Category: All
Why isn’t there any curvature of the horizon in a horizontally oriented picture?
That depends on what you mean by “horizontally oriented”. But if we take it as meaning perpendicular to the radius and parallel to the surface of the Earth (or any other sphere), then no matter which “horizontal” direction you look, you will not be looking directly at the horizon and the angle of depression down to the horizon will be the same in all such directions. So the horizon should appear perfectly straight in any undistorted view of that sort. It will appear curved up at the edges if your line of sight is pointing upwards and down at the edges only if you are looking downwards (which is why all those pictures taken from airplanes claiming to refute “flat earthers” are so silly).
What does show the curvature of the Earth (albeit in a way that is very hard to notice even at the highest commercial airplane altitudes) is the fact that, when you look out “horizontally”, the horizon is below the midpoint of the picture (which is almost impossible to demonstrate convincingly with a photograph since there is no way of proving that the picture has not been cropped from one where the horizon was above the middle).
Of course, if you are far enough from the sphere that the only way any part of it can be in your field of view is by looking downwards (as in the various pictures of the whole Earth taken from space), then you will indeed always see the curvature if you can see any part of it at all. But at any altitude achievable by a commercial aircraft you will almost surely not.
In particular, even at an altitude of 54 km (which is well above most commercial airline cruising heights) we have the following:
An altitude of 54km is just a bit less than 1% of the radius of the Earth, so the angle of depression of the horizon is about arccos(0.99) (* – see first note below if you need an explanation), which is approximately 8 degrees. Since that is about 10% of the 80 degree viewing angle of a 24mm wide angle camera lens, a “horizontally” oriented camera with a wide angle lens at an altitude of 54 km should produce an image of a perfectly straight horizon line about 10% below the middle of the frame. (With a fisheye lens the viewing angle is 180 degrees so with such a lens the horizon should still be a straight line but just [math]8/180\approx 4.4%[/math] below the middle.)
If the camera is pointed down so that the middle of the horizon is at the centre of the frame then with the fisheye lens (if you could put it outside the window, or perhaps be working in the cockpit with a wrap-around canopy window) the horizon should appear as a half ellipse with the edges still about 4.4% down from the middle.
But for a normal wide angle lens, since the half-view width of 40 degrees [per https://expertphotography.com/wide-angle-lens/ ] is less than a half of the 90 degree sweep to the side of the camera, and since the equation of the ellipse is quadratic rather than linear, the edges of the part of the horizon in the picture from that lens will be less than [math]4%/2^2=1%[/math] of the image height below the middle.
And with a telephoto lens of viewing angle just 10–20 degrees, the horizon should only be completely straight if it is right near the bottom of the picture, but the curvature that comes from pointing the camera down to centre the middle of the horizon should be even less noticeable as the image only shows 5 to 10 degrees worth of the drop. (** – See the second note below for the actual calculations)
But at the typical 10km cruising altitude of a commercial airliner there is really no visible effect. The angle of depression of the horizon at that altitude is arccos(0.998) which is less than 4degrees, so in a square picture shot with a normal 50mm focal length showing just the middle 46 degrees of the 180 degree full sweep the drop from centre to edge of the horizon should be less than 0.0015 of the height of the picture (ie just barely one pixel down at the edges in a 640×640 image) – as shown below.
With a 28mm wide angle lens, showing the middle 75 degrees the amount only goes up to about .004 of the image height or less than 2.5 pixels in 640.
With the 94 degree sweep of a 20mm extreme wide angle we still only get a downturn of .0065 of the image height or about 4 pixels in that 640×640 image.
And even with the full 180 degree sweep of the unobstructed fisheye (whose picture compresses what would be in our peripheral vision down into the middle of our field of view and so gives a much more dramatic effect than what we actually see even from that panoramically windowed cockpit) the drop to the edges of the horizon line is only 2% of the image height.
So, while it may be possible for a pilot in a cockpit with good visibility out of front and side windows to get a sense of curvature from peripheral vision, no picture taken from a passenger seat can show curvature unless the camera is pointed downwards (which is a natural thing to do from a plane since most of the interesting stuff is below the horizon).
Here’s what a picture from an ordinary wide angle lens should look like if the camera was pointing ten degrees downwards:
But if you are inclined to see this as evidence of curvature then compare it with the following view of what the same camera would produce looking 10 degrees downwards on a perfectly flat Earth:
(*)-Note1: The picture below shows that the angle of depression, [math]\delta[/math], is the same as the angle subtended at the Earth’s centre between the observer and the point seen on the horizon, whose cosine is just the ratio [math]\frac{R}{R+a}[/math] where [math]R[/math] is the radius and [math]a[/math] is the altitude of the observer.
