newtonian mechanics - Why does a car engine not do work if the wheels don't slip?

I saw this mind boggling result that if the tires don't slip then the work done by an engine to move a car is zero. Why is this true? Moreover, what does this truly mean?

Update: Sorry about not being clear, but I was talking about an idealised case where air resistance is negligible and the tires are perfectly rigid and there is no internal friction at play.

Answer

The answer to your question depends on precisely how it is interpreted. In my opinion, the clearest way of understanding a car driving on the road does in fact have the engine doing work on the car, but it is possible to define the system involved such that this is is not so. However, under this interpretation, the engine does no work on the car regardless of whether or not the tires are slipping.

To explain this distinction, I'll first give an example of work, then talk a little about work on the driving car, then finally discuss no-slip tires. I think this is a fairly-tricky issue, so please excuse the length of my response.

At Anna's request, let's assume no air drag, no internal friction, and no slipping. If the car is moving at a constant velocity, then the engine can be turned off, allowing the car to coast. There is clearly no work done and no change in energy in this scenario, so we'll talk about what happens as the car accelerates.

Work

In physics, "work" is a way of transferring energy into or out of a system. We say that a force does work on a system, thus adding energy to it. Work is defined by the equation

$$\textrm{d}W = \vec{F}\cdot \textrm{d}\vec{x}$$

The left hand side, $\textrm{d}W$, means "a small amount of work done". On the right hand side, $\vec{F}$ is the force that is doing the work, and $\textrm{d}\vec{x}$ is the movement of the part of the system where the force is applied. This does not particularly look like a change in energy, but it is possible to show, using Newton's laws, that work, as defined above, turns out to be the same as the change in kinetic energy induced by the force. (see work-energy theorem)

We need to keep in mind for later that although work done on a system induces a change in its kinetic energy, not all changes in kinetic energy are due to work being performed. For example, if I have a gerbil in a cage, and the gerbil starts running, the cage didn't go anywhere, so $\textrm{d}\vec{x} = 0$ for the system (gerbil + cage), so the work done on the cage/gerbil system is zero. Nonetheless, it gained kinetic energy. This is an internal conversion of energy from chemical energy in the gerbil to kinetic energy.

For an example of work, suppose you put your car in neutral and I stand behind it and push. If I push with a force of $100$ newtons, I do $1$ joule of work for each centimeter the car moves. In this example, $\textrm{d}W = 1 \textrm{J}$, $\vec{F} = 100 \textrm{N}$, and $\textrm{d}\vec{x} = 1 \textrm{cm}$. The "$\cdot$" symbol means the vector dot product. As long as the force and the motion are in the same direction, this is the same as normal multiplication (but if the force is at an angle to the motion, the work goes down; we won't worry about this in the rest of the answer). This work is the change in the kinetic energy of the car; the car speeds up some while I push.

It's important to define the system you're doing work on. In the example above, I considered the system to be your car. Then I applied an external force with my body. In an alternative view, we could define a system that consists of the car and me together. Then I didn't do any work on the system at all since, by definition, only external forces can do work. Finally, we might consider the system to be just the back half of the car. In this case, I'm still the external force and I did the same amount of work, but all that work went into the back half of the car. The back half of the car then in turn did some work on the front half of the car. This tells us that when I push on the back of the car, the car compresses a little, creating a force between the front half and back half that's half the size of the force I'm pushing with (since half the work done on the back half goes into the front half, giving those two halves the same change in speed). This is really true. If you slice your car in half, then hold the two halves together with some slices of white bread, when I start pushing on the back of the car the white bread will get crushed. Thus, the internal stresses in a car are different for front-wheel-drive and rear-wheel-drive vehicles.

Net Work on the Car

Let's take our system to be the entire car as it drives.

The car might be going at $45 \textrm{mph}$ or $20$ meters/second. In that case, in one second the car travels $20 \textrm{m}$. If, in one second, the car goes from $20 \textrm{m}/\textrm{s}$ to $21 \textrm{m}/\textrm{s}$, and the car has a mass of $1000 kg$, then the change in the car's kinetic energy works out to $20,500 \textrm{J}$. Assuming this is work done by an external force (me pushing, for example, although it's unlikely!) the work done on the car is $20,500 \textrm{J}$.

No-Slip Tires

The above analysis is wrong - the assumption that work is done by an external force is incorrect. The car is accelerating, so there must be an external force on it. But it turns out that external force does no work.

The external force on the car is the force from the road on the tires. Because the tires are not moving with respect to the road, this force does not do work. Thus, the road does no work on the car. This is as it should be. Everyone knows the road is not making your car go. The road cannot do work on your car because the road has no energy to give up.

Similarly, there is a force equal and opposite from your tires to the road, but again this force does no work. This says that your car does not put any energy into the road. This allows a car engine to be more efficient than a jet engine. A jet engine works by pushing air backwards. This air is moving with respect to the jet, and so the jet does considerable work on the air. That means that lots of the energy used by the jet engine goes into the air, and is wasted from the viewpoint of making the jet go.

When I'm pushing the car, my feet don't move while in contact with the road, so again no work is done on us. In both the case of me pushing (as long as I'm part of the system) and the car driving, the car's increase in kinetic energy is not due to external work, but to a conversion of chemical energy to kinetic energy completely within the system.

The Engine

Since the engine is inside the car, if the system we consider is the car itself, the engine does no work on the car. A system can't do work on itself, by definition. Instead, the engine spins an axle that spins the tires, adding kinetic energy to the car, but not doing work. This is why it's possible to say the engine does no work, and it is not dependent on whether or not the tires are slipping.

Since the kinetic energy of the car is changing, we might learn something more by redefining the system. If we define two systems, one being the body of the car minus the engine, and the other being the engine, now work is indeed being done.

The engine does work on the car. It does this by spinning an axle. To spin the axle, the engine must exert force on the outside edge of the axle, and since the outside edge of the axle is moving, the engine does work. (As inflector pointed out in his answer, this work is more easily calculated in terms of the torque on the axle as it rotates, but the two descriptions are equivalent). All the energy the car gains comes through the engine, so in this scenario the engine does all the work to accelerate the car.

The car also does work on the engine in this case. The car exerts a force on the engine. This is a structural force that keeps the engine in place inside the car. The car is paying back a portion of the energy it gained when the engine did work on it.

Conclusion

Two different people (in the same reference frame) can observe the same events, but disagree on whether work is done during them, or on how much work is done. This is because work is a transfer of energy between systems, so that if the two people define their systems differently, they will calculate different values for work done (and for heat exchange).

In light of this ambiguity, it is correct, although incomplete, to state that the engine does no work. This is not dependent on the no-slip tires, and is an artifact of defining systems such that the engine and car are part of the same system.

No slip tires imply that the road does no work on the car, and that the car does no work on the road.

homework and exercises - The use of the commutators in quantum mechanics: explanations

Considering that I've never studied quantum mechanics before I have need to understand the operator commutator. My start is: $[A,B]=AB-BA \tag{a}$

Now, why must be

$$\left[\frac{\partial }{\partial x},x\right]\stackrel{?}{=}1 \tag{1}$$ I have thought, from the rule (a),

This identity $$\left[x,\frac{\partial }{\partial x}\right]=-1 \tag{2}$$ is easy because $[A,B]=-[B,A]$. I have not understood, also, (3) and (4) $$\left[i\hslash\frac {\partial}{\partial x},x\right]=i\hslash \tag{3}$$

$$[p_x,x]=i\hslash \tag{4}$$ where $p_x$ is the momentum on $x-$ axis.

Answer

Equations (a), (1), (2), (3) and (4) all are operator equations. Therefore you need to understand what an operator equation actually is.

Now, why must be $$ \left[\frac{\partial }{\partial x},x\right]\stackrel{?}{=}1 \tag{1}$$

That means, the operators on the left-hand-side and on the right-hand-side always yield the same result when applied to arbitrary functions.

Hence, here you must prove that $$ \left[\frac{\partial}{\partial x},x\right] \psi(x) = 1\cdot \psi(x) $$ for every function $\psi(x)$.

The proof is a long sequence of very elementary steps: $$\begin{align} &\left[\frac{\partial }{\partial x},x\right] \psi(x) \\ = &\left(\frac{\partial}{\partial x} x - x \frac{\partial }{\partial x}\right) \psi(x) \\ = &\frac{\partial}{\partial x} x \psi(x) - x \frac{\partial }{\partial x} \psi(x) \\ = &\frac{\partial x}{\partial x}\psi(x) + x \frac{\partial \psi(x)}{\partial x} - x \frac{\partial \psi(x)}{\partial x} \\ = &\frac{\partial x}{\partial x}\psi(x) \\ = &1\cdot \psi(x) \\ \end{align}$$

vision - Why can I look at the sun with my sunglasses but not at the solar eclipse?

