How does a particle's kinetic energy vary with a(t)?

In summary: Expand on "slower moving particles (non-relativistic)".I calculated that for a matter dominated universe it would vary as 1/a2, while the photon temperature varies as 1/a. Is this correct?Yes, this is correct.
  • #1
Buzz Bloom
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I am curious about how a particle's kinetic energy changes with the explansion of the unverse. I calculated that for a matter dominated universe it would vary as 1/a2, while the photon temperature varies as 1/a. Is this correct? Is this also true in general? If so, the implication would be that assuming equlibrium at the time of recombination, the temperature of the atomic hydrogen gas that dominates bryonic matter would about 0.003 oK. Would this be detectible by finding very narrow fraunhoffer lines of light passing through this gas?
 
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  • #2
Buzz Bloom said:
I am curious about how a particle's kinetic energy changes with the explansion of the unverse.

Kinetic energy relative to what?

Buzz Bloom said:
I calculated that for a matter dominated universe it would vary as 1/a2, while the photon temperature varies as 1/a. Is this correct?

I can't tell because I don't know what this kinetic energy is supposed to be relative to. What kind of particles are you assuming, what state of motion are they in, and what is the kinetic energy being measured relative to?

Buzz Bloom said:
the implication would be that assuming equlibrium at the time of recombination, the temperature of the atomic hydrogen gas that dominates bryonic matter would about 0.003 oK.

This is certainly not correct. The temperature of the universe at the time of recombination was about 3,000 K. Since up to that point matter and radiation were in thermal equilibrium, that was the temperature of both. But this temperature is not calculated based on calculation of kinetic energy of a particle; it is simply the observed temperature of the CMBR today, 3 K, times the redshift factor of the CMBR, which is about 1000.
 
  • #3
My bad. I should have said that the kinetic energy is relative to the comoving coordinates of the explansion. I am assuming the particles are about 75% hydrogen atoms, 23% He atoms, ans 2% all other baryonic matter paricles formed by nuclear fusion during the early expansion, but the same relationship between kinetic energy and a(t) should also apply to dark matter particles.
 
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  • #4
Buzz Bloom said:
I should have said that the kinetic energy is relative to the comoving coordinates of the explansion.

A comoving particle is at rest in the comoving coordinates of the expansion, so its kinetic energy in those coordinates is zero.

In our cosmological models, we don't model the matter or radiation or dark matter or dark energy as individual particles; we model it as a fluid, with an equation of state relating pressure and energy density; "matter", "radiation", and "dark energy" simply correspond to different equations of state. The key relationship for each kind of fluid is how its energy density ##\rho## changes with the scale factor ##a##: for matter (ordinary or dark), we have ##\rho \propto 1 / a^3##; for radiation, we have ##\rho \propto 1 / a^4##; and for dark energy, we have ##\rho = constant##. But this energy density is not "kinetic", because the fluid is at rest in comoving coordinates; it is rest energy density, i.e., the energy density in the fluid rest frame.
 
  • #5
Buzz Bloom said:
I am curious about how a particle's kinetic energy changes with the explansion of the unverse. I calculated that for a matter dominated universe it would vary as 1/a2, while the photon temperature varies as 1/a. Is this correct? Is this also true in general? If so, the implication would be that assuming equlibrium at the time of recombination, the temperature of the atomic hydrogen gas that dominates bryonic matter would about 0.003 oK. Would this be detectible by finding very narrow fraunhoffer lines of light passing through this gas?

I believe momentum falls off as 1/a. The kinetic energy of primoridal neutrinos for example would be falling off about the same way as the CMB temperature.

I could be wrong of course. You might look up the Cosmic Neutrino Background
http://en.wikipedia.org/wiki/Cosmic_neutrino_background
analogous to the CMB except that "recombination" happened earlier and the temp is just a bit lower, around 2K.

I'll try to check a little bit. I understand that Weinberg treats this in his cosmology textbook.
Expansion drains particles of their momentum.
Always relative to "comoving observers" of course. Momentum and kinetic energy as seen by observers at CMB rest.
 
