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We consider a particle, A receiving energy from a second one,particle B in a one dimensional collision. $$E^2=p^2+m_0^2$$
$$EdE=pdp$$ For particle A:

$$E_AdE_A=p_Adp_A{\;\;\;\;\;\;}(1)$$

For Particle B:

$$E_B dE_B=p_Bdp_B {\;\;\;\;\;\;}(2)$$ Now $$\mid dE_A\mid=\mid dE_B\mid{\;\;\;\;\;\;}(3)$$

[Since enegry lost by one particle is gained by the other]

From Conservation of linear momentum we have:

$$\vec{p_A}+\vec{p_B}=\vec{K}$$ where $\vec{k}$ is a constant vector. Now, $$d\vec{p_A}+d\vec{p_B}=0$$ Or, $$d\vec{p_B}=-d\vec{p_A}$$ Or, $$\mid d\vec{p_B}\mid=\mid d\vec{p_A} \mid {\;\;\;\;\;\;}(4)$$

Applying relations (3) and (4) to (1) and (2) we have: $$\frac{E_A}{E_B}=\frac{\mid \vec{p_A} \mid}{\mid \vec{p_B} \mid}{\;\;\;\;\;\;\;\;}(5)$$ A pair of particles cannot interact unless relation (5) is satisfied. Can we conclude that that relation (5)to be a restriction for 1D collisions?

Now let's move to the general type of 3d collisions between a pair of particles A and B.

A frame is chosen where the particle B is initially at rest in it.

We may write: $$E_BdE_B=\vec{p_B}\cdot d\vec{p_B}{\;\;\;\;\;}(6)$$ If the particle B is initially at rest the RHS of (6)is zero. But the LHS cannot be zero unless $dE_B=0$.

How does one get round this problem?

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This looks like a homework problem so I applied that tag... –  FrankH Sep 22 '12 at 9:09
    
In the relation $EdE=\vec{p}\cdot d\vec{p}$ we may allow the magnitude of $\vec{p}$ to tend to zero. The RHS of the first relation tends to zero but the LHS will not go off to zero –  Anamitra Palit Sep 22 '12 at 12:08
    
@AnamitraPalit please stop bumping your question. –  David Z Sep 25 '12 at 11:02
    
@DavidZaslavsky: The last editing to the question was done 11 hours ago and the last editing to one of the answers was worked out 8 hours ago. And suddenly(1 hour ago) you have accused me of bumping the question. That's quite strange. –  Anamitra Palit Sep 25 '12 at 12:06
    
Not really, that's just when I got around to handling the flags on your question. In any case, you've already posted two answers, neither of which were judged useful by the community, so I'm just reminding you not to keep doing that. –  David Z Sep 25 '12 at 16:34
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3 Answers 3

You have assumed all the changes $dE$, $dp$ etc. to be infinitesimal – infinitely small – because you have used the differential calculus. So you "theorem" doesn't apply to arbitrary collisions; it only applies to elastic collisions in which just an infinitesimal amount of energy and momentum is transmitted.

Indeed, for such collisions, your identity has to hold. One may also write it as $$ \frac{|p_A|}{E_A} = \frac{|p_B|}{E_B}$$ which is probably more insightful because it simply says that the two particles must have the same velocity. When it's so, they have the same rest frame. In that rest frame, one may transmit the momentum between the particles and the kinetic energy of both remains equal to the rest mass, up to negligible second-order terms.

However, you won't be able to find any finite collision of this sort because it would already violate the conservation laws: the second-order deviations would become important and couldn't be canceled.

If you lift the assumption that the changes of the energy and momentum are infinitesimal, you will be able to see that there are also nontrivial collisions in which the momentum and energy of both particles change by a particular finite amount. Those collisions are particularly comprehensible in the center-of-mass system in which the initial particles have the opposite values of the momentum and in this frame, the only change is that each particle reverts its sign of the momentum.

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Let's consider a particle having energy E and momentum p.It receives $\triangle E$ amount of energy and the magnitude of its momentum changes by an amount $\triangle p$

Initially, $$E^2=p^2+m_0^2{\;\;\;\;\;\;}(1)$$ Finally, $$(E+\triangle E)^2=(p+\triangle p)^2+m_0^2{\;\;\;\;\;\;}(2)$$

Subtracting (2) from (1) $$2E\triangle E+(\triangle E)^2=2pdp+(\triangle p)^2{\;\;\;\;\;\;}(3)$$ We choose an inertial frame $K$ where the particle is initially at rest.We can have several such frames

