Monday, June 10, 2013

Relativistic Doppler Effect

One of the assumption of Special Relativity is that the velocity of light is constant in every inertial frame of reference. Consider two frames: one at rest (t,x,y,z); and a second one (t',x',y',z') is in motion with velocity V with respect to the first one. We've omitted the z- and z'-axis as representation of 4 dimensions on a 2-D surface is not possible.



Now suppose a light was turn on, and our two observers would be looking at this beam of light spreading throughout space as a sphere. (See fig above)

In the rest frame we have:

(1) x2 + y2 + z2 = (ct)2

Similarly in the moving frame:

(2) x'2 + y'2 + z'2 = (ct')2

Rewriting both equations (1) and (2) as:

(3) -(ct)2 + x2 + y2 + z2 = 0

(4) -(ct')2 + x'2 + y'2 + z'2 = 0

Since both expressions are equal to zero, they are equal to each other. But more fundamental, these two expressions are suggesting that they represent the same fundamental quantity in two different frames. That there are equal means something doesn't change from one frame to the other. Let's represent this quantity as:

(5) s2 = -(ct)2 + x2 + y2 + z2

= -(ct')2 + x'2 + y'2 + z'2

To keep in line with convention, we will express these quantities as small differences. We use the symbol d, which is common use in calculus. We also set c =1, as it is a constant and we can always put it back if we need to do a calculation.

(6) ds2 = -dt2 + dx2 + dy2 + dz2

We define the metric as the coefficient of each of the terms in the above:

(7) η00 = -1, η11 = 1,η22 = 1,η33 = 1,and ηij = 0 for i≠j

We also define the proper time τ as,

(8) dτ2 = - ds2

= dt2 - (dx2 + dy2 + dz2)

= dt2 ( 1 - (dx2/dt2 + dy2/dt2 + dz2/dt2))

= dt2 ( 1 - v2)

Where we have use the definition of velocity,
v = (dx/dt,dy/dt,dz/dt).

Taking the square root on both sides,

(9) dτ = dt/γ

(10) where γ = (1 - v2)

Now, in Newtonian physics, it was assumed that time was different from space. But in Relativity, this separation cannot be upheld any longer. We can see in the definition of the proper time, that both time and space form one manifold. So we need to redefine our quantities with this new perspective.

Convention: we use latin indices for 3-dimensional objects. For instance, the velocity, vi, where i = 1,2,3. So a velocity's components which were written as v = (vx,vy,vz), now will be written as v = (v1,v2,v3).

We use Greek letters (α,β,γ,δ...) to denote objects with 4 components. They will take values 0,1,2,3 for t,x,y,z. So, for example, uβ =(u0,v1,v2,v3). Sometimes, we want to break up the components as temporal and spatial. So we write,uβ =(u0,vi), where it is understood, i = 1,2,3, or uβ =(u0,v)

We will measure the velocity with respect to the proper time τ, not the ordinary time t.

(11) uβ =dxβ/dτ

Using the chain rule and equation (10),

(12) uβ = (dt/dτ)dxβ/dt = γ(dt/dt,dx1/dt,dx2/dt,dx3/dt) = γ(1,v1,v2,v3) = γ(1,v)= (γ,γv) .

The dot product between two vectors is now defined with the metric (see equation (7)),

(13) u2 = u•u = ηαβuαuβ

Note that in u2, the 2 means squaring, u2 = u squared, not the 2nd component of u. When there's confusion, we will point out what is meant by an upper index.

Expanding the above into the temporal and spatial components, and using the metric in (7),

(14) u•u = η00u0u0 + ηijuiuj
= (-1)γ2 + γ2v2 = (-1)γ2 ( 1 - v2)

But from (10),

(15) γ2 = (1 - v2)-1

Therefore,

(16) u2 = u•u = -1

Energy and Momentum

Momentum is defined as mass x velocity,

(17) pβ = muβ

Similarly, we define a 4-vector momentum as,

(18) pβ =(p0,pi) = (p0,p)

An important result is to calculate p2, where the 2 means squaring, not the component 2. First using equation (17)

(19)p2 = muβmuβ = m2u2 = - m2

Then using equation (18) and the metric in (7),

(20) p2 = p•p = η00p0p0 + ηijpipj
= (-1)(p0)2 + (p)2

We define p0 = E/c = E , (using the convention, c=1). Putting this altogether, we get,

(21) p2 = - m2 = (-1)E2 + (p)2

Or,

(22) E2 = m2 + (p)2.

Putting c into the equation,

(23) E2 = m2c4 + p2c2.

Notice when the particle is at rest, p = 0, and we get, E = mc2.

