geordief 114 Posted October 17, 2020 Share Posted October 17, 2020 (edited) Hopefully an extremely simple scenario. Let's have an observer A with another relatively stationary observer B at 100 light seconds distance A emits a pulse of light in the direction of B and that pulse is reflected and captured again by A at a distance of 1000 metres to his or her right (just in his or her vicinity, really) So we have 2 events ; the emission and the recapture of the signal. Am I right to say that the spatial measurement component of the spacetime interval between the 2 events is the 1000 metres and the time component ( ie the ct component) is the 200 light.seconds taken by the signal to make the round trip and return to the vicinity of its emission? If I am right so far ,can I ask : How does observer B calculate this spacetime interval ? (It must be the same as for A ,no?) Also can this same scenario be used whereby B is moving relative to A? Does it show that B still measures the spacetime interval the same as when he or she was stationary with respect to A? Edited October 17, 2020 by geordief Link to post Share on other sites

md65536 383 Posted October 17, 2020 Share Posted October 17, 2020 (edited) Conventionally in SR "observer" refers to a frame of reference. If you used that convention (not that you have to, just that it can be helpful to think in these terms), then A and B are the same observer when they're not moving relative to each other. They measure times and distances the same. Local measurements differ, like the relative timing of perceived light from distant events, that each can see (ie. locally measure) in different orders, but that doesn't matter in your example. A and B measure the same as each other, the time between the two events, and the distance between the two events. They measure an interval with a length of negligibly less than two seconds. A moving observer (another reference frame) would measure a generally different time between the two events, and a different distance between the two events, but end up with the exact same interval length. The interval you're describing is a timelike interval (meaning a clock could travel between the events, and record its length as a proper time). The time component that A measures is simply c multiplied by the time on A's clock measured between the events. But of course that's the distance that A measures light traveling in that time, so you're right that it is 200 light seconds. The reflection point being 100 light seconds away doesn't really matter either, for the interval you're describing. Basically you're measuring the time between the events using a very big light clock that ticks just once between the events. A smaller light clock that reflects a light signal multiple times, can measure the same thing. Edited October 17, 2020 by md65536 Link to post Share on other sites

geordief 114 Posted October 21, 2020 Author Share Posted October 21, 2020 On 10/17/2020 at 10:54 PM, md65536 said: Conventionally in SR "observer" refers to a frame of reference. If you used that convention (not that you have to, just that it can be helpful to think in these terms), then A and B are the same observer when they're not moving relative to each other. They measure times and distances the same. Local measurements differ, like the relative timing of perceived light from distant events, that each can see (ie. locally measure) in different orders, but that doesn't matter in your example. A and B measure the same as each other, the time between the two events, and the distance between the two events. They measure an interval with a length of negligibly less than two seconds. A moving observer (another reference frame) would measure a generally different time between the two events, and a different distance between the two events, but end up with the exact same interval length. The interval you're describing is a timelike interval (meaning a clock could travel between the events, and record its length as a proper time). The time component that A measures is simply c multiplied by the time on A's clock measured between the events. But of course that's the distance that A measures light traveling in that time, so you're right that it is 200 light seconds. The reflection point being 100 light seconds away doesn't really matter either, for the interval you're describing. Basically you're measuring the time between the events using a very big light clock that ticks just once between the events. A smaller light clock that reflects a light signal multiple times, can measure the same thing. Thanks for explaining the convention. So ,in my example the spacetime interval is as per s^2=[ct]^2 -r^2 and I make that to be [200ct]^2 -1000^2 ,which is measured in metres. [200ct]^2 and 1000^2 seem to me to represent the time and the spatial components of the spacetime interval.... Can I think of those 2 components as 2 spatial distances ,the former vastly greater than the latter and the ratio between the lengths of those 2 lines could represent the time /space ratio of the spacetime interval? Link to post Share on other sites

md65536 383 Posted October 21, 2020 Share Posted October 21, 2020 (edited) It's meters or lightseconds or any distance units, squared. The use of distance units is apparently a convention, see https://physics.stackexchange.com/questions/519707/is-the-unit-for-spacetime-intervals-time-or-space-distance Yes, for a time-like interval, the time component will be greater than the spatial. For time-like (or light-like for that matter) intervals, the ratio of r/t is the constant speed of a particle that moves between the two events. ct/r would be the ratio of the distance that light travels between the two events (along any path that gets it there, like your 200 lightsecond example) to the straight-line spatial distance between the two events, in the given frame. This ratio is frame-dependent, and undefined in frames where r=0. (ct)^2/r^2... I'm not sure of any meaning to that. As squares, the equation of the interval s^2 (a constant) =(ct)^2-r^2 is that of a hyperbola, and relates to the pythagorean theorem. Edited October 21, 2020 by md65536 1 Link to post Share on other sites

