Are superluminal Lorentz contraction velocities real velocities?

Physics Asked by Pat Dolan on November 27, 2020

Starting with the equation for Lorentz contraction

$L = sqrt{1-(v/c)^2}L’$

and making the substitution $v=at$ (be sure to see a better substitution two paragraphs down) then taking the derivative of the Lorentz contraction equation with respect to time gives

$$frac{dL}{dt} = frac{-tL’(a/c)^2}{sqrt{1 – (at/c)^2}}$$

which can be thought of as the Lorentz contraction velocity; that is to say, a measure of how fast an object is Lorentz contracting from the point of view of an observer accelerating with respect to that object. Notice the units are m/s just like regular velocity and that this velocity is always opposite to the direction of acceleration.

Now repeat the derivation using a better substitution $v = ctanh(at/c)$ to get

$L = sqrt{1 -tanh(at/c)^2}L’$

which gives the correct Lorentz contraction for velocities near $c$ for proper time $t$ and proper acceleration $a$. Differentiating with respect to $t$ now gives

$$frac{dL}{dt} = -(a/c)tanh(at/c)operatorname{sec h}(at/c)L’$$

which also has units m/s.

$dL/dt$ is neither linear nor monotonic. It increases to a maximum value then approaches zero asymptotically. The behavior of $dL/dt$ can be seen to rise from zero at $t=0$, achieve a maximum velocity of $aL’/2c$ when $(at/c) = operatorname{arctanh(sqrt{2}/2)} = .881$, then fall to zero as $t$ goes to infinity (see This equation implies that a suitably large value $L’$ can produce an enormous Lorentz contraction velocity many millions of times the speed of light.

My question: Are superluminal Lorentz contraction velocities real velocities?

An example to consider:

Andromeda is approximately $2.5$ million light years away from the Milky Way. An observer in the Milky Way starts to accelerate towards Andromeda at a constant and comfortable $1g$. At the point of maximum Lorentz contraction velocity, 312 days into the trip as measured by the observer, Andromeda is seen to be approaching the observer at $3.866times 10^{14}$ m/s. That’s $1.29$ million times the speed of light. The component of this velocity produced by thrust is negligible compared to the component due to continuous, ongoing Lorentz contraction.

4 Answers

Suppose you are considering some event. At time $0$ you are using a coordinate system in which its coordinates are $(x_0,t_0)$. At time $1$ you are using a coordinate system in which its coordinates are $(x_0',t_0')$. Your "Lorentz contraction velocity" would be $(x_0'-x_0)/(1-0)= x_0'-x_0$. It would have units of distance over time, but it would not be a velocity. $x'_0$ and $x_0$ are numbers in different coordinate systems. $x'_0$ is how far along the $x'$ axis the event is, and $x_0$ is how far along the $x axis it is. They are numbers, not physical quantities. It doesn't make sense to subtract them, and the difference doesn't represent a physical quantity.

As analogy, suppose you're displaying Earth on your computer screen. Since Earth is a sphere and your computer screen is a flat plane, you need to use a map projection to display Earth on the screen. Now suppose you decide to change to a different map projection. And you decide to have your computer continuously morph from one to another. Each city will move across the screen as its coordinates in the map projection changes. You can take its change in position and divide it by the time it takes to change, and you will get distance over time, but this is not a real velocity.

Answered by Acccumulation on November 27, 2020

Note that if you keep accelerating at one gee, you will reach Andromeda in about 15½ years. Any continuous function with a value of $2.5text{ Mly}$ at the start of the trip and $0$ at the end has to have an average slope of about $-160000c$, regardless of the details.

There's no lower bound on the amount of proper time required to get to any reachable location in the universe. You could say that the universe Lorentz contracts to get you there in time, and there would be some mathematical sense in which you'd be correct. Personally, I don't like that sort of language because physics is not subjective in the way that it suggests. The world doesn't revolve around you just because you chose a coordinate system in which you're stationary at the origin. The reality is that you're just a tiny insignificant rocket ship in an uncaring cosmos.

Answered by benrg on November 27, 2020

An Over-Interpretation

The possibility of superluminal Lorentz contraction velocities are contained in, and can be validly derived from the second postulate. So to state that the second postulate always and everywhere forbids two material objects from ever acquiring a superluminal velocity with respect to one another is a mistaken over-interpretation.

A New Perspective on the Second Postulate

The second postulate correctly implies that there is no force great enough to accelerate a first object up to, or beyond the speed of light with respect to a second non-accelerating object.

But this state of affairs is not symmetrical. Because it can also be deduced from the second postulate that a non-accelerating object can acquire a superluminal Lorentz contraction velocity with respect to an accelerating object. Furthermore, if the distance is great enough between the two then the non-accelerating object MUST acquire a superluminal Lorentz contraction velocity no matter how small the acceleration. In an infinite universe there is no acceleration so small such that there isn’t some distance L from the accelerating observer, beyond which all objects acquire superluminal Lorentz contraction velocity when (at/c) = .881

Another Broken Symmetry

It is because of the second postulate that two observers in relative motion will disagree on each other’s length, mass and rate of time. If one of the observers is accelerating we can also add “each other’s velocity” and even “each other’s direction of velocity” to the list of things on which they disagree. The relative motion that is responsible for breaking symmetries in length, mass and time now turns out to break its own symmetry as well.

Answered by Pat Dolan on November 27, 2020

Are superluminal Lorentz contraction velocities real velocities?

This depends on what you mean by “real velocity”.

It is really a quantity with units of length per time. So if that is sufficient to qualify something as a “real velocity” then it is.

You can create a coordinate system where a physical object, like the Andromeda galaxy, has a rate of change of position that is equal to your Lorentz contraction velocity. So if that is sufficient to qualify as a “real velocity” then it is.

The coordinate system in the previous paragraph does not represent an inertial frame, which is why the Lorentz contraction velocity can exceed c. So if being defined in an inertial frame is necessary to qualify as a “real velocity” then it is not.

In a frame invariant sense that velocity does not represent the velocity with respect to the timelike vector of any tetrad, even a non-inertial one. So if that is necessary to qualify as a “real velocity” then it is not.

Personally, I would tend to use that final criterion as the main one since it is frame invariant, but the word “real” is rather poorly defined and many things that people consider “real” are frame variant.

Answered by Dale on November 27, 2020

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