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md65536

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Posts posted by md65536

  1. 3 hours ago, jamesfairclear said:

    One could envisage an experiment whereby a light source is set in motion at a constant speed S and then illuminated at a distance D from a relatively stationary detector. A clock at the detector measures the time it takes the Doppler redshifted light to arrive from distance D in order to establish its speed.

    That doesn't work unless you know the time at the light source, and you can't measure that from the detector. You theoretically can't measure the one-way speed of light, but you don't have to measure it since it is by definition equal to c. You can measure the 2-way timing of light, and find the one-way speed because literally by definition, the time that it takes for the light to go 1 way is the same as the time it takes to go the other way.

    You can however confirm that light from a distance D takes the same time regardless of the motion of the light source. For example, if you have 2 sources moving in opposite directions, and a signal from each of them when they're at the same location, you can verify that the 2 signals arrive at the same time. You don't have to know what time they're sent at, if all you care about is that they were sent at the same time, and you can make sure that happens by sending them from the same location.

  2. 35 minutes ago, 34student said:

    Since the particles are 1 meter apart and they travel along world lines through time, then doesn't that imply that their world lines are also 1 meter apart?

    Your line of questions shows you're not going to understand the answers. Why don't you start with simpler concepts first and understand them before talking about world lines?

    The world line of a particle is made up of all the events in the particle's entire lifetime. You're asking for the distance between two arbitrary lines. I think what you're really asking for is the distance between particular pairs of events on those world lines, but you're not specifying that unambiguously.

    If you're using coordinates where Bob's world line remains fixed at one x,y,z location and only varies in T, then (based on the situation you've implied), the distance between a given event at time T1 on the world line of the front of the train, and an event at the same time T1 on the world line of the back of the train, will be 1 m apart. If you use other coordinates, you'll get other answers.

    If you're talking about measurements in the block universe, you should specify what coordinates you're using. If you're talking about observations, you should make it clear what frame of reference you're using. If you're measuring the distance between 2 objects over their entire lifetimes (ie. world line) you should make it clear what time you're talking about. And if you're talking about a single moment across a distance, to even make sense of that requires the set of coordinates or frame of reference! If you want specific answers, your questions have to make sense, and they haven't been.

     

  3. 25 minutes ago, 34student said:

    So if there are 10 cars to the train, are you saying that there actually are 20 cars?

    Could you please stop taking what is said and twisting it to fit your existing misunderstanding? It's been repeated ad nauseam that that there's only one train.

    2 hours ago, 34student said:

    Now here is my question that will either put my issue to rest for me or keep me wondering.  If the particle is at x1 for Bob, and x2 for people on Earth, does the particle necessarily have to go through x1 and x2 (not necessarily in this sequence)?

    No, because those are 2 different coordinate systems. For example, the front of the train can be at rest at x2=0 for the coordinate system of an observer at the front of the train, while Bob might use a coordinate system where his ship is at the origin and the location of the front of the train is x1=several light years and changing. If the front of the train enters a tunnel, that's an event. Everyone agrees that the location of the front of the train coincides with the location of the entrance of the tunnel when it enters, but that could be at x2=0 and x1=several light years. Though it might make more sense with all the other basic stuff about relativity that I'm sure you've read, if you called them x and x' to denote that they're different coordinate systems.

  4. 25 minutes ago, 34student said:

    That is what I thought, and this is why I am so confused.

    Going back to the example in my OP, let's think about a particle sitting at one end of the train.  The train lies on the x axis.  The particle has one "String"/world line for Bob and another for Earth observers, (and possible an infinite more creating a plane).

    Which world line is real, or do they both exist?  If both particle locations exist, why can't one of the two observers detect both particles?    

    Did none of the rest of what I wrote make sense?

    The particle has only one world line. The world line is the events (the x and t locations) that the particle passes through over its lifetime. The world line is a fixed set of events, but the x and t values of those events change if you rotate it into other coordinate systems representing other observers.

    For example, put a stick on a grid with an x and t axis. If you align it with the t axis, it has the same x value at different t, representing the world line of a particle at rest. If you rotate the grid, the stick stays the same but now it has different x at different t, representing a different observer, for which the particle is moving.