geordief 114 Posted October 24, 2020 Author Share Posted October 24, 2020 On 10/21/2020 at 10:11 PM, md65536 said: It's meters or lightseconds or any distance units, squared. The use of distance units is apparently a convention, see https://physics.stackexchange.com/questions/519707/is-the-unit-for-spacetime-intervals-time-or-space-distance Yes, for a time-like interval, the time component will be greater than the spatial. For time-like (or light-like for that matter) intervals, the ratio of r/t is the constant speed of a particle that moves between the two events. ct/r would be the ratio of the distance that light travels between the two events (along any path that gets it there, like your 200 lightsecond example) to the straight-line spatial distance between the two events, in the given frame. This ratio is frame-dependent, and undefined in frames where r=0. (ct)^2/r^2... I'm not sure of any meaning to that. As squares, the equation of the interval s^2 (a constant) =(ct)^2-r^2 is that of a hyperbola, and relates to the pythagorean theorem. If I differentiate the spacetime interval formula ** with respect to time I think I get the expression dr/dt =c^2.t/r Am I correct in the math and what might be the physical significance or interpretation of that expression? To me it looks as if, when t and r are measured in infinitesimally small amounts they seem to "switch places" (dr being in the numerator and r being in the denominator in that expression) Is it simplistic to suggest that time and space might "switch places" when the measurements are completely local rather than extended? If that is ,as I expect wrong headed what physical situation might that dr/dt=c^2.t/r be describing mathematically? ** s^2 = [ct]^2 - r^2 0 =2c ^2.t -2r.dr/dt dr/dt=c^2.t/r Edit: hope I was right that d[s^2]/dt =0 Link to post Share on other sites

md65536 383 Posted October 24, 2020 Share Posted October 24, 2020 (edited) You're in over my head! Hopefully someone else can help? That's the derivative of a hyperbola. I don't see it saying anything about switching places. When y (or r) is small, it changes quickly. As y gets bigger, it approaches x (or ct), and the rate of change approaches constant; a unit hyperbola asymptotically approaches the line y=x. If you take the spacetime interval and make r a function of t, I think what that means physically is... It describes how the spatial distance of the interval changes as a function of the time component of the interval, as you go through different frames of reference. The infinitesimal changes in t for example would mean, if you change inertial frames by just a little (ie. with infinitesimal change of speed), the time and spatial distance components of the interval change like a hyperbolic function does. Or, an infinitesimal change in speed corresponds with an infinitesimal hyperbolic rotation of the spacetime interval. Edit: I'll leave that there but it's wrong! When both x and y are very large, an infinitesimal rotation (I think) can still mean a huge change in x and y. So to correct that: A small change in inertial frame involves an infinitesimal change in t and an infinitesimal change in r. However, as speed approaches c, a small change in speed (but huge change in rapidity) can involve a huge change in t and r. That makes sense with respect to velocity composition, right? Maybe it's correct to say "an infinitesimal change in rapidity corresponds with an infinitesimal hyperbolic rotation of the spacetime interval", but I might change my mind again after learning more... Edited October 24, 2020 by md65536 1 Link to post Share on other sites

