Four-velocity of a particle in hyperbolic motion

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 27.

[Griffiths’s approach to the relativistic four-velocity is similar to that of Moore, although rather confusingly, he uses different notation (as well as keeping factors of {c} in the equations rather than setting {c=1}). To keep the notation consistent with Griffiths, I’ll use his notation here, but anyone attempting to follow both books should beware…]

We can now return to a particle travelling on a hyperbolic trajectory, so its position (one dimensional, on the {x} axis) is

\displaystyle  x\left(t\right)=\sqrt{b^{2}+c^{2}t^{2}} \ \ \ \ \ (1)

Here, {x} and {t} are the position and time as measured by an observer at rest (so they are not proper time for the particle). We can find the particle’s proper time as a function of {t} by using the relation between time intervals:

\displaystyle  d\tau=\sqrt{1-u^{2}/c^{2}}dt \ \ \ \ \ (2)

We have

\displaystyle   u \displaystyle  = \displaystyle  \dot{x}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{c^{2}t}{\sqrt{b^{2}+c^{2}t^{2}}}\ \ \ \ \ (4)
\displaystyle  d\tau \displaystyle  = \displaystyle  \sqrt{1-\frac{c^{2}t^{2}}{b^{2}+c^{2}t^{2}}}dt\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{b}{\sqrt{b^{2}+c^{2}t^{2}}}dt \ \ \ \ \ (6)

Integrating both sides, we get, taking {\tau=0} when {t=0} (the integral can be done by software or looked up as it is a standard integral):

\displaystyle   \tau \displaystyle  = \displaystyle  b\int_{0}^{t}\frac{1}{\sqrt{b^{2}+c^{2}\left(t'\right)^{2}}}dt'\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{b}{c}\ln\left[\frac{1}{b}\left(ct+\sqrt{b^{2}+c^{2}t^{2}}\right)\right] \ \ \ \ \ (8)

We can write the position as a function of {\tau} by starting with 1 and using this last result:

\displaystyle   \tau \displaystyle  = \displaystyle  \frac{b}{c}\ln\left[\frac{1}{b}\left(ct+x\right)\right]\ \ \ \ \ (9)
\displaystyle  ct \displaystyle  = \displaystyle  \sqrt{x^{2}-b^{2}}\ \ \ \ \ (10)
\displaystyle  \tau \displaystyle  = \displaystyle  \frac{b}{c}\ln\left[\frac{1}{b}\left(\sqrt{x^{2}-b^{2}}+x\right)\right]\ \ \ \ \ (11)
\displaystyle  be^{c\tau/b} \displaystyle  = \displaystyle  \sqrt{x^{2}-b^{2}}+x\ \ \ \ \ (12)
\displaystyle  \left(be^{c\tau/b}-x\right)^{2} \displaystyle  = \displaystyle  x^{2}-b^{2}\ \ \ \ \ (13)
\displaystyle  x \displaystyle  = \displaystyle  \frac{b}{2}\left(e^{c\tau/b}+e^{-c\tau/b}\right)\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  b\cosh\frac{c\tau}{b} \ \ \ \ \ (15)

For the ordinary velocity, we have

\displaystyle   u \displaystyle  = \displaystyle  \frac{c^{2}t}{\sqrt{b^{2}+c^{2}t^{2}}}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  c\frac{\sqrt{x^{2}-b^{2}}}{x}\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  cb\frac{\sqrt{\cosh^{2}\frac{c\tau}{b}-1}}{b\cosh\frac{c\tau}{b}}\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  c\tanh\frac{c\tau}{b}\nonumber

The four velocity is defined as

\displaystyle  \eta^{i}=\frac{u^{i}}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (19)


so we have (I’m assuming that in part (c) of Griffiths’s problem, he meant to ask for {\eta^{i}} in terms of {\tau}, not {t}, as the latter doesn’t give anything particularly informative):

\displaystyle   \eta^{0} \displaystyle  = \displaystyle  \frac{c}{\sqrt{1-\tanh^{2}\frac{c\tau}{b}}}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  c\cosh\frac{c\tau}{b}\ \ \ \ \ (21)
\displaystyle  \eta_{x} \displaystyle  = \displaystyle  \frac{u}{\sqrt{1-\tanh^{2}\frac{c\tau}{b}}}\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  \frac{c\tanh\frac{c\tau}{b}}{\sqrt{1-\tanh^{2}\frac{c\tau}{b}}}\ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  c\sinh\frac{c\tau}{b} \ \ \ \ \ (24)

As a check, we note that

\displaystyle  \eta_{i}\eta^{i}=-c^{2}\left(\cosh^{2}\frac{c\tau}{b}-\sinh^{2}\frac{c\tau}{b}\right)=-c^{2} \ \ \ \ \ (25)

Four-velocity’s square is a universal constant

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 26.

