Spacetime diagrams: an example

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 23.

As an example of a spacetime diagram, suppose we have our usual two inertial frames with {\mathcal{S}} at rest relative to observer {A} and {\bar{\mathcal{S}}} moving at speed {\beta=\frac{3}{5}} in the {+x} direction, with the origins of the two systems coinciding as usual.

We’d like to plot the lines of constant {\bar{t}} and {\bar{x}} on a spacetime diagram. Using Lorentz transformations we have

\displaystyle ct \displaystyle = \displaystyle \frac{x}{\beta}-\frac{\bar{x}}{\beta\gamma}\ \ \ \ \ (1)
\displaystyle ct \displaystyle = \displaystyle \beta x+\frac{c\bar{t}}{\gamma} \ \ \ \ \ (2)

For various values of {\bar{x}} and {\bar{t}}, these two equations give two sets of parallel lines. The first equation gives lines with slope {1/\beta} and {ct}-intercepts of {-\bar{x}/\beta\gamma}, while the second equation gives lines with slope {\beta} and {ct}-intercepts of {c\bar{t}/\gamma}. We can plot a few lines from each set as shown:

Griffiths 12.23

The red lines are lines of constant {\bar{x}} with the top line corresponding to {\bar{x}=-3} and the bottom line to {\bar{x}=+3}, in steps of 1. The green lines are lines of constant {\bar{t}} with the bottom line corresponding to {\bar{t}=-3} and the top line to {\bar{t}=+3}, again in steps of 1.

The thick blue line represents the world line of an object that starts at {\left(\bar{t},\bar{x}\right)=\left(-2,-2\right)} and moves to {\left(\bar{t},\bar{x}\right)=\left(3,2\right)}. We can find its velocity in {\mathcal{S}} by taking its slope on the graph. Finding the exact values of {x} and {t} is difficult by eyeballing a graph, so we can ‘cheat’ a bit and use the Lorentz transformations to find the corresponding values. We get for the starting point:

\displaystyle ct_{1} \displaystyle = \displaystyle \gamma\left(c\bar{t}+\beta\bar{x}\right)\ \ \ \ \ (3)
\displaystyle \displaystyle = \displaystyle \frac{5}{4}\left(-2-\frac{3}{5}2\right)\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle -4\ \ \ \ \ (5)
\displaystyle x_{1} \displaystyle = \displaystyle \gamma\left(\bar{x}+\beta c\bar{t}\right)\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle -4 \ \ \ \ \ (7)

and for the end point:

\displaystyle ct_{2} \displaystyle = \displaystyle \gamma\left(c\bar{t}+\beta\bar{x}\right)\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle \frac{5}{4}\left(3+\frac{3}{5}2\right)\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle \frac{21}{4}\ \ \ \ \ (10)
\displaystyle x_{2} \displaystyle = \displaystyle \gamma\left(\bar{x}+\beta c\bar{t}\right)\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle \frac{19}{4} \ \ \ \ \ (12)

The velocity is then

\displaystyle v \displaystyle = \displaystyle \frac{\Delta x}{\Delta t}\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \frac{35}{37}c \ \ \ \ \ (14)

We can check this using the velocity addition formula. Its velocity in {\bar{\mathcal{S}}} is

\displaystyle \bar{v}=\frac{\Delta\bar{x}}{\Delta\bar{t}}=\frac{4}{5}c \ \ \ \ \ (15)

so

\displaystyle v=\frac{\frac{4}{5}+\frac{3}{5}}{1+\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)}c=\frac{35}{37}c \ \ \ \ \ (16)

so it checks out.

Spacelike intervals

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 22.

If two events are separated by a spacelike interval, neither event can affect the other, since different observers may disagree about the order of the events. Here are examples of a couple of misconceptions that sometimes arise about such intervals.