    This is an oversimplified analogy and it uses the wrong rotation, so don't draw too many conclusions from it. If you rotate the stick or grid, the stick's x and t components change, but the length of the stick stays the same. It's the hypotenuse in the Pythagorean theorem r^2 = x^2 + t^2. With the correct hyperbolic rotation, s^2 = x^2 - t^2 stays the same.

  5. 11 hours ago, 34student said:

    I always thought that the past was one 4 dimensional static block, literally a block of static particles that would look like like strings because of the dimension of time.

    Sure, that would be a block of events, and the "strings" are world lines of particles.

    Imagine you have a block of say wood, and you draw a small line on it to represent a world line. You can rotate the block in your hands, but the shape and length of the line remains invariant. Say that you rotate it through a fixed grid of coordinates in a room. Using just 2 dimensions for simplicity, you can make the line align with the x-axis, or the y-axis, or anything in between. By rotating it, you can change how long the line is in the x-dimension and how long it is in the y-dimension, without changing its length.

    But the block universe also has a time dimension, so instead of x and y, imagine it's t and x. The line could be a function x(t) representing how far x a particle moves in time t in a particular set of coordinates, and you get different x-lengths in different coordinates just by rotating the block.

     

    In this analogy the rotation is a spherical rotation which keeps r^2 = x^2 + y^2 invariant as you rotate it. With spacetime, the rotation is hyperbolic, which instead keeps s^2 = x^2 - t^2 invariant as you rotate it. So you can't just turn it upside down and reverse time etc.

  6. 1 hour ago, MigL said:

    The 'Block' Universe model is essentially devoid of time, as past, present and future are all present in the 'Block'. [...]

    The worldline of an event does not move; it is already there.

    Do you mean worldline of an object? An event is a single point in spacetime (and in the "block") and many worldlines can pass through it.

    Worldlines still exist in the block, and one's proper time is an invariant length of the world line. Different coordinate times can be defined (including just using T to represent time) and the coordinate times between distant events can be calculated in the block. I don't understand what you mean by it being "devoid of time", since all the measures of time are still there. Some philosophical "flow or time" or whatever might be taken out, but time in SR and GR is the measurement, not the concept.

    6 minutes ago, 34student said:

    Bob and Earth begin with the same t value.  When Bob returns, his t value is the same as Earth's t value when t was 2051.  For some of that year, according to Bob's clock, the train was 1 meter.  But from 2050 to 2051 on Earth, the train was never 1 meter.    

    Where's the problem? You're using coordinate time t to be the time in Earth's frame. You're stating that the length of the train is relative, and depends on the observer (ie. reference frame). That agrees with observation. That agrees with relativity. Nothing there implies illusion.

  7. 4 hours ago, 34student said:

    For every t (time) there is a corresponding function S(t) (static universe).  That is all there is in a block universe.  If you agree, continue reading.

    For some t's between 2050 and 2100 Bob sees the train as 1 meter.  But from Earth, people see the train as 100 meters for all t's.

    There cannot be two S(t)'s for the same t in a block universe in this example.

    Then S(t) is a hypersurface of events representing a single moment?

    How do you determine t? It sounds like you're implying it uses dates according to a clock on Earth? What if you have a block universe with all of the same events, but with a t value that corresponds with Bob's clock?

  8. On 10/26/2021 at 10:59 AM, MigL said:

    In a 'block' universe ( 3D+T ) nothing is moving.
    Events trace out worldlines within the 'block'.

    So why would you even expect to see length contraction in that model.

     

    Don't you have to choose your coordinates for such a block universe? Then you can measure the distance between points with the same T value.

    For example, choose coordinates for the block universe where T corresponds with a clock that is stationary relative to the train. Take a pair of events with the same T value, one at the front of the train and one at the back. You can find the proper length of this train, say 100 m.

    Or choose coordinates where T' corresponds with Bob's proper time and find 2 different events with the same T' and find that the distance between those 2 events is 1 m.

    The aliens would also have to choose a set of coordinates, and could measure the length contraction of the train relative to its proper length.

  9. 38 minutes ago, 34student said:

    Will they see a 100 meter train or a 1 meter train or something else?

