Lorentz transformation is the transformation of spatial and temporal co-ordinates(time) from one inertial frame of reference to another which is moving at a uniform velocity V with reference to the other. Lorentz transformation is a special case of Poincare transformation w.r.t. linear uniform motion & rotation of co-ordinates.
* Suppose one observer is in frame S and the other in frame S', the co-ordinate system being Cartesian in Euclidean space.
* Clocks of both frames are synchronized at t=0 i.e. when both the frames coincide.
* S' is moving at a uniform velocity V in the direction of positive x-axis w.r.t. S.We take V as +Ve for motion along positive x/y/z direction. For simplicity, we assume that velocity has no component in y & z direction. Vx= V and angle Ψ =0°, angle Φ =90°, angle θ =90°. However in general it can have Vy , Vz components.
* An event is recorded in frame S'. Its co-ordinates are (x', y', z', t').Co-ordinates of the origin of S' are s'(x), s'(y), s'(z) and s'(t)
* If the same event is recorded in S and is having co-ordinates (x, y, z, t) , what are the transformation laws between S and S' for x,y,z,t ?
* The transformation law that measures (x',y',z', t') in terms of (x, y, z, t) or Vice Versa between two inertial frames of reference is called the Lorentz transformation.
* Had this been Galilean transformation ,it would have been
x'= x±vt; ( +vt when v is -Ve and -vt when V is positive )
* The essence of Lorentz transformation lies in treatment of time. In Galilean transformation, events are simultaneous in both the frames and so t'=t.. Not so in Lorentz transformation because event can be perceived only when the signal (light or any communication signal) reaches the observer and the time of propagation of signal to each observer is not instantaneous but is a function of the distance between the observer and the event and hence in general time =f(x,y,z) for each observer. So time t' of an event perceived in S' frame is not the same as time t perceived in frame S but is a function of x, y, z and t.
In general x' =A11x +A12y+A13z+A14t ;
y' =A21x +A22y+A23z+A24t ;
z' =A31x +A32y+A33z+A34t ;
t' =A41x +A42y+A43z+A44t ;
i.e. Each co-ordinate can be expressed as the linear combination of all the co-ordinates in other frame. The combination is assumed to be linear because if it is in any other power of x,y,z or t,then x1'-x' which is the length of an event would be different in different frames which is not tenable. Hence vector representation is maintained.
Since there is no projection of y-axis on x-axis or other axes by the Orthogonality condition of the axes, A11,A22 , A33 ,A44 are not equal to zero and all other co-efficientare zero. Aij =0 for all i ≠j except A24 ,A34 , A42 & A43 i.e. co-efficients associated with t.
x' =A11x +A14t ;
y' =A22y+A24t ;
z' =A33z+A34t ;
t' =A41x +A42y+A43z+A44t ;
In the instant case, since we have restricted movement of S' frame in x-axis, A22= A33 =1 & A24= A42 =A43= A34=0
x' =A11x +A14t
y' =y ;
t' =A41x +A44t
Again x'=x-vt which implies that A14=-A11v ( this part has to be more elaborately explained?)
x' =A11(x -vt) .......(1)
y' =y ;
t' =A41x +A44t ......(2)
Since the event can be recorded only on communication of the signal, if velocity of signal is c ,then
x2+y2+z2=c2t2 ...... or x2+y2+z2 - c2t2 =0 (3)
x'2+y'2+z'2=c2t'2 ...... or x'2+y'2+z'2 - c2t'2 =0 (4)
= x2+y2+z2 - c2t2
or x'2- c2t'2
= x2 - c2t2 ..........(5)
Substituting the value of x' and t' from (1) & (2) respectively in (5) and rearranging we,get,
x2(A211 -A241c2) + t2( A211v2-A244c2) -2xt ( A211v+A41A44c2) = x2 - c2t2 ..........(6)
Comparing the co-efficients of x2,t2 and xt , we get
A211 -A241c2 =1 ........ (7)
A211v2-A244c2 = -c2.......(8)
A211v+A41A44c2 = 0 ...... (9)
There are 3 equations and 3 unknowns. Solving them we have
A11 = 1/√(1-v2/c2) ..............(10)
A41 = (- v/c2)* (1/√(1-v2/c2)) ..............(11)
A44 = 1/√(1-v2/c2) ..............(12)
Putting them in (1) and (2),
x' =1/√(1-v2/c2) *(x -vt) .......(1)
y' =y ;
t' =(t -vx/c2)*1/√(1-v2/c2) ......(2)
When v has x,y ,z component, A41 =- ((vx x+ vy y + vz z)/c2)* (1/√(1-v2/c2))
Some Important Observations on Lorentz Transformation
1) γ is known as the Lorentz factor. γ =f(v). When v=0, γ =1
when v-->c, γ--> infinity without bound
1/γ is called reciprocal Lorentz Factor.