Example 1 If two people are sitting a couple of metres apart then they are at rest relative to each other so they share the same inertial frame and will agree about all time and space measurements. At a particular instant of time, the interval separating the two people is spacelike, so it might seem that they could not communicate with each other. However, if we draw each person’s world line on a spacetime diagram, then in their own frame, each person’s world line is a vertical line (remember that {ct} is plotted on the ordinate (‘y axis’) and {x} on the abscissa (‘x axis’). If they are speaking to each other, the sound waves have world lines that travel diagonally upwards between the two vertical world lines representing the 2 people. If person {A} is at {x=0} and {B} is at {x=2}, then if {A} says something at {t=0}, the sound reaches {B} at {t=2/v} where {v} is the speed of sound, so the slope of the sound wave’s world line is

\displaystyle  c\frac{\Delta t}{\Delta x}=c\frac{2}{2v}=\frac{c}{v} \ \ \ \ \ (1)

As {c\gg v}, the sound’s world line is very nearly vertical but it does angle from {A}‘s vertical line over to {B}‘s line. Similarly, if {B} says something to {A}, the sound’s world line travels in the {-x} direction with the same speed, so its slope is {-c/v}.

The interval between the events of {A} saying something and {B} hearing it is

\displaystyle   \Delta s^{2} \displaystyle  = \displaystyle  -c^{2}\Delta t^{2}+\Delta x^{2}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  -c^{2}\frac{4}{v^{2}}+4\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  -4\frac{c^{2}}{v^{2}}\left(1-\frac{v^{2}}{c^{2}}\right) \ \ \ \ \ (4)

For {c\gg v}, this is (a very large) negative value, so the interval between the two events is definitely timelike.

Example 2 Suppose that faster than light travel is possible, but that light signals still travel at {c}. In that case it would be possible for an object to travel from {A} to {B} such that the interval between the events of leaving {A} and arriving at {B} is spacelike. Since different observers can disagree on the order in which such events occur, it is possible for some observers to say that the object arrived at {B} before it left {A}.

However, if the object then returned from {B} to {A} (also faster than light, say), all observers would agree that the object arrived back at {A} after it left {A}. This is because the interval between the two events (leaving {A} and arriving back at {A}) is timelike (since they occur at the same place in {A}‘s frame), so they must be separated by a positive time interval in every inertial frame.

The invariant interval: some examples

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 20-21.

The invariant interval in special relativity is the scalar product of the interval between two events with itself:

\displaystyle  \Delta s^{2}\equiv\left(\Delta x\right)_{i}\left(\Delta x\right)^{i} \ \ \ \ \ (1)

Since {\Delta x} is the difference of two four-vectors, it too is a four-vector so the invariance under Lorentz transformations follows from that fact.

Because the 0 term is negative and the other three terms are positive, {\Delta s^{2}} can be negative, zero or positive. This gives three possible types of pairs of events:

  1. Timelike: If {\Delta s^{2}<0}, then it is possible to find a frame in which the two events occur at the same spatial point, but at different times, since it is the time component {-\left(\Delta x^{0}\right)^{2}} which is negative.
  2. Lightlike: If {\Delta s^{2}=0} then {c^{2}\left(\Delta t\right)^{2}=\Delta x^{2}} (if the motion is along the {x} axis; the argument is similar for arbitrary directions), so the events can be connected by a light signal.
  3. Spacelike: If {\Delta s^{2}>0}, then it is possible to find a frame in which the two events occur at the same time but at different places. Different observers may disagree about which event occurs first.

Example 1 In system {\mathcal{S}}, an event {A} happens at {\left(ct,x,y,z\right)=\left(15,5,3,0\right)} and {B} happens at {\left(5,10,8,0\right)}. The interval between them is

\displaystyle  \Delta s^{2}=-100+25+25+0=-50<0 \ \ \ \ \ (2)

so the interval is timelike. There is no frame in which {A} and {B} occur simultaneously. However, there is a frame where they occur at the same point. To find this frame, it’s easiest to orient the coordinates so that the {x} axis is along the line joining the events, which points in the direction {\hat{\mathbf{x}}+\hat{\mathbf{y}}}, and to redefine the origin so that {x=\bar{x}=0} and {t=\bar{t}=0} when {A} occurs. In {\mathcal{S}}, the events are separated by a distance of {5\sqrt{2}}, so in the new coordinate system (at rest relative to {\mathcal{S}}) we have

\displaystyle   A \displaystyle  = \displaystyle  \left(0,0,0,0\right)\ \ \ \ \ (3)
\displaystyle  B \displaystyle  = \displaystyle  \left(-10,5\sqrt{2},0,0\right) \ \ \ \ \ (4)

We then need to find {\beta} such that in the frame {\bar{\mathcal{S}}}, {\bar{x}_{B}=0}, so using a Lorentz transformation, we have

