Calculating the energy in an EM wave

  • #26
fisher garry
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Why in the world are we dealing with radiation pressure in the first place? Why do you think it has any relevance in describing Faraday's law? Why haven't you answered that?

Zz.
I think I understand. Let me change the scenario. You have a circular current coil and apply an external B field. I will rewrite the relations of the kinetic energy produced:
1578085321055.png

1578085348158.png

1578085623778.png

When the coil has turned 90 degrees it reaches equilibrium and deaccelerates to 0 velocity. Lets assume hypotetically that all this energy is released as one photon. Now I will try to introduce radiation pressure again:
1578085783111.png

So we hypotetically want the electron that has been deaccelerated to 0 velocity (I will ignore the current direction) to obtain the same kinetic energy and velocity. We want it to absorb a photon to do so. If we look at one half cycle of an EM wave that would be half of the energy of one photon if all of the energy of the photon is contained in one cycle. The EM wave exert radiation pressure as described above. It is also said that E=pc where p is radiation pressure density and E is energy density.

Over a half cycle of an EM wave the avearge cycle can be obtained the following way:
1578086451844.png

In order to calculate the amplitude of the EM wave I try the following:
First I calculate the kinetic energy from the 90 degrees turn of the circular coil
1578086611499.png

Then I try to find the amplitude of the EM wave and uses the theory for the radiation pressure above along with E=pc:

1578086392219.png



1578086422219.png


Is this also a wrong usage of the radiation pressure?
 

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  • #27
ZapperZ
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When the coil has turned 90 degrees it reaches equilibrium and deaccelerates to 0 velocity. Lets assume hypotetically thst all this energy is released as one photon.
LET'S NOT!

This is silly. There is nothing here that indicates that this is valid. Do you think an electron oscillating up and down and just performed half a cycle produces ONE photon? Says who? What physics are you using to justify this?

You are making all of this up.

Zz.
 
  • #28
fisher garry
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LET'S NOT!

This is silly. There is nothing here that indicates that this is valid. Do you think an electron oscillating up and down and just performed half a cycle produces ONE photon? Says who? What physics are you using to justify this?

You are making all of this up.

Zz.
Ok. Thanks for the answer. Yes most of this is something I have made myself so a reference is of course not available. I did assume that half of the energy of one photon was absorbed from a half cycle. Perhaps that is wrong. I also wondered when I used the theory of the radiation pressure from Tipler and Mosca why a higher velocity of the electron that the EM wave excerted radiation pressure on would lead to higher radiation pressure.
 
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  • #29
davenn
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Ok. Thanks for the answer. Yes most of this is something I have made myself so a reference is of course not available.
Did you forget about the bit in the forum rules that comments about no personal theories ?
best you have a reread before commenting further :smile:

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Perhaps that is wrong
Just stop thinking and talking about photons. You are learning classical electrodynamics, not quantum.
 
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  • #31
fisher garry
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I have been working a bit more and this time I ended up with some relations that could have the right numbers. So hopefully it is ok that I post this as I am not sure where else I would post.
1583424852043.png

1583424879530.png

in the theory in the top of this post there is also a volume integral which starts the derivation below:
1583425086435.png
1583425155856.png


The volume needed to make this calculation work seems to be very close to the volume of the 1s of the hydrogen atom. It also seems to denote V=constant for an energy outlet of an electron since the classical electron radius is denoted as the radius where electrons emits photons. Can anyone explain why this is wrong or not?
 
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  • #32
ZapperZ
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The volume needed to make this calculation work seems to be very close to the volume of the 1s of the hydrogen atom.
How can you tell? Your work was completely devoid of units! A omission like this in my intro class will incur points deduction.

And what is the "volume" of the 1s hydrogen that you are using? And why should this be of significance other than a broad coincidence (assuming that your calculation is valid and correct, which is a big assumption)?

It is still a puzzle what exactly you are obsessing... er... calculating for.

Zz.
 
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  • #33
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Seriously, learn to use TeX. Posting microscopic pictures is disrespectful of the time and effort people have put in trying to help you. Use units, like Zz said. Making people guess is disrespectful of the time and effort people have put in trying to help you. Finally, explainnwhat the heck you are trying to do rather than just dumping a big-wall-O-calculations. Doing otherwise is disrespectful of the time and effort people have put in trying to help you.
 