Is the problem with seeing an eclipse that you have a sudden change between penumbra and regular sun? If that's the case, why can I look to the sun with my sunglasses on but I can't do it in an eclipse because it would be harmful? Even if my sunglasses have an UV filter it would be bad for my eyes to see the eclipse with them?

newtonian mechanics - A large rock falls on your toe. Which concept is most important in determining how much it hurts?

A) The mass of the rock B) The weight of the rock C) Both the mass and the weight are important. D) Either the mass or the weight, as they are related by a single multiplicative constant, g.

Similarly, if the large rock merely sits on your toe, which concept is most important in determining how much it hurts?

For a falling rock, to me it seems that mass would be most important as the pain level would be proportional to the change in momentum, or the impulse, on your foot.

For a stagnant rock, it seems either weight or mass would be the same.

Am I correct?

rotational kinematics - Instantaneous axis of rotation and rolling cone motion

Suppose a cone is purely rolling (no slipping) around a fixed axis. I mean, it is revolving around a fixed axis perpendicular to the ground and passing through its vertex and also rotating, so the vertex is stationary. (sorry this might be a bit confusing but I hope you understand what I mean). Something like this, but rolling on plane surface instead of another cone: rolling cone http://upload.wikimedia.org/wikipedia/commons/thumb/4/43/Rolling_cone.pdf/page1-1024px-Rolling_cone.pdf.jpg

Now the instantaneous axis of rotation (IAR) of the cone is the 'line' that is touching the ground right? So how do you find the velocity of any other point using that? I mean, in the rolling wheel, you multiply the angular velocity by the distance from the IAR to get the velocity. Is it the same here?

If it is, then consider the center of base of the cone. If the height of cone is $h$ then its distance from the IAR is clearly $h\sin x$ where $x$ is the cone's half apex angle. So its velocity should be $ah\sin x$, where $a$ is the angular velocity with which the cone is rotating. Is this right?

Now we can also analyse the cone's motion by considering it in two parts: rotation + revolution, right? So again considering the center of base of the cone, it has no velocity by virtue of rotation (since the cone is rotating about an axis through the center), right? And by virtue of it rotating in a circle (of radius $h\cos x$) around the axis passing through its vertex, it has velocity $bh\cos x$ , where $b$ is the angular velocity with which the cone is revolving.

Now these two must be same, so we get $b=a \tan x$.

But Wikipedia states here that the ratio is $\sin x$.

And at the same time, this video (which I found in the external links section of the Wikipedia page) states that $a=b\cot x$ which is same as what I got.

So I'm really confused. Is everything that I did right? If not please correct me. Thank you.

Edit: OK so as carl commented, I'm also confused about how the instantaneous linear velocity of the center of the cone's base is different from the velocity of the center of a rolling disk.

Edit 2: How to find velocity of any point on the cone? There ought to be two approaches , one using IAR and another by considering motion as rotation + revolution but I'm not able to do it.

optics - Could the sky on a planet theoretically be any color?

The sky on the Earth is blue. Could the color of the sky on a planet with an atmosphere be of any color theoretically?

Which colors are the most likely?

I think it would be really awesome to have a planet with a green sky, or a red sky. What conditions would have to be in place for this to happen?

By the way I suspect its impossible for the sky to be white, black, grey, or brown, but maybe this is possible if not at all times during the day, a part of the day?

Does the answer depend on whether the planet is a gas planet or a rocky one? Initially I had in mind a rocky planet, but I suspect a gas planet allows you to cheat a little because gases can be a variety of colors (red even?).

Answer

This is an extremely hard question to answer definitely; its answer is heavily dependent on three things (1) the makeup of the atmosphere in question (2) the density of the atmosphere in question and (3) the surface temperature, and therefore the spectrum of output light, of the star in question.

If the atmopsphere's gasses are themselves are coloured (e.g. $N\,O_2$), then such gasses can clearly shift an atmosphere's colour greatly. In low density atmospheres, such as that of Mars, dust and matter swept up from the surface dominates, giving the sky a "butterscotch" colour. We clearly need to know what's floating around, so we can't say much more about point (1), so I shall now assume in this answer that the gasses in the atmosphere themselves are uncoloured.

In this case, the dominant effect setting the colour is Rayleigh Scattering. This is the diffraction of light from smaller-than-wavelength inhomogeneities in the atmosphere. When you "see" the atmosphere, you are seeing scattered, not direct light from the sun in question, and Rayleigh scattering happens most strongly at shorter wavelengths. It is a powerful effect: in the small inhomogeneity limit, it varies as the inverse fourth power of the wavelength. So the process always favours shorter wavelengths.

This means that for a sun of similar spectrum to ours, and in an atmospheric density like ours, the sky will be blue, like ours, or like Jupiter's (in the upper layers). However, for very high density atmospheres, the direct light reaching the surface has its shorter wavelengths heavily attenuated by Rayleigh scattering, so that only longer wavelengths reach the surface. It is for this reason that pictures taken by Russia's Venera landers on Venus suggest that the sky has an orange glow.

If the planet is orbitting a red giant, I am guessing this would probably mean a greenish or yellow sky for a planet with an atmospheric density like ours, as the spectrum of the star's light contains only longer wavelengths than that from our Sun. However, the inverse fourth power dependence means that now the glow from the atmospheric gasses themselves is much weaker, so the sky's colour is more likely to be dominated by things like dust, as in the case of Mars.

riddle - I'm not obsessed with this treasure, I promise!

So, yesterday I received a treasure, can you guess what it is?

For me to receive this treasure, I had to wait for quite some time,
A time I did pass with leisure, many obstacles I had to climb.

Never ceasing, I continued day by day.
Always piecing, many foes I did slay.

To receive a prize, one most precious indeed.
I widened my eyes, until I did succeed.

Can you tell me what I've gotten? It's the first of it's kind I've received.

Answer

Did you get

A gold badge on PSE?

For me to receive this treasure, I had to wait for quite some time, A time I did pass with leisure, many obstacles I had to climb.

Sure enough, you have had to answer and ask a lot of questions for quite some time, 100 days to be specific, before you received this!

Never ceasing, I continued day by day. Always piecing, many foes I did slay.

Yes, we have been seeing your challenging puzzles as well as piecing together your answers beating foes like @El-Guest :P

To receive a prize, one most precious indeed.

Gold badge is definitely the highest and a precious badge indeed!

I widened my eyes, until I did succeed.

I'm guessing you learnt a lot while asking and answering questions!

I only just saw @El-Guest's part of the answer that says:

FANATIC. We know you are a puzzling fanatic now!
And I just saw that the badge you received is FANATIC! :)

And Congratulations! :D

harmonic oscillator - Why don't tuning forks have three prongs?

I was reading Why tuning forks have two prongs?. The top answer said the reason was to reduce oscillation through the hand holding the other prong.

So if having 2 prongs will reduce oscillation loss, surely a 3-pronged tuning fork would be even more efficient.

Why don't you see more 3-pronged tuning forks?

Answer

The reason for having two prongs is that they oscillate in antiphase. That is, instead of both moving to the left, then both moving to the right, and so on, they oscillate "in and out" - they move towards each other then move away from each other, then towards, etc. That means that the bit you hold doesn't vibrate at all, even though the prongs do.

You might ask why it is that they do that, instead of oscillating in the same direction as one another. The answer is that at first they oscillate in both ways at the same time, but the side-to-side oscillations are rapidly damped by your hand, so they die out quickly, whereas the in-and-out ones are not damped this way, so they ring on long enough to hear them. An excellent illustration of this can be seen in this video of a FEM model of a two-pronged fork, which shows you all the vibrational modes separately. (Hat tip to ghoppe, who posted this video in a comment.)

Having a third prong wouldn't help very much with reducing damping. There are (at least) three different ways a three-pronged fork could vibrate: one with all three vibrating side-to-side in phase with one another, and two where one of the prongs stays still and the other two vibrate in and out. (At first I thought there would be three of this latter type of mode, but the third can be formed from a linear combination of the other two: $(1,0,-1) - (1,-1,0) = (0,1,-1)$.) The vibrational mode in which everything moves in the same direction would be damped by your hand, and some combination of the other two would continue to sound for a while. As Ilmari Karonen pointed out in a comment, there would also be a "transverse" $(1,-2,1)$ mode, where prongs vibrate out of the plane of the fork. This mode wouldn't necessarily have the same frequency as the primary in-and-out mode, so it's probably something we'd want to avoid.

