THE    GEOMETRY OF THE SPECIAL  RELATIVITY

 

This text is for people that knows the elements of the special relativity.

 

In the formulas of the special relativity, often  it   appears  expressions like:

v / c and             being c the light velocity and v the speed of the inertial system with respect the reference frame.

For instance, in  the Lorentz transformation  tensor :

 

               being      a= 1/  and    b = - a v/c

 

Remembering  trigonometry, it is possible to identify both  expressions with  functions in the way:  v/c = sin w    and therefore   (1- v²/c²) ^1/2 = cos w.  Since in relativity   must be  v £ c ,  v/c varies  between  0 and 1, the same that  sin w.

Then   a and b are transformed to

 

a= 1/cos w   y  b = - a v/c = -  sen w / cos w   = - tg w

 

 Now, the coordinates  transformation  with the Lorentz tensor is as follows:

 

 

being x’’,y’’,z’’,t’’ the coordinates of a point in the inertial system at velocity  v  relative to the reference frame and x’,y’,z’,t’  that coordinates in the reference frame .

Multiplying by  cosw both members of the equation:

 

 

       

 This pattern of tensor is projecting the coordinates of the inertial  system over an axis that forms with it an angle w. On will return on this item  later.

 

Developping the tensor we get:

 x’’cos w = x’-ct’sinw      y’’= y’         z’’= z’   ct’’cos w = ct’ – x’sinw

that are the same that the classic transformation formulas.. We can see the simplification on notation:

                         y’’=y’              z’’=z’              

 

The expressions of the  length  contraction  and time dilation are:

 

l = l₀ cos w  and    t  = t₀ cos w         that are projections over an axis at angle w.  And for the mass:         

m= m₀ /cos w       At speed c ,  cos w=0  and the mass observed  from the reference frame becames ¥.

 

For the  determination of speeds composition and in the cases  when differentiation is necessary, the calculations are  simplified using the trigonometric functions.

The speed’s composition if v is parallel  to the x’ axis (v_y =0, v_z =0) should be, differenciating  respect to t’’:

 

 

Being u’’ the speed of a point P with respect the inertial frame x’’- y’’ which is moving at speed v with respect the reference frame x’-y’. We look for the speed u’ that an observer standing at the reference frame sees for P, or that is the same, the adition of v and u’’. . In Galilean geometry  u’=u’’+v. But not in relativity.

From the last row  doing u’=dx’/dt’  and u’’ = dx’’/dt’’

c cos w = (- u' sin  w +c) dt'/dt".    

Dividing by c  and clearing :

dt'/dt" = cosw /(1- u'/c.senw)   but u'/c  as all the speeds can be  expressed in functión  of a sinus.   Let be  u'/c =sen a   (a  is not the constat of the Lorentz tensor now ).Then:

dt'/dt" = cos w/(1-sen a.sen w).  Doing the sustitution of dt’/dt’’,  dividing both terms  by cos w and passing the denominator  to the first member:

 

 

doing now u"/c = sen b  and   dividing by c in both members we get:

 

 

 And now, developping the firs line:

sen b = (sen a  - sen w) / (1 - sen w.sen a)    (1)    or as we look for sen a:

sen a= (sen w + sen b) / (1 + sen w.sen b )   (2)    or

 

sen w.sen b sena  =sen w + sen b - sen a     

 

This is an interesting expression for the velocities composition.
Doing the substitution of  the sinus by the speeds in (2) we get:

    and multiplying by c :            that is the classical presentation of the speeds composition.  In apendix we give a graphical way for the composition based in the use of  our method.

 

Going back to the beguining of the text  with  v/c = sin a  and remembering that  sin² a + cos² a = 1  we have  .   Being  cosa  a function that varies between 0 and 1, it is possible to identify it to a speed in the way  u/c = cosa  getting the expression    v²/c²  +  u² /c²  =1     or

                   v² +u² = c²

We have now a right triangle with leg s  v and u  and hypotenuse  c, being u  the projection of  c over an axis with the angle  a.

But what could represent the speed u?

Putting the expression in differencial way we have:

 

v=dx/dt           u = dw/dt           (dx)² + (dw)²  = c² (dt)²        or    (dw)²  = c² (dt)² - (dx)²

 

but dw is what classicaly has been defined as the invariant “universe interval”  ds.

 Let’s see now the speeds triangle

The universe is only energy, space and time. Then  we can assume  only two coordinates: space and time. If v is the speed on space, we can assume that u is a speed  on time. But, how can the time to be consideresd orthogonal to the space?

To answer this, we can  imagine the time as something flowing from each point of the space as a spherical  shape or wave, being in this way always orthogonal to a any system of cartesian axes that we can consider in the space with the center in the point.

Then, if the v direction is the space axis, and u direction the time axis, one can think as  u being a speed with respect the time. So, if v=0 then u=c, or the  standing observer looks the time in its point travelling at speed c. But if v=c, then u=0  that is, the inertial frame is going at speed c with respect to our reference frame but the speed with respect to the time is  zero because both are moving at the same speed: it is moving with our time and for this reason at the light speed the lapse of time with respect to an standig observer does not go ahead. For instance : in a photon .

That is too the result of  t₀ = t cos a  for v= c  being sina =1, a= 90º  and cos a=0, t=µ.

