# Symmetrical distribution of energy.

1. Dec 24, 2007

we know that during inelastic collision momentum remain conserved but kinetic energy of system does not
some part of K.E in converted to other forms like light ,sound ,internal energy etc.

let us consider two subatomic particles undergoing one dimensional inelastic collision(just as case)
in this collision light is emitted to the surrounding.

looking at the collision in newtonian way equations run like this
P1+P2=p1+p2 P-initial momentum p-final momentum (law of conservation of momentum)
initial K.E=final K.E+light energy.............law of conservation of energy.
but when i look at the collision in einstienian way i will assign some momentum to light
as light is just a stream of photons
momentum of photon is =energy/c
thus rewriting the equations i will write
P1+P2=p1+p1+{momentum of photons}
but this destroys our concept inelastic collisions.
therefore to preserve the concept i figured that momentum of photons must cancel each other
this is possible if identical photons(equal energy) are emitted in opp directions.

but we see that this case can be generallised to all inelastic collisions ,where energy is emitted to surrounding(dissipation takes place)
like in sound,as energy can be transfred to the surrounding only by the interaction of the system with the surrounding.but this destroys conservation of momentun of the system in this cases.
in order to explain it we need to assume that energy is given out in equal amount in opposite directions(symmetrically)( like sound wave spreading in spherical shell form.)
as this cancels out all the forces acting on system during dissipation of energy and conserves momentum of system in newtonian way.

but the correctness of my arguments and deduction is doubtfull
any views or corrections will be appreciated.................................thank you.

Last edited: Dec 24, 2007
2. Dec 24, 2007

### Shooting Star

In inelastic collisions in the macroscopic world, KE is lost because it is converted into mainly heat energy of the colliding bodies, also energy to deform the bodies and maybe into sound etc. That energy will not be available to be able to contribute to the KE of the system again.

In subatomic particles, there is no way of converting the KE into thermal energy, so every collision will be elastic.

In atomic collisions, suppose during a collision between two atoms or a photon and an atom, the electron in an atom is raised to a higher energy level. Then the KE may not be conserved, but it is not dissipated into mostly unrecoverable thermal energy. Or two atoms may stick together to make a molecule. Again, it has not been dissipated into thermal energy. Whether you call these cases inelastic is a matter of choice, I think.

3. Dec 24, 2007

when we consider a particle system of two when cannot define thermal energy on the basis of motion of two particles as it require much larger and random systems.like a ball
which is made of no. of atoms
defining thermal energy on the basis motion of electrons or other sub atomic particles
withen the colliding particles makes no sense other wise we would have assigned temperature to an atom at rest. i dont understand what do ment by thermal energy in sub atomic or atomic collisions.

I am sure their that their must be some collisions which might emit light ,then take it in
their case.
not only light when a collision emit sound ,Particles of air move that is they gain momentum ,the monentum must come from somewhere (if we dont consider waves moving out symmetrically),which is tranferred through a wave,take it in this case.

question is not on dissipation but tranfer of energy to surrounding
which definitly takes place in a no. of collisions.I think

as far as i have studied collisions are of only two kind along with their
equations and relations established on newtonian physics and not on thermodynamics or such other feilds.

Last edited: Dec 24, 2007
4. Dec 24, 2007

### Shooting Star

Actually, now I am not very sure about exactly what you are trying to say. I read your first post again an let me quote you:

>
thus rewriting the equations i will write
P1+P2=p1+p1+{momentum of photons}
but this destroys our concept inelastic collisions.
therefore to preserve the concept i figured that momentum of photons must cancel each other
this is possible if identical photons(equal energy) are emitted in opp directions.

So, you are trying to preserve the concept of inelastic collisions, by emitting two identical photons in opp directions? I hope that you have taken into account that the magnitudes of the momenta of the emitted photons are different in different frames?

Once you answer this, we can wrap up this thread, I think. (Also, for the record, I never tried to ascribe thermal energy to single atoms.)

Last edited: Dec 24, 2007
5. Dec 24, 2007

### i_island0

In this i want to make a comment and ask something. As i am also having a lot of confusion in this. I read a question which i am not able to comprehend conceptually.

A neutron moving with a speed v strikes a hydrogen atom in the ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision might take place. [take mass of neutron and hydrogen to be same].
Here, i wrote KE as: mv^2 + E(1) = mvo^2 + E(n)
E(1) and E(n) refers to the energy of electron in 1st and nth state.

Here, after writing this energy equation, will i call it elastic or inelastic collision.

Moreover, the electron in the hydrogen atom is in nth state. Assume that electron falls back to the ground state, then according to Bohr's theory, a photon will be emitted. And that photon cannot be recovered in the system. Then will i call it elastic or inelastic collision.

And for the overall event, will the collision be called elastic or inelastic.

6. Dec 24, 2007

### Shooting Star

First let me say that the following statement which I had written was wrong.
>
In subatomic particles, there is no way of converting the KE into thermal energy, so every collision will be elastic.

What I had meant to say was that this would be the case if the particles didn’t make any new particle or didn’t break up, but just remained the same particles as before.
---------------

In analogy to collisions in classical mechanics, if the KE of the colliding particles are different before and after, it's termed as an inelastic collision. If the two stick together, it will be called completely inelastic.

