Where is Contraction Point in Special Relativity?

[MrBlank]

TL;DR

Where is the contraction point associated with length contraction in special relativity?

Generally speaking, when a simple contraction occurs there is a contraction point. Length contraction in special relativity appears to be a simple contraction, and hence there should be a contraction point. Where is this contraction point located?

 

[Orodruin]

There is no such thing.

 

[AdirianSoan]

[User has been asked not to post misinformation like this in the future -- Mods]

Terrell-Penrose rotation will solve the issue you are fighting. It isn't really contracted, it is rotated with respect to time and the axis of travel. The contraction is just measuring how much of the spatial axis has been displaced by time.

Consider a light on the back of a traveling spaceship, moving towards you (off to the side) at relativistic speeds (relative to you). Terrell-Penrose rotation implies you will see it, which makes sense if you think about when the various light will arrive to yoy, and their paths along the way. The back of the ship appears rotated towards you.

What you won't see, except as reflections, is the light passing through the rest of the ship.

If you imagine a clock on the front and back of the ship, synchronized in ship time, the clock on the back of the ship will show an "older" time than the clock on the front. The ship has been rotated in time and space.

 

[Ibix]

Terrell-Penrose rotation will solve the issue you are fighting. It isn't really contracted, it is rotated with respect to time and the axis of travel.

This is not correct, as has been pointed out to you before. Terrel-Penrose rotation is an optical effect; when you correct for light travel time (as you need to if you wish to consider inertial frames), what is left is length contraction.

You appear to be confusing this rotation with the hyperbolic "rotation" of axes normally called a Lorentz boost. They are not the same phenomenon.

Where is this contraction point located?

There isn't one. Or, to put it another way, you can choose any point on (or, indeed, outside) an object and declare that this point is the centre of its contraction.

If the object is being accelerated in some specified way then there will typically be a point on it whose acceleration you know exactly (e.g. the exhaust vent of a rocket). It is probably convenient to regard this as the centre, but it is certainly not obligatory to do so.

 

[MrBlank]

There isn't one. Or, to put it another way, you can choose any point on (or, indeed, outside) an object and declare that this point is the centre of its contraction.

I don’t understand. How can the contraction point be everywhere and nowhere?

 

[Ibix]

I don’t understand. How can the contraction point be everywhere and nowhere?

You get to pick the point. But it's a completely arbitrary choice, so pick whatever you like. There's no physics based reason to pick any particular one. But that freedom of choice means that you could argue (as @Orodruin presumably did) that the lack of a physical criterion for picking a particular point implies that there's no real meaning to the question.

The point is that with something like a bar expanding under heating, you can sit it on a lab bench, paint marks along its length and see which mark doesn't move relative to the bench (which is also arbitrary if you get right down to it, but at least there's a reason to think in the lab frame here). But with length contraction you have a moving bar whose length only changes it it changes speed - so there's no real way to pick a fixed point on the bar and say "this is the point that didn't move". They all moved at varying speeds.

 

[MrBlank]

Alice and Bob are both at rest at the origin. A rocket starts at the origin and accelerates in the positive x-direction. Alice arbitrarily decides that the contraction point is on the negative x-axis at x = -10 billion light-years. Bob arbitrarily decides that the contraction point is on the positive x-axis at x = 10 billion light-years. If the rocket rapidly accelerates then Alice will observe the rocket move backwards towards x = -10 billion light-years. Bob will observe the rocket rapidly moving forward. However, given that they are both at rest at the origin, they should observe the same thing. How is it possible to arbitrarily choose the contraction point, and yet get the same result?

 

[FactChecker]

These are not the type of contraction that you are thinking of. You are thinking of a contraction where there is no dispute about the instant when the distance between two distinct points is measured. The very definition of measuring distance is more complicated in SR. In SR, the times at two ends of any distance can not be called "simultaneous" in all inertial reference frames. If there is a lot of relative motion between two reference frames, they do not determine the same endpoints of a distance at the same time.

Suppose a very long, fast rocket is flying past and I have asked the people in my reference frame to mark the position of both ends at exactly 12:00. They do that and determine the length of the rocket. But the people on the rocket disagree. They say that my people did not mark the locations of both ends at the same time and that there was a lot of travel in the time between the two end markings. That gives the length contractions of SR. Because the "contraction" is due to a disagreement in the timing of "simultaneous" events, there does not need to be any "center of contraction"; the timing disagreement can go on forever, getting greater and greater as the length increases in the direction of relative motion.

 

[Orodruin]

A rocket starts at the origin and accelerates in the positive x-direction.

Length contraction describes the length of an object moving at constant speed relative to its rest length. As soon as you introduce acceleration, things get much more complicated. Thus, as soon as you say ”acceleration” you need to be much more careful and you cannot blindly use the length contraction formula.

 

[PeroK]

I don’t understand. How can the contraction point be everywhere and nowhere?

What is a contraction point? How are you defining it?

 

[Ibix]

Alice and Bob are both at rest at the origin.

And they remain at rest. The only thing length contracted (from their perspective) is the ship. So no, they do not expect it to move rapidly backwards or anything.

