The h Bar

fqq

01:36

@ACuriousMind I've taken string theory with him and agree completely :)

Nihar Karve

04:19

Is the spinor-helicity formalism useful for QED?

FakeMod

04:41

yesterday, by FakeMod

Do we implicitly assume homogeneity of space when we talk about translational symmetry of Newton's laws of motion?

Slereah

05:51

No, the laws are symmetric

It doesn't mean the matter content is homogeneous

123

06:34

Hi @JohnRennie

John Rennie

@123 hi :-)

123

In EM wave "Direction of Electric Force on particle is same Direction of Magnetic Force on particle".

John Rennie

The direction of the force an electric field exerts on a particle is in the same direction as the electric field vector.

123

I get this concept from equation of electric force from Electric field and coulombs Force and Magnetic force formula

John Rennie

The magnetic field exerts no force on a stationary particle.

123

06:37

@JohnRennie Yes aye this is exactly i want to say about electric force.

Yes i agree magnetic force only exert on moving particle

John Rennie

So if for example an EM wave interacts with a stationary electron only the electric field affects the electron.

123

But direction of magnetic force is perpendicular to magnetic field and EM wave direction. which is same direction as electric force.

John Rennie

The direction of the force due to the magnetic field depends on the direction of the electron's velocity, nut just on the direction of the magnetic field. So it is not necessarily the same as the direction of the electric field.

123

Yes but direction of magnetic force is always perpendicular to electric force?? right or wrong.

If $\vec{B}$ is always perpendicular to $\vec{E}$ . Then $\vec{F}_B$ become same direction as $\vec{F}_E$

John Rennie

No. In an EM wave the direction of the magnetic field is alway normal to the direction of the electric field, but the direction of the magnetic force is not necessarily normal to the direction of the electric force.

i.e. $\mathbf B$ is normal to $\mathbf E$, but $\mathbf B \times \mathbf v$ is not necessarily normal to $\mathbf E$.

123

06:47

Aaaaah... I see.... I think this is what i am missing.

but $\bf{v \times B}$ is $\bf{F_B}$ which is parallel to $\bf{F_E}$ and $\bf{E}$

This is what i am confusing @JohnRennie Sir.

John Rennie

No, the direction of $\mathbf v \times \mathbf B$ depends on the direction of $\mathbf v$ where $\mathbf v$ is the velocity of the electron not the velocity of the light ray.

So the direction of the magnetic force is not necessarily the same as the direction of the electric field.

123

Ookay.. Let's take an example of free moving particle in vacuum in straight line but in wire $\bf{E}$ is zero.

Isolated particle velocity $\bf{v}$ has direction parallel to ground take as x-axis. So direction of $\bf{E}$ is $\perp$ perpendicular to x-axis as fixed polarization $\bf{E}$ has y-axis direction. Is it true???

John Rennie

Do you mean the light ray is travelling along the x axis and it is polarised in the xy plane i.e. the electric field lies in the xy plane?

123

Yes??? Electron is moving along x-axis creating electric field in the xy-plne

In $\bf{E}$ is creating with particle along xy-plane. Then magnetic field $\bf{B}$ always along z-axis. Is that right?

If it is the case $\bf{v}$ along x-axis , $\bf{E}$ along y-axis and $\bf{B}$ along z-axis (or xz-Plane).

John Rennie

07:03

Are you asking me about the magnetic field created by the moving particle, or the field of the light wave?

123

$\bf{v \times B}$ is pseudovector is along y-axis which is $\bf{F_B}$ . So y-axis is also direction of $\bf{E}$ and $\bf{F_E}$

First Sir take it as moving particle then i want to understand in EM waves.

John Rennie

A moving charge is like a current, so the magnetic field created by a charge moving along the x axis will be concentric rings centred on the x axis.

123

Also particle velocity is constant in time and direction. As DC Current.

What about $\bf{E}$ in moving particle. It's along y-axis or x-axis???

undefined

Does the speed of a photon have any influence on what happens when this photon hits a mirror? (Assuming the speed of the photon is less then c)

123

If an free isolated electron is moving along x-axis. What is the direction $\bf{E}$ and $\bf{B}$?

John Rennie

07:11

The electric field of the particle is just the usual 1/r² field of a point charge. It is unaffected by the velocity.

undefined

hello John, haven't seen for a long time. Hope you are alright

John Rennie

@undefined you have to be careful about using the concept of a photon where it is not applicable. Light is not photons and it not a wave. It is a quantum field that can behave like a photon in some cases and behave like a wave in other cases. So when you talk about "a photon hitting a mirror" this is not generally a useful way of describing the physics.

123

@JohnRennie Great Sir. It means point particle as electron is moving along x-axis create $\bf{E}$ in radially inward (For vector notation) in all direction with magnitude $\frac{1}{r^2}$. Does not depend on particle velocity.

