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Posts posted by md65536


5 hours ago, Halc said:
I'm just pointing out that these three observers are going to measure different times between the events, so they can't all have measured the frameinvariant interval between the two events. If one of them is by definition correct, then which?
Sure, they can measure different coordinate times. Those times can be a component of the invariant spacetime interval, without being invariant themselves. Different observers have different components that combine to the same interval.
It's the proper time that is the invariant length of a spacetime interval. Everyone agrees on the time that a clock measures on its world line between two events. But to different observers, the clock's path with have different coordinate times, and different coordinate lengths. Consider an infinitesimal line element of such a world line. For some observers, the clock can be stationary over that line element, with spacial components of 0 and an infinitesimal time component. For other observers the clock is moving, and the line element contains infinitesimal spacial component. They disagree on the components, but agree on its invariant length that results each set of those components.
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38 minutes ago, Halc said:
It cannot be done 'all at once' since different parts of the ruler need different levels of proper acceleration.
[...] The only rule was that at no point could there be either stress or strain.
You're right, though it could but not really but actually it could, but not really. "All parts accelerating equally and simultaneously" in a single frame basically describes Bell's paradox, but kind of the inverse because you're slowing it instead of speeding it. Definitely you can't do it as a rigid object. If you stop a ruler all at once, you change its rest length. Its former lengthcontracted length becomes its new proper length, it must be compressed. To do this would require an opposite rule, that strain or stress doesn't affect the ruler's behavior, so it could be arbitrarily compressed.
I guess a normal ruler would disintegrate if it was forced to stop quickly from relativistic speeds all at once, or if slowed gradually enough, it would have internal forces pushing it to resist being compressed, which would cause different parts of it to have different speeds. A realistic material ruler could survive not being perfectly Born rigid.
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3 hours ago, joigus said:
Namely: That no perfectlyrigid bodies can exist, for the simple reason that they would contradict relativistic causality. The reason being that one tip of the ruler would have to stop instantly and it would take some time for the local perturbation to reach the other end.
True... if Earth alone stopped the ruler and the ruler approached the limit of perfect rigidity, it would take at least 2 years for Earth to see part of the ruler come to rest 1 LY away, twice as long as if the ruler stopped all at once. But that's why I specified that the entire ruler was "made to stop" simultaneously in a given frame, rather than that it just stopped. I avoid worrying about practical details because SR isn't limited to what's practical. But it would mean this is a poor example to use as a thought experiment to explain what can be expected in real experiments. Not that anyone's argued that exactly!
Practically, you could stop a very long ruler all at once (in a given frame), eg. with a very long synchronized brake.
But, yikes... what happens if the traveler's 1 LYlong ruler is stopped by the traveler, and the rest of it stopped like a wave propagating across it at a speed of c? Before the traveler moving at .6c stops, it sees the ruler behind it at rest, 1 LY long, and Earth is .8 LY away (1/gamma) just before stopping, but appears .5 LY away (Doppler, receding). Then just after the traveler stops, but before the moving ruler stops, the ruler appears 2 LY long (Doppler, approaching). But it's lengthcontracted to .8 LY which means the far end has already passed the Earth (not yet seen) and really is still traveling at .6c... The closing speed of the far end of the ruler and the "stop" wave is 1.6c which means half a year for it to stop, it stops .2+.3 LY = halfway between Earth and the stopped traveler, as measured in the Earth's frame. Which means that stopping an object at its front would physically compress it by the Doppler factor (or more)! This agrees with what Earth would see: If the moving ruler stopped as if by a light signal traveling from its far end, it would appear to stop all at once, and it would appear .5 LY long (Doppler, receding) before that happened.