    This is meant to support the argument that length contraction is an illusion and not actually happening.

    In your aliens' view of the universe, is Bob's ship moving? Is the train moving? If so, how fast? Does this question support the argument that movement is an illusion and not actually happening?

  10. No, you don't need two. You can have the two, but there's nothing that stops working or making sense with just the frame-dependent definition of simultaneity.

    For order of events, causality is enough, and that doesn't depend on frames of reference or simultaneity. There is no "god's eye view" needed. You can add it in, but it doesn't explain any observations that can't be explained without it. Therefore it's likely more misleading than helpful for explaining. It would be like saying "To understand life and death, we have to first understand ghosts." Nothing requires that ghosts aren't real, but everything observed can be explained without ghosts.

    For order of events, if two distant events are simultaneous, they're not causally linked (one doesn't cause the other). Their order is frame-dependent, and there's no problem with that because their order has no bearing on causality and on what other events are effected, and thus no bearing on what can be observed (since observations can be treated as a set of events). If the god's eye view is necessary, it's for something other than what's been observed or what's predicted to be, ie. metaphysical. The "universal now" is simply not needed for the universe to function.

  11. I bet reddit.com/r/Whatisthis/ or reddit.com/r/whatisthisthing/ would figure it out within an hour.

    I figured it might be part of a tool, like a plane depth adjust, or a proportional divider knob.

    The closest I've seen is on a horological tool, like in the top left:

    p1130017.jpg

    The knob slides and tightens to lock. A string might fit in the groove, and the part is used to set the tension on it. This one looks like it spins freely though.

  12. 4 hours ago, geordief said:

    Are you saying that ,far from any mass it is possible for there to be an object with the independent means to rotate about itself and to create an artificial gravity?

    I wouldn't say that exactly. I'd also call it "inertia" instead of artificial gravity. But objects do tend to have some rotation, and inertia does contribute to forces acting on things (eg. the Earth bulges around the equator, along with a feeling of less pull of gravity than at the poles, at a fixed distance from the centre of the Earth), and for a static object that inertia does not require energy to maintain.

    My point is that if you choose a force that requires constant use of energy for one observer, and another force that doesn't for another observer, that's not a difference that concerns the equivalence principle. The principle doesn't say that any 2 different things you choose should be the same, and if there's a detectable difference (eg. in energy use) then the principle doesn't hold. Effectively it says if you make everything else the same, gravity is not detectably different from acceleration.

  13. On 10/9/2021 at 5:54 AM, mistermack said:

    Yes, but that's basically what I said in my first post. The more local you look, the more equivalent are the two phenomena. And the less local, then the less they are equivalent. So it's a useful approximation, only actually true for a time interval of zero. 

    In the case of my seventy years, if I was free floating in space, it would take an infinite amount of energy to maintain a force of 200lbs on me for seventy years. On earth it needs none at all.

    It wouldn't take infinite energy and you wouldn't reach a speed of c relative to anything. If you accelerated away from Earth with a constant proper acceleration for seventy years, the coordinate acceleration of Earth away from you would decrease and approach zero as its speed approached c, because of the velocity addition or composition law of SR.

    I don't think the difference in energy use is related to localness of effects, or whatever. You could hover over the Earth on a stationary platform that uses energy to create 200 lbs of thrust for 70 years. The equivalence principle still applies. You can't tell whether you're accelerating away or overcoming gravity. You could also simulate gravity by being in a rotating ship in freefall for 70 years, and use no energy, and experience no gravitational acceleration. The equivalence principle might not apply since that would be a rotating frame of reference.

  14. On 9/24/2021 at 6:09 PM, AbstractDreamer said:

    If you zoom OUT far enough, and assuming you zoom out at the same rate for the X and Y axes, then y=x^2 approaches a straight line.  Does this mean anything?

    Well, the curvature of the parabola decreases with increasing positive x. There's an analogy for what you're talking about. "Local" effectively means at small enough distances that the effects of distance don't matter, and that's not a fixed size. At small x, you have to zoom in to a smaller area to make the parabola appear flat, and at large x, it can appear flat over a larger range. Analogously, near to a gravitational mass "local" might be very small, but very big much farther away. If you're free-falling near a black hole and being spaghettified, spacetime is still locally Minkowskian but at distances much smaller than a human.

