Recent questions in Spread

College StatisticsAnswered question

ntaraxq 2022-07-10

Spread and direction of the cosmic background radiation

Something I can never understand is that where the cosmic background radiation spreads?

If I know well, the cosmic background radiation is actually the light of the Big Bang. If it happened exactly in the same time, it must have spread into the theoretic center of the universe. Which would mean that it already reached every parts of it, in case if it happened in the same time with the Big Bang.

It be possible that it spreads to this direction, but universe expands faster. In this case, the radiation approaches but also moves off - just like everything in the universe. But it's possible only if the universe expands faster than the speed of light. Is this possible?

If it would spread to the direction of the "edge" of the universe, we shouldn't be able to know about its existence, because it would never reach us.

Also, these theories are true with an important conclusion: universe has a beginning in time, which means that once upon a time, it started to expand - consequently it must have a size limit. The reason is that universe has 4 dimensions: length, width, height, and time. One of them (time) is not infinite, so none of the others can be infinite.

Something I can never understand is that where the cosmic background radiation spreads?

If I know well, the cosmic background radiation is actually the light of the Big Bang. If it happened exactly in the same time, it must have spread into the theoretic center of the universe. Which would mean that it already reached every parts of it, in case if it happened in the same time with the Big Bang.

It be possible that it spreads to this direction, but universe expands faster. In this case, the radiation approaches but also moves off - just like everything in the universe. But it's possible only if the universe expands faster than the speed of light. Is this possible?

If it would spread to the direction of the "edge" of the universe, we shouldn't be able to know about its existence, because it would never reach us.

Also, these theories are true with an important conclusion: universe has a beginning in time, which means that once upon a time, it started to expand - consequently it must have a size limit. The reason is that universe has 4 dimensions: length, width, height, and time. One of them (time) is not infinite, so none of the others can be infinite.

College StatisticsAnswered question

Frederick Kramer 2022-07-10

$\hslash \approx 0$ and the spread of QM wave function

Is there a direct mathematical method to show that if a quantum wave funtion is initially sharply localized, then it will stay sharply localized if $\hslash \approx 0$? In that case the Ehrenfest theorem implies the transition from quantum mechanics to classical mechanics.

Of course, we are dealing with the propagation of a wave function, but let's not mess with path integrals. Thus, the structure of the general solution of Schrödinger equation should imply the result - if possible.

Is there a direct mathematical method to show that if a quantum wave funtion is initially sharply localized, then it will stay sharply localized if $\hslash \approx 0$? In that case the Ehrenfest theorem implies the transition from quantum mechanics to classical mechanics.

Of course, we are dealing with the propagation of a wave function, but let's not mess with path integrals. Thus, the structure of the general solution of Schrödinger equation should imply the result - if possible.

College StatisticsAnswered question

myntfalskj4 2022-07-08

Can quantum particles spread out over large distances?

While trying to understand quantum mechanics I was wondering about this: since free quantum particles naturally spread out until the wave function collapses (if I understand correctly); does there exist an abundance of extremely spread out particles in outer space where interaction with other particles is rare or do the particles collapse before this happens?

To be more specific:

1.Does it occur often that particles in outer space reach macroscopic spreads of let's say multiple kilometers? Or does quantum decoherence occur before this happens? With spreading I mean ${\sigma}_{x}$ or the uncertainty in position.

2.If a particle reaches such a big spread, does this accelerate or inhibit wave function collapse?

A spread out particle covers more area making it interact with more matter but at the same the probability amplitude per area decreases making the chance of interaction smaller.

While trying to understand quantum mechanics I was wondering about this: since free quantum particles naturally spread out until the wave function collapses (if I understand correctly); does there exist an abundance of extremely spread out particles in outer space where interaction with other particles is rare or do the particles collapse before this happens?

To be more specific:

1.Does it occur often that particles in outer space reach macroscopic spreads of let's say multiple kilometers? Or does quantum decoherence occur before this happens? With spreading I mean ${\sigma}_{x}$ or the uncertainty in position.

