PHYS10121 - Quantum and Relativity

Relativity: What are the Galilean Transformations? When can they be used?

x=x^{\prime}+vt (or for y or z if velocity has y or z component)

t=t^{\prime}

• These are trivial from diagrams/suvat equations and can be used to prove that Newton’s first law holds in all inertial frames of reference (i.e by showing object moving parallel to y axis in S’ also moves in a straight line in S).

• However, (as we can see) it assumes there is universal time i.e time is unchanged between different reference frames. It also suggests that the length of an object will be the same in all frames (as if there are two positions at each end of rod, x1 and x2, we can see from equations x2 - x1 = x2’ - x1’)

• Special relativity shows us that these suggestions are not true. therefore these transforms can only be used when v « c , which will give answers that will equal the true answers to a ridiculously good accuracy/precision so do a good enough job

Relativity: Einstein’s first postulate and it’s consequences

• Einstein was influenced by Maxwell’s equations which predicted the existence of EM waves which travel at speed c, where c=\frac{1}{\sqrt{\mu0\epsilon0}}

• Einstein’s first postulate proposed that identical isolated experiments in different inertial frames deliver identical results.

• For example, experiments to do with Newton’s first law of motion (which Galilean transformations “showed” were the same in all frames) and experiments in electricity and magnetism:

F=\frac{q1q2}{4\pi\epsilon0r^2} , B=\frac{\mu0}{2\pi r} (B is magnetic field due to current in wire)

• This suggests that, if Newton’s first law of motion gives the same results in all inertial frames, there should be no way to experimentally prove that one object is “universally moving” - we can only declare it as “moving” with reference to another inertial frame.

• This also suggests that \mu0 and \epsilon0 would be the same in all inertial frames (as they must be so that experiments in electricity and magnetism are consistent with his first postulate). Therefore, IF Maxwell’s equations are correct, then the speed of light, c, must also be the same in all inertial frames (leading to Einstein’s second postulate…)

Relativity: Einstein’s second postulate and it’s consequences

• The speed of light in a vacuum is the same in all inertial frames. (as predicted by his first postulate, assuming Maxwell was correct)

• This has some strange consequences - i.e, imagine 2 observers, one in a frame stationary to a light source and one in a frame moving towards the light source at speed v. the observer in the stationary frame measures light approaching the other observer at c + v while the observer in the moving frame measures the speed of light as just c. This leads ideas of time dilation and length contraction, showing that time is not actually universally constant!

Relativity: Time dilation - light clock thought experiment

• Imagine light moving between two mirrors of separation d, set up in frame S’. the time for one ‘tick’ is given by \Delta t^{\prime}=\frac{2d}{c}

• Now imagine another frame, S, in which S’ has a constant velocity, v (see diagram). In S, the time for one ‘tick’ will be \Delta t, which we can see is given by \Delta t=\frac{2\sqrt{\left(\frac12v\Delta t\right)^2+d^2}}{c} (bit in sqrt is total distance back and forth between mirrors using pythagoras). note: we have used Einstein’s second postulate here by assuming speed of light is same in all frames.

• This can be rearranged to give \Delta t=\frac{2d}{c}\cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\Delta t^{\prime}

Therefore, we can see

\Delta t=\gamma\Delta t^{\prime} , where \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c2}}}

Time is NOT the same in all inertial frames! (it just seems to be when v « c as gamma becomes very very close to 1). This is also the same for all clocks (not just light clocks) - note that if it wasn’t and clocks in different frames drifted out of sync, we would be able to tell they were moving, which violates Einstein’s first postulate.

Relativity: Length contraction - ruler and clock thought experiment

• Consider a ruler of length L’ as measured in it’s frame, S’. Now imagine S’ is moving with a constant speed (in the x-axis) of v relative to another frame, S. What would the length be according to an observer in frame S (L)?

• We can use a light clock to measure the length of the ruler. The time dilation formula can therefore be used to express L in terms of gamma and L’

• Position one mirror (of the light clock) at one end of the ruler and position the other mirror at the other end (see diagram).

• IN S FRAME: On the outbound journey (of the light ray), we see the distance traveled is equal to c\Delta tout=L+v\Delta tout. On the return journey, the distance traveled will be c\Delta tback=L-v\Delta tback. this gives a total time (in S frame) of \Delta t^{}=\frac{L}{c-v}+\frac{L}{c+v}

• IN S’ FRAME: There is no relative motion between the ruler and the reference frame, so outbound and return time is the same and the total time in this frame can simply be found with suvat: \Delta t^{\prime}=\frac{2L^{\prime}}{c}

Using the formula for time dilation (\Delta t=\gamma\Delta t^{\prime}) we can see

\frac{L}{c-v}+\frac{L}{c+v}=\frac{\gamma2L^{\prime}}{c}

L=\frac{L^{\prime}}{\gamma} , where \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c2}}}

Length is NOT the same in all inertial frames!! Objects that are moving relative to a frame appear smaller in that frame than they appear in their rest frame (frame where they have no relative motion).