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  • #6
https://www.physicsforums.com/threads/expansion-drains-momentum-from-massive-objects.249589/
An alternative to looking it up in Steven Weinberg's textbook:

http://arxiv.org/abs/0808.1552
Note on the thermal history of decoupled massive particles
Hongbao Zhang
(Submitted on 11 Aug 2008)
"This note provides an alternative approach to the momentum decay and thermal evolution of decoupled massive particles. Although the ingredients in our results have been addressed in [Weinberg's new Cosmology text], the strategies employed here are simpler, and the results obtained here are more general."

Yes this confirms that momentum falls off as 1/a

So it would seem that for slower moving particles (non-relativistic) the kinetic energy would fall off as 1/a2 as you conjectured!
 
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  • #7
Buzz Bloom said:
My bad. I should have said that the kinetic energy is relative to the comoving coordinates of the explansion. I am assuming the particles are about 75% hydrogen atoms, 23% He atoms, ans 2% all other baryonic matter paricles formed by nuclear fusion during the early expansion, but the same relationship between kinetic energy and a(t) should also apply to dark matter particles.
The average kinetic energy of the particles in the co-moving frame is given by the temperature. When self-gravitation becomes important, there's no simple formula for how this temperature varies over time. When self-gravitation isn't important (i.e., before any bound objects form), the change in temperature should be an adiabatic process.
 
  • #8
I was not assuming the particles moved with the comoving coordinates. I assumed that a particle's velocity was relative to these coordinates.

This is how I calculated the change in velocity.

Assume that at time t1 a particle is at some point P1 in space and, P1 moves with the comoving coordinates, and at time t2 it is at another such point, P2. If v is the average velocity between P1 and P1, the distance the particle has moved is (t2-t1) v.

During the period of time a has increased from a(t1) to a(t1).

For relatively small t2-t1, h(t2) ~= (a(t2)-a(t1)) / (a (t2-t1)).

The Hubble redshift velocity between these two points, corresponding to the distance (t2-t1) v, is H(t2) v (t2-t1).

For a matter dominated universe, a = const t2/3, t = const a3/2,
and H(t) = 2/(3t).

Therefore the Hubble redshift velocity is 2/(3 t2) v (t2-t1). This represents the reduction in the particle's velocity v(t1)-v (t2) .

Let t = t2, dt = t2-t1. Also let dv = v(t2)-v(t1).

(1/v) dv = -(2/3) 1/t dt

Integrating:
log(v) = log(const) - 2/3 log(t) = log (const t-2/3) .

Exponetiating:
v = const t-2/3 = const/a .

So, if the velocity (or momentum) varies inversely with a, the kinetic energy would vary inversely with a2.
 
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  • #9
marcus said:
Yes this confirms that momentum falls off as 1/a

So it would seem that for slower moving particles (non-relativistic) the kinetic energy would fall off as 1/a2 as you conjectured!

Hi marcus:

Thank you for your post agreeing with my conclusion. I have downloaded the Zhang paper and plan to study it.

Do you know if astronomers could detect this phenomenon by finding extremely small widths for Fraunhoffer lines in light passing through a relatively small regions of primordial gas?
 
  • #10
marcus said:
I believe momentum falls off as 1/a. The kinetic energy of primoridal neutrinos for example would be falling off about the same way as the CMB temperature.

Hi again marcus:

If a neutrino has rest mass, then its toral energy should be reduced to an extremely small amount more than its rest mass if its kinetic energy varies with a2.

Do you know what assumptions are made (for example: a nutrino has zero rest mass an travels at light speed) when the current density for neutrinos is given as about the same value as for photons? (Please see https://en.wikipedia.org/wiki/Cosmic_neutrino_background .)
 
  • #11
Chalnoth said:
The average kinetic energy of the particles in the co-moving frame is given by the temperature. When self-gravitation becomes important, there's no simple formula for how this temperature varies over time. When self-gravitation isn't important (i.e., before any bound objects form), the change in temperature should be an adiabatic process.

Hi Chalnoth:
As I have come to understand (with much uncertainty) an adiabatic process assumes equilibrium, and the universe had not been in a state of equilibrium since at least the time of recombination, and possibly not since nuclear synthesis. Am i wrong about this?
 
