15b. Einstein addition
As another application of the vector products, let us discuss now the speed addition in relativity. Based on experiments by Fizeau, then Michelson-Morley and others, and some physics by Maxwell and Lorentz, Einstein came upon the following principles: \begin{fact}[Einstein principles] The following happen:
- Light travels in vacuum at a finite speed, [math]c \lt \infty[/math].
- This speed [math]c[/math] is the same for all inertial observers.
- In non-vacuum, the light speed is lower, [math]v \lt c[/math].
- Nothing can travel faster than light, [math]v\not \gt c[/math].
\end{fact} The point now is that, obviously, something is wrong here. Indeed, assuming for instance that we have a train, running in vacuum at speed [math]v \gt 0[/math], and someone on board lights a flashlight [math]\ast[/math] towards the locomotive, an observer [math]\circ[/math] on the ground will see the light travelling at speed [math]c+v \gt c[/math], which is a contradiction:
Equivalently, with the same train running, in vacuum at speed [math]v \gt 0[/math], if the observer on the ground lights a flashlight [math]\ast[/math] towards the back of the train, then viewed from the train, that light will travel at speed [math]c+v \gt c[/math], which is a contradiction again:
Summarizing, Fact 15.10, while physically true, implies [math]c+v=c[/math], so contradicts classical mechanics, which needs a fix. In the classical case, to start with, we have:
The classical speeds add according to the Galileo formula
This is clear indeed from the definition of speed, and very intuitive.
In order to find the fix, we will first discuss the 1D case, and leave the 3D case, which is a bit more complicated, for later. We will use two tricks. First, let us forget about absolute speeds, with respect to a given frame, and talk about relative speeds only. In this case we are allowed to sum only quantities of type [math]v_{AB},v_{BC}[/math], and we denote by [math]v_{AB}+_gv_{BC}[/math] the corresponding sum [math]v_{AC}[/math]. With this convention, the Galileo formula becomes:
As a second trick now, observe that this Galileo formula holds in any system of units. In order now to deal with our problems, basically involving high speeds, it is convenient to change the system of units, as to have [math]c=1[/math]. With this convention our [math]c+v=c[/math] problem becomes [math]1+v=1[/math], and the solution to it is quite obvious, as follows:
If we define the Einstein sum [math]+_e[/math] of relative speeds by
This is obvious indeed from our definition of [math]+_e[/math], because if we plug in [math]u=1[/math] in the above formula, we obtain as result:
Thus, we are led to the conclusion in the statement.
Summarizing, we have solved our problem. In order now to formulate a final result, we must do some reverse engineering, by waiving the above two tricks. First, by getting back to usual units, [math]v\to v/c[/math], our new addition formula becomes:
By multiplying by [math]c[/math], we can write this formula in a better way, as follows:
In order now to finish, it remains to get back to absolute speeds, as in Proposition 15.11. And by doing so, we are led to the following result:
If we sum the speeds according to the Einstein formula
We have two assertions here, which are both clear, as follows:
(1) Regarding the first assertion, if we are at low speeds, [math]v_{AB},v_{BC} \lt \lt c[/math], the correction term [math]v_{AB}v_{BC}/c^2[/math] dissapears, and we are left with the Galileo formula, as claimed.
(2) As for the second assertion, this follows from the above discussion.
The Einstein summation formula, while looking very simple, is in fact quite subtle, and must be handled with care. Indeed, getting back to [math]c=1[/math] conventions, we have:
The Einstein speed summation, written in [math]c=1[/math] units as
- [math]u,v \lt 1[/math] implies [math]u+_ev \lt 1[/math].
- [math]u+_ev=v+_eu[/math].
- [math](u+_ev)+_ew=u+_e(v+_ew)[/math].
- However, [math]\lambda u+_e\lambda v=\lambda(u+_ev)[/math] fails.
All these assertions are elementary, as follows:
(1) This follows from the following formula, valid for any speeds [math]u,v[/math]:
(2) This is clear too, coming from the following computation:
(3) We have indeed the following computation:
(4) This is clear too, with the remark however that the formula [math]\lambda u+_e\lambda v=\lambda(u+_ev)[/math] works at [math]\lambda=-1,0,1[/math], or when [math]u=0[/math], [math]v=0[/math], or [math]u+v=0[/math].
