general relativity - Apple falls for which of these 2 reasons?

Needles to say I am a visitor here. I do not belong to the science world;)

But I have read both of these things before:

1. Apple falls to the ground because curved spacetime pushes it there (same force as keeps moon in orbit)



2. Apple 'falls' to ground because the ground is rushing up to meet the apple (which is actually suspended in space) because of Earth's acceleration through space.

I don't think these can both be true. I'd appreciate any clarification - thank you.

Answer

The difficulty here is one of definitions.

• In the pre-Einstein-ian view there is a privileged frame (typically taken as that of the "fixed" stars), and whichever body is experiencing the smallest acceleration with respect to that frame (which is to say the Earth by a large margin) would have a stronger claim to being "still" Thus the apple accelerates and the ground is still.


• In general relativity it is inertia frames that are special, and you can tell if you are in one by setting a test mass next to and letting go of it. If it stays there you are in an inertia frame. In that view, the apple is still and the ground comes rushing up to it. Note that the Earth as a whole is free falling and is therefore in an inertial frame, but object on the surface are not.

At the kinds of energies that apply to falling apples on Earth you can do physics correctly in either view. We teach the former in physics 101, but the latter has a pretty strong claim to being more fundamental.

knowledge - Recondite key locked another puzzle... - Clue Twenty Nine!

<sub><<---First clue

<---Previous clue</sub>

---

_{A note from @Mithrandir: I have given explicit permission for @Techidiot to post this, and gave him the answer to encode. If you want to post one, ping @Mithrandir in chat and we'll talk.}

---

The door slides open, revealing a dimly-lit - you can't see how - small room, with a short wooden desk and a plastic swivel chair. On the desk is a keypad and several pieces of paper. You pick one of them up. It appears to be a crossword:

ACROSS

1. Los Angeles, operating system(4)
2. Back, if around climbing unfinished odd juice(1)

3. Most of the men within Japanese currency(19)
4. Hollow, idea covering starts newly designed internet(16)
5. Weird, crane following after(11)
9. Hidden strategy, pteropodans follow!(10)
10. Crocodile clipped impala's ends after a time(9)
12. Thousand glasses cover first sample of virus(12)
14. Inside each puzzled trier(2)
15. Ring man(13)
16. Queen and Rook, captured aces capturing time(5)

DOWN

1. First, I became yours in chuckles without hugs(18)
4. It always saw left(8)
6. Completely empty page rust endlessly(3)
7. Endless tale going westwards met en route(17)
8. Gut less month, before unfinished trial(15)
9. Part of ten glands!(6)
11. Incomplete, army starts eliminating novices with one area(7)
13. Clumsy final destination, ends back(14)

Interesting, you think. You pick up the other papers. They appear to be the next stage of the puzzle:

A key for the next part is lost in the crossword above..Find it and go for the next part...

 Nbznv. Jbvknu whtg. Jxqz - Vmpmurjz.

And then...

_ _ _ _ 

---

_{Next clue--->}

Answer

Solutions to the clues

ACROSS

1. Los Angeles, operating system(4)

LA+OS $\rightarrow$ LAOS

2. Back, if around climbing unfinished odd juice(1)

F(IJ)I (Backwards if is fi around, odd juice, unfinished ji, climbing ij) $\rightarrow$ FIJI

3. Most of the men within Japanese currency(19)

YE(ME)N $\rightarrow$ YEMEN

4. Hollow, idea covering starts newly designed internet(16)

I(NDI)A $\rightarrow$ INDIA

5. Weird, crane following after(11)

F+RANCE $\rightarrow$ FRANCE

9. Hidden strategy, pteropodans follow!(10)

strategy, pteropodans $\rightarrow$ EGYPT

10. Crocodile clipped impala's ends after a time(9)

CRO+aT+IA $\rightarrow$ CROATIA

12. Thousand glasses cover first sample of virus(12)

K+O(SOV)O $\rightarrow$ KOSOVO

14. Inside each puzzled trier(2)

E(RITRE)A $\rightarrow$ ERITREA

15. Ring man(13)

O+MAN $\rightarrow$ OMAN

16. Queen and Rook, captured aces capturing time(5)

Q(A(T)A)R $\rightarrow$ QATAR

DOWN

1. First, I became yours in chuckles without hugs(18)

L(IBY)A (Laughs-hugs=LA) $\rightarrow$ LIBYA

4. It always saw left(8)

IT+ALY(always-saw) $\rightarrow$ ITALY

6. Completely empty page rust endlessly(3)

CY(completely empty) + P+RUS $\rightarrow$ CYPRUS

7. Endless tale going westwards met en route(17)

LAT(TAL westwards) + VIA(en route) $\rightarrow$ LATVIA

8. Gut less month, before unfinished trial(15)

AUS(AUGUST-GUT) + TRIA $\rightarrow$ AUSTRIA

9. Part of ten glands!(6)

ten glands $\rightarrow$ ENGLAND

11. Incomplete, army starts eliminating novices with one area(7)

ARM+E+N+I+A $\rightarrow$ ARMENIA

13. Clumsy final destination, ends back(14)

FINLA+ND(DN back) $\rightarrow$ FINLAND

Here's the crossword:

