Ask the Physicist!

QUESTION:
I was reminded today about the old canard, "What weighs more a pound of feathers or a pound of lead?" i'd always accepted that a pound is a pound. I even accepted that weight isn't mass. I was confronted in my own mind today about the fact that gravity would affect a pound of feathers differently that lead if they were both tossed into a swimming pool. At what point is buoyancy considered in weight. A ballon full of helium has mass. In a vacuum its weight on Earth would be the same. The Earth abhors a vacuum and the balloon rises despite its weight.

ANSWER:
The confusion that arises here is one of nomenclature, what does the word "weight" mean. in everyday life, weight is what a scale reads. But no self-respecting physicist would take this as what weight is (although I once saw an introductory physics text which defined weight as what a scale reads). In physics, weight is the force which the earth exerts on an object. Your supposition that a buoyant force should matter is totally correct. Another example is an astronaut in orbit who is referred to as being weightless; but the earth is exerting a force on her so her weight is not zero.

QUESTION:
Assume there is a sphere of mercury 1 million km in diameter, with uniform density. If 1/8 of its volume is removed instantly from one edge of the sphere, how long would it take for the opposite edge to react?

ANSWER:
Why choose a liquid? This would greatly complicate trying to do any calculations. Why such a huge fraction to remove? Again, this complicates calculations and rules out perturbation methods. Also, exactly what happens would be determined by the shape of the removed volume. Since anything like a reasonable answer would involve a gigantic computational effort, I will just give a brief overview. First, find the speed of sound in mercury; that is the fastest that information can travel from one side or another. I can imagine surface waves to propogate from the removal site to the opposite side but have no idea how to calculate their velocities. The likeliest thing you would see is, instead of a static sphere, a dynamic shape would emerge and would vibrate, gradually losing amplitude until it settled down to a static smaller sphere (but maybe rotating). But what I have been imagining is that I have something smaller, maybe basketball size. Your sphere is so big that the pressure would be huge deep under the surface which would really complicate things.

QUESTION:
If electrons need a very specific energy amount to change levels, what happens when a photon hits them which is very close to the amount of energy needed? Does uncertainty help them interact? For instance if we need precisely 680 nm wavelength, and a photon appears with a wavelength of 681, or 680.01, or 679.99 Is it possible that wavelength is this discrete or is it uncertain? Does the difference in energy exist and does it need to be made up somewhere?

ANSWER:
What you are describing is generally discussed as "line broadening". What you are describing, a photon being absorbed by a level in an atom, is the time reversed process of a level in an atom decaying by emitting a photon. The two processes involve precisely the same physics; and the decay is much easier to understand than excitation; also, experiments to observe state decay by photon emission are much easier to perform. Therefore I will describe the state decay rather than what you are asking.

Now, consider an excited state in an atom which decays. Out pops a photon with a specific energy. A photon, being a stable particle, lives forever if it suffers no interactions. This is an uncertainty principle example: since we could observe it for infinite time, its energy is not uncertain at all, it has one energy which is the energy of the state from which it came. The only uncertainty in the experiment is due to how well we can measure the energy. Now we repeat this experiment thousands or millions of times and what we find is that the state is not a single energy but has a distribution of possible energies which can be observed as long as our photon measurement is accurate enough (which it is).

So maybe somebody says, "hey, the gas of atoms in this experiments is very hot, I wonder what would happen if it were cooler." What you find is that the colder it the photon spectrum for the emission from the state gets narrower and narrower. So you say, "Aha the state can't really be many possible energies because we are just seeing the Doppler effect from the motion of the atoms."

But here is the catch: as you make it colder and colder, it eventually stops getting broader. This now is the physics you allude to in your question, the uncertainty principle stipulates that ΔE≤ℏ/Δt. So the longer a state lives, the sharper its spectrum will be. The only atomic states which have zero widths are the ground states. In nuclear physics lifetimes of states are often measuring their spectra.

QUESTION:
I'm currently reading a book that involves physics (the name of the book is House of Leaves). I'm having a bit of a problem calculating a distance that an objects falls through, specifically a coin (a quarter). I tried using Newton's simplified formula to measure the distance that the coin falls, just as the book suggests. The quarter falls for 50 minutes, and (I don't understand why), the measurement I have to use for gravity is 32ft so that I can calculate the distance the coin travels downward.

[Here the questioner shows a bunch of arithmatic calculations ending with his answer of about 27 thousand miles. I have deleted all that.]

And my question is, is there another more concrete/specific/accurate way to calculate the distance this coin travels for 50 minutes as it falls? Because the book says there's a more accurate way to do it, yet it shows none. I don't know why the book (or the formula) uses 32 ft as gravity acceleration, but if there's another way to calculate this, does it exist? I've been trying to find a similar (or better) way to calculate this for a long time now because I just find it impossible that a coin can travel a distance of 27,272.73 miles in 50 minutes even though the coin is not moving at a constant speed, but it is accelerating, which probably means every second adds more speed.

ANSWER:
The first thing I must note is that the calculation you did assumes that the gravitational field is uniform (g=32 ft/s² the whole time) and that there is no air drag. In fact, gravity gets weaker and weaker the farther up you go to drop the coin; and, for a coin, the air drag gets very important at high speeds. Therefore, the simple calculation you did, h=½gt², is essentially meaningless. I much prefer working in SI units (meters, kilograms, seconds) and will convert to Imperial units at the end; so now g=9.8 m/s²≈10 m/s² and t=50 min=3000 s. I will discus three situations:

The simple calculation you did: h=½gt²=5x3000²=45x10⁶ m=27,962 miles. I like to think of distances on this scale in terms of earth radii, h=7 earth radii, more than a tenth of the distance to the moon. So,your calculation was spot on. Also, of interest is the speed v when it hits the ground. v=gt=10x3000=30,000 m/s=67,108 mph .
A calculation where we ignore air drag but now the force on a particle of mass m is not a constant but given by
    F=ma=-mMG/r²
where M=6x10²⁴ kg is the mass of the earth, the universal gravitational constant is G=6.67x10^-11 N·m²/kg², and R=6.4x10⁶ m is the radius of the earth. Since the acceleration is dv/dt,
    dv/dt=-MG/r²,
rather than dv/dt=-g. Now, the mathematics to get from here to an expression for t as a function of h is tricky and tedious, so I will jump right to the answer,
    t=∫dr/√[2GM((1/r)-(1/(R+h))];
the integral has limits R+h to R. r is the variable of integration and is the distance from the center of the earth. This integral can actually be expressed in closed form but it involves the inverse hyperbolic tangent function which I did not want to wade through. So, I turned it over to AI (ChatGPT) and asked for the numerical answer and a graph showing the time to fall as a function of h. Your answer is h=9.27x10⁶ m=5760 miles=1.45 earth radii.
Finally, let me say a few words about air drag. It turns out that its terminal velocity is about 28 mph=12.5 m/s. Keep in mind that this is what it is at sea level—it gets larger as you go higher. But, as a rough estimate let's just assume it is constant. When dropped the terminal velocity is probably acquired quickly, so I will just assume that it falls with a constant speed for 3000 s. The distance will be about h=3.75x10⁴ m=23 miles=0.006 earth radii.

