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md65536

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Posts posted by md65536

  1. 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: ______ ?

  2. 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.

  3. 37 minutes ago, Janus said:

    No.  The detected frequency would be 1.732 times the source frequency.  The red or blue shift is due to a combination of both The changing light propagation times and relativistic time dilation.

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

  4. 6 hours ago, geordief said:

    What about my scenario of an observer traveling at c/2    at the mid point of A and B in the direction of B (using your terminology)?

    This is flat spacetime ,I think but the frequency of  the light coming from B  is much higher than it would be if O (observer) was at rest wrt B and A.

    If many photons are emitted  from B towards O and he or she detects them on two detectors (as per your suggestion) will he measure the speed  as c?

     

    Lastly if O is accelerating towards B will this be the same as a gravitational  scenario? (Will he or she measure the speed of light as greater than c using the  2 detector method?)

    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.

  5. 3 hours ago, drumbo said:

    Ok. It appears I understand even less about relativity than I thought. I'll take a few days to read an introductory book.

    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.

  6. 10 hours ago, drumbo said:

    I see, and if he was going 0.9c would his travel time from the Earth's reference frame be 11.11... years?

    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.

    12 hours ago, drumbo said:

    In John's reference frame he records that it takes 11.11... years to complete the trip implying that the average velocity for the trip was 0.9 light years per year. Meanwhile back on Earth we have been observing John's spaceship. Due to John's high average velocity of 0.9c we experience time dilation which implies that by the time John reaches his destination more than 11.11 years will have passed here on Earth, and therefore if we calculate John's average velocity for the trip from our reference frame it will be greater than 0.9 light years per year.

    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.

  7. 11 hours ago, geordief said:

    If the observer chooses 2 points separated by 1 metre and records the time of detection at each point this should allow him or her to evaluate the speed ,but if the light is first detected at the first point then is not the photon absorbed by that detector  and retransmitted to the second detector?

    That doesn't matter because you can use a short pulse of light, ie. simply use many many photons instead of one, and they'll all behave the same with respect to speed. Some of the photons can be detected at the first point, and others at the second.

    11 hours ago, geordief said:

    So if the observer evaluates the speed of light between those 2 points is that actually the same as the speed of light between the Sun and the first detector?

    Yes and no. Assuming flat spacetime, yes: It is assumed that the speed of light is the same everywhere, and that agrees with experience. With GR, not really. Measuring the speed of light from a distance, at a different gravitational potential, isn't really meaningful. The local speed of light is invariantly c. Some might say "the coordinate speed of light in a gravitational well isn't necessarily c" but "coordinate speed" might not be an accepted scientific definition.

    11 hours ago, geordief said:

    How does that observer directly measure the speed of the light it detects at that moment?

    There's not really any theoretically accepted way to directly measure the one-way speed of light without something like the definition below. If you're using measurements of timing made at two different locations, you need a way to relate those two measurements, for example using synchronized clocks. Einstein established the definition we use to say that sync'd clocks read the same time. The definition he used in his 1905 paper on SR is (translated):

    Quote

    We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A.

    Using this you can measure the one-way speed of light between A and B because it's by definition the same as the two-way speed of light, which can be directly measured, and also because you can sync clocks at A and B, etc.

    Obviously, the one-way speed of light between A and B depends on the time it takes light to go from A to B, and the time at the different locations is defined. There's no way to independently measure the time at A and B without knowing that the one-way speed of light is the same as two-way, or to measure the one-way speed of light without knowing the time at A and B. So the time of a light signal is defined to be the same in either direction.

     

    (Of course... you could use alternative definitions as well. You could eg. put an observer C equidistant to A and B, and observe that one-way signals from A to B do take the same time as from B to A. But then you'd define (or assume) that the light from A to C takes the same as from B to C. Einstein did it right, defining instead of assuming that the time is the same, because if you used some alternative way to sync clocks in a consistent way, then you could make different assumptions on the timing of light that are true using your alternative definition of time.)

  8. https://www.fallacyfiles.org/redefine.html

    https://www.logicallyfallacious.com/logicalfallacies/Definist-Fallacy

    1. Define "heaven" to be something that's "scientifically supported." (Or maybe something simply vague enough that it can't be scientifically refuted?, ie. "not even wrong"?)