\displaystyle   \bar{x}_{B} \displaystyle  = \displaystyle  0=\gamma\left(5\sqrt{2}+10\beta\right)\ \ \ \ \ (5)
\displaystyle  \beta \displaystyle  = \displaystyle  -\frac{\sqrt{2}}{2} \ \ \ \ \ (6)

Therefore the velocity of {\bar{\mathcal{S}}} relative to our original frame {\mathcal{S}} is

\displaystyle  \mathbf{v}=-\frac{c}{2}\left(\hat{\mathbf{x}}+\hat{\mathbf{y}}\right) \ \ \ \ \ (7)

Example 2 Now we take {A=\left(1,2,0,0\right)} and {B=\left(3,5,0,0\right)}. The interval is

\displaystyle  \Delta s^{2}=-4+9=5>0 \ \ \ \ \ (8)

so the interval is spacelike. Since both events are already on the {x} axis, to find a frame in which the events occur at the same time, we change the origin to event {A}, giving

\displaystyle   A \displaystyle  = \displaystyle  \left(0,0,0,0\right)\ \ \ \ \ (9)
\displaystyle  B \displaystyle  = \displaystyle  \left(2,3,0,0\right) \ \ \ \ \ (10)

We now use a Lorentz transformation on the time to find {\beta} such that {\bar{t}_{B}=0}:

\displaystyle   \bar{t}_{B} \displaystyle  = \displaystyle  0=\gamma\left(t_{B}-\beta x_{B}\right)\ \ \ \ \ (11)
\displaystyle  \beta \displaystyle  = \displaystyle  \frac{t_{B}}{x_{B}}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{3} \ \ \ \ \ (13)

Example 3 The previous example is easily generalized to the case where {A=\left(t_{A},x_{A},0,0\right)} and {B=\left(t_{B},x_{B},0,0\right)}. We redefine the origin to be at {A}, giving {B} coordinates of {B'=\left(t_{B}-t_{A},x_{B}-x_{A},0,0\right)}. Assuming the interval is spacelike, the velocity of the frame in {A} and {B} are simultaneous is

\displaystyle   \bar{t}_{B} \displaystyle  = \displaystyle  0=\gamma\left(t_{B}-t_{A}-\beta\left(x_{B}-x_{A}\right)\right)\ \ \ \ \ (14)
\displaystyle  \beta \displaystyle  = \displaystyle  \frac{t_{B}-t_{A}}{x_{B}-x_{A}} \ \ \ \ \ (15)

Rapidity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 19.

An alternative way of writing the Lorentz transformations is to define a quantity called the rapidity:

\displaystyle  \theta\equiv\tanh^{-1}\beta \ \ \ \ \ (1)

Using this definition, we have

\displaystyle   \gamma \displaystyle  = \displaystyle  \frac{1}{\sqrt{1-\beta^{2}}}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{1-\tanh^{2}\theta}}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{\cosh\theta}{\sqrt{\cosh^{2}\theta-\sinh^{2}\theta}}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \cosh\theta \ \ \ \ \ (5)

since {\cosh^{2}\theta-\sinh^{2}\theta=1}.

Also

\displaystyle  \gamma\beta=\cosh\theta\tanh\theta=\sinh\theta \ \ \ \ \ (6)

so

\displaystyle   \Lambda \displaystyle  = \displaystyle  \left[\begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \left[\begin{array}{cccc} \cosh\theta & -\sinh\theta & 0 & 0\\ -\sinh\theta & \cosh\theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (8)

This is similar to a rotation through an angle {\theta} in 3-d space, except both the sinh terms are negative.

The velocity addition formula becomes

\displaystyle   \bar{u} \displaystyle  = \displaystyle  \frac{u+v}{1+uv/c^{2}}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{\beta_{u}+\beta_{v}}{1+\beta_{u}\beta_{v}}c\ \ \ \ \ (10)
\displaystyle  \beta_{\bar{u}} \displaystyle  = \displaystyle  \frac{\tanh\theta_{u}+\tanh\theta_{v}}{1+\tanh\theta_{u}\tanh\theta_{v}}\ \ \ \ \ (11)
\displaystyle  \tanh\theta_{\bar{u}} \displaystyle  = \displaystyle  \tanh\left(\theta_{u}+\theta_{v}\right) \ \ \ \ \ (12)

where in the last line we’ve used the formula for the tanh of a sum of two arguments.

