Category: Quora Answers

Simultaneity and Synchronization

Identical clocks in relative motion can only be synchronized from the point of view of an observer relative to whom they both have the same speed. To anyone else they seem to be progressing at different rates and so are always unsynchronized and there’s no “becoming” about it. There is a sense in which the relativity of simultaneity forces the clocks to be unsynchronized, but it’s generally easier to understand the other way around so I’ll look at that first.

Imagine that, when you pass by me at 86% of the speed of light, we set our clocks to both read, say, t=0. Then they will both read the same time at that event but will not be synchronized because each will see the other as running at just half speed.

and that I also use light signals to set a clock that is fixed in my frame one light year away (in the direction towards which you are headed) to also read t=0 at what seems to me to be the same time.

Then when you reach that clock it will read t=1/.86 years but your own clock will read just 0.5/.86 years (so your clock and my remote clock are not synchronized). I will attribute that difference to your clock running slow but from your point of view the time on your clock is correct because the distance which I saw as 1 light year appears to you to be only half of that.

But from your point of view it has seemed that my clocks were running slow. So according to you, the clock you see as reading 1/.86 years must have started at a time 2/.86 years ago which is long before the time when we passed one another. Or in other words the clock-setting event I thought was simultaneous with our meeting was not simultaneous according to you.

Now what I have shown here is just that relativity of simultaneity is a mathematical consequence of the symmetric nature of the asynchronization of the clocks. But despite the lack of “becoming” there is also a logical connection the other way.

Imagine that I send a message to reach you when I think you are 1 year away from reaching me (ie at a distance of 1/.86 light years when travelling at 0.86c) and ask you to set your clock at t=-1 at that time and that I set my own clock at t=-1 at the moment I think that message reaches you. But relativity of simultaneity says that you will think I made a mistake and that those two clock settings did not happen at the same time. You will actually think that the time when I asked you to set your clock was after I had set my own and this will cause you to expect that when you reach me my own clock would have got to t=1 if running at the same speed as yours but yours will just be at t=-0.5. And since my clock will actually be reading t=0 at our meeting this will lead you to conclude that it must be running slowly (and so that they cannot be synchronized).

Source: (1001) Alan Cooper’s answer to The simultaneity of relativity tells us that inertial observers in relative motion disagree on whether 2 spatially separated events are simultaneous, but how does this lead to the observers’ clocks becoming unsynchronised? – Quora

Black Hole Singularities

A Quoran asks: Do black holes have a singularity? (A point inside the event horizon which is infinitely dense)?
YES, black holes are theoretical objects in the theory of General Relativity which are usually associated with singularities of space-time. But NO such a singularity is NOT “a point inside the event horizon”. As a point in space-time the black hole’s singularity is actually an event rather than a particular point or location in space; and in fact for every point inside the event horizon, the singularity event is in the future of any particle that is ever at that point. In terms of the way it looks to an observer inside the event horizon, there is no longer any sense of any “centre” but oneself, and the theoretically predicted experience is not of falling inwards but rather of getting crushed by the collapse of everything around you. ….or something like that!

Source: (1001) Alan Cooper’s answer to Do black holes have a singularity? (A point inside the event horizon which is infinitely dense)? – Quora

Length Contraction Inherent in Maxwell Equations?

There are two senses in which length contraction is “inherent” in the Maxwell/Heaviside equations.

One sense is that in some cases length contraction follows from Maxwell’s equations. FitzGerald and Lorentz showed that if the structure of matter is determined by electromagnetic forces then the equilibrium spacing of stationary particles due to purely electrostatic forces would become contracted when they are in motion due to the presence of additional magnetic forces. [And one effect of this contraction, together with a related slowing and desynchronization of clocks, would be that moving observers (using their own contracted measuring rods and slowed clocks) would not notice any effect of their own motion on the apparent speed of light (or in fact on any electromagnetic phenomena).]

Another sense is that length contraction is necessary to preserve the form of Maxwell’s equations. Poincare had already noted that if the coordinates used by a moving observer were related to those of one who is stationary by a standard Galilean transformation (ie by just a progressive shift of position with no change of length and time scales) and if Maxwell’s equations applied to (say) the stationary one then they would have to be modified in order to make correct predictions for the other. And he showed that the only kinds of coordinate transformation that leave the form of Maxwell’s equations unchanged are those involving length contraction and time dilation.

