Thales et al

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madmac, There are clocks (with observers) at every theoretically possible point in the “original observer’s” inertial frame of reference. And all of these clocks, in the inertial frame of reference of the original observer, are synchronized. (Perhaps we can call him the “OG observer.”) All of the “clock/observers” are requested to report the “event” of the part of the square that is right next to him at one specified time (such as “12:00:00”). And so, when 12:00:00 rolls around on all of the synchronized clocks, one of the clock/observers will say the top right corner of the square is next to him and another will say that at 12:00:00 the lower left corner is next to him and so on (and not just for the four corners but for every point along each side there will be a clock/observer at 12:00:00 who can report what “event” is next to him at this time). And this way we can find out the “height” of the length contracted square in the inertial frame of reference of the “original observer.” (If the square is moving at a constant relative speed this this same “height” will be found no matter what agreed upon time we use (11:59:59 or whatever).) And the same thing is done for the “height” (the y displacement) of the vertically moving rectangle. This is why I don’t think “Relativity of Simultaniety” is the solution here. (I’ve tried but I still don’t see it.) The clocks that are relevant to the measurements here of the square and rectangle are in the rest frame and so are not un-synchronized. (But it’s possible that when we switch to the square’s rest frame or to the rectangle’s rest frame this “relativity of simultaneity” with the square or the rectangle then pops up from this switch and saves the day (paradox-wise) … but I can’t see it yet.) However … After reading your link Mike-from-the-Bronx … I may be able to start seeing this other way to the solution. I’m definitely not there yet … but I think I’m beginning to understand the basics of the kind of switching (double-switching) you are talking about. I’m still thinking … I can start to see something. Thank you. And thank you all!

Okay … maybe I am wrong. If a square is moving diagonally across an inertial frame of reference is that square’s height knowable? Is the square’s height one thing if we look at it one way and is the square’s height a different thing if we look at it a different way (at the same time in the same inertial frame of reference)? If that’s what I’m hearing that doesn’t seem right to me. It seems like the square’s height and the rectangle’s height should be knowable and solely based on their direction of motion relative to the given inertial frame of reference. And so, it seems to me that in the inertial frame of reference of the observer the height of the diagonally moving square should stay the same and the height of the vertically moving rectangle should decrease. But … that would mean a paradox. I’ll keep studying! Thanks for the links!

In the observer’s frame of reference, the square’s motion is diagonal. However, we can talk about the vertical component of the square’s velocity and the horizontal component of the square’s velocity. It is my understanding of the Special Theory of Relativity that length contraction only occurs in the moving body’s actual direction of motion. It is my understanding of the Special Theory of Relativity that length contraction does not occur in a moving body’s “component” directions. No? The square is moving diagonally and so it should contract diagonally. The square has a component of its motion moving vertically, but it does not (… or so I believe …) contract vertically. Nor should it contract horizontally. “Moving bodies contract in their direction of motion.” Am I right? "The key I would focus on is that a square contracted diagonally in 2 steps does not have the same shape as a square contracted diagonally in 1 step. That's how it still fits in the rectangle." Can we do length contraction like this? Am I wrong that length contraction occurs in the direction of overall motion of a body?

Alright … first things first … the most important question first … are you really from the Bronx? “Switch reference frames from initial setup to Third Observer. Rectangle now moving diagonally(with additional contraction). Square now moving diagonally (with contraction). Third Observer at rest.” In the third observer’s inertial frame of reference the rectangle is moving vertically (not diagonally) but the square is moving diagonally. In the third observer’s inertial frame of reference the rectangle is contracted vertically and the square is contracted diagonally. Well, at least that’s what I tried to say. More questions? And thank you for taking my idea here seriously. If there is a way to use “relativity of simultaneity” to resolve this “paradox,” I can’t see it. 1. Clearly in the square’s rest frame the square remains within the rectangle. 2. And clearly in the rectangle’s rest frame the square remains within the rectangle. And so, It must be that in the observer’s rest frame the diagonally moving square also remains in the vertically moving rectangle (… for there not to be a paradox). And so I suppose you could say that the bottom of the diagonally moving square is inside the vertically moving rectangle but at an earlier time and the top of the diagonally moving square is also inside the vertically moving rectangle but at a later point in time? But this leads to the contradiction of the diagonally moving square being both longer and shorter (in its y dimension) at the same time (in the inertial frame of reference of the observer). Using relativity of simultaneity in this way would mean contradicting basic relativistic length contraction. As far as I can tell, in the inertial frame of reference of the observer the length contracted square is outside the bounds of the length contracted rectangle. ?

In the ladder/barn/pole "paradox" the ladder and the barn doors only encounter each other at a specific moment in time. The ladder moves through the one door and then the ladder moves through the other door. Here, time is not necessarily a factor. This thought experiment could be modified to say "the square and rectangle are moving like this relative to one another over 1 billion year" or "the square and rectangle are moving like this relative to one another over an infinite amount of time" If there is a way to get the square to break through the rectangle in the square's rest frame and if there is a way to get the squrae to break through the rectangle in the rectangle's rest frame with "time" and with "non-synchronous events" I haven't been able to do that yet. But, perhaps you seem to have made it work. I don't know.

