2024-01-16 Reference Frames

Inertial Reference Frames

• Reference frames are used to describe the motion and position of…

… fill in later…

• Given a point $p$ @ $\vec{r}=\vec{r}(x,y,z)$, this same point can be described by ${\vec{r}}^{\prime }={\vec{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$.
• We can transform between these two coordinate systems using the vector $\vec{R}$, the vector between the two origins
• Time is the same in all frames for Galilean systems (absolute)
  • $t=t^{\prime }$
• To find velocity, take time derivative

$$\frac{d}{dt}(\vec{r})=\frac{d}{dt}({\vec{r}}^{\prime }+{\vec{R}}^{\prime })$$ $$\frac{d\vec{r}}{dt}=\vec{v}=\frac{d{\vec{r}}^{\prime }}{dt}+\frac{d{\vec{R}}^{\prime }}{dt}$$ $$v={\vec{v}}^{\prime }+\vec{V}$$

• Where $\vec{V}$ is the rate of change (velocity) of the distance between the two origins.
• For acceleration

$$\frac{d}{dt}(\vec{v})=\frac{d}{dt}({\vec{v}}^{\prime }+{\vec{V}}^{\prime })$$ $$\frac{d\vec{v}}{dt}=\vec{a}=\frac{d{\vec{v}}^{\prime }}{dt}+\frac{d\vec{V}}{dt}$$ $$\vec{a}={\vec{a}}^{\prime }+\vec{A}$$

• Acceleration in unprimed coordinate system is equal to acceleration in primed reference frame + $\vec{A}$
• An inertial reference frame is one in which the reference frame is not acceleration relative to another. $\vec{A}=0$

If Newtons laws of motions are invariant, (not changing) under a coordinate transform, then the frames are inertial.

$${\vec{F}}^{\prime }=\vec{F}$$

Starting in unprimed frame

$$\vec{F}=m\vec{a}$$

Transformation

$$\vec{F}=m({\vec{a}}^{\prime }+{\vec{A}}^{\prime })$$ $$=m{\vec{a}}^{\prime }+m\vec{A}$$ $$\vec{F}={\vec{F}}^{\prime }+m\vec{A}$$ $$m\vec{a}=m{\vec{a}}^{\prime }\text{ if}\vec{A}=0$$

These frames can still move relative to one another as long as velocity is constant.

Special Relativity

• We must use relativistic mechanics when the velocities involved are significant fractions of $c$
• The speed of light is a universal constant, the same in all reference frames
  • $c=299,792,458\frac{m}{s}$
  • Fine to use $3E8$
• This important fact means that our rule for velocity addition in different frames is incorrect
• The Galileo transformation violates $c=\text{constant}$
  • $c^{\prime }\ne \vec{v}+c$
• Time is relative $t\ne t^{\prime }$
• Our original rules are still valid for an approximation for $v\ll c$
• New set of coordinate transformations are needed which are in agreement with Einstein’s insight that light moves at constant speed

Lorentz Transformation

$$x^{\prime }=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$$

rOtAtIoN MatRiX

$$y^{\prime }=y$$ $$z^{\prime }=z$$ $$t^{\prime }=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$

also sometimes written with $\beta =\frac{v}{c}$

$$t^{\prime }=\gamma t-g\frac{vx}{c^2}$$

• v is the relative speed of the two reference frames. Reference frames cannot be traveling faster then the speed of light to each other.

1. Classical mechanics take place in 3 dimensional space (euclidean space) with a separate time $t$
2. For special relativity we use 4-dimensional Minkowsky spacetime (4D spacetime) $x_1=x_1,x_2=y_1,x_3=z_1,x_4=ct$ 4 vectors:

$$\vec{x}=\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right)$$

• Mathematically, 1 and 2 are equivalent

The Lorentz Transformation

• Expressed as a 4x4 matrix, represented by $\Lambda$ such that ${\vec{x}}^{\prime }=\Lambda \vec{x}$, with $\Lambda$ represented as

$$\Lambda =\left[\begin{array}{c}\gamma ,0,0,-\beta \gamma \\ 0,1,0,0\\ 0,0,1,0\\ -\beta \gamma ,0,0,\gamma \end{array}\right]$$

• “Standard-boost” movement along x-axis
• So transforming position from the 4-vector gives

$${\vec{x}}^{\prime }=\Lambda \vec{x}=\left[\begin{array}{c}\gamma ,0,0,-\beta \gamma \\ 0,1,0,0\\ 0,0,1,0\\ -\beta \gamma ,0,0,\gamma \end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right]=\left[\begin{array}{c}\gamma x+0y+0z-\beta \gamma ct\\ 0x+1y+0z+0ct\\ 0x+0y+1z+0ct\\ -\beta \gamma x+0y+0z+\gamma ct\end{array}\right]=\left[\begin{array}{c}\gamma x-\beta ct\\ y\\ z\\ -\beta \gamma x+\gamma ct\end{array}\right]$$ $$=\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\\ z^{\prime }\\ c{t}^{\prime }\end{array}\right]$$ $$x^{\prime }=\gamma x-\gamma vt$$ $$y^{\prime }=y$$ $$z^{\prime }=z$$ $$c{t}^{\prime }=-\frac{\gamma vx}{c}+\gamma ct$$

We can also transform the velocities The v in the equations above represents the speed between reference frames

$$v^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}$$

use primed position and time

$$d{x}^{\prime }=\frac{\partial x^{\prime }}{\partial x}dx+\frac{\partial x^{\prime }}{\partial t}dt$$ $$d{t}^{\prime }=\frac{\partial t^{\prime }}{\partial x}dx+\frac{\partial t^{\prime }}{\partial t}dt$$

Access partials

$$\frac{\partial x^{\prime }}{dx}=\gamma$$ $$\frac{\partial x^{\prime }}{\partial t}=-\gamma v$$ $$\frac{\partial t^{\prime }}{dx}=-\frac{\gamma v}{c^2}$$ $$\frac{\partial t^{\prime }}{\partial t}=\gamma$$ $$d{x}^{\prime }=\gamma dx+(-\gamma v)dt$$ $$d{t}^{\prime }=\left(-\frac{\gamma v}{c^2}\right)dx+\gamma dt$$ $$v^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}=\frac{dx-vdt}{-\frac{v}{c^2}dx+dt}$$ $$v^{\prime }=\frac{\frac{dx}{dt}-c}{-\frac{v}{c^2}\frac{dx}{dt}+1}=\frac{\frac{dx}{dt}-v}{1-\frac{vdx}{c^{2}dt}}$$

V is the speed of the x,y,z frame compared to x’,y’,z’

$$\frac{dx}{dt}=u$$

Where u is speed measured in x,y,z, frame with t clock

$$v^{\prime }=\frac{u-v}{1-\frac{uv}{c^2}}$$

This is the velocity transformation!