John Rennie

Yes you're correct

undefined

@JohnRennie I had a single photon in my mind, that's why I was referring to 'a photon'

John Rennie

07:16

Right, but that isn't necessarily a useful way to describe the interaction between the light and the mirror.

123

Grea Sir @JohnRennie . And at the same time we take magnetic field $\bf{B} \perp$ $\bf{E}$ at every point in space. Or anything else?

John Rennie

As a general rule photons are only a useful concept when the light is exchanging energy with something.

When you are describing the propagation of the light it is more useful to approximate the light as a wave.

@123 $\mathbf B \perp \mathbf E$ only for EM waves. It does not apply in general.

123

@JohnRennie Ookay. So what about moving particle for $\bf{B}$

John Rennie

I answered that already. A moving particle has a magnetic field that is similar to the field from a current flowing in a wire.

@undefined it's important to understand that both the photon and the wave are different approximate ways to describe how light behaves. Which is the better approximation depends on the circumstances.

undefined

I thought photons do exchange energy with a reflective surface/mirror upon interacting/hitting it

John Rennie

07:21

No, or not in the sense that light can excite electrons in atoms.

undefined

how do I say it correctly when I'm referring to 'the wave of a single photon'? Or is that wrong from the beginning?

John Rennie

The light becomes entangled with the electrons in the mirror, but it does not exchange energy with them like an electronic excitation exchanges energy.

123

I think if i take $\bf{B}$ for moving particle with respect to radial direction of $\bf{E} \perp \bf{B}$ it might be concentric ring. I don't know. Pls explain. I also want to simulate this behavior

By the way Sir you give me best explanation for my question. But part of magnetic field remains how it is creating concentric ring. whether we accept it as natural phenomenon there must be some specific pattern and behavior. Thank you

John Rennie

To really understand why electric and magnetic fields behave as they do requires you to understand the treatment of EM fields in special relativity, and that's a complicated subject that you wouldn't do until the second year of a physics degree.

123

I have read the ideas of and little math of relativity.

This exactly what i am thinking this time and you said the same. I thought about what happen if another charge is moving he sees magnetic field of stationary particle with respect to ground.

But first i want to understand simplest phenomenon then complicated . Now my target relate math with my understanding.

undefined

07:33

I thought/read that the atoms in the mirror absorb the light and then re-emit it. I learned that all quantities are (energy, momentum, spin, charge) conserved but that momentum and polarization are altered in a predictable and well-determined way. I thought that this is all true if the photon/the wave is traveling close to c. And the question I was thinking about is, if any of those known properties will change if the photon/the wave is traveling/propagating slower

John Rennie

07:45

What happens is the light and the electrons become entangled and together they form a system in which the average charge density inside the metal of the mirror oscillates in time. Note that this is not a single electron receiving energy from the light. Instead is a collective oscillation i.e. a collective motion of many electrons.

undefined

on which circumstances does the frequency of the oscillation and the area/number of the collective forming electrons depend?

but one side question, is it just me or is this whole 'reflective stuff' really complicated to understand in detail? I can't really wrap my head around it

123

As $\bf{E}$ change/depend on charge $q$ and $\frac{1}{r^2}$. What $\bf{B}$ depend and formula?

If linear moving particle creates $\bf{B}$ as concentric ring. How magnetic dipole define in this situation?

Nihar Karve

08:14

@123 I suggest you read Griffiths' Electrodynamics to get a solid grounding in this stuff

123

@NiharKarve Thanks for suggestion. I have Griffiths i must read.

Secret

09:18

The Black Hole Information Loss Problem is Unsolved. And Unsolvable.

►

Normally Hossenfeder's youtube video do not caught my attention

But this one...

feels particularly dense for some reason

123

Why Coulomb's Force read in electrostatic it also work on electrodynamics when both source and test charge are in motion.

Secret

But what she said is valid, maths alone don't claim anything, you need experimental data

The DFT literature also suffer from similar problems of overreliance on theory

And finally, coloumb equations works in classical electrodynamics because you can derive that from the electromagnetic tensor

and electrodynamics only adds some extra terms that deals with charges in moving frames with the help of the maxwell equations

123

09:38

Does the coulomb force work on both stationary and moving charge????

Secret

yes, but you need to correct for the motion of the moving charges

Liénard–Wiechert potential

The Liénard–Wiechert potentials describe the classical electromagnetic effect of a moving electric point charge in terms of a vector potential and a scalar potential in the Lorenz gauge. Built directly from Maxwell's equations, these describe the complete, relativistically correct, time-varying electromagnetic field for a point charge in arbitrary motion, but are not corrected for quantum-mechanical effects. Electromagnetic radiation in the form of waves can be obtained from these potentials. These expressions were developed in part by Alfred-Marie Liénard in 1898 and independently by Emil Wiechert...