This seems like an even worse "impractical example", sorry! Is it pointless to even consider such a thing? (an object in which the speed of sound equals c)
Back to this thread........... the above example would be what a traveler might see if a ruler accelerated along with the traveler. However, using long rulers as representing lengths in different inertial frames, the rulers would remain inertial; they'd remain in their respective inertial frames, and no part of any ruler would be seen moving at a different speed than the rest of the ruler. A moving ruler would have different parts appear distorted differently (compressed moving away from you, stretched moving toward you). Using these inertial rulers, it is obviously (with hand waving) paradoxical to measure an inertial Earth appearing to move closer to you after you've come to a rest relative to it.
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8 hours ago, Markus Hanke said:
The spacetime interval is defined as [...]
This quantity transforms as a rank0 tensor (a scalar), so it is covariant under changes in reference frames. All observers will agree on it. If this weren’t so, then observers would disagree on l both the geometric length of world lines, as well as on the geometry of spacetime itself  which would be problematic.
So instead of saying the spacetime interval isn't invariant in curved spacetime, I should have said the interval defined for Minkowski spacetime, ie. the quantity (ct)^2  r^2 isn't invariant in GR. I was going to ask if the spacetime interval in GR is a local thing only, that applies only to intervals between nearby events, but if it implies world line lengths are invariant, it might apply to any arbitrarily separated events? Oh but then, there can be multiple world lines between two events in GR, so the spacetime interval is local only??? and a world line's length depends on local variances in spacetime curvature. Conversely, in Minkowski spacetime there is only one straight line between any 2 points, so the spacetime interval equation for SR is invariant no matter how far apart the events are. Am I on the right track?
The interval in GR is based on a set of values of g for each pair of the 4 space and time dimensions??? Different observers would have different components for the 4 dimensions, but the interval itself would be invariant. Can that be paraphrased as, "The spacetime interval in GR is invariant because curved spacetime is locally Minkowskian", so that any variations in the interval's components are like those seen in SR's interval, or is that wrong or incomplete?
8 hours ago, Markus Hanke said:The notion of gravitational potential can only be meaningfully defined in spacetimes that admit a timelike Killing vector field, and which are asymptotically flat. Schwarzschild spacetime is one such example. A counter example would be spacetime in and around a binary star system.
I don't understand. If you have two events in spacetime in a binary star system, where you might carry a mass from one event to the other, isn't there a fixed gravitational potential energy difference between the masses at the two events? Doesn't that imply a meaningful notion of gravitational potential?
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10 hours ago, Bufofrog said:
Michel has to literally see a clock go in slow motion or see a meter stick shrink, anything other than that is fake news to him.
The funny thing is, he's already accepted that a Doppler effect is acceptable in his definition of reality, and it's an easy modification of a twin paradox setup to make neither twin inertial and make it truly symmetric. Then just say "that's what they both see."
It's also funny because for me, seeing the asymmetry in the Doppler analysis of the twin paradox is probably what fully sold me on the predictions of SR, and I never doubted the resolution of the paradox after that, even though I still would have struggled with "the Earth's clock jumps forward with the traveler's change in inertial frame."
Then if you cherry pick some predictions of SR, you can get something that fits Michel's reality and doesn't add up (which is not a problem for Michel). For example, if you let the traveling twin have a lightyearlong ruler attached behind it, and you make it so the entire ruler stops simultaneously in Earth's frame, then you can see something like this: Say v=.6c, from Earth the receding ruler appears compressed by the Doppler factor of 1/2. Then when the 1 LY mark on the ruler reaches Earth, that part of the ruler stops, but the traveler appears to keep moving until it reaches 1 LY rest distance. All along the ruler, a "wave" of successive lengths of the ruler being seen coming to stop and returning to normal length spreads down the ruler, the wave moving at an "apparent rate" of c, so that it takes 1 year to see the traveler and the end of the ruler coming to a stop 1 LY away. That's something SR predicts and sounds similar to what Michel has described the traveler seeing (instead of SR's prediction of the traveler seeing Earth's entire ruler appearing normal instantly, when the travelernot Earth's rulerstops and comes to rest in Earth's frame). I'm not positive I got the details right.
