Lorentz contraction and time dilatation can be deduced without Lorentz transformation. Can you deduce also the theorem of addition of velocities


without Lorentz transformation? Using just the constancy of light speed.


3 Answers 3


Since the Lorentz transformations are a consequence of the postulate of the constancy of the speed of light, together with some homogeneity and parallel postulates, it is a little difficult to make precise the request for a Lorentz-transformation-free demonstration.

But I will interpret the question as asking for a synthetic proof of the addition of velocities. A synthetic proof is a proof in the same style as the Euclidean geometry proofs you do in grade school. It gives you a picture and an intuitive way to understand the formula.

First, to start, you should understand that a velocity is the ratio of distance traveled over time elapsed, it is the analog of the tangent of an angle in geometry. In geometry, there is a law of addition of tangents:

$$ \tan(a+b) = {\tan(a) + \tan(b)\over 1 - \tan(a)\tan(b)} $$

The easiest way to understand relativity is that it is Lorentz geometry--- a geometry where the Pythagorean theorem has a minus sign, and perpendicular lines don't look perpendicular, but have the same slope relative to the 45-degree line (on opposite sides), where the distinguished 45-degree line is the light ray.

In Lorentzian geometry, the addition formula is the law of addition of hyperbolic tangents:

$$ \tanh(a+b) = {\tanh(a) + \tanh(b) \over 1 + \tanh(a)\tanh(b)}$$

If you interpret the angles as the rapidities (this is just the definition of the Lorentzian analog of angle), and the hyperbolic tangents as the velocities, this is the law of addition of velocities.

Synthetic proof of both laws of addition of tangents

First review these two answers: Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman and: What are the mechanics by which Time Dilation and Length Contraction occur?. You need a little intuition for the geometry.

Consider the following diagram:

The geometric addition of tangents

Where the length of segment AB is 1, the length of segment BC is u, and angle ABC is right. This implies that the angle CAB has a tangent of u, by definition, and that the length of AC is $\sqrt{1+u^2}$.

If I want the angle CAE to have tangent v, then I need the ratio of EC to CA to be v. This determines the length of EC to be $v\sqrt{1+u^2}$, and from this, you learn that the length of CD is v and the length of DE is uv (since triangle ABC and CDE are similar). You can fill in all the lengths with a pen, my drawing program didn't allow square roots.

Therefore, the tangent of the sum of the two angles is the ratio

$$ {EF\over AF} = {u + v \over 1 - uv} $$

And I hope the diagram makes it self-evident without any complicated formulas.

For relativity, you do the same thing in space-time. The analogous diagram is given below.

alt text

The relativistic addition of velocities--- time axis horizontal

In triangle ABC, AB is along the time axis (I drew it horizontally to make the diagram as similar as possible to the preceding one), and it has a length of 1. BC has length u so that the velocity of the line AC is u.

From this, you find the length of AC using the relativistic version of the Pythagorean theorem (with a minus sign). CE is then drawn relativistically-perpendicular to AC (that's how it looks--- get used to this), and triangle CED is similar to ABC (for the same reason as in Euclidean geometry) so that the lengths are proportional. From this you learn that DE has length v, and CD has length uv (just as before).

Now the total velocity is given by the ratio of EF to AF, as before, and it is now:

$$ {EF\over AF} = {u+v\over 1+uv}$$

To persuade yourself that this is really ok, you need to get comfortable with the rotations of relativity.


I endorse Ron's answer – it's the systematic way to proceed. The velocity $v/c$ may be written as $\tanh \eta$ where $\eta$, the rapidity or whatever, is the hyperbolic (Minkowski) counterpart of the (Euclidean) angle. The addition of velocities then boils down to an addition formula for $\tanh(\eta_1+\eta_2)$ because the rapidities just add additively.

Let me offer an elementary derivation without any fancy rapidities. Imagine that an object moves by the speed $u$ to the right, another object moves by $v$ to the left with respect to our frame. What is their relative speed?

The world line of the first observer is a straight line containing the points $(0,0)$ and $(1,u)$; the coordinates are $(t,x)$. The other object has a world line connecting $(0,0)$ with $(1,-v)$. Now, let's imagine that we tranform the situation into the rest frame of the second observer, i.e. boost it by the velocity $v$. How will the first observer's world line tilt?