  • #12
Buzz Bloom said:
Hi Chalnoth:
As I have come to understand (with much uncertainty) an adiabatic process assumes equilibrium, and the universe had not been in a state of equilibrium since at least the time of recombination, and possibly not since nuclear synthesis. Am i wrong about this?
To a very good approximation, the universe prior to the emission of the CMB was in thermal equilibrium at each point in time. This early expansion was approximately adiabatic. I'm not completely sure how to extend this to non-interacting particles (e.g. neutrinos, dark matter), but I doubt it changes too much.

This approximation breaks down pretty dramatically at later times, when gravitationally-bound systems start to form.
 
  • #13
marcus said:
The kinetic energy of primoridal neutrinos for example would be falling off about the same way as the CMB temperature.

But this is because neutrinos are relativistic--they're not quite massless, but they're close to it, so their equation of state is very similar to that of radiation. That means their energy density falls off like that of radiation, i.e., as ##1 / a^4##. But the energy density of non-relativistic matter falls off as ##1 / a^3##.

marcus said:
momentum falls off as 1/a

So it would seem that for slower moving particles (non-relativistic) the kinetic energy would fall off as 1/a2

I understand the argument for momentum falling off as ##1 / a##, but I'm still trying to reconcile that, and the corresponding conjecture that kinetic energy should fall off as ##1 / a^2##, with the behavior of the energy density of matter and radiation cosmological fluids, which I described above.
 
  • #14
Chalnoth said:
This approximation breaks down pretty dramatically at later times, when gravitationally-bound systems start to form.

Hi Chalnoth:

I am currently reading a very well written book: An Introduction to Cosmology by Derek Raine and Ted Thimas (2001). This source says that a very large majority of the universe's baryonic matter remains primoridal and not influenced by the gravity of galaxies, clusters of galaxies and super clusters. Also, the expansion of the universe does not effect these structures, For structures larger than super cluseters, the expnasnsion pushes these structures apart, and the greatest mass of primordial stuff is among these 4th order structures.
 
  • #15
PeterDonis said:
But this is because neutrinos are relativistic--they're not quite massless, but they're close to it, so their equation of state is very similar to that of radiation. That means their energy density falls off like that of radiation, i.e., as ##1 / a^4##. But the energy density of non-relativistic matter falls off as ##1 / a^3##.
The kinetic energy for non-relativistic particles is a negligible component of their total energy. So if you're interested in how the kinetic energy of non-relativistic particles changes over time, it's not useful to examine their total energy.
 
  • #16
Buzz Bloom said:
Hi Chalnoth:

I am currently reading a very well written book: An Introduction to Cosmology by Derek Raine and Ted Thimas (2001). This source says that a very large majority of the universe's baryonic matter remains primoridal and not influenced by the gravity of galaxies, clusters of galaxies and super clusters. Also, the expansion of the universe does not effect these structures, For structures larger than super cluseters, the expnasnsion pushes these structures apart, and the greatest mass of primordial stuff is among these 4th order structures.
I'm not sure what the ratios are, but I believe it.

One issue is that within clusters, most of the baryonic matter consists of the diffuse cluster gas whose temperature is essentially set by the depth of the gravitational well. That temperature is so vastly higher than the temperature of the gas between clusters (it emits in the X-ray range) that the impact of expansion on temperature for such gas is negligible.

It might be reasonable to use this sort of analysis to get an idea of the temperature of the matter that remains between galaxies and galaxy clusters, but for most purposes that temperature is going to be essentially zero. You can probably get a rough idea by assuming adiabatic expansion, though.
 
  • #17
Just bear in mind that the intercluster medium is now ionized (and has been since enough stars turned on at about z=20 or so), so it should interact with the CMB. That complicates things a bit.
 
  • #18
Chalnoth said:
Just bear in mind that the intercluster medium is now ionized (and has been since enough stars turned on at about z=20 or so), so it should interact with the CMB. That complicates things a bit.
Actually, now that I think of it, the ionization of the intercluster medium indicates that it absorbs quite a bit of energy from stars, meaning its temperature is likely much higher than that of the CMB, or than you would naively expect from looking at expansion alone. I wasn't able to find any decent sources on this with a quick search, but my bet is that getting at the details of how temperature of baryonic matter has changed over time is going to be really, really difficult. For dark matter, the difficulty there is that we don't know the temperature at early times, so it's hard to say how that changes (in the simplest model, which seems to fit the data pretty well, the temperature is identically zero).
 
