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5-1

Chapter 5 The Loedel Spacetime Diagram

5.1 The Loedel Spacetime Diagram – Contravariant Components We will now show that the spacetime diagram of figure 4.9 contains all the concepts of relativity within it. This new spacetime diagram will be called the Loedel spacetime diagram. As you can see in figure 4.9, all the scales from which numbers are read are parallel to their respective axes. Hence, all components read on these axes are contravariant components, and the Loedel diagram is a contravariant diagram of spacetime. Also notice from the diagram, the angle between the x’- and x’’-axis is equal to θ, and similarly the angle between the τ’- and τ’’-axes, is also equal to θ.

Let us consider a person at rest in the S’ frame of reference at the position x1’. The snap of his fingers marks the event E1 that occurs at the time τ1’ = 0 in figure 5.1. He snaps his fingers a second time at the same position x1’ but now at the time τ2’. An observer in the S’’ frame of reference sees event E1 occurring at the position x1’’ at the

Figure 5.1 The Loedel spacetime diagram.

time τ1’’ and event E2 occurs at x2’’ at the time τ2’’. Notice that in triangle ABC in figure 5.1 that

(τ2’’ − τ1’’) cos θ = L (5.1) While in triangle DFG

(x2’’ − x1’’) cos θ = K (5.2) and in triangle E1E2H

L cos α = K (5.3)

Substituting equation 5.3 into equation 5.2 gives

(x2’’ − x1’’) cos θ = L cos α (5.4) However, as can be seen in figure 5.1

α = 900 − θ and

cos α = cos(900 − θ) = cos 900 cos θ + sin900 sin θ but cos 900 = 0 and sin 900 = 1, hence

cos α = sin θ (5.5)

Replacing equation 5.5 into equation 5.4 gives

(x2’’ − x1’’) cos θ = L sin θ (5.6)

θ

θ

θθ

1’’

2’’

1’

’τ2

τ1’’’τ1

τ2’’

2E

1E

θL

L

KK

HG

B

C

D

F

A

Chapter 5 The Loedel Spacetime Diagram - Contravariant components

5-2

Upon substituting for L from equation 5.1 we get

(x2’’ − x1’’) cos θ = [(τ2’’ − τ1’’) cos θ ] sin θ Therefore

sin θ = (x2’’ − x1’’) (τ2’’ − τ1’’)

But τ’’ = ct’’, hence sin θ = (x2’’ − x1’’) (5.7)

c(t2’’ − t1’’) but

(x2’’ − x1’’) = v’’ (t2’’ − t1’’) the velocity relative to the S’’ frame. But since the person is at rest in the S’ frame, v’’ = v the velocity of the S’ frame with respect to the S’’ frame. Therefore equation 5.7 becomes

sin θ = v

(5.8) c

Equation 5.8 can also be interpreted as v = c sin θ (5.9)

That is, the velocity of the S’ frame of reference with respect to the S’’ frame of reference is simply c sin θ. Equation 5.8 also tells us the angle θ that the S’ frame of reference makes with respect to the S’’ frame of reference on the spacetime diagram. Equation 5.8 may seem to be a little confusing because in Chapter 2 we used the relation tanθ = v/c. The difference is that in Chapter 3 the angle θ referred to the angle between the S - S’ frames of references, while here it refers to the angle θ between the S’ - S’’ frames of references.

Equation 5.8 has an important consequence, in that it shows us the relationship between v and c. That is, since the sin θ is equal to the ratio of the opposite side of the triangle to the hypotenuse, figure 5.2 is the result. Notice from figure 5.2 that

2 2

cos c vc

θ −=

and hence 2 2cos 1 /v cθ = − (5.10)

We will use this relationship frequently in what follows

Figure 5.2 Visualization of equation 5.8, sin θ = v/c.