[Einstein then noted that if all the laws of physics, written in terms of the natural coordinates of an observer, have a form that is independent of the state of motion of the observer, then there is in fact no way to tell which of two relatively moving observers is “truly” stationary and whose clocks are “truly” synchronized. He therefore sought to express all the laws of mechanics (eventually also including gravity) in a form which used only the actually measured coordinates of each observer rather than those of some assumed fixed “rest”(or “aether”) frame (which would have been more complicated to do if possible, but was not actually possible to do properly because no observers could actually tell whether or not they were actually in motion).]

Source: (1001) Alan Cooper’s answer to Is length contraction inherent in the Maxwell/Heaviside equations? They do have some asymmetry. I remember reading something to this effect but can’t find it and could be wrong. – Quora

Superposition Independent of Space and Time

The spin observables for an elementary particle can be studied without any reference to the particle’s position. They are then modeled as operators on a finite dimensional Hilbert space of spin states that is essentially independent of the infinite dimensional Hilbert space of “wave” functions on position space (which identify those aspects of the particle’s state that are related to its position observables). And in this context for example the eigenstates for any spin component of a spin 1/2 particle are superpositions of eigenstates for other components.

Now you might (correctly) think that this talk of spin components means that we must still be thinking about directions in physical position space. But in fact the study of any two-valued observable (with values “Yes” and “No” or “True” and “False”) forces us to also consider other observables whose eigenstates are superpositions of the Yes and No eigenstates and whose relationship to the original observable are mathematically equivalent to those between spin components in different directions despite not actually having any connection with directions in physical space. And in quantum computing, although spin directions for a single particle are often used as a conceptual model, the choice of Yes/No observable might in practice be something different.

Of course, to study the progress of a quantum calculation, while disregarding space we still need to consider the time evolution of the system; but some aspects of the relationships between possible outcomes can be studied without reference to time in a similar way to how the time-independent Schrodinger equation can be used to study stationary states and energy levels of an atom or molecule. And in these types of analysis the question of whether and how some states are superpositions of other states is still relevant even though there is no reference to either space or time involved.

Source: (1001) Alan Cooper’s answer to Would quantum superpositions function in the absence of time and space? – Quora

Black Hole Fantasy

A black hole has been spotted heading towards Earth, and we have 200 years before it arrives. Does our species have a chance of survival?

Perhaps. But not on Earth. (Assuming any kind of astronomically plausible black hole, there is no way of avoiding substantial perturbation of the Earth’s orbit followed by tidal distortion and probable destruction of the Sun.)

However, it may be possible to colonize some asteroids and by small manipulation of their orbits ensure that they get slingshotted so as to end up at sufficient distance from the accretion disc (into which the sun and planets will be converted by tidal forces) so that the body of the asteroid is sufficient to shield its residents from the radiation.

Since the colonies will of necessity be small, most humans will be stuck on Earth and not survive; but if sufficient genetic diversity is brought along in the form of germ cells and/or frozen zygotes or blastocysts then perhaps the species itself might survive and adapt to the new low gravity environment.

Source: (1001) Alan Cooper’s answer to A black hole has been spotted heading towards Earth, and we have 200 years before it arrives. Does our species have a chance of survival? – Quora

Less Time for Same Distance? 

For the traveller it’s not the same distance; due to length contraction it’s a smaller distance, and so takes less time. For the observer who sees the distance as a full light year it appears to take more than a year, but the traveller’s clocks appear slowed down and so will advance by less than that. (The time experienced by the traveller will still be more than a year unless the travel speed  exceeds , ie just over 70% of the speed of light.)

Source: (1001) Alan Cooper’s answer to It takes light a year to travel a light year, but why would it take a person less time to travel the same distance due to time dilation? – Quora

Timelike vs Spacelike

If two distinct events are such that there is any inertial frame in which they have zero spatial distance between them, then there is no frame in which they are simultaneous and so they are said to be “timelike separated”. This is because the frame in which they have zero spatial separation corresponds to an observer who sees them both happening at the same place one after the other; and for any other inertial observer, the time between them is also nonzero (since for any v<c the Lorentz contraction factor is never zero).

On the other hand, any two events which are seen as simultaneous by some inertial observer (which is different from being seen simultaneously by that observer!) are said to be “spacelike separated”. But the appearance of simultaneity is relative to the observer and only happens in one particular frame. Other inertial observers won’t see the events as simultaneous but all will agree that it would take faster than light travel to see them both at the same place – eg to actually be present at both of them.

