I argue there are two other concepts physics generally hasn't considered. These are:
Linear Force; is a force exerted on an object with a constant acceleration with a linear trajectory.
Linear Momentum; is the momentum a body has with a constant acceleration with a linear trajectory.
I define the math as:
$$\hat{F}=ad$$
where $$\hat{F}$$ is the linear force. The linear force is given by acceleration times distance.
key equations
[1] $$d=\frac{1}{2}at^2$$
[2] $$a =\frac{v}{t}$$
This gives us
$$\hat{F}=\frac{v}{t}\frac{1}{2}at^2$$
solving gives us with solving (1)
$$\hat{F}=\frac{1}{2}atv$$
which is the Linear Force equation derived. The Liner Force equation has a constant acceleration in a given distance. Linear Momentum is derived as:
$$d(mv)=p(\hat{F}-a)=\hat{p}$$
$$p(\frac{1}{2}at^2)=d(mv)$$
$$=mv\frac{1}{2}\frac{v}{t}t^2$$
$$=\frac{1}{2}mv^2 t$$ [1]
where $$d(mv)$$ is linear momentum $$\hat{p}$$. At the point [1], we find an equation very similar to the kinetic energy equation, so my concept of linear momentum is the same magnitude of a variation of kinetic energy in some time. So the value of linear momentum is simply a multiple of $$\frac{1}{2}Mv^2$$.
If we now use the concept of a linear momentum given by $$d(mv)=\hat{p}$$, then we can now put this into relativistic terms,
$$F=\frac{\partial \hat{p}}{\partial t}=\frac{\partial(mvd)}{\partial t}$$
$$m\frac{\partial(vd)}{\partial t}+\frac{\partial m}{\partial t}(vd)$$
From here we can derive:
$$m\frac{\partial (vd)}{\partial t}=F-\frac{F.v}{c^2}(vd)$$
For the resultant force.
(1) - $$F=ad$$ so
$$ad=\frac{v}{t}\frac{1}{2}at^2$$ gives
$$ad=\frac{1}{2}atv$$
Linear Force; is a force exerted on an object with a constant acceleration with a linear trajectory.
Linear Momentum; is the momentum a body has with a constant acceleration with a linear trajectory.
I define the math as:
$$\hat{F}=ad$$
where $$\hat{F}$$ is the linear force. The linear force is given by acceleration times distance.
key equations
[1] $$d=\frac{1}{2}at^2$$
[2] $$a =\frac{v}{t}$$
This gives us
$$\hat{F}=\frac{v}{t}\frac{1}{2}at^2$$
solving gives us with solving (1)
$$\hat{F}=\frac{1}{2}atv$$
which is the Linear Force equation derived. The Liner Force equation has a constant acceleration in a given distance. Linear Momentum is derived as:
$$d(mv)=p(\hat{F}-a)=\hat{p}$$
$$p(\frac{1}{2}at^2)=d(mv)$$
$$=mv\frac{1}{2}\frac{v}{t}t^2$$
$$=\frac{1}{2}mv^2 t$$ [1]
where $$d(mv)$$ is linear momentum $$\hat{p}$$. At the point [1], we find an equation very similar to the kinetic energy equation, so my concept of linear momentum is the same magnitude of a variation of kinetic energy in some time. So the value of linear momentum is simply a multiple of $$\frac{1}{2}Mv^2$$.
If we now use the concept of a linear momentum given by $$d(mv)=\hat{p}$$, then we can now put this into relativistic terms,
$$F=\frac{\partial \hat{p}}{\partial t}=\frac{\partial(mvd)}{\partial t}$$
$$m\frac{\partial(vd)}{\partial t}+\frac{\partial m}{\partial t}(vd)$$
From here we can derive:
$$m\frac{\partial (vd)}{\partial t}=F-\frac{F.v}{c^2}(vd)$$
For the resultant force.
(1) - $$F=ad$$ so
$$ad=\frac{v}{t}\frac{1}{2}at^2$$ gives
$$ad=\frac{1}{2}atv$$