    2. Therefore, heaven is scientifically supported. QED.

     

    1. Define something as whatever I want to argue.

    2. Therefore, whatever I want to argue.

  9. 8 hours ago, swansont said:

    What will make a difference is the length of travel, and if expansion is an issue, that the v of a massive object will diminish.

    You should still be able to apply vt to the problem, but v will now be a variable.

    A 10^-4 fractional difference in the speed means 10^-4 difference in arrival time over some distance. If that's a light year, then it's 10^-4 of a year, or 3.15 x 10^3 seconds. That's less than an hour. At 100,000 LY, it's about a year.

    Time t will vary along with v, but also with d (length of travel), which itself should vary, and depends on time and expansion. You're giving an example where the differences in v and t are small, and difference in d is negligible? (Due to being nearby and/or negligible rate of expansion? Or assuming a fixed distance of expanded space, instead of being sent from a common source that is receding over time?)

    However if we were talking about an object at the cosmological horizon, light from it would arrive after infinite time. For any speed less than c, there should be another nearer horizon, from which an object traveling at that speed would take an infinite time to arrive. For a speed like .999999c, if someone is in between that speed's horizon and the cosmological horizon, then light from them would arrive after finite time, but a rock at .999999c would never arrive.

    So you can have "nearby" cases where expansion adds less than a second to the time between arrival of light and arrival of neutrinos, and the extreme distant case where the neutrinos take infinitely longer to arrive.

    8 hours ago, swansont said:

    For a photon the wavelength shift is the is the same factor as the energy shift, since energy is hc/lambda - twice the wavelength is half the energy. But for a massive object, twice the wavelength is half the momentum, not half the energy.

    That would imply expansion does slow down an object, if the object is redshifted. If light from a distance source was redshifted to twice the wavelength by expansion, would a massive object launched from the source and traveling at near c (and arriving nearly at the same time as the light) also arrive with nearly twice the wavelength it was launched with?

  10. 4 hours ago, Halc said:

    That was sort of my thinking at the OP,  but I wasn't sure. Energy is conserved, so where does it go?  The 'expansion' doesn't seem like a form of energy that can receive it. Maybe all the slowing down stuff powers the dark energy, translated into acceleration of expansion.

    There's this: https://www.forbes.com/sites/startswithabang/2018/07/28/ask-ethan-where-does-the-energy-for-dark-energy-come-from/

    Including:

    Quote

    When you have particles interacting in a static background of spacetime, energy is truly conserved. But when the space through which particles move is changing, the total energy of those particles is not conserved.

    [...]

    There was a paper written in 1992 by Carroll, Press, and Turner, which dealt with this exact issue. In it, they state:

        …the patch does negative work on its surroundings, because it has negative pressure. Assuming the patch expands adiabatically, one may equate this negative work to the increase of mass/energy of the patch. One thereby recovers the correct equation of state for dark energy: P = – ρc^2.

    Unfortunately I can't find anything about how expansion affects the energy of moving objects. I'd assumed it would be the same as with light.

  11. On 8/27/2020 at 12:49 PM, Halc said:

    Suppose I launch a rock at half light speed.  The proper distance between it and some object 7 GLY away is constant, making it sort of stationary relative to it in a way. But over time, the Hubble 'constant' goes down, and that distant object is no longer stationary relative to our rock, so the rock starts gaining on it.  Given that logic, perhaps it will not only eventually get there, but it will pass by it half light speed.

    I don't think that's right. I think the rock would have to lose energy to the expansion, which means it would arrive at a lower speed relative to its destination than its speed relative to the source when launched.

    Suppose a photon and a neutrino were sent, with the same energy. Say they arrived roughly at the same time, and the photon was red-shifted to half its energy. Wouldn't the neutrino need to have roughly half the energy it was sent with?

    In this example, the neutrino's speed is so close to c that it can lose half its energy to a decrease in speed and still be moving at a speed very close to c. In the case of a slower rock, I think it'd lose even more energy since it would take longer to travel.

    Since the expansion wouldn't affect the rest mass, I suspect it's a fraction of only the kinetic energy (of the rock or neutrino or photon) that is lost, not total energy.

