Special relativity question

First off, I am not attempting to undermine Special Relativity.

I merely want an explanation of questions raised by a friend.

It is my understanding that observers moving relative to each other both observe the other's distances to be shorter. Id est: A sees B's meter stick as being less than a meter & vice versa.

Based on the above, my friend came up with a paradox/anomaly which I cannot explain. I think I have seen this before with an explanation.

Thought experiment

In A's reference frame is a thin sheet with a circular hole one meter in diameter. In B's frame is a thin circular disk which is one meter in diameter.

Now assume the disk moves at some fraction of light speed & approaches the hole in the thin sheet.

From one frame, the hole is too small to allow the disk to pass though. From the other frame, the disk is contracted & small enough to pass though the hole.​

If the direction of movement is such that the disk will meet the hole, what actually happens?

 

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Hi Dinosaur,
Length contraction is only in the line of travel. So if the disk and the sheet are parallel, and they approach each other perpendicularly, then...
- in the reference frame with the sheet at rest, the disk will be thinner, but still a 1m circle
- in the reference frame with the disk at rest, the sheet will be thinner, but still with a 1m circular hole.

Does that answer your question?
Pete

 

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This is a slight twist of the pole-in-the-barn paradox or the ladder-in-the-garage paradox.

The point is that special relativity gives rise to a number of effects, including
1) length contraction -- in the direction of movement,
2) relativity of simultaneity (which is two say moving observers can diagree when separated events happen at "the same time" or not),
3) it is illegal to assume materials are infinitely rigid, which we can ignore unless your plan was to make the disk take some non-inertial path,
4) time dilation, which I think we can ignore for this.

At zero speed, the disk fits in the hole if the front edge and the back edge of the disk just meet with the corresponding edges of the hole and we can reasonable say these conditions happen at the same time.

At finite speed, observers differ as to what that last phrase means, so we ignore it. It's clear if the disk is going through the hole, it is either perpendicular (as Pete points out there will be no length contraction) or at an angle. I assume you mean a very slight angle to maximize the amount length contraction.

The hole observer sees the length-contracted disk flying in at an angle and diving into the hole. No problem
The disk observer sees the front edge of the disk enter the length contracted hole before the back edge of the disk gets near to position. This is how mail gets through a mail-slot. Again no problem.

Similar:
http://en.wikipedia.org/wiki/Ladder_paradox#Bar_and_ring_paradox

 

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I am no physicist, but I will have a partial go at answering.

Whenever two objects come into contact they are in the same
frame of reference, where there will be no discrepancy in the
units by which they are measured.

As to what would happen if a disc exacly one meter diameter
encountered a perfectly aligned hole of exactly one meter diameter
I am not sure, but the answer does not depend on consideration
of Special Relativity.

There might also different answers delending on whether space
is assumed to be euclidean, elliptic, or hyperbolic.

If one or both such objects are moving at close to the speed of light
they will undergo contraction in the direction of travel. If their surfaces
are parallel then it will not matter if one is going faster when they enter
a common frame of reference because the width of the hole and the
width of the disc, being perpendicular to the direction of travel, will
remain an identical one meter.

If one or both objects are moving at different speeds close to the
speed of light and their surfaces are not parallel, then I will have
to hope a physicist shpws up here to explain how the variables are
to figure in calculating a solution.

 

This reads like the "Meterstick and the Hole Paradox" As suggested, it is a variation of the "Pole in the Barn Paradox" but it is more complicated. It involves velocities in two directions.

I wrote this tutorial at the URL below specifically to address this paradox. Hope it helps.

http://mysite.verizon.net/mikelizzi/Tutorials/TutorialMeterstickAndHole.html

P.S.
I can't provide the exact link because I am not a fully entitled member. [Moderator note: I have allowed it.]
I hope I don't get in trouble by offering this URL.
But take out the apostrophies.

Oh. Just noticed the URL offered by rpenner also describes this example. So now you have two.

 

Yes.

None of the images are where they need to be for viewing.

The href in the jnlp file is invalid, because no codebase is specified.
From the JNLP spec: "An href element can either contain a relative URL or an absolute URL as shown above. A relative URL is relative to the URL given in the codebase attribute of the jnlp root element."

 

Oh my.

Thanks for the feedback. Both the Microsoft word document and the Java applet work fine on my Windows 7 machine and on the Windows machines in my local library.

So now to figure out why. I will check with some other folks too.

 

Actually I have a similar question.

I did the Lorenz transform for a thought experiment of my own.

I have 3 observers : A B and C. Observer A and B are 4 ly apart initially and have a constant (though 4 year old ) view of each-other and C who is 2 ly from each. Observer C is moving at the constant rate of his galaxy (i.e. << c) A and B both build ships to travel to meet each other. The ships accelerate rapidly to .95 c, this takes 1 ly to accomplish) Starting the calculations at the point at which they reach maximum velocity:

When I do the calculations, from the point at which they achieve the top speed, I keep getting some really weird results. And that is that the time at which observer A thinks he passes C is greater than the time at which he thinks he is passing observer B. Even though he's passing them at the same time.

