All – Page 14 – Physics Notes

In conventional spacetime diagrams (where inertial worldlines and lightlines are all represented by straight lines) the worldlines of accelerated observers must be bent (and so cannot be used as cartesian coordinate axes). If the acceleration of an observer is small enough and brief enough for velocity changes to be small relative to the speed of light, then the experience of that observer can be usefully approximated by treating it as inertial. (But how small the velocity changes must be kept depends on how accurately we are able to measure things, and the limiting factor is that the accelerations must be small and brief enough that relativistic effects are smaller than the experimental errors.)

In the case of a traveler passing an inertial observer at high speed and then decelerating and returning (as pictured in the diagram from your comment),

the world lines of the traveler and the inertial observer cannot both be represented by straight lines from the first meeting to the second since they intersect at two points and are not the same. In this example the velocity change is fixed and substantial, so even if the acceleration is very small then the time has to be very long and relativistic effects of acceleration may accumulate to noticeable amounts even if they are not noticeable over smaller time scales.

So, since a conventional spacetime diagram requires all inertial world lines to be straight we cannot draw such a diagram for the whole story in which the traveller’s world line is the t-axis.

We can however draw such diagrams for short time intervals in which the traveller’s velocity does not change too much.

And by piecing together such diagrams we can get a picture of what the story looks like in terms of the traveler’s coordinates.

(I might call this a “spacetime diagram” from the point of view of the traveler but since it results in a bent world line for the inertial observer it is NOT a spacetime diagram in the conventional sense.)

But note that the resulting picture is NOT a “mirror image” of the diagram drawn in terms of the inertial observer’s coordinates!

Now let’s see in more detail how the above diagram comes about.

For concreteness we might consider the case of a traveller which passes by the Earth at a speed of [math]0.866c=0.866ly/y[/math] (which gives a gamma factor of 2) and appears to an Earth-based observer to be subject to a constant acceleration back towards the Earth of [math]1g=9.8m/s^2=1.03ly/y^2[/math]. This brings the traveller back to pass by the Earth again (at the same speed in the opposite direction) after about 2(0.866)/1.03=1.64 years (ie about 20 months) have passed on Earth. Of course, this scenario of constant acceleration relative to the Earth would, if extended in time, lead to impossible situation of exceeding light speed. So although it might be possible to maintain such a constant Earth-relative acceleration for the duration of the trip it would be more realistic to consider the case of constant acceleration as perceived by the traveler. We’ll look at both cases case for now (and probably just go with whichever turns out to be easier to analyse).

In order to plot the two diagrams above, we’ll consider the situation as seen at several times according to the clock of the observer whose coordinates are being used.

For the Earth-based observer’s view of the situation with constant Earth-based acceleration this is easy as the path of the traveler is defined in terms of Earth-based coordinates. At time t years on the Earth clock the traveler is at a distance, in light years, of [math]x=0.866t-1/2(1.03)t^2[/math] – which gives the usual simple parabolic plot.

But to figure out how the Earth’s position [math]x'[/math] in traveller-coordinates depends on the traveller’s time [math]t'[/math] is a more interesting exercise.

First we note that because of the changing velocity we can’t just apply a single Lorentz transformation. For example at the event whose Earth-based coordinates are [math](t,x(t))[/math], the traveller’s clock appears to the Earth-based observer to be advancing at the rate [math]1/\gamma=\sqrt{1-v(t)^2}[/math] where [math]v(t)=0.866-1.03t[/math] is the velocity as seen from the Earth at time [math]t[/math]. So to find the time showing on the traveller’s clock we just have to integrate to get [math]\begin{align}t’&=\int_0^t \sqrt{1-(0.866-1.03\tau)^2}d\tau\\&=-1/1.03 (1/2)(\arcsin(0.866-1.03\tau)\\&+(0.866-1.03\tau)\sqrt{1-(0.866-1.03\tau)^2})|_0^t\end{align}[/math]…….

On the other hand, if we are given the acceleration as felt by the spaceship then we can analyse the spaceship’s view as follows:

