Special Relativity with Geometric Algebra - Spacetime Rotors

Velocity addition close to the speed of light

To motivate spacetime rotors, we need more trains (these are very popular in relativity apparently).
Figure 1 - Left: Alice is on a train throwing a ball and sees Bob outside the train, Right: Bob looks at Alice throwing a ball in her moving train
Figure 1 - Left: Alice is on a train throwing a ball and sees Bob outside the train, Right: Bob looks at Alice throwing a ball in her moving train
Consider Alice is now on a train. On the train, she is throwing a ball with speed0.4c(yes, yes... very contrived). Outside of the train is Bob who can see the train moving at0.8c. Alice, on the train, sees herself standing still but sees Bob passing by her at0.8c(ie. same speed but in the opposite direction of the train). We can now ask the question of how fast is the ball moving from the point of view of Bob.
If we just add up the speed of the train as seen by Bob and the speed of the ball as seen by Alice we get0.8c+0.4c=1.2cbut this can't be as nothing can exceed the speed of light. Let's recall some things:
  • From the paths section we know that the angle of paths is related to their speed
  • We also know that resting observers have path velocities proportional toet
  • Claim: We can use our new rotors to rotate between path velocities to find the correct velocity from Bob's perspective
Okay this might sound far-fetched first. Rotate between spacetime velocities? Let's look at this sketch.
Figure 2 - Diagonal yellow: Light. Left: Alice's view, Alice (blue) is at rest, Bob (purple) moves left, Ball (green) moves right. Right: Bob's view, Alice (blue) moves right, Bob (purple) is at rest, Ball (green) is moves right.
Figure 2 - Diagonal yellow: Light. Left: Alice's view, Alice (blue) is at rest, Bob (purple) moves left, Ball (green) moves right. Right: Bob's view, Alice (blue) moves right, Bob (purple) is at rest, Ball (green) is moves right.
Alice's view (left)
Alice (blue) is at rest so her path velocity vector points in heretdirection. From Alice's view Bob (purple) is moving left so his velocity vector points left. The rotation from Bob's to Alice's velocity vector is done by applying the rotorR1to Bob's velocity vector. The ball moves to the right, so its velocity vector is rotated to the right. The rotation between Alice and the ball is done with the rotorR2.
The rotations between these vectors are all active transformations because the three velocity vectors are actually different from each other and it is not just Alice changing her perspective on a single unchanging vector.
Bob's view (right)
Bob's velocity vector (purple) is at rest, ie. points in hisetdirection. Alice's velocity vector is rotated right withR1and the ball is rotated relative to Alice's velocity vector withR2.
Velocity Rotors
Now the key observation: the rotors in the left and the right diagram are identical. The values in them are the same (unlike for the vectors!). So if we know them in one diagram we can use them in the other. We knowR1from both views, but we only knowR2from Alice's view. But if we calcualteR2in Alice's view, we can just apply it to Alice's velocity vector in Bob's view.
So that's the idea. How to actually do this and how the angle relates to the speed comes in the next section.