2.If a particle reaches such a big spread, does this accelerate or inhibit wave function collapse?

A spread out particle covers more area making it interact with more matter but at the same the probability amplitude per area decreases making the chance of interaction smaller.

College StatisticsAnswered question

Palmosigx 2022-07-07

Do cracks in solids spread at a characteristic speed?

I have a feeling this is actually a pretty complicated problem in detail as I know a tremendous amount of research is done on the behavior of materials under stress, and I think the way materials fracture/crack is a useful diagnostic tool.

As a starting point, waves travel at the speed of sound in a material, which is to say they are mechanical waves. Do cracks also propagate at the speed of sound?

Another related question is, do the cracks in a solid move at a constant speed? Obviously it is possible for a crack to form and stop propagating, so this might imply a dissipation of some vibrational amplitude until the energy is below the point of being able to break bonds/fragment atoms (sheets for some solids). It also might imply that the crack does not behave like a wave and spreads at a non-constant speed. Which is right?

Lastly, do all solids follow the same rules for the speed at which they fracture? I'm thinking of glass, which probably is not a good place to start on this question because it is an amorphous solid (some people say incredibly viscous liquid but this feels like semantics). Obviously some glass will crack into large pieces and some will fracture like a spiderweb. Are these processes fundamentally the same but with different microscopic details? And how does fracturing in glass relate to something simpler like the breaking of a single atom solid? I haven't seen this happen, but surely it is possible.

Obviously there were quite a few question there, but they were only meant to give a general idea of the types of questions would be nice to have answered. Maybe the question is actually quite simple and one answer will suffice, or maybe it's complicated and there are multiple good answers.

I have a feeling this is actually a pretty complicated problem in detail as I know a tremendous amount of research is done on the behavior of materials under stress, and I think the way materials fracture/crack is a useful diagnostic tool.

As a starting point, waves travel at the speed of sound in a material, which is to say they are mechanical waves. Do cracks also propagate at the speed of sound?

Another related question is, do the cracks in a solid move at a constant speed? Obviously it is possible for a crack to form and stop propagating, so this might imply a dissipation of some vibrational amplitude until the energy is below the point of being able to break bonds/fragment atoms (sheets for some solids). It also might imply that the crack does not behave like a wave and spreads at a non-constant speed. Which is right?

Lastly, do all solids follow the same rules for the speed at which they fracture? I'm thinking of glass, which probably is not a good place to start on this question because it is an amorphous solid (some people say incredibly viscous liquid but this feels like semantics). Obviously some glass will crack into large pieces and some will fracture like a spiderweb. Are these processes fundamentally the same but with different microscopic details? And how does fracturing in glass relate to something simpler like the breaking of a single atom solid? I haven't seen this happen, but surely it is possible.

Obviously there were quite a few question there, but they were only meant to give a general idea of the types of questions would be nice to have answered. Maybe the question is actually quite simple and one answer will suffice, or maybe it's complicated and there are multiple good answers.

College StatisticsAnswered question

babyagelesszj 2022-07-04

Why do radio waves spread out while higher frequency waves travel in beams?

Why is it that radio waves spread out in proportion to the square of the distance, while higher frequency electromagnetic waves, like microwaves, infrared waves, light, etc are able to propagate as beams? What fundamental property allows higher energy waves to travel differently than lower energy?

Why is it that radio waves spread out in proportion to the square of the distance, while higher frequency electromagnetic waves, like microwaves, infrared waves, light, etc are able to propagate as beams? What fundamental property allows higher energy waves to travel differently than lower energy?

College StatisticsAnswered question

Sylvia Byrd 2022-07-02

How can the energy of a magnetic field spread at speed of light?

I would like to solve one question. Suppose that we have a point charge. Then we apply a force on it, and as it accelerates, it emits electromagnetic waves and is pushed back due to Abraham-Lorentz force. But then, when we drop it and it moves at constant speed, it starts to create a magnetic field that spreads at speed of light. So as the field occupies more volume, it increases its energy but we are not giving any energy as we have dropped it. So, where does this energy come from?

