md65536

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? 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?

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 lightyear-long 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 traveler---not Earth's ruler---stops 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.

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.)

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: space-like, light-like, and time-like? 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.

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 curved-spacetime 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.

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 non-relativistic v will change the speed of light in the medium by a LOT less than v.

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?

Adding to that: A ruler doesn't measure "accumulated" distance, but an odometer does. While a twin makes a round-trip 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.)

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. 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.

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 made-up 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.

Thanks for the interesting visuals, they're rarely shown. From what I understand, the only(?) differences between the length-contracted 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 1-side of the dice still appear to follow an undistorted straight line, and the bottoms of the 1s follow another, and same with the 6-side 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.

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!

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. You've answered your own question! 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.

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 2-hour round trip? Could it be that you were right? If you answer this, it will prove that I'm a fool.

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 space-like or time-like 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 light-like intervals being invariant?). You have a knack for showing no interest in explanations, but continuing to get people to put effort into giving them.

re. "how can he have his own image so close and measure that the same image goes away from him at c? " 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?

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:

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 length-contraction 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.

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.

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. 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: ______ ?

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.

Oops, right! It's the relativistic Doppler factor, not the Lorentz factor. 🤕 brain damage

The frequency increases (blue shifts) the faster that you move toward the light source. At c/2 gamma is about 1.15, the frequency would be 1.15 times what was emitted, if moving toward the source (or redshifted to 1/1.15 times the emitted frequency if moving away). Everyone will measure the speed of light as c in their inertial frame. The two detectors I mentioned I was referring to A and B. Acceleration complicates things. The equivalence principle implies that you can set up a scenario where an accelerating O measures the same things as an O in a uniform gravitational field. So...... using 2 detectors depends on a lot of stuff (which direction O's accelerating, etc). One problem is that if O is changing speeds, and the 2 detectors must detect the light at different times (the events "A detects light signal" and "B detects same light signal" are separated by a light-like interval, so there is no reference frame where they are simultaneous) then effectively you're talking about 2 different measurements made in 2 different reference frames. O changes speed in the time that it takes light to travel the distance between A and B. You'd get a "speed of light other than c" if O treats the two measurements as if they were made in a single inertial reference frame, which they weren't.

Sure! It's "length contraction". In case I wasn't clear, I was talking about two different cases, depending on what the 10 LY refers to. Either the 10 LY is the distance from Earth to destination as measured by Earth, and John measures that as length-contracted to a shorter distance. Or 10 LY is the distance that John measures, and Earth measures a longer distance that John measures as length-contracted to 10 LY.

You said the trip was 10 light years, but didn't say what frame that's measured in. It sounds like you intended that to be in the Earth's reference frame, so Earth would measure the trip taking 11.11 years. In John's reference frame, the distance to the destination is length-contracted, so even though the destination approaches John at 0.9c, it arrives at John sooner than 11.11 years. Gamma is ~ 2.29, the contracted travel distance is 10 LY/gamma = 4.36 LY, taking 4.84 years at .9c, measured by John. This could be true (until the last sentence) if it was 10 LY as measured by John. If that were true, then it's true that he'd measure 11.11 years and that Earth would measure more (11.11 years * gamma = 25.49 years), but Earth would also measure the distance traveled as much greater too (10 LY * gamma = 22.94 LY), and the speed would still be .9 c. Earth would not experience time dilation in those measurements, it would measure 25.49 years as normal. The only time dilation it would experience is that a moving clock ticks slower, and it agrees that John's clock ticks only 11.11 years during Earth's 25.49 years.