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sharad
Oct 7, 2010, 12:15:37 PM10/7/10
to RSI-C 2010
1. How can we prove the homogeneity and Isotropy of space?
2. Why do we consider only forward envelope in Hygen’s Wave theory-?
3. Why are there always 6 deg of freedoms for rigid bodies?
4. What is the expression for momentum of a photon?
2. Why do we consider only forward envelope in Hygen’s Wave theory-?
3. Why are there always 6 deg of freedoms for rigid bodies?
4. What is the expression for momentum of a photon?
sushruth
Oct 8, 2010, 8:50:27 AM10/8/10
to RSI-C 2010
hey man....
1. i dont know much....
but i guess it is something that is observed...and i am not sure if it
is properly accounted for yet...
homogeneous means you can move from one random point in the universe
to another and everything looks exactly the same. Look at the sky at
night with all the stars and galaxies that dot the sky. If you moved
to any other random place in the universe, the odds are it would look
very similar.
isotropy means you could turn any direction and the universe looks
exactly the same. Look at the sky in the Northern, then Southern
Hemisphere and statistically they look nearly identical.
that is all i know of isotropy and homogeneity...i guess to understand
u require much deeper knowledge in physics with a better mathematical
background of relativity etc etc....
2.This is bcoz all other points of the secondary wavelets cancel out
and u see only the forward envelope in the wavelets bcoz of
interference.....also it is mathematically derives that the amplitude
is proportional to (1+cos(theta)) {i do not know the mathematical
derivation for this but it is so...}...so if u say a backward envelope
theta=pi...so amplitude becomes 0...
3.degrees of freedom are the total number of independent coordinates
required to specify its position in space...
in a rigid body u just need 6 degrees of freedom to specify its
position(and so it also means the no. of coordinates it can vary
i.e,move)....
so if there are 'n' particles so it will be obvious that u require 3
coordinates to specify the position of each particle and so the DOF
will be 3n....but now if u apply k constraints then the DOF becomes 3n-
k...
in rigid bodies with n particles u wuld expect for nC2 constraints
i.e,put a constraint between each and every particle.
but some constraints become redundant bcoz even without those
constraints the body wuld still be rigid ....
for example in a 5 particle system u expect 10(5C2) constraints but
one constraint becomes redundant and so u can
get a rigid 5 particle body with just 9 constraints....so the
DOF=6...check it out for greater no. of particles for urself if u have
the time..:)...
what are these 6 coordinates that u require....?...
3 coordinates (x,y,z) to specify the position of any one
particle(translational)
once u know that u will know the distances for the rest of the
particles form that particles as the distance is always that constant
constraint which u applied..
u still need to specify its angular orientation of the rigid body...so
u require three angles (wt xy ,yz and zx planes)to specify the
orientation of the plane of the body.... so now u have 6....
this is not inclusive of vibrational dof's.....
4.p= hv/c = h/lambda...
i dont know the proper derivation or explanation for this..but i am
guessing ...(so if i am wrong i am requesting anyone who knows better
to correct me...)
if u r thinking m=0 and so momentum is zero i guess u r wrong ...
roughly speaking, i guess the inertial characteristic of the photon is
accounted through its energy even though u say its mass is 0...so u
could roughly say that...m=E/(c^2) {from E=m(c^2)}..i guess it is this
E/(c^2) which gives inertial characteristics or "mass".....we know
E=hv....so i guess u get momentum will be p = [E/(c^2)]*c = E/c = hv/c
= h/lambda......i guess it is that E/(c^2) which causes the photon to
be influenced by gravity inspite of it having no mass....
1. i dont know much....
but i guess it is something that is observed...and i am not sure if it
is properly accounted for yet...
homogeneous means you can move from one random point in the universe
to another and everything looks exactly the same. Look at the sky at
night with all the stars and galaxies that dot the sky. If you moved
to any other random place in the universe, the odds are it would look
very similar.
isotropy means you could turn any direction and the universe looks
exactly the same. Look at the sky in the Northern, then Southern
Hemisphere and statistically they look nearly identical.
that is all i know of isotropy and homogeneity...i guess to understand
u require much deeper knowledge in physics with a better mathematical
background of relativity etc etc....
2.This is bcoz all other points of the secondary wavelets cancel out
and u see only the forward envelope in the wavelets bcoz of
interference.....also it is mathematically derives that the amplitude
is proportional to (1+cos(theta)) {i do not know the mathematical
derivation for this but it is so...}...so if u say a backward envelope
theta=pi...so amplitude becomes 0...