Course, you'd have to sell your soul to argue that the predictions of SR describe reality and show that SR is wrong.
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1 hour ago, geordief said:
I struggle with the simple maths (and the implications thereof) of basic special relativity . So I can't with confidence say that it makes good sense or bad sense. (I will try to reread it a few times and it may become clearer)
The Pythagorean theorem seems to crop up a lot, and if you move one of the terms to the other side it looks like an equation for a hyperbola. Often you can see that in diagrams, where you have eg. a length in the x dimension, and one in ct, and the hypotenuse is meaningful in some way. Anyway I'm still figuring out things about that.
The numbers I used were just an interval s^2=1, with (ct)^2=1,r^2=0 in one frame, and (ct)^2=4, r^2=3 in another. A common speed used in examples is approx 0.866 c ie. sqrt(3)/2, because the Lorentz factor in that case is a simple factor of 2. You might try setting gamma to 2 and solve gamma=1/sqrt(1(v/c)^2) for v, if you can, to see why those numbers come about. Or get Wolfram Alpha to solve it for you if not. Otherwise, you can get away with following a lot of examples just using v=.866c, gamma=2.
(v=.6c, gamma=1.25, Doppler factor=2 is also common.)
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4 hours ago, geordief said:
I wasn't aware it was easy to synchronize clocks like you said and that is why I assumed there would have to be signal capture at E and M.
It's more about the coordinates (an inertial frame is 3 Euclidean space dimensions and a time that is the same everywhere within the frame), the clocks just represent a measure of the frame's time at different points. If you never had to consider different frames, you could use a single clock to represent time everywhere.
I'll just keep talking because I hope more people talk about the meaning of the spacetime interval being invariant! But with respect to that, do you know the 3 types of interval: spacelike, lightlike, and timelike? The type is invariant, and depends on if s^2 is respectively negative, zero, or positive (in the sign convention you used, s^2 = (ct)^2  r^2). For any spacelike interval, there's a set of inertial frames where the two events are simultaneous. For a timelike interval, there's an inertial frame where the two events occur at the same place.
I think the interval being invariant means that if S1 and S2 are more distant in M than E, then they must also be farther apart in time in M than E. An example is that a clock at rest ticks faster (smallest time between ticks) than if measured from any frame in which it is moving. If you have an interval where r=0, ct=1 (ie. a clock at rest ticking once, with appropriate choice of units) then s^2=1. In some other frame where it takes ct=2 for the moving clock to tick once, s^2 also must equal 1 = 4  3, so r must equal sqrt(3) units of distance. In that frame, there's more time between the two events, and more distance. And indeed, in such a frame, gamma=2, v=(sqrt(3)/2)c, and the moving clock moving for 2/c units of time moves a distance of sqrt(3).
Does that make sense? I'm not sure I got it right because I have no experience dealing with intervals.
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While waiting for a better answer...
Having the events on the sun unnecessarily complicates things because of spacetime curvature. You're measuring distances from afar, in a different gravitational potential, so there's not one "correct" way to measure those distances. I don't think curvedspacetime intervals are invariant. However, if you're treating the sun as just a location in flat spacetime, that's fine.
You wouldn't directly compare the arrival time of light signals from the events, you'd want the time those signals were sent. Basically you'd subtract the travel time of light to find that.
Edit: I think I'm misunderstanding "time the difference in signal capture between S(1) and S(2)". Signals aren't really a necessary part. You could use any clock(s) synchronized with the observer's, to measure the events' times. Typically a clock at S1 and one at S2 would be used, but in this example a single clock at E and then compensate for light travel time would be more practical. Then M in a different reference frame would measure its different set of times, either using a different set of clocks or by transforming the times from another reference frame.
To measure distance, you can imagine all of space being covered in a lattice of rulers, at rest in an observer's inertial frame. Another observer would use a different lattice of rulers. Then the events are located at some place on those rulers, and you can measure the distance between them. Simply knowing the position of the sun in the observer's coordinates, or timing distances using light signals, or other ways, tells you distances and/or the locations of the events.
