# Momentum of photon

1. Sep 12, 2015

### exmarine

Is there any experimental evidence that photons can transmit negative momentum? After all, there is also a negative root: E^2 = (pc)^2 + (mc^2)^2

2. Sep 12, 2015

### vanhees71

What do you mean by "negative momentum"? Momentum is a vector! In scattering processes, involving photons (e.g., Compton scattering, pair creation, bremsstrahlung,...), total momentum is conserved.

Or do you mean "energy"? Then you should know that photon energies are always positive and that this is one of the reasons, why one uses relativistic quantum-field theory to describe relativistic quantum phenomena. The free electromagnetic field is represented by a quantum field. In radiation gauge it reads
$$\hat{A}^{\mu}(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{\sqrt{(2 \pi)^3 2 \omega(\vec{k})}} \sum_{\lambda \in \{-1,+1 \}} \left [\hat{a}(\vec{k}, \lambda) \epsilon^{\mu}(\vec{k}) \exp[-\mathrm{i} \omega(\vec{k}) t+\mathrm{i} \vec{k} \cdot \vec{x}]+\text{h.c.} \right].$$
As you see, there are modes with positive and with negative frequency (the latter are hidden in the short-hand notation "h.c." which stands for "hermitean conjugated").

The crucial point is that the $\hat{a}(\vec{k},\lambda)$ discribe the annihilation of a photon with three-momentum $\vec{k}$, energy $\omega(\vec{k})=|\vec{k}|>0$, and helicity $\lambda$ while the $\hat{a}^{\dagger}(\vec{k},\lambda)$ describe the creation of a photon with three-momentum $\vec{k}$, energy $\omega(\vec{k})=|\vec{k}|>0$, and helicity $\lambda$. Nowhere are photons with "negative energy" involved, and that's, as stressed before, one of the reasons why the so far one and only successful attempt to define a physically consistent relativistic quantum theory is in terms of local, microcausal relativistic quantum-field theory.

3. Sep 12, 2015

### exmarine

OK, how about this. Can a photon ever transmit momentum in the opposite direction of its propagation?

4. Sep 12, 2015

### ZapperZ

Staff Emeritus

This is rather puzzling because the direction of the momentum is DEFINED by the direction of its propagation!

Zz.

5. Sep 13, 2015

### lightarrow

Maybe if the body hit by the photon had negative mass (the body would accelerate in the opposite direction of the force on it), but this is just speculation because negative mass doesn't exist.

--
lightarrow

6. Sep 13, 2015

### PeroK

If you forget about photons for a moment. If a small fast particle collides with a larger slower particle (both moving in the same direction), then the small particle will bounce off the large particle, which will be accelerated somewhat, and the small particle will be given an impulse in the opposite direction. You could describe this as the larger particle "transmitting" momentum in the direction opposite to its motion.

7. Sep 13, 2015

Staff Emeritus
I'm still trying to envision a photon (or, for that matter, anything) moving in one direction, but its momentum pointing opposite.

Two guys are walking, and one is knocked down by an unseen force. He gets up, puzzled, and the other says "Must have been a heck of a photon moving over yonder."

8. Sep 13, 2015

### vanhees71

It doesn't make sense at all. What do you mean by "a photon is moving in one direction". I cannot make sense of this sentence, because a photon has no position. So what do you mean when you say, it's "moving in a direction". The only thing we know about the photon is that it is described by a one-photon state. It can have a pretty well determined momentum (and then it also has a pretty well determined energy) and has a perfectly well determined polarization. The association of a state to a photon then describes everything you can know about it, and this are all kinds of probabilities of its detection or about its scattering with other quanta or many-body systems, no more no less.

The idea that a photon is a little massless bullet which is moving like such a macroscopic particle, is misleading and never needed in modern physics!

9. Sep 13, 2015

### Strilanc

Maybe the hypothetical being described is more like... two balls meet head on, but instead of bouncing off each other they mutually accelerate through each other. This doesn't violate conservation of momentum, but does violate conservation of energy, and is a bit like the change in momentum being negated.

10. Sep 14, 2015

### WhatIsGravity

Just to be mischievous... a photon passing through an event horizon will have a negative momentum... opposite it's velocity?

But I guess at that point, what is velocity?

Last edited: Sep 14, 2015
11. Sep 14, 2015

### WhatIsGravity

Can negative momentum be expressed as Hawking radiation?

