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(416) Adam Lantos’s answer to What are the three best reasons that support string theory? How do we know that this is a valid area of research? – Quora

Source: (416) Adam Lantos’s answer to What are the three best reasons that support string theory? How do we know that this is a valid area of research? – Quora

(416) Adam Lantos’s answer to What are the three best reasons that support string theory? How do we know that this is a valid area of research? – Quora

Source: (416) Adam Lantos’s answer to What are the three best reasons that support string theory? How do we know that this is a valid area of research? – Quora

(416) David Kahana’s answer to Can the Dirac Equation be simplified into the Schrodinger Equation in the non-relativistic case? – Quora

Source: (416) David Kahana’s answer to Can the Dirac Equation be simplified into the Schrodinger Equation in the non-relativistic case? – Quora

Can we draw a spacetime diagram from the POV of the travelling twin in the twin paradox?

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 $0.866 c = 0.866 l y / y$ (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 $1 g = 9.8 m / s^{2} = 1.03 l y / y^{2}$ . 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 $x = 0.866 t - 1 / 2 (1.03) t^{2}$ – which gives the usual simple parabolic plot.

But to figure out how the Earth’s position $x^{'}$ in traveller-coordinates depends on the traveller’s time $t^{'}$ 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 $(t, x (t))$ , the traveller’s clock appears to the Earth-based observer to be advancing at the rate $1 / γ = \sqrt{1 - v (t)^{2}}$ where $v (t) = 0.866 - 1.03 t$ is the velocity as seen from the Earth at time $t$ . So to find the time showing on the traveller’s clock we just have to integrate to get $\begin{aligned} t^{'} & = \int_{0}^{t} \sqrt{1 - (0.866 - 1.03 τ)^{2}} d τ \\ = - 1 / 1.03 (1 / 2) (\arcsin (0.866 - 1.03 τ) \\ + (0.866 - 1.03 τ) \sqrt{1 - (0.866 - 1.03 τ)^{2}}) |_{0}^{t} \end{aligned}$ …….

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 $t_{s}$ in the spaceship’s coordinates, if the Earth appears to be at a position relative to the spaceship given by $x_{s e} (t_{s})$ with its clock showing a time of $t_{e} (t_{s})$ then $\frac{d x_{s e} (t_{s})}{d t_{s}} = v_{s} (t_{s})$ , and if the spaceship is unaccelerated then at a time $Δ t_{s}$ after $t_{s, 0}$ the Earth’s new position will be at $x_{s e} (t_{s, 0}) + v_{s} (t_{s, 0}) Δ t_{s}$ , but if the spaceship is feeling an acceleration of $a (t_{s})$ the new inertial tangent worldline will be given by $x_{s} = a (t_{s, 0}) Δ t_{s} (t_{s} - (t_{s, 0} + Δ t_{s}))$ and the corresponding simultaneity space will be given by $t_{s} = t_{s, 0} + Δ t_{s} + (a (t_{s, 0}) Δ t_{s}) x_{s}$ so $t_{s} - t_{s, 0} = Δ t_{s} + (a (t_{s, 0}) Δ t_{s}) x_{s}$

which intersects the Earth’s worldline (given by $x_{s} = x_{s} (t_{s, 0}) + v_{s} (t_{s, 0}) (t_{s} - t_{s, 0})$ ) at $x_{s} = x_{s} (t_{s, 0}) + v_{s} (t_{s, 0}) (Δ t_{s} + (a (t_{s, 0}) Δ t_{s}) x_{s})$ so $x_{s} - v_{s} (t_{s, 0}) (a (t_{s, 0}) Δ t_{s}) x_{s} = x_{s} (t_{s, 0}) + v_{s} (t_{s, 0}) (Δ t_{s})$

so $x_{s} = \frac{x_{s} (t_{s, 0}) + v_{s} (t_{s, 0}) Δ t_{s}}{1 - v_{s} (t_{s, 0}) a (t_{s, 0}) Δ t_{s}}$ , $t_{s} = \frac{a (t_{s, 0}) x_{s} (t_{s, 0})}{1 - v_{s} (t_{s, 0}) a (t_{s, 0}) Δ t_{s}}$ .

