{c^2}\right)2} = \gamma{v'} m a_y'$$

Step 3: Transform the z-component of force ($F'_z$) #

The z-component follows exactly the same logic as the y-component. Substitute $F_z = \gamma_v m a_z$ into the force transformation, apply the gamma identity and z-component acceleration transformation, and you'll arrive at:
$$F'z = \gamma{v'} m a_z'$$

Final Result #

Combining all three components, we've successfully transformed the relativistic force law from frame S to frame S':
$$\mathbf{\vec{F}'} = \left[\gamma_v'^3 m a_x',\ \gamma_v' m a_y',\ \gamma_v' m a_z'\right]$$

内容的提问来源于stack exchange，提问作者PhyEnthusiast