If you don't know what terminal velocity is you can easily google it. If the math is over your head, don't worry about it—the graph illustrates what all the math gets you in the end.

I haven't said anything about impact velocities here, but it is pretty easy to talk about what they are qualitatively and comparatively. The uniform field (unphysical) quarter will be screamin' fast; the physical quarter without friction will be quite a bit slower; the most realistic, the quarter which also experiences air drag, will be a puny 28 mph.

QUESTION:
If mass bends space. The structure of space gets warped or bent. That is my understanding of gravity. How then does it relate to quantum mechanics? Is it because mass interacts with the fabric of space (quantum fields)?

ANSWER:
See an earlier answer. Only the second paragraph is of interest to you. Unfortunately, the essay referred to is no longer on line.

QUESTION:
I'm currently reading, "Surely You Are Joking Mr. Feynman." There is a lot of stuff I don't understand (I only took two college level physics classes), but one thing has me very curious. He talked about an early experiment he was doing that involved "shaking" electrons. What does it mean to shake an electron and how is it done?

ANSWER:
This is simply a reference to classical electricity and magnetism (E&M). The way students are generally introduced to electromagnetic radiation (light, radio, x-ray, etc.) is to imagine an electric charge in empty space oscillating, maybe like being attached to a spring which is vibrating; hence Feynman's "shaking". The result of this shaking is electromagnetic waves. What he was interested in was that if another charge was in the vicinity, it would feel forces because of the radiation and it would start shaking itself which meant that the first charge would feel the radiation from the second. In other words the first would feel forces which it originally caused. What he wanted was an explanation for another effect and it didn't pan out.

QUESTION:
We are in motion in many ways. The earth rotates and simultaneously revolves around the sun, while the solar system revolves around the Milky Way galaxy, etc. I have seen the question “Where are we going?” asked and answered, but I have never seen an answer to “What effect does all of this motion have on the time dilation predicted by Einstein’s theory of relativity?”. In other words, if all of this motion ceased, how much faster would time pass?

ANSWER:
Time dilation with respect to whom? For example, suppose you are on earth and Elon Musk is on Mars. The details of why you and he happen to be moving like they are is not important. Relative to you he will have some velocity; you would observe his clock to be running more slowly than yours as determined by using that velocity to calculate the time dilation. For most applications this is purely academic because the speeds are all very small relative to the speed of light.

QUESTION:
According to Einstein's theory of relativity, nothing, not even light, can escape from inside a black hole due to its immense gravitational pull. However, I have also read that when two black holes collide with each other, they release tremendous amounts of energy, primarily in the form of gravitational waves. My question is, how can this energy escape from the merger of two black holes if, in normal circumstances, nothing should be able to escape from a single black hole? Shouldn't all the energy collapse inside the black hole instead of being released?

ANSWER:
Here is a nice video of two black holes colliding. Each black hole has an event horizon which defines the region, inside of which, nothing can escape. As the two spiral into each other, their event horizons merge into one. Nothing inside can escape, but much of the huge energy release will be outside.

QUESTION:
Imagine you have an automatic baseball batting machine, stationary, which consistently swings the bat at a ball on a tee at 30 mph. Now, imagine that the batting machine is set on rails/tracks running perpendicular to home plate, so that it can travel forward as it swings (still set to swing at 30 mph). If the batting machine is moving forward past home plate at 10 mph when it swings and strikes the ball off the tee, is the swing speed now 40 mph? In other words, is the result swing speed plus forward motion? Does it make a difference if the forward motion is less than the 30 mph swing speed? What if it is moving forward at 30 mph? 50 mph?

ANSWER:
I presume that when you say "set to swing at 30 mph" the speed of the bat at the point where it hits the ball is 30 mph relative to its own rest frame. The only thing that matters is the relative speed of the ball and bat; so yes, machine speed =20 mph, relative speed=40 mph; machine speed =30 mph, relative speed=60 mph; machine speed =10 mph, relative speed=40 mph; machine speed=5 mph, relative speed=35 mph; etc.

QUESTION:
My understanding of a quantum state is it is given as a probability. It can not be known until it is measured. I never thought that a quantum particle could be in both states at the same time, only that its state was unknowable until measured.

ANSWER:
When you solve a linear algebraic equation you get an answer. When you solve a quadratic algebraic equation you get two answers. The thing in quantum mechanics which tells you the nature of state of some system is the wave function which is the solution(s) of the equations of quantum mechanics. These are partial differential equations which often will have more than one solution. So you are faced by a problem: Which is the appropriate solution for the situation? The answer is that the most general solution is a linear superposition* of all possible solutions. You may, for a given situation, be able to determine what the constants of the superposition by applying constraints. The square of the total wave function must be 1 because it is the probability of the system being in the state being in a given configuration determined by the As.

*Super position means: if ψ_n is the wave function for the n^th state possible, the form of the total wave function is Ψ=A₁ψ₁+A₂ψ₂+A₃ψ₃+….

QUESTION:
I skateboard down hills, but not steep ones. If I go down a hill at a 45 degree slope, and reach 20 mph, then the hill starts to flatten out to 15 degrees. Will I slow down, maintain or increase my speed?

ANSWER:
Seriously? A 45° slope sounds pretty steep to me! The only reason you would slow down would be friction, and it is hard to do a calculation without knowing lots of things you don't mention. I will later try to add in some reasonable variables to estimate the effects of friction. For starters, let's do the calculation assuming no friction. In the picture I have drawn, you are going down a hill of angle θ . The weight of you plus your board is denoted by W. This force is doing two things; the component of W parallel to the surface (x), -Wsinθ, is accelerating you down the hill and the component perpendicular to the surface (y), -Wcosθ, is pressing you into the surface. Newton's third law says that there is a force which you exert on the road, the road exerts an equal and opposite force on you; this is denoted by the force N, often called the normal force. Newton's second law says that the sum of all the forces equals the product of the acceleration and the mass m. Because there is no acceleration normal to the surface, N=Wcosθ and, for your situation, N=Wcos(15°)=0.97W . But, in the x-direction, there is an unbalenced force, -Wsinθ=ma where a is your accleration down the hill. Without going into further detail, a=-9.8sin(15°)=-2.54 m/s². To make this easier to relate to, the acceleration is 5.7 mph per second—your speed down the hill increases at the rate of 5.7 mph as every second ticks by.