5.2 The Lorentz Transformation by a Loedel Diagram If this new Loedel spacetime diagram is to represent the many consequences of relativity, it must first contain the Lorentz transformation in it. Let us show that it does. Consider the event E that occurs in spacetime, that has coordinates x’, τ’ in the S’ frame of reference and x’’, τ’’ in the S’’ frame of reference, as shown in figure 5.3. The S’ frame is moving to the right (+x’) at the velocity v with

θ

c v

v2c2−


5-3

Figure 5.3 Loedel diagram for the Lorentz transformations. respect to the S’’ frame of reference. What is the relationship between the coordinates of this event in both frames? Notice in figure 5.3 that the length

OB = OC + CB (5.11) But in triangle OAB

OB = x’’ cos θ while in triangle BCE

CB = τ’ cos α Notice that

OC = x’

Substituting these values into equation 5.11 gives

x’’ cos θ = x’ + τ’ cos α (5.12) But as we showed in equation 5.5

cos α = sin θ (5.5)


x’’ cos θ = x’ + τ’ sin θ (5.13) Solving for x’’ gives

x’’ = x’ + τ’ sin θ cos θ

Substituting for cos θ and sin θ from equations 5.10 and 5.8 respectively yields

τ ’cos α

cosαx’

E

FD

C

B

A


5-4

2 2

' '''

1 /

vxcx

v c

τ+=

−

Since τ = ct , this can be written as

2 2

' '''

1 /

vx ctcx

v c

+=

−

or

2 2

' '''1 /x vtx

v c+

=−

(5.14)

Equation 5.14 is of course the Lorentz transformation for the space coordinates. It gives the position x’’ in the S’’ frame of reference in terms of its position x’ and the time t’ in the S’ frame of reference, which is moving to the right at the velocity v with respect to the S’’ frame of reference. Notice that equation 5.14 is equivalent to equation 1.49, in chapter 1, that had the S frame of reference at rest and the S’ frame of reference moving to the right with the velocity v. Also note that the coordinate equation for the position reduces to the Galilean transformation when the velocity v of the reference frame is small compared to c. That is, if v2/c2 ≈ 0, then,

2 2

' ' ' ''' ' '1 01 /

x vt x vtx x vtv c+ +

= = = +−−

Thus, the Lorentz transformation equation as obtained from the Loedel spacetime diagram, reduces to the classical Galilean transformation equation, which it must, when the relative speed between the observers is small as compared to the speed of light.

To determine the Lorentz transformation equations for the time coordinate we notice that in figure 5.3 the length

OF = OD + DF (5.15) Notice also that

OF = τ’’ cos θ and

OD = τ’ while in triangle DEF

DF = x’ cos α Replacing these values into equation 5.15 gives

τ’’ cos θ = τ’ + x’ cos α (5.16)

But as we showed in equation 5.5, cos α = sin θ. Hence, replacing equation 5.5 into equation 5.16 gives

τ’’ cos θ = τ’ + x’ sin θ Solving for τ’’ gives

τ’’ = τ’ + x’ sin θ cos θ


2 2

' '''

1 /

vxc

v c

ττ

+=

− (5.17)


5-5

However, τ = ct, and equation 5.17 can be written as

2

2 2

' '''

1 /

vt xct

v c

+=

− (5.18)

Equation 5.18 is the Lorentz transformation for the time coordinates. It gives the time t’’ for the time observed in the S’’ frame of reference in terms of the time t’ and its position x’ in the S’ frame of reference, which is moving to the right at the velocity v with respect to the S’’ frame of reference. Notice that equation 5.18 is equivalent to equation 2.10, in chapter 2, that had the S frame of reference at rest and the S’ frame of reference moving to the right with the velocity v.