Then, taking the highlighted letters:

In the order given by the 'enumerations' we get -

FIJI F
ERITREA I

CYPRUS R
LAOS S
QATAR T
ENGLAND L
ARMENIA E
ITALY T
CROATIA T
EGYPT E
FRANCE R
KOSOVO S

OMAN O
FINLAND F
AUSTRIA T
INDIA I
LATVIA T
LIBYA L
YEMEN E

FIRSTLETTERSOFTITLE $\rightarrow$ FIRST LETTERS OF TITLE

Applying that, we can get

Using question title: RKLAP, sentence starting with WRONG.
Using Crossword title: WTTGC: 'Right. Nicely done. Next - Tqwtopng.'

The next step is

to visit , recognize that this represent the G4 Nations (thanks @Sp3000) and take the letters at the specified points of their names. This yield the final answer -
Brazil
Germany
Japan
India

The final answer is - LEAD

electromagnetism - CGS Units for Magnetism

Why does the formula for magnetic field force include the speed of light in the denominator in cgs units? Where does the extra $c$ go in SI units?

Answer

The natural way to write it in this notation is

$$F = q(E + \beta \times B)$$

where $\beta$ is the velocity measured in natural units - the velocity as a fraction of the speed of light.

In the CGS system, we instead write $\beta = \frac{v}{c}$ and the equation becomes

$$F = q(E + \frac{v}{c} \times B)$$

That's not silly enough, though, so we go really crazy and invent two new constants and give them obscure names: "permittivity of free space" ($\epsilon_0$) and "permeability of free space" ($\mu_0$) and relate them so $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

If that's where it ended the equation would be

$$F = q(E + v \sqrt{\mu_0 \epsilon_0} \times B)$$

But that would still not be silly enough. Instead, we then decided to invent a new magnetic field given by $B' = \sqrt{\mu_0/4\pi} B$ and a new electric field given by $E' = \frac{E}{\sqrt{4\pi \epsilon_0}}$. This would give us the new formula

$$F = q(\sqrt{4\pi \epsilon_0}E' + v \times \sqrt{4\pi \epsilon_0} B')$$

but we need to be more sillier, so we define a new measure of charge by $q' = q\sqrt{4\pi\epsilon_0}$. At last we get

$$F = q'(E' + v \times B')$$

and there you have the SI Lorentz force law. So the answer is that the speed of light appears only because of complicated ways of changing the units.

It's a bit instructive to look at the energy densities of the fields. In the original units they're just $\frac{E^2}{8\pi}$ and $\frac{B^2}{8\pi}$. But we made a bunch of units changes, so these become $\frac{4\pi \epsilon_0 E'^2}{8\pi} = \frac{\epsilon_0 E'^2}{2}$ and $\frac{4\pi B'^2}{8 \pi\mu_0} = \frac{B'^2}{2\mu_0}$

So the cost of making the speed of light go away in the Lorentz force law is that we pick up these strange $\mu_0$ and $\epsilon_0$ constants that hide the speed of light and flop around through all the subsequent formulas.

event horizon - Photons hitting the Black Hole Singularity

I was watching one of Dr. Susskind's Lectures on String theory and he was talking about black hole singularities (see the 40 minute mark of this lecture http://www.youtube.com/watch?v=SkEvsxg5Tu4&list=PL3E633552E58EB230). I'm curious, given that the singularity becomes a 'time' and not a 'place', if at the moment you cross the horizon, you emit a photon, according to the explanation given, does this mean that it would take an infinite amount of time for the photon to hit the singularity? And this would also mean that the faster you move after crossing the horizon, the longer it would take to hit the singularity (which seems kind of cool)? I'm just curious to know if I'm interpreting the explanation correctly.

Answer

I'm curious, given that the singularity becomes a 'time' and not a 'place', if at the moment you cross the horizon, you emit a photon, according to the explanation given, does this mean that it would take an infinite amount of time for the photon to hit the singularity?

Time is relative in general relativity, so questions like this do not automatically have well-defined answers.

If you want to know how much time a material object like a rock takes to hit the singularity, the most natural way of defining this is to ask how much time will elapse on a clock attached to the rock. The answer is then a finite number that depends on the size of the black hole. It isn't infinite.