Obviously I am doing something wrong here....

(Wishing I could link my mathcad sheet into this browser)

But basically GammaCA = 3.203 GammaAB = 19.513

Apparent velocity of approach from A to B is .999 c while the Approach on A to C is .95 (initial conditions)

So observer C would see them pass at the same time, observer A (or B) sees Observer B (or A) Pass long before observer C (3.239x10^6 s vs 1.037x10^7 s)

What have I done wrong here?
Did I do anything wrong?

Gamma = 1/SQRT(1-v^2/c^2)

t' = Gamma(t-v*x/c^2)

I took t to be the time from observer C who has a v^2/c^2 ~ 0

 

You are not accounting for the Relativity of Simultaneity.

When A reaches velocity with respect to C, he doesn't conclude that B does the same at the same instant. Imgine there are two clocks, each an equal distance from C and in sync with each other according to C. An observer at C notes that A and B each pass one of these clocks the instant they reach coasting speed( IOW each clock reads exactly the same when its ship passes it.)

However, according to A, these two clocks are not in sync and one actually shows a later time than the other. (IOW, A passes his clock and notes its time. Then taking into account the speed of light, the observed distance between himself and the other clock, and relative speed between himself and the other clock, calculates what time it is on the other clock at that moment. He will find that it does not read the same as the clock he passed.)

So the upshot is that you are assuming that A agrees with C that B started the same instant he himself did, whereas in reality A does not agree with this.

 

In my original thought experiment they were able to coordinate launch times, so even though they didn't see each other leave, they knew the other one was at a predetermined time, i.e. "When C fires off a nuke that we can see as a flare". Since the nuke is equidistant from either of them, they can begin from C's perspective at the same time.

I understand that A doesn't see B reaching top speed at the same time he sees C doing it. I think I understand now, Because at the time A launches and reaches .95 c, B is shown to him as NOT yet at .95 c, the numbers will balance? What I am wondering is how this is different than if we simply took two observers and gave them initial distance, and velocity of .95c with respect to observer C, I still get numbers that suggest different ordering of events (Which I accept is a consequence of the theory) but for events that all happen (from C's Perspective) at the same time. The question becomes if I managed to slow down the ships of the observers when they see that they have reached observer C they would either be past him, or think they were past the other observer and yet not be. (assume infinite negative acceleration to each ones own vector so we don't have to deal with reversing the dialation effects.)

 

Mikelizzi, I've tried your applet and it works just fine. I'm running Vista with AMD 64 chip. I would recommend folks update their Java. I didn't play with the applet much but at first glance it looked incredible.

 

Mikelizzi, I've continued to play with your Java applet and it truly is impressive. Did you write all the code yourself? You should create an import/export function that would allow people to share scenes...

 

RJBeery

Thanks for the positive feedback. After James R and others had problems with the program, I did some debugging. I guess it worked.

And thanks for the suggestion of an import/export function. I was thinking I needed something like that. I will put it on my "to do" list

 

Staring at the same time in C's perspective is not the same as starting at the same time in A or B's perspective.

I think you are still struggling with the concept of the Relativity. One thing all three will agree to is that A B and C all intersect at the same instant, and what times each one's individual clocks read at that instant. They will not agree upon whether and B left at the same time or which one left first.

For example. Assume you have a long stick and an observer traveling along side it. At the instant the observer passes the midpoint of the stick, a flash of light is emitted from that point.

According to someone at rest with respect to the stick (it does not matter where is is located with respect to the stick.) the flash of light starts at the midpoint of the stick traveling at a constant speed, and reaches the ends of the stick at the same time.

According to our observer traveling along the stick( or anyone else at rest with respect to him), something else happens. The light is emitted from his position and expands outward at a constant speed in all direction relative to himself . The stick is also moving relative to him, so one end is while one end is rushing to meet the light, the other end is making the light chase after it. Thus the light gets to one end before the other.

So we have a situation where one frame two events happened simultaneously, and other frame they don't.

So let's spread some clocks around and see what happens. We'll give our observer a clock and we'll put clocks at the ends and midpoint of the stick.

The clocks on the stick are set so that they read the same time according to the rest frame of the stick. T he observer clock it set so that it reads the same as the midpoint clock at the instant they pass each other and the flash originates.

Again, according to the stick frame, the light reaches the ends at the same time and thus when the all the stick's clocks read the same.