Except for air drag, just about all frictional forces are proportional to the normal force. This means that f=μN where f is the frictional force and μ is called the coefficient of friction. Looking at some physics of skateboarding sites I found that μ is generally around 0.002-0.005 for skateboards. Taking the normal force to be about 200 lb and using the biggest μ, f=1 lb. But Wsinθ is about about 52 lb. So friction due to everything but air drag is pretty negligible sort of at about the 2% level. Finally let us consider air drag which depends basically on the square of the speed v and the area A presented to the onrushing wind. If you use SI units a pretty good estimate is f=¼Av². Using A=1 m² and v=20 mph I find f=11 lb. So this is not negligible! But in your case, starting the 15° slope going 20 mph, the net force is still down the hill, about 52-11=41 lb. Eventually, as your speed continues to increase, you will reach the speed where the air drag is 52 lb and you will then continue at that speed with no further acceleration. (This is called the terminal velocity.)

QUESTION:
As a scientist who considers many questions, I am wondering why it is still considered that the Sun will definitely destroy the earth. My understanding is that gravity is proportional to an objects mass, so the Sun is losing mass all the time. Would it be possible to believe that the Suns influence on the planets etc, will be less as it loses mass and the orbits increase (planets move away gradually). Therefore as the sun changes to a red giant and increases volume (during this transition period the planets maybe moving further away and maybe the further objects moving away exponentially,) will destroy less of the solar system orbiting objects.

ANSWER:
This is an academic question! The sun, from its birth to death, will lose 0.034% of its original mass; this includes both the solar wind and the loss due to fusion. The effect on planetary orbits of this change would be negligible.

QUESTION:
What happens when you have a 10eV positron colliding with a 5eV electron? Do you get a 5eV positron and 5eV photon? Or a 10eV photon.

ANSWER:
The problem here is that you do not know relativistic kinematics. The whole thing can be understood, but it is a little complicated. Before I outline how to do this I want to note a very important conservation law, conservation of electric charge. The total charge before the collision is zero so it must be zero after; your first guess, a photon and a positron has charge so could never happen. If you had a single photon after the collision, charge would be conserved but, as I will show you, this is not possible for another reason.

First, note that there is no definite result of an annihilation; the results depend on the energies of the incident particles and could be massive particles other than photons. And I believe that there could be more than two photons. For illustration purposes I will assume that there are two photons after the annihilation. In the little sketch I have included, the red vectors are photons, the blue vectors are the electron and positron. You specify suppose the electron, e.g., has an energy of 5 eV; but you must do the calculation relativistically because massless photons are involved, so you need to know the total energy E of each of the particles. For particles with mass E=√[(mc²)²+(pc)²]; for photons E=hf; here c is the speed of light, f is the frequency of the photon, m is the rest mass of the particle, p is the magnitude of the linear moment of the particles with mass, and h is Planck's constant. Linear momentum for a particle with mass is p=mv/√[1-(v/c)²]; for massless particles (photons) it is p=E/c. Momentum must also be conserved, total momentum before and after the annihilation must be the same. But, lest we forget, momentum is a vector and, in my picture, each momentum vector will have x and y components, so to conserve momentum you can separately conserve their components. So now, supposing that we know the energies and momenta of the electron and positron, there are three equations (conservation of energy, conservation of x and y momentum) but six unknowns (the energy of each photon, the x and y momentum component of each photon); so you cannot know what the photon energies are nor know the magnitudes and directions of the photons; there are constraints, e.g., the total photon energies are the sum of the photon and electron energies. In this situation you would probably measure the energy of one of the photons and, since p=E/c, you could infer the momentum components from where you have placed the detector. Now you would have three unknowns (the other photon) so you could predict where the second photon went and its energy.

I said I would say why you can't have a single photon after the annihilation. The reason is that it is impossible to conserve both the energy and the momentum. A simple case is the case of the two particles both at rest when they annihilate: the photon would have to have an energy equal to the sum of the rest mass energies of the two particles and no momentum, not possible.

QUESTION:
I know that nothing can go faster than the speed of light. But what exactly stops objects from reaching the speed of light? For example, I know that the Large Hadron Collider accelerates particles to 99.9999991% the speed of light, and achieves collision energy levels of 14TeV, and, I know that future accelerators are being planned which will have collision energy levels higher than that. How can future colliders with vastly higher energy levels achieve those energy levels without breaking the speed of light?

ANSWER:
The simplest way to understand this is that the inertia (mass) m of an object increases as the velocity v increases. The equation describing this is m=m₀/√(1-(v²/c²)) where m₀is the mass if v=0 and c is the speed of light. Note that if v=c the mass is infinite meaning it would take infinite energy to get there. I have always maintained that accelerators are poorly named because they aren't doing any significant acceleration; they would be better named energizers.

QUESTION:
I am working my way through Hawking's 1974 paper about what eventually became named Hawking radiation and black hole evaporation. Being a mathematician I lack certain fundamental physical intuitions and I have no astrophysicists to badger about my questions. Firstly, having read the paper, I now understand Hawking's argument about radiation arising because of the interaction between the Heisenberg Uncertainty Principle and time dilation in the gravity well, and that is much more satisfying than the virtual particle/antiparticle pair explanation we usually get. However I'm left with two fundamental questions:

Since this all occurs within the framework of General Relativity, and General Relativity notoriously does not conserve energy (photons in an expanding spacetime lose frequency and become less energetic and that energy goes nowhere, dark energy expands the universe and adds more dark energy, etc.), why do we insist on believing that the emission of radiation from the vicinity of the black hole must necessarily diminish the mass-energy of the black hole? From my understanding of Hawking radiation's close relative Uhru radiation the same thermal bath is expected to be experienced by an accelerated observer and yet (as far as I can discern) that energy isn't being subtracted from anything. It's just being created. Couldn't the same apply to Hawking radiation?
Regarding black hole evaporation, even if Hawking radiation is subtracting (mass-)energy, since at the horizon time is infinitely dilated to a standstill, how could a remote observer ever witness any evolution (evaporation) of the black hole? Isn't an infinitely time-dilated region essentially frozen in time and therefore unable to evolve from the standpoint of an observer at infinity?

ANSWER:
First, the site clearly states that I do not do astronomy/astrophysics/cosmology. In addition, highly specialized, technical questions are discouraged. I will speak to a couple of your points, but not give anywhere close to a complete answer.