(**)- Note2: If the edge of a circle is at an angle of [math]\delta[/math] degrees below eye level in all directions, then tilting the view down to bring the part directly in front up to the middle of the picture will not change the fact that the parts off at 90 degrees in the left and right directions are still depressed by an angle of [math]\delta[/math] degrees (and if we had eyes in the back of our head they would be looking up so that the part of the circle behind us would be further down at an angle of [math]2\delta[/math] degrees). Since the projection of a circle is an ellipse, the image of the half-circle in front of us would be given analytically (in terms of the image coordinates [math]x[/math] and [math]y[/math] being the numbers of degrees to the right and above the centre of the image) by the part of [math](\frac{x}{90})^2+\frac{y+\delta}{\delta})^2=1[/math] with [math]y>-\delta[/math] (or parametrically by [math]x(t)=90cos(t), y=\delta sin(t)[/math] for [math]0<t<\pi[/math]). A fisheye lens would produce an image in which the [math]x[/math] and [math]y[/math] values range from -90 to +90, but with a lens of view width [math]w[/math] degrees the ranges in the picture would just be from [math]-w/2[/math] to [math]w/2[/math], and the images shown above were created by just zooming in on a graph of the ellipse so that the relevant ranges of [math]x[/math] and [math]y[/math] fit into the area of the picture.
640*4(1-sqrt(1-(94/180)^2))/90
90cos(t);4sin(t)-4;0,3.14
Source: Alan Cooper’s answer to Why isn’t there any curvature at the horizon when we look out if we live on a ball? – Quora
Lorentz Expansion!
A Quora question asks:
Two guns on Earth D metres apart fires simultaneously at a metal plate moving by at close to light speed. Is the distance between the two bullet holes in the plate length contracted?
It depends on who thinks the guns fired simultaneously.
If, as seems most likely, the question means that those Earth-based guns were synchronized by someone on Earth with them, then the distance between the two holes, as measured by any observer stationary with respect to the Earth, is (and remains) exactly equal to the distance between the two guns. Since the plate appears contracted to these Earth-based observers, if it was marked with units of length in its own rest frame those markings would appear closer together to the Earth-based observers and so there would be more of them between the holes than the number of length units measured on Earth.
In other words the distance between the holes would appear to be
greater from the point of view of someone travelling with the plate.
It may seem puzzling that this happens despite the fact that from the traveller’s point of view the distance between the guns is “length contracted” and so appears to be less than that measured between them on Earth.
The puzzle is resolved by the fact that from the point of view of the traveller the guns did NOT fire simultaneously. The one making the front hole appears to the traveller to have gone off earlier. (And it is a worthwhile exercise for anyone seeking to learn about relativity to work through the calculation needed to show that the delay is by exactly the right amount for the forward movement of the plate to create the observed bigger distance between the holes).
Alternatively, if the traveller thinks the guns went off simultaneously, then the Earth-based observers think there is a delay. (And again, working out the details is a worthwhile exercise for any beginning student of the subject.)
P.S. The question of whether or not a length or object is “length contracted” does not really make sense without any mention of which observer is doing the measurement.
What does it mean to say that there is a distance between two events in time? (Another Quora Question)
It probably means that the speaker is taking a Galilean approach to physics.
In modern relativistic physics, the property of having a time-like separation between two events is independent of observer, but the magnitude of that separation depends on the observer. And two events which are spacelike separated, while having no time difference for some observers, will still appear to have a non-zero time difference for others.
So the concept of a time (or space) “distance” (ie a specific value of the difference) between two events in space-time does not make sense without reference to an observer.
However in the case of two time-like separated events the time difference is nonzero for all observers, and if we restrict to inertial observers it has a nonzero minimum (which corresponds to the time difference as seen by an observer who experiences both events directly without any intervening acceleration or gravitational field gradient). But although this minimum is in principle computable by any observer it does not correspond to the time difference actually “seen” by that observer.
Are Bayesian and Frequentist probability really inherently completely different, or is there a false dichotomy between them?
I would say that they are inherently “somewhat” different even though they are both used for similar purposes and obey the same mathematical axioms (of a countably additive non-negative measure on some space of “outcomes” in which the measure of the set of all outcomes is 1).
Of course there are many probability measures in the mathematical sense which have nothing to do with any actual probabilities in the real world, so having the same mathematical structure does NOT preclude their being “completely different”.
But probabilities in the real world are supposed to have something to do with our expectations regarding what will happen when we perform an experiment which could in principle be repeated an unlimited number of times. And since any concept of probability is intended as a measure of how “likely” we expect physical events to be, if both kinds are useful they can’t really be all that different in what they lead us to expect.