Relativistic Doppler Effect

In the last blog, Einstein's Derivation of the Famous Equation, E=mc2 , we were given the energy of the photon as,

(24) γB+ = ½ E(1 + (V/c)cosΦ)(1 – V2/c2) for the incoming photon

γB- = ½ E(1 - (V/c)cosΦ)(1 – V2/c2) for the outgoing photon



The energy is,

(25) Etotal = - pβ uβ = - (η00p0u0 + ηijpiuj)

= - (-1)Eγ - (pcosθ, psinθ, 0)(-γV,0,0)

= Eγ + (pcosθ)γV = Eγ(1 + (V/c)cosθ),

Where the last step , we use p = E/c.

Each photon released will carry half the energy in opposite direction. For the incoming photon,

(26) γB+ = ½ Etotal = ½ E γ(1 + (V/c)cosΦ)
= ½ E(1 + (V/c)cosΦ)(1 – V2/c2)

Where we use equation (10) in the last step.

For the outgoing photon, we get an extra minus sign in the momentum,



(27) pi = (-pcosθ, -psinθ, 0)

Giving,

(28)γB- = ½ E(1 - (V/c)cosΦ)(1 – V2/c2)

Notice that for the energy of the incoming photon is greater than the energy of the outgoing photon. This is the Doppler Effect, as a light coming towards you will appear blueshifted, while a photon moving away from you will be redshifted.



Appendix

From here on, the speed of light is c.

We will use the following conventions:

(A1) β = v/c

(A2) γ = (1 - v2/c2) = (1 - β2)

For a wave, it is proportional to ei(k∙x - ωt),where,

(A3) k = |k| = 2π/λ, ω = 2πf, and ω = kc

Just like we can define a 4-vector position, xμ = (ct,xi), we can also define a 4-vector wavenumber, kμ = (ω/c,ki). The phase factor, (k∙x - ωt), can now be written as,

(A4) k∙x - ωt = ημνxμkν, where ημν is the Minkowski metric tensor with signature (-1,1,1,1).

The importance of the phase factor, which basically counts the number of peaks and troughs of the wave, must be frame-independent, that is, a Lorentz scalar.

(A5) kμ → k'μ = Lμνkν

Where Lμν is the Lorentz transformation (See diagram below).



In particular, under a Lorentz boost in the +x direction,

(A6) k'x = γ(kx - βω/c)

(A7) ω'x = γ(ω - βckx)
= γ(ω - βckcosθ)


Where θ is the angle between the boost direction in the +x direction and the direction of the wave propagation k.

Using ω = kc (equ. A3), and γ = (1 - β2) (equ. A2) then equation A7 becomes,

(A8) ω' = γ(ω - βckcosθ) = ω(1 - βcosθ)(1 - β2)

Note: if θ = π/2, then ω' = 0. There is no Doppler shift in the transverse direction.
if θ = 0, then ω'/ω = [(1- β)/(1+β)]½
For low velocity, v < c, then

(A9) ω'/ω ≈ (1- β)

Or,

(A10) Δω'/ω ≈ -v/c

Saturday, June 01, 2013

Einstein's Derivation of the Famous Equation, E=mc2

Here’s a quick rundown on one of the most famous equation in physics, E = mc2. Einstein knew from experiments previously done that a particle could decay and release gamma rays. He reasoned that when this happened, the particle would lose kinetic energy, and this could only be accounted by a loss of mass. So how did he come to that conclusion? He analysed the situation both in a rest frame and in a moving frame.



The value of γ was already known, but we can derive it from Relativistic Doppler Effect (to be the topic for another blog A).

By the law of conservation of energy:

The energy before decay = the energy after decay

(1) In the rest frame:

A0 + KA0 = A1 + KA1 + ½ E + ½E = A1 + KA1 + E

(2) In the moving frame:
B0 + KB0 = B1 + KB1 + ½ E(1 – (v/c)cosΦ)(1 – v2/c2)
+ ½ E(1 + (v/c)cosΦ)(1 – v2/c2)

= B1 + KB1 + E(1 – v2 /c2)

Now taking a look at the energy difference of the particle in each of the frame:

(3) In the rest frame:

A1 – A0 = KA0 – E – KA1 = – ΔKA – E

(4) In the moving frame:

B1 – B0 = KB0 – KB1 – E(1 – v2 /c2)
= – ΔKB – E(1 – v2 /c2)

Whether the observer is at rest or moving with respect to the particle, the energy difference should be the same.

(5) – ΔKA – E = – ΔKB – E(1 – v2 /c2)

Calculating the difference in kinetic energy:

((6) ΔK = ΔKA – ΔKB = E(1 – v2 /c2) – E

= E ((1 – v2/c2) – 1)

≈ E ((1 + ½v2/c2) – 1)

= ½ E (v2/c2)

Einstein had reasoned that if the kinetic energy of the particle is smaller by ½ E (v2/c2), the only way this can happen is that the particle must lose mass when emitting radiation.

By definition the kinetic energy is,

(7) K = ½ mv2

If the particle was initially at rest,

(8) ΔK = K = ½ mv2

Equating (6) and (8)

½ mv2 = ½ E (v2/c2)

(9) Therefore E = mc2.