But ultimately there's little reason why, in a three-pronged fork, the vibrations would ring on longer than just two prongs - there would be the same amount of vibrational energy as in a two-pronged fork, but just shared between two or three modes of vibration instead of one.

electromagnetism - Discretization of the induction equation using the Finite Volume Method (FVM)

I am trying to discretize the induction equation using the finite volume method. The equation with the quasi-static assumption is given as

$$( \mathbf{B} \cdot \nabla)\mathbf{u} - (\mathbf{u} \cdot \nabla)\mathbf{B} + \frac{1}{\sigma \mu} \nabla^2 \mathbf{B} = 0.$$

My first guess is rewriting the equation in such a way that the Gauss divergence theorem can be applied. Like this

$$( \mathbf{B} \cdot \nabla)\mathbf{u} = \nabla \cdot ( \mathbf{B} \mathbf{u} ).$$

But this is actually wrong. How can I discretize these equation?

PS: The Laplacian isn't a problem

Answer

The confusion lies in writing the formula as you have. If you instead started with Faradays law, $$ \frac{\partial \mathbf B}{\partial t}=-\nabla\times\mathbf E\tag{1} $$ under the assumption of perfect conductivity, (1) becomes $$ \frac{\partial \mathbf B}{\partial t}=\nabla\times\left(\mathbf u\times\mathbf B\right)\tag{2} $$ You, of course, have a diffusion term added to (2), which comes from $\mathbf J=\eta\nabla\times\mathbf B$ (cf. this post of mine). Combining these in the steady-state approximation & using some vector calculus, you can get the form you have in your first equation.

In terms of discretizing the PDE, it's easier to work with Eq. (2) above than your first equation, so along the $x$ direction, you'll have $$ \frac{\partial}{\partial x}\left(\mathbf u\times\mathbf B\right)_x=\frac{\partial}{\partial x}\left(\begin{array}{c}0 \\ v_xB_y-v_yB_x \\ v_xB_z - v_zB_x\end{array}\right) $$ The $y$ and $z$ directions are pretty similar; applying FVM/FDM to these terms should be straight forward.

Note that, while the $x$ term here is zero (required for satisfying $\nabla\cdot\mathbf B=0$), numerical instabilities can still exist and drive inaccuracies in your model; fixing this requires ''divergence cleaning'' methods. There are some tricks in either the discretization scheme or the algorithm to employ that have been discovered over the years to do this cleaning: Balsara & Kim (2003) and Tóth (2000) cover the many of the schemes available (Tóth's paper is paywalled, but the Balsara & Kim paper is on arXiv).

special relativity - Is acceleration relative?

A while back in my Dynamics & Relativity lectures my lecturer mentioned that an object need not be accelerating relative to anything - he said it makes sense for an object to just be accelerating. Now, to me (and to my supervisor for this course), this sounds a little weird. An object's velocity is relative to the frame you're observing it in/from (right?), so where does this 'relativeness' go when we differentiate?

I am pretty sure that I'm just confused here or that I've somehow misheard/misunderstood the lecturer, so can someone please explain this to me.

Answer

I find the phrase "acceleration need not be relative anything" to be awkward, but I can see where it comes from.

For the moment restrict our consideration the Galilean relativity (just to keep the math simple). Consider two frames of reference one ($S$) in which the body is at rest and another ($S'$) in which it moves with velocity $\vec{v'_i} = \vec{u} = u \hat{z}$.

So we have the initial velocity of the body in frame $S$ as $v_i = 0$, and $v' = v + u \hat{z}$

Now assume that the the body accelerates from time $t$ at acceleration $\vec{a} = a \hat{Z}$ resulting in a velocity in frame $S$ of $\vec{v_f} = a t \hat{z}$.

Compute the final velocity in frame $S'$ as $v'_f = v_f + u \hat{z} = (u + a t)\hat{z}$, and from that the acceleration in the primed frame as $a' = a$.

So the acceleration is the same in all frames (you can check the cases for $a \not\parallel u$ yourself), and it is reasonable to say that accelerations are not relative anything.

All of this is a consequence of the simple form of the transformation between frames:

$$ \vec{x'} = \vec{x_0} + \vec{u} t $$ $$ t' = t $$

So what about Einsteinian relativity?

Here the transformation between frames is more complicated, and the math is much more complicated resulting in observers in different frames seeing different accelerations, but they will all agree on the acceleration as measured in the body's own frame. In my opinion "the acceleration need not be relative" risks causing unnecessary confusion on these points. The magnitude and direction measured will measure in depend on the frame of the observer, which is often what is meant when people say "it's relative".

astrophysics - Where is all the Dark Matter? Theoretical Question

In my Physics Class we had to look into possible areas where Dark Matter could be "hiding." Such as Black Holes and so on. Dark matter really can be "seen" through its gravitational effects, and we still come up short.

Why does the matter need to be inside our universe? I like to imagine the Universe as a sort of atom or even star. I mean multiple universes acting on each other. If they are contracting or expanding similar to our universe could these multiple universes act out forces upon each other. This would explain the lack of matter through forces acting on the outside of the universe pushing against the expansion. I do not have the background in physics to explain it in more than a simplistic theoretical way.

Answer

There are several reasons to believe that dark matter is a particle. The most widely accepted alternative explanations for the different phenomena that led us to conjecture dark matter in the first place, can collectively be labeled "we don't understand gravity well enough".

But no matter what, the effects of dark matter are sort of "localized". The rotation curves of galaxies suggest that there's an "extra" pull from matter inside the galaxies. That is, either there's more matter than we can see, or the matter that we can see pulls more than we expect. The same can be said about the interpretation of the velocity dispersion of galaxies in clusters of galaxies, and of massive objects acting as gravitational lenses. Also, in order to explain how structure has the time to form before the Universe expanded too much, something must have made matter collapse faster than expected.

If these forces were applied from an "outside" agent, you would have to come up with a very special way for the forces to act. The forces we know of all depend on the distance between bodies, either increasing or decreasing with distance. But if an outside force were to explain e.g. galaxy rotation curves, it seems it would have to repeatedly "switch sign", i.e. alternate between attracting and repelling.

I should say that I don't really know much about multiverse theories, but for such a theory to explain the observables, it seems there would be the need for many extra add-ons and ad hoc-explanation. The theory of dark matter (and to some extend those of modified gravity, I suppose) offer the most simple explanation.

And we like simple explanations.

homework and exercises - Is there any difference between relative motion and motion itself

If relative motion is the change in position of a body with reference to another body, and motion is the change in position of a body with time . Then why is relative motion not associated with the types of motion

Answer

Every object which is in motion is actually in relative motion. Suppose, you found an object which is in motion but not in relative motion. Ask yourself how do you know that the object is in motion? You will realize that the object is in motion with respect to you, which means the object is in relative motion.

I am not sure if you got the concept correctly or you did not frame your doubt properly. You seem to be confused with relative motion with motion.

Motion, as you said, is change in position with respect to time but the definition of relative motion you gave is not really correct. Relative motion is the change in position with respect to a point in space with time. In fact, relative motion and motion are exactly one and the same.

Any time you talk about motion , its relative motion. You say something is in motion because it is moving with respect to you which means, say what, it is in relative motion with respect to you.

There is no such thing as absolute motion.

Will there be translation + rotational motion?

If a rod is on a frictionless plane, and a force is applied on one of it's end, will there be both, translation + rotation motion? Also, if only a single force is applied on a body that does not pass through Centre of Mass, will it always produce rotation + translation?

quantum mechanics - Double slit experiment - how to see an electron going through a slit?

This is my first question here. Hope it not too dull to you guys :p

I found this video on youtube. After 3:24 it says if there is an observer detecting which slit the electron goes through, there won't be an interference pattern. Has there ever been any experiment (not thought experiment) demonstrating this phenomenon really using microscopes or other detectors to see where the electron goes?

My physics teacher don't know the answer, but he took an alternative route, said if we suddenly close one of the slits, we know the electron can only go through the other, and it turns out no interference pattern shown. It sounds convincing to me and the experiment is much easier to perform. But if one can really shine a light and see the electron, he or she can alter the intensity or energy of the light to find a critical value below which the interference pattern won't be destroyed. Sounds even cooler :)

Answer

This or a similar video was topic here already:

Is Dr Quantum's Double Slit Experiment video scientifically accurate?

This: ""But if one can really shine a light and see the electron, "" is the problem, in practice one had to "shoot" electrons at the electrons to "see" them. Or on could use gamma rays (Compton effect), but always this would disturb the experiment. Ther is no "eye" in the quantum world to "watch" those balls without interaction.

quantum field theory - Is the world $C^infty$?