An event has a time and a location in any reference frame's coordinates, ie. a place and a time on a lattice of rulers and synchronized clocks. Then r is the spatial distance between the two events, and t is the time between the two events.
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The speed would change in different reference frames, exactly as the speed of an object would. To see this, imagine light through some medium, and an object traveling along with it at the same speed, so that they both pass through the same set of events. In another frame, if their speeds were no longer equal, they wouldn't pass through the same events, which would be a paradox. As swansont mentioned, because of the velocity addition (composition) formula, changing the observer's speed by nonrelativistic v will change the speed of light in the medium by a LOT less than v.
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2 hours ago, michel123456 said:
They are observing THE SAME THING.
This is literally your thread now, I'm curious how this adds up in your examples, without relativity.
You said that when traveling twin A stops, it doesn't immediately see Earth as stopped because of the delay of light. Relativity says Earth doesn't see A stop immediately, and you're saying A sees Earth appearing the same.
For example, suppose Earth is at rest at the 0 mark on a ruler, and A travels to a 1 LY mark, and then stops. The whole time while traveling, A sees that Earth appearing to stay at the 0 mark, agreed?
When A stops at the 1 LY mark, how far away does it see the Earth? I think it should be 1 LY. It sees Earth at 0 and itself at 1 LY and the distance is "normal", not contracted, and it is 1 LY. If A sees Earth continuing to move away, how is it appearing to move? Does it appear to move farther than 1 LY away? How do those numbers add up, in your view?
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8 hours ago, Eise said:
you assert that in the so called 'twin paradox', there is an asymmetry, because the effect of the time dilation stays (the traveler has not grown older so much as the homestayer), but the length contraction has gone (the twins are still equally sized). Truth is that you comparison is wrong. After arriving back home the twin's clocks tick at the same rate, so the time dilation itself is gone, just as the length contraction.
Adding to that: A ruler doesn't measure "accumulated" distance, but an odometer does. While a twin makes a roundtrip at constant speed, if its clock records half the time Earth's does, it will measure that it traveled half the distance Earth measured it traveling. Its odometer retains the accumulated effects of length contraction.
(The trip's clock and odometer dicrepancy ratios would generally differ is the speed wasn't constant.)
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9 hours ago, michel123456 said:
Quest1. Why is the distance d2 larger than the distance d1? Shouldn't it be contracted?
It is, see image (b). Image (c) is what the dice look like when you receive (at the same time) light that left the dice at different times and traveled different distances. Image (b) is what you see if you remove the differences due to delay of light.
In fact, if you put points of light all over the dice and sync'd them to flash at a single moment in the observer's frame, what you see would look exactly like (b). This assumes persistence of vision or exposure time somewhat proportional to d2, since the light wouldn't all arrive at the same time.
You could figure out so many mysteries if you learned SR.
1 hour ago, swansont said:Is the “length” moving? No. So it’s not contracted.
Do you mean the lengths between the dice? Those lengths are contracted, see (b).
Here's a thought experiment to show that the spaces between objects are contracted the same as objects themselves: Consider the dice in their rest frame. Put an enclosure around each die, and connect them with sticks. In the moving frame, everything contracts, and the dice never leave the enclosures. The distance between the dice must contract the same as an object of the same length.
Edit: After catching up to the rest of the conversation I see my explanation isn't necessary. I think that if you showed the dice say 1s apart (big dice) in a Newtonian frame I guess??? in (a), and showed them "at the same events" 1/gamma s apart in (b), it would look the same as it does now. If (b) showed them 1s apart in the new frame, they'd be spaced farther.
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5 minutes ago, The victorious truther said:
Not only that. . . are you just going to ignore the meat of my previous posts. . . anything to say. . .