Relativity: Muons as an example of both time dilation and length contraction

When measured on Earth (so the muon is at rest in Earth’s reference frame), muons have a half-life of just 2.2 microseconds. They are created in cosmic ray collisions ~20 km above Earth’s surface, yet can be detected by detectors on the ground, suggesting they are traveling faster than the speed of light!?

• Explanation with time dilation: The created muons move very close to the speed of light, let’s say with v = 0.9999c. From Earth’s frame, the muons are time dilated, so their half life (according to earth) is longer than the 2.2 microseconds measured in muons on Earth (= 2.2 microseconds * gamma), explaining why (according to earth) they can travel 20km while not moving faster than the speed of light.

• Explanation with length contraction: from the muon’s frame, the Earth is moving towards them at 0.9999c, so the atmosphere has been length contracted by a huge amount (= 1/gamma) allowing them to travel through all of it in (according to them) just 2.2 microseconds.

This shows us that time dilation and length contraction are basically just different ways of showing the same thing (and that they are both needed as if only one existed without the other to explain phenomena from the other observer’s point of view, it would literally break every law of physics ever and cause lots of confusion and mess.)

Relativity: Doppler effect for light - time interval between clock ticks for observer moving relative to clock

• Let observer be in S frame, clock in S’ frame which is moving with speed v (x-axis) relative to S. Time interval between ticks for observer = \Delta T .

• At start of tick, clock at distance d from observer. at end of tick, clock at distance v\Delta t from observer (see diagram). Note the time that has passed does NOT equal the time interval.

• Assume the light from the start of the tick arrives at the observer at t = 0. The light emitted at end of tick therefore arrives \frac{v\Delta t}{c}+\Delta t later. Therefore:

\Delta T=\Delta t\left(1+\frac{v}{c}\right)=\gamma\Delta t^{\prime}\left(1+\frac{v}{c}\right) (due to time dilation equation)

\Delta T=\frac{1+\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}\Delta t^{\prime}=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}\Delta t^{\prime}

• We can use this to get an expression for the doppler effect including relativistic effects: T=\frac{1}{f}, therefore for S, T=\Delta T and for S’, T=\Delta t^{\prime}

Therefore:

\frac{1}{f}=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}\cdot\frac{1}{f^{\prime}} (f is frequency in S, f’ in S’)

f=\left(\frac{1-\frac{v}{c}}{1+\frac{v}{c}}\right)^{\frac12}f^{\prime}

Using f=\frac{c}{\lambda} (which is usable in all inertial frames as c is constant) can be used to find the doppler shift in wavelength for light emitters moving at relativistic speeds towards/away from observers.

Relativity: Lorentz transformations:

Lorentz transforms are equations that relate the coordinates of an event in one inertial frame to the coordinates of the same event in a second inertial frame (i.e Galilean transforms but actually account for relativity)

Assuming velocity in x direction:

• x^{\prime}=\gamma\left(x-vt\right) (LT1)

• t^{\prime}=\gamma\left(t-\frac{vx}{c^2}\right) (LT2)

• y^{\prime}=y (LT3)

• z^{\prime}=z (LT4)

Reverse transforms:

• x^{}=\gamma\left(x^{\prime}+vt\right) (LT5)

• t^{}=\gamma\left(t^{\prime}+\frac{vx^{\prime}}{c^2}\right) (LT6)

These can also be used directly for coordinate differences (\Delta x=x2-x1 , \Delta t=t2-t1) by simply subbing in x=\Delta x and t=\Delta t (and vice versa for prime versions)

Relativity: Deriving Lorentz transforms

• We can assume position in one frame will be related to position AND time in the other. Same assumption for time. We then assume that the transformations are linear, i.e is the same everywhere (true for special relativity… but not general!) Therefore in simplest terms,

x^{\prime}=ax+bt (1)

t^{\prime}=dx+et (2)

We now find coefficients a, b, d, and e (note we didn’t use c as this denotes speed of light):

• State that origin O’ (where x’ = 0) moves along x axis with speed v where x = vt. Sub into (1):

0=avt+bt , -\frac{b}{a}=v (3)

• Similarly, state O (x = 0) moves along x’ axis at speed -v (duh) where x’ = -vt’. Sub into (1):

x^{\prime}=bt

t^{\prime}=et

-\frac{b}{e}=v (4)