5.3 The Inverse Lorentz Transformation by a Loedel Diagram The inverse transformations can be found by replacing x’ with x’’, t’ with t’’, and v with -v, in

equations 5.14 and 5.18 to yield

2 2

'' '''1 /x vtx

v c−

=−

(5.19)

and

2

2 2

'' '''

1 /

vt xct

v c

−=

− (5.20)

Let us, however, derive these inverse Lorentz transformations directly from the Loedel spacetime diagram in figure 5.4, which is essentially the same diagram as in figure 5.3 but with an emphasis on different parts of the diagram. Notice in figure 5.4 that the length

EB = EC + CB (5.21)

Also notice that in the large parallelogram OAEBO that

OA = x’’ = EB while in triangle CDE

EC = x’ sin α and triangle OCB gives

CB = τ’’ sin θ Replacing these values into equation 5.21 gives

x’’ = τ’’ sin θ + x’ sin α or

x’ sin α = x’’ − τ’’ sin θ (5.22) But as can be seen in the diagram

α = 900 − θ


5-6

Figure 5.4 The inverse Lorentz transformation equation for space coordinates by a Loedel spacetime diagram.

Hence

sin α = sin(900 − θ) But

sin(A − B) = sinA cosB − cosA sinB or

sin(900 − θ) = sin900 cos θ − cos900 sin θ However, sin900 = 1 and cos900 = 0, therefore

sin α = sin(900 − θ) = cos θ (5.23)

Replacing equation 5.23 back into equation 5.22 gives

x’ cos θ = x’’ − τ’’ sin θ Solving for x’ gives

x’ = x’’ − τ’’ sin θ cos θ


2 2

'' '''

1 /

vxcx

v c

τ−=

−

Since τ = ct, this can be written as

2 2

'' '''

1 /

vx ctcx

v c

−=

−

E

αx’ sin

sin

B

D C

A


5-7

or

2 2

'' '''1 /x vtx

v c−

=−

(5.24)

Equation 5.24 is the Inverse Lorentz transformation for the space coordinates. It gives the position x’ in the S’ frame of reference in terms of its position x’’ and the time t’’ in the S’’ frame of reference, which is moving at the velocity −v with respect to the S’ frame of reference. Notice that equation 5.24 is equivalent to equation 2.14, in chapter 2, that had the S’ frame of reference at rest and the S frame of reference moving to the left with the velocity −v.

Let us now derive the inverse Lorentz transformations for time from the Loedel spacetime diagram in figure 5.5, which is essentially the same diagram as in figure 5.3, and 5.13 but with an emphasis on different parts of the diagram.

Figure 5.5 Inverse Lorentz transformation for time coordinates by a Loedel spacetime diagram.

Notice from the diagram in figure 5.5 that the length OA is given by

OA = OB + BA (5.25) where

OA = τ’’ and in triangle OCB

OC = τ’ and

OB = τ’ cos θ and in triangle ABE

BA = x’’ cos α Substituting these values back into equation 5.25 yields

τ’’ = τ’ cos θ + x’’ cos α (5.26)

τ’coscos αx’’

E

B

C

A


5-8

But as we showed in equation 5.5, cos α = sin θ. Hence, replacing equation 5.5 into equation 5.26 gives

τ’’ = τ’ cos θ + x’’ sin θ Solving for τ’ yields

τ’ cos θ = τ’’ − x’’ sin θ τ’ =

cos θ τ’’ − x’’ sin θ


2 2

'' '''

1 /

vxc

v c

ττ

−=

−

Substituting for τ = ct, this becomes

2

2 2

'' '''

1 /

vt xct

v c

−=

− (5.27)

Equation 5.27 is the inverse Lorentz transformation for the time coordinates. It gives the time t’ for the time observed in the S’ frame of reference in terms of the time t’’ and its position x’’ in the S’’ frame of reference, which is moving at the velocity −v with respect to the S’ frame of reference. Notice that equation 5.27 is equivalent to equation 2.14, in chapter 2, that had the S’ frame of reference at rest and the S frame of reference moving to the left with the velocity −v. 5.4 Length Contraction by a Loedel Diagram Consider a rod of length L0 at rest in the S’ frame of reference as shown in figure 5.6. A line drawn from the tip of L0, parallel to the τ’-axis represents the world line of the tip of L0. If we extend the world line backward into the S’’ frame of reference we see that where it intersects the x’’-axis represents the length L of the rod as observed in the S’’ frame. That is, the length of the rod L in S’’ is the projection of L0 onto the x’’-axis giving