[Griffiths’s approach to the relativistic four-velocity is similar to that of Moore, although rather confusingly, he uses different notation (as well as keeping factors of {c} in the equations rather than setting {c=1}). To keep the notation consistent with Griffiths, I’ll use his notation here, but anyone attempting to follow both books should beware…]

The four velocity is defined as

\displaystyle  \eta^{i}=\frac{u^{i}}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (1)


so the square is

\displaystyle   \eta_{i}\eta^{i} \displaystyle  = \displaystyle  -\frac{c^{2}}{1-u^{2}/c^{2}}+\frac{\sum_{i=1}^{3}\left(u^{i}\right)^{2}}{1-u^{2}/c^{2}}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{-c^{2}+u^{2}}{1-u^{2}/c^{2}}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  -c^{2} \ \ \ \ \ (4)

This is the equivalent of the result we derived earlier in Moore’s book, except with {c=1} so {\eta_{i}\eta^{i}=-1}. It is a universal invariant for all four-velocities.

Four-velocity versus ordinary velocity: an example

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 25.

[Griffiths’s approach to the relativistic four-velocity is similar to that of Moore, although rather confusingly, he uses different notation (as well as keeping factors of {c} in the equations rather than setting {c=1}). To keep the notation consistent with Griffiths, I’ll use his notation here, but anyone attempting to follow both books should beware…]

Here’s an example of calculating the ordinary and four-velocities of an object. Suppose the object moves with ordinary velocity (as measured in the ground frame) {\frac{2}{\sqrt{5}}c} in the {45^{\circ}} direction (towards the upper right; note that there is a typo in Griffiths’s statement of the problem as he states that the velocity is {2\sqrt{5}c} which is, of course, greater than {c}). This resolves into components:

\displaystyle  \mathbf{u}=\sqrt{\frac{2}{5}}c\hat{\mathbf{x}}+\sqrt{\frac{2}{5}}c\hat{\mathbf{y}} \ \ \ \ \ (1)

The four velocity is defined as

\displaystyle  \eta^{i}=\frac{u^{i}}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (2)


so the components are

\displaystyle  \eta_{x}=\eta_{y}=\sqrt{2}c \ \ \ \ \ (3)

We now introduce another frame moving in the {x} direction with ordinary speed {\sqrt{\frac{2}{5}}c}. The ordinary velocity of the object transforms using the velocity addition formulas

\displaystyle   \dot{\bar{x}} \displaystyle  = \displaystyle  \frac{\dot{x}-v}{1-\dot{x}v/c^{2}}\ \ \ \ \ (4)
\displaystyle  \dot{\bar{y}} \displaystyle  = \displaystyle  \frac{\dot{y}}{\gamma\left(1-\dot{x}v/c^{2}\right)} \ \ \ \ \ (5)

so we get, using {\gamma=1/\sqrt{1-\frac{2}{5}}=\sqrt{\frac{5}{3}}}:

\displaystyle   \bar{u}_{x} \displaystyle  = \displaystyle  0\ \ \ \ \ (6)
\displaystyle  \bar{u}_{y} \displaystyle  = \displaystyle  \frac{\sqrt{2/5}}{\sqrt{5/3}\left(1-2/5\right)}c\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{2}{3}}c \ \ \ \ \ (8)

As {\eta^{i}} is a four-vector, it transforms using Lorentz transformations, for which we need

\displaystyle  \eta^{0}=\frac{c}{\sqrt{1-u^{2}/c^{2}}}=\sqrt{5}c \ \ \ \ \ (9)

so

\displaystyle   \bar{\eta}_{x} \displaystyle  = \displaystyle  \gamma\left(\eta_{x}-\beta\eta^{0}\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{5}{3}}\left(\sqrt{2}c-\sqrt{\frac{2}{5}}\sqrt{5}c\right)=0\ \ \ \ \ (11)
\displaystyle  \bar{\eta}_{y} \displaystyle  = \displaystyle  \eta_{y}=\sqrt{2}c \ \ \ \ \ (12)