According to the observer, the light reaches one end before the other. However, he does agree that the clocks at each end read exactly the same time when the light reaches them. So for him, one clock is ahead of the other. In addition, one clock is ahead of the midpoint clock and the other is behind it. This means that the midpoint clock cannot read the same at the end clocks when the light reaches the ends, and, in fact, will show a different time when one light reaches its end than when the other light reaches it.

And conversely, anyone at rest with the stick will say that the light reached the ends when the observer clock read a single particular value, while the observer will say that his clock read two distinctly different values when each light reached its end.

If we continue the experiment until the observer is even with one of the ends of the stick, everyone will agree as to what value is on the observer clock and the clock at that end when this happens. (they will not, however, agree as what time it is according to the clock on the other end or the midpoint. )

The upshot is that while all observers, regardless of their relative motion, will agree to what events happen at a given point in space-time. (Such as the readings on clocks when they meet), they will not agree upon order of events that are separated in space-time. (Such as the time on two clocks at opposite ends of a stick.)

 

I understand that, what I *AM* struggling with is the following, perhaps I should have explained this in my post.

From any frame of reference we can say that all other frame of reference are moving with respect to that frame. i.e. Saying that A is moving in the direction of C is as valid as saying that C is moving in the direction of A.

Now if A and B and C are separated as before, and moving as before, According to this idea, It should take, from A's perspective, just over a year from .95c point for C to reach him, and just over 2 for B to reach him. The numbers though give a different result entirely, that of longer for C than B. And if they are closing together like this they should reach all at the same time.

I realize that A will see the distance between him and B (and therefor C) shrink as he approaches this velocity. But why do the times come out backwards doing the math? And if he sees the distance shrink, does he at some point see observer B approach him faster than possible?

Did I put the numbers in backwards?
(is there an online application for this calculation?)

 

You should get something like represented in the following space-time diagrams.

The first shows events according to C (blue line). A(green line) and B(red line) start off at rest with each other(vertical and parallel to each other at the bottom of the image).
Then A and B start heading towards c at equal speeds and finally meet at C.

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Now here's how things occur according to the inertial frame of A after it starts moving towards C. (the other part of the green line is now vertical)

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Note that, in this frame, B started moving towards A before A changed its own velocity.

A a result, at the moment after A changes its velocity, B is not twice the distance from A as C is, and is actually closer to C than it is to A. C will approach at 0.95c and B at ~0.999c, and given their relative distance at that instant, will arrive at A at the same time.

Given the fact that in you state that C would take just over a year to reach him and that B would take a little over 2, That despite your claim otherwise, you do not understand what I was talking about in my last post. Because you still are holding on to the idea that B and C would start towards A at the same time according to A.

 

Thanx --- The ladder & barn scenario is easier to understand than a circular disk & circular hole, although both are really the same situation.

When I started this thread, I knew that SR was not disproven by this apparent paradox, but long ago had forgotten the explanation.

 

SR isn't disproven by the pole-and-the-barn paradox. However the explanations offered above do not actually deal with the issue of whether length contraction is real. Imagine you're in space in a gedanken spaceship. You're motionless with respect to a star, and it appears spherical to you. Then you accelerate rapidly so that you're moving towards the star at a relativistic speed. The star now looks flattened, resembling a discoid rather than a sphere. People say that in your frame of reference the star is length-contracted, but in truth that star hasn't changed one jot. Nor has the universe. You will see the stars, galaxies, and indeed the whole universe as length-contracted. You will see the CMBR blueshifted in front of you and redshifted behind. But that's because you're moving through the universe, not because it's changed. Instead you’ve changed.

As to how, step out of your spaceship holding a two-metre pole in one hand, pointing in the direction of travel. Hold a one-metre butterfly net in the other hand. I'm at rest with respect to the star, and you're about to pass me by. I have a similar pole and a similar butterfly net. Can you scoop your net over my pole whilst I scoop mine over yours? The answer is no. See Simple inference of time dilation due to relative velocity as to why. Look at the illustrations. The observer at rest with respect to the light pulse sees it follow a path like this ║ whilst the observer with relative motion sees it like this /\. Now go from a single-shot pulse to a light beam an inch wide turned on for a short period of time. The observer with relative motion doesn't see a vertical beam length-contracted to half an inch wide. Instead he sees the beam as being "smeared out" in the direction of motion. Draw a series of /\s, each one shifted a little to the right to picture this. There's a symmetry here, pair production tells us that electrons can be literally made from light, and you are made of electons etc. So apply the same thinking to you, your net and pole, and your spaceship. When you move fast through the universe, you are smeared out through the space of the universe. You don't notice it locally because other things are similarly affected. But you see the star, the universe, and me as length-contracted. If you and I have no stars or CMBR to go on, then we can't say who's moving. We each look length-contracted to on another, because we are each smeared out with respect to one another.

All this might sound like it isn't special relativity, but see The Other Meaning of Special Relativity by Robert Close. SR is right, and this paper gets tells you why it's right.