Geometrical presentation of special relativity (based on the …Special Relativity“ by Sander Bais) Nándor Bokor, BUTE, 2013. Postulates: 1. In all inertial frames in all directions

Geometrical presentation of special relativity (based on the book „Very Special Relativity“ by Sander Bais) Nándor Bokor, BUTE, 2013.

Postulates: 1. In all inertial frames in all directions the speed of light is measured to be c. 2. The laws of physics are the same in all inertial frames. Figure 1: The concepts of event and worldline.

Figure 1

Figure 2: Synchronization of clocks in frame K. Two observers (a and b) at rest with respect to one another synchronize their clocks. When a receives the light signal, his clock reads t2. He sends this data to b, who sets the reading of her own clock so that (retroactively) t1 is one half of t2. This also means that the time of event A (an event on a’s worldline) is

!

t22

= t1

⇒ the broken line is the line of simultaneity.

Figure 2

Figure 3: Synchronization of clocks in frame K′. The same kind of synchronization in a frame K′ which is in relative motion with respect to K (here the two persons with clocks are called a′ and b′): The light signal emitted by a′ is depicted by a 45° straight line, just like the light signal emitted by a was! The line of reasoning is the same as before ⇒ the time of event Aʹ′ (an

event on the worldline of a′) is

!

t'22

= t'1 ⇒ for a′ and b′ the line of

simultaneity is the slanted broken line. The „primed“ observers measure distances along the line of simultaneity (and the unprimed observers measure distances along their line of simultaneity) ⇒ the x′-axis is parallel to the slanted broken line! SIMULTANEITY IS RELATIVE.

Figure 3

Figure 4: How are things in newtonian kinematics? Galilean transformation. Time is absolute (there is only one kind of lines of simultaneity in the figure, the horizontal broken lines). ⇒ the speed of light is necessarily relative.

the speed of a light pulse emitted by a is:

!

c =xt

"

# $

%

& '

according to a′:

!

c'= x "#xt

= c" #xt

= c" v ,

where v is the speed of a′.

........................................................................... Galilean transformation (see event B in bottom Figure):

!

x'B = xB "#xB = xB " vtB

!

t'B = tB ...........................................................................

Figure 4

Figure 5: Back to Einstein. The order of events. The order of events can be different in K and Kʹ′:

!

tA < tB , but

!

t'A > t'B This phenomenon can be explained by noting that K and Kʹ′ have different lines of simultaneity in the figure (and that, in turn, can be explained by the fact that K and Kʹ′ measure the same speed of light). But: there can be no causal relationship between events A and B. Events that can be influenced by A and B lie within the green and red regions, respectively. (These regions are the so-called future lightcones.) As seen, for both A and B, one lies outside the lightcone (outside the region of influence) of the other.

Figure 5

Figure 6: Velocity transformation.

!

u =xDtD

= < because the green triangles are similar > =

⇒

!

=x'A + x'Dt'A

(1)

⇒

!

=ct'A "ct'D

x'A# c (2)

(1) ⇒

!

x'D = ut'A " x'A

(2) ⇒

!

t'D = t'A "u #x'Ac2

Dividing the last two equations:

!

u'= x'Dt'D

=ut'A " x'A

t'A "ux'Ac2

=u "

x'At'A

1" uc2

x'At'A

=u " v

1" uvc2

Remark: we have not calibrated the primed axes to the unprimed axes yet, but so far this hasn’t caused any problems, because in every ratio that

appears above (e.g.

!

xDtD

, (1), (2)) the numerator is calibrated in the same

reference frame as the denominator (either both are calibrated in the Kʹ′ or both are calibrated in K).

Figure 6

Figure 7: Inverse velocity transformation. (Of course, the inverse velocity transformation formula can also be obtained algebraically, by solving the velocity transformation formula above for u.)

!

u'= x'Dt'D

= < because the green triangles are similar > =

⇒

!

=xD " xAtA

(1)

⇒

!