We can check this using 2:

\displaystyle   \bar{\eta}_{x} \displaystyle  = \displaystyle  \frac{\bar{u}_{x}}{\sqrt{1-\bar{u}^{2}/c^{2}}}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  0\ \ \ \ \ (14)
\displaystyle  \bar{\eta}_{y} \displaystyle  = \displaystyle  \frac{\bar{u}_{y}}{\sqrt{1-\bar{u}^{2}/c^{2}}}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \frac{\sqrt{2/3}c}{\sqrt{1/3}}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{2}c \ \ \ \ \ (17)

so it checks out.

Four-velocity again

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 24.

Griffiths’s approach to the relativistic four-velocity is similar to that of Moore, although rather confusingly, he uses different notation (as well as keeping factors of {c} in the equations rather than setting {c=1}). To keep the notation consistent with Griffiths, I’ll use his notation here, but anyone attempting to follow both books should beware…

The four-velocity (or proper velocity as Griffiths calls it) is defined using the symbol {\eta} (which is used by Moore for the flat space metric) as the derivative of the four-position with respect to proper time:

\displaystyle  \eta^{i}\equiv\frac{dx^{i}}{d\tau} \ \ \ \ \ (1)

Ordinary velocity (which Griffiths calls {u}; Moore uses {u} for the four-velocity. As I say, it gets confusing.) is defined as the derivative of the four-position with respect to the time component of the four-position:

\displaystyle  u^{i}=\frac{dx^{i}}{dx^{0}}c \ \ \ \ \ (2)

where the factor of {c} is there to cancel out the {c} in {x^{0}=ct}. Since proper time and coordinate time are related by

\displaystyle  d\tau=\sqrt{1-\frac{u^{2}}{c^{2}}}dt \ \ \ \ \ (3)

we have

\displaystyle   dx^{0} \displaystyle  = \displaystyle  c\; dt\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{c\; d\tau}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (5)

where {u^{2}=\sum_{i=1}^{3}\left(u^{i}\right)^{2}}, that is, the square magnitude of the spatial ordinary velocity. Therefore

\displaystyle  \eta^{i}=\frac{u^{i}}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (6)

In particular

\displaystyle   \eta^{0} \displaystyle  = \displaystyle  \frac{u^{0}}{\sqrt{1-u^{2}/c^{2}}}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{c}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (8)

Note that the components {\eta^{i}} can be larger than {c}. This is OK because the four-velocity is the derivative of distance in one coordinate system with respect to the time coordinate in another system, so we’re not actually calculating the speed of an object as measured in one specific frame (unless that frame is the object’s own rest frame, in which case {u=0} and all components {\eta^{i}} are safely less than {c}).

If we sum up the squares of the spatial components, we get

\displaystyle   \eta^{2}\equiv\sum_{i=1}^{3}\left(\eta^{i}\right)^{2} \displaystyle  = \displaystyle  \frac{u^{2}}{1-u^{2}/c^{2}}\ \ \ \ \ (9)
\displaystyle  u^{2} \displaystyle  = \displaystyle  \frac{\eta^{2}}{1+\eta^{2}/c^{2}}\ \ \ \ \ (10)
\displaystyle  \mathbf{u} \displaystyle  = \displaystyle  \frac{\boldsymbol{\eta}}{\sqrt{1+\eta^{2}/c^{2}}} \ \ \ \ \ (11)

where bold-face symbols represent 3-d (spatial) vectors.

We can express {\eta} in terms of rapidity {\theta}, defined as

\displaystyle  \frac{u}{c}\equiv\tanh\theta \ \ \ \ \ (12)

We get, assuming the motion is in the {x} direction (using {\cosh^{2}\theta-\sinh^{2}\theta=1} in the denominator):

\displaystyle   \eta^{2} \displaystyle  = \displaystyle  \frac{c^{2}\tanh^{2}\theta}{1-\tanh^{2}\theta}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  c^{2}\sinh^{2}\theta\ \ \ \ \ (14)
\displaystyle  \eta \displaystyle  = \displaystyle  c\sinh\theta \ \ \ \ \ (15)

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)

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