All the above is very nice, but remember, takes place in 1D. So, time now to get seriously to work, and see what all this becomes in 3D. Expect of course a lot a vector calculus, as usual in relation with 3D problems, and in the hope that you love that. As a main goal, we must review the Einstein speed summation formula: \begin{question} What is the correct analogue of the Einstein summation formula
in [math]2[/math] and [math]3[/math] dimensions? \end{question} In order to discuss this question, let us attempt to construct [math]u+_ev[/math] in arbitrary dimensions, just by using our common sense and intuition. When the vectors [math]u,v\in\mathbb R^N[/math] are proportional, we are basically in 1D, and so our addition formula must satisfy:
However, the formula on the right will not work as such in general, for arbitrary speeds [math]u,v\in\mathbb R^N[/math], and this because we have, as main requirement for our operation, in analogy with the [math]1+_ev=1[/math] formula from 1D, the following condition:
Equivalently, in analogy with [math]u+_e1=1[/math] from 1D, we would like to have:
Summarizing, our [math]u\sim v[/math] formula above is not bad, as a start, but we must add a correction term to it, for the above requirements to be satisfied, and of course with the correction term vanishing when [math]u\sim v[/math]. So, we are led to a math puzzle:
\begin{puzzle}
What vanishes when [math]u\sim v[/math], and then how to correctly define
as for the correction term [math]\gamma_{uv}[/math] to vanish when [math]u\sim v[/math]? \end{puzzle} But the solution to the first question is well-known in 3D. Indeed, here we can use the vector product [math]u\times v[/math], that we met before, which notoriously satisfies:
Thus, our correction term [math]\gamma_{uv}[/math] must be something containing [math]w=u\times v[/math], which vanishes when this vector [math]w[/math] vanishes, and in addition arranged such that [math]||u||=1[/math] produces a simplification, with [math]u+_ev=u[/math] as end result, and with [math]||v||=1[/math] producing a simplification too, with [math]u+_ev=v[/math] as end result. Thus, our vector calculus puzzle becomes:
\begin{puzzle}
How to correctly define the Einstein summation in [math]3[/math] dimensions,
with [math]w=u\times v[/math], in such a way as for the correction term [math]\gamma_{uvw}[/math] to satisfy
and also such that [math]||u||=1\implies u+_ev=u[/math], and [math]||v||\implies u+_ev=v[/math]? \end{puzzle} In order to solve this latter puzzle, the first observation is that [math]\gamma_{uvw}=w[/math] will not do, and this for several reasons. First, this vector points in the wrong direction, orthogonal to the plane spanned by [math]u,v[/math], and we certainly don't want to leave this plane, with our correction. Also, as a technical remark to be put on top of this, the choice [math]\gamma_{uvw}=w[/math] will not bring any simplifications, as required above, in the cases [math]||u||=1[/math] or [math]||v||=1[/math]. Thus, certainly wrong choice, and we must invent something more complicated.
Moving ahead now, as obvious task, we must “transport” the vector [math]w[/math] to the plane spanned by [math]u,v[/math]. But this is simplest done by taking the vector product with any vector in this plane, and so as a reasonable candidate for our correction term, we have:
Here [math]\alpha,\beta\in\mathbb R[/math] are some scalars to be determined, but let us take a break, and leave the computations for later. We did some good work, time to update our puzzle:
\begin{puzzle}
How to define the Einstein summation in [math]3[/math] dimensions,
with the correction term being of the following form, with [math]w=u\times v[/math], and [math]\alpha,\beta\in\mathbb R[/math],
in such a way as to have [math]||u||=1\implies u+_ev=u[/math], and [math]||v||\implies u+_ev=v[/math]? \end{puzzle} In order to investigate what happens when [math]||u||=1[/math] or [math]||v||=1[/math], we must compute the vector products [math]u\times w[/math] and [math]v\times w[/math]. So, pausing now our study for consulting the vector calculus database, and then coming back, here is the formula that we need:
As for the formula of [math]v\times w[/math], that I forgot to record, we can recover it from the one above of [math]u\times w[/math], by using the basic properties of the vector products, as follows:
With these formulae in hand, we can now compute the correction term, with the result here, that we will need several times in what comes next, being as follows:
The correction term [math]\gamma_{uvw}=(\alpha u+\beta v)\times w[/math] is given by
According to our vector product formulae above, we have:
Thus, we are led to the conclusion in the statement.
Time now to get into the real thing, see what happens when [math]||u||=1[/math] and [math]||v||=1[/math], if we can get indeed [math]u+_ev=u[/math] and [math]u+_ev=v[/math]. It is convenient here to do some reverse engineering. Regarding the first desired formula, namely [math]u+_ev=u[/math], we have:
Thus, with the parameter choice [math]\alpha=1,\beta=0[/math], we will have, as desired:
In what regards now the second desired formula, namely [math]u+_ev=v[/math], here the computation is almost identical, save for a sign switch, which after some thinking comes from our choice [math]w=u\times v[/math] instead of [math]w=v\times u[/math], clearly favoring [math]u[/math], as follows:
Thus, with the parameter choice [math]\alpha=0,\beta=-1[/math], we will have, as desired:
All this is mixed news, because we managed to solve both our problems, at [math]||u||=1[/math] and at [math]||v||=1[/math], but our solutions are different. So, time to breathe, decide that we did enough interesting work for the day, and formulate our conclusion as follows:
When defining the Einstein speed summation in [math]3{\rm D}[/math] as
- When [math]u\sim v[/math], we recover the previous [math]1{\rm D}[/math] formula.