It is, by no means a settled issue that in gravitational redshift the energy simply "goes nowhere".

The virtual-pair model is as deep as my knowledge of Halking radiation goes. A summary of my understanding can be found in the two links in the question following this one.

Although the question of energy conservation for the universe is not definitively answered, energy in a local, relatively compact volume is certainly conserved.

It is important to keep in mind, if you are studying physics, that what we know about the laws of physics can be assumed to only be valid for about 5% of the universe. There are no theories which describes the physics of dark energy or dark matter.

QUESTION:
What exactly is hawking radition is it electro magnetic wave or what?

ANSWER:
Hawking radiation can be anything which results from a particle-antiparticle pair; for example, if an electron-positron pair is created just outside the event horizon, one will be captured and one not captured, the not captured one being the radiation. The nature of the vacuum outside is such that such pairs are constantly popping into and out of existence; although you might think that would violate energy conservation, the uncertainty principle allows it if the pair exists for a short enough time before annihilating back to the vacuum. Everything I know about Hawking radiation can be seen here and the earlier answer linked to there. (I remind you that according to site ground rules I usually do not answer questions on astronomy/astrophysics/cosmology.

QUESTION:
Massive objects, such as the sun or better yet, the black hole at the center of the galaxy, all have gravity. Here on Earth I feel the gravity of Earth, but not that of the sun or black holes. Why is that?

ANSWER:
All the objects you talk about are spherically symmetric spheres, which means that the density depends only on how far you are from the center. The gravitational force from an object is proportional to its mass; i.e., if the mass of the earth were twice as large as it is, your weight would be twice as big. But, the force is also inversely proportional to the square of the distance from the center of the object; i.e., if the radius of the earth were twice as large but its mass the same as it is, your weight would be one fourth as large. The sun's mass is hugely bigger than the earth's; but your distance from the center of the sun squared is hugely bigger than your distance from the center of the earth squared. The effect from the distance falls off as distance increases much faster than the effec from the mass increases as mass increases. The net result is that the gravitational fields of very distant objects at points on the earth are neglibly small for most purposes.

So, as an example I thought I should do a calculation of the force on you here on earth due to the sun. To my surprise, I found that the acceleration you would experience due to this force would be about 5.9 m/s²! This is about 60% what your acceleration would be due to the earth, 9.8 m/s², hardly neglegible?! But I quickly realized that you would be completely unaware of this force; you, like the earth, are moving in an almost circular orbit around the sun and this force, pointing toward the center of the sun, is the force keeping you in orbit. This is just like the "weightlessness" experienced by astronauts in orbits around earth where they don't feel the gravity even though it is there. If you were placed at rest somewhere in space on the earth's orbit but without the earth, you would accelerate toward the sun with an acceleration 5.9 m/s².

QUESTION:
I have a rather interesting Physics question. If the normal force balances gravity when an object is at rest, then why doesn't it balance gravity when an object is lifted up at constant speed?

ANSWER:
It does! When moving with constant velocity an object is in equilibrium just as if it were sitting still; i.e., all forces on it must add to zero. So whatever is keeping moving upward is exerting an upward force (normal) equal to the object's weight. Not "rather interesting"!

QUESTION:
my question is regarding pry bar surface area: 4 inch surface area vs a 2 inch surface area - both are lifting 2000lbs does the 4 inch surface area help pry an object easier than a 2 inch surface area

ANSWER:
Assuming that everything else but the pry bars is identical, you get no lifting force advantage with a larger area. However, the force (2000 lb) is distributed over a bigger area. So if what you are lifting is breakable or deformable you might be better with a larger area. (By the way, area is measured in square inches, non inches.)

QUESTION:
How is it that a flashlight powered by a battery or two can instantly propel light to 186,000 miles per second?

ANSWER:
See a recent answer.

FOLLOWUP QUESTION:

Thanks, but I was directed to the answer below, and it doesn’t answer my question. At least I don’t understand it. It seems to me that the flashlight batteries energize the bulb, and then the bulb instantaneously emits photons at the speed of light. How can two humble flashlight batteries produce that much energy?

ANSWER:
Okay, let's try a different approach to answering your question. Suppose we have a 10 Watt light bulb. That means that this light bulb, when on, is consuming 10 Watts of energy per second, 10 Joules. This is the rate at which energy is being taken from the batteries, a very modest amount. The battery is a storage for energy and when all that stored energy is used up the bulb will stop emitting light. The energy of a car depends on its speed, but it can go at many different speeds so if you go faster you have more energy; but a car also has mass and its energy depends on its mass also. But light has no mass and only one speed, so it's a whole different can of worms. What you can get from the earlier answer is that that the light doesn't need to be "speeded up" to 186,000 mps, it is already there when emitted.

QUESTION:
When we look at galaxies that are 12 billion light years away, we see them has they appeared 12 billion years ago. I can understand that because it took the light that long to reach us. But then I did a thought experiment. Let's pretend we have a super powerful telescope that can not just see a galaxy 12 billion light years away, but one that can zoom up on a planet in that galaxy and then on the actual surface of that planet and let's pretend this planet has life that we can see. Doesn't have to be intelligent life. Let's say it is a planet with dinosaur type creatures and our telescope can peer across the cosmos and see these creatures as they moved around 12 billion years ago. Now here is the thought experiment.....these creatures, unknowingly, are being viewed by intelligent life from 12 billion light years away. Since we are seeing them as they appeared 12 billion years ago, from their perspective are they are being viewed by people from the future that technically do not even exist yet? Am I correct with this line of thought? And if so, would that mean a race of intelligent beings that do not even exist yet could view us from the future?

ANSWER:
Since your creatures are participating in your thought experiment, it would be more illuminiating to let them be sentient creatures who also have the same fantastic telescope model as yours. We have established that these creatures existed 12 gigayears ago. So what you are seeing is the distant past of this planet. I am bothered by your statement "…from their perspective are they are being viewed by people from the future that technically do not even exist yet?" They have no "perspective" because they aren't viewing you at all, they would be viewing your world 12 billion years ago in their world. If they thought about a bit they would say, "That place we are looking at might be looking at us in 12,000,000,000 years in their future if evolution allows them to be clever enough to make a telescope as cool as ours is."

QUESTION:
How long does it take for a photon/wave/particle to achieve light speed when an atom absorbs enough energy to emit. It seems to me that 0 to light speed cannot be instantaneous and adhere to basic laws of motion.