     

  12. 11 hours ago, dimreepr said:

    How do you do that?

    It's a fictional example. Everything has their own clock, and they can't all be stopped "at the same time". In fiction where "time stops" in some contrived way, some observers' clocks stop while other observers keep observing, because if there's nothing to observe that some clocks have stopped, the stop is meaningless and doesn't look good on tv etc. When they use "time is stopped" for things like time travel, it's some made up way with most likely no basis in reality.

  13. 6 hours ago, studiot said:

    "If the pilot and spaceship observe the universe contracted to 1m, how small does that make the observed size of the spaceship/pilot from the point of view of the universe"

    I roughly calculated an order of magnitude somewhere between a Planck length and a quark. It depends on how big you say the universe is. There are many orders of magnitude between Planck length and quark though; it would be around some billionths the size of a quark, and some billions of Planck lengths.

  14. 45 minutes ago, empleat said:

    I thought everyone have their own clocks. I don't even know what to imagine behind that. Than how could be some object measured by other clocks, which have stopped?!

    Sure, everything has their own clocks. I'm giving an example the breaks the laws of physics. For example, say you have two observers A and B, a light year apart, and A sends a message to B. Then you stop time for A and B, but let the message keep going. Then start time so that A and B measure the message taking say only an hour, as if it instantly jumped across space while they and their clocks were stopped.

     

    But then suppose there's another observer C moving at high speed relative to A and B. Normally, if A and B, in their rest frame, are a light year apart and there's an event at A and another at B one hour later, then C can be moving such that the event at B happens before the one at A, which is normally fine because the events aren't causally related. But if A can send a message to B, that is causal relation, and if C can observe B receiving it before A sends it, causality is broken. By ignoring the laws of physics, I made an example that doesn't make sense, and is a paradox. For C, "time starts again" first, before B receives the message, then later "time stops" after A sends it. As long as you're ignoring the laws of physics, you could make up any number of ways to resolve this paradox, and if you're just making things up, then yes literally anything could be imagined.

     

  15. 1 hour ago, michel123456 said:

    However length contraction & time dilation go hand by hand, you cannot have the one without the other.

    And of course relativity of simultaneity, which is directional. A pair of clocks placed on a line perpendicular to their direction of motion remain synchronized with each other.

    Please work through an example of this on paper or numbers. Put it into geometry or use equations. You can reiterate the same philosophical questions for 10+ years and still not be any closer to a philosophical understanding of the answers.

    For example, suppose you have a stick 1 light second long, and you send light from one end (event A) to the other (B) and back (C). In the rest frame of the stick, the time between A and C is two seconds. In a frame where the stick is moving perpendicular to its length, A and B can be very far apart (many light seconds or even light years) because the stick moves, even without the stick's length changing. It can take much longer for the light to go from A to C, yet a clock on the stick only records 2 seconds during that time. There you have time dilation without length contraction. Then add a stick moving parallel to its length and see how length contraction is now needed. Draw this out on paper and if it still makes no sense, show the numbers that you're having trouble adding up.

  16. 1 hour ago, phyti said:

    U will measure the A ship as length contracted due to its motion.

    A will measure the world as length contracted due to his perception.

    The two frames are equivalent, the same reasons for length contraction apply equally to both of them.

    Motion is relative. If A is in motion relative to U, then U is in motion relative to A.

    Only moving things* are length contracted. If A measures "the world" as length contracted, then the world is moving relative to A. Any part of the world not moving relative to A, won't be length contracted according to A.

     

    * Or, trying to be more accurate: "the lengths of, and distances between, objects as measured in their rest frame, are contracted in frames in which they're moving".

  17. I can't see how you could possibly break causality.

    Time is relative, and time is what clocks measure. Time, or a clock, will stop on a black hole event horizon according to an observer outside the horizon. That doesn't break causality, which is perfectly fine with that type of "time stopping".