The rapidities therefore simply add, giving a simpler measure of relativistic velocity:

\displaystyle  \theta_{\bar{u}}=\theta_{u}+\theta_{v} \ \ \ \ \ (13)

Compound Lorentz transformations

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 18.

The Lorentz transformations can be written in matrix form as

\displaystyle  \Lambda_{x}=\left[\begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (1)

where the 0 ({ct}) component is the first row and first column, followed by the 1, 2, and 3 directions in order. This matrix is for relative motion along the 1 axis.

The Galilean transformations can be written as a matrix as well, where the first coordinate is just {t} rather than {ct}:

\displaystyle  \Gamma=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ -v & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (2)

or if we want to use the same symbols as in the Lorentz case, where the top row of {\Gamma} is a {ct} coordinate, we can write

\displaystyle  \Gamma=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ -\beta & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (3)

The Lorentz transformation along the 2 ({y}) axis is obtained by putting the transformation terms in row and column 2:

\displaystyle  \Lambda_{y}=\left[\begin{array}{cccc} \gamma & 0 & -\beta\gamma & 0\\ 0 & 1 & 0 & 0\\ -\beta\gamma & 0 & \gamma & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (4)

If we apply a Lorentz transformation first in the {x} and then in the {y} direction (with different relative velocities), we get the compound matrix:

\displaystyle   \Lambda_{y}\Lambda_{x} \displaystyle  = \displaystyle  \left[\begin{array}{cccc} \gamma_{y} & 0 & -\beta_{y}\gamma_{y} & 0\\ 0 & 1 & 0 & 0\\ -\beta_{y}\gamma_{y} & 0 & \gamma_{y} & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} \gamma_{x} & -\beta_{x}\gamma_{x} & 0 & 0\\ -\beta_{x}\gamma_{x} & \gamma_{x} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \left[\begin{array}{cccc} \gamma_{y}\gamma_{x} & -\gamma_{y}\gamma_{x}\beta_{x} & -\beta_{y}\gamma_{y} & 0\\ -\beta_{x}\gamma_{x} & \gamma_{x} & 0 & 0\\ -\beta_{y}\gamma_{y}\gamma_{x} & \gamma_{y}\gamma_{x}\beta_{x}\beta_{y} & \gamma_{y} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (6)

Note that although {\Lambda_{x}} and {\Lambda_{y}} are both symmetric, their product is not. This means that applying the transformations in the opposite order gives a different result.

\displaystyle   \Lambda_{x}\Lambda_{y} \displaystyle  = \displaystyle  \Lambda_{x}^{T}\Lambda_{y}^{T}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \left(\Lambda_{y}\Lambda_{x}\right)^{T}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \left[\begin{array}{cccc} \gamma_{y}\gamma_{x} & -\beta_{x}\gamma_{x} & -\beta_{y}\gamma_{y}\gamma_{x} & 0\\ -\gamma_{y}\gamma_{x}\beta_{x} & \gamma_{x} & \gamma_{y}\gamma_{x}\beta_{x}\beta_{y} & 0\\ -\beta_{y}\gamma_{y} & 0 & \gamma_{y} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (9)

Invariance of scalar product under Lorentz transformations

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 17.

Although the time and position of an event can change under Lorentz transformations, the scalar product of any two four-vectors is an invariant under a Lorentz transformation. That is

\displaystyle  \bar{a}_{i}\bar{b}^{i}=a_{i}b^{i} \ \ \ \ \ (1)

where for motion along the 1-axis ({x} axis) the transformations are

\displaystyle   \bar{a}^{0} \displaystyle  = \displaystyle  \gamma\left(a^{0}-\beta a^{1}\right)\ \ \ \ \ (2)
\displaystyle  \bar{a}^{1} \displaystyle  = \displaystyle  \gamma\left(a^{1}-\beta a^{0}\right)\ \ \ \ \ (3)
\displaystyle  \bar{a}^{2} \displaystyle  = \displaystyle  a^{2}\ \ \ \ \ (4)
\displaystyle  \bar{a}^{3} \displaystyle  = \displaystyle  a^{3} \ \ \ \ \ (5)

and

\displaystyle   a_{0} \displaystyle  = \displaystyle  -a^{0}\ \ \ \ \ (6)
\displaystyle  a_{j} \displaystyle  = \displaystyle  a^{j} \ \ \ \ \ (7)

for {j=1,2,3}. We can see this directly by calculation, using {\gamma^{2}=1/\left(1-\beta^{2}\right)}:

\displaystyle   \bar{a}_{i}\bar{b}^{i} \displaystyle  = \displaystyle  -\gamma^{2}\left(a^{0}-\beta a^{1}\right)\left(b^{0}-\beta b^{1}\right)+\gamma^{2}\left(a^{1}-\beta a^{0}\right)\left(b^{1}-\beta b^{0}\right)+a^{2}b^{2}+a^{3}b^{3}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \gamma^{2}\left[a^{0}b^{0}\left(-1+\beta^{2}\right)+a^{1}b^{0}\left(\beta-\beta\right)+a^{0}b^{1}\left(\beta-\beta\right)+a^{1}b^{1}\left(-\beta^{2}+1\right)\right]+a^{2}b^{2}+a^{3}b^{3}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -a^{0}b^{0}+a^{1}b^{1}+a^{2}b^{2}+a^{3}b^{3}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  a_{i}b^{i} \ \ \ \ \ (11)

The twin paradox analyzed using Lorentz transformations

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 16.

The twin paradox can be analyzed using Lorentz transformations to get an idea of what each twin perceives at key points in the journey. Suppose we have a brother Andy and his twin sister Betty. At {t=t'=0} (which corresponds to the twins’ 21st birthday) and {x=x'=0} in both systems, Betty leaves on a spaceship travelling at {\frac{4}{5}c} heading towards a star {X}. Upon arriving at {X}, Betty immediately transfers to another spaceship and heads back to Earth, also at {\frac{4}{5}c}, arriving on her 39th birthday (according to her own reckoning).

Andy sees Betty as moving in both directions, so to him, Betty’s clock runs slow. Using the time dilation formula, Andy sees a time interval of

\displaystyle   \Delta t_{A} \displaystyle  = \displaystyle  \gamma\Delta t_{B}\ \ \ \ \ (1)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\times18\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  30\mbox{ years} \ \ \ \ \ (3)

Thus Andy will be {21+30=51} years old when Betty arrives home.

According to Andy, Betty’s trip took 15 years each way at a speed of {\frac{4}{5}c}, so star {X} is at a distance of

\displaystyle  d_{X}=\frac{4}{5}\times15c=12c\mbox{ light years} \ \ \ \ \ (4)

In Andy’s frame, the coordinates of Betty’s jump between spaceships is

\displaystyle  \left(x,t\right)=\left(12,15\right) \ \ \ \ \ (5)

In Betty’s outgoing frame, we can apply Lorentz transformations to get her coordinates

\displaystyle   x' \displaystyle  = \displaystyle  \gamma\left(x-vt\right)\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\left(12c-\frac{4}{5}15c\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  0\ \ \ \ \ (8)
\displaystyle  t' \displaystyle  = \displaystyle  \gamma\left(t-\frac{xv}{c^{2}}\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\left(15-\frac{4}{5}12\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  9\mbox{ years} \ \ \ \ \ (11)

These values are consistent, since Betty doesn’t move relative to her own frame, so we’d expect {x'=0}, and since her clock runs slow relative to Andy’s clock, the time interval is shortened by a factor of {1/\gamma} giving {t'=9\mbox{ years}}.

Now consider the frame (indicated by a double prime) fixed to Betty’s returning spaceship. We’ll fix the coordinates so that {t=t'=t"=0} and {x=x'=x"=0}. Its velocity relative to Andy is {-v=-\frac{4}{5}c}, so in this frame, the jump occurs at

\displaystyle   x" \displaystyle  = \displaystyle  \gamma\left(x+vt\right)\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\left(12c+\frac{4}{5}15c\right)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  40\mbox{ light years}\ \ \ \ \ (14)
\displaystyle  t" \displaystyle  = \displaystyle  \gamma\left(t+\frac{xv}{c^{2}}\right)\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\left(15+\frac{4}{5}12\right)\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  41\mbox{ years} \ \ \ \ \ (17)

Thus for Betty to adjust her clock so that it agreed with the {\mathcal{S}"} frame, whe would have to advance it by 32 years just after making the jump. In this frame, the coordinates of her arrival back on Earth are given by transforming Andy’s coordinates of {\left(x,t\right)=\left(0,30\right)}, so we have

\displaystyle   x" \displaystyle  = \displaystyle  \gamma\left(x+vt\right)\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\left(0+\frac{4}{5}30c\right)\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  40\mbox{ light years}\ \ \ \ \ (20)
\displaystyle  t" \displaystyle  = \displaystyle  \gamma\left(t+\frac{xv}{c^{2}}\right)\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{5}{3}\left(30+0\right)\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  50\mbox{ years} \ \ \ \ \ (23)

Note that {x"} doesn’t change as Betty travels home, again because this is her own rest frame. Also, the time interval for her to get home is again 9 years in her own frame, the same as for the outbound journey.