123

@Secret What is meant by that moving charge?

Secret

electric fields still exists when dealing with moving charges, the coloumb force of the source charge then exerts onto the moving charge

Also the moving source charge may be brought into motion by some process earlier

You do not need to constantly exert a force to keep something in motion

123

Yes. But i want to know if in both case stationary and moving charge has coulomb's force on another charge. How do we know the force act on another particle is due to electric or magnetic force?

Secret

they will always felt an electric field between each other. What matters then is whether you have a magnetic field and hence by faraday law, you have an electromotive force that is felt by the charges. The composition of the force that acts on the particle depends on your frame of reference, because force is frame dependent

If you are in the rest frame of both charges, the force is entirely columb

But if you are in a moving frame, there can be contribution from the magnetic force due to the motion of the charges

Under electrodynamics, magnetic fields are kinda just electric fields in a moving frame

123

10:27

Thanks @Secret

AccidentalBismuthTransform

12:38

I get a strange result about the energy flux entering and leaving a material in steady state

in the case of a temperature difference is maintained at the 2 ends of a wire for instance, the energy flux are equal, which is fine and makes sense. If we turn on Joule heat by passing a current in the wire, we also get that the fluxes are equal, which again makes sense

however if I introduce thermoelectricity, then I do not get this anymore. I get that they differ... in steady state... how is this possible?

user2723984

12:50

@Secret it's in her usual controversial style, but as usual, she does have a point

B. Brekke

13:43

I am trying to understand an old article (1981) using notation from the book \emph{Properties of the Thirty-Two Point Groups}. In order to understand the labeling I think I need to read the book somehow, but I don't know how to get hold of it. It is not in my university library, I don't think it is sold anymore and I can't find it online. Does anyone have any idea of what I should try next?

bolbteppa

@B.Brekke the full thing is here

B. Brekke

@bolbteppa Wow thanks! Seems like I really need to work on my googling skills...

Emilio Pisanty

14:14

@ACuriousMind @JMac I'm rather befuddled about this conversation about national origins, taking place with extremely scant evidence.

but, that said

I'm somewhat confused about what happened with user 26076

the Wayback Machine puts that user with a location in Australia

did I miss something getting announced?

ACuriousMind

@EmilioPisanty I don't think that user announced anything about their rather sudden change in their online presence, no

Emilio Pisanty

OK, thanks

none of my business, I think.

ACuriousMind

ah, but speculating about things that are none of one's business is so fun :P

user2723984

it's none of your business what happens at the Planck scale

JMac

14:29

@EmilioPisanty That whole conversation was weird. Someone took a sample of top 11 users to suggest that EU was over-represented on physics SE, based on a few locations. I mostly was just pointing out that you could get a totally different conclusion if you took a different sample size (top 11 seemed pretty obviously chosen to get the result they wanted); and that it was all assumptions about location anyways.

Emilio Pisanty

@JMac no worries.

@ACuriousMind anyways, it's good to see that there's a familiar PSE face in town, and to see that the PSE gender representation at leaderboard level is not as dire as it was previously

I guess I should update my location and profile soonish

(significant re: assumptions)

Nihar Karve

16:19

@JMac huh, what a weird thing to do

Jim

17:04

@JMac For future reference, QMechanic was born from a quantum fluctuation in the SE vacuum. He draws energy from editing physics posts

Nihar Karve

17:29

@Qmechanic actually, what is your mojo while editing posts (specifically the links)?

I know you shorten the question links and change references to doi - what else?

Qmechanic

17:48

@NiharKarve : In general to replace with permalinks as a protection against link rot.

ACuriousMind

18:32

my roommate ordered a mulled wine dispenser and now we have to drink what we put into it for a test run...I don't think I'll do much else this evening

3

Charlie

A kinda simple question asked on the site earlier made me wonder because I didn't know the answer, how does an object like $V^\mu\partial_\mu$ behave under lorentz transformations? Does some weird behaviour result from the fact that it's an operator?

ACuriousMind

that's just a vector :P

Charlie

oh yeah

ok ignore me i retire from physics

ACuriousMind

kthxbye

Charlie

there's some things you just can't come back from

ACuriousMind

18:42

(but I guess this goes back the the conversations we've have about whether or not a full vector (and not the components) is "invariant" "covariant" or something else)

Ryan Unger

19:50

what does covariant mean

ACuriousMind

bolbteppa

22:22

Covariant notation/derivations makes everything better/easier

Stupid question inc

22:42

Is it true

that

A force is field if it depends on position the particle?

Ryan Unger

22:59

I prefer contravariant things

bolbteppa

@ACuriousMind from today: "In this note, while revisiting Haag’s theorem within standard QFT, we provide a fresh look into its consequence ... All these proofs require a substantial amount of effort to get familiar with the formalism used In this note, we proof and clarify the theorem in the language of standard QFT."

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