L = L0 cos θ (5.28) But we showed in equation 5.10 that

2 2cos 1 /v cθ = − Replacing equation 5.10 into equation 5.28 gives

2 2

0 1 /L L v c= − (5.29)

which, of course, is the same length contraction formula we derived in chapter 2, equation 2.20, from the Lorentz transformation equations. Here we see that the contraction is built right into our spacetime diagrams.

The length contraction is of course reciprocal and a rod of length L0 at rest in the S’’ frame of reference is shown in figure 5.7. The projection of Lo onto the x’-axis gives the length L that would be observed in the S’ frame of reference. From the diagram we see that

L = L0 cos θ (5.30)

and again using equation 5.10 for the cos θ yields


5-9

Figure 5.6 Length Contraction by a Loedel spacetime diagram - Object at rest in S’.

2 2

0 1 /L L v c= − (5.31)

Equation 5.31 gives the length contraction of a rod L0 at rest in the S’’ frame of reference as it would be observed in the S’ frame of reference, while equation 5.29 gave the length contraction for a rod at rest in the S’ frame of reference. Note that the contraction results are the same in either coordinate system, as they must by the first postulate of special relativity. Equation 5.31 is also the same length

Figure 5.7 Length Contraction by a Loedel spacetime diagram - Object at rest in S’’.

contraction formula we derived in chapter 2, from the Lorentz transformation equations. Again we see that the contraction is built right into our spacetime diagrams.

L

L0

World line

LL0

World line


5-10

5.5 Time Dilation by a Loedel Diagram Consider a clock at rest at the position x’ in the S’ frame of reference as shown in figure 5.8. The event E1 occurs at the time τ1’ in the S’ frame of reference and at the time τ1’’ in the S’’ frame of reference. A second event E2 occurs at the same position x’ in the S’ frame of reference but at the time τ2’ in the S’ frame of reference and at the time τ2’’ in the S’’ frame of reference as is also shown in figure 5.8. The time interval in the S’ frame of reference is given by

Figure 5.8 Time dilation by a Loedel spacetime diagram – clock at rest in the S’ frame of reference.

∆τ0’ = τ2’ − τ1’

We have again used the subscript 0 on ∆τ0’ because this is the time on a clock at rest in the S’ frame of reference, which is of course, what we called the proper time. The time interval on the clock in the S’’ frame of reference is given by

∆τ’’ = τ2’’ − τ1’’

From the diagram in figure 5.8 we see that ∆τ0’ = ∆τ’’ cos θ (5.32)

But again, as we showed in equation 5.10 2 2cos 1 /v cθ = −


2 2

0 ' '' 1 /v cτ τ∆ = ∆ − (5.33) Upon solving for ∆τ’’ we get

0

0

0


5-11

02 2

'''1 /v c

ττ

∆∆ =

− (5.34)


02 2

'''1 /

ttv c

∆∆ =

− (5.35)

Equation 5.35 is the same time dilation formula we found in chapter 1, equation 1.64, and gives the time interval ∆τ’’ observed in the S’’ frame of reference in terms of the time interval ∆τ0’ observed in the S’ frame of reference.