Relation (3) reduces to: $$2E\triangle E+(\triangle E)^2=(\triangle p)^2$$ $$2E\triangle E=(\triangle p)^2-(\triangle E)^2{\;\;\;\;\;}(4)$$ $$2E\triangle E=(\triangle p+\triangle E)(\triangle p-\triangle E)$$ Or, $$E=0.5(\frac{\triangle p}{\triangle E}+1)(\triangle p-\triangle E){\;\;\;\;\;\;}(5)$$ If $\triangle p>>\triangle E$ we have, $$E=0.5\frac{\triangle p}{\triangle E}\times \triangle p{\;\;\;\;\;\;}(6)$$

Let's use relation (4) to understand the transfer[reception] of a finite amount of energy $\triangle E$ by the particle.For relation (4) to work the energy should be transferred in a single "burst" in the form of a packet.If the same energy was to flow in a continuous manner the particle will no more be at rest wrt the frame K after receiving a small amount of total anticipated energy transfer and we will not be able to apply our foregoing calculations. A continuous transfer may be broken up into a stream of packets

Rest frame considerations: [Multiple Transfers in the form of a Stream of Packets of Energy]

When the particle receives a small amount of energy it accelerates wrt K(inertial). So the rest frame is not an inertial frame if energy absorption is considered.For each transfer we consider an inertial frame $K_i$ which is in uniform motion wrt K and which coincides with the rest frame of the particle for the initial moment of each transfer. Then one has to subdivide $\triangle E$ and $\triangle p$ into appropriate packets(keeping in mind the frames $k_i$) to work out the same transfer in the form of a string of packets/quanta.You may apply relation (4) (5) or (6) remembering that E represents the amount of energy with the particle at the current state wrt to the frame $K_i$

Discrete transmission of Energy is favored

It is also to be noted that the transferred packets do not lie on the mass shell

For the emitting particle elation (4) from its own rest frame becomes: $$-2E\triangle E=(\triangle p)^2-(\triangle E)^2{\;\;\;\;\;}(7)$$ The emitted energy must exceed the emitted momentum for the emitter.But the rest frames of the two particles in general are different(rest frame of one particle at the time of emission and the rest frame of the other at the time of reception are not identical).That may resolve the issue.

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For any type of energy reception $\triangle E<<\triangle p$ . Discrete transfer favors such a condition.For an apparently continuous transmission each "packet " should comply with relation (4) (5) or (6) wrt to the frame $K_i$ concerned –  Anamitra Palit Sep 24 '12 at 8:59
    
$\Delta E$ is certainly not much smaller than $\Delta p$ in the ultra-relativistic limit (they are manifestly of the same magnitude). If your conclusion depends upon a fact which is not Lorentz invariant then it is highly suspect before you even get started. –  dmckee Sep 25 '12 at 4:28
    
@dmckee:The transmitted packet is "off the mass shell".The relation $E\approx pc{\;\;}Or,\triangle E\approx c\triangle p$---(A) is valid for high speed real particles especially those having speed close to the value of light(and less than it). Such particles lie on the mass shell. Relation (A) does not apply to packets of energy transfer(virtual particles).Virtual particles,incidentally, do not lie on the mass shell.The relation coming from my calculations relates to virtual particles and are quite justified. –  Anamitra Palit Sep 25 '12 at 6:43
    
Going off the mass shell does not help your case because the mass-squared of exchange particles can have either sign or be zero. –  dmckee Sep 25 '12 at 16:45
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Lets analyze the transfer of energy and momentum from a particle B to a particle A. Observation is carried out from some arbitrary inertial frame K. Initial States of the particles(Wrt K):

Particle $A:(E_A,\vec{p_A})$ $$E_A^2=p_{Ax}^2+p_{Ay }^2+p_{Az}^2+m_0^2{\;\;\;\;\;\;\;}(1)$$ Particle $B:(E_B,\vec{p_B})$ $$E_B ^2=p_{Bx}^2+p_{By }^2+p_{Bz}^2+m_0^2{\;\;\;\;\;\;\;}(2)$$

Final states after the interaction:

Particle $A:(E_A+E,\vec{p}_A+\vec{p})$

$$(E_A+E)^2=(p_{Ax}+p_x)^2+(p_{Ay }+p_y)^2+(p_{Az}+p_z)^2+m_0^2{\;\;\;\;\;\;\;}(3)$$

Particle $B:(E_B+E,\vec{p}_B+\vec{p})$

$$(E_B-E)^2=(p_{Bx}-p_x)^2+(p_{By }-p_y)^2+(p_{Bz}-p_z)^2+m_0^2{\;\;\;\;\;\;\;}(4)$$

The amount of energy transferred from B to A : $E$

The amount of momentum transferred from B to A:$\vec{p}:(p_x,p_y,p_z)$

By subtracting (3) from (4) and applying (1) and (2) on the result thus obtained , we have,

$$E(E_A+E_B)=\vec{p}.(\vec{p}_A+\vec{p}_B)$$

Or, $$EE_{total}=\vec{p}\cdot\vec{p}_{total}{\;\;\;\;\;\;}(5)$$ $$EE_{total}=\mid\vec{p}\mid\vec{p}_{total}\mid\cos\theta{\;\;\;\;\;\;}(6)$$