Michel has established that he's spent 20 years denying relativity. There are many, many hijacked threads over the decades, any that have certain keywords (in this case, "time, direction"), turned into 6+ pages of failed attempts to get him personally to accept something about SR. He's stated in this thread that he's not interested in relativity, and of all the hundreds and hundreds of answers to his repeated questions, not a single one of them he acknowledges as an answer to his questions. The only time I've seen any calculation or attempt to work through a problem, is when he's twisting his beliefs and madeup definitions into a nonsensical answer that confirms that relativity is wrong. Trying to understand relativity is destructive to his goals, and completely avoided. So the answer is yes, anything that shows relativity working will be ignored. If you're enjoying explaining it, that's good. If you're hoping Michel will learn something, ... I wouldn't.
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4 hours ago, The victorious truther said:
Hmmmm. . . I found it! There is a difference between length contraction (measured) and what is observed which respectively would be (b) versus (c) as well as what you would expect with naive classical physics (e).
(a) A row of dice at rest moving from left to right in a single file at 95% of the speed of light. (b) The moving dice are length contracted, so that one might (wrongly) expect them to look as here. (c) If you actually observe the dice, however, they will appear rotated. (d) But when some perception in depth is provided, you’d see them as sheared rather than rotated. (e) Shown here is the predicted “classical” appearance of the dice, with no length contraction. You can view a short film of part c online here. (Courtesy: U Kraus 2008 Eur. J. Phys. 29 1)
Thanks for the interesting visuals, they're rarely shown.
From what I understand, the only(?) differences between the lengthcontracted measured (b) and the seen (c) are abberation of light (where the orientation of a ray of light is different in different frames of reference, due to the time that it takes light to travel from source to destination while those points are moving) as well as different parts of the scene being seen in different positions due to delay of light from different distances.
The rotation seen must be a rotation in 4 dimensions, nothing is rotated in 3? To demonstrate, if you add rails that the dice slide on (say 4 rails that 4 edges of each die slide on), such that the rails are at rest relative to the observer, you'll see of course that the rails do not appear distorted, and that the dice edges never appear to leave the rails. For example, the tops of the 1side of the dice still appear to follow an undistorted straight line, and the bottoms of the 1s follow another, and same with the 6side at the back. When moving in the x direction, an objects's parts don't rotate off their yx coordinates, but instead appear stretched between them. That can also be seen in the black and white animation you posted earlier.
Edit: I just noticed, the (d) image shows this... Instead of rails, have the dice slide along lines in the floor. It looks like the checkered floor is at rest relative to the observer, and each die has the same width and proper length as a floor tile. The sides of the dice appear remaining aligned with the straight floor tiles. The description also describes them as sheared, not rotated, but the shearing is how the 4D rotation appears in 3D?
Edit2: No wait... I think there are 2 separate things. A hyperbolic rotation of lengths and times, between the x and t dimensions (y and z are not affected at all if there's no motion in those directions) and the superficial appearance of the objects appearing rotated. The observable effect of the actual rotation is length contraction. The distortion that appears similar to a rotation isn't a rotation at all, but shearing. The entire front (in direction of travel) of a die passes the parallel floor lines simultaneously, both in the die's frame and this observer's frame, but appears not to because the die face isn't all the same distance to the camera.
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10 minutes ago, michel123456 said:
You have been fooled.
Absolutely! You are absolutely correct! It was all worth it for the laugh.
Sounds good. Your example has a traveler moving away for one hour, and then coming back in half an hour. What does Galilean relati ah forget it. Thanks for the laugh!
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1 hour ago, michel123456 said:
The half and double rate are observed from the Earth. From this FOR, the outbound & inbound travel will not be observed the same.
I suspect that the same goes for the traveling twin, he will not measure the same time for the outbound & for the inbound, because of the Doppler shift taking place only in 1 direction (toward the Earth).
No symmetry then? Abandoned the thought experiment the very second it didn't show what you wanted? Thus proving me a fool for even trying despite this obvious outcome.