The time dilation is of course reciprocal and a clock at rest at the position x’’ in the S’’ frame of reference is shown in figure 5.9. The event E1 occurs at the time τ1’’ in the S’’ frame of reference and at the time τ1’ in the S’ frame of reference. A second event E2 occurs at the same position x’’ in

Figure 5.9 Time dilation by a Loedel spacetime diagram - clock at rest in the S’’ frame of reference. the S’’ frame of reference but at the time τ2’’ in the S’’ frame of reference and at the time τ2’ in the S’ frame of reference as is also shown in figure 5.9. The time interval in the S’’ frame of reference is given by

∆τ0’’ = τ2’’ − τ1’’

We have again used the subscript 0 on ∆τ0’’ because this is the time on a clock at rest in the S’’ frame of reference, which is of course, what we called the proper time. The time interval on the clock in the S’ frame of reference is given by

∆τ’ = τ2’ − τ1’

From the diagram in figure 5.9 we see that ∆τ0’’ = ∆τ’ sin α (5.36)

But as we showed in equation 5.23 sin α = cos θ (5.23)

0

0

0


5-12

Replacing equation 5.23 back into equation 5.36 gives

∆τ0’’ = ∆τ’ cos θ (5.37) But again, as we showed in equation 5.10

2 2cos 1 /v cθ = − Replacing equation 5.10 into equation 5.37 gives

2 2

0 '' ' 1 /v cτ τ∆ = ∆ − Upon solving for ∆τ’ we get

02 2

'''1 /v c

ττ

∆∆ =

− (5.38)


02 2

'''1 /

ttv c

∆∆ =

− (5.39)

Equation 5.39 gives the time interval ∆t’ observed in the S’ frame of reference in terms of the time interval ∆t0’’ observed in the S’’ frame of reference where the clock is at rest. Equation 5.39 is the same time dilation formula we found in equation 5.35 where there the clock was at rest in the S’ frame of reference, while here the clock is at rest in the S’’ frame of reference. Note that the two equations have the same form, as they must by the first postulate of special relativity. 5.6 Simultaneity by a Loedel Diagram Consider the two events A and B that occur at the same time (τ’) in the S’ frame of reference as shown in figure 5.10. These simultaneous events in the S’ frame of reference are not simultaneous in the S’’ frame of reference as can be seen from the diagram. Event A occurs at the time τA’’ and event B occurs at the time τB’’ in the S’’ frame of reference.

This same problem of simultaneity occurs in the S’’ frame of reference. Consider now the two events A and B that occur at the same time (τ’’) in the S’’ frame of reference as shown in figure 5.11. These simultaneous events in the S’’ frame of reference are not simultaneous in the S’ frame of reference as can be seen from the diagram. Event A occurs at the time τA’ and event B occurs at the time τB’ in the S’ frame of reference.

Notice that in all these Loedel diagrams, the components of every quantity were found by dropping lines that were parallel to the appropriate axes, and hence all these diagrams were examples of using contravariant components. In the next chapter we will look at the same physical phenomena but in our analysis we will use the covariant components.


5-13

Figure 5.10 Simultaneity by a Loedel spacetime diagram - two events simultaneous in S’ frame of reference.

Figure 5.11 Simultaneity by a Loedel spacetime diagram - two events simultaneous in S’’ frame of reference.

A

B

AB

A

B

AB


5-14

5.5 The Invariant Interval in a Spacetime Diagram We saw in chapter 3 that the quantity

(ds)2 = c2(dt)2 − (dx)2 (3.8)

which we called the invariant interval, was a constant in spacetime. All observers, regardless of their state of motion, agreed on this value in spacetime. An observer in the S frame of reference found that

ds2 = c2(dt)2 − (dx)2 While the observer in the S’ frame of reference found that

ds2 = c2(dt’)2 − (dx’)2

Hence ds2 = c2(dt’)2 − (dx’)2 = c2(dt)2 − (dx)2

Thus the quantity c2(dt)2 − (dx)2 as measured by the S observer was equal to the same quantity c2(dt’)2 − (dx’)2 as measured by the S’ observer. The invariant interval, was the same in all inertial systems. Using the invariant interval we plotted a family of hyperbolas for a series of values of x and τ. Where the x’ and τ’ axes intersected the family of hyperbolas determined the length scale along the x’ and τ’ axes. The final result of the invariant interval was to establish an S’ frame of reference which was a skewed coordinate system with the scales on the x’- and τ’-axis that were not the same as the scales on the x- and y-axis. Those skewed coordinate system became the basis of our Loedel and Brehme diagrams of spacetime. Even though they were derived on the basis of the invariant interval, it is instructive to see this effect geometrically in these diagrams.