In your example of a photon, we don't have this option. We can't duct-tape a clock to a photon, so it's not possible to define the amount of time experienced by a photon. Therefore the answer to your question is fundamentally undefined. It's not that we don't know the answer -- the question just isn't meaningful. It's like asking whether happiness is blue.

quantum mechanics - Is there something behind non-commuting observables?

Consider a quantum system described by the Hilbert space $\mathcal{H}$ and consider $A,B\in \mathcal{L}(\mathcal{H},\mathcal{H})$ to be observables. If those observables do not commute there's no simultaneous basis of eigenvectors of each of them. In that case, in general if $|\varphi\rangle$ is eigenvector of $A$ it will not be of $B$.

This leads to the problem of not having a definite value of some quantity in some states.

Now, this is just a mathematical model. It works because it agrees with observations. But it makes me wonder about something. Concerning the Physical quantities associated to $A$ and $B$ (if an example helps consider $A$ to be the position and $B$ the momentum) what is really behind non-commutativity?

Do we have any idea whatsoever about why two observables do not commute? Is there any idea about any underlying reason for that?

Again I know one might say "we don't care about that because the theory agrees with the observation", but I can't really believe there's no underlying reason for some physical quantities be compatible while others are not.

I believe this comes down to the fact that a measurement of a quantity affects the system in some way that interferes with the other quantity, but I don't know how to elaborate on this.

EDIT: I think it's useful to emphasize that I'm not saying that "I can't accept that there exist observables which don't commute". This would enter the rather lengthy discussion about whether nature is deterministic or not, which is not what I'm trying to get here.

My point is: suppose $A_1,A_2,B_1,B_2$ are observables and suppose that $A_1$ and $B_1$ commute while $A_2$ and $B_2$ do not commute. My whole question is: do we know today why the physical quantities $A_1$ and $B_1$ are compatible (can be simultaneously known) and why the quantities $A_2$ and $B_2$ are not?

In other words: accepting that there are incompatible observables, and given a pair of incompatible observables do we know currently, or at least have a guess about why those physical quantities are incompatible?

spacetime - What is the curvature of this shape? Is it closed?

This is a follow up question to this issue. I have the following metric:$$\mathrm{d}s_s^2=[1-4\phi^2T^2]\mathrm{d}T^2-\phi^2T^4\mathrm{d}\Theta^2-\phi^2T^4\sin^2\Theta \mathrm{d}\Phi^2$$ Where $\phi$ is a constant in units of $km$ $s^{-2}$ (length per square time), $\Phi$ and $\Theta$ are 3D polar coordinates and $T$ is time. What is the curvature of this space at time, $T$?

If I travel in a straight line in the space defined by this metric and travel long enough, will I end up back where I started (is it closed)?

general relativity - how many times have the first photons lapped the universe?

I'm not really interested in the actual number of times, but rather an explanation of a closed Universe. How do we still measure the background radiation of the Big Bang? Those electromagnetic waves should have all passed this point in space a long time ago. Unless the Universe is closed and those Electromagnetic waves pass over the same region of space over and over again (like a bug running around the hemisphere of an expanding ball). If I recall correctly, at this point in time, the (circumference) of the Universe is increasing faster than the speed of light so it would be impossible for any current photons to ever complete another lap, but was this always so? Is the background radiation also continuously lowering in frequency as it red-shifts through ever expanding space?

Answer

Yes, the cosmic microwave background frequency has been dropping since its birth 380,000 years after the Big Bang, and it will continue to do so forever.

And yes, it was always impossible for the photons to complete a "lap". In fact, it was even more impossible a short time after the birth of the CMB about 380,000 years after the Big Bang, because the expansion was making the task of completing a lap increasingly hard.

This fact may be seen on the Penrose "causal" diagram of our Universe: its height is actually finite and smaller than the width. This fact is the source of the so-called "horizon problem": because the photons don't have enough time for a round trip, it's hard to understand why all places in the Universe agreed upon a common temperature. Cosmic inflation solves this problem.

In the text above, I assumed that there are no "spooky periodicities" in the visible Universe. It had been proposed that the shape of the Universe was an orbifold defined by a dodecahedron. However, such an assumption implies predictions – we should be able to see certain objects many times in the skies – and no such periodic multiple images have been found which imposes strong constraints on such models.

Finally, I need to disagree with a sentence of yours:

Those electromagnetic waves should have all passed this point in space a long time ago.

That isn't so. The CMB was created at every point of the Universe about 380,000 years after the Big Bang. Right now, we are observing photons that were created exactly at the right place so that they needed those "13.7 billion years minus 380,000 years" to get from the place of birth to the point where our gadgets are located now. As the Universe is getting older, we are observing photons that were created at increasingly distant places of the Universe, but whatever the age of the Universe is, we always find some places with the right distance so that the time needed for the photons to get here exactly agrees with (any) age of the Universe.