    But you probably mean if all of time stops at once? That doesn't make physical sense in the universe as we know it (ie. described by General Relativity) so you'll have to specify what you mean by it. However, if you came up with an arbitrary definition of simultaneity that made physical sense for a particular abstract observer, and stopped all physical processes at the same time, then "later" (still letting time pass for the non-physical observer, whether that's even meaningful) they all continued as they were before and at the same time, then not only would causality not be broken, I don't think any physical thing would be able to measure any difference having happened at all... I think. So whether everything stopped and started all together, or didn't, doesn't even have any bearing on reality, which I think would make it basically a philosophical exercise rather than a scientific question. Even though things wouldn't all stop at the same time according to others, if everything stopped and started in the same way, I don't think they could possibly notice.

    However, you might be able to contrive some meaning of "stopping time" that breaks causality. For example if you temporarily stop time for some clocks but locally let a physical object keep moving as if time wasn't stopped, that object can move faster than light as measured by the stopped clocks, and that can break causality.

     

  18. 6 hours ago, Winterlong said:

    If we, on the other hand, want to keep the fabric-of-something idea. Well, then it should have a length for a particular observer, whatever it is, that can be contracted up to 1 m
    [...]

    From the early "yeah, but the universe expandes at c" to the last views of "spacetime, curvature? fabric?" or your calculations about the speed

    It sounds like you've moved on from the length-contraction aspects, but if a bubble is expanding at a rate of c in the bubble's frame, it should also expand at a rate of c in the ship frame. If it's a given size before you accelerate, I don't think it's possible to make it length-contract any smaller than that size by accelerating, if it's expanding at c.

    As a very rough look at this, suppose you have a particle moving away from you at c, and is "now" at location x, a billion light years away. Now say you accelerate so that the distance to x contracts to 1m. But due to relativity of simultaneity (think of the Andromeda paradox, or the twin paradox) the clock at x is now advanced a great time relative to your clock (almost a billion years), and the particle is not at x "now" but has long ago moved past it.

    Anyway there are a lot of interesting details related to this, a puzzle to figure out, if I say more I'll probably get it wrong.

  19. 1 hour ago, Winterlong said:

    [...] This is different from saying that the ladder is "outside of space"

    [...] I  don't disagree, but it is commonly accepted that space-time is kind of... fabric?

    It doesn't mean cloth, just like if someone talks about the makeup of spacetime, they don't mean it's made of cosmetics.

    In SR there's no single universal frame of reference that space is "in". The ship frame and the huge bubble frame are equal, there's no such thing as "one frame contains spacetime and the other doesn't". You might wrap your head around that by taking your example and having half your universe moving relative to the other half and vice-versa. Which is the universe? Or consider a universe made only of two identical ships moving relative to each other. Each, in its own frame, is at rest, not moving through space.

    SR has no problem with an object bigger than another in one frame, being completely inside it in another. Also, just like you can't contract a bubble to make the back of the ship stick out, you also can't use length-contraction alone to make it stick out the front. Either the end of the ship moves through the edge of the bubble (an event that happens in all frames, just with different timing), or the ship contracts along with the bubble. The seemingly paradoxical aspects of SR are resolved in SR, and what you're left with is a question that amounts to "What happens if the universe has an edge and something moves past that edge, where is it?" SR doesn't answer that.

    Thinking of it like the "fabric" has a rest frame and contracts to a finite size in another frame, is like supposing the universe is a finite bubble of Ether, with a rest frame, and then something leaves that bubble. SR doesn't imply that at all, but it also has no problem with that nor with an object leaving that bubble. In SR the frames of reference aren't finitely sized.

  20. On 8/9/2020 at 12:47 PM, Winterlong said:

    We apply Lorentz-contraction and, for whatever length is considered, it is just a matter of to speed up close enough to c to contract it to 1 m. I imagine this contraction in the direction of the movement, like two walls closing on the ship from the front and rear.

    The "closing in" is a change of the length contraction factor, which only happens during acceleration. But combining this idea with this:

    17 hours ago, Markus Hanke said:

    There's another issue here as well - if the observable universe is somehow contracted to 1m, while at the same time having the topology of a closed manifold, then the pilot will not measure his ship to be 10m long any longer.