=ctD " ctA

xA# c (2)

(1) ⇒

!

xD = u' tA + xA

(2) ⇒

!

tD = tA + u'" xAc2

Dividing the last two equations:

!

u =xDtD

=u' tA + xAtA + u' xA

c2=

u'+ xAtA

1+u'c2

xAtA

=u'+v

1+u'vc2

Figure 7

Figure 8: Time dilation.

!

t'A : „proper time“ between O and A (a′ measures it on his wristwatch).

!

tA : time between O and A as measured by a. To do this measurement, she has to use the line of simultaneity (i.e. she has to use the data registered on the wristwatch of b).

!

tB : „proper time“ between O and B (a measures it on her wristwatch).

!

t'B : time between O and B as measured by a′. To do this measurement, he has to use the line of simultaneity (i.e. he has to use the data registered on the wristwatch of d′). From symmetry:

!

tA = " # t'A and

!

t'B = " # tB , where the two

!

" ’s are the same. ⇒

!

t'B = " # tB = " # tA = " 2t'A ⇒

!

" >1. Proper time is always the shortest time interval between two events. Anyone else will measure a longer time interval than the proper time. TIME DILATION.

Figure 8

Figure 9: Time dilation (cont’d). The green and red triangles are similar, so:

!

v =xAtA

"

# $

%

& ' =

ct'B (ct'Ax'B

) c

Using the result on the previous page

!

t'A =1" 2t'B

⇒

!

v =

c2t'B 1"1# 2

$

% &

'

( )

x'B=

c2 1" 1# 2

$

% &

'

( )

v (since

!

x'Bt'B

= v)

⇒

!

v2

c2=1" 1

# 2

⇒

!

" =1

1# v2

c2

This is the multiplying factor that appears in the time dilation formula.

Figure 9

Figure 10: About calibration. The calibration of (i.e. the location of the 1 meter markings on) the ct-axis must be the same as the calibration of the x-axis, because this is the only way that the worldline of light will be a 45° straight line. For the same reason the calibration of the ct′-axis must be the same as the calibration of the x′-axis (i.e. on the spacetime diagram the separation between the 1 meter markings on the ct′-axis must be the same as the separation between the 1 meter markings on the x′-axis). (Note that we haven’t discussed yet how the primed axes should be calibrated relative to the unprimed axes!)

Figure 10

Figure 11: Length contraction: the length of a rod at rest in Kʹ′. The ratio between

!

x'E and

!

xE must be the same as the ratio between

!

t'B and

!

tB (see previous page about calibration). Figure 8:

!

t'B = " # tB ⇒

!

x'E = " # xE The proper length („rest length) of the rod is

!

l0 = x'E (measured in Kʹ′) The length of the moving rod is

!

l = xE =x'E"

= l0 1# v2

c2 (measured in K)

LENGTH CONTRACTION.

Figure 11

Figure 12: Length contraction: the length of a rod at rest in K. Figure 8:

!

tA = " # t'A ⇒

!

xF = " # x'F The proper length („rest length“) of the rod is

!

l0 = xF (measured in K) The length of the moving rod is

!

l = x'F =xF"

= l0 1# v2

c2 (measured in Kʹ′)

Remark: from a naive, euclidean interpretation of the Figure we would draw the conclusion that

!

xF < x'F , but as we see from the formula above,

!

xF > x'F ! This tells us that the calibration of the primed axes with respect to the unprimed axes will not follow euclidean geometry.

Figure 12

Figure 13: Calibration. In Figure 10 we discussed how the xʹ′-axis should be calibrated with respect to the ctʹ′-axis (and how the x-axis should be calibrated with respect to the ct-axis). Now the question is: how should we calibrate the ct-axis with respect to the ctʹ′-axis (and to the ctʺ″-axis, etc.)? For example:

!

ctA = cT (measured in meters). Where is event G for which

!

ct'G = cT?

!

t'G =T = tA = " # t'A > t'A ⇒ G is „towards the top“ in the figure. The coordinates of G in K:

!

ct = "ct'G = "cT (1)

!

x = vtG = "vT (2) Let’s express the relation between x and ct by eliminating v. Our conclusion will then be valid not only for event G, but for events Gʺ″, etc. as well.