  • #19
Chalnoth said:
The kinetic energy for non-relativistic particles is a negligible component of their total energy.

Ah, of course; the ##1/a^3## is just the number density being diluted by the expansion.
 
  • #20
marcus said:
http://arxiv.org/abs/0808.1552
Note on the thermal history of decoupled massive particles
Hongbao Zhang

I hope someone can help me understand this. I read trhe Zhang article recommended by marcus. Here is a quote:

Whence we know that for a freely traveling massive particle in
an expanding FLRW universe, it is its momentum rather than energy that goes like
p ∝ 1 / a (2.7).


p here is momentum. Later the article says:

So the number density of massive particles at the time t with momentum between p and p + dp would be

Zhang1.PNG

Zhang then discusses the use ot the Fermi-Dirac and Bose-Einstein distributions in the "second step", and observes:

Therefore the form of the Fermi-Dirac and Bose-Einstein distributions are preserved for the thermal evolution of decoupled
massive particle,
...

The d subscripts represent variable values at the latest time of equilibriium. The e subscripts a later time when the particles are no longer in equilibrium. The article also shows the derived temperature realtion below.

Zhang2.PNG


This seems to say that the temperature of a collection of particles not in equilibrium varies inversly with a. Near the end of the article, the following is said:

... although the spectrum has still kept the form of the Fermi-Dirac and Bose-Einstein distributions
since decoupling, it is not the thermal spectrum with the effective temperature and
chemical potential since the effective mass is not equal to the static mass.


What I think all this comes to, although it is not said explicitly (and I am not at all clear in my mind that I have it right) is the Fermi-Dirac and Bose-Einstein distributions form are preserved assuming an artificial temperature, while the real, Maxwell-Boltzmann distibution of the energy (in terms of the velocity-squared of the collection of particles) would correspond to a temperature that varied inversely with a2.

Also, as I understand it, the Fermi-Dirac and Bose-Einstein distributions relate to leptons and bosons rather than to atoms, but the conclusions should apply to atoms also with respect to the Maxwell-Boltzmann distibution.

I hope someone will comfirm my understanding if it is right, and correct it if I have it wrong.
 
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  • #21
I think it's probably correct. Just bear in mind that it ignores a lot of the messy details of reality (mentioned above).
 
  • #22
Hi Chalnoth:

Thank you very much for you response.

I now how some new questions regarding this understanding. I will soon ask these questions as new threads.
 
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  • #23
If momentum scales as 1/a then combining this with ## E^2=m^2c^4+p^2c^2 ##, I get for a population of particles moving wrt the comoving frames ## \rho=\sqrt{(\frac{\rho_m^0}{a^3})^2+(\frac{\rho_p^0}{a^4})^2} ## which does work for either light or comoving matter - is it correct in general ? I had been wondering about that case.
 
  • #24
I have been confused for a long time about the relationship between entropy and expansion. From this very interesting thread, and from the Zhang paper I am back to it. I can imagine the universe as a closed Hamiltonian (Louiville) phase space. But it can't be called an "isolated" system can it, If temperature, kinetic-energy is bleeding away from the standpoint of any observer of a moving massive particle?What's the relevant distinction between a closed system and an isolated system here? And why is it one and not the other, and the is the phase space open or closed? Or are you all already in agreement that the direct expansion related "thermodynamic" effect you all are discussing mean that the SLOT has to be tied mechanically to expansion somehow, not just as a potentially incidental, anthropocentric "observed tendency for the arrow of time"?

And just to make sure I'm asking an egregious number of questions, :woot: How does this relate to something like Verlinde's entropic gravity?
 
  • #25
Jimster41 said:
I can imagine the universe as a closed Hamiltonian (Louiville) phase space.

I think this is only true if the universe is spatially closed. If it's spatially infinite, I don't think the phase space is closed.

Jimster41 said:
it can't be called an "isolated" system can it,

The entire universe is, yes; there's nothing outside it for it to exchange heat and work with.

Jimster41 said:
If temperature, kinetic-energy is bleeding away from the standpoint of any observer of a moving massive particle?

This just means that the moving massive particle, taken by itself, is not isolated. It doesn't mean the universe as a whole is not isolated.
 
  • #26
Those are helpful answers. Gotta think about them.