    As an example, suppose the ship has its back against a brick wall, and then the entire ship accelerates away from the wall in some coordinated way (it can't keep accelerating "simultaneously" according to the pilot because the ship never shares a single inertial frame while accelerating), it doesn't matter for this example how. If the pilot accelerates fast enough, SR says that the wall can contract toward the pilot faster than the pilot moves away from the wall. But the wall can never get closer than the back of the rocket is, to the pilot. (The wall won't length-contract through other stuff.) With that high acceleration, the back of the ship must also length-contract toward the pilot. This demonstrates that there must be an acceleration limit to Born rigidity. It also means the back of the rocket will never stick out the back of the contracting region it was in when it accelerated forward.

  21. 7 hours ago, Winterlong said:

    I don't see why we cannot use the SR and the SR alone, as their premisses are fulfilled (hence, not considering acceleration)

    Sure, but it's best not to refer to it as "the universe" because 1) you're intentionally avoiding real properties of the universe that require GR, and 2) you're unintentionally treating your simple model as if it should be like the real universe.

    So let's call it a "bubble", say a spheroid region one billion light years in diameter, with a boundary and some stuff in it at relative rest, in an inertial frame. Then it has an inertial ship passing through it from one edge through the middle to the other edge. The ship has a proper length of 10m. The ship is traveling fast enough that the bubble is length-contracted to 1m in its own frame.

     

    Inside the bubble, the ship is length-contracted so it is flattened in the direction of its travel to less than the diameter of a quark. It takes just over a billion years for it to pass through the bubble, during which it ages only about 3.3 nanoseconds.

    In the rest frame of the ship, the bubble is a disk that is 1 m thick and has a diameter of one billion light years, which passes (thick-wise) through the ship at near the speed of light. It takes about the time it takes light to travel 1 m (about 3.3 nanoseconds) for any point on the ship to enter and then exit the bubble. That's one way to know that a point on the ship only ages 3.3 ns in the bubble's frame.

    [You probably don't want to know, but it's interesting that it take about 36.7 ns for the bubble to move 11 m and pass completely through the ship, and yet it only ages 3.3 ns according to an observer in the bubble. This is because of relativity of simultaneity. Clocks on the front and end of the ship that are in sync in the ship's frame, are out of sync by about 33.4 ns in the bubble's frame. In the bubble, if a clock at the front of the ship reads 0 on entry, a clock at the rear simultaneously reads 33.4 ns on entry, and then 36.7 ns on exit. The ladder paradox explains how the observer in the bubble can say the entire ship is inside the bubble, while an observer on the ship disagrees.]

     

    Okay so back to your "paradox". In the ship's frame, only the bubble is length-contracted. When you say "the entire universe" is length-contracted, you might be imagining all of space ie. all of the measurements of space around the ship, are contracted too, but the ship's inertial frame's measurements, or its space, doesn't get contracted. Where is the ship? It's in its own inertial frame, with all of its rest clocks and rulers behaving completely normally. If you suppose there's nothing else at rest in that frame except the ship, then all of the real clocks and rulers are on the ship, but the spacetime surrounding it would still be measured as normal, according to SR. Since it's only the bubble that's moving, that's all that gets length-contracted. Spacetime as measured in the ship's inertial frame, is not moving, and is not contracted. The spacetime isn't "stuff", I think it's nothing more than the measurements.

  22. 1 hour ago, md65536 said:

    I think a pilot in the middle of the ship could consistently conclude, "I'm in the middle of the "universe" which is 1m wide and is smaller than my ship would be at rest (which it currently is not, it doesn't share a single inertial frame), but the back of my ship has already passed through the front edge of the universe has not yet reached the same speed as me and the universe is not yet contracted for them."??? That's confusing, I doubt I got it right. However, relativity of simultaneity does resolve this part of the paradox if you do it right.

    I got this wrong. There is a paradox if the ship starts in the middle of the simplified universe and accelerates, ending up sticking out both ends. It's resolved by Born rigidity https://en.wikipedia.org/wiki/Born_rigidity

    Quote

    Already Born (1909) pointed out that a rigid body in translational motion has a maximal spatial extension depending on its acceleration, given by the relation b < c^2 / R, where b is the proper acceleration and R is the radius of a sphere in which the body is located, thus the higher the proper acceleration, the smaller the maximal extension of the rigid body.