As Orodruin points out, acceleration is complicated and you cannot use naive intuition. While it is undergoing constant acceleration (and making reasonable assumptions about how the acceleration is applied), each point on the rocket follows a hyperbolic worldline. You can pick anyone (even extending the family of worldlines beyond the end of the ship) and regard it as "the center of contraction". That is, the ratio of distance from any such point to the nose and tail of the ship should remain constant. Although I was slightly wrong to say that you can pick any point as this center - it can't be more than $c^2/a$ behind the ship if the proper acceleration of its tail is $a$.

If you are talking about comparing inertial frames then you can pick any event as the origin. The linearity of the Lorentz transforms guarantees that lengths measured will be independent of the choice of origin.

 

[jbriggs444]

Alice and Bob are both at rest at the origin. A rocket starts at the origin and accelerates in the positive x-direction. Alice arbitrarily decides that the contraction point is on the negative x-axis at x = -10 billion light-years. Bob arbitrarily decides that the contraction point is on the positive x-axis at x = 10 billion light-years. If the rocket rapidly accelerates then Alice will observe the rocket move backwards towards x = -10 billion light-years. Bob will observe the rocket rapidly moving forward. However, given that they are both at rest at the origin, they should observe the same thing. How is it possible to arbitrarily choose the contraction point, and yet get the same result?

This seems to assume things about rigid acceleration that are not true. If you accelerate an object (or a frame of reference, for that matter) so that it remains rigid (Google "born rigidity"), you will not be accelerating all of its pieces at the same rate.

 

[jartsa]

Here's a thought:

We can measure how much an accelerating rod is contracted at some moment. From that we can calculate a velocity that corresponds to that amount of contraction. That velocity would be the velocity of the whole rod at that moment, right?

Then we can find a point on the rod that has exactly that velocity.

Then we can say that that one point is moving at the velocity of the rod. And the other points are moving at different velocities than the rod because of the contraction motion of the rod. So there was one point with no "contraction motion".

 

[jbriggs444]

We can measure how much an accelerating rod is contracted at some moment.

Certainly. For instance, by computing the difference in x coordinate values for the rod's endpoints from our chosen inertial rest frame.

From that we can calculate a velocity that corresponds to that amount of contraction.

Certainly. We can express the coordinate difference computed above as a function of time and differentiate once.

That velocity would be the velocity of the whole rod at that moment, right?

No. It is the difference in the coordinate velocities of the endpoints at some instant (as judged from our chosen inertial frame).

The "velocity of the rod as a whole" would correspond more naturally to the average velocity of the endpoints (their sum divided by two) rather than to their difference.

 

[PeterDonis]

Then we can find a point on the rod that has exactly that velocity.

You're assuming that your procedure will always give the same point on the rod. I'm not sure that's true.

 

[Ibix]

I tried calculating the "length contraction" of a rocket undergoing constant proper acceleration so that each point is a Rindler observer with common foci. If the rocket is of length $l$ in an inertial reference frame where it's instantaneously at rest at $t=0$, and if the tail of the rocket experiences proper acceleration $a$, then I make its "length contraction" in the initial frame at time $t$ (i.e. its time $t$ length divided by its time 0 length) to be:$${{\sqrt{a^2t^2+a^2l^2+2al+1}-\sqrt{a^2t^2+1}}\over{al}}$$You can equate this to $1/\gamma$ and solve for $v$. You get something of a mess, which doesn't seem to me to match the functional form of any particular point on the rocket. I haven't played around with it very much, though.

So I tend to agree with @PeterDonis and @jbriggs444 - you can certainly perform the procedure, but I don't think it gives you a unique point on the rocket.

 

[pervect]

Let's try and answer the question in terms of a rigid ruler. The first question one need to ask is "is there such a thing?" The answer is yes, more or less, with some interesting limitations that I won't get into much. The concept is a rather advanced one, called "Born Rigidity". See for instance the <wiki link>.

Given this rigid object, our "born rigid" object, we can describe how the object accelerates. In it's own instantaneous inertial frame, the object always has a constant length. But as it accelerates, it appears to shrink. How does this happen? That's a question we can answer.

Basically, the front of the object, in order to maintain rigid motion, accelerates less, a lower "proper acceleration", than the front of the rocket.

This is closely related to Bell's spaceship paradox, but the details are different. Bell's spaceship paradox covers the case of two rockets with the same proper acceleration, and discusses how the proper distance between them increases with time.

The rule for the proper acceleration reading is that the product of the distance from the so-called "Rindler horizon" and the proper acceleration is constant. So the acceleration vs distance is not a linear curve, it's a hyperbola.

So, if we have a point accelerating at 1 light year/ year^2, which is approximately 1 Earth gravity, (whcih I'll henceforth cal 1G), the Rindler horizon is 1 light year behind the object.

The acceleration needed for an object to keep up in rigid motion approaches infinity at this Rindler horizon, 1 light year behind the point that is accelerating at (roughly) 1G. And it drops to half a G 2 light years away from the horizon, which puts it 1 light year ahead of the part that's accelerating at 1G.

Again, this is fairly advanced stuff, not really a good idea for an intorduction to SR, but the wiki entry on the Rindler coordinates and rindler horizon is at <wiki link>.

I would suggest a different approach than the whole "shrinking rulers" idea to learn special relativity. My general advice is to focus on things that remain the same (this includes proper time, proper distance, and proper acceleration) rather than things that vary with different observers. Probably the place to start is to learn about the difference between proper time and coordinate time, and go on from there. Exactly where to learn about prop;er time is a good question, though - I'm not sure of the best resource here.