2024-01-19 Special Relativity

Special Relativity

Velocity addition

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

u=c example shows c is constant If c is constant, scalar product of the 4-vector with itself is invariant under Lorentz transform

$${\vec{x}}^{\prime }\cdot {\vec{x}}^{\prime }=\vec{x}\cdot \vec{x}$$ $$\vec{x}\cdot \vec{x}=x_1^2+x_2^2+x_3^2-c^2{t}^2$$

Tensors are matrices Minkowski Tensor

$${\mu }_{\nu }=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right)$$ $$\vec{x}\cdot \vec{x}=x{\mu }_{\nu }{x}^{\nu }$$

Test:

$$x_1^{\prime }=\gamma x_1-\gamma \beta ct$$ $$x_2^{\prime }=x_2$$ $$x_3^{\prime }=x_3$$ $$c{t}^{\prime }-\gamma \beta x_1+\gamma ct$$ $$x_1^{\prime 2}=(\gamma x_1-\gamma \beta ct{)}^2={\gamma }^2{x}_1^2+{\gamma }^2{\beta }^2{c}^2{t}^2-2{\gamma }^2\beta x_{1}ct$$ $$x_2^{\prime 2}=x_2^2$$ $$x_3^{\prime 2}=x_3$$ $$c^2{t}^2=(-\gamma \beta x_1+\gamma ct{)}^2={\gamma }^2{\beta }^2{x}_1^2+{\gamma }^2{c}^2{t}^2-2{\gamma }^2\beta x_{1}ct$$ $${\vec{x}}^{\prime }\cdot {\vec{x}}^{\prime }=x_1^{\prime 2}+x_2^{\prime 2}+x_3^{\prime 2}-c^2{t}^2$$ $$={\gamma }^2{x}_1^2+{\gamma }^2{\beta }^2{c}^2{t}^2-2{\gamma }^2\beta x_{1}ct+x_2^2+x_3^2-{\gamma }^2{\beta }^2{x}_1^2-{\gamma }^2{c}^2{t}^2+2{\gamma }^2\beta x_{2}ct$$ $$={\gamma }^2{x}_1^2(1-{\beta }^2)+x_2^2+x_3^2-{\gamma }^2{c}^2{t}^2(1-{\beta }^2)$$ $$\to =x_1^2+x_2^2+x_3^2-c^2{t}^2=\vec{x}\cdot \vec{x}$$

Invariant!

Relativistic Mechanics

Newtons laws of motions

1st law of motion:

Object at rest stays at rest Object in motion stays in motion If no net external forces $\vec{F}=0$, $\vec{v}=\text{constant}$.

• Also valid in relativity

2nd law of motion

$\vec{F}=m\vec{a}=\frac{d}{dt}\vec{p}$ If constant $\vec{F}$ then constant $\vec{a}$ $v=at\to \infty$ as $t\to \infty$ This is not allowed! $v\le c$ must be true

Requires momentum modification:

$$p=mv\to p=\gamma mv$$

This is relativistic momentum. Mass is function of velocity now:

$$m(v)=\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}$$

$m(v=0)$ is the "rest mass " of the object

As speed increases, mass functionally increases As it approaches the speed of light, mass reaches infinity if you were able to accelerate at $9.8\frac{m}{s^2}$ in perpetuity, youd reach c in about a year

$$\vec{F}=\frac{d\vec{p}}{dt}=m\frac{d}{dt}\frac{\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$$

In the limit of $v\ll c$, should get $F=ma$ Denominator goes to $\approx 1$, so you get $\vec{F}=m\vec{v}\frac{d}{dt}$ which is equivalant to

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

Newtons 3nd law

Third law pairs Equal and opposite forces This is valid for contact forces Friction, impact etc etc This is not valid for non-contact forces at a distance Gravity The force of gravity doesn’t travel faster then the speed of light If something moves at a significant portion of the speed of light, its gravitational potentials may move slow enough to cause discrepancies

• Takes time for things to propagate and update and space is big

Relativistic Kinetic Energy

$$T=m{c}^2(\gamma -1)=m{c}^2\gamma -m{c}^2$$

Derived from work-energy theorem using relativistic momentum $p=\gamma mv$ Rest Energy:

$$E_0=m{c}^2$$

Total energy of the system:

$$E=\gamma m{c}^2$$ $$E=T+m{c}^2=m{c}^2\gamma -m{c}^2+m{c}^2$$

Multiply both sides of relativistic by c and square

$$c^2{p}^2={\gamma }^2{m}^2{v}^2{c}^2={\gamma }^2{m}^2{c}^4\frac{v^2}{c^2}$$ $$g^2=\frac{1}{1-\frac{v^2}{c^2}}$$ $$\frac{v^2}{c^2}=1-\frac{1}{{\gamma }^2}$$ $$c^2{p}^2=g^2{m}^2{c}^4\left(1-\frac{1}{{\gamma }^2}\right)$$ $$={\gamma }^2{m}^2{c}^4-m^2{c}^4$$ $$c^2{p}^2=E^2-E_0^2$$ $$E^2=c^2{p}^2+E_0^2$$

For a mass-less particle:

$$E=cp$$ $$E=h\nu =\frac{hc}{\lambda }$$ $$\frac{h}{\lambda }=p$$

This agrees with quantum physics!

