- #1
SiennaTheGr8
- 493
- 193
In classical mechanics, there's some debate about whether to define force as ##\mathbf{f} = m\mathbf{a}## or as ##\mathbf{f} = \dot {\mathbf{p}} = d (m\mathbf{v})/dt##. The former is much more popular I think, and it has the virtue of always having a Galilean-invariant magnitude ##f##, whereas the latter has the virtues of emphasizing a conserved quantity (momentum) and smoothing the transition to SR but leaves ##f## frame-dependent in a variable-mass situation (##\mathbf{f} = \dot{m} \mathbf{v} + m\mathbf{a}##).
In SR, we have the four-vector relation ##\mathbf{F} = d\mathbf{P} / d\tau = d(m\mathbf{V})/d \tau##, which also equals ##m \mathbf{A}## if ##m## is constant. But here's my sanity check: even if ##m## isn't constant, the magnitude ##F## of the four-force is Lorentz-invariant, right? ##\mathbf{F} = (dm / d \tau) \mathbf{V} + m \mathbf{A}##, but here ##V## is Lorentz-invariant, and ##\mathbf{V} \cdot \mathbf{A} = 0##, so everyone should get the same value for ##F##.
I'm 99.9% sure about this, especially since by other means I can calculate that ##F## must be invariant (##F^2## equals the difference of the squares of the proper power and the proper force). Mostly just looking for confirmation that the logic above is sound.
There's something rather pleasing about this result, IMO.
In SR, we have the four-vector relation ##\mathbf{F} = d\mathbf{P} / d\tau = d(m\mathbf{V})/d \tau##, which also equals ##m \mathbf{A}## if ##m## is constant. But here's my sanity check: even if ##m## isn't constant, the magnitude ##F## of the four-force is Lorentz-invariant, right? ##\mathbf{F} = (dm / d \tau) \mathbf{V} + m \mathbf{A}##, but here ##V## is Lorentz-invariant, and ##\mathbf{V} \cdot \mathbf{A} = 0##, so everyone should get the same value for ##F##.
I'm 99.9% sure about this, especially since by other means I can calculate that ##F## must be invariant (##F^2## equals the difference of the squares of the proper power and the proper force). Mostly just looking for confirmation that the logic above is sound.
There's something rather pleasing about this result, IMO.