So the expanding universe, is a closed phase space if we expect it to stop expanding at some point. In other words there was always a metric limit with a countable number of states. So the bath the massive particle is coupled to is spacetime with a locally expanding metric, which is "bleeding" it's energy. Is that correct? I guess I'm confused about the "location" of the bath, or it's gradient, the massive particle is coupled with, that it is exchanging energy with. It is the same everywhere in space (not counting other particles) but not the same everywhere in time?
 
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  • #27
wabbit said:
If momentum scales as 1/a then combining this with ## E^2=m^2c^4+p^2c^2 ##, I get for a population of particles moving wrt the comoving frames ## \rho=\sqrt{(\frac{\rho_m^0}{a^3})^2+(\frac{\rho_p^0}{a^4})^2} ## which does work for either light or comoving matter - is it correct in general ? I had been wondering about that case.
I don't think so. You have to use conservation of stress-energy to determine how energy density scales with expansion.
 
  • #28
Ah OK thanks it was too easy then :)

Edit: did not try the stress-energy, but following a stream of particles flowing past comoving observers for an infinitesimal interval and using relativistic velocity addition I get ##\frac{\delta v}{v}=-(1-v^2)H\delta t,\frac{\delta p}{p}=-H\delta t,\frac{\delta V}{V}=3H\delta t## and I still find that p scales as 1/a and that the energy density of the stream follows that formula. Is there something wrong in this method?
 
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  • #29
I started working through the equations, and I'm pretty sure this doesn't work. I might have made a mistake, but it really looks like there probably isn't an analytical solution. It's not easy to do the math correctly, though, as the special relativistic equations describe a single particle at a time, while the fluid equations describe the behavior of a collection of particles.

What I did was assume an ideal fluid, for which the conservation of stress energy equation is:

[tex]\dot{\rho} + 3{\dot{a} \over a}\left(\rho + p\right) = 0[/tex]

Typically the way this equation is solved is to convert it into a differential equation in terms of a, by using the substitution:

[tex]{d\rho \over dt} = {d\rho \over da}{da \over dt}[/tex]

Using this and simplifying, the equation becomes:

[tex]{d\rho \over da} + {3 \over a}\left(\rho + p\right) = 0[/tex]

For the case where ##p = w\rho## with constant ##w##, this is a very easy differential equation to solve. But it doesn't look to be so easy when you make use of the relativistic pressure of an ideal gas, which I believe is given by:

[tex]p^2c^2 = {1 \over 3}\left(\rho^2 - \left({N_0 m c^2 \over a^3}\right)^2\right)[/tex]

Here ##N_0## is the number density of the particles when ##a = 1##, ##m## is the mass of the individual particles, ##\rho## is the energy density, and ##p## is the pressure. Perhaps I've made a mistake in translating momentum to pressure, but the prospects do not look good.
 
  • #30
Hmm I am not following here - but I wasn't thinking of solving the Friedmann equation here, more of the evolution of such matter as a marginal component of the universe, i.e. assuming a given FRW background - such as the cooling of a neutrino gas. The calculation I did was actually for a (local) stream of particles initially sharing a single common comoving velocity, though I thought it would generalize to a collectively comoving pressureless gas.
 
  • #31
marcus said:
http://arxiv.org/abs/0808.1552
Note on the thermal history of decoupled massive particles
Hongbao Zhang
(Submitted on 11 Aug 2008)
"This note provides an alternative approach to the momentum decay and thermal evolution of decoupled massive particles. Although the ingredients in our results have been addressed in [Weinberg's new Cosmology text], the strategies employed here are simpler, and the results obtained here are more general."

Buzz Bloom said:
I hope someone can help me understand this. I read trhe Zhang article recommended by marcus.

This seems to say that the temperature of a collection of particles not in equilibrium varies inversly with a. Near the end of the article, the following is said:

... although the spectrum has still kept the form of the Fermi-Dirac and Bose-Einstein distributions
since decoupling, it is not the thermal spectrum with the effective temperature and
chemical potential since the effective mass is not equal to the static mass.


What I think all this comes to, although it is not said explicitly (and I am not at all clear in my mind that I have it right) is the Fermi-Dirac and Bose-Einstein distributions form are preserved assuming an artificial temperature, while the real, Maxwell-Boltzmann distibution of the energy (in terms of the velocity-squared of the collection of particles) would correspond to a temperature that varied inversely with a2.