So what do you feel we should call it in the case of a photon being emitted? And the overall event?

7. Dec 24, 2007

QUOTE=Shooting star;1551096]Actually, now I am not very sure about exactly what you are trying to say. I read your first post again an let me quote you:

>
thus rewriting the equations i will write
P1+P2=p1+p1+{momentum of photons}
but this destroys our concept inelastic collisions.
therefore to preserve the concept i figured that momentum of photons must cancel each other
this is possible if identical photons(equal energy) are emitted in opp directions.

So, you are trying to preserve the concept of inelastic collisions, by emitting two identical photons in opp directions? I hope that you have taken into account that the magnitudes of the momenta of the emitted photons are different in different frames?

well i think i can put this up in a better way
taking the question to atomic level is creating a lot of confusion

try figuring a collision between two metal ball
the collision can be elastic ,inelastic ,or somewhere in between 0<coefficient of restitution<1.

what we assume is that system remain isolated i.e net external force is 0.
& total mass &state remain same( no disintegration takes place)

but if we go by the definition then inelastic collision is such a collision
in which K.E is not conserved but is transformed to other forms but it is not made clear
,to what forms and whether energy remains withen the system or not.

in the above collision sound in emitted ,but sound is such an energy which is transferred to surrounding , energy cannot be converted to such forms like sound ,without the application of forces by surrounding(reaction by air molecules during vibrations).then how can we think of system being isolated.

in the case of sound it can be shown in many cases that though energy is emitted but net external forces on system are 0,(if you yourself not apply forces in some specific direction on the system ) .

this can be shown like a vibrating ball (COM is at rest ) does not start moving if placed in air ,though air does apply forces on ball during sound production.and it can be shown that sound produced is symmetric(you know what it mean)...

similiarly if it can be shown on sound (net force in 0),can it be shown on other forms
like light which takes energy out of the system.

can net force be 0.

if not in all cases then why it is written in that way that energy is converted to other formes like light.........

so what i mean to say that in the forms of energy which move out of system like sound ,light .can we consider system to be isolated.if yes then how ????????

Last edited: Dec 24, 2007
8. Dec 25, 2007

Last edited: Dec 25, 2007
9. Dec 26, 2007

### i_island0

i want to comment on things i wrote earlier, pls check if its correct..
I haven't read relativity yet, so i cant comment on that.
A neutron moving with a speed v strikes a hydrogen atom in the ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision might take place. [take mass of neutron and hydrogen to be same].
Here, i wrote KE as: mv^2 + E(1) = mvo^2 + E(n)
E(1) and E(n) refers to the energy of electron in 1st and nth state.

Here, after writing this energy equation, will i call it elastic or inelastic collision.
here i can call the collision elastic, as i assuming that no sound is emitted and no energy is lost by the system

Moreover, the electron in the hydrogen atom is in nth state. Assume that electron falls back to the ground state, then according to Bohr's theory, a photon will be emitted. And that photon cannot be recovered in the system. Then will i call it elastic or inelastic collision.
in this case energy is lost by the system in form of radiation. If somehow we can recover that energy then it can be called elastic, else inelastic.

And for the overall event, will the collision be called elastic or inelastic.[/QUOTE]
If energy remains unrecoverable, then its inelastic. else overall its elastic.
In collision of gases, subatomic particles are not colliding, so relating these collisions to them is a mistake.

10. Dec 26, 2007

### Shooting Star

A partially inelastic collision can happen when part of the initial KE is used to excite the H atom. I think in this case your equation, i.e., classical mechanics may be valid, because the change in energy is low. This is definitely partially inelastic.

A completely inelastic collision in this case will mean that the N and H stick together, forming a Deuterium atom. You have to use relativistic treatment for this case. The KE is converted to binding energy.

After emission of the photon, the overall collision may be considered to be inelastic, since the initial KE of the N and H is different from their final KE. But note that in this case, unlike classical mechanics, even the final momentum is different from the initial, because photons carry away momentum. The terminology is a matter of choice, but I believe this is termed in nuclear Physics as inelastic, especially when it’s not the atomic electron which is sent to a higher energy state, but the nucleus itself, which afterwards emits a high energy photon to come back to ground state. In nuclear reactions, the total energy and the total momentum are always conserved.

11. Dec 26, 2007

I also have not studied relativity as an chapter but only a bit as an interest that to two years ago in 9th standard

but still as far as i know when photon is not emitted then also the collision is partly or completly inelastic

as K.E is not conserved and in this case else everything is perfectly conserved .
energy is not moving out..
but equations need to be written in relativistic manner(dont bother)

in the case of emission of photon it is neither elastic nor inelastic if you consider on n,h.
as system is losing mass and momentum ........

what do you call to such collisions where system breaks???????

12. Dec 26, 2007

Really,
what if total energyof both is utilised in just exiting the electron.
neutron ,h will lie next to each other without any attraction,just at rest
this doest mean it will form deutrium
well for it exeptionaly large amount of energy is required far greater than what is needed to exite electron

13. Dec 26, 2007

### Shooting Star

EDIT: Deleting previous post, because I may have made a mistake. Will repost after correction.