There is a square moving at a constant speed across an inertial frame of reference. Relative to an observer at rest in the inertial frame of reference the square is moving diagonally. The moving square is length contracted in its direction of motion. (A diagonally moving square is diamond in its direction of motion. Contracted, it becomes a rhombus.) And so, from the perspective of the observer, the height of the diagonally moving and length contracted square is the same when it is at rest or when it is in diagonal motion. There is also a rectangle moving at a constant speed across this same inertial frame of reference. Relative to the same observer the rectangle is moving vertically. The moving rectangle is length contracted in its direction of motion. And so, from the perspective of the observer, the height of the vertically moving rectangle is less when in vertical motion than when at rest. From the perspective of the observer, in the inertial frame of reference of the observer, the height of the diagonally moving square in motion is the same as when it is at rest and the height of the vertically moving rectangle in motion is less than when it is when at rest. The square is inside the rectangle. When at rest, the height of the square is slightly less than the height of the rectangle. The square moves diagonally down and to the right. The rectangle moves straight down. The vertical component of the square’s velocity is the same at the rectangle’s vertical velocity. The moving square contracts in its direction of motion. The moving rectangle contracts in its direction of motion. (The velocity of the square is more and so it contracts more in its direction of motion.) The height of the square remains the same when at rest. And the height of the rectangle decreases. And so, with the height of the square only slightly less than the height of the rectangle when at rest, then, assuming the velocities relative to the observer are considerable, the length contracted square will be beyond (break through, depending on the materials) the bounds of the length contracted rectangle. The Special Theory of Relativity predicts, in the inertial frame of reference of the observer, the square will not remain within the walls of the rectangle. 1. From the perspective of the square at rest, where the rectangle and the observer are in motion, does the square remain within in the rectangle or does the square break through the rectangle? 2. From the perspective of the rectangle at rest, where the square and the observer are in motion, does the square remain within the rectangle or does the square break through the rectangle? 3. Does the square break through the rectangle in one inertial frame of reference while the square remains with the rectangle in other inertial frames of reference? 4. Is this a paradox?
10. Ives-Stilwell

This thread ended a long time ago. But I found a page that includes a pretty good description of what is physically happening in the Ives-Stilwell experiment. http://spiff.rit.edu/classes/phys314/lectures/doppler/doppler.html And so I figured I'd post it here if anyone is interested. Cheers!
11. Ives-Stilwell

Okay. I think I'm starting to get it. Thank you all for helping me out!
12. Ives-Stilwell

zztop: This is my current drawing: In my drawing the positive ion moves from the container on the left to the spectrometer on the right. However, you said: “1. Light coming directly from the ion is redshifted because the source (the ion) is moving AWAY from the receiver 2. Light going into the mirror has the same frequency as the light reflected from the mirror (due to the energy conservation). So, light reflected by the mirror into the receiver is blueshifted because the source (the ion) is moving TOWARD the mirror.” And so I’ve changed my drawing: (I changed the arrow on the moving ion from left to right to right to left) Is this now right? If so, then this clears up my red-shift/blue-shift confusion. Cool. But, then … if this is correct … my basic understanding that the ion comes out of the container of gas … and so would then move from left to right is all askew. My question (simple, like all of my previous ones): how does the ion end up moving from right to left? (This is so simple, I know it must be dumb.) Thank you.
13. Ives-Stilwell

Cool. More progress. Okay, in my drawings the grate and the spectrograph are left pretty much untouched/unexplained. I want to get to their description, but I still have a question about what has been said up to this point. I have a question about what is in my drawings in post #17: I show the light coming from the ion and to the receiver as red-shifted and I show the light coming from the ion and to the mirror (and then to the receiver) as blue-shifted. But shouldn’t it be the other way around? Shouldn’t it be that the light coming from the ion and moving in the same direction as the ion is blue-shifted and the light coming from the ion and moving in the opposite direction as the ion is red-shifted? No? Thank you!
14. Ives-Stilwell

zztop,I reread post #13, and if I’m reading it correctly now, that means I misread it before. This is what I’m now thinking. Please correct me where I’m wrong. --- Positive ions move from the gas filled cathode towards the spectrometer. The ions emit electromagnetic waves (light). Some of these move towards the spectrometer. And some of these move in the other direction towards the mirror. The waves that have come directly from the moving ion (the waves that were emitted in the same direction as the moving ion) arrive at the spectrometer red shifted. And the waves that first bounce off of the mirror (the waves that were emitted in the opposite direction as the moving ion) arrive at the spectrometer blue shifted. The amounts of red and blue shift are measured and compared. The amount of red and blue shift from the Doppler effect from the velocity of the moving source can be calculated. If this calculated amount is removed from the measured amounts, then what we are left with is the amount of additional red shift due to relativistic time dilation. --- Do I have this right? Please let me know. And please correct me where I’m wrong. And thank all of you for helping me with my problem.
15. Ives-Stilwell

Line a: “-the ions emit light DIRECTLY towards the receiver” “-the light arrives redshifted at the receiver:” Line b: “-the ions also emit light AWAY from the receiver, towards a mirror located at the opposite end from the receiver in the cathode tube that emitted the ions in the first place” Line c: “-the light is reflected by the mirror and arrives at the receiver blueshifted” Correct? Am I correct so far? (I realize there are things missing from my drawing right now, like the interaction with the grating, but first things first.) --- You might really think I’m dumb when I ask this question, but I’ve reread the Wikipedia article, and I still don’t know what physical element is “in motion” in this experiment in order for there to be “time dilation due to relative motion.” I assume the container of gas itself is not in motion. Or is it? Is it the moving ions themselves that are the “moving” element? I don’t know. Thank you for helping me. (zztop, I realize I only address a fraction of what you responded with. Thank you for all of the information. I will need to work through it all. Right now, I’m just trying to figure out the very basic basics.)