Lorentz Transformation
As in the previous lecture about time dilation, consider two inertial reference frames:
frame α with Cartesian coordinates X, Y, Z and time T
and frame β with Cartesian coordinates x, y, z and time t.
So, all capital coordinates (X, Y, Z) and time (T) will be related to α-frame, events and experiments occurring in it will be referred to as α-events and α-experiments, and observer at rest in it will be referred to as α-observer.
Similarly, all lower case coordinates (x, y, z) and time (t) will be related to β-frame, β-events, β-experiments and β-observer.
Assume that at time T=t=0 both frames coincide, that is their origins are at the same point in space and their corresponding axes overlap, but frame β is moving relative to α along the X-axis with constant speed v, so the velocity vector of β-frame in α-coordinates has components {v,0,0}.
Since the movement is relative, we can say at the same time that frame α is moving relative to β along its x-axis with speed −v, so the velocity vector of α-frame in β-coordinates has components {−v,0,0}.
We still rely on two main principles:
(a) Principle of Relativity that states that all the Physics laws must be the same, if expressed quantitatively, using corresponding coordinates and time, in all inertial reference frames.
(b) Speed of light is constant and the same in all inertial reference frames in empty space, we will use a symbol c for this universal constant.
Our current task is purely mathematical - express the transformation of space and time coordinates from one inertial reference frame to another that moves relative to the first, preserving the integrity of the above principles.
At this point we would like to remind that this type of task is not unusual in Math.
One of the elementary tasks in coordinate geometry was to transform the Cartesian coordinates if XY-axes are rotated by some angle around the origin.
In this case the transformation from coordinate system {X,Y} to {x,y}, where axes are rotated by angle φ around origin is
x = X·cos(φ) − Y·sin(φ)
y = X·sin(φ) + Y·cos(φ)
Consider now our case of the transformation of space and time coordinates between two inertial reference frames moving relatively to each other as described above:
α-frame {X,T} and
β-frame {x,t} that uniformly moves relatively to α-frame along the X-axis with constant speed v, preserving the parallelism of corresponding axes, provided at time T=t=0 both reference frames coincide.
Since the direction of β-frame movement is along X-axis, coordinates y and z will be the same as, correspondingly, Y and Z. This simplifies our task, reducing it to transformation of only two coordinates - X in space and T in time into x and t.
We will look for a simple linear transformation from {X,T} system to {x,t} system of the form
x = p·X + q·T
t = r·X + s·T
where p, q, r and s are four unknown coefficients of transformation, which we are going to determine.
We should not add any constants into above transformations since coordinates (X=0,T=0) should transform into (x=0,t=0).
Equation 1
Since β-frame moves along X-axis of α-frame with speed v, its origin of space coordinate (point x=0) must at any moment of α-time T be on a distance v·T from the origin of coordinates of a stationary α-frame.
Hence, if X=v·T then x=0 for any T.
From this and the first transformation equation
we derive:
0 = p·v·T + q·T or
0 = (p·v + q)·T.
Since this equality is true for any time moment T,
p·v + q = 0
and, unconditionally,
q = −p·v
This is the first equation for our unknown coefficients.Equation 2
Consider a ray of light issued at time T=t=0 from the point of coinciding origins of both reference frames in the direction of positive coordinates X and x.
Since the speed of light c is the same in both systems {X,T} and {x,t}, according to Principle of Relativity, an equation of the motion of the front of the light wave in the α-frame must be X=c·T and in the β-frame it is x=c·t.
Therefore, if X=c·T, then x=c·t.
Put X=c·T into both equations of transformation of coordinates.
We get
Substitute these expressions into x=c·t:
p·c·T + q·T = r·c²·T + s·c·T.
Reduce by T,
p·c + q = r·c² + s·c
This is the second equation for unknown coefficients.Equation 3
Consider a ray of light issued at time T=t=0 from the point of coinciding origins of both reference frames in the direction of negative coordinates X and x.
Repeat the logic of a previous paragraph for the light moving now in the opposite direction with a speed −c.
Therefore, if
Put X=−c·T into both equations of transformation of coordinates.
We get
Substitute these expressions into x=−c·t:
−p·c·T + q·T = r·c²·T − s·c·T.
Reduce by T,
−p·c + q = r·c² − s·c
System of Equations
So, this is the system of three linear equations for four unknown coefficients of transformation p, q, r, s:
(a) q = −p·v
(b) p·c + q = r·c² + s·c
(c) −p·c + q = r·c² − s·c
Usually, we need four equations for four unknown, but additional considerations will help to resolve this situation.
Solution
From (b) and (c), adding and subtracting these equations, we get:
2q = 2r·c², therefore q = r·c²
2p·c = 2s·c, therefore p = s
Now the system of equations is
q = −p·v
q = r·c²
p = s
Let's express q, r and s in terms of only one unknown p and known constants c and v.
q = −p·v
r = q/c² = −p·v/c²
s = p
The original system of equations now looks like
x = p·X − p·v·T
t = −p·(v/c²)·X + p·T
In a matrix form the transformation from α-frame {X,T} to β-frame {x,t} looks like
| x |
| t |
| = |
| p | −p·v |
| −p·v/c² | p |
| · |
| X |
| T |
Symmetrically, the motion of system {X,T} relative to system {x,t} is a uniform motion with speed −v along x-axis.
That means that expression of {X,T} coordinates in terms of {x,t} should look similar to above with the only difference of the sign of the speed v
X = p·x + p·v·t
T = p·(v/c²)·x + p·t
In a matrix form the transformation from β-frame {x,t} to α-frame {X,T} looks like
| X |
| T |
| = |
| p | p·v |
| p·v/c² | p |
| · |
| x |
| t |
Since the latter matrix of transformation is an inverse of the one that represents transformation from α-frame {X,T} to β-frame {x,t}, the product of these two matrices must produce a unit matrix.
| 1 | 0 |
| 0 | 1 |
Therefore, multiplying two matrices of transformation, we should have satisfied the following equations for all four elements of the unit matrix
p² − p²·v²/c² = 1
p²·v − p²·v = 0
−p²·v²/c² + p²·v²/c² = 0
−p²·v²/c² + p² = 1
The first equation is the same as the fourth and produce the value for coefficient p
p²·(1 − v²/c²) = 1 and
p = 1/√1−v²/c² .
The second and the third equations above are identities that we can disregard.
Consequently, the values for other coefficients of transformation of coordinates are
q = −p·v = −v/√1−v²/c²
r = −p·v/c² = (−v/c²)/√1−v²/c²
s = p = 1/√1−v²/c²
Traditionally, factor v/c is replaced with Greek letter β, which results in formulas:
x = (1/√1−β² )·X +
+ (−v/√1−β² )·T
t = ((−v/c²)/√1−β² )·X +
+ (1/√1−β² )·T
One more simplification is usually done by introducing Lorentz factor γ equaled to
x = γX − γv·T = γ(X − v·T)
t =−γv·X/c²+γT = γ(T−v·X/c²)
The final form of Lorentz transformation from inertial α-frame coordinates {X,T} to inertial β-frame coordinates {x,t}, when β-frame is uniformly moving with speed v along α's X-axis, in the Special Theory of Relativity is:
| x = |
|
| t = |
|
The final transformation looks exactly like in Einstein's article On the Electrodynamics of Moving Bodies (1905), but the derivation is mathematically rigorous based on two postulates - Principle of Relativity and the universality of the constant speed of light in all inertial reference frames in vacuum.
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