PeterDonis said:
That's not quite correct. The correct statement is that temperature varies inversely with ##a## for relativistic particles (which includes radiation), and inversely as ##a^2## for non-relativistic particles (which includes all the ordinary matter and dark matter in our present universe). At the time of neutrino decoupling, the neutrinos are still highly relativistic, so their temperature will still vary as ##1 / a##. Whether or not the neutrinos become non-relativistic at some point after decoupling depends on the neutrino masses; for small enough neutrino masses, they could still be relativistic even today.

There seems to be a misunderstanding here. The temperature scaling of a decoupled species is determined by properties at the time of decoupling. If a species was non-relativistic at the time of decoupling, then, after decoupling, its temperature is inversely proportional to a^2. If a species was (ultra)relativistic at the time of decoupling, then, after decoupling, its temperature is inversely proportional to a. The latter is true even when a species that was relativistic at the time of decoupling is cooled by expansion to non-relativistic speeds.

I think this is what Chalnoth means by

Chalnoth said:
Also, let me add one other thing. This statement isn't true:

Temperature varies inversely as ##1/a(t)## as long as the neutrinos remain relativistic. They don't need to interact with anything to maintain this temperature scaling because they'll retain their thermal distribution without interactions.

I spent a pleasant morning looking at the quantitative details of Zhang's paper, a couple of whose points seem to be somewhat well-known before publication of the paper. Right now, I have to pick up my daughter from school, and take her to a piano lesson and soccer practice. I will try to make an effort tomorrow to fill in the quantitative details of how Zhang's paper relates to the above.
 
  • #32
George Jones said:
The latter is true even when a species that was relativistic at the time of decoupling is cooled by expansion to non-relativistic speeds.

Hm, I wasn't aware of this. Is there a quick heuristic explanation of why it's true? I'm not getting anything from the Zhang paper that is helpful in that regard.
 
  • #33
George Jones said:
There seems to be a misunderstanding here. The temperature scaling of a decoupled species is determined by properties at the time of decoupling. If a species was non-relativistic at the time of decoupling, then, after decoupling, its temperature is inversely proportional to a^2. If a species was (ultra)relativistic at the time of decoupling, then, after decoupling, its temperature is inversely proportional to a. The latter is true even when a species that was relativistic at the time of decoupling is cooled by expansion to non-relativistic speeds.

I think this is what Chalnoth means by
Chalnoth said:
Also, let me add one other thing. This statement isn't true:

Temperature varies inversely as 1/a(t) as long as the neutrinos remain relativistic. They don't need to interact with anything to maintain this temperature scaling because they'll retain their thermal distribution without interactions.
No, that's not what I meant. What I was getting at is that the thermal distribution that they had at the time of decoupling remains thermal from there on. That is, the expansion reduces their temperature but doesn't make them non-thermal. This is certainly a true statement as long as the neutrinos remain relativistic. I am not quite certain it's true when the neutrinos transition from relativistic to non-relativistic.

The calculations I was trying to do later were using the assumption that you could deal with all particles of a given energy independently, and track their behavior forward in time.
 

1. How does a particle's kinetic energy change with acceleration?

The kinetic energy of a particle is directly proportional to its acceleration. This means that as the acceleration increases, so does the kinetic energy of the particle.

2. Does a particle's kinetic energy increase or decrease with time?

A particle's kinetic energy can either increase or decrease with time, depending on the direction and magnitude of its acceleration. If the particle is accelerating in the same direction as its velocity, its kinetic energy will increase. However, if the particle is decelerating or changing direction, its kinetic energy will decrease.

3. How does the mass of a particle affect its kinetic energy?

The mass of a particle does not directly affect its kinetic energy. However, a particle with a larger mass will require more energy to accelerate to a certain velocity compared to a particle with a smaller mass.

4. Can a particle have kinetic energy without acceleration?

No, a particle cannot have kinetic energy without acceleration. Kinetic energy is the energy that a particle possesses due to its motion, and acceleration is necessary for the particle to be in motion.

5. How does the velocity of a particle affect its kinetic energy?

The kinetic energy of a particle is directly proportional to the square of its velocity. This means that as the velocity of a particle increases, its kinetic energy will increase exponentially.

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