3.degrees of freedom are the total number of independent coordinates
required to specify its position in space...
in a rigid body u just need 6 degrees of freedom to specify its
position(and so it also means the no. of coordinates it can vary
i.e,move)....
so if there are 'n' particles so it will be obvious that u require 3
coordinates to specify the position of each particle and so the DOF
will be 3n....but now if u apply k constraints then the DOF becomes 3n-
k...
in rigid bodies with n particles u wuld expect for nC2 constraints
i.e,put a constraint between each and every particle.
but some constraints become redundant bcoz even without those
constraints the body wuld still be rigid ....
for example in a 5 particle system u expect 10(5C2) constraints but
one constraint becomes redundant and so u can
get a rigid 5 particle body with just 9 constraints....so the
DOF=6...check it out for greater no. of particles for urself if u have
the time..:)...
what are these 6 coordinates that u require....?...
3 coordinates (x,y,z) to specify the position of any one
particle(translational)
once u know that u will know the distances for the rest of the
particles form that particles as the distance is always that constant
constraint which u applied..
u still need to specify its angular orientation of the rigid body...so
u require three angles (wt xy ,yz and zx planes)to specify the
orientation of the plane of the body.... so now u have 6....
this is not inclusive of vibrational dof's.....
4.p= hv/c = h/lambda...
i dont know the proper derivation or explanation for this..but i am
guessing ...(so if i am wrong i am requesting anyone who knows better
to correct me...)
if u r thinking m=0 and so momentum is zero i guess u r wrong ...
roughly speaking, i guess the inertial characteristic of the photon is
accounted through its energy even though u say its mass is 0...so u
could roughly say that...m=E/(c^2) {from E=m(c^2)}..i guess it is this
E/(c^2) which gives inertial characteristics or "mass".....we know
E=hv....so i guess u get momentum will be p = [E/(c^2)]*c = E/c = hv/c
= h/lambda......i guess it is that E/(c^2) which causes the photon to
be influenced by gravity inspite of it having no mass....
sharad
Oct 8, 2010, 12:44:55 PM10/8/10
to RSI-C 2010
1 How did we even comment about the homogenity and isotropy of the
universe by simple observing it around our close vicinity?
3. Consider am octagon with each vertex with an atom. Let it be a
rigid body.
degrees of freedom = 3*8 - 8 = 16
There are no redundant constraints.
How are there still only 6 degrees of freedom?
universe by simple observing it around our close vicinity?
3. Consider am octagon with each vertex with an atom. Let it be a
rigid body.
degrees of freedom = 3*8 - 8 = 16
There are no redundant constraints.
How are there still only 6 degrees of freedom?
vba...@physics.iitm.ac.in
Oct 9, 2010, 12:03:15 AM10/9/10
to rsi-c...@googlegroups.com
> 1 How did we even comment about the homogenity and isotropy of the
> universe by simple observing it around our close vicinity?
>
> universe by simple observing it around our close vicinity?
>
The observations are not restricted to our 'close vicinity'.
> 3. Consider am octagon with each vertex with an atom. Let it be a
> rigid body.
> degrees of freedom = 3*8 - 8 = 16
> There are no redundant constraints.
> How are there still only 6 degrees of freedom?
An octagon in which the bonds are just the outer edges of the polygon is
not rigid. You have to include a sufficient number of diagonals (bonds
joining atoms that are not nearest neighbours) to make the system rigid.
The point is easier to envisage in the case of a square:
A square is not rigid. (You can bend the square about a diagonal without
changing the side lengths.) If you add a diagonal to the square, you can
still bend it about that diagonal. But if you add both diagonals, the
object becomes rigid. You now have (3*4) - 6 = 6 independent degrees of
freedom. Similar arguments hold good for larger collections of particles.
B.
Message has been deleted
sharad
Oct 9, 2010, 12:15:07 PM10/9/10
to RSI-C 2010
1. How was homogeneity observed/verified?
2. A photon has a constant velocity in a medium. Still we don't
assign
a constant momentum (i.e., one independent of frequency). Is this
because it is travelling at c or is it because we cannot ignore wave-
like properties at the quantum level?
3. Given a single pulse of light from a point source, how does the
backward envelope of the secondary wavelets cancel?
Q. How do particles at 0K vibrate if they have no thermal energy? Is
there another form of energy they use?
2. A photon has a constant velocity in a medium. Still we don't
assign
a constant momentum (i.e., one independent of frequency). Is this
because it is travelling at c or is it because we cannot ignore wave-
like properties at the quantum level?
3. Given a single pulse of light from a point source, how does the
backward envelope of the secondary wavelets cancel?
Q. How do particles at 0K vibrate if they have no thermal energy? Is
there another form of energy they use?
Venkataraman Balakrishnan
Oct 10, 2010, 2:41:40 AM10/10/10
to rsi-c...@googlegroups.com
On Sat, Oct 9, 2010 at 9:40 PM, sharad <olorin_...@yahoo.com> wrote:
1. A photon has a constant velocity in a medium. Still we don't assign
a constant momentum (i.e., one independent of frequency). Is this
because it is travelling at c or is it because we cannot ignore wave-
like properties at the quantum level?