While it is quite common to use piecewise constant functions to describe reality, e.g. the optical properties of a layered system, or the Fermi–Dirac statistics at (the impossible to reach exactly) $T=0$, I wonder if in a fundamental theory such as QFT some statement on the analyticity of the fields can be made/assumed/proven/refuted?

Take for example the Klein-Gordon equation. Even if you start with the non-analytical Delta distribution, after infinitesimal time the field will smooth out to an analytical function. (Yeah I know, that is one of the problems of relativistic quantum mechanics and why QFT is "truer", but intuitively I don't assume path integrals to behave otherwise but smoothing, too).

Answer

This is a really interesting, but equally beguiling, question. Shock waves are discontinuities that develop in solutions of the wave equation. Phase transitions (of various kinds) are non-continuities in thermodynamics, but as thermodynamics is a study of aggregate quantitites, one might argue that the microscopic system is still continuous. However, the Higgs mechanism is an analogue in quantum field theory, where continuity is a bit harder to see. It is likely that smoothness is simply a convenience of our mathematical models (as was mentioned above). It is also possible that smooth spacetime is some aggregate/thermodynamic approximation of discrete microstates of spacetime -- but our model of that discrete system will probably be described by the mathematics of continuous functions.

(p.s.: Nonanalyticity is somehow akin to free will: our future is not determined by all time-derivatives of our past!)

general relativity - How does classical GR concept of space-time emerge from string theory?

First, I'll state some background that lead me to the question.

I was thinking about quantization of space-time on and off for a long time but I never really looked into it any deeper (mainly because I am not yet quite fluent in string theory). But the recent discussion about propagation of information in space-time got me thinking about the topic again. Combined with the fact that it has been more or less decided that we should also ask graduate level questions here, I decided I should give it a go.

---

So, first some of my (certainly very naive) thoughts.

It's no problem to quantize gravitational waves on a curved background. They don't differ much from any other particles we know. But what if we want the background itself to change in response to movement of the matter and quantize these processes? Then I would imagine that space-time itself is built from tiny particles (call them space-timeons) that interact by means of exchanging gravitons. I drawed this conclusion from an analogy with how solid matter is built from atoms in a lattice and interact by means of exchanging phonons.

Now, I am aware that the above picture is completely naive but I think it must in some sense also be correct. That is, if there exists any reasonable notion of quantum gravity, it has to look something like that (at least on the level of QFT).

So, having come this far I decided I should not stop. So let's move one step further and assume string theory is correct description of the nature. Then all the particles above are actually strings. So I decided that space-time must probably arise as condensation of huge number of strings. Now does this make any sense at all? To make it more precise (and also to ask something in case it doesn't make a sense), I have two questions:

1. 
   

   In what way does the classical space-time arise as a limit in the string theory? Try to give a clear, conceptual description of this process. I don't mind some equations, but I want mainly ideas.

   
   


2. 
   

   Is there a good introduction into this subject? If so, what is it? If not, where else can I learn about this stuff?

   
   

Regarding 2., I would prefer something not too advanced. But to give you an idea of what kind of literature might be appropriate for my level: I already now something about classical strings, how to quantize them (in different quantization schemes) and some bits about the role of CFT on the worldsheets. Also I have a qualitative overview of different types of string theories and also a little quantitative knowledge about moduli spaces of Calabi-Yau manifolds and their properties.

Answer

First, you are right in that non-Minkowski solutions to string theory, in which the gravitational field is macroscopic, it should be thought of as a condensate of a huge number of gravitons (which are one of the spacetime particles associated to a degree of freedom of the string). (Aside: a point particle, corresponding to quantum field theory, has no internal degrees of freedom; the different particles come simply from different labels attached to ponits. A string has many degrees of freedom, each of which corresponds to a particle in the spacetime interpretation of string theory, i.e. the effective field theory.)

To your question (1): certainly there is no great organizing principle of string theory (yet). One practical principle is that the 2-dimensional (quantum) field theory which describes the fluctuations of the string worldsheet should be conformal, i.e. independent of local scale invariance of the metric. This allows us to integrate over all metrics on Riemann surfaces only up to diffeomorphisms and scalings, which is to say only up to a finite number of degrees of freedom. That's an integral we can do. (Were we able to integrate over all metrics in a way that is sensible within quantum field theory, we would already have been able to quantize gravity.) Now, scale invariance imposes constraints on the background spacetime fields used to construct the 2d action (such as the metric, which determines the energy of the map from the worldsheet of the string). These constraints reduce to Einstein's equations.

That's not a very fundamental derivation, but formulating string theory in a way which is independent of the starting point ("background independence") is notoriously tricky.

(2): This goes under the name "strings in background fields," and can be found in Volume 1 of Green, Schwarz and Witten.

thermodynamics - Why can I touch aluminum foil in the oven and not get burned?

I cook frequently with aluminum foil as a cover in the oven. When it's time to remove the foil and cook uncovered, I find I can handle it with my bare hands, and it's barely warm.

What are the physics for this? Does it have something to do with the thickness and storing energy?

Answer

You get burned because energy is transferred from the hot object to your hand until they are both at the same temperature. The more energy transferred, the more damage done to you.

Aluminium, like most metals, has a lower heat capacity than water (ie you) so transferring a small amount of energy lowers the temperature of aluminium more than it heats you (about 5x as much). Next the mass of the aluminium foil is very low - there isn't much metal to hold the heat, and finally the foil is probably crinkled so although it is a good conductor of heat you are only touching a very small part of the surface area so the heat flow to you is low.

If you put your hand flat on an aluminium engine block at the same temperature you would get burned.

The same thing applies to the sparks from a grinder or firework "sparkler", the sparks are hot enough to be molten iron - but are so small they contain very little energy.

quantum mechanics - What is the difference between $|0rangle $ and $0$?

What is the difference between $|0\rangle $ and $0$ in the context of $$a_- |0\rangle =0~?$$

Answer

$|0\rangle$ is just a quantum state that happens to be labeled by the number 0. It's conventional to use that label to denote the ground state (or vacuum state), the one with the lowest energy. But the label you put on a quantum state is actually kind of arbitrary. You could choose a different convention in which you label the ground state with, say, 5, and although it would confuse a lot of people, you could still do physics perfectly well with it. The point is, $|0\rangle$ is just a particular quantum state. The fact that it's labeled with a 0 doesn't have to mean that anything about it is actually zero.

In contrast, $0$ (not written as a ket) is actually zero. You could perhaps think of it as the quantum state of an object that doesn't exist (although I suspect that analogy will come back to bite me... just don't take it too literally). If you calculate any matrix element of some operator $A$ in the "state" $0$, you will get 0 as a result because you're basically multiplying by zero:

$$\langle\psi| A (a_-|0\rangle) = 0$$

for any state $\langle\psi|$. In contrast, you can do this for the ground state without necessarily getting zero:

$$\langle\psi| A |0\rangle = \text{can be anything}$$

quantum mechanics - Why is lens flare doubled if odd number of blades but equal to number of blades if even?

An article in Picture Correct describes how the number of blades in a particular lens correlates to the number of starburst points associated with lens flare. The number of starbursts is double the number of aperture blades if there are an odd number of blades and equal to the number of blades if the lens contains an even number of blades. Here's a jpeg of the article should the above link ever break:

What property of light could cause this phenomenon?
_{This seems more appropriate here on the Physics page than it does on the Photography SE community, since the question is about the physics behind this and not whether or not it's true.)}

Answer

A straight edge causes diffraction in the direction perpendicular to that edge. When two straight edges are parallel to each other their diffraction patterns will overlap (point along the same line) - making it look like there is just one.

That happens when there is an even number of blades.

Note - I believe the diagram for the six pointed blade is wrong - it shows the flare aligned with the corners of the hexagon. I am pretty sure they align with the centers of the blades instead. (Note - for an odd number of blades the distinction cannot be made).

statistical mechanics - Proving that the Boltzmann entropy is equal to the thermodynamic entropy

I've been trying to understand how we can equate the Boltzmann entropy $k_B \ln \Omega$ and the entropy from thermodynamics. I'm following the approach found in the first chapter in Pathria's statistical mechanics, and in many other texts. Many other questions on stackexchange come close to addressing this problem, but I don't think any of the answers get at my specific question.