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  • #34
fisher garry
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I guess I could write it down in tex:smile:

##\textbf{F}\cdot d \textbf{l}=q (\textbf{E} +\textbf{v}\times\textbf{B})\cdot \textbf{v}dt=q \textbf{E} \cdot \textbf{v}dt##

current density is defined as ##\textbf{J}=\rho \textbf{v}##

##\textbf{F}\cdot d \textbf{l}=\frac{q}{\rho} \textbf{E} \cdot \textbf{J}dt##

Maxwell-Ampere's law:

##\textbf{E} \cdot \textbf{J}=\frac{1}{\mu _0}\textbf{E} \cdot (\nabla \times \textbf{B})- \epsilon _0 \textbf{E} \cdot \frac{\partial \textbf{E}}{\partial t} ##

the identity

##\nabla \cdot (\textbf{E} \times \textbf{B})= \textbf{B} (\nabla \times \textbf{E}) - \textbf{E} \cdot (\nabla \times \textbf{B})##

We also have
##\textbf{B} \cdot \frac{\partial \textbf{B}}{\partial t}= \frac{1}{2} \frac{\partial }{\partial t} B^2##

##\textbf{E} \cdot \frac{\partial \textbf{E}}{\partial t}= \frac{1}{2} \frac{\partial }{\partial t} E^2##

##\textbf{E} \cdot \textbf{J}= \frac{1}{2} \frac{\partial }{\partial t}(\epsilon _0 E^2+\frac{1}{\mu _0} B^2)-\frac{1}{\mu _0} \nabla \cdot (\textbf{E} \times \textbf{B})##

The first term to the right in the last equation is the energy density. We want to look at the energy transported out which is the second term. Since we want what goes out we change the sign

##\textbf{F}\cdot d \textbf{l}=\frac{q}{\rho} \textbf{E} \cdot \textbf{J}dt=\frac{q}{\rho} \frac{1}{\mu _0} \nabla \cdot (\textbf{E} \times \textbf{B})dt=\frac{q}{\rho} \frac{1}{\mu _0} \nabla \cdot cB^2dt##

I omitted the direction vector above because I am only looking for a quantification of the energy. Above we still have the units Joule.

Then we look at an EM-wave

##B=\textbf{B}_{0}sin[2\pi ft]=\textbf{B}_{0}sin[2\pi f\frac{x}{c}]##

##\nabla \cdot (B^2)=\textbf{B}_{0} 4 \pi f \frac{1}{c} cos[2\pi f\frac{x}{c}] ##

##\frac{q}{\rho} \frac{1}{\mu _0} \nabla \cdot cB^2=\frac{q}{\rho} \frac{1}{\mu _0} \textbf{B}_{0} 4 \pi f \frac{c}{c} cos[2\pi f\frac{x}{c}]=\frac{q}{\rho} \frac{1}{\mu _0} \textbf{B}_{0} 4 \pi f cos[2\pi ft]##

We integrate over a quarter of a cycle since cos is positive in the first quarter and multiply by 4 afterwards

##4\int_{0}^\frac{T}{4}cos[2\pi ft]dt=4\frac{1}{2\pi f}(sin[2\pi f\frac{T}{4}]-sin[2\pi f 0])=2\frac{1}{\pi f}##

##E=\frac{q}{\rho} \frac{1}{\mu _0} \textbf{B}_{0} 4 \pi f 2\frac{1}{\pi f}=8 \frac{V}{\mu _0} \textbf{B}_{0} ##

We still have units joule above

If all the energy was sent out from an electrons perspective instead

##\frac{dW}{dt}=\textbf{E} \cdot \textbf{J} d \tau##

By using the rewriting above and again assume that all the energy is sent out and therefore the energy density is 0:

##\frac{dW}{dt}=\frac{1}{\mu _0} \int c \cdot B^2 d \tau##

divergence theorem:

##\frac{dW}{dt}=\frac{c}{\mu _0} \int \nabla B^2 d A=\frac{c}{\mu _0}B^2 4 \pi r^2##

Again we want all the energy to be preserved in one wave

##W=4 \pi r^2\frac{c}{\mu _0} B_{0}^2 \int_{0}^T sin^2[2\pi ft] dt##

##sin^2[2\pi ft] =\frac{1}{2}- cos[4\pi ft]##

##W=4 \pi r^2\frac{c}{\mu _0} B_{0}^2 \int_{0}^T \frac{1}{2}- cos[4\pi ft] dt##

the lower limit leads to 0 since the integral of cos is sin and we are left with

##W=4 \pi r^2\frac{c}{\mu _0} B_{0}^2 ( \frac{T}{2}- sin[4\pi fT])= 2\pi r^2\frac{c}{\mu _0} B_{0}^2\frac{1}{f}##

The units are still joule

By equating the two energy relations

##W=8 \frac{V}{\mu _0} \textbf{B}_{0}=2 \pi r^2\frac{c}{\mu _0} B_{0}^2\frac{1}{f}##

