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what is a virtual work and how could it be possible.

thaks for ur valuable time.

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- Thread starter jaan
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what is a virtual work and how could it be possible.

thaks for ur valuable time.

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HallsofIvy

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Okay, I'll give one example: According to Newton's law of gravity, together with his other laws (which I am more inclined to attribute to Gallileo than Newton), the orbit of any planet around the sun should be an ellipse (ignoring effects from other planets). The planet mercury is close enough to the sun that the effect of other planets is neligible but its orbit is

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BJ

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Also an example from Special relativity is, the equation F=ma is not correct when the velocity is near the speed of light.

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jtbell

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yu_wing_sin said:Also an example from Special relativity is, the equation F=ma is not correct when the velocity is near the speed of light.

However, the statement [itex]\vec F = d \vec p / dt[/itex], which is closer to Newton's own statement of the Second Law,

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jtbell said:However, the statement [itex]\vec F = d \vec p / dt[/itex], which is closer to Newton's own statement of the Second Law,isstill correct in special relativity.

A good overview of Newtonian Mechanics should show that at molecular and atomic levels, Newtonian mechanics breaks down, for instance various axioms of Newtonian gravitation, like the Sphere Theorem, and also the Centre of Gravity 'method'. Newton's approaches are quite accurate for distances and scales around the size of the Solar System.

However, the Sphere Theorem fails at atomic levels, because it is an approximation based upon the continuous distribution of 'matter'. At these sizes, matter turns out to be discretely distributed, and so 'smooth' approximations like the Sphere Theorem fail.

The Centre of Mass 'Method' fails for slightly different reasons: Here the idea has been known since its proposal by Newton to be an approximation, but quite accurate enough when the distances sufficiently dwarf the radius of the objects under consideration (e.g., planets and stars). At close distances where objects interact with each other at distances approaching their effective radii, the CM method fails and forces become underdetermined.

Finally, (according to SRT and current field theories) it is now believed that all exchanges of energy and actions of forces must involve particle exchanges and/or fields, which can store and transmit energy, however at the cost of limits to the speed of propagation of forces and effects.

The original Newtonian formulation of gravity in contrast poses 'instantaneous action at a distance', which while it may have been a necessary fiction for Newton, to organize what was known of physics in his time, is considered inaccurate today.

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:rofl: my question was 'how can a particles momentum change without the application of force'.i know that generalised coordinates were the advantage of legrangian mechanics which made mechanics easier.

just quote me an example where we cannot apply newtons laws and go for legrangian mechanics ( i know that in theory of relativity it fails )

thanks a lot

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HallsofIvy

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I confess that I was sorely tempted to say "In south Philadelphia" but I fought the temptation!

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This is one of the two claims in Grifith's "Introduction to electrodynamics" that I wish he would elaborate on further. The other claim is that "the magnetic field does no work".

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jaan said::rofl: my question was 'how can a particles momentum change without the application of force'.i know that generalised coordinates were the advantage of legrangian mechanics which made mechanics easier.

just quote me an example where we cannot apply newtons laws and go for legrangian mechanics ( i know that in theory of relativity it fails )

thanks a lot

Lagrangian mechanics says that the rate of change of a particle's generalized momentum is proportional to the generalized force. So it won't do what you are asking. It's rather unclear why you are asking, anyway.

I.e. if we let q be a coordinate and [itex]\dot{q}[/itex] be the velocity, the rate of change of a coordinate with time we can write

Genearlized momentum

[tex]

p = \frac{\partial L}{\partial \dot{q}}

[/tex]

Generalized force

[tex]

F = \frac{\partial L}{\partial q}

[/itex]

Lagrange's equation

[tex]

\frac{dp}{dt} = F

[/tex]

This is just using the definitions of generalized momentum and generalized forces in the standard equations, i.e.

http://scienceworld.wolfram.com/physics/LagrangesEquations.html

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jtbell

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Crosson said:This is one of the two claims in Grifith's "Introduction to electrodynamics" that I wish he would elaborate on further.arun_mid said:

Consider a positive charge A which (at a particular instant of time) is traveling along the x-axis, in the +x direction, and is located at x = 0; and a positive charge B which (at that instant) is traveling along the y-axis, in the +y direction, and is located at some y > 0.

The magnetic field produced by A runs in circles around the x-axis. If A is positively charged, the field that it produces at the location of B is in the +z direction. This field exerts a magnetic force in the +x direction on B.

The magnetic field produced by B runs in circles around the y-axis, and is zero at all points on the y-axis. Therefore there is no magnetic force on A.

The other claim is that "the magnetic field does no work".