  15. On 9/16/2021 at 2:52 PM, studiot said:

    The curvature of the rubber sheet (manifold) is extrinsic.

    I think it also has intrinsic curvature in the usual setup.

     

    22 hours ago, Markus Hanke said:

    However, there are also scenarios where the effects of gravity are in some sense ‘relative’. Consider a hollow shell of matter, like a planet that has somehow been hollowed out (not very physical of course, but I’m just demonstrating a principle here). Birkhoffs Theorem tells us that spacetime everywhere in the interior cavity is perfectly flat, ie locally Minkowski. There’s no geodesic deviation inside the cavity. Now let’s place a clock into the cavity, and another reference clock very far way on the outside, so both clocks are locally in flat Minkowski spacetime. What happens? Even though both clocks are locally in flat spacetime (no gravity), the one inside the cavity is still gravitationally dilated with respect to the far way one! This is because while both local patches are flat, spacetime in between them is curved - if you were to draw an embedding diagram, you’d get a gravitational well with a ‘Mesa mountain’ at the bottom; and the flat top of that mountain sits at a lower level than the far away clock, thus the time dilation. So in this particular case one could reasonably say that gravitation effects are ‘relative’ between local patches. Or you can put it like this: both regions are Minkowski, but one is more Minkowski than the other :)

    Sure, but the gravitational time dilation should depend only on their relative gravitational potential (right?) and not how space between them curves to produce that difference in potential. For example if you have 2 locations at different potentials in a constant gravitational field, you have 2 locally Minkowskian regions and no curvature anywhere, but you still get gravitational time dilation.

    Yours is an interesting example because the regions are intuitively flat. I guess the Riemann tensor is zero at those locations? But for example a region just outside the hollow shell is also locally Minkowskian in the coordinates of a particle in free fall, and locally measurably "flat" to such a particle (but spacetime is not actually flat there and the Riemann tensor isn't zero).

    I think I'm too stuck on confusion about the meaning of "locally Minkowskian," and trying to visualize it as horizontal on a diagram showing local coordinates, but then don't know how distant curved space could be shown.

    Now I remember you mentioned the covariance of tensors to me before, and I looked up the components of the Riemann tensor and couldn't make sense of it (same as now!). I think I need to learn more basics before understanding curvature in different observers' coordinates.

  16. 12 hours ago, Markus Hanke said:

    There is no detectable information about this at any one point on the Earth’s surface. This is because the geometry concerns relationships between points, so what you do is take measurements of path lengths, areas, or angles. For example, you’ll find that the sum of the angles in a triangle on Earth’s surface is no longer exactly 180 degrees - it’s possible to directly measure this deviation. But you can’t do it at a single point, you need to measure across some distance. That’s because the effects of a non-flat metric are accumulative - mathematically, you integrate components of the metric to obtain path lengths.

    Tying this in with the rubber sheet analogy: The height on the sheet or on Earth corresponds (assuming constant g) with gravitational potential, proportional to how much energy it takes to lift an object to a specific height. The derivative of that, the slope of the sheet or ground, corresponds with gravitational force---how quickly a marble would accelerate as it rolls along that point. The derivative of that represents curvature. If we imagine the rubber sheet to be semi-rigid, and you actually have to bend it into a curved shape, the severity of the bending at a given point corresponds with curvature. You can tilt a sheet, and that corresponds with a constant gravitational force but no curvature.

    (Just to complete the analogy, curvature corresponds with tidal forces. If you have a toy car where the front and back are separate and connected by a spring, and roll it down a hill with constant slope, the spring doesn't stretch or compress. If you put it on a curve, eg. on top of a ball, the front and back are on different slopes and can pull away from each other.)

    The "local flatness of spacetime" implies that if you're looking only at a single point, you can't detect curvature, but you also can't detect gravitational force (is that right???). So for the Earth analogy, using it only to show what you can detect locally, and not how it affects the motion of objects, we could take the real Earth and modify it so that "up" is always normal to the surface you're standing on. Like in some cartoons where if you're standing on the side of a mountain, your body is tilted so your feet remain "flat" against the slope. Then if you can only look at the ground beneath your feet, you can't tell if you're on a slope or not.