    If the ship started "inside" the universe and accelerated quickly enough that its 10m rest length would stick out both ends, that ship could not maintain a 10m length during that acceleration, and it would not stick out both ends. If the ship starts outside the simplified universe and is already inertial and 10m before the back of the ship enters the 1 m contracted universe, it could stick out both ends of the universe.

  23. 2 hours ago, Winterlong said:

    I see a paradox, and I don't see how the simultaneity can solve it. For the sake of simplicity, the pilot does studiot's calculations (assuming that he knows the true length of the universe) and concludes that the length of the ship is bigger than the length of the universe, without even opening the windows of the ship to look outside.

    There's no paradox. You're talking about a finite spacetime that seems to be flat. Or basically an object with a proper length of over ten billion light years. You're calling that "the universe", which is fine as long as you avoid attaching meaning to that label, like that everything else you specify must be within that.

    Not everything in the (flat, toy) universe gets length contracted. Only the stuff that is moving relative to you does. The pilot isn't moving relative to the ship. If the ship is part of the universe, the universe won't be contracted to 1m, only the stuff moving relative to it. If the ship's not part of the universe, the universe can be like a 1m-thick wall traveling past the ship at near c.

    If you want to talk about the ship being at rest inside the flat universe, and then accelerating "instantly" to near c, then simultaneity is important if the universe is not static. It sounds like the ship is implicitly Born rigid, and clocks on parts of it would become out of sync with each other (by billions of years??). I think you would see the far edge of the approaching 'wall' appearing to age over twenty billion years in the nano seconds it takes to pass you, due to the relativity of simultaneity and relativistic Doppler shift.

    Actually, that idea's more complicated than I thought. Say the ship starts in the middle of a toy universe, and instantly accelerates to near c. Ignoring simultaneity, you might conclude that the universe contracts to a wall in the middle of the ship, with the ship sticking out both ends. But that's impossible because the back of the ship never travels backward. No part of the ship ever enters the "back" half of the universe. But with relativity of simultaneity, the different parts of the (Born rigid) ship travel through the front half of universe at different times... eek is that right? I think a pilot in the middle of the ship could consistently conclude, "I'm in the middle of the "universe" which is 1m wide and is smaller than my ship would be at rest (which it currently is not, it doesn't share a single inertial frame), but the back of my ship has already passed through the front edge of the universe has not yet reached the same speed as me and the universe is not yet contracted for them."??? That's confusing, I doubt I got it right. However, relativity of simultaneity does resolve this part of the paradox if you do it right.

     

    22 hours ago, studiot said:

    Winterlong, the fly in your ointment argument is that the pilot does not see a contracted universe, that is the view of some observer travelling relative to him.

    As the pilot masures things, not only does the ship have its normal 10m length, but the universe has its 'normal' length, whatever that is (we don't have a good number for this).

    It sounds like you're referring to the spacetime as 'the universe' and others are referring to all the moving stuff in it as the universe? If all the stuff was moving, the pilot would measure it as length-contracted.

  24. On 8/3/2020 at 5:21 PM, Janus said:

    This leads our bike rider to conclude that while he decelerates to a stop and then accelerates back up to speed back towards the source, the source clock runs very fast and ticks off the time between  12:34-12:35 and 13:44, over a very short period by the bike rider's clock.

     

    5 minutes ago, Janus said:

    The source clock runs slower (not backwards), if you are accelerating away from it( by your measure).

    These two statements aren't consistent with each other.

    If the bike rider accelerates, ie. changes inertial frame, there is a shift in relative simultaneity with respect to the distant source clock. If the rider accelerates towards it, the coordinate time of the distant clock can change from 12:34 to 13:44 in a small local time, as in your example. But if the bike rider then accelerates away and returns to its former reference frame, the coordinate time can change from 13:44 to 12:34, or even earlier if it accelerates away more. If the first change in relative simultaneity is "the source clock runs very fast", how is the second not running backwards?

    It's because of confusion like this that I don't like the phrasing that a change in relative simultaneity means a clock is running fast, because it also implies a clock can run backwards. They're both just changes in relative simultaneity, but one is accepted as intuitively reasonable but the other isn't, meaning it isn't an intuitive way to describe it.

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