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.

(405) What is a tensor product in simple words? – Quora

For any two vector spaces $V$ and $W$ , the tensor product $V ⨂ W$ is the space of bilinear functions on $V \times W$ .

If $V$ and $W$ are inner product spaces then for any $v \in V$ and $w \in W$ we can define the pure tensor $v \otimes w$ in $V ⨂ W$ by $v \otimes w (v *, w *) = (v \cdot v *) (w \cdot w *)$ also often written by physicists as $⟨ v | v * ⟩ ⟨ w | w * ⟩$ . But there are also elements of $V ⨂ W$ that are not of the pure tensor form.

For example if $v_{1} \otimes w_{1} + v_{2} \otimes w_{2}$ could be written in the form $(a_{1} v_{1} + a_{2} v_{2}) \otimes (b_{1} w_{1} + b_{2} w_{2})$ , then for all $v \in V$ and $w \in W$ , we’d need $\begin{aligned} (a_{1} v_{1} + a_{2} v_{2}) \otimes (b_{1} w_{1} + b_{2} w_{2}) (v, w) \\ = (a_{1} v_{1} \cdot v) (b_{1} w_{1} \cdot w) + (a_{1} v_{1} \cdot v) (b_{2} w_{2} \cdot w) \\ + (a_{2} v_{2} \cdot v) (b_{1} w_{1} \cdot w) + (a_{2} v_{2} \cdot v) (b_{2} w_{2} \cdot w) \\ = (v_{1} \cdot v) (w_{1} \cdot w) + (v_{2} \cdot v) (w_{2} \cdot w) \end{aligned}$ .

But this is only true if $a_{1} b_{1} = a_{2} b_{2} = 1$ and $a_{1} b_{2} = a_{2} b_{1} = 0$ , but if one of $a_{1}$ or $b_{2}$ is zero then one of $a_{1} b_{1}$ or $a_{2} b_{2}$ must be also.

Source: (405) What is a tensor product in simple words? – Quora

(354) Is a system of two entangled particles a pure state, and are its subsystems in a mixed state? – Quora

Not necessarily (but the pure state case is often what we study), and not exactly (but kind of).

WITH REGARD TO THE FIRST QUESTION:

I don’t think it is necessary to restrict the concept of entanglement to pure states. One could do so, but it would be odd to describe a state with a 50% chance of being in an entangled pure state as “not entangled”.

I would therefore say that a composition of two systems can be said to be in an entangled state whenever that state is represented by a density operator that includes at least one term which is not a pure tensor product of state vectors for the corresponding component systems.

WITH REGARD TO THE SECOND QUESTION:

In order to make sense of this question we need so decide what is meant by the state of a subsystem of a composite system. The standard approach is to use the “relative state” which is implemented mathematically by a “partial trace” which gives “marginal probability” distributions of observed quantities. And if one does that, then yes, the relative state of a subsystem may indeed be mixed state even for a pure state of the combined system.

Before going on I should point out that it is important to understand both the distinction between a statistical mixture and a linear superposition, and the fact that for a composite system the states that are “pure tensors” are just a subset of all the pure states.

First, with regard to mixtures vs superpositions:

For any system, mixed states (which are classical statistical mixtures of pure states) can be represented by so-called density matrices which are operators of the form $ρ = Σ p_{i} | ψ_{i} ⟩ ⟨ ψ_{i} |$ where $p_{i}$ is the probability of being in pure state $ψ_{i}$ and the operator $| ψ_{i} ⟩ ⟨ ψ_{i} |$ is just the projector onto the subspace spanned by the state vector $| ψ_{i} ⟩$

The expectation of observable $O$ in state $ρ$ is then given by the trace $⟨ O ⟩ = T r (ρ O) = Σ p_{i} ⟨ ψ_{i} | O ψ_{i} ⟩$ , which is just the overall expectation from a process which gives the expected value for state $| ψ_{i} ⟩$ with probability $p_{i}$ . The case of a pure state $ψ$ corresponding to a single vector $| ψ ⟩$ can also be represented by a density matrix in which the sum has only one term and so the density matrix is a one dimensional projector (ie of rank one), and the trace formula gives $⟨ O ⟩ = T r (| ψ ⟩ ⟨ ψ | O) = ⟨ ψ | O ψ ⟩$ which is just the usual form for the expectation value.

