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

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

Block Universe Nonsense

A Quora question asks: Is it true that the past still exists and the future is predetermined? Is the ‘’block universe’’ correct, since Special relativity is scientifically proven? – Quora

No. It’s not true that in the context of Special Relativity “the past still exists”.

In fact, according to Special Relativity (which does appear to be correct for situations where the effects of gravity are negligible – though nothing outside of mathematics is ever “scientifically proven”), the whole idea of the (unique) past is meaningless since there are events that you consider past that are not in the past for a remote relatively moving observer who is at an event which you consider simultaneous with your present.

Also, again according to Special (and maybe* General) Relativity, the word “still” is also meaningless as its use at any time refers to events that are simultaneous with the statement, but that criterion will correspond to different sets of events for different observers.

*- I have added the word “maybe” since Viktor Toth# has mentioned that in General (but not Special) Relativity it is possible to identify a preferred set of inertial frames (in which the redshift due to expansion is the same in all directions), and these do share a common concept of simultaneity.

But of course, even in a Galilean universe, there is also the question of what you mean by saying that the past “exists”. Yes it exists as the past (and even in SR the past of my worldline exists as my past) but that does not mean it is happening now. And the same applies to the future. Even in a non-deterministic universe the future “exists”, as what will turn out to have happened, regardless of the fact that we have no way of predicting it.

#PS I am not linking to Viktor Toth’s answer because, despite having some good bits, I feel that it is marred by the inclusion of some very misleading (and quite unnecessary) statements about Quantum Mechanics.

Source: (1003) Alan Cooper’s answer to Is it true that the past still exists and the future is predetermined? Is the ‘’block universe’’ correct, since Special relativity is scientifically proven? – Quora

Another Quora Question

A Quora question asks: “How does quantum physics know that if a system were not measured it would be in multiple possible states (without measuring it), and that when measuring it collapses into one definite state?

Quantum theory doesn’t “know” anything. All it does is describe what we know and makes predictions about what we may find out in future. What we call the “state” of a system is just a summary of what we know about it, and a system on which we can gain no new information without losing some of what we already have is said to be in a “pure” state.

An example is the case of a single electron whose position we are ignoring (so it can be considered fixed) and whose only measurable property is the direction of its spin. If we first measure the spin component in the vertical direction (say that of the z-axis), then we will always find that the spin is pointing straight up or straight down; and if we repeat that measurement we will see that the direction is unchanged. But if we then follow that with a measurement in any perpendicular direction then we have a 50% chance of finding that the spin is now pointing in that new direction and 50% chance of its opposite. And if we now return to the original direction we find equal chances for pointing up or down.

Here, the spin up state is a pure state because we cannot determine the sideways component without losing information about the vertical component, but being in the up state is not the same as being in both left and right states at the same time, and it is also not the same as being in a statistical mixture (ie in one or other of those two states but we just don’t know which).

P.S. In either quantum or classical physics, a system that is not measured or which we have only measured incompletely may be in any one of several pure states with different probabilities, and we call the resulting state a statistical mixture; but that is not the same as being in multiple pure states at once.

Source: (1002) Alan Cooper’s answer to How does quantum physics know that if a system were not measured it would be in multiple possible states (without measuring it), and that when measuring it collapses into one definite state? – Quora