(1), (2) ⇒

!

ct = cT " 1

1# x2

c2t 2

⇒

!

c2t 2 " x2 = c2T 2 We obtained the equation of a hyperbola (green curve in the Figure). This is the so-called „calibration hyperbola“.

Figure 13

Figure 14: Calibration 2. It follows from our reasoning in Figure 8 that the calibration between the x-, xʹ′-, xʺ″-, etc. axes is done in the same way, using a calibration hyperbola the equation of which is

!

x2 " c2t 2 = X 2 (this is the equation of the hyperbola that specifies the X meter markings on the axes).

Figure 14

Figure 15: Doppler-effect of light. O, A, B, C, ...: emission events O*, A*, B*, C*, ...: detection events

!

ctA* = ctA + xA = ctA + vtA = ctA 1+

vc

"

# $

%

& ' =

ct 'A

1( v2

c2

1+vc

"

# $

%

& ' = ct'A

c+ vc( v

Taking the inverse of both sides:

!

fobserved = femittedc" vc+ v

Remark: the formula above is valid when the source and the observer are moving away from each other.

Figure 15

Figure 16: Lorentz-transformation. Our goal is to determine

!

x'P ,t'P( ) from

!

xP ,tP( ). „Auxiliary events“: A, B. The green and red triangles are all similar (triangles with the same color are congruent too).

!

x'P = x'B "#x'PB = x'B " x'A = x'B "vt'A =xP

1" v2

c2

" v tP

1" v2

c2

,

where we used the formulas for length contraction (see Figure 11) and time dilation (see Figure 8). The Lorentz transformation formula for xʹ′ is thus

!

x'P =xP " vtP

1" v2

c2

. (L1)

Similarly:

!

ct'P = ct'A "ct'B =ct'Bx'B

=vc

= ct'A "vcx'B = c tP

1" v2

c2

"vc

xP

1" v2

c2

.

The Lorentz-transformation formula for tʹ′ is thus

!

t'P =tP "

vc2xP

1" v2

c2

. (L2)

Figure 16

Figure 17: Inverse Lorentz-transformation. (Remark: the inverse transformation formulas can of course also be obtained by algebraically solving the set of equations (L1), (L2) for x and t.) „Auxiliary events“: D, E.

!

xP = xE + "xEP = xE + xD = xE + vtD =x'P

1# v2

c2

+ v t'P

1# v2

c2

where we used the formulas for length contraction (see Figure 12) and time dilation (see Figure 8). The (inverse) Lorentz transformation formula for x is thus

!

xP =x'P +vt 'P

1" v2

c2

. (IL1)

Similarly:

!

ctP = ctD + ctE =ctExE

=vc

= ctD +vcxE = c t'P

1" v2

c2

+vc

x'P

1" v2

c2

.

The (inverse) Lorentz-transformation formula for t is thus

!

tP =t'P +

vc2x'P

1" v2

c2

. (IL2)

Figure 17

Figure 18: Spacetime interval between two events. „Spacetime displacement“ vektor between two events: - in frame K:

!

dx,dy,dz,cdt( ) - in frame Kʹ′:

!

dx',dy',dz',cdt'( ) K and Kʹ′ do not agree on the values of separate coordinate differentials (e.g.

!

dx " dx'). This is in fact what the Lorentz-transformation is about. However:

!

c2dt'2 "dx'2 = c2dt 2 " 2v

c2dxdt +

v2

c4dx2

1" v2

c2

"dx2 " 2vdxdt + v2dx2

1" v2

c2

= c2dt 2 " dx2

That is, the quantity

!

c2d" 2 # c2dt 2 $ dx2 is invariant. This is the spacetime interval („spacetime distance“) between two events. The meaning of

!

d" : PROPER TIME (the time lapse measured in a frame (xʺ″,ctʺ″) where A and B occur at the same location).