Another quite different way of combining states is by way of linear combination of state vectors to create what is called a superposition. This is different from the classical mixture because the expected average value of an observable $O$ in the superposition state $| ψ ⟩ = c_{1} | ψ_{1} ⟩ + c_{2} | ψ_{2} ⟩$ is given by $| c_{1} |^{2} | ⟨ ψ_{1} | O ψ_{1} ⟩ + c_{1}^{*} c_{2} ⟨ ψ_{1} | O ψ_{2} ⟩ + c_{2}^{*} c_{1} ⟨ ψ_{2} | O ψ_{1} ⟩ + | c_{2} |^{2} | ⟨ ψ_{2} | O ψ_{2} ⟩$ and the cross terms represent the fact that, if $| ψ_{i} ⟩$ are not both eigenstates of $O$ , then observation of $O$ has a mixing effect which produces interference between them.

Second, with regard to composite systems:

If the systems A and B have pure state vectors of the form $| a_{α} ⟩$ and $| b_{β} ⟩$ in Hilbert spaces $H_{A}$ and $H_{B}$ , then any (normalized) linear combination of pure tensors corresponds to a pure state of the combined system.

The special feature of pure tensors is that they represent states in which the properties of the two subsystems are statistically independent, sometimes called separable states. But in most states (which are formed by taking linear combinations of pure tensors) the properties are correlated (and we say that in such other states, which are not represented by pure tensors, the two systems are “entangled”).

In this setting any pure tensor corresponding to a state vector of the form $| a_{α} ⟩ \otimes | b_{β} ⟩$ has a rank one density matrix $ρ = (| a_{α} ⟩ \otimes | b_{β} ⟩) (⟨ a_{α} | \otimes ⟨ b_{β} |) = | a_{α} b_{β} ⟩ ⟨ a_{α} b_{β} |$ and is not entangled.

But neither is any classical statistical mixture of such states with density matrix $ρ = Σ p_{i} | ψ_{i} ⟩ ⟨ ψ_{i} |$ with each contributing $ψ_{i}$ being a pure tensor of the form $ψ_{i} = | a_{i} ⟩ \otimes | b_{i} ⟩$ .

On the other hand an entangled state might be pure state represented by a vector of the form $c_{1} | a_{1} ⟩ \otimes | b_{1} ⟩ + c_{2} | a_{2} ⟩ \otimes | b_{2} ⟩$ (which is a linear superposition of pure tensors ), but it might also be a classical mixture in which one or more of the possible $ψ_{i}$ is of that form.

In order to make sense of the second question we need so decide what is meant by the state of a subsystem of a composite system.

For an unentangled (pure tensor product) pure state it is natural to take the corresponding factor in the tensor product. For an entangled state it is less obvious what to do, but it makes sense to think of the relative state of system A as giving any observable $O_{A}$ the expectation value that results from observing the combined state but ignoring the state of system B. This amounts to observing the identity operator in system B so the corresponding observable on the combined system would correspond to the operator $O_{A} \otimes I_{B}$

For the pure entangled state $ψ = c_{1} | a_{1} ⟩ \otimes | b_{1} ⟩ + c_{2} | a_{2} ⟩ \otimes | b_{2} ⟩$ , if $⟨ b_{1} | b_{2} ⟩ = 0$ , this gives the expectation $\begin{aligned} ⟨ ψ | (O_{A} \otimes I_{B}) ψ ⟩ \\ = | c_{1} |^{2} ⟨ a_{1} | O_{A} a_{1} ⟩ ⟨ b_{1} | I_{B} b_{1} ⟩ + c_{1}^{*} c_{2} ⟨ a_{1} | O_{A} a_{2} ⟩ ⟨ b_{1} | I_{B} b_{2} ⟩ \\ + c_{2}^{*} c_{1} ⟨ a_{2} | O_{A} a_{1} ⟩ ⟨ b_{2} | I_{B} b_{1} ⟩ + | c_{2} |^{2} ⟨ a_{2} | O_{A} a_{2} ⟩ ⟨ b_{2} | I_{B} b_{2} ⟩ \\ = | c_{1} |^{2} ⟨ a_{1} | O_{A} a_{1} ⟩ + | c_{2} |^{2} ⟨ a_{2} | O_{A} a_{2} ⟩ \end{aligned}$