Figure 5.12 is the Loedel spacetime diagram that shows the invariant interval in spacetime. Consider the length L in spacetime between the two events E1 and E2. By drawing lines parallel to the appropriate axes we get the contravariant components of the length L. Notice from the two triangles that

L2 = (∆x’)2 + (∆τ’’)2 = (∆x’’)2 + (∆τ’)2

Placing all the prime terms on one side of the equation, and all the double primes on the other side gives

(∆x’)2 − (∆τ’)2 = (∆x’’)2 − (∆τ’’)2 (5.40)

Figure 5.12 The Invariant Interval of Spacetime in a Loedel Diagram.

But τ’ = ct’ and τ’’ = ct’’, therefore

(∆x’)2 − c2(∆t’)2 = (∆x’’)2 − c2(∆t’’)2 (5.41) Hence, as we see in our diagram, the quantity (∆x’)2 − c2(∆t’)2 in the S’ frame of reference is equal to (∆x’’)2 − c2(∆t’’)2 in the S’’ frame of reference. Therefore it must be an invariant because it has the same value in either frame of reference. We, of course, designate this invariant quantity as ∆s, and we write it as

L


5-15

(∆s2) = (∆x’)2 − c2(∆t’)2 (5.42)

and (∆s2) = (∆x’’)2 − c2(∆t’’)2 (5.43)

Equations 5.42 and 5.43 can be written in terms of differential quantities as

(ds2) = (dx’)2 − c2(dt’)2 (5.44)

and (ds2) = (dx’’)2 − c2(dt’’)2 (5.45)

If the other two space dimensions are included, the invariant interval in four-dimensional spacetime becomes

(ds)2 = c2(dt)2 − (dx)2 − (dy)2 − (dz)2 (5.46) As we have just seen, the invariant interval is built into the Loedel diagram of spacetime.

We will see in the next chapter that it is also built into the Brehme diagram, and either diagram can be used, because they will both give the same results.

Summary of Basic Concepts

The Loedel Spacetime Diagram. In a Loedel diagram, all the scales from which numbers are read are parallel to their respective axes. Hence, all components read on these axes are contravariant components, and the Loedel diagram is a contravariant diagram of spacetime.

Summary of Important Equations

Lorentz transformation for space coordinates

2 2

' '''1 /x vtx

v c+

=−

(5.14)

Lorentz transformation for the time

coordinates. 2

2 2

' '''

1 /

vt xct

v c

+=

− (5.18)

Inverse Lorentz Transformation for space

coordinates 2 2

'' '''1 /x vtx

v c−

=−

(5.19)

Inverse Lorentz Transformation for time

coordinates 2

2 2

'' '''

1 /

vt xct

v c

−=

− (5.20)

Inverse Lorentz transformation for the space coordinates.

2 2

'' '''1 /x vtx

v c−

=−

(5.24)

Inverse Lorentz transformation for the time coordinates.