• Combining (3) and (4) gives e = a , b = -av. Sub into (1) and (2):

x^{\prime}=ax-avt , t^{\prime}=dx+at (5)

Note here that if t = t’, we recover the Galilean transformations! (as d = 0, e = a = 1)

• We now use Einstein’s second postulate to eliminate x / x’: c is same in all inertial frames - Consider a photon emitted at O and O’ when they are coincident, i.e when t’ = t = 0. Photon travels along the x and x’ axis so that x = ct, x’ = ct’. Sub into equations (5):

ct^{\prime}=act-avt , t^{\prime}=dct+at

d=-\frac{av}{c^2}

• From combining this with (1) and (2), we are now left with:

x^{\prime}=a\left(x^{}-vt^{}\right) , t^{\prime}=a\left(t-\frac{vx}{c^2}\right) (6)

Note that equations (6) must take same forms in all inertial frames so we can simply swap the primed and unprimed coordinates and swap the sign of v to find the same equations just reversed:

x^{}=a\left(x^{^{\prime}}+vt^{^{\prime}}\right) , t^{}=a\left(t^{\prime}+\frac{vx^{\prime}}{c^2}\right) (7)

• To solve for a, we substitute x’ and t’ from (6) into equations (7):

x=a\left(ax-avt+avt-\frac{av^2x}{c^2}\right)=a^2x\left(1-\frac{v^2}{c^2}\right)

cancel for x and rearrange:

a=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma !!!!!!!!!

• Now we have all unknowns, giving us our Lorentz transforms!

Relativity: When deriving Lorentz transformations, why do we assume they are linear transformations?

• 1) Spacetime is same everywhere (under Special relativity): we could choose origin of coordinate system to be anywhere and the equations will have the same forms

• 2) Newton’s first law: We proved N1 is valid in all inertial frames under Galilean transforms so we assume the same will be true for Lorentz transforms

• 3) (best reason kinda) Einstein’s first postulate: We could say Lorentz transforms must be linear so that Einstein’s first postulate, that laws of physics are same in all inertial frames, is true, as if they were not linear, N1 would not be the same in all inertial frames so would break Einstein’s first postulate

tl;dr, Einstein’s first postulate states N1 must be true in all frames, N1 is true because spacetime is same everywhere, therefore Lorentz transforms must be linear so that spacetime is the same everywhere.

Relativity: Deriving time dilation with Lorentz transforms

• consider clock at rest in S’ at position x0’. One tick of clock starts at t1’ and ends at t2’, giving \Delta t^{\prime}=t2^{\prime}-t1^{\prime}. How long is this tick in S? (\Delta t)

We have two events: start of tick, end of tick. We can use LT6 (t^{}=\gamma\left(t^{\prime}+\frac{vx^{\prime}}{c^2}\right)) to transform these events in S’ into S.

Event 1: (x0’, t1’) in S’, in S, t1^{}=\gamma\left(t1^{\prime}+\frac{vx0^{\prime}}{c^2}\right)

Event 2: (x0’ , t2’) in S’, in S, t2=\gamma\left(t2^{\prime}+\frac{vx0^{\prime}}{c^2}\right)

\Delta t=t2-t1=\gamma\left(t2^{\prime}-t1^{\prime}+\frac{vx0^{\prime}}{c^2}-\frac{vx0^{\prime}}{c^2}\right)

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

Relativity: Deriving length contraction with Lorentz transforms

• Consider ruler in S’ with one end at x1’ and other at x2’, \Delta x^{\prime}=x2^{\prime}-x1^{\prime}. What is length in S? (\Delta x)

We have two events: Measurement of one end (of ruler), measurement of other end. Both events occur at the same time in S (= t0) by definition; we want length of ruler measured by one observer at rest in S.

We can use either LT1 or LT5 - LT5 directly transforms x’ into x, but using LT1 actually looks neater due to how the length contraction formula looks in the end*:

Coordinate (/event) 1: x1^{\prime}=\gamma\left(x1-vt0\right)

Coordinate (/event) 2: x2^{\prime}=\gamma\left(x2-vt0\right)

\Delta x^{\prime}=x2^{\prime}-x1^{\prime}=\gamma\left(x2-x1+vt0-vt0\right)

\Delta x^{\prime}=\gamma\Delta x,

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

Relativity: Velocity transformations

• Consider an object moving with velocity v’ (vector with both vx’ and vy’ components) in S’. An observer in S sees the S’ reference frame moving away from them with speed u (purely in x direction). What is the velocity, v, of the object in the S frame?

We want 2 equations: one for vx, and one for vy. We can do this using Lorentz transforms as:

vx=\frac{dx}{dt} and vy=\frac{dy}{dt}

Just like we subbed in \Delta x=x and \Delta t=t into Lorentz transforms before, we can do the same with dx and dt.