Rotors in the Spacetime Algebra

To continue, we need to look at what the rotors in the Spacetime Algebra actually do. We will only need the rotors for bivectors squaring to+1, ie.etx,ety,etz. Let's stick withetxfor now. How do we get rotors from bivectors? We stick them in the exponential function with an angleφ, usually negated and halved because we need to apply a two-sided product.
(1)R(φ)=eφ2etx
Can we expand this further? Usually we get something involving sine and cosine for performing ordinary rotations, but that is because the bivector in the exponential squared to1. It turns out in the case of+1we get hyperbolic sine and hyperbolic cosine. You can get this result by using the taylor series for the exponential function and comparing it with the taylor series of the hyperbolic sine and cosine functions. Anyway, we now get
(2)R(φ)=eφ2etx=cosh(φ2)+sinh(φ2)etx
What does this rotor do? It's not ordinary rotation, it is hyperbolic rotation! What is hyperbolic rotation? I promised you earlier we could use these rotors to rotate velocity vectors and use them for velocity addition. Let's apply them to the rest velocity vectoretand see what we can infer from that
(3)R(φ)etR~(φ)=cosh(φ)et+sinh(φ)ex
The steps between the left-hand side and right-hand side are left as an excercise. This will require some identities involving the hyperbolic sine and cosine functions, or you can just throw them into Wolfram Alpha to see what the parts simplify to.
We can see the result is some mix ofetandex.
Figure 3 - Yellow: Light. Dark blue: Velocity of object at rest. Light blue: Velocity of moving object. Orange: Rotor between object at rest and moving object.
Figure 3 - Yellow: Light. Dark blue: Velocity of object at rest. Light blue: Velocity of moving object. Orange: Rotor between object at rest and moving object.
To get some more intuition let's look at the slope of the result. Theetpart is on the y-axis and theexpart is on the x-axis so we have
(4)slope(φ)=sinh(φ)cosh(φ)=1tanhφ
So the slope is equal to one over the hyperbolic tangenttanh. If you're familiar with neural networks you probably already know how thetanhfunction looks (and perhaps other fields I'm not familiar with). Here's a sketch
Figure 4 - Hyperbolic tangent function, zero at phi equal to 0, one towards positive and negative infinity.
Figure 4 - Hyperbolic tangent function, zero at phi equal to 0, one towards positive and negative infinity.
The second part we need for plotting what the result looks like is the length of the resulting vector. The length in our graph is the euclidean length. This is of course not the length / distance we use in the algebra, but in our drawings, if theexandetcomponents get bigger, then the vector we draw always gets bigger too. So we also have
(5)graphlength(φ)=cosh2(φ)+sinh2(φ)
Now we take a deeper look at the slope result we got earlier for different rotor anglesφ.
Ifφ=0, then the slope is(vertical) and the graph length is1. This makes sense as rotating by0should leave the input unchanged. Furthermore, forφ, we can see thatslope1. As we increase the angle of the rotor, the slope tends towards1. A slope of1is a 45° angle. Recall that light travels at 45° angles, so this is good! If we increase the angle, our velocity vector rotates towards the speed of light. Using these rotors we can never exceed (or even reach) 45°. Finally, the length also tends towards infinity, meaning if we rotate the rest velocity vector, its length will always increase.
Relation between velocity and angle
Given what we learnt, we can try to figure out the relationship between the angleφand the velocityv. We know that the slope for light is1, sotanh(φlight)=1. We also know that the rest velocity has slope infinity (vertical), ie.tanh(0)=0. Based on these two data points, what do you think the relation is? The correct answer that also fits both data points is
(6)tanh(φ)=vc
We will use this relation in the next section to finally solve our velocity addition problem.

Velocity addition close to the speed of light: revisited

Figure 5 - Diagonal yellow: Light. Left: Alice's view, Alice (blue) is at rest, Bob (purple) moves left, Ball (green) moves right. Right: Bob's view, Alice (blue) moves right, Bob (purple) is at rest, Ball (green) is moves right.
Figure 5 - Diagonal yellow: Light. Left: Alice's view, Alice (blue) is at rest, Bob (purple) moves left, Ball (green) moves right. Right: Bob's view, Alice (blue) moves right, Bob (purple) is at rest, Ball (green) is moves right.
And we're back. This time equipped with the power of rotors that can rotate velocities and the relation between angles and velocities! We want to figure out the velocity of the ball from Bob's perspective (which we don't know yet). For that we look at Bob's frame where he is at rest and apply the two rotors (which we do know). Let me show you one more thing (that you probably already knew!) about rotors before we tackle the problem.
When we compose two rotors in the same plane we get (no matter what the bivector squares to)
(7)R(φ1)R(φ2)=eφ12etxeφ22etx=eφ1+φ22etx=R(φ1+φ2)
When composing two rotations the angles just add. This is still the case even with our hyperbolic rotations. It turns out to solve the problems we don't even need to use rotors directly. First we calculate the two angles corresponding to the two velocities we were given. To get the angle from velocity we solve(6)forv. Then we add the angles. Finally we need to convert the resulting angle back to a velocity using(6), and that will be the velocity of the ball as seen from Bob.
(8)vball=ctanh(φ1+φ2)=ctanh(tanh1(0.8)+tanh1(0.4))0.9c
The ball moves at around0.9cfrom Bob's perspective.

Conclusion

We started by noticing that adding velocities close to the speed of light fails with ordinary addition because the speed of light can be exceeded. We then saw how we can relate velocity vectors using rotors with bivectors that square to+1. We noticed that these rotors rotate velocity vectors towards the speed of light but never exceed it. We also saw that the rotors don't change with passive transformations so we could compose them to solve our velocity addition problem. The result was that we can add the rotor angles for rotors in the same plane and relate them to velocities using the formulatanh(φ)=vc.

Formulas

  • Relation between angle and velocity:tanh(φ)=vc
  • Addition of angles:φa+b=φa+φb
  • Addition of velocities:va+b=ctanh(φa+b)=ctanh(tanh1(vac)+tanh1(vbc))

Up next

Next we will look at how we can take measurements in Spacetime.

Special Relativity with Geometric Algebra - Measuring in Spacetime