    Is that a fair analogy to spacetime, or is it only curvature that is locally undetectable, but not gravitational force?

     

     

    Also I have a feeling I've asked a similar thing but still don't get it: Is the curvature relative, so it actually is locally zero but a different value from a distant? Or does local flatness merely mean, like you suggest, that the local value of the curvature only has measurable meaning across some distance? I think it's the latter??? Can curvature be called a scalar field, and is it invariant in a static universe?

  17. On 9/12/2021 at 12:32 PM, studiot said:

    Here is a better one, soor no pretty dagram at the moment.

    Say you are a road runner that can only travel along the road.

    Now say you are standing at 5 West Street on a grid pattern of roads where the sides are completely built up with buildings.

    And say you want to get to 5 North street.

    Well you can't cut through the corner of the buildings, you have to go first along West street to the intersection of North Street and West Street.
    And then you have to go up North Street until you reach number 5.

    These are the 'rules' of this grid pattern of points.

    Furthermore this route is the shortest possible route for a road runner.

    This is an example in 2D, that works without invoking the 3rd dimension at all.

     

    Now suppose we scale this up to 3D.
    Again we have an arrangement of point, just now in 3D.
    And we have rules either of travel between these points or equivalently the way these points are laid out.
    This time there is no need to invoke a 4th dimension (ignoring time for this).

     

    The (mathematical) rules are pretty complicated, but that basically how General Relativity works.

    Can you please explain the statement that I bolded? You implied that I should back up assertions with detail, and when I looked closer at your analogy that you claim is a "better one", I find that it makes no sense at all. Can you explain what this is meant to show about GR, especially related to the topic of how curvature in space-time is shown?

    2 hours ago, studiot said:

    No I don't think you did answer either my questions, nor those of swansont.

    I don't see any questions asked by swansont. I agree with all the points he made.

  18. 9 hours ago, studiot said:

    Everyone is entitled to thir opinion, including those that can't or won't back up their assertions with detail.

    I already justified the opinion, your model doesn't show the paths of objects bending in a curved space, and the rubber sheet analogy does. But I don't see what your analogy is even trying to say. What I get from it is you're saying that objects can only travel along gridlines through space?

    On 9/12/2021 at 12:32 PM, studiot said:

    Well you can't cut through the corner of the buildings, you have to go first along West street to the intersection of North Street and West Street.

    And then you have to go up North Street until you reach number 5.

    These are the 'rules' of this grid pattern of points.

    Are the streets representative of dimensions? Are they both spatial, or is one meant to be time? (I guess spatial, since you said "ignoring time", but then I don't understand MigL's comment that it "considers the space-time interval".) You can only travel in one dimension at a time, one "first" and then another? And that should give someone an idea of how gravity works?

  19. 2 hours ago, studiot said:

    These misunderstandings show why I dislike this analogy. [...]

    1) It is meant to represent the curvature of a space (just not real space), and the paths of objects through that space. What I meant was that I don't think the sheet's curvature is intended to actually model real spacetime curvature. I don't think they curve in the same way. But that's fine, it is an analogy of real spacetime curvature, not a model of it. My problem is that it is not clear how closely it is analogous, or even what properties exactly it is representing. Usually with a physical analogy you can clearly see the differences between the analogy and the real thing, and you don't confuse them. Here, the "fabric of spacetime" is such an elusive abstract thing that people see the sheet as a model of that "fabric".

    2) I don't understand. It doesn't represent the real path of an object through spacetime (see (1)), yet a marble initially at rest on the trampoline does follow the line to the COG of the mass.

    4) Why can't a 1D line curve? Can't you map one 1D space onto another different 1D space?

    5) Again it's an analogy. You can't practically show the curvature of a sheet due to mass alone, the thing that causes the curvature in the sheet is only an analogy to the thing that causes curvature in spacetime.