- When [math]||u||=1[/math], speed of light, we have [math]u+_ev=u[/math].
- However, [math]||v||=1[/math] does not imply [math]u+_ev=v[/math].
- Also, the formula [math]u+_ev=v+_eu[/math] fails.
Here (1) and (2) follow from the above discussion, with the following choice for the correction term, by favoring the [math]||u||=1[/math] problem over the [math]||v||=1[/math] one:
In fact, with this choice made, the computation is very simple, as follows:
As for (3) and (4), these are also clear from the above discussion, coming from the obvious lack of symmetry of our summation formula.
Looking now at Proposition 15.20 from an abstract, mathematical perspective, there are still many things missing from there, which can be summarized as follows: \begin{question} Can we fine-tune the Einstein speed summation in [math]3{\rm D}[/math] into
with [math]\lambda\in\mathbb R[/math], chosen such that [math]||u||=1\implies\lambda=1[/math], as to have:
- [math]||u||,||v|| \lt 1\implies||u+_ev|| \lt 1[/math].
- [math]||v||=1\implies||u+_ev||=1[/math].
\end{question} All this is quite tricky, and deserves some explanations. First, if we add a scalar [math]\lambda\in\mathbb R[/math] into our formula, as above, we will still have, exactly as before:
On the other hand, we already know from our previous computations, those preceding Proposition 15.20, that if we ask for [math]\lambda\in\mathbb R[/math] to be a plain constant, not depending on [math]u,v[/math], then [math]\lambda=1[/math] is the only good choice, making the following formula happen:
But, and here comes our point, [math]\lambda=1[/math] is not an ideal choice either, because it would be nice to have the properties (1,2) in the statement, and these properties have no reason to be valid for [math]\lambda=1[/math], as you can check for instance by yourself by doing some computations. Thus, the solution to our problem most likely involves a scalar [math]\lambda\in\mathbb R[/math] depending on [math]u,v[/math], and satisfying the following condition, as to still have [math]||u||=1\implies u+_ev=u[/math]:
Obviously, as simplest answer, [math]\lambda[/math] must be some well-chosen function of [math]||u||[/math], or rather of [math]||u||^2[/math], because it is always better to use square norms, when possible. But then, with this idea in mind, after a few computations we are led to the following solution:
Summarizing, final correction done, and with this being the end of mathematics, we did a nice job, and we can now formulate our findings as a theorem, as follows:
When defining the Einstein speed summation in [math]3{\rm D}[/math] as
- When [math]u\sim v[/math], we recover the previous [math]1{\rm D}[/math] formula.
- We have [math]||u||,||v|| \lt 1\implies||u+_ev|| \lt 1[/math].
- When [math]||u||=1[/math], we have [math]u+_ev=u[/math].
- When [math]||v||=1[/math], we have [math]||u+_ev||=1[/math].
- However, [math]||v||=1[/math] does not imply [math]u+_ev=v[/math].
- Also, the formula [math]u+_ev=v+_eu[/math] fails.
This follows from the above discussion, as follows:
(1) This is something that we know from Proposition 15.20.
(2) In order to simplify notation, let us set [math]\delta=\sqrt{1-||u||^2}[/math], which is the inverse of the quantity [math]\gamma=1/\sqrt{1-||u||^2}[/math]. With this convention, we have:
Taking now the squared norm and computing gives the following formula:
But this formula can be further processed by using [math]\delta=\sqrt{1-||u||^2}[/math], and by navigating through the various quantities which appear, we obtain, as a final product:
But this type of formula is exactly what we need, for what we want to do. Indeed, by assuming [math]||u||,||v|| \lt 1[/math], we have the following estimate:
Thus, we are led to the conclusion in the statement.
(3) This is something that we know from Proposition 15.20.
(4) This comes from the squared norm formula established in the proof of (2) above, because when assuming [math]||v||=1[/math], we obtain:
(5) This is clear, from the obvious lack of symmetry of our formula.
(6) This is again clear, from the obvious lack of symmetry of our formula.
That was nice, all this mathematics, and hope you're still with me. And good news, the formula in Theorem 15.22 is the good one, confirmed by experimental physics.
General references
Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].