# (Yet another) relativity conundrum

#### zanket

##### Human
Valued Senior Member
Two cars stopped on different streets begin accelerating identically. Thus after a given duration of time on street clocks, the cars have traveled the same street distance respectively.

Let the cars accelerate identically starting at the same time on street clocks, along the same street, with one car following distance x behind the other. Just as they did when they traveled independently, after a given amount of time on street clocks the cars travel the same street distance respectively. Having traveled the same street distance by any particular moment on the street clocks, any bystander along the cars’ route should measure the distance between the passing cars to be x. The bystander-measured distance between the cars should remain constant as the cars accelerate.

But special relativity says that a bystander should measure the distance between the cars to be less than x; the original distance x should be length-contracted. The distance between the cars should be length-contracted because the two cars can be considered a single object that is moving past the bystander. (If you have a hard time imagining them as a single object then put a rope between the cars to make them a single object. The car-rope-car object should be length-contracted.)

Can anyone resolve this conundrum?

i think the answer is in your post itself...the car-rope-car is A SINGLE OBJECT!!

Zanket. I am certainly no supporter of SR, but I don't see your conundrum? What do you see as a dilemma according to SR? To begin with, an accelerating frame is a
non-inertial frame and not addressed by SR. The acceleration can be broken down
into minute time intervals where no acceleration is taking place I suppose, and an
inertial frame postulated. In an inertial frame of reference, it would not matter if two
cars were behind each other, on separate streets, or if there were ten cars. If their
speeds were identical in relation to a bystander on the street, the cars would be in
the same frame of reference and the bystander would be in a different FoR. Whether
ten cars or one car, they would be length contracted AND time dilation (simultaniety)
would be part of the effects of two different frames of reference, the cars in one, the
bystander in the other. But at non-relativistic speeds, the differences between the
two frames of reference would be almost infinitesimal, too small to really measure.
Have I misunderstood your question?

It is a conundrum - I have seen it before, but can't remember the resolutions (I think there was one, but could be mistaken).

The way I remember it, there was a rod that was uniformly accelerated in the direction of its length. Does a stationary observer see the rod contract? If so, why does the trailing end accelerate faster than the leading end?

Edit:
Just found a page addressing this paradox in the Usenet Relativity FAQ:
Bell's Spaceship Paradox

I do not understand the explanation.

Another curly SR issue seems to be the Rigid Rotating Disk

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Thank you Pete! I took a peek at the link. I’ll digest it and report back here in layman’s terms. This one’s been giving me a headache for 2 days now. I’m glad someone else thought of it before!

ecclesiastes said:
i think the answer is in your post itself...the car-rope-car is A SINGLE OBJECT!!

The link Pete posted puts the conundrum this way:

Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other. The ships start out at rest in some coordinate system (the "lab-frame"). Since they have the same acceleration, their speeds should be equal at all times (relative to the lab-frame) and so they should stay a constant distance apart (in the lab-frame). But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it?

It doesn’t matter whether you consider them as two ships or as a single ship where the bow and stern of the ship accelerate identically. It’s a conundrum either way. It just might be easier to visualize that the rope must length-contract when it’s a car-rope-car or ship-rope-ship object.

If they are connected, I imagine you'll run into some variant of the "rigid objects can't exist" problem, which also arises in the rotating disc problem.

2inquisitive said:
Zanket. I am certainly no supporter of SR, but I don't see your conundrum? What do you see as a dilemma according to SR?

Take a look at the link Pete posted. The conundrum or paradox is that SR seems to demand that the distance between the ships stays the same as it was before the ships accelerated, and also demands that the distance length-contract.

To begin with, an accelerating frame is a non-inertial frame and not addressed by SR.

Actually SR handles constant acceleration just fine, as noted by The Relativistic Rocket. These equations, which are derived from SR, were published after GR was published, as far as I know.

If their speeds were identical in relation to a bystander on the street, the cars would be in the same frame of reference and the bystander would be in a different FoR. Whether ten cars or one car, they would be length contracted AND time dilation (simultaniety) would be part of the effects of two different frames of reference, the cars in one, the bystander in the other.

Agreed. The conundrum is that, because the cars accelerate identically, the distance between them must stay constant; that is, the distance must not length-contract.

I’ll digest the link and report back here.

This is a tough one for me. It seems that Bell’s Spaceship Paradox addresses only half of the paradox, by explaining how the lab-frame people observe the distance between the rockets as length-contracted due to failure of simultaneity. I understand this, but still don’t understand how it is resolved that the lab-frame people do not observe the distance between the rockets as not length-contracted (the paradox). Since both rockets accelerate identically, they should traverse the same lab-frame distance in any given lab-frame elapsed time; that is, the same experiment should yield the same result. The link doesn’t fully resolve the paradox as far as I can tell. Nor do related links. I'll have to think about it some more.

Those moving would measure the distance between their veachles as unchanged and distances outside their frame of reference i.e. evrything else as contracted. Siimilary everyone else assuming that they were at rest relative to the veachles would measure the distance as contracted and everythig in their frame as unchanged. The simple fact is; although the difference would be negligable neither frame would measure the other as unchanged, where's the conumdrum?