  20. The mass on a trampoline shows how paths of objects bend in a curved space. studiot, I don't see how your model shows that. The curved sheet doesn't technically need gravity to show this; distort the sheet some other way, and run a "straight" line of tape over the curve and the path will bend (analogous to a null geodesic).

    Yes the analogy has problems. The mass represents mass, but the curvature is not representative of spacetime curvature, which I think is 0 at the center of the mass?* Showing the Earth resting on the sheet incorrectly suggests that it's the volume of matter displaces spacetime. I think this fails Einstein's "as simple as possible, but no simpler" criterion. Instead of saying "fabric", it could be called a manifold made up of events---would that stop people from asking what it's made of? Maybe the rest of the analogy could be fixed by labeling things similarly abstractly, instead of using concrete things like an Earth. But I don't know how you'd label it because I don't know what the curvature of the trampoline is actually meant to represent. Is it gravitational potential? Or is it just a toy example of an abstract curved space? I think the trampoline model could be set up and described differently, "no simpler than possible", so that it would both be clearer what it's meant to show, and not suggest other things. At the very least, I feel it should make people think something like "curved spacetime bends the paths of objects" and not "gravity pulls on the fabric of space" or whatever.

    Speaking of space vs. spacetime, the inclusion of time in the curvature is what makes masses at rest gravitate toward each other (is that right? along with constancy of 4-velocity magnitude?) but I don't see how that could be represented on a curved sheet.

     

    * Edit: now I'm confused because the curvature of the trampoline is also zero under the center of the mass, so maybe it does fairly represent curvature?

  21. 17 hours ago, Dagl1 said:

    if they aren't at rest (both moving along line A-B at the same speed so that the distance between them does not change),

    Maybe this was already explained better but I didn't see it. An inertial object is only moving relative to something else (ie. an observer). If they're inertial and their distance isn't changing, then they're relatively at rest. If you then say that they're moving at the same speed but still at rest relative to each other, then that's relative to another frame, which you haven't mentioned.

    In non-relativistic physics, you might assume you're talking about some universal frame, but probably no one else here would do that; if you only mention frames A and B and then a speed, I think everyone would assume you mean A's speed relative to B and vice versa. You would have to specify a third frame ("Earth frame" for example) for people to get what you mean.

     

    Also, an observer is a frame of reference in SR. https://en.wikipedia.org/wiki/Observer_(special_relativity) This is because you observe ie. measure the same distances and times no matter where you are in a given inertial frame, so you don't have to distinguish different viewpoints in the same frame as different observers.

     

    17 hours ago, Dagl1 said:

    If I move towards a star at 80% the speed of light, and I measure the distance to be 6LY, then if I moved very slowly towards it it would have been 10LY? And measured from the star, I too am 6 LY away when moving at at 0.8c, right?

    The first part sounds right........ if the star is moving towards you at 80% of the speed of light and is 6 LY away, then you accelerate instantly so it is at rest, it should now be 10 LY away (not "would have". It still is 6 LY as measured in your first frame, and is 10 LY in your second frame).

    The second sentence is kinda wrong and this is where it gets fun! You could ignore this until you get the rest of the replies in the thread. Because of relativity of simultaneity, you and the star don't measure the distance between you as 6 LY at the same time. For example if the star is 6 LY away, approaching at 0.8c, it will take 6LY/.8c = 7.5 years for it to arrive. But as observed by the star, your clock is ticking at a rate of 0.6x its own. So while you measure 7.5 years to reach the star (from the event where you measure the star being 6 LY away), the star measures 12.5 years in its own frame, during which you travel 12.5 y * 0.8c = 10 LY. The star measures you as 10 LY away when (according to you!) you measure the star being 6 LY away.

    You also measure the star's clock ticking at .6x your own. You and the star are symmetric: each of you measure the other as 6 LY away at the moment (in your own frame) that the other measures being 10 LY away.

    Or another way to see it is: at the moment (according to you) that the star passes the 6 LY mark on your rulers (which are at rest relative to you), the star's rulers are length-contracted by a factor of 0.6, and you are passing the 10 LY mark on its ruler. At the event where you say "we're 6 LY away", the star measures you at the mark that's 10 LY away.