ANSWER:
First, since a photon has no mass there is nothing to prevent it from instantaneously going from zero to c. However, since a photon is not a point particle, it will not be created instantaneously but rather over the time of the decay of the atom to a lower state. This time depends on the time it takes the excited atom to decay; the photon is "created" but created already traveling c, (This time, the time for the transition to occur, is not be confused with the half-life of the excited state.)

QUESTION:
The rate of expansion of our Universe seems to be accelerating due to some unknown dark energy they say, can it be that a part of dark energy is just photons or electromagnetic radiation. Photons seem to have momentum so when they hit they transfer their energy and we seem to have a shit ton of radiation travelling all around the universe

ANSWER:
As clearly stated on the site, I do not do astronomy/astrophysics/cosmology. I think I can answer this question, though. First, it is probably safe to guess that, over large volumes, the photons move in random directions. Therefore, the net force on anything due to photons would be expected to be near zero. Second, it is unimaginable to me that any swarm of photons would exert sufficient force to move an entire galaxy or massive black hole.

QUESTION:
I have a conceptual question that touches the edge between physics and metaphysical belief. Some traditions hold that prayer or focused intention can, in some way, positively influence people or events at a distance--even without direct interaction. The idea is not necessarily that energy is transmitted, but that there's some kind of deeper interconnectedness that allows such effects to happen. From the standpoint of physics, would this idea directly contradict core principles like causality or conservation of energy--or could it be viewed instead as something that simply lies beyond what current physics can model or test? I'm not asking whether there's scientific evidence for this--I understand there isn't. I'm just curious whether physics as we know it would say "this is impossible," or if it would simply say "this is outside our scope."

ANSWER:
There are three different constituents of our known universe:

Normal matter: This is the stuff we see and can directly measure or observe. It consists of mass and force fields. Modern physics describes, pretty well, the universe we observe.
Dark matter: In recent decades there have been some observations which are not well-described by modern physics. To account for these anamolies the presence of matter which which interacts with the rest of the universe only via gravity has been hypothecized. We have never directly observed dark matter and there is virtually nothing known about it. It is estimated to be 27% of the universe.
Dark energy: More recently it has been observed that distant galaxies are accelerating away from us. The only way, at least as described by modern physics, that this could be is that there is an unknown repulsive force which is pushing the universe apart while gravity alone would be pulling so as to eventually pull it all back together. Like dark matter, we are mostly ingnorant of what this dark energy is or where it comes from. It is estimated to be 68% of the universe.

So the bottom line for modern physics is that it describes, pretty well, 5% of our universe! It would therefore be pretty foolish to say that the effectiveness of prayer is impossible; at least it would not be described by physics. I personally do not believe in the effectiveness of prayer, but I acknowledge that we do not know everything.

QUESTION:
There's a debate where even the dictionaries can't agree. The word is opaque. Some definitions say it's the same as translucent. I contend that this is incorrect.

ANSWER:
The meaning of translucent is obvious: light passes through. The meaning of opaque, in physics at least, is no light passes through. Their meanings are complete opposites.

QUESTION:
I’m am having a debate with a friend about flat earth. I know it’s a globe but I can’t explain. Can you give a quick rundown how water sticks to a spinning ball?

ANSWER:
It's quite easy: the force of gravity is holding it from being thrown off. However, if it were spinning much faster the gravity would not be able to hold the water down.

If you want to go into more detail, I can do an example of water at the equator; Think of a small mass m of water on the surface of the ocean. It has a weight of mg and a centrifugal force of mv²/R where g=9.8 m/s²is the accelation due to gravity, R=6.4x10⁶ m is the radius of the earth, and v=463 m/s is the speed of m as it travels around the equator. Then the centrifugal force is mv²/R=0.034m kg·m/s² whereas the weight is 9.8m kg·m/s². Comparing the two forces, W/C=288. The force holding the water to the surface is almost 300 times a great as the force trying to throw it off. For water to be thrown off, the centrifugal force would have to be larger than the weight; if you do the arithematic you will see that when the speed v is about 17 times larger than it is today, W/C=1 and water would start coming off the surface. This would be when the length of the day would be 17 times smaller, 1 day would be 24/17=1.4 hours long. It is more complicated at latitudes other than the equator, but the same ideas lead hold sway.

QUESTION:
Why does it take more energy to split a particle than to fuse one? (Fusion and fission reactions don't require the same amount of energy; fusion requires more energy to initiate than fission, but produces significantly more energy per unit of fuel.)

ANSWER:
Heavy nuclei tend to be unstable the heavier they get because of the nature of the forces which hold nuclei together. For example, a uranium nucleus has many possible modes of decaying to lighter elements; one possible mode of decay is spontaneous fission (SF) where an isolated uranium nucleus spontaneously breaks into two much lighter nuclei and several neutrons. So in this case, it takes zero energy input for fission to get energy out. However, this is a rare decay mode; for example, one gram of ²³⁵U will have about 14 SF/day. So even though you get a lot of energy compared to fuel "burned", the energy out is puny. You probably know that you can induce fission by having a very slow neutron absorbed by the nucleus. So if you flood a fissile material with slow neutrons you still put a relatively small amount of energy in but get a comparatively large amount out.

Now consider nuclear fusion. You must push two nuclei close to each other for the nuclear force to be felt. As long as the final nucleus is lighter than iron, the result will be the release of much more energy per amount of fuel "burned" than for fission. The key here is that in order to fuse two positively-charged nuclei, you must bring them very close to each other. But there is a repulsive electrical force between them which means that you have to do a lot of work (energy input) just to get them close enough for the nuclear force to kick in.

QUESTION:
Lets say I have a ship with a velocity of zero begin accelerating at 1G away from an observer.

What I'm interested in is the perspective of those on board. Lets say they need to emit X particles of reactant per second in order to maintain this acceleration. As they approach the speed of light, even though from their perspective they are still pushing out the same number of particles per second, from an outside perspective this decreases over time, for example the outside observer would see X/2 particles emitted per second at 87% the speed of light, while the passengers wouldn't notice any difference in their emission rate. This continues as the lorentz factor increases, with fewer and fewer particles emitted per second resulting in slower and slower acceleration from the perspective of the outside observer, while for the passengers it "feels" like their acceleration is constant and if they looked back at the observer they would see that their clock is ticking faster.

What this means in my estimation is that eventually you reach a point where from the passenger's perspective, if they could emit one more particle, they would reach the speed of light (after approx .98 years by my math, from their perspective). However due to time dilation, that particle will never actually be emitted... the observer could sit outside and watch the ship for an infinite amount of time, and it will never emit that last particle... implying that time, for those on board time has been effectively paused due to intense time dilation, just as they've approached the limit to their velocity, they've also approached the limit to the amount of time they can experience.