I would like to solve one question. Suppose that we have a point charge. Then we apply a force on it, and as it accelerates, it emits electromagnetic waves and is pushed back due to Abraham-Lorentz force. But then, when we drop it and it moves at constant speed, it starts to create a magnetic field that spreads at speed of light. So as the field occupies more volume, it increases its energy but we are not giving any energy as we have dropped it. So, where does this energy come from?

College StatisticsAnswered question

Ximena Skinner 2022-07-02

When does the wave function spread over the volume of a box?

I have heard colloquially that for any initial state, a particle enclosed in some volume $V$ will spread itself relatively evenly over that volume after large time, so that $|\psi (\overrightarrow{x}){|}^{2}\approx 1/V$ may be used as the position probability density in the box.

This seems reasonable but of course it is not true, for example, of energy eigenstates. A sine wave with a wavelength fitting into the box will remain a sine wave for all times; this is the case for any energy eigenstate. However one might argue that an exact energy eigenstate is an idealization.

Does anyone know of a condition on the initial wave function ${\psi}_{0}(\overrightarrow{x})$ which is sufficient to guarantee that for large times,

$|\psi (\overrightarrow{x}){|}^{2}\approx 1/V\text{}?$

Or else, is there a counterexample of a more realistic state which does not fit the above approximation? My hunch is that any initial state whose fourier transform is supported on all of ${\mathbb{R}}^{3}$ might give the approximation above, but I don't know of any such result. Energy eigenstates, for example, have a delta-function fourier transform.

I have heard colloquially that for any initial state, a particle enclosed in some volume $V$ will spread itself relatively evenly over that volume after large time, so that $|\psi (\overrightarrow{x}){|}^{2}\approx 1/V$ may be used as the position probability density in the box.

This seems reasonable but of course it is not true, for example, of energy eigenstates. A sine wave with a wavelength fitting into the box will remain a sine wave for all times; this is the case for any energy eigenstate. However one might argue that an exact energy eigenstate is an idealization.

Does anyone know of a condition on the initial wave function ${\psi}_{0}(\overrightarrow{x})$ which is sufficient to guarantee that for large times,

$|\psi (\overrightarrow{x}){|}^{2}\approx 1/V\text{}?$

Or else, is there a counterexample of a more realistic state which does not fit the above approximation? My hunch is that any initial state whose fourier transform is supported on all of ${\mathbb{R}}^{3}$ might give the approximation above, but I don't know of any such result. Energy eigenstates, for example, have a delta-function fourier transform.

College StatisticsAnswered question

Brock Byrd 2022-07-01

Photoelectric effect: Why does monochromatic radiation produces photoelectron with a spread of velocities?

Recall that in photoelectric effect, $V=\frac{hf}{e}-\frac{Wo}{e}$. The incident photon with frequency f produces an electron with energy eV. This should result in a single velocity, why is there a velocity spread?

Recall that in photoelectric effect, $V=\frac{hf}{e}-\frac{Wo}{e}$. The incident photon with frequency f produces an electron with energy eV. This should result in a single velocity, why is there a velocity spread?

College StatisticsAnswered question

juanberrio8a 2022-06-29

What does "spread of momentum" actually mean?

I was reading Feynman's lecture in which Feynman invoked his own way of explaining the uncertainty principle using single-slit experiment.

There I found:

To get a rough idea of the spread of momentum, the vertical momentum ${p}_{y}$ has a spread which is equal to ${p}_{0}\mathrm{\Delta}\theta $, where ${p}_{0}$ is the horizontal momentum . . .

What is he talking of? What does "spreading" actually mean? And how did he measure it?

I was reading Feynman's lecture in which Feynman invoked his own way of explaining the uncertainty principle using single-slit experiment.

There I found:

To get a rough idea of the spread of momentum, the vertical momentum ${p}_{y}$ has a spread which is equal to ${p}_{0}\mathrm{\Delta}\theta $, where ${p}_{0}$ is the horizontal momentum . . .