    ... But then, you might also see, if you're at the 10 LY mark on the star's ruler and it's your ruler that's length-contracted according to the star, then when you're at the star's 10 LY mark, the star is at a 16.666 (repeating of course) LY mark on your ruler! According to you, the event of you passing the star's 10 LY mark is simultaneous with the event of the star passing your ruler's 0.6 LY mark. According to the star, the event of you passing its 10 LY mark, is simultaneous with it passing your 16.666 LY mark. This is no problem because the relative simultaneity of distant events is different for different frames of reference.

    There are a lot of ways to describe this, I edited it to try to simplify, others probably have clearer and simpler ways to say it.

  22. I'm assuming flat spacetime (no mass, SR only), and inertial motion unless specified.

    12 hours ago, Dagl1 said:

    If I have a map with the 'position' of different stars around the galaxy, or different galaxies in the observable universe, and we assume these bodies are moving at high speeds relative to each other, does the distance between two galaxies depend on the specific galaxy I measure these from?

    Generally yes. Also "when are you talking about?" matters and is more complicated than a Newtonian description. The positions of things on a map are coordinates within a coordinate system, and those are different for different observers. You could have a map where the Milky Way is at a fixed location and Andromeda is moving, or one where Andromeda is fixed and the Milky Way is moving. Those correspond to the coordinate systems of 2 observers at rest in the respective galaxies.

    12 hours ago, Dagl1 said:

    If so then is there any 'absolute' distance between any points? Does it make sense to say that a star is x light years away, if by speeding up or slowing down, that distance is changed.

    Yes, the distance to the star is different in the different frames of reference. There are invariant measures of distance, eg. the "proper length" of a 1 m stick at rest is always 1 m and everyone will agree on that, even if the stick is moving relative to some observer and is length-contracted ie. has a coordinate length less than a meter in that observer's coordinates.

    12 hours ago, Dagl1 said:

    I suppose the reason is that because galaxy B is in motion, we cannot say it is twice as far away from us, as we can only measure things from one frame of reference.

    We can say that. We can measure the distances to both galaxies using one frame of reference (eg. the one in which we're at rest), and you can measure the motion of objects using the coordinates of that frame.

    Consider the map analogy. The spatial coordinates can be represented by a grid drawn on the map. The same grid coordinates can be shown by putting a lattice of rulers throughout space. In our own frame of reference, our rulers are not moving and so they don't length-contract. An object light years away can wobble at speeds near c, but yet stay near one place in the grid of rulers.

    Meanwhile, that distant wobbling object is moving relative to our lattice of rulers, and our rulers do length-contract, in its frame of reference. For simplicity consider two different inertial frames of reference F1 and F2 that the wobbling object switches between. Each of those frames has its own set of rulers making up a lattice throughout space, each at rest and not length-contracted in its own rest frame. Say I'm at 1 LY from Earth, as measured by Earth, and I'm wobbling relative to Earth. I stay near the 1 LY mark on Earth's set of rulers, but those rulers are contracted by different amounts in F1 vs F2. For example, in F1 Earth might be 0.8 LY away from me and the 1 LY mark, and only 0.6 LY away in F2's frame. The reason that the distance as measured by Earth isn't changing much, and the distance measured by me is changing drastically, is that I'm switching between different frames of reference. The distance between Earth and the 1 LY mark, which has a proper distance of 1 LY, is length-contracted by different amounts to different observers, depending on their relative speed.

  23. 51 minutes ago, Ant Death said:

    Duration is as recorded by clocks and unique to each observer dependant upon change of position (distance) within the universe irregardless of the rate of change (velocity).

    [...]

    An inertial motion frame has a distance value greater than zero so duration values are less than time values.

    [...]

    Inertial light frame velocity = 299792458m/0s = infinite metres / zero seconds

    You're making up definitions but it sounds like you're using "duration" to describe proper time.

    You're not talking about velocity here. Velocity is a measure of distance/time as measured by a single observer (aka inertial frame). There is a measure of rate of motion called proper velocity or celerity that instead of measuring time using the observer's clock, it uses proper time as measured by the moving clock. Celerity approaches infinity as velocity approaches c. It's not a measure of velocity because you're measuring distance in one frame and time in another.

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