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| bio | website | |
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| location | Kolkata,India[91-33-25514464] | |
| age | ||
| visits | member for | 1 year, 6 months |
| seen | Mar 27 at 12:06 | |
| stats | profile views | 225 |
Author/Teacher from India. Interested in General Relativity and other areas of physics
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Sep 25 |
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Interaction between a Pair of Particles @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. |
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Sep 25 |
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Interaction between a Pair of Particles 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. $Relation{\;} (10){\;\;}incidentally{\;\;} remains{\;\;\;\;} unchanged{\;\;\;}!$ |
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Sep 25 |
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Interaction between a Pair of Particles 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 |
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Sep 25 |
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Interaction between a Pair of Particles @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) |
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Sep 25 |
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Interaction between a Pair of Particles @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. |
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Sep 25 |
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Sep 25 |
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Sep 25 |
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Sep 25 |
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Interaction between a Pair of Particles edited tags |
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Sep 25 |
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Interaction between a Pair of Particles added 1 characters in body |
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Sep 24 |
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Interaction between a Pair of Particles added 306 characters in body |
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Sep 24 |
answered | Interaction between a Pair of Particles |
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Sep 24 |
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Interaction between a Pair of Particles 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 |
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Sep 24 |
answered | Interaction between a Pair of Particles |
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Sep 22 |
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Interaction between a Pair of Particles 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 |
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Sep 22 |
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Sep 22 |
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Sep 22 |
asked | Interaction between a Pair of Particles |
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Sep 21 |
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Newton's Law of Gravitation, Gauss Law and GR (in continuation)The speed of light is independent of its source. The idea of speed in the said formulation is different from the spatial part of four velocity ie from proper speed (or celerity).Proper speed can exceed the speed of light without hurting or violating relativity. But the three velocity concept plays an important role in the construction of relativity itself though it is not of a covariant form |
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Sep 21 |
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Newton's Law of Gravitation, Gauss Law and GR Well,the four acceleration of a test particle following a geodesic is zero[all components are zero].This is in conformity with the fact that gravity is not a force.But this four acceleration is different from the acceleration we perceive in the physical world,for example the acceleration of an apple falling from a tree.The non covariant form of acceleration is important for understanding physics.But the covariant form has a different type of elegance in so far as the transformation rules are concerned.An interesting analogy would be the concept of the classical three velocity |
