4-Force
We define 4-force F ⃗ \fv F F 4-momentum :
F ⃗ = d p ⃗ d τ . \fv F = \frac{\d \fv p}{\d \tau}. F = d τ d p . It’s often more useful to use the 3-force since that’s what we can
actually measure. We’ll often need to convert between 3- and 4-force
depending on the situation. Let’s first consider how the 3-force
transforms, then how to relate it to the 4-force.
We will use the classic example of a train moving through a station
with velocity v = v e x \tv v = v \tv e_x v = v e x p S = γ ( u ) m u \tv p_S = \gamma(u) m \tv u p S = γ ( u ) m u E S E_S E S F S \tv F_S F S Lorentz transform
to find
( F T ) x = d p T x d t T = γ ( d p S x − v d E s / c 2 ) γ ( d t S − v d x S / c 2 ) = ( F S ) x − ( v / c 2 ) ( d E s / d t S ) 1 − v ( u ) x / c 2 .
\begin{align*}
(\tv F_T)^x &= \frac{\d p_T^x}{\d t_T} = \frac{\gamma(\d p_S^x - v\d E_s/c^2)}{\gamma(\d t_S - v \d x_S / c^2)} \\
&= \frac{(\tv F_S)^x - (v/c^2)(\d E_s/\d t_S)}{1-v(\tv u)^x / c^2}.
\end{align*}
( F T ) x = d t T d p T x = γ ( d t S − v d x S / c 2 ) γ ( d p S x − v d E s / c 2 ) = 1 − v ( u ) x / c 2 ( F S ) x − ( v / c 2 ) ( d E s / d t S ) . We simplify further by recalling that E 2 = p 2 c 2 + m 2 c 4 E^2 = p^2c^2 + m^2 c^4 E 2 = p 2 c 2 + m 2 c 4
E s d E S d t S = p s ⋅ d p S d t S c 2 γ m c 2 d E S d t S = γ m u ⋅ d p S d t S c 2 d E s d t S = F S ⋅ u .
\begin{align*}
E_s \frac{\d E_S}{\d t_S} &= \tv p_s \cdot \frac{\d \tv p_S}{\d t_S} c^2 \\
\gamma mc^2 \frac{\d E_S}{\d t_S} &= \gamma m \tv u \cdot \frac{\d \tv p_S}{\d t_S} c^2 \\
\frac{\d E_s}{\d t_S} &= \tv F_S \cdot \tv u.
\end{align*}
E s d t S d E S γm c 2 d t S d E S d t S d E s = p s ⋅ d t S d p S c 2 = γm u ⋅ d t S d p S c 2 = F S ⋅ u . Which tells us
( F T ) x = ( F S ) x − ( v / c 2 ) F S ⋅ u 1 − v ( u ) x / c 2 .
\begin{align*}
(\tv F_T)^x &= \frac{(\tv F_S)^x - (v/c^2)\tv F_S \cdot \tv u}{1-v(\tv u)^x / c^2}.
\end{align*}
( F T ) x = 1 − v ( u ) x / c 2 ( F S ) x − ( v / c 2 ) F S ⋅ u . We compute the other components to find
( F T ) y , z = ( F S ) y , z γ ( 1 − v ( u ) x / c 2 ) .
\begin{align*}
(\tv F_T)^{y,z} &= \frac{(\tv F_S)^{y,z}}{\gamma(1-v(\tv u)^x / c^2)}.
\end{align*}
( F T ) y , z = γ ( 1 − v ( u ) x / c 2 ) ( F S ) y , z . This looks a lot like velocity addition . It’s
a remarkable symmetry.
We can also easily relate the 3-force to the 4-force. Considering
first the spacial components,
F i = d p i d τ = γ ( u ) d p i d t = γ ( u ) ( F ) i .
\begin{align*}
F^i &= \frac{\d p^i}{\d \tau} = \gamma(u) \frac{\d p^i}{\d t} = \gamma(u) (\tv F)^i.
\end{align*}
F i = d τ d p i = γ ( u ) d t d p i = γ ( u ) ( F ) i . Then the time component,
F 0 = d p 0 d τ = γ ( u ) d d t ( E c ) = γ ( u ) c d E d t .
\begin{align*}
F^0 = \frac{\d p^0}{\d\tau} = \gamma(u) \frac{\d}{\dt} \left(\frac Ec\right) = \frac{\gamma(u)}{c} \frac{\d E}{\dt} .
\end{align*}
F 0 = d τ d p 0 = γ ( u ) d t d ( c E ) = c γ ( u ) d t d E . We have shown that d E / d t = F ⋅ u \d E/\dt = \tv F \cdot \tv u d E / d t = F ⋅ u
F 0 = γ ( u ) c F ⋅ u .
\begin{align*}
F^0 = \frac{\gamma(u)}{c} \tv F \cdot \tv u.
\end{align*}
F 0 = c γ ( u ) F ⋅ u .