geordief Posted November 9, 2022 Share Posted November 9, 2022 If any point in the spacetime model is specified wrt any reference point is it inevitably the case that that point can only ever be an approximation to any physical activity that actually takes place there? If so is this because of theories like the Uncertainty Principle or does it simply follow from the spacetime model itself, because it models both position and time ,as well as (I imagine) that at the most detailed level that all things are in relative motion no matter how we try to set up any scenario that might illustrate processes at rest to each other?** **Not completely sure if that is completely accurate and I think Studiot recently disagreed with Joigus as to whether "Panta Rhei" was as fundamental a proposition as I have always taken it to be. Link to comment Share on other sites More sharing options...

exchemist Posted November 9, 2022 Share Posted November 9, 2022 2 hours ago, geordief said: If any point in the spacetime model is specified wrt any reference point is it inevitably the case that that point can only ever be an approximation to any physical activity that actually takes place there? If so is this because of theories like the Uncertainty Principle or does it simply follow from the spacetime model itself, because it models both position and time ,as well as (I imagine) that at the most detailed level that all things are in relative motion no matter how we try to set up any scenario that might illustrate processes at rest to each other?** **Not completely sure if that is completely accurate and I think Studiot recently disagreed with Joigus as to whether "Panta Rhei" was as fundamental a proposition as I have always taken it to be. No. The uncertainty principle comes from QM, not relativity, and relates to non-commuting operators or, in the wave formalism, to certain pairs of properties being related to each other through Fourier transforms, whereby increasing precision in one leads to loss of precision in the other. QM is applied to physical systems - objects, if you like - rather than to spacetime. 1 Link to comment Share on other sites More sharing options...

studiot Posted November 9, 2022 Share Posted November 9, 2022 8 minutes ago, exchemist said: No. The uncertainty principle comes from QM, not relativity, and relates to non-commuting operators or, in the wave formalism, to certain pairs of properties being related to each other through Fourier transforms, whereby increasing precision in one leads to loss of precision in the other. QM is applied to physical systems - objects, if you like - rather than to spacetime. Very nicely put. +1 I would add that there are observable and measureble effects in real physical systems. as a result of of this precision tradeoff. Link to comment Share on other sites More sharing options...

geordief Posted November 10, 2022 Author Share Posted November 10, 2022 (edited) On 11/9/2022 at 6:20 PM, exchemist said: No. The uncertainty principle comes from QM, not relativity, and relates to non-commuting operators or, in the wave formalism, to certain pairs of properties being related to each other through Fourier transforms, whereby increasing precision in one leads to loss of precision in the other. QM is applied to physical systems - objects, if you like - rather than to spacetime. Is Minkowski spacetime not just basically (ignoring the c) plotting x against dx/dt? A bit like plotting a quantity against a property of itself Are there other examples where a function can be plotted against it's own derivative wrt time in a precise way? I don't have the physics or the maths to judge but it feels to me like there could be some kind of self reference in the two quantities being plotted against each other where they should be independent. Edit:apologies. I see I am quite wrong(confusion** set in ) about x being plotted against it's own derivative wrt time in the Minkowski diagram Feel free to disregard this contribution (can't see the "embarrassed" emoticon) **I must have confused c ,which is a speed with dx/dt which is also a speed but which is not being plotted on the Minkowski diagram) Edited November 11, 2022 by geordief Link to comment Share on other sites More sharing options...

