How does a photon not "feel" electromagnetism?

  • #1
Marshall2389
3
1
TL;DR Summary
A table from Sean Carroll's 'The Particle at the End of the Universe' says that photons don't feel electromagnetism. I'm wondering how this is so.
I've attached a picture of a table in Sean Carroll's The Particle at the End of the Universe. It says that photons don't "feel" electromagnetism, but gluons feel the strong force, the W and Z bosons feel the weak force, and gravitons feel gravitation. How is this so?

(I have no formal quantum physics training, just a fair amount of classical as a mechanical engineering PhD)
 

Attachments

  • IMG_0080.JPG
    IMG_0080.JPG
    88.5 KB · Views: 77

Answers and Replies

  • #2
33,674
11,244
Photons have no charge.
 
  • Like
Likes bhobba, Marshall2389, vanhees71 and 1 other person
  • #3
Marshall2389
3
1
Alright, that's what 'feel' means. That makes sense. I can't really make sense of how these bosons are vibrations in their respective fields but some are and some aren't affected by the forces arising from that field. Do you recommend any particular introduction to quantum mechanics textbook that might give me a deep enough understanding to get rid of my confusion here?
 
  • #4
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
29,560
15,006
Do you recommend any particular introduction to quantum mechanics textbook that might give me a deep enough understanding to get rid of my confusion here?

Probably not.

This was covered in my second year of graduate school. You are three or so textbooks behind. If I told you "it's because electromagnetism is a U(1) and QCD is an SU(3)" it would be true, but that assumes years of background.
 
  • Like
Likes dextercioby, bhobba, PhDeezNutz and 2 others
  • #5
Marshall2389
3
1
What’s a U(1)?

Just kidding. Thank you for the clear answer.
 
  • #6
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
29,560
15,006
What’s a U(1)?

It';s a force where the carriers have no charge! (See the problem?)
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,012
12,921
That's really some deep mathematical stuff called "gauge theory". As @Vanadium 50 says, you need quite a while in the university curriculum to get there. The first time you encounter the notion of a gauge theory is in the 3rd semester of the theory course, when classical electrodynamics is taught (and you are lucky and your professor treats classical electromagnetism in such a modern way ;-)).
 
  • Like
Likes dextercioby, bhobba and Marshall2389
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,745
15,354
Do you recommend any particular introduction to quantum mechanics textbook that might give me a deep enough understanding to get rid of my confusion here?
If you want to dive in at the deep end:

Introduction to Elementary Particles by David Griffiths.

Photons don't feel the EM force because photons themselves have no charge and QED is relatively simple. Gluons, on the other hand, have color and QCD becomes monumentally complicated.

That ultimately stems from the dimension of the symmetry that nature chose in each case: a simple ##U(1)## symmetry for QED and an ##SU(3)## symmetry for QCD.
 
  • Like
Likes bhobba, Marshall2389 and vanhees71
  • #9
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,745
15,354
  • Like
Likes PhDeezNutz and vanhees71
  • #10
40,580
18,269
What’s a U(1)?

A slightly more informative answer than the one @Vanadium 50 gave is that U(1), the gauge group of electromagnetism, is an abelian group--any two elements of the group commute--while SU(2) (weak force) and SU(3) (strong force) are non-abelian--not all group elements commute. And it turns out that if the gauge group is abelian, the force carriers don't carry any charge, but if the gauge group is non-abelian, they do.

Of course, understanding why all that is the case will still take the years of study that have been referred to.
 
  • Like
  • Informative
Likes Klystron, vanhees71, bhobba and 4 others
  • #11
10,166
3,312
As @Vanadium 50 says, you need quite a while in the university curriculum to get there.

At the undergrad level I have always found the following a good starting point:
https://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html

This shows how the vector potential comes about and gives the first two of Maxwell's Equations which is the divergence of B and the curl of E. But we would also like the curl of B and the divergence of E. Let p, by definition, be the divergence of E. We then define the electromagnetic field tensor Fuv:
https://quantummechanics.ucsd.edu/ph130a/130_notes/node451.html

We define Ju = ∂v Fuv as per the above link, (easier to do if we use units where c, the speed of light, is 1) and note the zero component is p as defined above. Hence Ju is the 4 current of p, if p is a density of something, that something we call charge, q. This gives the curl of E and the divergence of B. These are known as Maxwell's equations. To get the Lorentz Force Law we have to delve into the Lagrangian Formulation of Maxwell's Equations, which is done by Lenny Susskind here:




Added Later: I could have been sneakier. When adding the counter-term to the Schrodinger equation, associated the q introduced there to ensure local global invariance, with the charge defined above, and you get the Hamiltonian of the equations of motion and hence the Lorentz Force Law. But I think it is a bit too sneaky and assumption laden.

An even simpler 'derivation', not using gauge invariance, can be found here:
https://arxiv.org/pdf/1507.06393.pdf

Then there is the one that uses Coulomb's Law and Relativity:
http://cse.secs.oakland.edu/haskell/Special Relativity and Maxwells Equations.pdf.

Interesting question for further investigation - why does it fail for Gravity? Part of the answer is of relevance to your question. Hint - the source of Maxwell's Equations is charge. EM Fields do not have charge. But gravitational fields have energy - which is a source of gravity.

Personally, as Schwinger does in his textbook on EM, the above, or something similar, (Schwinger uses something similar) is IMHO the way to begin EM:
https://www.amazon.com/dp/0738200565/?tag=pfamazon01-20

When you have read a few of these 'derivations' or 'justifications' it is fun and instructive coming up with your own. But Jackson in his standard textbook on EM thinks they are silly. I disagree.