So here's the actual question:

Assuming that I've reasoned my way through all of this appropriately, does that mean that effectively if the ship is to try to accelerate continuously at 1G, that the passengers will only ever experience .98 years of time from their perspective, because the next tick of the clock will never actually come? If I'm wrong, where in my reasoning did I go wrong, and if I'm right, that's pretty cool and thanks for being a good sport and reading my hopefully not too rambling question.

ANSWER:
Here is the problem: acceleration is not at all a useful quantity when special relativity is important. Why is it important in Newtonian mechanics? Because no matter what your speed is, a force will cause the same acceleration. Therefore, Newton's second law, F=ma, will be true. That is not true if your speed is comparable to the speed of light. I have answered several questions in the past with very long answers which would help you understand this issue. Look at the faq page, in particular "Questions involving acceleration, force, linear momentum, and Newton's second law in special relativity". It is pretty tough going but important if you really want to understand this stuff!

QUESTION:
A tree branch is in equilibrium so all forces acting on it are balanced out. What forces are acting on it? There is the weight which is balanced by the vertical component of tension in the branch. There is horizontal component of tension, also. What forces are acting is balancing that? If it were spinning, I would say centrifugal force but it's not spinning.

ANSWER:
First, what are all the forces which act on the branch? There are only two, its weight W and the force which the tree trunk exerts on the branch. W acts at the center of gravity of the branch but we do not know where the force by the tree trunk should be drawn because the actual force is distributed all over the area of contact of the branch in some way we cannot know. However, you can allow for that by imagining two forces at the top and bottom of the contact area; I have shown the horizontal and vertical components of these two unknown forces. So we now have four unknowns and we will get three equations, translational equilibrium in the horizontal and vertical directions and rotational equilibrium. The translational equations are simple and yield H₁=H₂ and V₁+V₂=W.

I will just talk qualitatively now about rotational equilibrium (sum of all torques about some axis=0) because I don't have any numbers given. Suppose I choose my axis about which to calculate torques to be at the bottom where green forces are drawn. About this axis W causes a torque tending to make the branch rotate in the counterclockwise direction and the only other force exerting a torque is H₁ because all the other three have zero moment arms; this torque tends to make the branch rotate clockwise and could therefore hold the branch in rotational equilibrium. (Since H₁ has a much smaller moment arm than W, my drawing is not accurate because H₁ needs to be quite a bit larger than W for their torques to be the same magnitude.) If I chose the top of the branch to be the axis, only H₂ would exert a torque, still opposing the torque due to W.

When there is an ice storm, branches break off because the weight of the branch gets bigger than the ability of the trunk to hold on. Once in early summer a really big branch of a pecan tree in my back yard had nuts growing on it; this gradually increased the weight of the branch resulting in its breaking with a resounding bang in the middle of the night.

QUESTION:
Try as I might I cannot get my head round the reasons for the basic laws of gravity. I have thought up the following scenario - (Using rounded figures) The gravitational pull of earth is 10 m/s/s The gravitational pull of Jupiter is 25 m/s/s So an object the mass of Jupiter would fall to earth at a rate of 10 m/s/s as would an object of any other mass. But also an object the mass of earth would presumably fall to Jupiter at a rate of 25 m/s/s. I cannot reconcile these two seemingly contradictory conclusions. Surely there are no "ups and Downs" in space? I'm presuming the answer is something to do with relativity?

ANSWER:
First, I want to clear up some nomenclature you used.

When you say that "the gravitational pull of earth is 10 m/s/s", you should have said "the acceleration due to gravity is approximately 10 m/s/s provided that the accelerating mass is very small compared to the mass of the earth and that the position of the mass is close to the surface of the earth. Similar statments can be applied to Jupiter. When you say "gravitational pull" you are suggesting that 10 m/s/s is a force (we think as a push or a pull as a force); as I am sure you know, it is an acceleration.

Now , what causes acceleration? Force! So we must find the force which each experiences. Newton's third law says that bothe experience the same magnituce force but in opposite directions. Both forces are pointed toward the other. Newton's universal law of gravity is F=GMm/r²where G is Newton's universal gravitational constant and r is the distance between the centers of the two planets. Now Newton's second law says that the acceleration is the force divided by the mass, so a_earth=GM/r² and a_Jupiter=Gm/r². All this has nothing to do with relativity.

Incidentally, it is lucky that these are spheres Also they must be spherically symmetric masses also which means the density only can depend on the distance from the center; you can't have a lot more mass in the northern hemisphere than the southern. Newton himself proved that you may do this calculation (as I did) assuming all the mass of the planet may be treated as if it is all in a point mass in the centers of the planets.

QUESTION:
I am writing a classy science fiction and want the science to be believable. So, I got a lot of things covered and can understand much of the research I do, but there are things that no one has had to think about because they only relate to my book. Is there a name for an orbit that would circle the earth only once a year like in the figure attached? Could this work without a ridiculous amount of fuel? Would that shape of orbit work as well? It is meant for an Earth ship to dock once a year to resupply the observatory, so the orbit shape is based on the need to dock during the slowest speed of its orbit. Would that even be a concern?

ANSWER:
I have shown just a part of the figure you attached. No, that orbit is impossible unless you you burn a "ridiculous amount of fuel"; the natural orbit of one object orbiting another is (Kepler's first law) an ellipse with the object which is being orbited (earth in your case) located at the focus of the ellipse as shown in my figure. The figure shows the semimajor (a) and semiminor (b) axes which would be necessary for the observatory to be close to earth at some point in the orbit. Kepler's third law relates the period T (one year for your case) of the orbit the a: T²=(4π²/(GM)a³, where M is the mass of the earth and G is Newton's universal gravitational constant. It is important to note that every orbit with semimajor axis a has the same period. It is instructive to look at a circular orbit of radius a; I find that if T is one year, a=2.5x10¹⁰ m. It is easier to visualize if we express this distance in earth radii, R_E=6.4x10⁶ m and therefore a=3900 R_E, almost four thousand earth radii; for comparison, the distance to the moon is about 60 R_E. Unfortunately, the velocity at the closest distance to earth is the fastest anywhere on the orbit, the slowest is when the observatory is farthest away (Kepler's second law), just the opposite of what you need to dock conveniently. The animation below shows this. The red line is how the radius in a circular orbit would move; note that both the red and blue lines take the same time to go around once.