At a time [math]t_s[/math] in the spaceship’s coordinates, if the Earth appears to be at a position relative to the spaceship given by [math]x_{se}(t_s)[/math] with its clock showing a time of [math]t_e(t_s)[/math] then [math]\frac{dx_{se}(t_s)}{dt_s}=v_s(t_s)[/math], and if the spaceship is unaccelerated then at a time [math]\Delta t_s[/math] after [math]t_{s,0}[/math] the Earth’s new position will be at [math]x_{se}(t_{s,0})+v_s(t_{s,0})\Delta t_s[/math], but if the spaceship is feeling an acceleration of [math]a(t_s)[/math] the new inertial tangent worldline will be given by [math]x_s=a(t_{s,0})\Delta t_s(t_s-(t_{s,0}+\Delta t_s))[/math] and the corresponding simultaneity space will be given by [math]t_s=t_{s,0}+\Delta t_s+(a(t_{s,0})\Delta t_s)x_s[/math] so [math]t_s – t_{s,0}= \Delta t_s+(a(t_{s,0})\Delta t_s)x_s[/math]

which intersects the Earth’s worldline (given by [math]x_s=x_s(t_{s,0})+v_s(t_{s,0})(t_s – t_{s,0})[/math] ) at [math]x_s=x_s(t_{s,0})+v_s(t_{s,0})( \Delta t_s+(a(t_{s,0})\Delta t_s)x_s)[/math] so [math]x_s – v_s(t_{s,0})(a(t_{s,0})\Delta t_s)x_s =x_s(t_{s,0})+v_s(t_{s,0})( \Delta t_s)[/math]

so [math]x_s=\frac{x_s(t_{s,0})+v_s(t_{s,0}) \Delta t_s}{1-v_s(t_{s,0})a(t_{s,0})\Delta t_s}[/math] , [math]t_s=\frac{a(t_{s,0})x_s(t_{s,0})}{1-v_s(t_{s,0})a(t_{s,0})\Delta t_s}[/math].

Source: (95) Alan Cooper’s answer to Can you draw a spacetime diagram from the POV of the travelling twin in the twin paradox with very slow turnaround (VSTTP paradox)? The travelling twin is approximately in an inertial frame. See the comment at the question for details. – Quora

(508) Alan Cooper’s answer to There is no twin paradox if, when resolving, one is assuming that the Earth is at rest. How do the diagrams compare with the Earth(and earthbound twin)making the voyage and the shuttle/travelling twin at rest? – Quora

“If we measure a distance x at time t from Earth to the spaceship, then we must measure a distance of -x from the spaceship to the twin.” Who is “we” here? If “we” are both using the Earth’s reference frame it is true. But those distances “at time t” are between the Earth and spaceship at the same time from the point of view of the Earth. And when someone moving with the spaceship measures the distance they get a smaller value because what the Earth thinks of as at the same time as when Earth’s clock said time t is to the spaceship actually not at the same time but somewhat earlier or later (depending of direction of motion and due to the different corrections they make for light travel time by using the same speed of light despite the fact that they are moving relative to one another). When the outbound spaceship thinks it is at the same time as the Earth clock showing time t it is actually earlier so the distance between the Earth’s clock showing time t and the spaceship at what it thinks is the same time is less than the Earth’s idea of that distance.

Source: (508) Alan Cooper’s answer to There is no twin paradox if, when resolving, one is assuming that the Earth is at rest. How do the diagrams compare with the Earth(and earthbound twin)making the voyage and the shuttle/travelling twin at rest? – Quora

(425) What is quantum entanglement explained as simple as possible? – Quora

Source: (425) What is quantum entanglement explained as simple as possible? – Quora

Entanglement is a relationship between two or more systems such that observations of one provides information about another.

Perhaps the first thing one needs to know about quantum entanglement is how it compares to the classical entanglement that we are all familiar with (and which no-one gets particularly excited about). So let’s start with the classical case.

If we put two distinct items (eg black&white balls, left&right gloves, or Bertelmann’s socks) in identical boxes, and send them far apart after making a random choice to decide which box goes in which direction, then an observer who opens a box and sees one of the items knows immediately what’s in the other.

Before the observation, the state of the system is not known precisely but instead can be described as a probabilistic mixture which “collapses” as soon as the observation is made. But of course we all understand that the probabilistic nature of the situation depends on the observer, and the collapse seen by the first observer doesn’t happen for the second until either the second box is opened or a message from the first observer travels between them.

In the classical case, all of the observer-dependent randomness can be attributed to a lack of knowledge, but in the quantum situation at least some of the randomness appears to be essential in that sometimes we cannot determine one quantity without introducing randomness in another. It’s as if the boxes contained a pair of socks such that if one looks red then the other looks green, but if we look in different light the possible colours are yellow and blue. But a sock that looks either yellow or blue is equally likely in the original light to appear red or green.

Actually by cleverly packing random pairs of socks it is possible to reproduce those (red-green vs yellow-blue) quantum predictions in a perfectly classical way, but there are also intermediate situations, analogous to lightings in which the colours are orange and turquoise, and what quantum theory predicts about how an observation of either orange or turquoise affects the probabilities of red vs green and yellow vs blue is slightly outside the range of anything that could have been caused by any process of selecting from a random selection of pre-packed sock pairs.