joigus 505 Posted October 24, 2020 Share Posted October 24, 2020 (edited) Mmmm. Interesting problem, but some technical difficulties I can see there. The first thing I see is that going from event 1 (emission) to event 2 (reflection), and from there to event 3 (reception) involves three relevant events, IMO. This implies two 4-vector translations, and their sum. Let's call them d_{12}, d_{23}, d_{13}. They satisfy: d_{13}=d_{12}+d_{23}. But unfortunately, \[s_{13}^{2}\neq s_{12}^{2}+s_{23}^{2}\] If you write down the correct expression, you get, \[s_{13}^{2}=s_{12}^{2}+s_{23}^{2}+\left(x_{2}-x_{1}\right)^{\mu}\left(x_{3}-x_{2}\right)_{\mu}=\] \[=s_{12}^{2}+s_{23}^{2}+c^2\left(t_{2}-t_{1}\right)\left(t_{3}-t_{2}\right)-\left(\boldsymbol{x}_{2}-\boldsymbol{x}_{1}\right)\cdot\left(\boldsymbol{x}_{3}-\boldsymbol{x}_{2}\right)\] You can see very clearly in this expression that how much it takes for the signal to reach the reception point depends on the relative positions (including orientation) of the whole "triangulation." It is not specified well enough by Euclidean distances only. This may be at the root of the problem that @md65536 sees. In your second post it looks like you're trying to get a differential equation that relates r and t. There's a problem there too. In the words of @md65536: 10 hours ago, md65536 said: Edit: I'll leave that there but it's wrong! When both x and y are very large, an infinitesimal rotation (I think) can still mean a huge change in x and y. So to correct that: A small change in inertial frame involves an infinitesimal change in t and an infinitesimal change in r. However, as speed approaches c, a small change in speed (but huge change in rapidity) can involve a huge change in t and r. I would re-word it as: your problem is under-determined, which I think is what md65536 is saying. The back and forth rays belong to different hyperbolas*. In fact, from the differential equation that you're trying to get at, which is correct: \[\frac{dr}{dt}=c^{2}\frac{t}{r}\] You get, not only the outgoing ray r=ct, but also the return trip r=-ct, plus the unwanted free gift of an infinite set of constant accelerated motions, \[r^{2}-c^{2}t^{2}=K\] with K being an arbitrary constant. So going back and forth implies jumping from one inertial system to another (in the case of the light rays, one relative sign to another in r, t). Sorry I took the whole thing to the language I understand better. I hope that helps, and I hope it has some bearing on the problem. *In the case of the light rays, they're "degenerate hyperbolas", meaning straight lines in Minkowski space. Corresponding to K=0. Edited October 24, 2020 by joigus Link to post Share on other sites

md65536 383 Posted October 24, 2020 Share Posted October 24, 2020 2 hours ago, joigus said: It is not specified well enough by Euclidean distances only. This may be at the root of the problem that @md65536 sees. No, I didn't see that as a problem. The original spec is, "So we have 2 events ; the emission and the recapture of the signal," and "200 light.seconds taken by the signal to make the round trip." It's not specified where the reflection point is, but I don't think that matters because it's only used to establish the time between the two events, and that's given. For me the light reflection path is irrelevant. It's only used here as a clock, and any stationary clock would do. Yes, there are light-like intervals between the reflection point and each of the 2 events, but I wasn't thinking of those. I think I've completely confused the meaning of hyperbolic angle, which I tried to relate to the derivative dr/dt. With a given invariant spacetime interval, a change in the t and r parameters doesn't involve moving along a world line between the two events. It involves rotating the fixed interval through different frames of reference, to vary the t and r components that make up the same fixed interval. If you do consider a particle moving along such a world line, it's moving through different points along that line, ie. different events, each of which makes a different spacetime interval between it and the initial event. (Though, in the case of light-like paths, all of those intervals are 0! But they can still be rotated so that different observers measure light between the two events traveling a different distance during a different time. Lol I'm sure there's a simpler way to look at this.) Link to post Share on other sites

joigus 505 Posted October 24, 2020 Share Posted October 24, 2020 (edited) 11 minutes ago, md65536 said: No, I didn't see that as a problem. The original spec is, "So we have 2 events ; the emission and the recapture of the signal," and "200 light.seconds taken by the signal to make the round trip." It's not specified where the reflection point is, but I don't think that matters because it's only used to establish the time between the two events, and that's given. But it does matter. If the mirror is on the same line where the reception point is, but in the same direction, the signal's delay will be minimal. On the contrary, if it's on the same line but in opposite direction, the delay will be maximal for the given distance, corresponding to the extreme values of the cosine in, \[\left(\boldsymbol{x}_{2}-\boldsymbol{x}_{1}\right)\cdot\left(\boldsymbol{x}_{3}-\boldsymbol{x}_{2}\right)\] The reception event is described only in terms of its Euclidean (spatial) distance to the emission in the given frame. The reflection point being at certain distance does not determine the time it takes for the photon to get to the reception point. It's not a "home-run," but an open trajectory in Minkowski space in which we only know the radial distance from the emission point. In what directions? The problem is under-determined. Maybe I misunderstood something... Edited October 24, 2020 by joigus Link to post Share on other sites