Or, $$\cos\theta=\frac{E}{\mid\vec{p}\mid}\frac{E_{total}}{\mid\vec{p}_{total}\mid}{\;\;\;\;\;\;}(7)$$

The above relation signifies that the net transfer of momentum(its magnitude) should be much greater than the amount of energy transferred. Suppose we write the transferred momentum in the form : $$\vec{p}=\vec{p_1}+\vec{p_2}+......\vec{p_n}$$

We have by applying relation (5) to each transfer,

$$\Sigma E_i E_{total}=\Sigma (\vec{p_i}\cdot\vec{p}_{total})$$

Or,

$$\Sigma E_i E_{total}=\Sigma (\mid\vec{p_i}\mid \mid\vec{p}_{total}\mid\cos\theta_i){\;\;\;\;\;\;}(8)$$

From relations(6) and (8) we have, $$\mid\vec{p}\mid\vec{p}_{total}\mid\cos\theta=\Sigma (\mid\vec{p_i}\mid \mid\vec{p}_{total}\mid\cos\theta_i){\;\;\;\;\;\;}(9)$$ If the different $\vec{p_i}$ s have different directions we have a picture similar to some sort of a multiple scattering event with one particle in the field of the other.

If the frame $K$ is chosen such that the emitting particle ie, B is initially at rest in it, we have for the initial transfer: $$2E_B E'=E'^2-\mid\vec{p'}\mid^2{\;\;\;\;\;\;}(10)$$

$E'$ and $\vec{p'}$ are the initial transfers of energy and momentum

Incidentally $$E'^2>\mid\vec{p'}\mid^2$$

[ It would be better to choose the inertial frame K in such a manner that the ratio of energy transmitted to magnitude of momentum transmitted in the initial stage (in relation to particle B of course)is greater than one.This may be done since in the rest frame of B the same ratio is greater than one]

to keep the RHS of (10) positive

A comparison between relations (7) and (10) suggests that the ratio of energy to momentum for transferred packets should change from greater than one to less than one even if the $\vec{p_i}$ s ($\Sigma \vec{p_i}=\vec{p}$) are collinear for the total transfer.The change of the said ratio implies an increase in momentum for the packet in a manner consistent with the conservation principles

The other option would be that some energy from the "packet" flows out in the form of radiation or perhaps it gets stored in some hidden dimension.But finally it should reach the target particle(A in this case) in a manner consistent with conservation of momentum principle.

There is also an indication towards discretization.

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Again, you have made assertions about relative magnitudes that manifestly fail in the ultra-relativistic case. Which is a shame because you are getting close to several basic relationships in high energy scattering. If you want to do relativistic mechanics you have to keep a close eye on the possibility of $m \ne m'$. –  dmckee Sep 25 '12 at 4:46
    
@dmckee: In case you have implied the interaction between a pair of particles having different rest masses there is no problem. The mass terms on the RHS of (1) and (2) are different. The same different mass terms occur on the RHS of (3) and (4). when you subtract (3) from (4) and use the relations (1) and (2) the individual mass terms do cancel and relations (5) (6) etc remain unchanged.Relation (1) considers the release of energy (and momentum) by the particle B. The mass term is not there(it cancels out) –  Anamitra Palit Sep 25 '12 at 7:55
    
Whatever you do in particle physics is in consistency with the notions of Relativity.So relativity by itself should be in a position strong enough to predict or rather encompass in its broadest perspectives the results of scattering phenomena in particle physics.These results are consistent with the formula $E^2=p^2+m_0^2$ and of course with the laws of energy and momentum conservation. Nothing should go against the said formula and Laws having the veto power –  Anamitra Palit Sep 25 '12 at 8:07
    
A slightly Generalized Case in absence of annihilation of mass:Initial States:Particle $A:(E_A,\vec{p}_A,m_A);Particle B:(E_B,\vec{p}_B,m_B)$. Final States:$A:(E_A+E,\vec{p}_A+\vec{p},m_A+m);Particle B:(E_B-E,\vec{p}_B-\vec{p},m_B-m)$.Total rest mass is assumed to be constant;"m" may be positive or negative.Instead of relation (5) in answer we have:$EE_{tot}=\vec{p}.\vec{p}_{tot}+mm_{tot}$;where $m_{tot}=m_A+m_B$.The transferred "packet" is not necessarily on the mass shell. $Relation{\;} (10){\;\;}incidentally{\;\;} remains{\;\;\;\;} unchanged{\;\;\;}!$ –  Anamitra Palit Sep 25 '12 at 9:15
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