QUESTION:
Im not asking for help on homework; my teacher claims that a boat moving east at 5m/s and a water current north at 3m/s will travel 10 meters in 2 seconds but wouldnt the current have a effect on the time? i should use a resultant vector because i am not moving at 5m/s i am moving around 5.8 m/s north east if u use Pythagorean theorem

ANSWER:
The main problem you have here is that you misunderstand the question. "…a boat moving east at 5m/s…" means exactly what it says: in spite of the north-moving current of 3 m/s, the boat is moving east at 5 m/s. So 5 east is not what the boat would have in still water as you have assumed. In some sense this is a trick question because what is given is V_BGwhich is the boat with respect to the ground which is given as 5 m/s east, so of course the boat will go 10 m in 2 s. We might as well go on and find out what we might want to know, how would the boat be moving in still water so that it would end up going east with a speed of 5 m/s. We have to know the velocity addition theorem which is V_BG=V_BW+V_WG. Here, Vij means the velocity of i with respect to j. So we want V_BW=V_BG-V_WG. The first figure shows what you are given, the velocities of the boat and water as seen from the ground. The second figure shows the application of the velocity addition theorem applied to find what the velocity of this boat has relative to the water and this is how the boat would move if there were no current.

QUESTION:
If you had a very long indestructible bar and spun from the center of the bar at the speed of light, wouldn't the edges be moving faster than the speed of light because they are covering more distance. In the same way that when a record or disk is spinning, the center is moving slower than the edges. And if nothing can move faster than the speed of light, what happens?

ANSWER:
First, site ground rules specify that questions should not contain physically impossible conditions, viz. "indestructible bar". But the answer to your question doesn't doesn't depend on the indestructibility of the bar. Second you specify how fast something something is rotation by an angular velocity, e.g., RPM (revolutions per minute); in physics we usually specify angular velocity as radians/second and denote it as ω. This question has been answered before.

QUESTION:
i am a high school student from Scandinavia and have a upcoming assignment where i choose to research something of my choice. I visited the local university some time ago and saw a cloud chamber for the first time. I wondered if there are any temperature differences between the paths of an electron and a positron in a cloud chamber. At first thought i would believe that there would be no temperature difference, but after some research i discovered that there is a lot of unknown phenomena when it comes to positrons. Do you think that this is a topic worth researching with a DIY cloud chamber?

ANSWER:
I don't understand what you mean by 'temperature differences". We don't think of there being a temperature you can assign to a single particle, rather temperature is a statistical property describing the average kinetic energy of a very large number of particles. I can tell you that a positron and an electron have very similar (practically identical tracks in a cloud chamber if they have equal kinetic energies; they have equal masses and equal magnitudes of charges. The only thing different is that they have opposite charges. The way you can tell them apart is by making them pass through a magnetic field. If you see one track curve left, the other will curve right. You should, of course, have access to both β⁺ and β^- radioactive sources.

QUESTION:
I am an AP Physics teacher in NY. I was hoping you could help me explain conceptually why the electric potential of a charged conducting sphere changes when you surround it concentrically with a conducting shell.

ANSWER:
I found, when teaching, that students have considerable difficulty understanding what potential is, they are more comfortable thinking about potential energy and, even better, work. Before the concentric shell is brought in, how much work does it take it take to move a charge q from infinity to the surface of the sphere? If the total charge on the sphere is Q, the force on q when it is a distance r from the center is kqQ/r²; the total work is -kqQ/R. If q and Q are of opposite sign, the work you do is positive since you have to push the charge in the same direction it is moving. Now, put the spherical shell, inner and outer radii R_i and Ro. The total charge on the sphere is Q and the total charge on the shell is zero. But the field from the sphere at r=R_i will cause an induced surface charge on inside of the shell but also the negative of that surface charge on the outer surface. It is important here to note that all the charge on a conductor resides only on the surfaces, there is zero field inside a conductor. So the field seen by q, which you are imagining pushing in, is exactly the same for ∞<r<R_o and R<r<R_i as it was before the shell was brought in. But when R_i<r<R_o, there is no force and therefore no work is done. So the total work done is different from that done before the shell was added by the amount

±|(1/R_o)-(1/R_i)|

You could say all this by just saying that because there is zero field inside any conductor there is no work done on q in the conductor which means the electric potential inside the shell will be different from what it was without the shell.

QUESTION:
I'm reading a book by Sean Carroll about Quantum Physics. It is an introduction to the subject for non-physicists. There, it says that if you put a detector at one (or both I guess) of the slits in the double-slit experiment, the interference pattern disappears. The reason would be that the detector and the electron’s wave function become entangled. I have two questions:

Is it practically possible to build such a detector? Would it sense the electric field of one single electron? If so it would need to be extremely sensitive. Also, it would be impossible to sense the electric field without affecting the electron's path, wouldn't it? Which would ruin the experiment.
From the electron's point of view, what is the difference between the detector and the slit arrangement? Doesn't the electron become entangled with the slit setup as well? Thus the interference pattern wouldn't appear even without the detector. Yet it does, why?

ANSWER:
I was going to begin with a statement about how this experiment could not be done as described because the wavelengths of the electrons are too small at reasonable speeds and the slits would have to be much narrower and closely placed that we would not have the ability to fabricate them. But Richard Feynman, around 1960, said it much better than I could: “We choose to examine a phenomenon which is impossible, absolutely impossible, to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery. We should say right away that you should not try to set up this experiment. This experiment has never been done in just this way. The trouble is that the apparatus would have to be made on an impossibly small scale to show the effects we are interested in. We are doing a 'thought experiment', which we have chosen because it is easy to think about. We know the results that would be obtained because there are many experiments that have been done, in which the scale and the proportions have been chosen to show the effects we shall describe”. But Davisson and Germer, around 1927, showed interference of electrons from scattering from the surface of single crystal nickel where the atoms played the role of the slits and were separated by a small enough distance to deal with the small wavelengths of the electrons. But, the ability to actually fabricate extremely narrow and close together slits had already been done and Feynman was apparently unaware. It is done by milling gold plated on a very thin substrate of silicon e.g., with an atomic beam.

Now, I have no idea what "the detector and the electron’s wave function become entangled" means. I believe that is much easier to say that if a detector detects it, it is all there, i.e., it is behaving like a particle, not a wave. In order for the interference pattern to happen, the electron must interfere with itself, i.e., pass through both slits, something only a wave can do. Regarding whether or not it is possible to actually have such a detector, I would say that I am quite confident that is would be straight forward to detect the presence of a single electron

QUESTION:
What happens when ice at zero degree Celsius is dropped into water at zero degree Celcius?

ANSWER:
Nothing will happen. There is no temperature difference so there will be no heat flow. The ice cube will not melt, the water will not freeze.

QUESTION:
Let's say that the entire observable universe was compressed to be about the size of a softball. How dense would that be and how much energy would it take to do that?

