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Cosmic Background Radiation and the Electromagnetic Field


geordief

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(Hope this is the right subforum)

 

Really a question about waves,particles and fields wrt electromagnetism.

 

I understand that light propagates as a wave in the Electromagnetic Field and think I am also coming to realise  that the Cosmic Microwave Background (CMB) may well be such a field and a primordial one at that.

So,this is what I am wondering (and my question)

When light propagates through the vacuum as a wave in the EM field can we say that it is propagating through this CMB(field) ?

Is the vacuum (in this instance)  the CMB field?

 

As a second related question,how does one detect the CMB field if it is the same in all directions ?

What apparatus might one employ to measure its value and how would its needle move if it didn't know what direction to move in?(if that makes sense)

Edited by geordief
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Let me clarify what the CMB is.

The universe started off very hot.
For a couple of hundred thousand years, the radiation that permeated the whole universe was hot enough that, if an electron were to attach to a proton ( hydrogen nucleus ) or a Helium nucleus, a gamma, or x, ray would soon come along and hit it with enough energy to cause it to escape. The whole universe was permeated by this plasma, where Hydrogen and Helium nuclei, as well as electrons,  are kept from combining in this sea of high energy radiation. This is like conditions on the Sun, where temperatures keep elements in a plasma state, and atoms dissociated; you cannot see through plasma, the glow of scattering makes it opaque.

But the universe kept expanding and cooling ( according to gas laws ), and at a certain point ( about 30000 K, temperature of last scattering ), the radiation was cool enough that it could no longer knock electrons loose, after they formed an atom. The glow of the plasma stopped, and the universe became transparent because radiation was no longer scattered, but free to travel to any observer ( even future ones like us ).

Over the last 13 1/2 billion years, the universe has expanded by about a factor of 1040 times, and cooled by the same factor ( again according to gas laws ), which means that we would expect to see the afterglow of that plasma radiation about 1040 times cooler ( see Gamow ), and sure enough, but purely accidentally, two Bell Labs employees detected this much weaker afterglow ( see Penzias and Wilson ) at 2.70 K.
 
That is the farthest we can see back in time electromagnetically, as before that the universe was opaque, and, we see it in all directions because it permeated the whole universe. Actually it was 'heard' in all directions because it is shifted to 1040 times longer wavelength/ lower frequency, by a square horn antenna in the 50s.

Edited by MigL
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Quote

Is the vacuum (in this instance)  the CMB field?

No.

Quote

 

As a second related question,how does one detect the CMB field if it is the same in all directions ?

You detect the photons coming from a particular direction. And notice that it’s the same as any other direction.

Quote

What apparatus might one employ to measure its value and how would its needle move if it didn't know what direction to move in?(if that makes sense)

As MigL said, it was a square horn antenna, which is a directional antenna, so it detects a signal from some (small) solid angle of sky.

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So,I am getting the feeling now  that the CMB is not ,as I  imagined an example of an electromagnetic field per se.

Is it ,though  or can it be described as a field of some kind (a set of measurements over space)?

 

Also ,if  the CMB is not an electromagnetic field what would be the first (earliest) known example of an electromagnetic field and  would that field have perpetuated itself (in a continuously changing form) in perpetuity down to the present day?

Edited by geordief
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19 minutes ago, geordief said:

So,I am getting the feeling now  that the CMB is not ,as I  imagined an example of an electromagnetic field per se.

There’s more than one sense of “field” and you may be conflating different ones 

You can talk about the concept of an electric field, and that field would be everywhere, but also that the value of the electric field is zero, so there is no field in a region.

So “light propagates as a wave in the Electromagnetic Field” is that first sense, but not in the second, because the EM wave is the field. (I’m not prone to phrase things as “light propagates as a wave in the Electromagnetic Field” because I take the second approach: if E=0, there is no field)

 

19 minutes ago, geordief said:

Is it ,though  or can it be described as a field of some kind (a set of measurements over space)?

The CMB can be described as a photon field, but nothing propagates in it. That makes it sounds (to me) like it’s a medium, and it isn’t

 

 

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17 minutes ago, swansont said:

because I take the second approach: if E=0, there is no field)

When does E=0?

Not,surely because to detecting apparatus is too insensitive?

Are there points in the region of an electric charge when E is equal to precisely zero?

Is ,perhaps the field not continuous?

You can't have a  perfect shield against the electric field can you?