What is he talking of? What does "spreading" actually mean? And how did he measure it?

College StatisticsAnswered question

sedeln5w 2022-06-29

How do ripples form and why do they spread out?

When I throw a rock in the water, why does only a small circular ring around the rock rises instead of the whole water body, and why does it fall outwards and not inwards or why fall out in any direction instead of just going back down like ball which after getting thrown up falls back to its initial position?

When I throw a rock in the water, why does only a small circular ring around the rock rises instead of the whole water body, and why does it fall outwards and not inwards or why fall out in any direction instead of just going back down like ball which after getting thrown up falls back to its initial position?

College StatisticsAnswered question

Dayami Rose 2022-06-27

Spread of the energy levels and sharp energy eigenvalues of the Schrodinger equation of the H-atom

Solving the Schroedinger equation for the H-atom (or any other system, say a particle in a box, or harmonic oscillator or anything), we obtain the energy eigenvalues are sharp with no spread. However, in reality, there must be a spread for the excited states which follows from uncertainty principle, and the spread of an energy level $\mathrm{\Delta}E\sim \frac{\hslash}{\tau}$ where $\tau $ is the lifetime of the particle in that state. Why is this effect not captured in the calculation of energy levels by solving the Schroedinger equation?

Is there any other way to compute the spread other than using the uncertainty principle and match them?

Solving the Schroedinger equation for the H-atom (or any other system, say a particle in a box, or harmonic oscillator or anything), we obtain the energy eigenvalues are sharp with no spread. However, in reality, there must be a spread for the excited states which follows from uncertainty principle, and the spread of an energy level $\mathrm{\Delta}E\sim \frac{\hslash}{\tau}$ where $\tau $ is the lifetime of the particle in that state. Why is this effect not captured in the calculation of energy levels by solving the Schroedinger equation?

Is there any other way to compute the spread other than using the uncertainty principle and match them?

College StatisticsAnswered question

Fletcher Hays 2022-06-25

Light of laser doesn't spread?

How exactly can a combination of monochromaticity and coherent light make a light not want to spread out? When it's all one wavelength and it's all amplified it should make the light brighter and thus more energetic. I can't seem to understand why the light doesn't spread normally.

How exactly can a combination of monochromaticity and coherent light make a light not want to spread out? When it's all one wavelength and it's all amplified it should make the light brighter and thus more energetic. I can't seem to understand why the light doesn't spread normally.

College StatisticsAnswered question

Manteo2h 2022-06-22

Does accelerating cosmological expansion increase beam spread?

In the standard textbook case, a transmitter of diameter $D$ can produce an electromagnetic beam of wavelength $\lambda $ that has spread angle $\theta =1.22\lambda /D$. But what happens in an expanding cosmology, especially one that accelerates so that there is an event horizon? Does $\theta $ increase with distance?

Obviously each photon will travel along a null geodesic and after conformal time $\tau $ have travelled $\chi =c\tau $ units of co-moving distance. The distance between the beam edges would in flat space be growing as $\delta =2c\mathrm{sin}(\theta /2)\tau $. Now, co-moving coordinates are nice and behave well with conformal time, so I would be mildly confident that this distance is true as measured in co-moving coordinates.

But that means that in proper distance the beam diameter is multiplied by the scale factor, $a(t)\delta $ (where $t$ is the time corresponding to $\tau $), and hence $\theta $ increases. However, the distance to the origin in these coordinates has also increased to $a(t)(ct)$, so that seems to cancel the expansion - if we measure $\theta (t)$ globally by dividing the lengths.

But it seems that locally we should see the edges getting separated at an accelerating pace; after all, the local observers will see the emitter accelerating away from them, producing a wider and wider beam near their location since it was emitted further away. From this perspective as time goes by the beam ends up closer and closer to $\theta =\pi $ (and ever more red-shifted, which presumably keeps the total power across it constant).

Does this analysis work, or did I slip on one or more coordinate systems?