Andy’s age, as viewed by Betty, depends on which spaceship she is on. Just before making the jump she is in the {\mathcal{S}'} frame, so according to her, the time back on Earth at that point is given by taking {t'=9}, {x=0} in the transformation

\displaystyle   t' \displaystyle  = \displaystyle  \gamma\left(t-\frac{xv}{c^{2}}\right)\ \ \ \ \ (24)
\displaystyle  9 \displaystyle  = \displaystyle  \frac{5}{3}\left(t-0\right)\ \ \ \ \ (25)
\displaystyle  t \displaystyle  = \displaystyle  \frac{27}{5}=5.4\mbox{ years} \ \ \ \ \ (26)

That is, Betty thinks Andy is actually 3.6 years younger than she is when she reaches the star. This is because to her, it is Andy who is moving so his clock runs slow by a factor of {1/\gamma} relative to her 9 years. How can we reconcile this with the fact that Andy thinks that Betty takes 15 years to get to the star, so according to him, he is actually 6 years older than Betty when she reaches the star? The confusion arises because of differences in simultaneity as perceived by Andy and Betty. Andy says that the two events (Andy on Earth at age {21+15=36} years, and Betty at the star at age {21+9=30} years) are simultaneous, but Betty disagrees with this, saying that the two events (Andy on Earth at age {21+5.4=26.4} years, and Betty at the star at age {21+9=30} years) are simultaneous. In general, two observers can agree about two events being simultaneous only if these events happen at the same location in both systems.

Just after the jump, Betty can now calculate Andy’s age in the {\mathcal{S}"} system by using {t"=41\mbox{ years}} and {x=0}:

\displaystyle   t" \displaystyle  = \displaystyle  \gamma\left(t+\frac{xv}{c^{2}}\right)\ \ \ \ \ (27)
\displaystyle  41 \displaystyle  = \displaystyle  \frac{5}{3}\left(t+0\right)\ \ \ \ \ (28)
\displaystyle  t \displaystyle  = \displaystyle  \frac{123}{5}=24.6\mbox{ years} \ \ \ \ \ (29)

Thus Betty now thinks that Andy is {21+24.6=45.6\mbox{ years}} old. Andy’s age, of course, hasn’t changed as Betty jumps between ships; only her perception of it has changed.

In her own frame, Betty says the return trip to Earth takes 9 years, so by the same logic as in 26, she says that 5.4 years elapse for Andy, making him {45.6+5.4=51} years old when she arrives home. Thus both Andy and Betty agree on Andy’s age.

The difference between the two twins is, of course, that Betty undergoes deceleration and then acceleration in the reverse direction when she jumps between ships at the star. A more realistic treatment would have Betty gradually decelerating as she approached the star and then gradually accelerating as she got onto the other ship for the journey home. During this process, Betty is continuously changing reference frames resulting in her ‘seeing’ Andy age rapidly (but continuously, rather than the jump in age from 26.4 to 45.6 that happens here) in the process. Thus the paradox is real, in the sense that different amounts of time actually do pass for the two twins.

Velocity addition: chasing space pirates viewed in four reference frames

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 15.

As an example of using the relativistic velocity addition formula, we can revisit the problem of the space police chasing some pirates. In the problem, space pirates are fleeing from the solar system police in a spacecraft that is moving at {\frac{3}{4}c} (relative to the Earth). The police’s spaceship is travelling at only {\frac{1}{2}c} but in an attempt to stop the pirates they fire a torpedo at them. The torpedo’s velocity, relative to the police’s ship, is {\frac{1}{3}c}.