The difference, as I see it, is as follows:
The frequentist probability of an “Event” (which is just the jargon for a subset of the space of all possible outcomes) is assumed to be the proportion of “all” experimental trials in which the outcome is consistent with the event. Despite its rather fuzzy definition (in terms of whatever we mean by “all” trials), this is something that is assumed to be an actual property of the experiment, albeit one that can never be determined by experimental tests (because any finite set of trials might fail to be representative). Frequentist statisticians often try to choose between different possible sets of assumptions by using each set of assumptions for computing the probability of what they have actually seen and choosing the assumptions that lead to higher probability, but they generally do so with the mindset that there is one set of assumptions that is really true.
Bayesian probability, on the other hand, is something that is always acknowledged to depend on the observer (through both initial choices and subsequent acquired knowledge). Given those choices and experience, the Bayesian probabilities are perfectly well known because the Bayesian approach provides an explicit rule for modifying one’s assumptions in the light of experiments. But because it depends on the the observer’s assumptions and experience, the Bayesian probability is not a well-defined property of the experiment and outcome alone.
The difference is subtle though, because one cannot do any frequentist analysis without making an assumption (usually about what constitutes a set of “equally likely” outcomes). But though one can make and change the assumptions, the Frequentist attitude remains that they are assumptions about something that exists and are either true or false, while the Bayesian does not necessarily ever claim that there is one set that is really “true”.
However there are theorems which tell us (roughly speaking) that if some frequentist assumption is in fact correct, then no matter how bad the initial assumptions of the Bayesian are, after a lot of experiments it is very probable (in the frequentist sense) that the Bayesian probabilities become close to the “correct” ones.
So in the end both approaches end up giving the same advice about what one “should” expect (though neither gives anything I can understand as a non-circular definition of what that means!) and whether the difference in attitude is a “false dichotomy” is something I think we each have to decide for ourselves.
Given that the Lorentz transformation is symmetrical with respect to interchange of space and time, how does it lead to length contraction but time dilation?
This is a question that I am surprised to not have seen before (especially since I have had to remind myself of the answer more than once – including, I suspect but can’t be sure, from way back before I entered my dotage).
It is true that in one space dimension the transformation equations
[math]x’=\gamma(x-\beta t)[/math] and [math]t’=\gamma(t-\beta x)[/math]
are completely symmetrical with respect to interchange of [math]x[/math] with [math]t[/math] and [math]x’[/math] with [math]t’[/math].
(and in the case of three space dimensions the same applies if [math]x[/math] and [math]x’[/math] are the coordinates in the same direction as the relative velocity, so it’s not got anything to do with the dimension).
So what is the difference?
Well here it is in a nutshell.
When we measure the length of a moving measuring rod, we look at both ends at the same time and so are looking at the spatial distance between two events at the same time in our frame of reference.
But when we measure the time between two ticks of a moving clock we are looking at the time difference between two events that are NOT at the same spatial position in our frame.
So the nature of the two measurements is not symmetrical with respect to interchange of space and time.
I may add some more explanation and diagrams to show how this does lead to contraction for the rod length and dilation for the tick interval, but I wanted to get this off my chest right away – and also to address a couple of natural follow-up questions.
Namely, what kind of measurements would give the symmetrical outcome? Are there situations in which these others might be relevant? And why do we instinctively prefer the ones we do?
So, for example, what kind of time measurement would be symmetrical compared to our usual rod length measurement (and so would give a “time contraction” rather than the usual time dilation)?
Since the rod length involves looking at both ends at the same time in our frame, the corresponding time measurement would involve looking at the interval between two ticks at the same place. But how can we do this if the clock is moving? Well we could if the clock was extended in space, and if we have a long train of clocks that are synchronized in their own frame, then you can easily check that observers who look at the time between the ticks right in front of them will actually see a shorter interval than that measured by the travelling system – ie a time contraction.
And going the other way, what kind of measurement would give a length dilation? Well that would have to be the symmetric version of our usual clock measurement. And corresponding to our usual measurement of the time interval between two ticks at the same place in the moving clock’s frame, interchanging space and time would have us measuring the spatial distance between events where the two ends of the rod are at the same time in the rod’s frame. For example the managers of the rod might set off flares at both ends in a way that they, travelling with the rod, perceive as simultaneous. If we measure the distance between where we see those two flares then it will indeed appear dilated relative to the length of the rod in its own frame.
So now we come to the final question. Is there anything really “wrong” about these alternative kinds of measurement? If so what is it? Or is there just something about us which makes us think of what we do as natural and the alternative as somehow, if not actually wrong, then at least rather odd?