##4 V= \pi r^2c \textbf{B}_{0}\frac{1}{f}##

##f= \pi r^2c \textbf{B}_{0}\frac{1}{4 V}##

If we assume that we have the ionization energy of hydrogen electron

##E_{ion}=2.1 \cdot 10^{-18}##

We try to detrmine the correpsonding magnetic field of the ionization energy:

##8 \frac{V}{\mu _0} \textbf{B}_{0}=2.1 \cdot 10^{-18}##

But we have two unknowns the volume and the B field. If we instead start with E=hf and see if we can get something that makes sense

##8 \frac{V}{\mu _0} \textbf{B}_{0}=hf##

##8 \frac{V}{h\mu _0} \textbf{B}_{0}=f##

##f4 \frac{V}{c\pi r^2}= \textbf{B}_{0}##

##E=hf=8 \frac{V}{\mu _0} \textbf{B}_{0}=8 \frac{V}{\mu _0} f4 \frac{V}{c\pi r^2}##

It is said that electrons emit photons around the classical electron radius

##r_{classical}=2.82 \times 10^{-15}##

##h=8 \frac{V}{\mu _0} 4 \frac{V}{c\pi r^2}=6.6 \times 10^{-34}##

##V^2=\frac{\mu _0}{32}c\pi r^2 6.6 \times 10^{-34}=\frac{1.26 \times 10^{-6}}{32}3 \times 10^8 \pi (2.82 \times 10^{-15})^2 6.6 \times 10^{-34}##

##V^2=19.5 \times 10^{-62}##

##V=4.419.5 \times 10^{-31}##

This volume seems to be close to the 1s volume.

In general

##h=2.7 \times 10^{-2} \frac{V^2}{r^2}=6.6 \times 10^{-34}##

The second energy relation obtained above:

##2 \pi r^2\frac{c}{\mu _0} B_{0}^2\frac{1}{f}=E##

## r^2B_{0}^2 =\frac{f\mu _0}{c2 \pi}E=\frac{E}{2.7 \times 10^{-2} \frac{V^2}{r^2}}\frac{\mu _0}{c2 \pi}E##

## V^2B_{0}^2 =\frac{E}{2.7 \times 10^{-2} }\frac{\mu _0}{c2 \pi}E=0.0255 \times 10^{-12}E^2##

## VB_{0} =1.59 \times 10^{-5}E##

and the first energy relation obtained above:

##8 \frac{V}{\mu _0} \textbf{B}_{0}=E##

##V \textbf{B}_{0} =\frac{\mu _0}{8}E=1.58 \times 10^{-5}E##
 
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  • #35
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It is said that electrons emit photons around the classical electron radius
What? By whom?
 
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  • #37
ZapperZ
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But do you know what a scattering "cross section" actually means? This is the "effective size" of an electron, by itself.

And what does this "r" have anything to do with the 1s hydrogen atom? You appear to be mixing two different quantities together that have nothing to do with one another.

But again, you refused to indicate what all of this is all about. Go read a physics publication. Read the abstract. It tells the readers what you intend to do and WHY. This is something you have consistently neglected to show here, even after repeated queries!

Zz.
 
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  • #38
fisher garry
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But do you know what a scattering "cross section" actually means? This is the "effective size" of an electron, by itself.

And what does this "r" have anything to do with the 1s hydrogen atom? You appear to be mixing two different quantities together that have nothing to do with one another.

But again, you refused to indicate what all of this is all about. Go read a physics publication. Read the abstract. It tells the readers what you intend to do and WHY. This is something you have consistently neglected to show here, even after repeated queries!

Zz.
the first energy relation deals with an electron revolving with velocity in the 1s shell and has energy given as

##8 \frac{V}{\mu _0} \textbf{B}_{0}##

The second energy relation deals with the same energy being given out from the electron and instead of a revolving electron in an orbital we look at the electron in its rest frame. This energy formula becomes

##2 \pi r^2\frac{c}{\mu _0} B_{0}^2\frac{1}{f}##

After that I equate those two. I guess a question would be can you equate those. And if not why not?
 
  • #39
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The word "emit" is nowhere to be found in Libert's message or the Quora thread.
It's one thing to misunderstand what someone else said. It's another to misrepresent what they said.
 
  • #40
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If we instead start with E=hf and see if we can get something that makes sense
You can't. ##E=hf## relates the energy and frequency of a photon so is completely irrelevant to any problem involving electromagnetic waves. It is always possible to torture unrelated equations into superficial equalities, and that's all you're doing here.

At this point it is time to close this thread.
 
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