Originally, we worked out the velocity of the torpedo relative to Earth as {\frac{5}{7}c} and saw that the torpedo would not catch the pirates. We can use the velocity addition formula to work out the velocity of each component in the problem relative to every other component and check that the pirates escape no matter how we look at it. If an object has velocity {\bar{u}} in system {\bar{\mathcal{S}}} which is moving at speed {v} relative to system {\mathcal{S}}, then the object’s velocity {u} as measured in {\mathcal{S}} is

\displaystyle  u=\frac{\bar{u}+v}{1+\bar{u}v/c^{2}} \ \ \ \ \ (1)

The velocity of the torpedo relative to the Earth is

\displaystyle  v_{r}=\frac{\frac{1}{2}c+\frac{1}{3}c}{1+\frac{1}{6}}=\frac{5}{7}c=\frac{20}{28}c \ \ \ \ \ (2)

The speed {v_{pp}} of the pirates relative to the police is found from the condition that the velocity {v_{pe}} of the pirates relative to Earth is

\displaystyle  v_{pe}=\frac{v_{pp}+\frac{1}{2}c}{1+\left(\frac{1}{2}c\right)v_{pp}/c^{2}} \ \ \ \ \ (3)

Solving for {v_{pp}} we get

\displaystyle   v_{pp} \displaystyle  = \displaystyle  \frac{v_{pe}-\frac{1}{2}c}{1-\left(\frac{1}{2}c\right)v_{pe}/c^{2}}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{\frac{3}{4}c-\frac{1}{2}c}{1-\left(\frac{1}{2}c\right)\left(\frac{3}{4}c\right)/c^{2}}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{5}c \ \ \ \ \ (6)

Similarly, the velocity {v_{tp}} of the torpedo relative to the pirates is

\displaystyle   v_{tp} \displaystyle  = \displaystyle  \frac{\frac{5}{7}c-\frac{3}{4}c}{1-\left(\frac{5}{7}c\right)\left(\frac{3}{4}c\right)/c^{2}}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{13}c \ \ \ \ \ (8)

In summary, the following matrix gives the velocity of the object named in the column header relative to the object in the row header:

Earth Police Pirates Torpedo Escape?
Earth 0 {\frac{1}{2}c} {\frac{3}{4}c} {\frac{5}{7}c} Yes
Police {-\frac{1}{2}c} 0 {\frac{2}{5}c} {\frac{1}{3}c} Yes
Pirates {-\frac{3}{4}c} {-\frac{2}{5}c} 0 {-\frac{1}{13}c} Yes
Torpedo {-\frac{5}{7}c} {-\frac{1}{3}c} {\frac{1}{13}c} 0 Yes

The matrix is antisymmetric ({A_{ij}=-A_{ji}}) since the velocity of {A} relative to {B} is just the negative of the velocity of {B} relative to {A}. In all cases, {v_{torpedo}<v_{pirates}} so the pirates always escape.

Velocity addition formulas for all 3 directions

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 14.

The relativistic velocity addition formula is most easily derived from the Lorentz transformations:

\displaystyle   \bar{x} \displaystyle  = \displaystyle  \gamma\left(x-vt\right)\ \ \ \ \ (1)
\displaystyle  \bar{t} \displaystyle  = \displaystyle  \gamma\left(t-\frac{xv}{c^{2}}\right) \ \ \ \ \ (2)

As usual, system {\bar{\mathcal{S}}} is moving to the right relative to system {\mathcal{S}} with velocity {v}. If an object is moving in the {\mathcal{S}} system with velocity {u=dx/dt}, then we can take differentials on both sides of the above equations to get

\displaystyle   d\bar{x} \displaystyle  = \displaystyle  \gamma\left(dx-v\; dt\right)\ \ \ \ \ (3)
\displaystyle  d\bar{t} \displaystyle  = \displaystyle  \gamma\left(dt-\frac{v}{c^{2}}dx\right)\ \ \ \ \ (4)
\displaystyle  \bar{u} \displaystyle  = \displaystyle  \frac{d\bar{x}}{d\bar{t}}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{dx-v\; dt}{dt-\frac{v}{c^{2}}dx}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{\left(dx/dt\right)-v}{1-\frac{v}{c^{2}}\left(dx/dt\right)}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{u-v}{1-uv/c^{2}}\ \ \ \ \ (8)
\displaystyle  \dot{\bar{x}} \displaystyle  = \displaystyle  \frac{\dot{x}-v}{1-\dot{x}v/c^{2}} \ \ \ \ \ (9)

By symmetry (or algebra if you don’t trust the physical argument) this relation can be inverted to give