So, we're considering two isolated systems 1 and 2, which are brought into thermal contact and allowed to exchange energy (let's assume for simplicity that they can only exchange energy). On the thermodynamic side of the problem, we have the necessary and sufficient condition for thermal equilibrium

$$T_1=\frac{\partial E_1}{\partial S_1}=T_2=\frac{\partial E_2}{\partial S_2},$$

where the temperatures $T_1$ and $T_2$, the internal energies $E_1$ and $E_2$, and the entropies $S_1$ and $S_2$ are all defined appropriately in operational, thermodynamic terms. On the other hand, we can show that the necessary and sufficient condition for equilibrium from the standpoint of statistical mechanics is given by

$$\beta_1 \equiv \frac{\partial \ln \Omega_1}{\partial E_1}= \beta_2 \equiv \frac{\partial \ln \Omega_2}{\partial E_2}.$$

Here, $\Omega_1$ and $\Omega_2$ are the number of microstates associated with the macrostate of each system. Now, since both of these relations are necessary and sufficient for equilibrium, one equality holds if and only if the other also holds. My question is: How can we proceed from here to show that $S=k_B \ln \Omega$, without limiting our scope to specific examples (like an ideal gas)? In Pathria's text and in other treatments, I don't see much explanation for how this step is justified.

My possibly wrong thoughts are: It seems like we first need to show that $\beta$ is a function of $T$ alone (and indeed the same function of $T$ for both systems), and then show that the form of this function is in fact $\beta \propto T^{-1}$. But I'm not sure how to prove either of those claims.

pattern - What is a Scientific Word™

If a word has a certain property, I call it a Scientific Word™.

In each of the short, unrelated sentences below, only one of the choices is a Scientific Word™.

The sentences do not affect whether a word is a Scientific Word™. Case, font, and letter/word appearance do not matter.

1. The child's mother gave him a (apple/banana) as a healthy snack.


2. When queried with the question about a raise, Dave's boss said (no/yes).


3. Billy's breakfast consisted of (bacon/beans), eggs, and toast.


4. The snake's sharp fangs contained a deadly (poison/venom) ready to be inserted into your bloodstream at the slightest movement.


5. Jerry's computer had gotten infected with (a) (virus/malware).


6. The soldier died out in (war/battle).


7. After Riley had died so many times on his video game, he was filled with (anger/rage).


8. The same series of events had happened years (ago/before), and the result was not good.



9. Joe has such a smart (brain/mind).


10. To be nice, Sarah offered her umbrella to an old (lady/woman).

What makes a word a Scientific Word™?

Answer

I think a Scientific Word™ is one that

that scientists around the world can use without any ambiguity, simply because these words are composed entirely from the symbols of elements in the periodic table.

1.

Ba Na Na (Barium Sodium Sodium)

2.

No (Nobelium)

3.

Ba Co N (Barium Cobalt Nitrogen)

4.

Po I S O N (Polonium Iodine Sulphur Oxygen Nitrogen)

5.

V I Ru S (Vanadium Iodine Rubidium Sulphur)

6.

W Ar (Tungsten Argon)

7.

Ra Ge (Radium Germanium)

8.

Ag O (Silver Oxygen)

9.

B Ra In (Boron Radium Indium)

10.

La Dy (Lanthanum Dyprosium)

All other words do not follow this Scientific Rule (patent pending).

visible light - Can something without mass exert a force?

I am something of a dilettante in physics, so please forgive me if the answer to this question is painfully obvious. The question is simple, can something that theoretically has no mass exert a force. I have been tossing around this and other similar questions in my head for a while now and have not really found any concrete answers to my inquiry. I am thinking about how light seems to be able to push objects but yet has no mass, however I expanded the question to be more encompassing in hopes of further learning.

Answer

Yes, photons can. See https://en.wikipedia.org/wiki/Radiation_pressure (and photons are certainly massless).

PS In fact, any massless particle has momentum(*) and if it is scattered on a body, it changes its own and the body's momentum, which is what a force does.

(*) $p = \hbar k = E/c$ where $E$ is its energy and $c$ is speed of light

special relativity - Why don't clocks on a train read the same time?

Two clocks are positioned at the ends of a train of length $L$ (as measured in its own frame). They are synchronized in the train frame.

The train travels past you at speed $v$. It turns out that if you observe the clocks at simultaneous times in your frame, you will see the rear clock showing a higher reading than the front clock. By how much?

The solution to this exercise is given in the book, and it denotes that we put "a light source on the train, but let’s now position it so that the light hits the clocks at the ends of the train at the same time in our frame." As in this figure:

I don't understand:

1. Why is there a difference in the clocks in the first place?


2. Why are the fractions are multiplied by half? i.e. what is the origin of the $2$ in $$\frac{L(c+v)}{\textbf{2}c}\text{?}$$

Note: This one is taken from David Morin textbook for mechanics.

optics - Why does light change direction when it travels through glass?

This was explained to me many years ago, by a physics teacher, with the following analogy:

"If someone on the beach wants to reach someone else that is in the water, they will try to travel as much as they can on the beach and as little as possible on the water, because this way they will get there faster."

I'm paraphrasing of course, but this is as accurate as I recall it.

This explanation makes no sense to me. Was he telling me the light knows where it is going? It wants to get there faster? It chooses a different direction?
(No need to answer these questions, this was just me trying to understand the analogy.)

My attempts to clarify the issue were without success and many years later I still don't know.

Why does light change direction when it travels through glass?

mathematics - Ali Baba and the 10 thieves

EDIT: This puzzle originates from the "International Mathematics Tournament of the Towns", and was published in the book "S.M.A.R.T circle overview" by Professor Andy Liu. The author has (albeit a little retroactively) granted us the permission to use the puzzle.

Ten thieves, ranked A to J, are trying to cross a river in a boat requiring two rowers. Unfortunately, if the ranks of any two in the boat differ by more than 1, those two will refuse to stay in the boat. This constraint means they can’t get across the river. Their leader, with a rank of A, asks Ali Baba for help and Ali Baba replies, “If you give me a rank of A, equal to yours, we can all cross the river.” The leader agrees.

How many one-way crossings are the least required to get Ali Baba and the 10 thieves across the river?

NOTE:

More than 2 people can accumulate in the boat. Boat is operated by 2 rowers. So a single person cannot row it on his own.

Answer

Here’s my quadruple-checked, optimality-guaranteed solution with

33 crossings.

(Glorfindel’s initial answer had the same crossing count earlier, but UselessInfoMine discovered an error in that method. The updated version of that method works, but uses 4 crossings more than the optimal method.)

You can get any 2 consecutive thieves over like this

+Aab, -ab, +bc, -Ab, +Aab, -bc, +XY, -Aa (8 moves)

Repeat four times, always bringing the two lowest ranked thieves over. Bring the rest over with the final move.

If there’s a quicker way to do it, it must be Very Clever Indeed, since

The only move that adds a thief on the opposite shore is the +Aab, which cannot happen more often than every fourth move. (Barring the silly -Aab waste of move, of course.)
Given that the final +Aab will bring 3 guys, a total of 9 +Aab’s are required, and in between them, there will have to be at minimum 8x3 other crossings. This sums up to a minimum of 33 moves.

Therefore, barring lateral thinking and other trick answers, this solution is optimal.

How do electrons get a charge?

Electrons belong to a group of elementary particles called leptons. There are charged and neutral leptons. And electron is the charged one. But how come it got charged?

The negative or positive charges were assigned by convention. But it is a fact that electrons are charged. My question is why electrons? and not neutrons?

Also while reading http://en.wikipedia.org/wiki/Electron, I saw that "Independent electrons moving in vacuum are termed free electrons. Electrons in metals also behave as if they were free. In reality the particles that are commonly termed electrons in metals and other solids are quasi electrons, quasiparticles, which have the same electrical charge, spin and magnetic moment as real electrons but may have a different mass ( or Effective mass - extra mass that a particle seems to have while interacting with some force )."

What does this mean?

Answer

Your question touches the question of ontology in particle physics. Historically we are used to be thinking of particles as tiny independent entities that behave according to some laws of motion. This stems from the atomistic theory of matter, which was developed some two thousand years ago from the starting point of what would happen if we could split matter in ever smaller parts. The old Greeks came to the conclusion that there had to be a limit to that splitting, hence the atom hypothesis was born.

This was just a philosophical idea, of course, until around the beginning of the 19th century we learned to do chemistry so well that it became obvious that the smallest chunks that matter can be split into seemed to be the atoms of the periodic table. A hundred years later we realized that atoms can be split even further into nuclei and electrons. What didn't change was this idea that each chunk had its own independent existence.

This idea ran into a deep crisis during the early 20th century when we discovered the first effects of quantum mechanics. It turns out that atoms and nuclei and electrons do not, at all, behave like really small pieces of ordinary matter. Instead, they are behaving radically different, so different, indeed, that the human imagination has a hard time keeping up with their dynamic properties.