md65536 383 Posted October 24, 2020 Share Posted October 24, 2020 1 hour ago, joigus said: The problem is under-determined. Maybe I misunderstood something... No, I think you're right. When OP wrote, On 10/16/2020 at 7:33 PM, geordief said: Am I right to say that the spatial measurement component of the spacetime interval between the 2 events is the 1000 metres and the time component ( ie the ct component) is the 200 light.seconds taken by the signal to make the round trip and return to the vicinity of its emission? you're basically saying that the 200 LS is imprecise or an assumption about where the reflection point is. I was treating it as though it was supplying the previously missing information, but that's not explicit. Link to post Share on other sites

joigus 505 Posted October 24, 2020 Share Posted October 24, 2020 4 minutes ago, md65536 said: No, I think you're right. When OP wrote, you're basically saying that the 200 LS is imprecise or an assumption about where the reflection point is. I was treating it as though it was supplying the previously missing information, but that's not explicit. Exactly, (either one of them, the reflection point or the reception point), and because the exact location of the reflection point is imprecise (we only know the Euclidean distance in that FOR), when the reception takes place is unknown. You can only take one of them as defining the x-direction with the emissor, if you will, but the other one could be at an angle. The three don't have to be collinear. I hope you agree. Link to post Share on other sites

geordief 114 Posted October 25, 2020 Author Share Posted October 25, 2020 @joigus @md65536I see now that my scenario was not water tight. I won't attempt to patch it up as I would probably make a bad job worse. Still I hope I may have learned a little overall (about 4-vectors, perhaps - or at least when a 3-vector does not do the job a of a 4-vector) Link to post Share on other sites

md65536 383 Posted October 25, 2020 Share Posted October 25, 2020 1 hour ago, geordief said: I won't attempt to patch it up as I would probably make a bad job worse. I think you should! If you think it's a bad job and don't know how to make it better, then we haven't done a good job in explaining it. There are aspects of this topic that I'm going to keep getting wrong until I see it in the right way. I know for myself it'll take repetition, to keep looking at it. Besides, I don't think that you did a bad job. Originally you didn't specify the two events of the interval precisely, but 1) it was good enough to understand what you implied, and 2) the imprecision only changes the 200 seconds value by +/- 3.3 microseconds, so imprecision is not a problem there. I'm not concerned with the exactness of the example interval, but I'm concerned about the meaning of it especially with respect to multiple frames of reference. Link to post Share on other sites

Markus Hanke 519 Posted October 25, 2020 Share Posted October 25, 2020 (edited) Could I perhaps suggest a very short (few minutes) video on the invariance of the spacetime interval: Edited October 25, 2020 by Markus Hanke Link to post Share on other sites

md65536 383 Posted October 25, 2020 Share Posted October 25, 2020 (edited) 12 hours ago, Markus Hanke said: Could I perhaps suggest a very short (few minutes) video on the invariance of the spacetime interval: Thanks for that. Is the repeated use of the word "true" in the video not a standard scientific term (even misleading)? If someone said that if a measure of something being length contracted isn't a true measure of length, I'd argue that's wrong, whereas saying it's not its proper length is using a scientific term. About exact specification of intervals: If you have 2 events and you don't care about their locations, only their relative separation, then you're talking about their spacetime interval. Further if you don't care about how they're measured in one particular frame of reference, then all you need to completely specify the interval is the one value, s^2. So in OP's example, suppose that s^2 is about 39999.9999999999889111 light seconds squared. This describes two events that, in one frame, are separated by say 1000 meters and 200 seconds. But it also describes the same two events that, in another frame, are in the same place and separated by 199.99999999999997227774 seconds. The latter is a measure of the proper time between the two events. (Sorry for those numbers, but if I don't use that many digits, the 1000 meters gets completely lost to rounding.) But also... that same s^2 describes the same 2 events separated by billions of light years and billions of years, in yet another frame. It also describes 2 completely different events a long time ago in a galaxy far far away that had the same separation relative to each other. This seems to imply that all light-like intervals are "the same." They all have s^2 = 0. What this means physically, is that if you have a light signal from A to B, no matter how near or far they are in your frame of reference, you can find other frames of reference where that light signal is arbitrarily short, and others where it is arbitrarily long. Those all describe the same thing, and there is no one frame in which the distance or timing of the light signal is "proper". An exception is the interval between an event and itself??? That seems to produce a valid interval where s^2 = 0, yet there are no frames of reference that can separate the events in time or space. I've never seen any mention of this. Is "spacetime interval" only defined for two different points? Edited October 25, 2020 by md65536 Link to post Share on other sites