Here’s what I think (at least for now). The thing that makes us prefer to measure lengths in terms of events at the same time in our frame but times in terms of events at the same place in the moving frame is the fact that we, as blobs of space time, are much more extended in time than in space. (This is evident in the fact that we live for many years but do not extend for many light years in our spatial extent – or equivalently that in units adapted to our own spatial and temporal extent the numerical value of c is very large.)
So here’s a follow-up question. Could we imagine an entity which was the other way around? (ie of brief duration but of great spatial extent) And from the point of view of such an entity would it make sense to define measurements differently (as suggested above to achieve the effect of time contraction and length dilation)?
OR is it more just a matter of causality?
P.S. This is a question and answer that I have been meaning to post for some time, but was prompted to do so by Domino Valdano’s excellent answer to another question (in which she covers pretty much the same ground with a slightly different way of expressing the ultimate reason for why we measure as we do – which I may yet end up deciding that I prefer to my own). Please do read that one too!
More TwinStuff from Quora
In the twin paradox where does the missing time go? If the twin turns back to Earth then turns away again their notion of now switches back to the past. What does this mean for the experience of the observer on Earth relative to the moving twin?
“In the twin paradox where does the missing time go?” I am not aware of any “missing” time. One twin experiences less time than the other but there is no gap where any time goes missing. (There is however an apparent speed-up of the Earth clock from the point of view of the traveller while turning around, and in the physically impossible case of an instantaneous turn-around that would look like a gap in the traveller’s understanding of what was happening “at the same time” on Earth; but in any possible actual scenario it would just be an apparent speed-up rather than a jump.)
“If the twin turns back to Earth then turns away again their notion of now switches back to the past. What does this mean…” Indeed! Does that sentence actually have any meaning at all?
Perhaps what that second sentence is referring to is the traveller’s idea of the time that is “now” back on Earth. It is true that when we accelerate away from something we infer a slowing down of its clock at a rate proportional to the distance. If the distance is great enough this effect can make the clock “behind” us appear to stop, but beyond that distance (called the “Rindler horizon”), rather than see it run backwards we actually just don’t see it at all. (And, yes, the Rindler horizon perceived by an accelerated observer is indeed related to the Event horizon surrounding a gravitational singularity.)
Yet Another Quora Question:
Is there experimental evidence of RoS (relativity of simultaneity) available? According to Carl Popper something that cannot be falsified is metaphysics. Can the ALT (absolute Lorentz transformation) and absolute simultaneity be used instead of RoS?
Yes. There is lots of evidence that IF observers define simultaneity of remote events by comparing the arrival times of light signals from those events and adjusting for the light travel times, then they will NOT agree on which pairs of events are or are not simultaneous.
What cannot be falsified (and so, according to Popper, is not worthy of consideration as scientific) is the claim that one particular set of “stationary” observers are “correct” and the rest are “wrong” about the “actual” simultaneity (with the observations of the “wrong” ones being explained by effects of their “motion” on their clocks and measuring rods).
The “metaphysical” question of whether or not there is such a particular set of “stationary” frames is generally resolved by reference to “Ockham’s Razor” which advises us to prefer an explanation which involves fewer arbitrary choices. To the extent that the identification of a particular preferred set of “stationary” frames is arbitrary, we are therefore advised to treat them all as equally valid and make no choice of what constitutes “absolute” simultaneity.
Of course, any particular feature of the universe (such as for example, the cosmic background microwave radiation (CBMR)) can, if we like, be taken as defining an “absolute” rest frame (and so an “absolute” sense of simultaneity). And if I were moving at a noticeably high speed relative to the CBMR, then it might indeed be presumptuous to declare my own synchronization as more fundamental, but it would not in fact lead to any difference in anyone’s prediction of any testable event.
Another silly Quora question
Is Special relativity based on a fundamental flawed claim of inertial reference frames without acceleration which do not exist anywhere in space?
NO. Special Relativity is not based on any “claim” at all.
It is based on the observation that Maxwell’s Equations (and the consequent value of the speed of light) appear to hold with the same constants in every freely falling reference frame, and is only expected to be valid in situations satisfying the simplifying condition that gravitation has negligible effects on the quantities of interest.
That simplifying condition of course limits the domain in which SR applies; but the existence of locally inertial frames is readily apparent, and for the purpose of measurements over a sufficiently small range of space and time SR has no difficulty dealing with accelerated frames as well.
Relativistic Dynamics
Some of Larry G King’s Quora answer to Why does an object’s mass increase as it accelerates towards light speed? should probably get included in my Relativistic Dynamics page.