Figure 18

Figure 19: Spacetime interval between two events 2. As we saw:

!

c2dt 2 " dx2 = c2dt'2 "dx'2 = c2dt"2 "dx"2 = ...= invariant For events A and B in Figure 18 this difference of squares is positive ⇒ we can define a proper time

!

d" between A and B. The spacetime interval between A and B is a „timelike interval“. ............................................. Between A and D:

!

c2dt 2 " dx2 = c2dt'2 "dx'2( ) = 0 The spacetime interval between A and D is a „lightlike interval“. ............................................. Between A and E:

!

c2dt 2 " dx2 = c2dt'2 "dx'2( ) < 0 The spacetime interval between A and E is a „spacelike interval“. In this case:

!

ds 2 " dx2 # c2dt 2 The meaning of

!

ds : PROPER LENGTH (the distance between the locations of A and E, measured in a frame (xʺ″,ctʺ″) where A and E occur at the same time).

Figure 19

Figure 20: Back to Newton; dynamics. Two particles undergo inelastic collision. Their momentum is such that after collision the combined particle is at rest in frame K. A: collision event. Kʹ′, Kʺ″, K: the instantaneous rest frames of the first, second and combined particles, respectively, at the moment of collision (i.e. immediately before and after the collision). (e.g.

!

"x'1 = 0 ,

!

"x"2 = 0) Newton ⇒ no time dilation ⇒ the calibration of the axes is such that

!

"t1 = "t'1 = "t"1. E.g. the speed of the first particle just before collision is

!

v1 ="x1"t1

!

v'1 ="x'1"t'1

= 0#

$ %

&

' (

In the same way we can interpret the following quantities (not shown in the Figure):

!

"x2,c"t2,c"t '2 ,c"t"2 ,

!

"x3,c"t3,c"t '3 ,c"t"3 . Consider the vicinity of event A. Let’s draw three vectors (related to the worldlines of the three particles) that obey the following rules: their

horizontal (i.e. x-) projection is

!

"ximi

"ti and their „vertical“ (i.e. t-)

projection is

!

c"timi

"ti. [It doesn’t make any difference whether we choose

primed or unprimed

!

"ti’s in the factor

!

mi

"ti, because they have the same

numerical value.]

Figure 20

Figure 21: Back to Newton; dynamics 2. The three new vectors thus obtained will be parallel to the three worldlines of Figure 20 (this follows from the construction rules). In addition, the lengths of the three new vectors (that point along the ct-, ctʹ′- and ctʺ″-axes, respectively) are

!

m1c,

!

m2c and

!

m3c, respectively [more precisely: the tip of vector 1 is at the „

!

m1c“ marking of the axis that is oriented along the ctʹ′-direction, the tip of vector 2 is at the „

!

m2c“ marking of the axis that is oriented along ctʺ″, and the tip of vector 3 is at the „

!

m3c “ marking of the axis that is oriented along ct]. As seen, we have to write p (and pʹ′and pʺ″) on the horizontal axis, because the horizontal projections of the vectors give their momenta as measured in frames K, Kʹ′ and Kʺ″ [The figure shows three of these measured momenta.] Newtonian mechanics, inelastic collision ⇒ ⇒ (1) conservation of momentum

!

p1 + p2 = p3( ) (2) conservation of mass

!

m1 + m2 = m3( )

Figure 21

Figure 22: Back to Newton; dynamics 3. In Figure 21 this means that (1) cons. of mom. ⇒ (the horizontal component of the green vector) +

(the horizontal component of the red vector) = (the horizontal component of the black vector)

(2) cons. of mass ⇒ (the „length“ of the green vector along tʹ′ [as well

as its vertical component]) + (the „length“ of the red vector along t [as well

as its vertical component]) = (the „length“ of the black vector along t [as well

as its vertical component]) From this it follows that the same relation holds in Figure 21 for the vectors themselves: the green vector + the red vector = the black vector --------------------------------------- Questions: 1. What physical quantity do Π, Πʹ′, Πʺ″ represent? (Our first thought is: they represent mass. But do they really?) 2. Our starting point (see Figure 20) is newtonian, i.e. not correct! How will the correct figures look? What are the correct formulas (that approach the newtonian formulas for

!

v1,v2,v3 << c)?