This is equivalent to  the standing observer  seing the inertial system  with the space and time  axes  turned an angle a as represented in the figure. If  the inertial system has speed c, its space axis being turned 90º  is coincident with our time axis.

 

In the figure, the space axes are e for the reference frame and e’  for the inertial system and the respective time axes,  t and t’; the right  triangle of speeds  OAB  (OA = v, AB = c.; OB=u) and OL is an initial length in the e axis, which after turning an angle a is seen by the  standing observer as OL’ = OL cos a.

In the time axis  OT = t₀ = m₀ = E₀  (time, mass and energy)  is the projection of t, m  and E equals to OT’ on the time axis of the inertial system, turned also an angle a, remaining space and time always orthogonal.

(To note that the Minkowsky interpretation of special relativity needs an imaginary time and hyperbolic trigonometric functions. Our way does not.)

Being OA = v and AB = c, sina = OA/AB,  OBA = AOL’ = BOD = a  and then  DO = DB =DA = c/2.

 

Let’s see now a way to draw graphicaly the coordinates change from x’’ to x’ and t’’  to t’.

The Lorentz tensor for two coordinates  (y=0, z=0)  is:

 

 

and developping:

x’’cosa = x’ –ct’sina   and  ct’’cosa = ct’ – x’sina

 

Going to the reference frame  e’-t’ in the following graphic, OA = vt’. The inertial system e’’-t’’ has moved  OA  at speed v but we look at it as  turned the angle a = A’OA defined by sin a = v/c. The coordinate x’’ =AB after turning is A’B’. In the galilean system we had that x’=vt’+x’’. Now x’=x’’cosa + ct’sina = x’’cosa + vt’

( As  v= c.sin. a).

In the triangle A’B’F’ is A’F’ =x’ cosa  and as OA’ = vt’ we have that 

x’=OA’ + A’F’=OF drawing a parallel  from F’ to AA’. And drawing a  perpendicular from F to OD  we get OC = x’sina.

Being ct’’=(ct’ – x’ sina)/ cosa  and ct’=OD,  CD = OD – OC = ct’ – x’ sina and then

CD/ cos a =CH = ct’’ . This way is possible to see the length contraction  OF<OB and the time dilation CD <DH.

.

 

If we use light-time to express the lengths, all is still simpler. Let be

 

x' = c t_x',   x'' =c t_x''        the transformation formula is now:

          

  

  and dividing both sides by c:

 

 and developping

       t_x''cos a =  t_x' - t' sin a                   t'' cos a= t' - t_x'sen a

or   t_x'' =  t_x' - t' tan a                             t'' = t' - t_x'tan a

The graphic construction is also simplified:

Now  OA =vt' =t'c sena   and in light-time OA = t' sen a

and         t_x' = t_x''cos a  + t' sen a = OA + A'F'  = OF            OD = OA/sen a=t'

OC=OF sen a    DC = t' - t_x'sen a          HD =DC/cos a = t''

 

 

Let us go now to the energy expressions. We have in the classical way:

E= total energy

   being  the rest mass  and m the mass at speed v as observed from the reference frame.

Using our conventions:

 E= m₀ c² /cosa = mc²   and then as seen before m₀ =m.cos a

When introducing the momentun  p, the classical  fomula is:

 

   or   E² = E₀²  + (pc)²     being E₀  the rest energy.

This can be easily deduced from our right triangle of sides u,v,c

which  acomplishes:

 

v² +u² = c²

 

M ultiplying  each side of the triangle by mc we get a similar triangle with sides mcv,   mcu,  and mc²   that  form too a right triangle which hypotenuse is mc²  .  In this triangle we can write:

 

(mc²)² = (mcv) ² + (mcu) ²        but being     u =  c cos a    mcu = m c²cos a   but    mcosa= m₀     and   p=mv  we have then:

E² = (pc) ² + m²c²cos²ac²  =  (pc) ² +   ( m₀c² ) ²     being E= mc²  and   m₀c² =E₀      we get  finally:

 

 E² = E₀²  + (pc)²

 

The same that the classical formula.

We have used only the previous knowledge of   m.cos a= m₀     and E= mc²

Other calculations can be simplified using the method described .

 

Appendix

In the attached drawing it is described the velocities composition based on the trigonometrical method.

In a circumference of diameter OM = c = 1 we trace OA = v/c =sina  and OB =u/c = sinb. Drawing OA’ = OA on OB  and projecting  this on OM, the segment OP=OA’sinb =  sen asenb.  Drawing now MN=OP at the prolongation of the diameter OM, we have ON = 1+ sen a senb. Adding OA  in the direction OB we get OB’ = sin a + sin b. Now,  drawing a circumference of diameter ON passing at O and center in OM  and tracing  OC’ = OB’ we see that  OC’/ ON = = (sina + sin b) /(1 + sin asin b)  = sin g. Finally, a parallel from M to MC’  gives the angle OMC= g  where OC/OM = sin g   or   u’/c =OC/OM and being c=OM=1, OC = w, the composition of the two velocities  u and v  as seen from the reference frame.

In the same way it is possible to compose the speeds  with respect the time, cos a  and cos b  giving as result

 MC =  cos g = ( cos a cos b) / (1 + sen a sen b).

 

 

 

 

 

 

 

 

 

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