In the standard textbook case, a transmitter of diameter $D$ can produce an electromagnetic beam of wavelength $\lambda $ that has spread angle $\theta =1.22\lambda /D$. But what happens in an expanding cosmology, especially one that accelerates so that there is an event horizon? Does $\theta $ increase with distance?

Obviously each photon will travel along a null geodesic and after conformal time $\tau $ have travelled $\chi =c\tau $ units of co-moving distance. The distance between the beam edges would in flat space be growing as $\delta =2c\mathrm{sin}(\theta /2)\tau $. Now, co-moving coordinates are nice and behave well with conformal time, so I would be mildly confident that this distance is true as measured in co-moving coordinates.

But that means that in proper distance the beam diameter is multiplied by the scale factor, $a(t)\delta $ (where $t$ is the time corresponding to $\tau $), and hence $\theta $ increases. However, the distance to the origin in these coordinates has also increased to $a(t)(ct)$, so that seems to cancel the expansion - if we measure $\theta (t)$ globally by dividing the lengths.

But it seems that locally we should see the edges getting separated at an accelerating pace; after all, the local observers will see the emitter accelerating away from them, producing a wider and wider beam near their location since it was emitted further away. From this perspective as time goes by the beam ends up closer and closer to $\theta =\pi $ (and ever more red-shifted, which presumably keeps the total power across it constant).

Does this analysis work, or did I slip on one or more coordinate systems?

College StatisticsAnswered question

Winigefx 2022-06-21

How can an electron be fired at a target when uncertainty principle says it will spread out around axis of motion?

Consider an electron fired at a target. Taking the axis of motion to be $x$, and position to be $(x,y,z)$ then

$\mathrm{\Delta}y=\mathrm{\Delta}z=0$

Therefore by the uncertainty principle

$\mathrm{\Delta}{p}_{y}=\mathrm{\Delta}{p}_{z}=\mathrm{\infty}$

So the electron will spread out in $y$ and $z$ and never hit its target (unless that target was very close i.e. on the same scale as the initial effective size of the electron). How can it hit a target in reality?

I realise there is a possible duplicate to this question: How can particles travel in a straight line? The accepted answer for the duplicate is "because $\hslash $ is really small so the particles don't spread out much before hitting the target". I'm not sure this stands though. The time-dependent Schrodinger equation is:

$i\hslash \frac{\delta}{\delta t}\psi (r,t)=[-\frac{{\hslash}^{2}}{2\mu}{\mathrm{\nabla}}^{2}+V(r,t)]\psi (r,t)$

In the absence of external potential $V$ and dividing both sides by $\hslash $

$i\frac{\delta}{\delta t}\psi (r,t)=-\frac{\hslash}{2\mu}{\mathrm{\nabla}}^{2}\psi (r,t)$

So if we change $\hslash $ (or $\mu $), sure the particle will spread out less as a function of time, but won't it also move slower in exactly the same proportion to its lack of spreading out, so it still won't hit the target?

Consider an electron fired at a target. Taking the axis of motion to be $x$, and position to be $(x,y,z)$ then

$\mathrm{\Delta}y=\mathrm{\Delta}z=0$

Therefore by the uncertainty principle

$\mathrm{\Delta}{p}_{y}=\mathrm{\Delta}{p}_{z}=\mathrm{\infty}$

So the electron will spread out in $y$ and $z$ and never hit its target (unless that target was very close i.e. on the same scale as the initial effective size of the electron). How can it hit a target in reality?

I realise there is a possible duplicate to this question: How can particles travel in a straight line? The accepted answer for the duplicate is "because $\hslash $ is really small so the particles don't spread out much before hitting the target". I'm not sure this stands though. The time-dependent Schrodinger equation is:

$i\hslash \frac{\delta}{\delta t}\psi (r,t)=[-\frac{{\hslash}^{2}}{2\mu}{\mathrm{\nabla}}^{2}+V(r,t)]\psi (r,t)$

In the absence of external potential $V$ and dividing both sides by $\hslash $

$i\frac{\delta}{\delta t}\psi (r,t)=-\frac{\hslash}{2\mu}{\mathrm{\nabla}}^{2}\psi (r,t)$

So if we change $\hslash $ (or $\mu $), sure the particle will spread out less as a function of time, but won't it also move slower in exactly the same proportion to its lack of spreading out, so it still won't hit the target?