Figure 22

Figure 23: Relativistic (i.e. correct) discussion of inelastic collision. For simplicity, we consider the collision of two identical particles with mass m. Frame K: where particle 1 is at rest before the collision. Frame Kʹ′: where the collision is symmetrical, i.e.

!

v'1 = "v'2 . This is the instantaneous rest frame of the combined particle 3. Frame Kʺ″: where particle 2 is at rest before the collision. In a given reference frame:

for different worldlines, if

!

"t is the same then each

!

"x (each x-projection) is proportional to the speed of the particle with the given wordline.

Figure 23

Figure 24: Momentum. Motivation: we want to define a quantity called „momentum“ so that it is a conserved quantity in any inertial frame. We want to plot momentum (based on the newtonian discussion) as the projection of a vector on the x-, xʹ′-, xʺ″-axes. Question: by what factor should we stretch the given spacetime displacement vectors in order to obtain vectors whose projections on the x-, xʹ′-, xʺ″-axes satisfy the law of conservation? Another question: on the so obtained diagram, what will be the physical meaning of the quantities on the axes? (we know that

!

" = p,

!

"'= p', etc., since we defined momentum that way, but how about the other axes?

!

" = ?,

!

" '= ? , etc.) ---------------------------------- Remark: Since each

!

c"t,"x( ) spacetime displacement vector is just stretched by a constant we can be sure that the new Π-, Πʹ′-axes will be mutually calibrated in the same way as the ct-, ctʹ′-axes were. The same is true for the calibration of Δ, Δʹ′ and x, xʹ′. ⇒

!

" ,#( ) and

!

" ',#'( ) will also be connected through the Lorentz-transformation!

Figure 24

Figure 25: The multiplying factor 1. It follows from elementary geometry that if we want the law of „addition“ (i.e. the law of conservation)

!

"1 + "2 = "3,

!

"'1 +"'2 = "'3 ,

!

""1 +""2 = ""3 , ..., to be valid for the Δ-projections in every inertial frame then the same law of addition must be valid for the vectors themselves! So the question is: what multiplying factor should we apply for the given spacetime displacement vectors so that the new vectors satisfy the law of vector addition? From the viewpoint of Kʹ′ the collision is symmetrical and particles 1 and 2 are completely identical ⇒ vectors 1 and 2 must have equal „lengths“ on the diagram (as Figure 25 shows). The „length“ of the vectors is denoted with k. This „length“ can be read on the Πʺ″ or Π axis along which the given vector is pointing.

Figure 25

Figure 26: The multiplying factor 2. Another line of reasoning: in Kʺ″ particle 2 is at rest ⇒ whatever „length“ particle 1 has in K, particle 2 must have the same „length“ in Kʺ″. So the two vectors point to the same markings (i.e. k) on the Π- and Πʺ″-axes, respectively.

!

" 1 = k

!

" "2 = k Lorentz-transformation:

!

"2 ="'2 +

v2c# '2

1$ v22

c2

=

v2ck

1$ v22

c2

We want Δ to represent momentum ⇒ we can obtain the physical meaning of k. The way we can do it is this: In the newtonian limit the above equation must give the newtonian formula for momentum: If

!

v2 << c then

!

"2 = m2v2

That is:

!

k v2ck

1" v22

c2

# k v2c

= m2v2

This gives us the value of k:

!

k = m2c In summary: the multiplying factor we were looking for is the mass of the given particle multiplied by the speed of light.

Figure 26

Figure 27: The relativistic expression of momentum. As we have just seen, the vector representing a given particle lies along the Π-axis of the instantaneous rest frame of that particle (for particles 1, 2 and 3, these frames are K, Kʺ″ and Kʹ′, respectively) and the tip of the vector is at the marking that gives the particle’s mass. Thus the new, relativistic formula of momentum can be obtained from the Lorentz-transformation equation for Δ and from our requirement that Δ (Δʹ′, Δʺ″, etc.) represent momentum in a given frame. The momentum in frame K of a particle at rest in frame Kʹ′ is thus

!

p =p'+ v

c" '

1# v2

c2

=0 +

vcmc

1# v2

c2

=m

1# v2

c2

v

Figure 27

Figure 28: Energy. Question: What is the physical meaning of the quantity we denoted with Π (Πʹ′, Πʺ″, etc.)? Applying what we know about the calibration of axes (or using the Lorentz-transformation) we get

!

