Kinetic Energy and Inelastic Collisions

[protonman]

Here is my understanding of the problem. What is being overlooked is the object's internal elastic potential energy. The incorrect assumption is that the objects involved in the collision are incompressable. If the objects were in compressable then the instant the blocks collide the force at the point of contact would instantaneously arise at all points along the object [in the direction of motion]. This would violate SR because the information [that there was contact] would travel faster that the speed of light [i.e. instantaneously].

The situation is like a compressed spring that once compressed does not return to equilibrium. The initial kinetic energy of both objects is stored in internal potential energy. In reality a perfect inelastic or perfectly elastic collision is impossible because it neglects friction. But in our model we were not concerned with the effects of friction. Even in [other] models that ignore friction KE is conserved. The key issue here is that a perfectly rigid [i.e. incompressable] body is in violation of SR.

Newtonian mechanics is not complete. It is only an approximation and therefore where it does not consider the postulates of SR it should not be assumed to be correct [in all cases].

 

[jcsd]

Here you imply that energy is not conserved in all reference frames.

Here you say energy is conserved in all reference frames.

Where did I imply that energy is not conserved in all reference frames? 4-momenta is always conserved, so energy is always conserved in all refernce frames, the two go hand-in-hand.

 

[jcsd]

My apologies, I thought that all sub-atomic collisions weren't (perfectly) inelastic?

jcsd: I don't get what you're arguing. You've been told that the energy goes to heat (from friction) and sound (as well as others, I'm sure...). Do you think the energy just disappears? Why aren't you accepting the answers you wanted?

Again where did I see the energy can't go to heat. I never asked any questions either.

 

[protonman]

Where did I imply that energy is not conserved in all reference frames? 4-momenta is always conserved, so energy is always conserved in all refernce frames, the two go hand-in-hand.

On Post #22

 

[robphy]

In my opinion, I think the original problem
"Why is KE not conserved in inelastic collisions?"
is not clearly posed since the original poster is obviously seeking for something deeper than the textbook definition [that an "inelastic collision" is one in which the total-KE is not conserved],

Can you please pose a clear question (possibly followed by what features you are looking for, or not looking for)?

[If "inelastic collision" means something other than the textbook definition, it would be helpful if you precisely define what YOU mean.]

 

[protonman]

In my opinion, I think the original problem
"Why is KE not conserved in inelastic collisions?"
is not clearly posed since the original poster is obviously seeking for something deeper than the textbook definition [that an "inelastic collision" is one in which the total-KE is not conserved],

Can you please pose a clear question (possibly followed by what features you are looking for, or not looking for)?

[If "inelastic collision" means something other than the textbook definition, it would be helpful if you precisely define what YOU mean.]

The textbook answer is wrong and/or incomplete.

A bit of advice. You don't know how much physics I have studied. I ask questions here to inspire discussion, not to get answers. This is the last place I would look for answers. If you want to test my understanding of the physics I have been discussing look at what I have said about SR. The best most people can do here is regurgitate their textbooks which is useless because anyone who can read can do the same. The whole point is to understand what the books are saying and then check if what they say is valid.

 

[robphy]

Physics Lesson One:

What is called the 4-momentum is not really "momentum" in the classical sense. If you differential the space-time 4-vector with respect to proper time you get what is called the momentum 4-vector.

I think you need to multiply by the rest mass to get what is called the momentum 4-vector.

Both photons and electrons have momentum. If they have an equal amount of momentum in opposite directions they will collide and both have a final velocity of zero.

Photons can't be brought to rest.

 

[protonman]

I think you need to multiply by the rest mass to get what is called the momentum 4-vector.



Photons can't be brought to rest.

WRT the first statement what I said initially is correct.

But they can be absorbed. The whole problem is that physicsts don't even understand what a photon is.

 

[robphy]

You need to read the post before you write.

They do have zero velocity because we are talking about inelastic collisions.

Is there anyone here who can think?

BTW, if you do the calculation out for a photon and electron moving with opposite momentums towards one another it does work out. Using the relativistic momenta.

(Energy of photon)/c - (gamma)(mass of electron)(velocity of electron) = 0

Just solve for the velocity of the electron in this case and see for yourself.

But they can be absorbed. The whole problem is that physicsts don't even understand what a photon is.

Let k be the null 4-momentum of the incident photon.
Let p be the timelike 4-momentum of the electron before the collision.
Let p' be the timelike 4-momentum of the electron after the collision.
On the assumption that the photon was "absorbed", there is no k'.

4-momentum conservation implies
k + p = p'

squaring
<br /> \begin{align*}<br /> k\cdot k + 2 k \cdot p + p \cdot p &amp;= p&#039; \cdot p&#039; \\<br /> 0 + 2 k \cdot p + m^2 &amp;= m^2 <br /> \end{align*}<br />
where m is the rest mass of the electron.

So,
k \cdot p = 0
Since k is null and p is timelike and nonzero, then k must be zero [that is, there is no such incident photon... the proposed collision is not possible.] (This supports Gokul43201's statement that "A free electron can not absorb a photon".)