Thanks
Bill
 
Last edited:
  • #12
andresB
588
336
While correct, it seems counterproductive to talk about gauge theory and the like when it is clear that the Op doesn't have the background to understand such answer and Maxwell theory is enough to give a good idea of what the book is talking about

@Marshall2389, direct a laser beam through a region of space where a magnetic and/or an electric field is present. Since classical electrodynamics is a linear theory, the laser beam will be unaffected by the magnetic (electric) field. You can even try to "collide" two laser beams and they will pass through each other completely unaffected. In that sense photons don't feel each other, aka, they don't feel electromagnetism.


=================================

As a twist of the story, photon DO feel other photons and electromagnetic fields in general. They do it indirectly since they don't have change, but the interaction is there and it might be relevant. The process is called photon-photon scattering
light-by-light.jpg

The image shows two photon "hitting" each other. But they do it indirectly using electrons/positrons as intermediaries.

This process is relevant only if the fields involved are strong enough. That's why we don't see it in classical electrodynamics. Everyday electric/magnetic phenomena involve fields far too weak for such non-linearity.

https://arxiv.org/abs/1101.3433
https://arxiv.org/abs/1012.1134

It is predicted that around magnetars the magnetic lensing can even dominate over the gravitational lensing of general relativity.
 
Last edited:
  • Like
  • Love
Likes dextercioby, vanhees71, PeroK and 2 others
  • #13
Sophrosyne
122
19
Summary:: A table from Sean Carroll's 'The Particle at the End of the Universe' says that photons don't feel electromagnetism. I'm wondering how this is so.

I've attached a picture of a table in Sean Carroll's The Particle at the End of the Universe. It says that photons don't "feel" electromagnetism, but gluons feel the strong force, the W and Z bosons feel the weak force, and gravitons feel gravitation. How is this so?

(I have no formal quantum physics training, just a fair amount of classical as a mechanical engineering PhD)

Photons ARE the force carrier for electromagnetism. They do not interact with each other (well except in very exceptional situations way out of the scope of the discussion here). They only interact with other charged particles.
 
  • #14
10,166
3,312
Magnetic fields do effect light as demonstrated in a famous experiment done by Faraday:
https://en.wikipedia.org/wiki/Faraday_effect

And yes, in special high energy processes photon-photon interaction is possible, but it is by an indirect route (eg gravity) not by direct EM interaction. You can look it up using Google.

Thanks
Bill
 
Last edited:
  • #15
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,012
12,921
The Faraday effect is due to the effect of the magnetic field on the matter.

The non-linearity and thus the self-interaction of photons in the vacuum is a higher-order effect due to radiative corrections, i.e., due to quantum fluctuations of the order ##\alpha_{\text{em}}^4##, i.e., very small (see the Feynman diagram in #12).
 
  • Like
Likes bhobba, Vanadium 50 and PeroK
  • #16
10,166
3,312
The Faraday effect is due to the effect of the magnetic field on the matter.

Damn - I was caught out. The Faraday effect will not work in a vacuum.

Thanks
Bill
 
  • #17
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,012
12,921
Well, it can work in principle, again because of the radiation corrections, i.e., the quantum fluctuations of the em. field and the charges. The diagram is the same as for light-by-light scattering but with two of the external legs meaning some (strong) classical electromagnetic field (in this case a very strong magnetic field). Then the diagram is a contribution to the vacuum polarization of the photons at presence of this field, and I'd expect that there is the corresponding "Faraday effect" contribution to the resulting photon propagator.

I've never heard about any experimental confirmation of such an effect though.

The Delbrück scattering itself has been recently observed in the strong Coulomb field of ultraperipheral lead-lead collisions at the LHC (by the ATLAS collaboration):

https://cerncourier.com/a/atlas-spots-light-by-light-scattering/

There's of course no doubt anyway that this effect really exists. Indirectly it's seen also in the very accurate measurement and calculated anomalous magnetic moment of the electron. For the muon anomalous magnetic moment the related contributions from the strong interaction (quarks/hadrons) are the largest theoretical uncertainty.
 
  • Like
Likes dextercioby
  • #18
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
29,560
15,006
In the spirit - nay, PF tradition, of A-level niggles to B-level questions, and in the hopes that we haven't completely lost the OP, but also in the hopes that this post will push him into endless confusion,

if you have a vacuum Faraday effect based on light-by-light scattering, in which direction does the polarization plate rotate? To the right or the left?
 
  • #19
andresB
588
336
Being pedantic, there is no Faraday effect in vacuum because Faraday effect is linear in B while the effect that comes from the Euler-Heisenberg Lagrangian is of higher order in the field strength. I don't remember the correct name at the moment, though I have to say that calling it vacuum Faraday effect is a somewhat common mistake.

Now, at the purely optical level, the vacuum magnetic birefringence and dichroism have proven to be too hard to observe.

https://arxiv.org/pdf/2005.12913.pdf
 
  • Like
Likes bhobba, vanhees71 and Vanadium 50
  • #20
dextercioby
Science Advisor
Homework Helper
Insights Author
13,264
1,551
It is only on PF that asking the simplest question pulls the most complicated (and obviously uncomprehensible to the OP) answers. I used to provide such answers in the past, mostly to show off, tbh.

So the only B-level explanation of this question is: Photons are quanta of electromagnetic energy and they ##\bf{do}## feel electromagnetism (i.e. they ##\bf{do}## feel other photons) at sufficient high energies (Delbrück scattering).
 

Suggested for: How does a photon not "feel" electromagnetism?

Replies
15
Views
2K
Replies
13
Views
570
  • Last Post
Replies
3
Views
425
Replies
11
Views
768
  • Last Post
Replies
32
Views
1K
Replies
3
Views
419
  • Last Post
Replies
2
Views
346
Replies
16
Views
484
Replies
1
Views
350
Replies
18
Views
812
Top