College StatisticsAnswered question

Adriana Ayers 2022-06-21

The shape characteristics of gravitational wells given different masses and spread of objects

I am curious as to research that calculates the shape of gravitational wells, and their limits, and affect on time, for different masses and spread of mass. The actual question is st the end.

For example: When the sun becomes a red giant the Earth's orbit is supposed to move out with the redistribution of mass, but what is the science behind this? Re-edit: Which I think was explained as the spreading density of the suns mass (it was a "TV" program. Another example, is it is said that the termination of the Sun's gravity field is 1.5 light years out. I am unaware of any distance studies to measure shape to specifically prove the theory.

The Question: Explanation of how the shape, time over distance, and extent of a gravitational well changes with the density of the mass, and the science behind this please? Re-edit: The curvature, research to verify theory, and actual simple graphical description of how the physical shape responds and changes based on contributing mass distribution. Say, does a more dense matter object cause the gravity well to more tightly curve to the surface of the matter, than to the surface of a cloud of gas of equal weight but magnitudes bigger. How does that look in physical shape over distance, how does the field terminate in shape. I'm interested in observational research on the profile. For instance does it continue the same decay equation or does it change/flatten out etc to a different equation at distance. This is more looking at verification/explanation of the conventional versus deviations. If we can only say so much at X distance from verified studies, that would be appreciated?

As we know there has been speculation based on deviations in observation of gravity on grander scales, such as across the galaxy. But I do not wish to go into those hypothesises, only the limits of what we have verified we know, which is a good starting point onto looking into this further.

I am curious as to research that calculates the shape of gravitational wells, and their limits, and affect on time, for different masses and spread of mass. The actual question is st the end.

For example: When the sun becomes a red giant the Earth's orbit is supposed to move out with the redistribution of mass, but what is the science behind this? Re-edit: Which I think was explained as the spreading density of the suns mass (it was a "TV" program. Another example, is it is said that the termination of the Sun's gravity field is 1.5 light years out. I am unaware of any distance studies to measure shape to specifically prove the theory.

The Question: Explanation of how the shape, time over distance, and extent of a gravitational well changes with the density of the mass, and the science behind this please? Re-edit: The curvature, research to verify theory, and actual simple graphical description of how the physical shape responds and changes based on contributing mass distribution. Say, does a more dense matter object cause the gravity well to more tightly curve to the surface of the matter, than to the surface of a cloud of gas of equal weight but magnitudes bigger. How does that look in physical shape over distance, how does the field terminate in shape. I'm interested in observational research on the profile. For instance does it continue the same decay equation or does it change/flatten out etc to a different equation at distance. This is more looking at verification/explanation of the conventional versus deviations. If we can only say so much at X distance from verified studies, that would be appreciated?

As we know there has been speculation based on deviations in observation of gravity on grander scales, such as across the galaxy. But I do not wish to go into those hypothesises, only the limits of what we have verified we know, which is a good starting point onto looking into this further.

College StatisticsAnswered question

Sonia Gay 2022-06-15

Oil spreading over water

The book says that oil spreads over water due to the greater surface tension of water as compared to oil, so the comparatively stronger water film stretches the oil surface and makes it spread... But if this is the case, doesn't that mean that the oil layer will go on spreading indefinitely(even if it gets one molecule thick)until it covers the whole surface, because at any point the surface tension of water is still greater than oil.... But this does not happen, oil spreads only until it makes a particular contact angle with the water surface.

So is the explanation given in the book a bit faulty or am I getting the whole thing wrong? Someone please help...