Alternatively,
evaluate the invariant k \cdot p = 0 in the rest frame of the electron (so, p = \left( \begin{array}{cc} m \\ \vec 0\end{array}\right) and k = \left( \begin{array}{cc} k_0 \\ \vec k \end{array}\right), where k_0=|\vec k|). Then 0=k \cdot p = m(k_0)-\vec 0\cdot \vec k=m(k_0). Since m \neq 0, then k_0=0 (and hence \vec k=\vec 0 and so the 4-vector k=0).

 

[protonman]

Let k be the null 4-momentum of the incident photon.
Let p be the timelike 4-momentum of the electron before the collision.
Let p&#039; be the timelike 4-momentum of the electron after the collision.
On the assumption that the photon was "absorbed", there is no k&#039;.

4-momentum conservation implies
k + p = p&#039;

squaring
<br /> \begin{align*}<br /> k\cdot k + 2 k \cdot p + p \cdot p &amp;= p&#039; \cdot p&#039; \\<br /> 0 + 2 k \cdot p + m^2 &amp;= m^2 <br /> \end{align*}<br />
where m is the rest mass of the electron.

So,
k \cdot p = 0
Since k is null and p is timelike and nonzero, then k must be zero [that is, there is no such incident photon... the proposed collision is not possible.] (This supports Gokul43201's statement that "A free electron can not absorb a photon".)

Alternatively,
evaluate the invariant k \cdot p = 0 in the rest frame of the electron (so, p = \left( \begin{array}{cc} m \\ \vec 0\end{array}\right) and k = \left( \begin{array}{cc} k_0 \\ \vec k \end{array}\right), where k_0=|\vec k|). Then 0=k \cdot p = m(k_0)-\vec 0\cdot \vec k=m(k_0). Since m \neq 0, then k_0=0 (and hence \vec k=\vec 0 and so the 4-vector k=0).

Your mathematics don't impress me. If you want to debate with adults you need to think for yourself. Just because the math works does not mean you have proved anything.

 

[jcsd]

On Post #22

No I didn't, read what I wrote. 4-momanta is always conserved so:

1) Free electronns can't absorb photons

2) energy is conserved in all referbce frames

 

[Gokul43201]

It looks like protonman has figured out where the lost KE of the inelastic collision goes - "the object's internal elastic potential energy". So, pending the question of photon absorption by a free electron and doubts about 'who said what, and when', I believe this question has reached a resolution.

 

[jcsd]

It looks like protonman has figured out where the lost KE of the inelastic collision goes - "the object's internal elastic potential energy". So, pending the question of photon absorption by a free electron and doubts about 'who said what, and when', I believe this question has reached a resolution.

There's nothing to resolve about whthere or not a free photon can absorb an electron it's basic physics, and it's hardly a big secret as it appears in enough physics textbooks.

 

[electronman]

No I didn't, read what I wrote. 4-momanta is always conserved so:

1) Free electronns can't absorb photons

2) energy is conserved in all referbce frames

Recall from post #22

God you're obnoxious. I didn't say momenta isn't conserved I said 4-momenta isn't conserved. For 4-momentum to be conserved, energy (which can be seen as the time coponent of 4-momentum) must also be conserved in all reference frames.

 

[electronman]

It looks like protonman has figured out where the lost KE of the inelastic collision goes - "the object's internal elastic potential energy". So, pending the question of photon absorption by a free electron and doubts about 'who said what, and when', I believe this question has reached a resolution.

And the interesting part is that no one even acknowledged my post. It seems here that when the bad guy is the right guy they ignore and eventually ban users.

 

[jcsd]

Recall from post #22

And if you recall the context, it's apparent that I meant 4-momentum isn't conserved in the example that he gave and hence the example is physically impossible.

 

[electronman]

And if you recall the context, it's apparent that I meant 4-momentum isn't conserved in the example that he gave and hence the example is physically impossible.

Ha Ha you lose!

 

[jcsd]

Ha Ha you lose!

What!? quit your babbling, what was the point to this thread? have you just come here to troll?

 

[electronman]

What!? quit your babbling, what was the point to this thread? have you just come here to troll?

This post ended at post #31 as far as I am concerned since no one raised any objection to my answer.

 

[jcsd]

The question was answered befopre then. You don't need to consider SR at all and the compressibilty of objects is irrelevant as even in SR objects can be point particles.

 

[kevin282819]

hi, new user to the forum, so sorry if i break any rules on thread necromancy (i checked the rules but i didnt see anything on it) I am just a physics 12 studen and i was also wondering the same thing in this question, protonman provided a good explanation about the objects being compressed and all, but there's a problem in it i don't quite get yet, maby its just because of lack of knowledge but, if the objects WERE to be completely elastic, the original question would have momentum conserved and EK not conserved, so if we turned EK into conserved, we would have to change the velocity, seeing as mass is unchangable, but if we were to do that, then wouldn't the momentum before and after be different while kinetic energy is conserved? (P.S. i would quote the post but i am typing this out from a ps3 and it has a limit to how much i can type out at once)