The trip that I described is realistic. If you travel outbound at one speed, and return at the same speed, it will take you the same time to make each leg of the trip. Galilean relativity even agrees with that.
2 hours ago, michel123456 said:Why the relativistic scenario does not end to be symmetric? While presenting a totally symmetric equation for frequencies? see my post above.
You've answered your own question!
8 hours ago, Bufofrog said:You seriously have been trying to understand special relativity for 20 years, and you still say things like this?
Maybe you should give up. Seriously, at some point you need to realize you're wasting your time.
Denial of science is a lifelong pursuit, you don't just give it up. The same reason that we would keep trying to explain relativity to someone with an amazing commitment to avoid understanding it, keeps someone else trying to show that relativity is wrong, even if he thinks the people he's trying to convince are "insane to believe such a thing."
Re. "at some point you need to realize you're wasting your time."... it's a battle of stubbornness, who will refuse for longer to give up? My bet's on Michel. You don't throw away 20 years invested in denying relativity, and risk it all by accepting something new now. You take it to the grave, regretting only that the insane people wouldn't listen.
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1 hour ago, michel123456 said:
Fine. Why the relativistic scenario does not end to be symmetric? While presenting a totally symmetric equation for frequencies? see my post above.
A sees B traveling 2 and 1/4 hrs while going away.
And you say that A sees B traveling back in 15 minutes? Or 2 and 1/4hrs divided by 3 = 45 min? Why do you divide 45 minutes (as observed by the traveling clock)? instead of the time observed by A?
I think you found something nobody considers; there is symmetry in the twin paradox! How can the maths possibly work out? You might convince me?
Okay, let's say that a traveling twin travels outward for one hour while seeing Earth's clock appear to tick at half the rate. Then it returns with a symmetric trip and for one hour, it sees Earth's clock appear to tick at double the rate.
Half the rate for one hour, and double the rate for one hour. What is the total time that it sees ticking on Earth's clock, during its own 2hour round trip? Could it be that you were right? If you answer this, it will prove that I'm a fool.
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10 hours ago, Markus Hanke said:
I am not entirely sure what you mean. Basically, you’d calculate the interval in one frame first; you then apply Lorentz transformations to the time and space parts in order to “go” into the other frame, and check what this does to the overall interval. What you’ll find is that the interval remains the same  Lorentz transformations are just hyperbolic rotations in spacetime, so when you go from one frame to another, you essentially rotate some portion of the space part into the time part, or vice versa. This leaves the overall quantity unchanged.
I think I misunderstood. "The spacetime interval is invariant" was used to explain several things, I think including (paraphrasing) "why are lengths and times different in different frames?" and "how is that consistent?" If you consider a single spacelike or timelike interval, it being invariant seems to demonstrate on its own that the measurements between frames are consistent, but doesn't give a reason why the measurements are different. I think what I missed is: it's that *all* intervals are invariant that explains why the measurements must be different (which can be demonstrated just by lightlike intervals being invariant?).
3 hours ago, michel123456 said:My quest is about time, not relativity.
You have a knack for showing no interest in explanations, but continuing to get people to put effort into giving them.
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re. "how can he have his own image so close and measure that the same image goes away from him at c? "
5 hours ago, Markus Hanke said:Because the spacetime interval is an invariant.
So basically, say in an Earth frame, with a rocket approaching at near c, the image of the rocket when it began its journey can be very close to the rocket itself, for nearly the entire journey, but in the rocket frame the image reaches Earth less than half way through the journey (since the Earth is approaching at near c, Earth and the outbound image will meet near the halfway mark).
The spacetime interval is invariant implies for example that... if you consider some pair of events, one that the "image" passes through and one that the rockets passes through, that are spatially near each other (separated by x) in the Earth frame, they're also temporally near each other (separated by t) in the Earth frame, and you get a small interval (ct^2  x^2). In the rocket frame, those events are far apart spatially, but also far apart in time, and you get the same small interval.