ANSWER:
This question has a number of possible answers, it is not unabiguous. If you mean the best estimate of the observable mass of the universe, it is about 1.5x10⁵³ kg. Taking the volume to be about 5x10^-4 m³, the density would be about ρ=3x10⁵⁸ kg/m³. There is estimated to be about 5 times more dark matter than normal matter in the universe, so that would increase the density to 1.8x10⁵⁹ kg/m³. (Of course, dark matter has never been directly measured, just inferred.)

You can ask how much energy it takes to assemble a sphere of radius R and mass M in classical mechanics. Because the force is gravitational, it would assemble itself because gravitation is attractive! But we should really ask how much work must be done such that each piece of mass we bring in arrives with zero kinetic energy; in that case the energy you supply is negative because the force you would have to exert is opposite the force of the gravity. E=-(3/5)GM²/R where G=6.67x10^-11 m³/(kg·s). You could go ahead and calculate this energy but it would be a purely academic exercise because matter can not exist at such high density. That's why black holes exist: when you get huge amounts of mass, like the remnants of a star which has burnt all its fuel, it will collapse into a neutron star or a black hole. You may be sure that this much mass in this small a volume would have collapsed into the most massive black hole anyone has ever seen. So technically the answer to your question is when you try to make your "softball" it will collapse long before it has all been assembled and the density will ever thereafter be infinite; the energy to do this will be zero.

Finally I should remind you that my site ground rules state that I do not do astronomy/astrophysics/cosmology and I usually discard such questions. I thought this one would be fun to play with. You may choose to believe that everything I have said is BS!

QUESTION:
In quantum entanglement, we can entangle two particles that can affect one another instantly even if they are galaxies apart (which in some ways appear to challenge our modern understanding of physics like the fact that nothing can travel faster than the speed of light). My question is. What would happen if we entangle two particles and then preserve one on Earth and the other we send it on a journey throughout the galaxy at supersonic speed, dilating time, and making it age only 1 year, but it would have passed 100 years on Earth? Would the entanglement between the two particles still work, and if so, could we use this entanglement to communicate with our past or future selves?

ANSWER:
This question has previously been answered. No information can be transmitted using entanglement.

QUESTION:
I am having a hard time trying to understand a simple DT fusion reaction (I am 72 years old). Since you start with one deuterium atom (one proton and one neutron) and one tritium atom (one proton and two neutrons) and they fuse to become a single helium atom and a free neutron, all the original particles are still acccounted for - exactly what mass is being converted into energy? I tried to read several explanations without satisfaction, but I found that the mass of an atom is less than the mass of its constituent particles, because when the atom is formed, some mass is converted into binding energy that holds the nucleus together (I thought that was gluons…), and it brought up a similar question - what mass is being converted to binding energy? Don’t all protons and neutrons have exactly the same mass, and isn’t it the same whether it is a free particle or part of an atom?

ANSWER:
Your problem can be understood from your sentence "exactly what mass is being converted into energy?" You are thinking that mass is something which gets converted into energy, but the fact is that mass is energy. I sometimes find that these kind of issues can be understood best from a simple gedanken (German for a thought experiment). Let's consider a deuteron. If you measure its mass you will find (as you seem to already know) it has a mass less than the sum of the masses of a proton+neutron. Now, these are bound together by a force or they wouldn't stay together. Now you wish to pull them apart; what do you have to do? You have to pull them apart, and this is positive work because you are pulling in the same direction the particles are moving. If you are doing positive work on the system that means that the energy of the system is increasing. Finally they are far enough apart that they no longer feel the force which formerly bound them, which need not be very far apart because the nuclear force is very short-ranged. (One micron might as well be infinity!) So now you weigh them and find that their masses are now to the masses these particles have when free, i.e., more than when they were bound. Your question: where did that mass come from? My answer: you gave it to them by doing positive work. Now, suppose you want to put them back together. So you push them together, but as soon as you get them close enough to feel the force you will have to hold them back, that is, doing negative work. The result is that the deuteron is going to have less mass once it is bound because you took it away by doing negative work. So bound particles in general do not have the same mass as free particles. And in a heavy nucleus, you can't say "oh, that proton over there must have less mass than it would outside"; what you can say is "that total collection of protons and neutrons has less mass than they would if they were all unbound."

QUESTION:
A 1kg sphere rolls at 1 m/sec along the X-axis. The center of mass of the sphere will have a fixed y value, but will pass through an infinite number of x values each second. However briefly, the ball will have each of these (infinite) x-coordinates. If the ball is traveling at 2m/sec in the X-direction, it will be 'located' at each of the infinite X-coordinates for half as long as when it was traveling at 1 m/sec. How should I think of this motion? Simple limits (as t > 0)? If so, I would apply the same limit rational for both the 1m/sec ball and 2m/sec ball?

ANSWER:
You are messing around with infinity. The mathematics of this number is not the same as usual numbers. You have stumbled on one quirky property when the you conclude that the time per point is half what it is when you double the speed. And think of this pecularity: there are an infinite number of 'points' in an inch but there are also an infinite number in a mile. The simple way to answer your question is that a point spends zero time on each of an infinite number; so if the path is one meter long the time spent will be 1 s=(0 s)x(∞ points) which is a perfectly fine equation because 0x∞ can be anything. Students often think that it is meaningful to ask what the probability of finding a point particle at some specific point in space at some particular time. In fact, it is zero. To get what you want you need to do a little calculus: consider an infinitesimal volume dV, apply the physics, take the limit as dV→0; this will give you a function showing the general behavior of probability as a function of location and time, but it will not give you what the probability is of finding the particle at any single point. You get a probability density as a function of time and you can find the probability of the particle being in a volume which you can make arbitrarily small.

QUESTION:
I have a question related to time dilation. I have a good idea on how it works and its principles, but I could not find any information regarding the question. Here it is:Since the closer you go to the speed of light, time passes more slowly for you, and quicker for the people you're travelling away from. What is (if there is one) the fastest speed a human could travel through space and come back without noticing any change?

ANSWER:
Actually, time passes perfectly normally for you in your frame of reference. I suggest that you read my answer on the twin paradox. Your question is qualitative and therefore unanswerable. The giveaway is the phrase "noticing any change"; change in what? And the change will depend on how far he goes. The satellites which support the GPS system have speeds of 8000 m/s which is about 0.00003 of the speed of light. Is that noticeable? It surely would be if you are trying to get somewhere and the system did not correct for time dialation. Another consideration, even if you defined what noticeable is, is that an 80 year old person could look 60 or 90 but, relative to more reliable clocks he carried along, he has actually aged 80 years.