" =mc

1# v2

c2

.

What sort of physical quantity is this? It is a quantity that is measured to be larger in K (in which the particle is moving) than in Krest (in which the particle is at rest). The amount by which it is measured to be larger in K than in Krest is

!

" #" rest =mc

1# v2

c2

#mc.

To find out the meaning of Π consider the expression above in the newtonian limit:

If

!

v << c:

!

" #" rest $ mc 1+v2

2c2%

& '

(

) * #1

+

, -

.

/ 0 =12m v2

c.

This is kinetic energy! (divided by c) ⇒ Inescapable conclusion: Π is the particle’s energy (divided by c)

⇒

!

" =Ec

=mc

1# v2

c2

⇒

!

E =mc2

1" v2

c2

Figure 28

Figure 29: Energy-momentum vector. In summary:

The momentum of a particle:

!

p =mv

1" v2

c2

(1)

The energy of a particle:

!

E =mc2

1" v2

c2

(2)

Combining (1) with (2) and eliminating the speed of the particle from we obtain the following relationship between E and p:

!

Ec

"

# $

%

& ' 2

( p2 = mc( )2 (3)

Thus for a particle

!

Ec

> p (see Figure). This is also easy to see if we

remember that the original worldline – which was our starting point when constructing the (E,p) vector – was timelike. We can construct a similar energy-momentum vector for light too. There our starting point is the 45° worldline of light, hence the energy-momentum vector for light will be a 45° line on the diagram.

⇒ for light:

!

Ec

= p

For light, formulas (1) and (2) cannot be applied (because

!

v = c), but formula (3) is legitimate to use (because the singularities of (1) and (2) have been eliminated from it). From (3) the mass of the photon turns out to be zero: for light:

!

m = 0.

Figure 29

Figure 30: Is mass conserved? Is mass conserved in an inelastic collision? Consider the inelastic collision depicted in Figure 23. As is immediately obvious from Figure 30, mass is not conserved!

!

M > 2m (because

!

M2

> m, as seen in the Figure)

Now the details: Applying our knowledge about calibration (or using Lorentz-transformation) we get

!

E '1c

=mc

1" v2

c2

(v: the speed of particle 1 in Kʹ′)

!

E '2c

=mc

1" v2

c2

(the speed of particle 2 is also v in Kʹ′)

Hence for particle 3:

!

E '3c

=2mc

1" v2

c2

(1)

On the other hand:

!

E '3c

= Mc (2)

(1), (2) ⇒

!

M =2m

1" v2

c2

Mass is not conserved in a collision.

But (and this follows directly from the fact that we added up vectors): energy (the Π-projection of vectors) does satisfy the law of conservation in every inertial frame.

Figure 30

Figure 31: Mass is invariant. Consider a particle that undergoes a spacetime displacement

!

cdt,dx( ) between two nearby events.

!

p =mv

1" v2

c2

= m dxdt

1

1" v2

c2

= m dxd#

,

where

!

d" is proper time, and we used the time dilation formula. On the other hand:

!

E =mc2

1" v2

c2

= mc2 dtd#

(again we used the time dilation formula)

The invariance of spacetime interval (see Figure 18) is expressed as

!

c2d" 2 = c2dt 2 # dx2 = c2dt'2 #dx'2 =invariant.

Multiply this equation by the (invariant!) quantity

!

m 2

d" 2.

This leads to:

!

mc( )2 =Ec

"

# $

%

& ' 2

( p2 =E 'c

"

# $

%

& ' 2

( p'2 =invariant.

Different observers measure different values for the particle’s energy and momentum, but they will agree on the particle’s mass.

Figure 31