Is that all there is to it? Does that sufficiently explain it? It seems to, but normally I'd have to figure out the separate time dilation, length contraction, and relativity of simultaneity to reconcile the different reference frames. Are those still needed, separately (or combined like in Lorentz transformation), to calculate the components of the interval? It seems like the interval being invariant isn't enough to know the x and t in another frame, without calculating one or the other separately?
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8 hours ago, swansont said:
The lack of understanding is perfectly acceptable; you are making an effort to learn.
I don't think that's true. I don't see a single reply here from michel123456 that relates to trying to understand any of the answers and explanations given. No asking for further details. No working through a solution. Every reply is a justification of not making an effort to learn, an argument of why the explanations can be ignored.
Literally 10 years ago he was asking about the same "problems" he had with relativity. 10 years from now, he'll have a similar list of "problems", after thousands of attempts by people to explain it to him, after 0 attempts to work through it.
What he's good at, is asking questions that makes one think he's interested in learning about relativity. But look at the replies. The only interest is in what doesn't make sense to him. Anything making sense of it is ignored. That's the only answer he's interested in: that it doesn't make sense. All his questions are phrased as if the answer he expects is that it can't make sense, never a question about how the resolution to the problems work out correctly. So I think he's soapboxing, getting much better responses by stating "relativity is nonsense" as a question.
Edit: To be fair, page 1 of this thread is full of counter examples to what I said, including asking about specific examples and numbers and their explanations. I don't know how we got from that on page 1 to page 2 with:
2 hours ago, michel123456 said:And when the twin comes back, he is not length contracted according to his position (upright or sitting) in the spacecraft. It is insane to believe such a thing. [...] The twin that comes back younger is like a train derailing at the horizon: it is the same flawed concept.
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24 minutes ago, michel123456 said:
Sorry but I am talking about what Relativity doesn't make sense to me:
[...](it would make more sense if length contraction happened in all directions. More sense if length contraction was a kind of illusion, a kind of perspective effect and not a real thing.
[...] What the other observer is measuring is a distorted image, it is not a 2nd reality.
[...]And there is something wrong in it.(and not endless conversations about accelerations in order to determine who is the traveler & who is at rest, that is not the problem)
That's not talking about relativity. Why not work through the explanations given, instead of ignoring them and always falling back to accepting some alternative, accepting that relativity is wrong?
All of your problems have been explained to you before. They all have examples that you *could* work through. You *could* work through them and see how you arrive at the answers that SR predicts, or find where you're getting hung up. You *could*!...
I gave you an example that *showed* the effect of time dilation without lengthcontraction perpendicular to the direction of travel. You expressed incredulity and moved on. Go through the example. We can show you what you're missing. Don't look at any of the details, and the answer's simple: You're missing everything.
However I think you're opposed to understanding relativity and are wasting people's time repeating the same questions about it while ignoring the answers. Go through the mathematical details of an example.
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22 minutes ago, michel123456 said:
i do have some problems with Relativity;
relativity is based on a principle that says that SOL is measured the same by all observers, no matter their state of motion. To me it does not go well with speed being relative. To me (but that must be me) it would be simplier to state that the phenomena that an observer observes moving at c is called "light". IOW that the observers will call "light' some different thing. As if light was an array of radiation traveling at all possible velocities.
Always I see the same pattern. Brush aside explanations and equations as if you didn't even read them, but always latch on to any idea that justifies a failure to understand relativity as if that's just another equally valid viewpoint. I think Asimov's "my ignorance is just as good as your knowledge" quote applies. You say your view is "simpler" but it's just a misunderstanding. It makes me think that people who put effort into trying to explain things to you over and over are just wasting their time. You ask questions as if you want to understand, and then reply to answers as if your questions were only meant to demonstrate what you see as "problems with Relativity" and you had no interest in understanding how they're resolved. If you were interested in understanding it, you'd spend more time talking about what relativity says that doesn't make sense to you, and less about how much sense an alternative makes.