Using LT5 and LT6:

For vx: vx=\frac{\gamma\left(dx^{\prime}+udt^{\prime}\right)}{\gamma\left(dt^{\prime}+\frac{udx^{\prime}}{c^2}\right)}=\frac{\left(\frac{dx^{\prime}}{dt^{\prime}}+u\right)}{\left(1+\frac{u}{c^2}\frac{dx^{\prime}}{dt^{\prime}}\right)} , vx^{\prime}=\frac{dx^{\prime}}{dt^{\prime}}

vx=\frac{vx^{\prime}+u}{1+\frac{uvx^{\prime}}{c^2}}

For vy: vy=\frac{dy^{\prime}}{\gamma\left(dt^{\prime}+\frac{udx^{\prime}}{c^2}\right)}=\frac{\frac{dy^{\prime}}{dt^{\prime}}}{\gamma\left(1+\frac{u}{c^2}\frac{dx^{\prime}}{dt^{\prime}}\right)}

vy=\frac{vy^{\prime}}{\gamma\left(1+\frac{uvx^{\prime}}{c^2}\right)}

Note: be careful with sign of u, and also remember u = relative speed of S’ as observed in S, NOT speed of object (that is v duh)

Relativity: 4D position vectors of Spacetime

• Spacetime is 4-dimensional: 3 direction dimensions and 1 time dimension. This gives us the equation for a position vector in spacetime:

X=\left(ct,x,y,z\right)

Note: c is included so that all dimensions have the same units. We can do this as c is invariant in spacetime so is the same always.

Relativity: The Spacetime interval - What is it?

• Defined as \Delta S where \left(\Delta S\right)^2=c^2\left(\Delta t\right)^2-\left(\Delta x\right)^2 and \left(\Delta S\right)^2=\left(\Delta S^{\prime}\right)^2.

• This last bit can be easily proven using Lorentz transforms and shows that it is invariant, i.e it is the same in ALL inertial frames, like the speed of light!

• The first bit is very interesting as it looks a lot like pythagoras but with a -ve? What does this suggest about \Delta S? To find out, we look at vectors of spacetime:

Consider 2 events:

Event 1: X1 =\left(ct1,x1\right)

Event 2: X2 =\left(ct2,x2\right)

(note that these events have y and z = 0, hence why they are not included)

Plot these events on the same spacetime diagram in the same inertial frame (see diagram). We can see:

\Delta X=X2-X1=\left(c\left(t2-t1\right),\left(x2-x1\right)\right)=\left(c\Delta t,\Delta x\right)

We know from maths that the length of a vector can be found by the dot product of itself square rooted (in 3D, this is just pythag! however, in 4D dot product works a bit differently..)

Remembering that \left(\Delta S\right)^2=\left(c\Delta t\right)^2-\left(\Delta x\right)^2 , let’s define the dot product in 4D as the same in 3D, but with a minus sign instead of a plus sign (just cause why not).

This gives us:

\Delta X\cdot\Delta X=\left(c\Delta t\right)^2-\left(\Delta x\right)^2=\left(\Delta S\right)^2

This suggests that \Delta S is the length in 4D spacetime of \Delta X, referred to as the interval between two events in spacetime (Event 1 and Event 2 in this case). The fact that \Delta S is invariant means that the interval (between two events in 4D spacetime) is the same in ALL inertial frames!! Obviously, Lorentz transforms show us that \Delta x and \Delta t do not equal \Delta x^{\prime} and \Delta t^{\prime}, but the interval (between these two events) does not change, no matter what reference frame you are observing it from!!!!!!!!

Relativity: Deriving time dilation using the spacetime interval (\Delta S)

• Imagine clock stationary in S’, S’ is moving with speed u relative to S.

• In S’, the spacetime interval between ticks of clock is \left(\Delta S^{\prime}\right)^2=\left(c\Delta t^{\prime}\right)^2-0 (as clock is stationary so \Delta x^{\prime}=0)

• In S, the spacetime interval is \left(\Delta S\right)^2=\left(c\Delta t\right)^2-\left(\Delta x\right)^2 and \Delta x=u\Delta t

• Spacetime interval is invariant, \left(\Delta S\right)^2=\left(\Delta S^{\prime}\right)^2, therefore:

\left(c\Delta t^{\prime}\right)^2=\left(c\Delta t\right)^2-\left(u\Delta t\right)^2

\Delta t^{\prime}=\sqrt{1-\frac{u^2}{c^2}}\Delta t

\Delta t=\gamma\Delta t^{\prime} (where \gamma=\sqrt{\frac{1}{1-\frac{u^2}{c^2}}} as per usual)