The book says that oil spreads over water due to the greater surface tension of water as compared to oil, so the comparatively stronger water film stretches the oil surface and makes it spread... But if this is the case, doesn't that mean that the oil layer will go on spreading indefinitely(even if it gets one molecule thick)until it covers the whole surface, because at any point the surface tension of water is still greater than oil.... But this does not happen, oil spreads only until it makes a particular contact angle with the water surface.

So is the explanation given in the book a bit faulty or am I getting the whole thing wrong? Someone please help...

College StatisticsAnswered question

Summer Bradford 2022-06-14

Gaussian State Spread

A measurement device which can be represented by a 1D quantum system (with canonical observables $X$ and $P$) 'is prepared in a Gaussian state with spread $s$'

$|\psi \u27e9=\frac{1}{({\pi}^{2}{s}^{2}{)}^{1/4}}\int \mathrm{exp}[-\frac{{x}^{2}}{2{s}^{2}}]\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x|x\u27e9$

Can somebody tell me what it means that its canonical variables are $X$ and $P$?

A measurement device which can be represented by a 1D quantum system (with canonical observables $X$ and $P$) 'is prepared in a Gaussian state with spread $s$'

$|\psi \u27e9=\frac{1}{({\pi}^{2}{s}^{2}{)}^{1/4}}\int \mathrm{exp}[-\frac{{x}^{2}}{2{s}^{2}}]\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x|x\u27e9$

Can somebody tell me what it means that its canonical variables are $X$ and $P$?

College StatisticsAnswered question

Lovellss 2022-06-14

How is finite energy spread wholly in the infinite universe

We all know that energy is never lost, but it transforms into another form.

Doesn't that mean that energy is not unlimited? I mean, why is it that way if it was infinite?

However, if energy is spread throughout the whole world, how can something finite fully fill something infinite?

Does that mean that energy isn't finite? Or the universe isn't infinite? Or maybe there are places without any energy at all?

Of course, there is probably something stupid I'm missing, but an answer would hopefully fix that.

Note: sorry if it's a dumb question, I just heard of it and wasn't able to answer it, nor find it on the web.

We all know that energy is never lost, but it transforms into another form.

Doesn't that mean that energy is not unlimited? I mean, why is it that way if it was infinite?

However, if energy is spread throughout the whole world, how can something finite fully fill something infinite?

Does that mean that energy isn't finite? Or the universe isn't infinite? Or maybe there are places without any energy at all?

Of course, there is probably something stupid I'm missing, but an answer would hopefully fix that.

Note: sorry if it's a dumb question, I just heard of it and wasn't able to answer it, nor find it on the web.

College StatisticsAnswered question

Celia Lucas 2022-06-14

Why do wave packets spread out over time?

Why do wave functions spread out over time? Where in the math does quantum mechanics state this? As far as I've seen, the waves are not required to spread, and what does this mean if they do?

Why do wave functions spread out over time? Where in the math does quantum mechanics state this? As far as I've seen, the waves are not required to spread, and what does this mean if they do?

College StatisticsAnswered question

Ezekiel Yoder 2022-06-13

How does positive charge spread out in conductors?

I know that when there are excess positive charges in a conductor, for example, a metal sphere, the positive charges will spread out over its surface. However, I am confused about how this excess charge spreads out over the surface, if protons cannot move and only electrons can move.

Can someone please inform me on how the excess positive charge spreads out over the surfaces of conductors?

I know that when there are excess positive charges in a conductor, for example, a metal sphere, the positive charges will spread out over its surface. However, I am confused about how this excess charge spreads out over the surface, if protons cannot move and only electrons can move.

Can someone please inform me on how the excess positive charge spreads out over the surfaces of conductors?

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The measures of spread Math are mostly used in finances and banking, accounting, and the marketing assignments, which is why you should take a look at the examples and the questions that already have the solutions. The answers that are provided will deal with the various use of spread, which is why you should seek help by starting with at least one spread Math example. If you are unsure about what your subject stands for, take a look at similar questions and check if there are equations and formulas that look close to what you may require to write down.