12. Sep 14, 2015

### vanhees71

Is the behavior of a photon at/beyond the event horizon well defined? I don't have a clue about this. QFT in curved (background) space times is not a simple topic in general, let alone what happens close to event horizons. Perhaps the experts here have references concerning this topic?

13. Sep 14, 2015

### bahamagreen

Does it count if the photon is right circularly polarized? Its spin angular momentum's vector points in the opposite direction of its propagation direction...

From Wiki; "Spin angular momentum of light" - When a light beam is circularly polarized, each of its photons carries a spin angular momentum of [plus or minus h] , where [h] is the reduced Planck constant and the sign is positive for Left and negative for Right circular polarizations (this is adopting the convention most commonly used in optics). This SAM is directed along the beam axis (parallel if positive, antiparallel if negative). The above figure shows the instantaneous structure of the electric field of left (L) and right (R) circularly polarized light in space. The green arrows indicate the propagation direction.

14. Sep 14, 2015

### vanhees71

I consider the notion of "spin" for the electromagnetic field as ill defined. This is also clear from the representation theory of the Poincare group, which is relevant to define the canonical observables of quanta. Only the total angular momentum of a photon (again it's a Fock state in QED, not some semiclassical concept, which is not applicable to the case of photons) has a clear defined physical sense. This also becomes clear in the Wikipedia article, where the spin is defined as a gauge-dependent quantity. Thus, it's a priori clear that this is an unobservable quantity.

15. Sep 14, 2015

### exmarine

You guys are way ahead of me, obviously. One of the things that provoked my question was Feynman rambling on and on about all the wondrous things that photons do. For example, “the nucleus keeps its electrons close by, by exchanging photons with its electrons”. I don’t feel like looking up an exact quote right now, but that’s pretty close. Well, doesn’t that require that those exchanging photons transmit negative momentum to produce attractive forces?

Thanks for all the responses.

16. Sep 14, 2015

### ZapperZ

Staff Emeritus
If this was the impetus for you to ask this question, you should have started with this and not simply throw a question which really isn't relevant at all!

What Feynmann is talking about is the physics of Quantum Field Theory, QFT, in which ALL interactions are mediated via an exchange of a virtual particle. He was demonstrating via a "cartoon picture" of the electromagnetic interaction, which is mediated by virtual photons. (The other interactions: Weak - W and Z vector bosons; Strong - gluons.)

There is no transmission of "negative momentum" here because what is relevant is the nature of the coupling between the particle and the quantum field. What you need to understand more fundamentally is the physics describing QFT.

Zz.

17. Sep 14, 2015

### exmarine

PS. Oh, and I can work my way through the derivation of Compton’s experiment with the momentum magnitude; p = –E/c instead of +E/c. I believe that produces his final equation with (1+cos) instead of (1-cos). The only “data” I have from his experiment is a crude sketch in one of my physics books (Alonso & Finn) which does not appear to support (1+cos). But I wonder what the actual data looks like. From long engineering experience, I know that authors can consciously or unconsciously discard data points that do not fit their understanding. So does anyone have access to his actual results, or to any later experiments of that type? Also, was he able to capture and characterize the ejected electrons, etc.?

18. Sep 14, 2015

### Guido Diforti

in Einstein's own words:

"[...]In every elementary process of absorption or emission a momentum of absolute magnitude $\frac{h \nu}{c}$ is transferred to the molecule "
The momentum is in the direction of the movement of the photon during the absorption and in the opposite direction during emission.

19. Sep 14, 2015

### bahamagreen

Do you think he meant that the momentum is applied at the center of molecular mass (to impart a linear motion) or applied to the electron as a lateral component of the molecule (to impart an angular motion)? In either case, how does pressing or pulling the electron make the whole molecule change momentum? Can the answer be the subsequent exchange of more particles which is not a kind of nested circular argument?

As far as the Op's question (How does particle exchange result in attraction?), doesn't the Uncertainty Principle suggest that the localization of the point of exchange receipt allows a wide margin for the attributes of the momentum, perhaps including its direction - even pointed back to the source of exchange?

20. Sep 15, 2015

### vanhees71

Again, photon energy and momentum are transferred according to energy-momentum conservation in the process in question. I don't know, where you got this quote by Einstein from. Please point us to the source, so that we can see, in which context it was made.