which corresponds to the mixed state for system A with probability $| c_{i} |^{2}$ of being in state $a_{i}$ (and density matrix $ρ = | c_{1} |^{2} | a_{1} ⟩ ⟨ a_{1} | + | c_{2} |^{2} | a_{2} ⟩ ⟨ a_{2} |$ ).

This procedure of mapping the density matrix $ρ = | ψ ⟩ ⟨ ψ |$ which is an operator in $H_{A} ⨂ H_{B}$ to an operator on just $H_{A}$ is often referred to as taking the partial trace.

Source: (354) Is a system of two entangled particles a pure state, and are its subsystems in a mixed state? – Quora

If photons travel at the speed of light in a vacuum, and are said to “slow” when they travel through different media but do they really “slow” or do they just bounce off more atoms and thus travel farther and seem to slow? – Quora

Photons don’t really “travel” at all. They are just units of energy exchange between the electromagnetic field and other fields such as that of the electron (or any other charged “particle”) and only “exist” at the events (positions and times) of energy transfer, by “emission”, into or, by “absorption”, out of the EM field.

It is only after an emission and absorption event have both been observed that we describe the process as corresponding to the motion of a photon, but really all that has happened might just as well be described by saying that a wave of changes in the EM field propagated from the emission event, and that at any point in spacetime (given the presence of a suitably excitable atom), the probability of an absorption event is proportional to the magnitude of the appropriate frequency component in the EM field at that point in space and time.

But the propagation of the EM field obeys Maxwell’s equations in empty space, and through a medium which can be approximated as consisting of classical particles it is governed by the same equations but with reflected waves from all the particles interfering with one another in such a way that the overall effect is like a wave moving more slowly. So there is something like the “bouncing of all the electrons idea” going on even in the classical case.

Also, the propagation of disturbances in the EM field from one point at one time to another point at a later time can be calculated by a process which amounts to considering all possible paths by which an imagined particle could end up at the right place and time (taking into account all of the ways that imagined particle could bounce off other things in its environment), and using something like a sum of contributions from all such paths to give the field values at the later event. So although there isn’t actually any particular path followed by any real photon, the effect we see is kind of like the average over all possible paths, and the presence of all those electrons allows longer paths to contribute in such a way that the overall effect is as if there was a photon travelling more slowly than it would have if it had only been able to go in a straight line.

Or something like that.

Source: (310) Alan Cooper’s answer to If photons travel at the speed of light in a vacuum, and are said to “slow” when they travel through different media but do they really “slow” or do they just bounce off more atoms and thus travel farther and seem to slow? – Quora

Do thought experiments really uncover new scientific truths? | Aeon Essays

Well, I’m late to the party here, as the article linked below is now five years old; but I do have two comments.

Source: Do thought experiments really uncover new scientific truths? | Aeon Essays

One is that Norton is right. ALL valid thought experiments are just logical deductions of consequences of things that we have learned from experiment seem to always be true. Case in point being Einstein’s deduction of Lorentz transformations from the assumption that speed of light is invariant which is not itself deduced but extrapolated from observations (either Michelson-Morley or tests of Maxwell’s equations). In fact anyone who designs a machine is doing the same thing – thinking about logical consequenses of the assumption that previously observed patterns will continue to apply. And if the assumptions have been well tested then we strongly expect the derived laws to apply in just the same way as we expect the machine to work.

My other point is that Galileo’s beautiful argument is NOT valid. It is basically circular in that there is no reason other than prior experience to expect that two cannonballs will not fall faster when close together than when far apart. In fact it is easy to imagine scenarios where combining objects (such as water droplets) does increase their rate of fall and there is no a priori reason why all space might not be filled with a permeable but resistive medium in which the drag effect is reduced by proximity (as for cyclists in a peloton for example).