To find the answer, note that by making the Lorentz boost which fixes the origin $(0,0)$, the Lorentzian inner product of the two vectors, $(1,u)$ and $(1,-v)$, will not change; I define the inner product of $(A,B)$ and $(C,D)$ as $AC-BD/c^2$ where the minus sign comes from the Lorentzian relativity twists and $c^2$ is the conventional conversion of length to time. Their length won't change, either. It also means that the inner product divided by the product of lengths won't change. In the original frame, it is equal to $$ \frac{(1,u)\cdot (1,-v)}{|(1,u)|\cdot |(1,-v)|} = \frac{1^2+uv/c^2}{\sqrt{1-u^2/c^2}\sqrt{1-v^2/c^2}} $$ Note that by taking the ratio, I canceled the absolute normalization of the two vectors $(1,u)$ and $(1,-v)$, so this normalization doesn't matter. However, this ratio must be the same in the new frame where the observers' vectors indicating the directions of the world line are $(1,0)$ and $(1,V)$ where $V$ is the total relative velocity. From those two vectors, the same ratio as above (which again cancels the normalization) is equal to $$ \frac{(1,0)\cdot (1,V)}{|(1,0)| \cdot |(1,V)|} = \frac{1}{\sqrt{1-V^2/c^2}} $$ Just to be sure, the ratios have to be equal and they may also be written as $\cosh \eta$ where $\eta$ is the total "hyperbolic angle" i.e. rapidity in between the two world lines, the same "angle" as discussed above. The formula for $\cosh$ is analogous to the high school formula for $\cos$ involving the inner product but you don't need to know anything from this paragraph to follow my derivation.

Now, we have $$ \frac{1}{\sqrt{1-V^2/c^2}} = \frac{1^2+uv/c^2}{\sqrt{1-u^2/c^2}\sqrt{1-v^2/c^2}}$$ Square it and invert it: $$ 1 - \frac{V^2}{c^2} = \frac{(1-u^2/c^2)(1-v^2/c^2)}{(1+uv/c^2)^2} $$ Expand the product in the numerator and subtract one from both sides: $$ - \frac{V^2}{c^2} = \frac{1 - u^2/c^2 - v^2/c^2 + u^2 v^2/c^4 - 1 - 2uv/c^2-u^2 v^2/c^4}{(1+uv/c^2)^2} $$ The numerator of the right hand side simplifies, two pairs of terms cancel: $$ - \frac{V^2}{c^2} = \frac{- u^2/c^2 - v^2/c^2 - 2uv/c^2}{(1+uv/c^2)^2} $$ Now, multiply both sides by $(-1)$ to get rid of the signs. And even I am able to compute the square root now: $$ \frac{V}{c} = \frac{u/c+v/c}{1+uv/c^2} $$ which we wanted to prove. Feel free to multiply it by $c$ again.

  • $\begingroup$ ok +1, this is a good way also, but possibly a little formal for OP. $\endgroup$
    – Ron Maimon
    Apr 12, 2012 at 14:08

David Mermin came up with a beautiful thought experiment, which can be found in his book Boojums all the way through:

Suppose that an observer in an inertial frame $S$ watches a train move at constant velocity $v$. The length of the moving train, as measured in $S$, is equal to $L$. Now, at time $t=0$, a photon and a massive particle (with speed $w < c$) start moving from the rear of the train to the front.

The photon reaches the front at a time $t=T$, and is then reflected back; meanwhile, the massive particle is still moving towards the front. At some later time $t=T+T'$, the particle and the reflected photon will meet each other at a specific location in the train. Let's express this location as a fraction $f$ of the total length $L$. Since this location is the same in every reference frame, we can state that $f$ is invariant (if we make the reasonable assumption that ratios between lengths do not change from one reference frame to another).

From this, we have $$ w(T + T') = c(T - T')\tag{1}. $$ Indeed, both the particle and the photon have travelled the same net distance during the total time $T+T'$, but the photon was moving in the opposite direction during the time interval $T'$. The distance that the photon moved during $T$ from the rear to the front is the length $L$ of the train + the distance that the train moved during this time: $$ cT = L + vT\tag{2}. $$ Similarly, the distance that the photon moved during $T'$ from the front to the meeting point $fL$ is $$ cT' = fL - vT'\tag{3}. $$ From equation (1), we find $$ \frac{T'}{T} = \frac{c-w}{c+w}\tag{4}, $$ and combining this with (2) and (3), we find $$ f = \left(\frac{c+v}{c-v}\right)\frac{T'}{T} = \frac{(c+v)(c-w)}{(c-v)(c+w)}.\tag{5} $$ Since $f$ is invariant, this relation remains true in the reference frame $S'$ that moves with the train. In this frame, $v=0$ and the massive particle will have a different velocity, say $u$. So we get $$ f = \frac{c-u}{c+u}.\tag{6} $$ If we now equate (5) and (6), we find $$ \frac{c+w}{c-w} = \left(\frac{c+u}{c-u}\right)\left(\frac{c+v}{c-v}\right),\tag{7} $$ and solving for $w$ leads to the familiar relativistic addition law: $$ w = \frac{u+v}{1 + uv/c^2}.\tag{8} $$ The only assumptions in this simple thought experiment are that the speed of light $c$ and ratios between lengths are invariant.


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