For a while we were in a limbo regarding our description of nature at the microscopic scale. It seemed like we could cling to some sort of "little weird billiard ball with mass, charge, spin etc. properties" kind of theory for electrons, but as time went by, this became ever more hopeless. Eventually we discovered quantum field theory, which does away with the particle description completely, and with that all the ontological problems of the past century have disappeared.

So what's the new way of describing nature? It is a field description, which assumes that the universe is permeated by ONE quantum field (you can split it up into multiple components, if you like). This quantum field has local properties that are described by quantum numbers like charge. This one quantum field is subject to a quantum mechanical equation of motion which assures that some properties like charge, spin, angular momentum etc. can only be changing in integer (or half integer) quantities (in case of charge it's actually in quantities of 1/3 and 2/3 but that's a historical artifact). Moreover, this field obeys symmetry rules that leave the total sum of some of these quantities unchanged or nearly unchanged. Charges in particular can only be created on this field in pairs such that the total charge remains zero.

So now we can answer your question in the language of the quantum field: the electron gets its charge by the field allowing to create one positive charge state and one negative charge state at the same time, leaving its total charge zero. This process takes some energy, in case of the electron-positron pair a little over 1MeV. Every other property that is needed to uniquely characterize an electron is created in a similar way and at the same time. The elementary particle zoo is therefore nothing but the list of possible combinations of quantum numbers of the quantum field. If it's not on the list, nature won't make it (at least not in form of a real particle state). Our list is, of course, at best partial. There are plenty of reasons to believe that there are combinations of quantum numbers out there that we have not observed, yet, but which are still allowed.

general relativity - black hole event horizon

Given gravitational time dilation, under what conditions will a test particle cross an event horizon before the black hole evaporates? Assume zero background radiation.

electrostatics - Coulomb potential

It is known that the Coulomb potential can be obtained by Fourier transform of the propagator from E&M. Is this because one of Maxwell's equations have the form $\nabla \cdot \mathbf{E}=\rho$?

pattern - Woody words, tinny words

Note: There are some rude words in this question.

There's an implicit puzzle in posed in this classic Monty Python sketch (you can also ready the script online).

Various words are described as being 'woody' or 'tinny', and one is described as being 'PVC'.

Here are some woody words:

• Gorn (possibly Gone)


• Sausage


• Seemly


• Prodding


• Vacuum



• Bound


• Vole


• Caribou


• Intercourse


• Pert


• Thighs


• Botty


• Erogenous Zone


• Ocelot


• Wasp



• Yowling

Here are some tinny words:

• Litter bin


• Newspaper


• Antelope


• Recidivist


• Tit


• Simpkins

This one is PVC:

So here's the question, what makes a word woody, and what makes it tinny? or even PVC? I'd prefer some kind of rule which can be used to categorise words into one of these categories.

Here's some words to test your rule on, tell me which category they belong to and why?

• Guildford


• Godalming


• Artichoke


• Trigonometry



• Brigand


• Fox


• Bookworm


• Gaggle

logical deduction - Heyawake: An Introductory Puzzle

This is a Heyawake ("divided rooms") puzzle.

Rules of Heyawake:

• 
  

  Shade some cells of the grid.

  
  


• 
  

  Shaded cells cannot be orthogonally adjacent; unshaded cells must be orthogonally connected.

  
  



• 
  

  There cannot be a horizontal or vertical line of unshaded cells that passes through two borders.

  
  


• 
  

  If a number is in a room, there must be exactly that many shaded cells in that room.

  
  

Answer

I think the answer is as follows

Reasoning

The first observation is that the box with the 3 near the top left hand corner can have its cells shaded in only one way. It must be a checkerboard pattern to avoid adjacent shaded cells and we must not trap against an unshaded cell against the edge

Now the cell with the 3 in the bottom left corner can immediately have its cells shaded alternately and no cell in the adjacent box can be shaded since it would either abut an already shaded square or trap an unshaded square. By rule 3 this means that any cell which is two over from the unshaded cells must be shaded like so.


From here, we can immediately colours the cells in the bottom 3-cell in a checkerboard pattern and the adjacent 2-cell must then have its 2nd and 4th cell shaded. Now, the cell above the 2 can be shaded (because below it cannot be). And by rule 3, the cell up-right and two down-right of that must also be. This gives us the following

Now look at the boxes with a 4 in them. The first column of the first must be completely unshaded leaving a 3 by 3 square to shade in a checkerboard pattern. This is the same for the other 4-cell. We cannot use the shading which surrounds an unshaded square so we must have all four shaded squares on the diagonals (the X-shape). Since the ends of these X-shapes are adjacent for the 4-cells, we can guarantee three of the shaded squares in each X. That is, we can shade as follows

We can shade some more squares on the right using rule 3. Then note that the square marked 'x' must be unshaded because, otherwise it would cause an unshaded section at the top to be trapped by shaded squares.

Using this fact and exploiting the 0-cell, we can shade in some more squares using rule 3.

Next we can look at which of the two shading options do we use in the remaining square of the 4-cell. Notice in this next diagram that none of the squares with an 'x' can be shaded. If we wish to satisfy rule 3, one of the two squares above this column must be shaded but if we take the shading choice as marked by the two blue 'A's, we will end up trapping an unshaded region on the left.


Hence, the shading must be as follows

From there, there is only one option for shading the remaining squares in the central 3-cell and we can use repeated applications of Rule 3, while ensuring that the unshaded section at the bottom doesn't become disconnected from the rest, to shade in the rest of the squares.

quantum field theory - Irrelevance of parastatistics for space dimension > 2

Consider a system of $n$ undistinguishable particles moving in $d$-dimensional Euclidean space $E^d$. The configuration space is $M=((E^d)^n \setminus \Delta)/S_n$ where $\Delta$ is the diagonal (subspace where at least 2 particles have coincidental positions) and $S_n$ is the group permuting the particles

Quantization of this system yields superselection sectors corresponding to unitary irreducible representations of $\pi_1(M)$: $S_n$ for $d > 2$, $B_n$ for $d = 2$. The trivial representation yields bosonic statistics, the sign representations yield fermionic statistics. For $d > 2$ there are no other 1-dimensional representations. For $d = 2$ there are other 1-dimensional representations in which switching two particles generates an arbitrary phase. These yield anyonic statistics.

What about higher dimensional irreducible representations? These correspond to parastatistics. It is said that for $d > 2$ we can safely ignore them because in some sense they are equivalent to ordinary bosons/fermions. However for $d = 2$ this is not the case. Why?

Why is parastatistics redundant for $d > 2$ but not for $d = 2$?

electromagnetism - Is the electric field zero inside an ideal conductor carrying a current?

By an ideal conductor, I mean one with zero resistance. Inside an ideal conductor with no current, the electric field is zero, but is the electric field still zero with the ideal conductor carrying a current?

Can x-ray radiation be compared to background radiation?

I've been trying to learn about the possible effects of x-ray radiation from dental x-rays and most of the resources I come across compare the exposure to that of natural background radiation.

Here's an example (not specific to dental x-rays): http://www.hpa.org.uk/webw/HPAweb&HPAwebStandard/HPAweb_C/1195733826941?p=1158934607708

I know electro-magnetic-radiation comes in different wavelengths and some are ionizing and some are not. Is this a valid comparison?

Update: Thanks for the helpful responses so far. Looking at the chart here from @anna v, is comparing x-rays with background radiation comparing EMR that is 10^-10 with wavelengths that are longer than infrared? That's what I still don't understand. Wouldn't those different types of waves have different properties?

Answer

First of all terminology:

When physicists speak of radiation they primarily speak of electromagnetic radiation. When health physicists speak of radiation they include radiations of other types, alpha and beta and neutrons in addition to gamma and xrays. They have developed a system where radiation is given in Becquerel ignoring the particular source.

So when somebody says that one xray is equivalent to background radiation, they compare the becquerels that one gets from one xray to the equivalent becquerels one would get from the ambient surroundings. This background comes from cosmic ray muons (about 1 per cm^2 per second) to natural radioactivity of stones and materials, to gases in the atmosphere released by volcanoes etc. Natural radiation is mainly non electromagnetic, since high energy photons produced by close by radioactive decays are easily absorbed by the materials intervening, and photons coming from the cosmos are absorbed or interact high in the atmosphere ( another important point for life, to have an atmosphere).

Cosmic background radiation is another story, and is not health physics related :it is photons left over from the Big Bang and has very little energy, it is in the microwave part of the electromagnetic spectrum( mm) whereas x-rays are of higher energy, in the range of nanometers.

The comparison is valid because it is the result of painstaking studies of calibration and measurements.

Response to updated question:

Wouldn't those different types of waves have different properties?

Matter responds differently to the different wavelengths of photons, due to the increasing energy they carry which is proportional to their frequency and inversely proportional to their wavelength.

