User:DominicPrice/sandbox

Informal treatment
In a two and three dimensional space, there is an ambiguity in taking an integral between two points as there are infinitely many paths that you could choose to get between the two points - apart from the straight line formed between the two points one could choose a curved path of greater length as shown in the figure. Therefore in general the value of the integral depends on the path taken. However, in the special case of a conservative vector field, the value of the integral is independent of the path taken which can be thought of as a large-scale cancellation of all elements dR which don't have a component along the straight line between the two points. To visualise this, imagine two people climbing a cliff; one decides to scale the cliff by going vertically up it, and the second decides to walk along a winding path that is longer in length than the height of the cliff, but at only a small angle to the horizontal. Although the two hikers have taken different routes to get up to the top of the cliff, at the top they will have both gained the same amount of potential energy due to gravity. This is because a gravitational field is conservative. As an example of a non-conservative field, imagine pushing a box from one end of a room to another. If you push the box in a straight line across the room, you will do noticeably less work against friction than if you pushed the box in a curved path covering a greater distance.

Equation of Motion
Velocity is defined as the rate of change of position with respect to time, i.e.


 * $$\boldsymbol{v} = \frac{d\boldsymbol{x}}{d\mathit{t}}$$

where v is velocity and x is the displacement vector. This will give the instantaneous velocity of a particle, or object, at any particular time t. Although the concept of an instantaneous velocity might at first seem counter-intuitive, it is best considered as the velocity that the object would continue to travel at if it stopped accelerating at that moment.

Although velocity is defined as the rate of change of position, it is more common to start with an expression for an objects acceleration and from there obtain an expression for velocity, which can be done by evaluating


 * $$ \int \boldsymbol{a}  d\mathit{t}$$

which comes from the definition of acceleration,


 * $$ \boldsymbol{a} = \frac{d\boldsymbol{v}}{d\mathit{t}}$$

Sometimes it is easier, or even necessary, to work with the average velocity of an object, that is to say the constant velocity, that would provide the same resultant displacement as a variable velocity, v(t), over some time period Δt. Average velocity can be calculated as
 * $$\boldsymbol{\bar{v}} = \frac{\Delta\boldsymbol{x}}{\Delta\mathit{t}}$$

The average velocity is always less than or equal to the average speed of an object. This can be seen by realizing that while distance is always strictly increasing, displacement can increase or decrease in magnitude as well as change direction.

In terms of a displacement-time graph, the velocity can be thought of as the gradient of the tangent line to the curve at any point, and the average velocity as the gradient of the chord line between two points with t coordinates equal to the boundaries of the time period for the average velocity.

Constant acceleration
In the special case of constant acceleration, velocity can be studied using the suvat equations. By considering a as being equal to some arbitrary constant vector, it is trivial to show that
 * $$\boldsymbol{v} = \boldsymbol{u} + \boldsymbol{a}t$$

with v as the velocity at time t and u as the velocity at time t=0. By combining this equation with the suvat equation x=ut+at2/2, it is possible to relate the displacement and the average velocity by
 * $$\boldsymbol{x} = \frac{(\boldsymbol{u} + \boldsymbol{v})}{2}\mathit{t} = \boldsymbol{\bar{v}}\mathit{t}$$.

It is also possible to derive an expression for the velocity independent of time, known as the Torricelli equation, as follows:
 * $$v^{2} = \boldsymbol{v}\cdot\boldsymbol{v} = (\boldsymbol{u}+\boldsymbol{a}t)\cdot(\boldsymbol{u}+\boldsymbol{a}t)=u^{2}+2t(\boldsymbol{a}\cdot\boldsymbol{u})+a^{2}t^{2}$$
 * $$(2\boldsymbol{a})\cdot\boldsymbol{x} = (2\boldsymbol{a})\cdot(\boldsymbol{u}t+\frac{1}{2}\boldsymbol{a}t^{2})=2t(\boldsymbol{a}\cdot\boldsymbol{u})+a^{2}t^{2} = v^{2} - u^{2}$$
 * $$\therefore v^{2} = u^{2} + 2(\boldsymbol{a}\cdot\boldsymbol{x})$$

where v=|v| etc...

The above equations are valid for both Newtonian mechanics and special relativity. Where Newtonian mechanics and special relativity differ is in how different observers would describe the same situation. In particular, in Newtonian mechanics, all observers agree on the value of t and the transformation rules for position create a situation in which all non-accelerating observers would describe the acceleration of an object with the same values. Neither is true for special relativity. In other words only relative velocity can be calculated.

Quantities that are dependent on velocity
Kinetic energy, momentum, lorentz factor

The kinetic energy of a moving object is dependent on its velocity by the equation
 * $$E_{k} = \frac{1}{2}mv^{2}$$

where Ek is the kinetic energy and m is the mass. Kinetic energy is a scalar quantity as it depends on the square of the velocity, however a related quantity, momentum, is a vector and defined by
 * $$\boldsymbol{p}=m\boldsymbol{v}$$

In special relativity, the dimensionless Lorentz Factor appears frequently, and is given by
 * $$\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

where γ is the Lorentz factor and c is the speed of light.

Escape velocity is the minimum speed a ballistic object needs to escape from a massive body like the earth. It represents the kinetic energy that, when added to the object's gravitational potential energy, (which is always negative) is greater than or equal to zero. The general formula for the escape velocity of an object at a distance r from the center of a planet with mass M is
 * $$v_e = \sqrt{\frac{2GM}{r}},$$

where G is the Gravitational constant. The escape velocity from Earth's surface is about 11 100 m/s.

For non collinear vectors
In the case of the acceleration and velocity vectors not being collinear, it is possible to rewrite the equations of motion in terms of vectors. The derivations are essentially the same as in the scalar case, however it is worth considering the derivation of Torricelli's equation, as this contains vector multiplication.
 * $$\begin{align}

\boldsymbol{v} & = \boldsymbol{a}t+\boldsymbol{v}_0 \quad [1]\\ \boldsymbol{r} & = \boldsymbol{r}_0 + \boldsymbol{v}_0 t + \frac{{\boldsymbol{a}}t^2}{2} \quad [2]\\ \boldsymbol{r} & = \boldsymbol{r}_0 + \left( \frac{\boldsymbol{v}+\boldsymbol{v}_0}{2} \right )t \quad [3]\\ v^2 & = v_0^2 + 2\boldsymbol{a}\cdot\left( \boldsymbol{r} - \boldsymbol{r}_0 \right) \quad [4]\\ \boldsymbol{r} & = \boldsymbol{r}_0 + \boldsymbol{v}t - \frac{{\boldsymbol{a}}t^2}{2} \quad [5]\\ \end{align}$$