2

2 2

'' '''

1 /

vt xct

v c

−=

− (5.27)

Length contraction formula 2 2

0 1 /L L v c= − (5.29) Time dilation formula

02 2

'''1 /

ttv c

∆∆ =

− (5.35)

02 2

'''1 /

ttv c

∆∆ =

− (5.39)


5-16

The invariant interval of Spacetime (∆s2) = (∆x’)2 − c2(∆t’)2 (5.42)

and (∆s2) = (∆x’’)2 − c2(∆t’’)2 (5.43)

The invariant interval of Spacetime in terms of differential quantities

(ds2) = (dx’)2 − c2(dt’)2 (5.44)

and (ds2) = (dx’’)2 − c2(dt’’)2 (5.45)

The invariant interval in four-dimensional spacetime

(ds)2 = c2(dt)2 − (dx)2 − (dy)2 − (dz)2 (5.46)

Questions for Chapter 5

1. Why can’t we just use orthogonal systems in our analysis of relativity?

2. What is a contravariant vector? 3. What is a covariant vector? 4. What is the difference between a Loedel

spacetime diagram and a Brehme spacetime diagram?

5. Is a unitary vector the same as a unit vector?

6. When using a product of two vectors, is it better to have two covariant vectors, two contravariant vectors, or one of each?

Problems for Chapter 5 5.1 The Loedel Spacetime Diagram

1. An observer in the S’ frame is moving at a speed of 0.45c with respect to an observer in the S’’ frame. (a) Find the angle θ between the τ’ and τ’’ axes. (b) Find the angle α between the x’ and τ’ axes. (c) Draw the Loedel spacetime diagram for this case.

2. Using the Loedel spacetime diagram from problem 1, pick an arbitrary event in spacetime with the coordinates (x’, τ’). (a) find the (x’’,τ’’) coordinates for this event graphically from the spacetime diagram. (b) Then find the x’’ and τ’’ coordinates using the Lorentz transformation equations.

3. Using the Loedel spacetime diagram from problem 1, pick an arbitrary event in spacetime with the coordinates (x’’, τ’’). (a) find the inverse (x’,τ’) coordinates for this event graphically from the spacetime diagram. (b) Then find the inverse x’’ and τ’’ coordinates using the Lorentz transformation equations.

4. Using the Loedel spacetime diagram from problem 1, draw a line L0 for a rod at rest in the S’ frame of reference. (a) Show graphically the length L in the S’’ frame of reference. (b) Find the length L using the length contraction equation.

5. Using the Loedel spacetime diagram from problem 1, draw a line L0 for a rod at rest in the S’’ frame of reference. (a) Show graphically the

length L in the S’ frame of reference. (b) Find the length L using the length contraction equation.

6. Using the Loedel spacetime diagram from problem 1, draw the world line for a clock at rest in the S’ frame of reference. Pick out two events. (a) Show graphically the times τ1’ and τ2’ in the S’ frame of reference. (b) Show graphically the corresponding times τ1’’ and τ2’’ in the S’’ frame of reference. (c) Show graphically the time interval ∆τ’ in the S’ frame and ∆τ’’ in the S’’ frame. (d) Find the time difference ∆τ’’ in the S’’ frame using the time dilation equation.

7. Using the Loedel spacetime diagram from problem 1, draw the world line for a clock at rest in the S’’ frame of reference. Pick out two events. (a) Show graphically the times τ1’’ and τ2’’ in the S’’ frame of reference. (b) Show graphically the corresponding times τ1’ and τ2’ in the S’ frame of reference. (c) Show graphically the time interval ∆τ’’ in the S’’ frame and ∆τ’ in the S’ frame. (d) Find the time difference ∆τ’ in the S’’ frame using the time dilation equation.

8. Using the Loedel spacetime diagram from problem 1, pick out two events that occur at the same time (τ’) in the S’ frame of reference. (a) Show graphically that these two simultaneous events in the S’ frame of reference are not simultaneous in the S’’ frame of reference.


5-17

9. Using the Loedel spacetime diagram from problem 1, pick out two events that occur at the same time (τ’’) in the S’’ frame of reference. (a) Show graphically that these two simultaneous events in the S’’ frame of

reference are not simultaneous in the S’ frame of reference.

To go to another chapter, return to the table of contents by clicking on this sentence.

http://www.farmingdale.edu/faculty/peter-nolan/pdf/TOCRel.pdf

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