The column on the far right gives the energy of the photon. A micron wavelength is in the electron Volt range and can affect molecular distances and cohesion and living matter. Below that the interaction with matter is in bulk, not individual molecules and cells after the Ultraviolet level. The electromagnetic radiation that can affect health is ultraviolet and smaller wavelengths. The smaller the wavelength the larger the possibility of destruction of living cells which is the study of health physics: by, as the frequency increases, heating in depth,breaking of chemical bonds, ionizing, and finally destroying complete cell structures when going to MeV energies.

homework and exercises - Light of a specific wavelength going through a prism

I won't post the entire question here since I would just like a bit of help getting started as I am quite lost. The question is essentially saying that a light ray parallel to the x axis with a wavelength of $\lambda$ is heading into an isosceles triangle with an index of refraction $n_2$ from air with an index of refraction $n_1$. I'm supposed to find the value of $n_1$ so that light of wavelength $>\lambda$ will exit the prism and light of wavelength $<\lambda$ will be reflected perpendicular to the x axis. I am given the value for $n_2$, $\lambda$ and the angles inside the prism.

The main thing I don't understand is how the wavelength is effecting whether or not the light exits the prism. I believe this depends on whether or not the light is experiencing total internal reflection but in this scenario that would be given by $sin\theta _c=\frac{n_2}{n_1}$ I also know that $n=\frac{c}{v}=\frac{c}{f \lambda }\Rightarrow \sin\theta_c=\frac{\lambda_1}{\lambda_2}$ since only wavelength changes in the different mediums. So the the critical angle is clearly affected by the wavelength of light in the different mediums but I just can't figure out how this relates to higher or lower frequencies exiting or staying in the prism.

Answer

To complement Samuel Weir's answer, look at this image of incandescent (white) light passing through a prism.

As can be seen violet light is refracted ('bent') more than red light. Violet light is more energetic than red light and we know that:

$\lambda=\frac{hc}{E}$. ($h$ is Planck's constant, $c$ the speed of light, $E$ is energy).

So violet light is of shorter wavelengths (larger $E$). Light of shorter wavelength is thus more refracted by a prism than light of longer wavelength.

Edit:

Following the asker's comments I understand his problem to be as in the following diagram:

For rays following path I, we get refraction:

$n_2\sin\theta_1=n_1\sin\theta_2$

Or: $\sin\theta_1=\frac{n_1}{n_2}\sin\theta_2$

For rays following path II we have total reflection:

Acc. Snell's Law, for total reflection, $\theta_2=90^0$, so $\sin\theta_2=1$.

So $\sin\theta_1=\frac{n_1}{n_2}$ and $n_1=n_2\sin\theta_1$.

Because the triangle is isosceles, $\theta_1=45^0$, so:

$\large{n_1=\frac{\sqrt{2}}{2}n_2}$

This the critical value for $n_1$, in reality to get total refection the condition:

$\large{n_1<\frac{\sqrt{2}}{2}n_2}$ must be met.

black holes - Bekenstein bound and an infinite universe

I have a question about the Bekenstein bound or the holographic principle.

I make the following assumptions:

1. The universe is infinite



2. The universe is homogeneous

These assumptions are at least considered plausible (on large scale). The Bekenstein bound (or holographic principle) say that the information contained in a ball of radius $R$ is at most $q A$ where $q$ is a proportionality constant and $A$ the area of the sphere. But if the universe is homogeneous, its density is constant and so the information contained in a sphere should be proportional to $R^3$; while the bound is proportional to $R^2$. That seems impossible, a big enough sphere will eventually exceed the bound.

A similar argument seems to imply that some sufficiently large portion of the universe should necessarily be a black hole.

What I am doing wrong here?

Answer

Bekenstein's bound does indeed fail when applied to an infinite universe, but then Bekenstein specified conditions for the validity of his bounds:

The system must be of constant, finite size and must have limited self-gravity, i.e., gravity must not be the dominant force in the system. This excludes, for example, gravitationally collapsing objects, and sufficiently large regions of cosmological space-times. Another important condition is that no matter components with negative energy density are available.

Various attempts have been made to extend the Bekenstein bound. The current front runner appears to be Bousso's covarient entropy bound - see also the summary on Wikipedia. This works for an infinite universe as well as for systems conforming to Bekenstein's original restrictions.

special relativity - Why does a sign difference between space and time lead to time that only flows forward?

Ever since special relativity we've had this equation that puts time and space on an equal footing:

$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2.$$

But they're obviously not equivalent, because there's a sign difference between space and time.

Question: how does a relative sign difference lead to a situation where time only flows forward and never backward? We can move back and forth in space, so why does the negative sign mean we can't move back and forth in time? It sounds like something I should know, yet I don't - the only thing I can see is, $dt$ could be positive or negative (corresponding to forwards and backwards in time), but after being squared that sign difference disappears so nothing changes.

Related questions: What grounds the difference between space and time?, What is time, does it flow, and if so what defines its direction? However I'm phrasing this question from a relativity viewpoint, not thermodynamics.

Answer

We can move back and forth in space, so why does the negative sign mean we can't move back and forth in time?

As illustrated in the answer by Ben Crowell and acknowledged in other answers, that relative sign doesn't by itself determine which is future and which is past. But as the answer by Dale explains, it does mean that we can't "move back and forth in time," assuming that the spacetime is globally hyperbolic (which excludes examples like the one in Ben Crowell's answer). A spacetime is called globally hyperbolic if it has a spacelike hypersurface through which every timelike curve passes exactly once (a Cauchy surface) [1][2]. This ensures that we can choose which half of every light-cone is "future" and which is "past," in a way that is consistent and smooth throughout the spacetime.

For an explicit proof that "turning around in time" is impossible, in the special case of ordinary flat spacetime, see the appendix of this post: https://physics.stackexchange.com/a/442841.

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References:

[1] Pages 39, 44, and 48 in Penrose (1972), "Techniques of Differential Topology in Relativity," Society for Industrial and Applied Mathematics, http://www.kfki.hu/~iracz/sgimp/cikkek/cenzor/Penrose_todtir.pdf

[2] Page 4 in Sanchez (2005), "Causal hierarchy of spacetimes, temporal functions and smoothness of Geroch's splitting. A revision," http://arxiv.org/abs/gr-qc/0411143v2

time - Hamilton equations from Poisson bracket's formulation

Referring to Wikipedia we have that the equation of motion for a $f(q, p, t)$ comes from the formula \begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} f(p, q, t) = \frac{\partial f}{\partial q} \frac{\mathrm{d}q}{\mathrm{d}t} + \frac {\partial f}{\partial p} \frac{\mathrm{d}p}{\mathrm{d}t} + \frac{\partial f}{\partial t} \tag{1} \end{equation} which is, un the Poisson bracket notation $$ \frac {\mathrm{d}}{\mathrm{d}t} f(p, q, t) = \{f, H\} + \frac{\partial f}{\partial t} \tag{2} $$ Now, many books say that if we want to get the Hamilton equations from here you just have to substitute, respectively for the first and the second equation ($k= 1, \dots, 2\cdot3N$ equations actually, for a system with $N$ particles and $3$ degrees of freedom) $f(q, p, t) = q(t)$ and $f(q, p, t) = p(t)$. So you should get the two equations: $$ \frac {\mathrm{d}}{\mathrm{d}t} q = \{q, H\} + \frac{\partial q}{\partial t}\tag{3} $$ and $$ \frac {\mathrm{d}}{\mathrm{d}t} p = \{p, H\} + \frac{\partial p}{\partial t}\tag{4} $$ So, in order to get back to the Hamilton equations we should have $$\frac{\partial p}{\partial t} = \frac{\partial q}{\partial t} = 0,\tag{5}$$ but why it is so? Why the partial time derivative is zero, if $q$ and $p$ are function of time?

Answer

The confusion stems from a bit of mathematical sloppiness. It's necessary useful sloppiness, because as you'll see in a moment, the full machinery is a pain, but it's helpful to see and keep in the back of your mind when confusions like this arise.

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Phase Space Trajectories

I won't go through the trouble of constructing the cotangent bundle to the configuration space and all of that mess - we can start from the intuitive notion of the phase space $\Omega$. A point $x\in\Omega$ can be labeled by the corresponding position $q$ and momentum $p$ (i.e. $x\equiv(q_x,p_x)$). It's important to note that $q$ and $p$ are not functions of time or anything else - they are just numbers which label a particular location in phase space.