geordief 114 Posted October 26, 2020 Author Share Posted October 26, 2020 @md65536 yes ,I suppose the looseness in my scenario was not such a big deal. I was trying to show a concrete example of this spacetime interval and it appears that I was along the right lines. You seem to ave a better feel for the ramifications of the issue than I do. @Markus Hanke thanks for that video.It dealt with a few things I had had on my mind for some time. I like the "5 minute from the shop " explanation I notice that ,as formulated the spacetime intervals has units of spatial distance.Is there a related formulation that would have units of pure time (even if convoluted)? I mean ,can the s^2= [ct]^2 -r^2 formula be rearranged or rejigged so that both [ct]^2 and r^2 are expressed in terms of time ( with c understood) ? Maybe T^2 =t^2 -[r/c]^2 where T is the temporal distance? Can I also ask whether you think that adopting the invariance of the spacetime interval as a postulate leads to the requirement of a universal and invariant speed limit ? I know you said you didn't think that it would work as a way to derive Relativity itself but might it be enough to require the existence of a "c" of some value? Link to post Share on other sites

Markus Hanke 519 Posted October 26, 2020 Share Posted October 26, 2020 9 hours ago, geordief said: Is there a related formulation that would have units of pure time (even if convoluted)? You can express it either as a distance in m, or as a time in s, by using the speed of light c as a conversion factor. You are ultimately completely free to choose whichever units work best for a given problem (so long as they are used consistently of course). 9 hours ago, geordief said: Can I also ask whether you think that adopting the invariance of the spacetime interval as a postulate leads to the requirement of a universal and invariant speed limit ? I don't think it does. You need to also specify a metric; in particular, the metric signature plays an important role here, namely that time and space have opposite signs within the metric. Link to post Share on other sites

md65536 383 Posted October 26, 2020 Share Posted October 26, 2020 (edited) 17 hours ago, geordief said: Can I also ask whether you think that adopting the invariance of the spacetime interval as a postulate leads to the requirement of a universal and invariant speed limit ? Here's a bit of a meander through how I understand this topic. There are other ways to look at it that you might prefer. First, if you move a light clock across a timelike interval in a particular frame, you can derive the time dilation factor for the moving clock using Pythagoras theorem. It looks quite similar to the scene you described in the initial post. I'd take a look at this if you've never seen it, I could post a video. If you repeat this for a bunch of different frames, you'll find you're looking at a bunch of different triangles that all have one of their sides in common: the side representing the proper time measured by the clock. Or working in the opposite direction: If you look at the time dilation factor with these equations: t/tau = 1/sqrt(1-v^2/c^2), and r = vt, you get (c tau)^2 = (ct)^2 -r^2, the spacetime interval. You can also consider space-like intervals, and either add a ruler (a proper length) instead of a light clock, and/or replace the proper lengths with times measured by light passing over those lengths, effectively swapping time and distance to get the same situation as above. From this you have a simple geometric picture of the spacetime interval components, in a triangle that gets stretched for different frames of reference, but has one side remaining invariant, and you can see the equation of the spacetime interval in the Pythagoras theorem. I get the sense that you're trying to say something like, "The spacetime interval is some natural measure of separation, and the fact that it's invariant must say something fundamental about relativity." The way I see it, the definition of the interval was chosen because it's something that is invariant---proper time---and as seen above, simply relates time and distance in different frames. Rather than starting by defining it and then assuming it is invariant, I think we start by defining it as something that is already assumed to be invariant. Rather than deriving relativity from it, I'd say the opposite is true; since it represents the measurements (eg. of time) in a particular frame, it shows that Newtonian time in a "rest frame" can be derived from special relativity. Like Markus's video suggests, it shows that not everything becomes relative when going from a Newtonian model to SR. The fact that proper time is invariant is not saying anything more (to me at least) than that 1) there's a single measure of time between a pair of events in a single frame (same as with Newtonian time). Or, different clocks sharing a rest frame don't measure time differently, and 2) while different observers measure time differently than each other, they all agree on the (proper) measurements that each other is making. Edited October 26, 2020 by md65536 Link to post Share on other sites

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