Special Relativity - Lorentz Transformations And Spacetime Flashcards
Deriving the values of A,B,C,D and fundamental effects using postulates, working out how we transform space and time in multiple reference frames.
Dropping deltas.
R.F S’ moving at constant v along x axis w.r.t frame S.
x = Ax’ + Bt’
t = Ct’ + Dx’
y = y’
z = z’
x = y(x’ + vt’)
x’ = y(x - vt)
t = y(t’ + vx’/c^2)
t’ = y(t - vx/c^2)
Think about an event that is synchronous in S but not in S’.
(Dx,Dt) = (Dx,0)
Feed these into transformations and you find length contraction and leading clockwise lag idea.
Think about two events that both occur in S’ such as successive clock ticks.
(Dx’,Dt’) = (0,Dt’)
Here, we directly see time dilation spat out at us. In both separate occasion we see opposing time dilation equations, they aren’t related though, as they consider scenarios where either frame is stationary in separate cases.
Finally to length. Consider a stick at rest in S’, length : l’. Simultaneously measured in S, space time separation (x,0). We then obtain x = x’/y. We see the same if the stick is stationary in frame S.
So we see the same equation for both scenarios? No, because in one t = 0 and the other t’ = 0, these are different, if they were the same, then v = 0 and Lorentz = 1. Not very interesting.
The 4 facts to determine the Lorentz coefficients?
- Physical setup - S’ moves at ‘v’ w.r.t S.
- The Principle of Relativity - SR1, S see’s things in S’ in the same way vice versa. Only difference is the sign of ‘v’.
- The speed of light Postulate - c is the same everywhere and to everyone.
- S moves at ‘-v’ w.r.t S’.
An object moves a ‘u’ in S, with S’ moving away at ‘v’. How do we find u’ (the objects speed in S’)?
First of all, physically, u’ should be smaller than u. Say event one occurs in origin for both S and S’ such that (0,0) = (0,0), and then event two, (x,t) and (x’,t’). Remember u’ = x’/t’.
Let’s divide the two transformations. u’ = (u - v)/(1 - uv/c^2)
State the invariance equation and how it relates to different variables in different frames.
s^2=c^2(t^2) - (x)^2
c^2(t^2)-(x^2) = c^2(t’^2)-(x’^2)
s: the spacetime interval in
This can be proven by substituting the Lorentz transformations in there.
For a photon. c=x/t, so s*2=0.
Given any events where s*2=b we can work it out for either the as or S’ frame.
With events separated by coordinates (Dx,cDt), this is known as the spacetime separation.
ct is another dimension, r*2 is an example of an invariant in the x-y plane.
s*2 is the rotational symmetry in 4d space, the Poincaré group of spacetime.
Run through the different types of spacetime separations.
Light-like separated: s^2=0 , x/t = c, this is frame independent due to s^2, and reiterates SRP2.
Impossible to find an S’ where events happen at the same place or time as it would need to travel at v=c.
Time-like separated: s^2>0 , x/t<c , makes sense. Consider Lorentz for x’, if x/t<c then there exists a v that makes x’ = 0.
Here it is possible for events to happen at the same place. (Travel from one to another)
Space-like separation: s*2<0 , x/t>c , here there is an x/t that exists to make t’ = 0. It is possible to find a frame S’ where events happen at the same time but not position. The absolute value of ‘s’, is then the distance x’ between events in the frame where events happen at the same time.
Deriving time dilation from invariance.
Say two events occur in the same position in S’ a time t’ apart. And the frame travels vt w.r.t S frame. Separations for each are S’ = (0,t’), S = (vt,t)
Now:
c2(t’2) - 0 = c2(t2) - (vt)2
t’2 = t2(c2-v2)/c2
t = yt’
Minkowski Diagrams for multiple reference frames.
A frame S’ is going to have its x’ and ct’ axis tilted inwards, still keeping the speed of light line symmetrical.
Event (x’,ct’) = (0,1)
For x=y(x’-vt’)
And t=y(t’-vx’/c*2)
(x,ct) = (yv/c,y) = (yB,y)
Now, imagine the graph, the angle between the axis is tan(O)=B
And the point mentioned has a distance to the origin of : y(sqrt(1+B^2)), so one ct’ unit/ one ct unit = sqrt((1+B^2)/(1-B*2))
This must always be greater than or equal to 1. This was for ct’ axis.
For x’ axis, everything is the opposite but the ratio and the angle is the exact same.
This shows that each axis is stretched and titled by the same factor.
Loss of simultaneity.
Seen through Minkowski, tilted simultaneous event line in S’ and horizontal line in S shows the lack of geometric equivalence. Events simultaneous in one frame are not in another.
When looking at a space-time interval in S of ct, and the one in the S’ frame, ct’ is longer from the S frame perspective, hence t=yto, and time dilates.
Supplemental: light below the ct axis within the fan can reach and influence the origin (the past). Light events above the ct axis in the fan can affect each other if the gradient between the two events is -1<m<1.
When analysing the twins paradox on the MD. We can see using equally spaced signals, that due to the slanted simultaneity line in S’, less signals are sent out compared to in vertical S, hence time dilation.
Loss of simultaneity.
Seen through Minkowski, tilted simultaneous event line in S’ and horizontal line in S shows the lack of geometric equivalence. Events simultaneous in one frame are not in another.
When looking at a space-time interval in S of ct, and the one in the S’ frame, ct’ is longer from the S frame perspective, hence t=yto, and time dilates.
Supplemental: light below the ct axis within the fan can reach and influence the origin (the past). Light events above the ct axis in the fan can affect each other if the gradient between the two events is -c<m<c.
When analysing the twins paradox on the MD. We can see using equally spaced signals, that due to the slanted simultaneity line in S’, less signals are sent out compared to in vertical S, hence time dilation.
Relativistic Doppler?
f’ = fs((v+vd)/(v-vs))
Imagine source as frame S and observer moving as S’. Time until next wavefront reaches observer is t = h/(c-v). In observer frame, t’ = t/y for next wavefront.
t= fh/f(c-v)
t= c/f(c-v)
t= 1/f(1-B)
f’=1/t’ = y/t = yf(1-B)
y= 1/sqrt(1-B2)= 1/sqrt(1-B)(1+B)
f’ = fsqrt(1-B)(1-B)/sqrt(1-B)(1+B)
f’ = f*sqrt((1-B)/(1+B))
Expanding spacetime on large scales causes redshift not Doppler.
Seeing green as blue?
Green = 530 nm
Blue = 470 nm
require wavelength to be a factor 470/530 = 0.887 shorter.
So frequency must be 1/0.887 =1.128 larger
(1-B)/(1+B)=1.272
B = -0.12
v = 12% of c