2) Elements of Lorentz transformation form a Group. Basically 4 conditions are to be satisfied for a set of objects to form a group. These are
a) If A, B are elements of the group, then C=A. B must also be an element of the group where . is a binary operation.
b) If A,B,C are elements of the group, then A(BC) =(AB)C must be true which is associative law.
c) There exists an identity element I such that AI=IA=A
d) Every element has an inverse which is also an element of the group.
Lorentz transformation obeys all these. The group contains 3 matrices, identity,forward,inverse transformation matrix.
3) Lorentz transformation occurs in Pseudo-Euclidean space because in x2+y2+z2 - c2t2 the last term is negative in stead of being positive and the same term is dependent upon space co-ordinates unlike in Euclidean space where all terms (x,y,z) are independent of each other. Here space, time are combined into a single manifold.
4) Next we consider the addition of velocities under Lorentz transformation. Suppose S-frame is the ground, S'-frame is the train moving with a uniform speed v w.r.t. ground in +ve x-direction ( for simplification) and both the frames coincided at t=0. A passenger moves with a uniform speed v' w.r.t. train in positive x-direction. What is the velocity of the passenger with respect to the ground?
x' =v't' = v'[t-vx/c2 ] /√(1-v2/c2)
but x' = (x -vt) / √(1-v2/c2)
Equating v'[t-vx/c2 ] /√(1-v2/c2) = (x -vt) / √(1-v2/c2)
or x = (v+v') t /(1+vv'/c2 )
But x=Vrel t
So Vrel = (v+v') /(1+vv'/c2 ) .......(13)
Had it been Galilean transformation, Vrel = (v+v')
We observe here that relativistic addition of velocities is not a mere addition of individual velocities.
Can we find some term which is additive of 2 other terms under Lorentz transformation & each term contains velocity and therefore can be aptly termed as a relativistic analog of velocities.
Now we recast equ (13). Vrel / c = ( v/c + v' / c) /(1+vv'/c2 ) ....(14)
The above expression is similar in structure to the following--
tan(A+B) = (tanA + tanB) / (1-tanAtanB) ....(15)
tanh(A+B) = (tanhA +tanhB)/(1+tanhA*tanhB) ....(16)
(14) looks analogous to (16).
So we put v/c =tanhA ; v'/c = tanhB and moreover tanhA or tanhB can vary between [-1,1] and v/c can also vary between [-1,1] as v can be between -c to c. Hence the above analogy is quite appropriate.
Vrel / c = (tanhA +tanhB)/(1+tanhA*tanhB) = tanh(A+B)
Or if A =tanh-1 ( v/c) (we call it rapidity of S' w.r.t. S)
B =tanh-1 ( v'/c) (we call it rapidity of passenger w.r.t. S')
then A +B = tanh-1 (Vrel /c) (we call it rapidity of passenger w.r.t. S)
Now if we define A,B as rapidities, then resultant rapidity is additive like Galilean velocities. Hence under Lorentz transformation, the counterpart of velocity is rapidity.
We conceptualize a term rapidity and define the Rapidity A,B is an hyperbolic angle that differentiates a moving frame of reference from a fixed frame of reference. At low speeds, U--> v/c
Since V is in the interval [-c,c]
β is in the interval [-1,1]
A= tanh-1 (v/c) or tanhA= v/c
tanh maps the real line from minus infinity to plus infinity onto the interval [-1,1].
Hyperbolic functions are analogs of the ordinary trigonometric or circular functions. As cosθ and sinθ form a circle with unit radius, coshθ and sinhθ form the right half of a equilateral or unit hyperbola i.e. cosh2θ - sinh2θ =1( of the form x2-y2=1 in Cartesian plane) and
coshθ =(ex+e-x)/2 and sinhθ =(ex - e-x)/2
Hyperbolic tangent addition, tanh(a+b)= (tanha +tanhb)/(1+tanha*tanhb)
tanh(a-b)= (tanha - tanhb)/(1-tanha*tanhb)
For hyperbolas, click 1,2,3 ,4
For quadratic equn, click
See also Lorentz - 2,3