A photon in free space (vacuum) has an energy E and momentum
vec(p), in the particle picture. In the wave picture, the attributes are an angular frequency omega and a wave vector vec(k). The basic relation between the frequency and the wave vector is called the dispersion relation. In free space this is simply
omega = c k,
where k is the magnitude of vec(k) and c is the fundamental velocity (a universal constant).
The Einstein-de Broglie relations connecting these wave and particle pictures are
E = h nu and vec(p) = h vec(k)/(2pi)
where h is Planck's constant.
In a material medium, while light has a specific speed, this is an effective speed. Individual photons themselves actually interact with the atoms of the medium, and are constantly absorbed and re-emitted by the atoms, and are also scattered by these atoms.
As a result, the effective relation between omega and k is generally nonlinear. Calculating the function explicitly from first principles is a nontrivial task, and requires a knowledge of the multiple interactions between atoms and photons that goes on in the medium.
vec(p), in the particle picture. In the wave picture, the attributes are an angular frequency omega and a wave vector vec(k). The basic relation between the frequency and the wave vector is called the dispersion relation. In free space this is simply
omega = c k,
where k is the magnitude of vec(k) and c is the fundamental velocity (a universal constant).
The Einstein-de Broglie relations connecting these wave and particle pictures are
E = h nu and vec(p) = h vec(k)/(2pi)
where h is Planck's constant.
In a material medium, while light has a specific speed, this is an effective speed. Individual photons themselves actually interact with the atoms of the medium, and are constantly absorbed and re-emitted by the atoms, and are also scattered by these atoms.
As a result, the effective relation between omega and k is generally nonlinear. Calculating the function explicitly from first principles is a nontrivial task, and requires a knowledge of the multiple interactions between atoms and photons that goes on in the medium.
2. Given a single pulse of light from a point source, how does the
backward envelope of the secondary wavelets cancel?
This (1 + cos theta) factor is a sort of fudge factor that is put in to kill the backward envelope in the elementary Huygen's theory of diffraction. In the more detailed Kirchhoff theory of dffraction, it emerges automatically.
Q. How do particles at 0K vibrate if they have no thermal energy? Is
there another form of energy they can use?
The "vibration" at 0 K is not a mechanical vibration in the classical sense. It must be regarded as a quantum fluctuation without a classical analogue. The whole point is that the classical idea of a particle with a definite trajectory in space is wrong in quantum mechanics, as that would violate the basic uncertainty principle.
B.
Message has been deleted
Venkataraman Balakrishnan
Oct 19, 2010, 9:24:01 AM10/19/10
to rsi-c...@googlegroups.com
On Tue, Oct 19, 2010 at 5:11 PM, sharad <olorin_...@yahoo.com> wrote:
4. Heisenberg's uncertainty principle is supposed to be applicable to
1. How do we know that gravity acts for objects in the middle
dimensions(i.e., where neither object is massive like a planet)? i.e.,
how did newton claim that every object attracts every other object?
The first experiment in this regard was carried out by Cavendish, after whom the famous Cavendish Lab at Cambridge is named. The "modern" experiment to verify the inverse square law of gravitation was done by Eotvos, a Hungarian physicist, around the period 1885-1905, using a torsion balance and two heavy masses. The experiment has been repeated several times with improved apparatus, and the inverse square law has been verified correct to 1 part in 10^10 or so.
Interestingly, what has NOT been done (and what is extremely difficult to do)
is to verify the law of gravity at length scales of a millimetre or less. The issue becomes important in recent theories involving extra spatial dimensions.
Interestingly, what has NOT been done (and what is extremely difficult to do)
is to verify the law of gravity at length scales of a millimetre or less. The issue becomes important in recent theories involving extra spatial dimensions.
2. If 2 bodies are light years away and the position of one of them is
changed. The force of gravitation should change much faster than light
can travel. Does this mean gravitons travel at near infinite speeds?
Wouldn't this imply infinite energy?
No, the inverse square law of gravitational attraction is what is known as a static appoximation or limit, just like Coulomb's Law between charges. Both laws are special cases of interactions mediated by fields (the gravitational and the electromagnetic fields, respectively). There is really no action at a distance. Influences propagate at the speed of light in vacuum, in both cases. A proper treatment shows how there is a retardation effect that takes into account the finite propagation speed of both electromagnetic and gravitational forces.
3. I read that virtual photons are responsible for electromagnetic
forces. What are they and how are they different from 'normal'
photons?
The answer to this question requires some quantum field theory, and it would take me too long to give even a qualitative account here.
4. Heisenberg's uncertainty principle is supposed to be applicable to
cannonical conjugate pairs. What are they? Are there other examples
other than momentum-position and energy-time?
Again, the answer involves some knowledge of classical mechanics at the level of the Hamiltonian formalism.
Other examples are angular momentum and angular coordinates, etc. By the way, energy and time constitute a slightly different knd of pair than all the others, because time is not, by itself, a dynamical variable in the usual sense. Rather, it parametrizes the variation of all the other dynamical variables.
B.
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