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41 minutes ago, michel123456 said:
You have kept velocity the same. why haven't you changed velocity and kept time & space as they are? Why is it so important to keep velocity the same for all observers?
It's only important if you want to be consistent with what we actually measure of reality.
It's a combination of the definition of speed being relative, and that measured speeds are consistent with that. If you have A moving at 0.8c relative to B, does it make sense that B is moving at a different speed relative to A? If you wanted that, you could define speed differently (eg. define speed to be absolute, and please call it something else), but you would end up with a system of measurements that is either inconsistent with measurement, or more cumbersome than what we have.
2 hours ago, michel123456 said:I assure you I am not deliberately twisting everybody's replies, so I am afraid the other option may be valid.
I think it's a 3rd option: I think you're determined not to accept relativity and so you're determined not to understand it. I think we could find out with a quiz! Do you think that a) you will accept relativity and understand it together, or b) you will eventually understand it, and then accept it after, or c) if you accept that it's correct first, that will make it easier to understand, or d) you will never accept it and it's more likely that you'll find a flaw in it before anyone convinces you that it's true. Or e) other: ______ ?
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11 hours ago, geordief said:
Thanks. That seems to solve the problem I had . So an observer accelerating towards a source of light needs 2 nearby detectors and to measure a short pulse of light at each.
The closer in time he or she collates measurements from the two detectors the closer the measurement of the light will come to c.
Do I have it?
I think so... but I'll nitpick. I wouldn't say the observer "needs" 2 separated detectors. For example Markus's method I think involves making only local measurements. Instead I would say, that if you *are* using 2 separated detectors, you have to coordinate them properly.
Not all measurements that rely on a separation of detectors will be the same as a local measurement, just by making the separation smaller. But in this case, by making the two detectors closer, you're minimizing the time that the observer accelerates, so minimizing the effects of difference in speed, and yes getting closer to what an inertial observer measures.
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michel123456's relativity thread (from Time dilation dependence on direction)
in Speculations
Posted
Credit where due, you seem to be making progress. Using conventions like time on the yaxis and light signals shown at 45 degree angles really helps in communicating ideas.
In cases where you start with bad assumptions and then show what happens, the conclusions aren't going to be useful. If you put "garbage in", you get "garbage out." Eg. the Doppler factor of 5 from assuming the clocks tick at the same rate, is not useful. You *can* assume the clocks tick at the same rate, but then you'll find that the speed of light isn't constant etc., which doesn't match reality.
Where you lined up "lines of simultaneity" by connecting moving clocks reading the same time, is not correct. That's not what simultaneity means.
If you want to make further progress, try this: In diag 2, where it shows a light signal from Planet X to Earth when the traveler turns around, draw a light signal from Earth to Planet X at that same time, and see where (and when) it reaches the traveler. You'll see that Earth and the traveler can't see the same thing, and there must be asymmetry. That diagram would show what each observer actually sees. Any other diagram would have to agree with it. Diag 5 does not show the same thing as diag 2. If diag 5 showed the same scenario but from a different perspective, it would still show that the traveler sees the same things that diag 2 shows the traveler seeing.
In my opinion, that interpretation is not incompatible with SR. But, however you interpret what is "really" happening, it still has to match the measurements, and the twins "really" do measure a difference in age.
Earth doesn't see that distance lengthcontracted, the clock does.
To wrap your head around it, you can ask "is the ruler that I'm using to measure this distance, moving?" If yes, length contraction will apply, otherwise it won't. For example if you measure the distance between Earth and Planet X is 1 LY according to Earth, that ruler is not moving. If something travels from Earth to X, the ruler doesn't move. Earth measures distances along that ruler without length contraction.
Meanwhile if a traveler going from Earth to X would see that same ruler connecting Earth and X moving along with them; the distance between Earth and X is lengthcontracted.