Source: (1001) Alan Cooper’s answer to Is there relativity in simultaneity for events that have distance between them from the prespective of one frame but don’t have distance between them from the prespective of another? – Quora

GR and Twin Paradox

General relativity theory does not “solve the twin paradox of special relativity”.

Despite being “paradoxical” in the sense of contradicting our intuition that the time ordering of separated events should be absolute, there is no “paradox” in the sense of internal contradiction in special relativity. Nor is it impossible to analyse the experience of an accelerated observer in special relativity; and in the case where one twin is turned back (eg by a rocket) this leads to the conclusion that both agree on the difference between their ages when reunited.

The only case in which general relativity is needed is when the acceleration is due to gravity (eg by slingshotting about a massive star) – and so does not lead to the feeling of applied force by the freely falling traveller. But as soon as gravity comes into the picture we are no longer talking about special relativity.

Source: (1001) Alan Cooper’s answer to How does the general relativity theory solve the twin paradox of special relativity? – Quora

Definite values and Eigenstates

A Quoran asks: Once we decide that physical states are represented as vectors in a Hilbert Space and observables as operators, can we deduce that states of definite values of an observable should be eigenstates of the corresponding operator?
 
Not in any conventional sense of the word “deduce”. (It is possible to “represent” physical states as vectors and observables as operators in all kinds of ways that have nothing to do with the representation that is useful in quantum theories.)
 
But if you are talking about the usual connection in quantum mechanics that we make between states and vectors and between observables and (self-adjoint) operators, then that includes also the condition that for the observable represented by operator [math]A[/math], the distribution of observed values, when it is measured in the state represented by vector [math]\psi[/math], is such that the probability of a measurement of [math]A[/math] being in any specified interval [math]L[/math] is given in terms of the spectral projector [math]E_{L}(A)[/math] by the inner product [math]\frac{\langle \psi |E_{L}(A) \psi\rangle}{\langle \psi |\psi\rangle}[/math] (or just [math]\langle \psi |E_{L}(A) \psi\rangle[/math] if [math]\psi[/math] is normalized with [math]\langle \psi |\psi\rangle=1[/math]).
 
And if you are requiring that we decide to adopt that representation, then it does follow (essentially immediately from that extra assumption) that a state in which the observable has a definite value is represented by an eigenvector of the corresponding operator.
 
PS The spectral projectors [math]E_{L}(A)[/math] of any observable [math]A[/math] also correspond to question-type observables giving the value 1 when [math]A[/math] is observed to give a value in [math]L[/math] and 0 otherwise, and the expectation value of any question is just the probability of its value being 1 (ie of the answer being “true”). So the assumption made above actually follows from the simpler version which just says that the expectation values of observables are always given by inner products of the form [math]\langle \psi |A \psi\rangle[/math] where [math]\psi[/math] is a corresponding normalized vector and [math]A[/math] is the corresponding observable (which includes the case of spectral questions about other observables).

Source: (1003) Alan Cooper’s answer to Once we decide that physical states are represented as vectors in a Hilbert Space and observables as operators, can we deduce that states of definite values of an observable should be eigenstates of the corresponding operator? – Quora

A Question About Geometrical Optics

How do you draw the image of the light source through the converging lens if the light sources are located higher than the top of the lens (it is not allowed to extend the lens)?

In a conventional diagram of the geometrical optics of any lens or mirror, the line through the source parallel to the axis of the optics does not necessarily represent an actual ray, but rather is included as a way of easily identifying where the image is located. It is used because the optics is assumed to have the property that all rays from the source to any part of the lens (or mirror) are bent in such a way as to converge on some other point (the image) and the mathematical form of the rule describing how this bending occurs makes sense for all points in the plane of the lens (or mirror) regardless of whether or not they are actually on the lens or mirror (and so regardless of whether or not the deflection actually occurs).

It is actually possible to calculate the angle of deflection for each ray (either by an algebraic formula or a geometric construction), and so to just use rays which pass through the lens (such as those through T and B, which I have shown as solid lines in the picture) without any reference to the imaginary ray (shown as a dashed line in the picture) which would bend at H if the lens was big enough; but the formula or construction is simplest for that imaginary ray and it works just as well as any other for determining the image point.

Source: (1003) Alan Cooper’s answer to How do you draw the image of the light source through the converging lens if the light sources are located higher than the top of the lens (it is not allowed to extend the lens)? – Quora