For case of non-relativistic speeds $(v\ll c)$, then relativistic kinetic energy should reduce to

$$\frac{1}{2}m{v}^2$$ $$T=m{c}^2(\gamma -1)$$ $$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

Taylor expansion of $\gamma$

$$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\approx \frac{1}{(1-x{)}^{1/2}}$$

First equation turns into

$$=1+\frac{1}{2}\frac{v^2}{c^2}+0\left(\frac{v^4}{c^4}\right)$$

Ignoring higher orders

$$T=m{c}^2\left(1+\frac{1}{2}\frac{v^2}{c^2}\dots \right)$$ $$T=\frac{1}{2}m{v}^2$$

Work energy theorem to get kinetic energy

$$\Delta T=w=\int Fdx$$ $$F=\frac{dp}{dt}\cdot \frac{dx}{dx}$$ $$F_{dx}=dpv$$ $$p=\gamma mv=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$dp=\frac{mdv}{\sqrt{1-\frac{v^2}{c^2}}}+\frac{mv\left(-\frac{1}{2}\right)\left(-2\frac{v}{c^2}dv\right)}{{\left(1-\frac{v^2}{c^2}\right)}^{3/2}}$$ $$dp=\gamma mdv+\frac{m{v}^2}{c^2}$$ $$=\gamma mdv\left(1+{\gamma }^2\frac{v^2}{c^2}\right)$$ $$=\gamma mdv{\gamma }^2$$ $${\gamma }^2=\frac{1}{1-\frac{v^2}{c^2}}$$ $${\gamma }^2\frac{v^2}{c^2}=\frac{v^2}{\left(1-\frac{v^2}{c^2}\right)c^2}$$ $$=\frac{v^2}{c^2-v^2}$$ $$1+{\gamma }^2\frac{v^2}{c^2}=1+\frac{v^2}{c^2-v^2}=\frac{c^2-v^2}{c^2-v^2}+\frac{v^2}{c^2-v^2}$$ $$=\frac{c^2}{c^2-v^2}\frac{\frac{1}{c^2}}{\frac{1}{c^2}}=\frac{1}{1-\frac{v^2}{c^2}}={\gamma }^2$$ $$dp={\gamma }^{3}mdv$$ $$\Delta T=\int vdp=\int m{\gamma }^{3}vdv$$ $$=m\int \frac{v}{{\left(1-\frac{v^2}{c^2}\right)}^{3/2}}dv$$ $$\to \frac{c^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

Integrate from 0 to T and from 0 to v

$$m\left[\frac{c^2}{\sqrt{1-\frac{v^2}{c^2}}}-\frac{c^2}{1}\right]$$ $$T=m{c}^2(\gamma -1)$$

Lorentz Transformation Derivation

If a light pulse is emitted from a flashbulb from a common origin of $k$ and $k^{\prime }$ when they are coincident then from postulate 2 (c is constant value in all frames) the wavefronts in each system is described by the following

$$\sum _{i=i}^3{x}_i^2-c^2{t}^2=0$$

These are eq1

$$\sum _{i=1}^3{x}_i^{\prime 2}-c^2{t}^2=0$$

This comes from the invariance. The rate of the pulse is $ct$ (dot product of 4-vector using the +++- conversion )

*insert picture here*

Recall from Galileo transforms

$$x_1^{\prime }=x_1-vt$$ $$x_2^{\prime }=x_2$$ $$x_3^{\prime }=x_3$$ $$t^{\prime }=t$$

and that using eq.1 in accordance with the postulates of relativity can not be reconciled

Einstein’s contribution was that Galileo transformation was approximately correct

We confined movement to the positive x direction for the k’ system meaning y’=y and z’=z

At t=t’=0 the flashbulb goes off, the position of the 0’ origin (of k’ system) is measured to be $x=vt$ (w.r.t the k system), or $x-vt=0$

In the k’ system the origin 0’ is at x’=0

So at t=t’=0 set them equal

$$x^{\prime }=x-vt$$

But we know that the Galileo transform is incorrect So we can add a coefficient to the transform. We could call it gamma… It would depend on v and is independent of x,x’,t,t’

$$x^{\prime }=\gamma (x-vt)$$

Eq. 2

We can describe motion in k with origin o in terms of k’

Postulate I demands that the laws of physics be the same in both reference frames

$$\gamma ={\gamma }^{\prime }$$

Relativistic Meter stick problem: How long is the meterstick? At a 45 degree angle

Continued on 1-23

$$x^{\prime }=\gamma (x-vt)$$ $$x=\gamma (x^{\prime }+v{t}^{\prime })$$

Sub top equation into above equation and solve for t

$$x=\gamma [\gamma (x-vt)+v{t}^{\prime }]$$ $$x={\gamma }^2(x-vt)+\gamma v{t}^{\prime }$$ $$x={\gamma }^2(x-vt)+\gamma v{t}^{\prime }$$ $$x={\gamma }^{2}x-{\gamma }^{2}vt+\gamma v{t}^{\prime }$$ $$\gamma v{t}^{\prime }=x-{\gamma }^{2}x+{\gamma }^{2}vt$$ $$t^{\prime }=\gamma t+\frac{x}{\gamma v}(1-{\gamma }^2)$$

The speed of light is the same in both frames

Position of flashbulb pulse:

$$x=ct$$ $$x^{\prime }=c{t}^{\prime }$$

Now solve for $\gamma$ using substitutions

$$\gamma =\frac{1}{\sqrt{\left(1-\frac{v^2}{c^2}\right)}}$$

Also:

$$t^{\prime }=\gamma \left(t-\frac{xv}{c^2}\right)$$

Time dilation

$$\Delta t=\gamma \Delta t^{\prime }$$

Length contraction

$$\Delta x=\frac{\Delta x^{\prime }}{\gamma }$$