The magnetic force is given by

[tex]{\vec F}_{mag} = q \vec v \times \vec B[/tex]

The cross product of two vectors is always perpendicular to both of those vectors. Therefore [itex]{\vec F}_{mag}[/itex] is always perpendicular to [itex]\vec v[/itex], and so it can do no work.

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The magnetic field produced by B runs in circles around the y-axis, and is zero at all points on the y-axis. Therefore there is no magnetic force on A.

I followed your example, and it was quite convincing. But it all depended on that precise moment of time (at any other time the force on A would be nonzero). Magnetic forces do not occur instantaneously, they transmit at the speed of light.

If you consider that the force on A in your example is due to B at B's former position (r1-r2)/c seconds ago, then the example is not so clear cut. Even if the velocities of the particles were much less then the speed of light, the force is still nonzero.

Consider a magnetic dipole positioned in a magnetic field. It has potential energy:

[tex]U = -\vec{\mu} \cdot \vec{B}[/tex]

Since work is the change in potential energy, I think it is pretty clear that the magnetic field is able to do non-zero work on magnetic dipoles.

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jtbell

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Crosson said:If you consider that the force on A in your example is due to B at B's former position (r1-r2)/c seconds ago, then the example is not so clear cut. Even if the velocities of the particles were much less then the speed of light, the force is still nonzero.

I can't check my copy of Griffiths right now because I'm at home, but I'm pretty sure it presents the solution for the fields produced by a point charge moving at constant velocity, in the chapter that deals with retarded potentials etc. If that's the case, it should be easy to check whether this "propagation effect" can account for the Third Law violation.

I'm skeptical that this is the case, because the generally accepted explanation is that the electromagnetic field itself carries momentum and energy. The sum of the momenta of the particles and the integrated momentum density of the fields is constant, which is the important thing.

Remember that in classical mechanics, we customarily derive conservation of momentum from the Third Law; but we can consider conservation of momentum to be actually more fundamental, precisely because of situations like this one.

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I laughed out loud when I read this. That was really quite clever. Designed to confuse students methinks.The magnetic field produced by B runs in circles around the y-axis, and is zero at all points on the y-axis. Therefore there is no magnetic force on A.

Of course every algebraic equation known to physics has asymptotes or discontinuities, due to the fundamental nature of ordinary algebra. For instance, anytime there is a multiplication, then rearrangement can give a division, which means the equation is undefined and meaningless when the denominator is zero. For instance, Ohm's Law:

[tex]I = \frac{V}{R}[/tex] is meaningless when R is set to zero.

Similarly, you have here poised us to consider the instant in time when the x-particle crosses the Y-axis, precisely when the equations are invalid. Well done.

The magnetic field example was equally cute.

Again, hats off.

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Crosson said:I followed your example, and it was quite convincing. But it all depended on that precise moment of time (at any other time the force on A would be nonzero). Magnetic forces do not occur instantaneously, they transmit at the speed of light.

If you consider that the force on A in your example is due to B at B's former position (r1-r2)/c seconds ago, then the example is not so clear cut. Even if the velocities of the particles were much less then the speed of light, the force is still nonzero.

Consider a magnetic dipole positioned in a magnetic field. It has potential energy:

[tex]U = -\vec{\mu} \cdot \vec{B}[/tex]

Since work is the change in potential energy, I think it is pretty clear that the magnetic field is able to do non-zero work on magnetic dipoles.

From what I know, magnetic fields don't really have a potential energy associated with them.I didn't do much on non-conservative fields...so correct me if I'm wrong here... because a magnetic field is ONLY produced by dipoles, all work done by one pole automatically cancels out work done by the other pole--at any instant! Hence the work done is zero. Work is done in a magnetic field, but not by the magnetic force...however I'm not clear on what force does the work then... can anyone clear this up?

And although magnetic fields don't travel instantaneously, as long as you can prove that the forces don't form an action reaction pair (ie that they are equal in magnitude and opposite in direction) that's enough to show that the third law fails there. maybe this is the case...perhaps the force on A is non-zero but still unequal to that on B. I haven't touched relativity yet, so I won't venture into the math of it.

btw, I don't understand Nam_Sapper's logic:

Why are the equations invalid? isn't the magnetic force proportional to the cross product of velocity and relative position? If the vectors are collinear the cross product becomes zero.Similarly, you have here poised us to consider the instant in time when the x-particle crosses the Y-axis, precisely when the equations are invalid. Well done.

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Goldstein makes a very similar point.

Goldstein constructs examples where angular momentum is not conserved when one ignores the field contributions, for instance.

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Gokul43201

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Quit the personal remarks. You know you're already treading on thin ice here !Nam_Sapper said:... Well done.

The magnetic field example was equally cute.

Again, hats off.

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