Markus Hanke Posted November 11, 2022 Share Posted November 11, 2022 (edited) 7 hours ago, geordief said: Is Minkowski spacetime not just basically (ignoring the c) plotting x against dx/dt? No, but I think you have picked up on that yourself already. The crucial feature of Minkowski spacetime is found in how it defines the separation between points. You might remember from your school days the Pythagorean theorem - if you have a pair of points in some Euclidean space, the squared separation between them is the sum of squares of coordinate differences: \[(\Delta s)^{2} =( \Delta x)^{2} +( \Delta y)^{2} +( \Delta z)^{2}\] In Minkowski spacetime, you have one additional dimension, being time - so the squared separation will involve four coordinates. However, unlike in Euclidean space, Minkowski spacetime does not simply add them; instead it ensures that time and space have opposite signs in the separation formula, like so: \[(\Delta s)^{2} =( \Delta t)^{2} - ( \Delta x)^{2} -( \Delta y)^{2} -( \Delta z)^{2}\] So the squared difference isn’t just the sum of (squared) spatial separations, but the difference between (squared) separation of time and space: (total separation)^2 = (separation in time)^2 - (separation in space)^2 Note that the choice of signs is arbitrary - I could have made time negative and space positive, without affecting the result. This is an example of hyperbolic geometry (as opposed to Euclidean geometry). What does this do physically? Well, having a difference rather than a sum enables you to make simultaneous changes to the space part and the time part in equal but opposite measure, without affecting the overall separation in any way. So you can trade a decrease in space for an increase in time (or vice versa), and still end up with the same overall separation. And that’s exactly what happens in Special Relativity - for example, if you are looking at a clock passing you by at relativistic speeds, you’ll find that the clock is time-dilated (meaning it takes longer for the clock’s hands to move, from your point of reference), while at the same time the clock itself will be length-contracted in its direction of motion, so its size becomes shorter (again, from your point of reference). So in this scenario, and from your point of reference, “space is traded for time”, in a manner of speaking. This happens in equal but opposite measures - the decrease in size is by the same factor as is the increase in time - which is why the ratio between them remains the same always, which physically means that the speed of light is always the same in any inertial frame. This is is purely a consequence of the hyperbolic geometry of Minkowski spacetime. A simple change in signs makes all the difference! The above is very simplified and not especially rigorous, but hopefully you get the central idea. Edited November 11, 2022 by Markus Hanke 1 Link to comment Share on other sites More sharing options...

md65536 Posted December 7, 2022 Share Posted December 7, 2022 On 11/10/2022 at 11:53 PM, Markus Hanke said: What does this do physically? Well, having a difference rather than a sum enables you to make simultaneous changes to the space part and the time part in equal but opposite measure, without affecting the overall separation in any way. So you can trade a decrease in space for an increase in time (or vice versa), and still end up with the same overall separation. And that’s exactly what happens in Special Relativity - for example, if you are looking at a clock passing you by at relativistic speeds, you’ll find that the clock is time-dilated (meaning it takes longer for the clock’s hands to move, from your point of reference), while at the same time the clock itself will be length-contracted in its direction of motion, so its size becomes shorter (again, from your point of reference). So in this scenario, and from your point of reference, “space is traded for time”, in a manner of speaking. This happens in equal but opposite measures - the decrease in size is by the same factor as is the increase in time - which is why the ratio between them remains the same always, which physically means that the speed of light is always the same in any inertial frame. I'm trying to make sense of this and can't. I think you have it backwards. s^2 is the spacetime interval between 2 events. I don't think that equation can directly represent length-contraction of an object, because the length of an object in a reference frame is the spatial distance between 2 simultaneous (in that frame) events. The spacetime interval is the measure between the same 2 events in different frames, while length contraction compares 2 different sets of events in different frames. I don't think you can put 2 lengths of an object (ie. proper and contracted) into the equation and get the same s^2, or in other words I don't see how you can express "longer t and shorter x" with that equation. The interval being invariant implies that the space and time coordinates actually change in the same way, not opposite. That means that in one frame, if 2 events are a certain time apart and distance apart, in another frame where they're a longer time apart, they're also a greater distance apart. I always confused that idea with the way that length contraction seems to say the opposite of that, so I'll just go through an example to show how relativity of simultaneity clears up the confusion. Say you have a train passing through a tunnel so that in the train's frame, the train is exactly the length of the contracted tunnel. The events are the front of the train passing the exit of the tunnel, and the rear of the train passing the entrance of the tunnel, simultaneously. The spacetime interval has (delta)t=0 and x=the proper length of the train = contracted length of tunnel, so s^2 is negative, a spacelike interval. In the tunnel frame, the tunnel is longer than it is in the train frame, while the train is length-contracted and fits entirely inside the tunnel. s^2 is the same since it's invariant, with x'=proper length of tunnel > x, but now t' is also larger than t because the train's rear enters the tunnel before the front of the train leaves. If I were to confuse things, I might say "let x' be the contracted length of the train, and it is smaller while t' is larger", but those statements are talking about 2 different spacetime intervals. Link to comment Share on other sites More sharing options...

Markus Hanke Posted December 7, 2022 Share Posted December 7, 2022 1 hour ago, md65536 said: The interval being invariant implies that the space and time coordinates actually change in the same way, not opposite. Yes. By “opposite” I simply meant the sign; the squared interval is the difference between the squared time part and the squared space part. The crucial bit is that whatever change occurs, it can only be such that the overall interval remains invariant. Link to comment Share on other sites More sharing options...

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