From here, we consider the notion of a trajectory through phase space. A trajectory $\gamma$ is a continuous map which takes a real number (the time) and maps it to a point in phase space:

$$\gamma:\mathbb{R} \rightarrow \Omega$$ $$t \mapsto \gamma(t)$$

If we feed $\gamma$ a time, it tells us the location of the system in phase space. As we move forward in time, the trajectory tells us how the state of the system evolves.

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Dynamical Variables

A dynamical variable $F$ takes a point in phase space and a value of time and maps them to a real number: $$F :\Omega\times \mathbb{R} \rightarrow \mathbb{R}$$ $$(q,p,t) \mapsto F(q,p,t)$$

Next we introduce the projection functions $\mathcal{Q}$ and $\mathcal{P}$, which map a particular point in phase space to the corresponding values of $q$ and $p$. $$\mathcal{Q}:\Omega\times\mathbb R \rightarrow \mathbb{R} $$ $$(x,t)\mapsto \mathcal{Q}(x,t)\equiv\mathcal{Q}\big(q_x,p_x,t\big) = q_x$$ and $$\mathcal{P}:\Omega\times\mathbb R \rightarrow \mathbb{R} $$ $$(x,t) \mapsto \mathcal{P}(x,t)\equiv\mathcal{P}\big(q_x,p_x,t\big) = p_x$$

Essentially, $\mathcal{Q}$ just takes a point in phase space and a time and tells you the position coordinate while ignoring the momentum coordinate and the time, while $\mathcal{P}$ takes a point in phase space and a time and tells you the momentum while ignoring the position coordinate and the time. Notice that these two functions are particular examples of time-independent dynamical variables, in the sense that $\frac{\partial \mathcal{P}}{\partial t} = \frac{\partial \mathcal{Q}}{\partial t}=0$.

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Dynamical Variables Along Phase Space Trajectories

Now we can combine these two concepts. Given a trajectory $\gamma$ and a dynamical variable $F$, we can combine them to form a map $F_\gamma$ which takes a single real number $t$ and returns the value of $F$ at time $t$ along $\gamma$:

$$F_\gamma : \mathbb{R} \rightarrow \mathbb{R}$$ $$ t \mapsto F\big(\gamma(t),t\big)$$

We can apply this definition to $\mathcal{Q}$ and $\mathcal{P}$. Notice that $$\mathcal{Q}_\gamma : \mathbb{R} \rightarrow \mathbb{R}$$ $$ t \mapsto \mathcal{Q}\big(\gamma(t)\big)$$ so $\mathcal{Q}_\gamma(t)$ is the position coordinate of the system at time $t$, while $\mathcal{P}_\gamma(t)$ is the momentum coordinate of the system at time $t$.

Notice that for a given dynamical variable $F$, we can also write that $$F_\gamma(t) = F\big(\gamma(t),t\big) = F\big(\mathcal{Q}_\gamma(t),\mathcal{P}_\gamma(t),t\big)$$

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Total Derivatives

Because such maps are functions of a single variable, it makes sense to take a total derivative with respect to time. This is the total rate of change of $F$ along the trajectory $\gamma$:

$$\frac{dF_\gamma}{dt} \equiv \frac{\partial F}{\partial q}\frac{d\mathcal{Q}_\gamma}{dt} + \frac{\partial F}{\partial p}\frac{d\mathcal{P}_\gamma}{dt} + \frac{\partial F}{\partial t}$$

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Hamilton's Equations

Hamilton's equations are the differential equations which govern phase space trajectories. Without delving into their derivation, they tell us that $$ \frac{d\gamma}{dt} \equiv \left(\frac{d\mathcal{Q}_\gamma}{dt},\frac{d\mathcal{P}_\gamma}{dt}\right) = \left(\frac{\partial\mathcal{H}}{\partial p},-\frac{\partial\mathcal{H}}{\partial q}\right)$$ where $\mathcal{H}$ is the Hamiltonian - yet another dynamical variable.

Once the Hamiltonian $$\mathcal{H}:\Omega \times \mathbb{R} \rightarrow \mathbb{R}$$ $$(q,p,t) \mapsto \mathcal{H}(q,p,t)$$ has been written down, then all possible phase space trajectories have been determined.

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Poisson Bracket

Using Hamilton's equations, we can rewrite the total derivative in the following way: $$ \frac{dF_\gamma}{dt} = \frac{\partial F}{\partial q}\frac{d\mathcal{Q}_\gamma}{dt} + \frac{\partial F}{\partial p}\frac{d\mathcal{P}_\gamma}{dt} + \frac{\partial F}{\partial t} = \left(\frac{\partial F}{\partial q}\frac{\partial \mathcal{H}}{\partial p} - \frac{\partial F}{\partial p}\frac{\partial \mathcal{H}}{\partial q}\right) + \frac{\partial F}{\partial t}$$

This motivates the definition of the Poisson bracket of two dynamical variables: $$ \{A,B\} \equiv \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p} \frac{\partial B}{\partial q}$$

at which point we can rewrite the total derivative formula one last time:

$$\frac{dF_\gamma}{dt} = \{F,H\} + \frac{\partial F}{\partial t}$$

All done! Notice that the right hand side makes no mention of the trajectory $\gamma$, for good reason - once we specify the Hamiltonian and our particular location in phase space, then there's no freedom left in the evolution of the system (and therefore no freedom left in the evolution of any dynamical variable).

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The Punchline

We're now equipped to answer your question. Consider the function $\mathcal{Q}$ (the function which takes a phase space point $(q,p)$ and returns its position coordinate $q$), as well as the associated function $\mathcal{Q}_\gamma$ which is associated to a phase space trajectory. We have that

$$\mathcal{Q}(q,p,t) = q$$ so $$\frac{\partial \mathcal{Q}}{\partial q} = 1$$ $$\frac{\partial \mathcal{Q}}{\partial p} = 0$$ $$\frac{\partial \mathcal{Q}}{\partial t} = 0$$

and therefore

$$\frac{d\mathcal{Q}_\gamma}{dt} = \{\mathcal{Q},\mathcal{H}\}+\frac{\partial \mathcal{Q}}{\partial t}$$ $$ = \left(\frac{\partial \mathcal{Q}}{\partial q}\frac{\partial \mathcal{H}}{\partial p} - \frac{\partial \mathcal{Q}}{\partial p}\frac{\partial \mathcal{H}}{\partial q}\right) + \frac{\partial \mathcal{Q}}{\partial t} $$ $$= \frac{\partial \mathcal{H}}{\partial p}$$

and similarly, $$\frac{d\mathcal{P}_\gamma}{dt} = -\frac{\partial \mathcal{H}}{\partial q}$$

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So there you have it. When we write everything out in excruciating detail, there is no ambiguity whatsoever. $q$ and $p$ are numbers, not functions, so differentiating them is meaningless. When we do physics, what we're actually differentating are the projection functions $\mathcal{Q}$ and $\mathcal{P}$, as well as their associated functions which are attached to the phase space trajectory $\gamma$ along which the system evolves. Again, it's crucial to notice that $\mathcal{Q}$ and $\mathcal{Q}_\gamma$ are associated with each other, but are not the same thing.

Of course, when I do physics, I don't write all this out - I differentiate $q$ and $p$ just like everybody else. But it's useful to be able to frame problems in this context when those little points of confusion arise.

newtonian mechanics - Why does rotation simulate gravity if motion is relative?

In Einstein's theory of relativity, if motion is truly relative, then why would somebody in a rotating space station experience (artificial) gravity? I mean, I get why they experience gravity IF the space station is rotating, but what I do not get is how can you say it is rotating in the first place? Would you not need to define its motion in relation to another point in space?

So let's say we have two space stations and in our hypothetical universe they are the only two things that exist in the entire universe.

• In space station 1, there is no sensation of gravity.


• In space station 2, there is a sensation of gravity.

So we know that space station 2 is rotating and space station 1 is not because of the sensation of gravity we feel in station 2.

To an observer in station 2, they would feel gravity and see station 1 appearing to orbit around them. An observer in station 1 would feel no gravity and see station 2 spinning but staying in one spot. I think everyone would agree with all of those observations thus far.

But here is where it breaks down, I think, you have to have some fixed reference point in space to be able to say that station 2 is spinning and it is not station 1 orbiting in order to get the sensation of gravity on station 2. Newton seemed to have an answer to this because he said there was a fixed "at rest" but according to Einstein, I don't think there is a fixed at rest?

Does my question make sense? My confusion also applies to acceleration, it seems you have to have some fixed reference frame to say anything is accelerating at all. We know you would feel the effects of acceleration if in a space ship in deep space, but why if all motion is relative. I mean, how can you even say you are accelerating and it is not everything else in the universe accelerating and you are actually standing still without a fixed “at rest” point in space?