Hello Everyone...

If equal and opposite forces 10N applied on the object it will go under compression (stress).

If we apply 10N from left side and 15N from right side, object will accelerate to the right. The 10N which are equal and opposite can create stress when object in motion?

yes. In fact, it will be a bit bigger because you will have the excess 5N spread out as a force distribution in the object

@naturallyInconsistent Is it possible to combine individual photon EM fields(quantized) to get the classical EM field which has like billions of photons?

I was wondering if there was a scheme to bring out the classical EM field by some sort of summing up of the individual photon EM fields that make up the classical field

I think that is necessarily possible, but it is probably conceptually fraught.

Surely some statistical thermodynamics trickery ought to directly get it, especially if you include the use of coherent states. After all, coherent states of photons already are the closest link between quantum wavefunctions and classical field configurations

Damn

I wish to understand the process some day : )

What do you mean? I think it is pretty understandable as it is. Not tooooo bad, especially compared to the other stuff in QM.

Anyway, miao miao just sent CIAAW and CODATA group an email suggesting that they update the mass of Hydrogen atom, protium, to a more precise value, about 1000x better, because the theoretical value has a much smaller uncertainty. Not sure if they will accept theoretical-only suggestions, though.

sure

👍

@naturallyInconsistent I meant I don't have the prereqs to understand it as of now : )

@Arjun Do you know what a Fourier transform is?

hi

@LuckyChouhan That's because I haven't written it yet

@JohnRennie Yes,I do!

@Arjun If you take any EM field you can Fourier transform it and you get the field as a sum of plane waves.

Then each plane wave can be written as a sum of identical photons.

But that doesn't mean an EM field really is a sum of photons. All we've done is to find a way of writing the field as if it was a sum of photons.

Hello @JohnRennie . How are you. You became silent user since 1 year in this group. We need your participation and knowledge sharing.

@JohnRennie I'm not sure what "sum of identical photons" means here,is there an electromagnetic field associated with each individual photon?Is this field classical or it comes from qed?

@Arjun I don't really know to be honest.

In QED the "photons" are modes of the photon field, where each mode represents a momentum eigenstate i.e. an infinite plane wave.

But I don't think a classical EM infinite plane wave is just a sum of modes.

@JohnRennie can we write work done general equation as $W(x(t),v(t), t ) = F(x(t), v(t), t) dx(t)$ ?

Because Knk above drived work done formula when force is the function of position, not for others. Pls clear that confusion. Force in work done formula can only be function of position or every possible things?

@123 No,the above result will hold even if F is a function of v(t),See that changing F(x) by F(x,v) in the first step is not going to affect the steps that follow it

@123 Force in the work energy theorem can be an explicit function of x,v,t and not just x

@Arjun Hello . Why from N2L they derived work energy theorem?

Are there other ways to drive the work done when F(v) or F(t)?

One of the reasons is to illustrate mechanical energy conservation for conservative forces,note that when forces are conservative their work is just the decrease in the potential energy function,this equals the increment in kinetic energy from WE theorem,hence Mech energy conservation.

Also you'll see when you solve problems how useful work energy theorem will prove as a computational tool,you can't simply solve Newton's laws in all cases,using energy conservations principle helps in such cases where obtaining solutions from Newton's laws is practically difficult

@123 Sorry! Define work? Isn't there a single definition for it in all cases,i.e the scalar product of force and the displacement of the point at which it acts?

@123 no matter what the force's form is we have the same definition

@Arjun Thanks

$2as = V_f^2 - V_i^2$ multiply with mass on both sides and divide by 2 gives us work formula.

Does deriving work energy theorem from above formula can only be explain when acceleration is constant or variable acceleration also?

@123 note that this approach that you took ,works only when the particle moves in a straight line,since the formula you used is only for constant acceleration straight line motion,you have to use integration to obtain the general result

@Arjun Hmm.. Nice thanks. Same as i think

It means general equation for work is $W(x, v, t) = \int F(x, v, t) dx$

Yeah..but in the true general thing you'd have the dot product between vector F and vector dr : )

@Arjun Yes

@JohnRennie I see;thanks to you I think I can actually phrase my question better now.I was wondering if classical EM field could be somehow brought up(not necessarily simply adding them up) from these "modes",you seem to suggest that it might not be the case..thank you : )

Note that the Work Energy Theorem is nonsense in modern physics. One part because forces are nonsense in quantum theory. Another part because in SR nobody ever considers the Work Energy Theorem; if you try to derive it, you will find that you have to first assume $E=+\sqrt{m^2+p^2}$, and then get back that, so that the argument is circular.

In actuality, in modern physics, we understand that momentum and energy are fundamental conserved quantities worth considering by themselves, and never involve yourself with mass and forces, and minimally with velocities, because they would take care of themselves if you take care of momentum and energy.

@JohnRennie It is actually perfectly represented as a single mode; you can use one single coherent state to describe any such single mode infinite plane waves. The issue is that actual real world light waves and so forth are never infinite plane waves, and so we try to approximate them somehow. Even if you use a wavepacket, it will be difficult, because real-world systems will have classical mixtures, i.e. not describable by one single wavefunction.

have to use density operators

@naturallyInconsistent Thanks for your input.

@Slereah you seem like interested in math, physics and computer science. Can you tell me something interesting in computer science or math? Any problem(s) or concepts(s) you enjoyed the most?

I really like the animation of Puffer Trains in your blogpost Cellular automata I : Conway's Game of Life.

scientificamerican.com/article/… it is about the quantum zeno effect

zeno said things shud remain frozen in time

in quantum zeno effect, the state of a system is frozen in time becuz of quick measurements

quantum zeno effect is also a problem with Rovelli's relational interpretation

but i doubt that that problem can be mathematically stated... it's just critics saying vague criticisms

for it to be a real problem, the measurements would have to be global and instantaneous and continuous... Relational interpretation does not assume any of that

Any tips pls

in the discussion of a periodic potential (in 1 dimension) of height $V_0$ and periodicity $a$ where the single potentials are centered at $na, n\in \mathbb{Z}$, denoting with $|n\rangle$ the ground state of the restriction of $V(x) in $x \in [x-a/2,x+a/2]$, namely where one of wells is located my professor writes the following

is there a reason why we must insert a minus sign is there when we consider one of the two nearest neighbors matrix entries?

the reason why we expect the nnb matrix elements to be nonzero is because of quantum tunnelling effect-induced "wavefunction leakage" outside of $[x-a/2, x+a/2]$

sorry for the insufferable wording :P

Hi @BinkyMcSquigglebottom

@RyderRude your surname is really 'Rude'?? Imagine your friend try to introduce you to his other friends by saying meet "Mr. Rude" hehe (not so funny :( )

@Claudio purely for convenience. We want the state with all the sites equally occupied, the simplest thing, to have minimum energy, so this choice makes that energy eigenstate have the energy eigenvalue $E-2\Delta$, where $\Delta>0$

otherwise, you would have to deal with $\Delta<0$

@naturallyInconsistent That makes sense. Thanks