 

Edited by geordief
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You can treat the EM field classically; then there is a particular configuration that is consistent and is made up of an E that gives rise to a B, that gives rise to an E, etc. without any charge density sustaining it. Those are so-called "vacuum solutions" of the Maxwell equations and correspond to radiation, any EM radiation. The CMB is made up of these EM waves in the vacuum. They are clearly distinguishable from static fields.

The quantum picture is one of photons, of course.

The CMB is an EM field, but it's completely thermalised. So it's not like the light from a light bulb, or from the Sun, or from a diffraction grid, let alone from a laser. It has no particular direction, and has a considerable dispersion in frequencies. But most of the photons are around,

\[\frac{\hbar\omega}{k_{B}T}\sim1\]

in frequency \( \omega \) for \( T = 2.7 K \) of absolute temperature, where \( k_B \) is Bolzmann's constant.

so it's very cold.

You can shield almost perfectly against EM fields by using a conductor.

Edited by joigus
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5 hours ago, geordief said:

So does an EM  wave propagating through the vacuum interact  with this CMB field in theory even if  the effects were too small to measure?

Sure it would in principle, those would be focalised beams of photons that would meet those completely thermal photons, here and there. But the chances of two photons scattering is completely negligible in QED at given temp. It is this process:

220px-Photon-photon_scattering.svg.pnghttps://en.wikipedia.org/wiki/Two-photon_physics

Edited by joigus
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The QED matrix element for this process "goes like",

\[\textrm{Tr}\left(A_{\mu}e\overline{\psi}\gamma^{\mu}\psi A_{\nu}e\overline{\psi}\gamma^{\nu}\psi A_{\rho}e\overline{\psi}\gamma^{\rho}\psi A_{\sigma}e\overline{\psi}\gamma^{\sigma}\psi\right)=e^{4}\textrm{Tr}\left(\cdots\right)\]

There are space-time integrals involved, but roughly speaking the probability for this process is controlled by the number,

\[e^{4}\sim\left(\frac{1}{137}\right)^{2}\simeq5.328\times10^{-5}\]

Which, again roughly speaking, means that this process of photon-photon scattering takes place 5 times every 100'000 times photons cross paths. The internal legs of the diagram represent two pairs of virtual e, e- pairs once appearing and once disappearing, and two processes of electron-photon and positron-photon scattering. At large enough scales what you see is photons scattering off each other every once in a blue moon.

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12 hours ago, geordief said:

When does E=0?

Not,surely because to detecting apparatus is too insensitive?

Are there points in the region of an electric charge when E is equal to precisely zero?

Is ,perhaps the field not continuous?

You can't have a  perfect shield against the electric field can you?

 

The field is zero inside of a conductor 

1 hour ago, joigus said:

The QED matrix element for this process "goes like",

 

Tr(Aμeψ¯¯¯γμψAνeψ¯¯¯γνψAρeψ¯¯¯γρψAσeψ¯¯¯γσψ)=e4Tr()

 

There are space-time integrals involved, but roughly speaking the probability for this process is controlled by the number,

 

e4(1137)25.328×105

 

Which, again roughly speaking, means that this process of photon-photon scattering takes place 5 times every 100'000 times photons cross paths. The internal legs of the diagram represent two pairs of virtual e, e- pairs once appearing and once disappearing, and two processes of electron-photon and positron-photon scattering. At large enough scales what you see is photons scattering off each other every once in a blue moon.

My recollection is that the process is energy dependent, so low energy photons are less likely to undergo the process (low energy means less likely to form virtual pairs, or the pairs last for a shorter time), which is a reason they use gammas in two-photon experiments, or that they only worry about the process under those conditions

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27 minutes ago, swansont said:

My recollection is that the process is energy dependent, so low energy photons are less likely to undergo the process (low energy means less likely to form virtual pairs, or the pairs last for a shorter time), which is a reason they use gammas in two-photon experiments, or that they only worry about the process under those conditions

You're absolutely right. I tried to convey that with,

4 hours ago, joigus said:

the chances of two photons scattering is completely negligible in QED at given temp.

The energy dependence you're referring to would indeed be reflected in the \( \int d^{4}xe^{ipx} \) integrals, to which you'd apply the cutoff \( \Lambda \) for the given energy range. So, as you say, it'd be something like,

\[e^{4}f\left(m/\varLambda\right)\]

I was kinda fixing this temp-dependent factor, or normalising it to 1, if you will. At that energy, the \( e^2 \) terms would totally dominate the photon-photon ones. Low-energy photons are never seen to scatter off each other at room temperatures, let alone at cosmic-background energies.

Edited by joigus
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