\displaystyle  u=\frac{\bar{u}+v}{1+\bar{u}v/c^{2}} \ \ \ \ \ (10)

We can also derive the formulas for transforming velocity components perpendicular to the motion. Although perpendicular distances don’t change under the Lorentz transformation, perpendicular velocities do, because they involve the ratio of perpendicular distance to time, and time does change from one system to another. We get, with a dot indicating a time derivative in the respective system:

\displaystyle   \dot{\bar{y}} \displaystyle  = \displaystyle  \frac{d\bar{y}}{d\bar{t}}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{dy}{\gamma\left(dt-\frac{v}{c^{2}}dx\right)}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{\dot{y}}{\gamma\left(1-\dot{x}v/c^{2}\right)}\ \ \ \ \ (13)
\displaystyle  \dot{\bar{z}} \displaystyle  = \displaystyle  \frac{\dot{z}}{\gamma\left(1-\dot{x}v/c^{2}\right)} \ \ \ \ \ (14)

Thus perpendicular velocities depend not only on the perpendicular velocity in the original system, but also on the horizontal velocity.

Example Suppose we have a spotlight mounted on the roof of our usual train, and the spotlight is aimed to point towards the rear of the train, making an angle of {\theta} with the roof of the train. If the train moves with speed {v}, what angle will a ground observer see the beam make with the train’s roof?

In the train’s system, the light’s velocity components are

\displaystyle   \dot{x} \displaystyle  = \displaystyle  -c\cos\theta\ \ \ \ \ (15)
\displaystyle  \dot{z} \displaystyle  = \displaystyle  c\sin\theta \ \ \ \ \ (16)

In the ground system, we get from 10:

\displaystyle   \dot{\bar{x}} \displaystyle  = \displaystyle  \frac{-c\cos\theta+v}{1-v\left(\cos\theta\right)/c}\ \ \ \ \ (17)
\displaystyle  \dot{\bar{z}} \displaystyle  = \displaystyle  \frac{c\sin\theta}{\gamma\left(1-v\left(\cos\theta\right)/c\right)}\ \ \ \ \ (18)
\displaystyle  \tan\bar{\theta} \displaystyle  = \displaystyle  -\frac{\dot{\bar{z}}}{\dot{\bar{x}}}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{c\sin\theta}{\gamma\left(c\cos\theta-v\right)} \ \ \ \ \ (20)

As a check, we can work out {\dot{\bar{x}}^{2}+\dot{\bar{z}}^{2}=c^{2}} (after simplifying). When the beam points directly up in the train’s frame ({\theta=\frac{\pi}{2}}), it points slightly forward in the ground’s frame. When {\cos\theta=\frac{v}{c}}, {\tan\bar{\theta}=\infty} so the beam points directly up in the ground’s frame.

Lorentz transformations and simultaneity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 13.

Here’s a simple example of using the Lorentz transformations. Suppose that two events occur simultaneously in the Earth frame, a distance 500 km apart. An observer in a (very fast) plane travelling at {\frac{12}{13}c} along the line joining the events passes the first event {A} such that their respective coordinate origins coincide at that event. At what time does the plane observer think event {B} occurs?

The Lorentz transformations are

\displaystyle   \bar{x} \displaystyle  = \displaystyle  \gamma\left(x-vt\right)\ \ \ \ \ (1)
\displaystyle  \bar{t} \displaystyle  = \displaystyle  \gamma\left(t-\frac{xv}{c^{2}}\right) \ \ \ \ \ (2)

Event {A} occurs at the same time {t=\bar{t}=0} in both systems. For event {B}, {x=500\mbox{ km}} and {t=0}, so

\displaystyle   \bar{t} \displaystyle  = \displaystyle  \gamma\left(-\frac{500v}{c^{2}}\right)\ \ \ \ \ (3)
\displaystyle  \gamma \displaystyle  = \displaystyle  \frac{13}{5}\ \ \ \ \ (4)
\displaystyle  \bar{t} \displaystyle  = \displaystyle  -\frac{13}{5}\left(\frac{12c}{13}\frac{500}{c^{2}}\right)\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1200}{c}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1200\mbox{ km}}{3\times10^{5}\mbox{ km s}^{-1}}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -4\times10^{-3}\mbox{ s} \ \ \ \ \ (8)

Thus the plane observer thinks that {B} occurs 4 milliseconds before {A}.

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