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location Kolkata,India[91-33-25514464]
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visits member for 2 years, 11 months
seen Mar 27 '13 at 12:06

Author/Teacher from India. Interested in General Relativity and other areas of physics


Dec
4
comment Can Parallel Transport always move a Vector Parallel to Itself?
You are not allowed to rotate the tangent planes wrt each other-----they are fixed on the curved surface at the two points concerned.
Dec
4
comment On Parallel Transport
Does "parallel transport" move a vector parallel to itself on a curved surface even in the infinitesimal sense? You may think of two adjacent tangent planes on a curved surface.Is it always possible to have parallel vectors at the points of contact(one vector being preassigned) even if the planes are awkwardly inclined?
Dec
4
comment On Parallel Transport
At the point N'(referring to the original posting)you may consider a second vector tangent to the latitude-line at N'. If it(2nd vector) is parallel transported along the latitude to the point N" on the meridian NB, it is no more a tangent wrt to the latitude at N". The angle this vector makes with the tangent at N" should be equal to the angle which the first vector(moved up from the equator) makes with the meridian at N". This angle is expected to be small
Dec
3
comment On Parallel Transport
Points to Observe:(1)The vector after going through a loop[by parallel transport] rotates by a negligible small angle though the enclosed area is large. You will find this in the example I have referred to in the comment after the question after Lubos Moti's comment.(2)An infinitesimally small area is not sufficient to warrant a flat space-time.The Christoffel symbols are point functions. They have non-zero value for curved spacetime.
Dec
3
comment On Parallel Transport
I am referring to the change between the initial and the final positions of the vector when it goes round a loop on a curved surface(by parallel transport). You may connect the point N'(in the original posting) with some point on the meridian NB by a $small{\;\;}$ curve so that the transported vector on landing on the meridian becomes tangential parallel or nearly tangential to the meridian NB.
Dec
3
comment On Parallel Transport
(in continuation) The above idea is embodied in the non-zero value of the Christoffel tensors in curved space.
Dec
3
comment On Parallel Transport
Let's's consider the vector components $A^\gamma(x^\alpha)$ and $ A^\gamma(x^\alpha+d x^\alpha)$. They have an infinitesimally small separation.This is not indicative of flat space-time over the infinitesimally small spacetime region concerned .Reason:For the purpose of calculating the derivative we have to parallel transport the vector-component $A(x^\alpha +d x^\alpha)$ to the location $x^\alpha$ and this vector definitely changes its orientation wrt to its initial position even though it has moved through an infinitesimally small distance
Dec
2
comment On Parallel Transport
You may consider a small curved line from N' to the meridian NB so that the vector becomes parallel to the meridian NB on reaching it. At A there is no turning all due to the exclusion of a small area.
Dec
2
comment On Parallel Transport
Quoting Ron Maimon:"It ends up in nearly exactly the same direction as it was when you start the parallel transport"---in such a situation if you move back to A the vector does not turn by alpha. It turns by a much smaller amount! On removing the triangle the vector in the initial and the final situation (at A) make a very small angle. If the triangle is there it turns by 90 degrees after looping round
Dec
2
comment On Parallel Transport
Only if you consider a geodesic the an infinitesimally think space round it and parallel to it is nearly flat--the tangent vector propagates parallel to itself
Sep
25
comment 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.
Sep
25
comment 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. $Relation{\;} (10){\;\;}incidentally{\;\;} remains{\;\;\;\;} unchanged{\;\;\;}!$
Sep
25
comment 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
Sep
25
comment 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)
Sep
25
comment 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.
Sep
24
comment 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
Sep
22
comment 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
Sep
21
comment 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
Sep
21
comment 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
Sep
7
comment Counterpart of the Klein Gordon Equation on the “Coordinate Shell”
$\frac{\partial E}{\partial t}=\frac{\partial E}{\partial \psi}\frac{\partial \psi}{\partial t}$Or,$\frac{\partial^2 E}{\partial t^2}=\frac{\partial^2 E}{\partial \psi^2}(\frac{\partial \psi}{\partial t})^2+\frac{\partial E}{\partial \psi}\frac{\partial^2 \psi}{\partial t^2}$Similarly:$\frac{\partial^2 E}{\partial x^2}=\frac{\partial^2 E}{\partial \psi^2}(\frac{\partial \psi}{\partial x})^2+\frac{\partial E}{\partial \psi}\frac{\partial^2 \psi}{\partial x^2}$ We finally have$\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}=0$ for points where mass=0.