I believe bonemeal and 2inquisitive nailed it. Its all relative (duh). Relative to the drivers of the cars, no contraction of the distance x is observed. Relative to the bystander, the contraction of x is observed. If the rope connecting the cars is regarded as a "rigid body" (assuming there is no slack in the rope) then the drivers observe no contraction because they themselves contract in proportion to the rope's contraction. The bystander of course would not regard the rope as a "rigid body with which to make spatial measurements" because the rope seems to shrink when compared to the rigid body (say a ruler) that the bystander is holding in his hypothetical hand.
As is the case with most of SRs apparent contradictions, they are only apparent contradictions.

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If the bystander measures the distance between the cars as contracted, then two identical experiments do not yield the same result. Or, if the bystander measures the distance between the cars as unchanged from the original distance, then it must be explained how the cars would length-contract in the bystander frame (SR is clear that they would) yet a rope stretched between them would not length-contract. That’s the conundrum, the paradox.

If both cars accelerate identically for the same elapsed time on bystander clocks then they should traverse the same distance in the bystander frame. If the distance between them contracts then the cars do not traverse the same distance.

Here is the relevant equation:

d = (sqrt((a * t)^2 + 1) – 1) / a

This equation returns the distance d in the bystander frame that a car traverses for a given acceleration a (acceleration the driver feels, in terms of c) for elapsed time t on bystander clocks. In the paradox a and t are the same for both cars, hence--because a and t are the only inputs to the equation--d should be the same, hence the distance between the cars should be constant in the bystander frame. If the distance between the cars contracts in the bystander frame then the equation is fiction; two experiments involving the same a and t do not yield the same d result.

Let both cars be initially at rest and one light year apart in the bystander frame, then accelerate to so close to c in one second on bystander clocks that the distance between them contracts to one meter in the bystander frame. Now the pursuing car has surpassed the speed of light in the bystander frame, beating its own image that travels toward the pursued car at c, leading to a causality paradox.

So for two important reasons, the distance between the cars in the bystander frame should not change as the cars accelerate.

I may have misread the convoluted web page about Bell’s Spaceship Paradox. The relevant quote may be:

This first picture interprets "two ships with the equal constant accelerations" to mean "constant for the co-moving observers, and equal in the lab-frame". Note that the lab-frame says that the accelerations are not constant, and the co-moving observers say the accelerations are not equal! (More precisely, any particular co-moving observer says this. The phrase "the co-moving observers" does not refer to a single frame of reference, unlike the phrase "the lab-frame".) The lab-frame says the ships maintain a constant distance from each other; the co-moving observers don't agree.

I think this is the resolution to the paradox: the bystanders measure the distance between the accelerating cars as unchanged. A rope between them would break (because the pursuing driver observes that the other car is pulling away--it's a given that the cars have the same constant acceleration so the rope doesn't affect that, but make the rope a thread if you wish the breakage to be more believable) and length-contract along with the cars.

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To fulfill my promise to explain the resolution to this paradox in laymen’s terms, I wish to elaborate on my last post:

In Bell’s Spaceship Paradox the rocket ships accelerate at the same constant rate. But in which frame of reference? If I tell you that a rocket accelerates at a constant 1g then you’ll probably expect the ship’s crew to feel a 1g acceleration. If the crew feels that then the 1g acceleration is not being measured in the rocket’s frame of reference. The acceleration accelerates the rocket’s frame of reference, so you have to be careful about measuring the acceleration in that frame. At any given moment the crew feels a 1g acceleration and in that moment the ship has a constant velocity. Acceleration is a change in velocity over time. Momentarily the velocity doesn’t change; it changes from one moment to the next. So the 1g acceleration applies to an inertial frame of reference that momentarily moves with the rocket at its constant velocity at that moment. This frame of reference is called a momentarily comoving inertial frame (MCIF) and is described more here. As the rocket accelerates, visualize a stream of successive MCIFs, one for each moment.

In my last post I noted how it is determined that the distance between the ships/cars does not change in the lab frame/bystander frame. Now visualize what the ships’ crews observe. Gravitational time dilation applies to the accelerating ships. The lead ship is effectively at a higher altitude and thus its clocks run faster than those of the trailing ship. With a higher rate of time and the same acceleration in the comoving observers’ frame (MCIF), the lead ship pulls away from the trailing ship. Suppose they both start accelerating at noon on their clocks (they are initially at rest in the lab frame, so their clocks can be synchronized). Later the trailing ship’s clock might read 1pm and the lead ship’s clock might read 1:01pm. With a later time and same acceleration, the lead ship must have traversed a greater distance than the trailing ship did, as measured from their respective starting points in the lab frame. If there is a thread connecting the ships, it will break at some point during a prolonged acceleration.

Rather than using gravitational time dilation you can visualize what the crews observe using failure of simultaneity. Having the same constant acceleration, both ships traverse the same lab-frame distance in the same respective proper time. From the crews’ perspective, clocks attached to mileposts further down the path (upon which the rockets are accelerating) show later times due to failure of simultaneity. These later lab-frame times correspond perfectly with the later proper time shown on the lead ship’s clocks as compared to the trailing ship’s clocks.

Now consider a string of ships that are loosely coupled. The ships are initially at rest in the lab frame. Then they begin to accelerate, with the same constant acceleration in their respective comoving observers’ frames, to relativistic velocity in the lab frame. The couplings on these ships will eventually fail. In the lab frame none of the ships will exceed c (naturally). The ships will individually length-contract in the lab frame; the distance between the head and tail of adjacent ships will increase but the distance between the heads of adjacent ships will remain constant.

It follows that if a rocket accelerates such that its head and tail accelerate at the same constant rate in the MCIF, at some point in a prolonged acceleration the rocket will break apart.