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I'm not a physicist, but my understanding is that electromagnetism (including attraction between opposite charges) is mediated by the photon, and gravity is probably (hypothetized to be?) mediated by the graviton.

I'm curious how that works from the point of view of the conservation of momentum. My naive imagination is, if a particle leaves A to the direction of B, doesn't that mean that A would have to change its momentum to the other direction (away from B)? And when B absorb this particle coming from A, shouldn't B now change its momentum to the same direction (away from A)?

How come is it that in the case of gravity and electromagnetism, A and B move towards each other as a result of this interaction?

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  • $\begingroup$ How I wish virtual particles stopped being a thing. $\endgroup$ – Javier Mar 21 '15 at 21:14
  • $\begingroup$ Required reading: profmattstrassler.com/articles-and-posts/… $\endgroup$ – Alfred Centauri Mar 21 '15 at 23:16
  • $\begingroup$ In classical electrodynamics you have to give to the fields a momentum and an energy, for the laws of conservation to hold. $\endgroup$ – Constantine Black May 11 '15 at 6:45
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If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear.

For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the equations of motion:

$$m_x \ddot x = F(x, y)$$ $$m_y \ddot y = -F(x, y)$$

For the time derivative of the total momentum we get:

$$\partial_t P = \partial_t (p_x + p_y) = \partial_t (m_x \dot x + m_y \dot y) = m_x \ddot x + m_y \ddot y = F(x, y) - F(x, y) = 0$$.

That is the total momentum is conserved.

If we consider that the fields causing the forces propagate (and thus the forces are not instantaneous) we have to consider the momentum of the fields and can write local equations for momentum conservation.

Now: Do not take the virtual particle thing too serious. They are in many ways just mathematical artefacts of how we compute things in quantum field theory (so called perturbation theory). Most importantly, do not confuse them with some macroscopic particle. Rather they are "packets" of waves. Furthermore each elementary process conserves momentum (techspeak: the momentum is conserved at all vertices of a Feynman diagram)! As they are a computational device, the virtual particles do not follow the usual rules of propagating particles, but even if a virtual particle starts from A with a moment away from the particle B, it still can reach B and there interact and give B the momentum carried away from A (thus conserving the total momentum).

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I will bring an example from classical electrodynamics.

In EM(electromagnetism) you have to consider that the fields(electrical and magnetic) also have energy and momentum. A classical example is to apply the third law of Newton(each action has an equal counter-action) to two moving charges. Then you'll conclude that the third law does not hold-thus the conservation of momentum does seems not to hold. To save the conservation law you have to assume that the fields also have momentum.

In a more mathematical viewpoint, for the momentum we begin by calculating the total electromagnetic force on volume V containing charges. We can prove that: $$f=ε_0 [(\nabla \cdot E)E + (E \cdot \nabla ) E ]+{1 \over μ_0}[(\nabla \cdot B)B + (B \cdot \nabla ) B ] - {1 \over 2}\nabla(ε_0 |Ε|^2 + {1 \over μ_0}|B|^2)-ε_0 {\partial \over \partial t}(E \times B) $$ where E,B the electric and magnetic field.By introducing the Maxwell electromagnetin tension tensor: $$T_{ij}== ε_0 (E_i E_j -{1 \over 2}δ_{ij}E^2)+{1 \over μ_0}(B_i B_j -{1 \over 2}δ_{ij}B^2) $$ we have: $$f= \nabla T -ε_0 μ_0 {\partial S \over \partial t} $$ and S is the Poynting vector $S={1 \over μ_0 }(\bar E \times \bar B) $

Why all this mathematics? We can show now from Newton's second law the conservation of momentum. The second law states that the force on an object is equal the time derivative of it's momentum, and we hsve just found the force of the electromagnetic field. Thus we have: $$f={d p_{mechanical} \over dt} \rightarrow {d p_{mechanical} \over dt}= -ε_0 μ_0 {d \over dt} \int_V Sdτ + \oint T \cdot da$$ Concluding, from here we have that the total momentum of the particles in the volume V is equal to the sum of the momentum saved in the EM field and the momentum over time unit coming out of the volume's surface. That is: $$p_{EM}=ε_0 μ_0 {d \over dt} \int_V Sdτ $$. Thus a change in the momentum of both the particles and the fields is equal to the momentum that the fields bring.

Note that the Poynting vector mentioned above is used to bring forth the conservation of energy(by saying that the field have also an energy.)Finally, even more abstract is the fact that we can define an orbital angular momentum for the fields and prove is a conservative.

Hope this helps.

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If particle A is attracted towards particle B because of an electric charge or gravity, then both will have a momentum towards or away from each other. Thus, momentum is conserved. Grab some paper and a pencil, go on, right now, I'll wait. Make a number line. Put A on -1 and B on 1. They will meet at 0. -1 + 1 = 0, and 1 - 1 = 0. Thus, A's velocity is positive and B's is negative. Momentum is conserved. Try the same with repulsion, it works.

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  • $\begingroup$ Yeah, I understand that, but my question is how does it work with the fact that there is an intermediate particle? $\endgroup$ – user69715 Mar 21 '15 at 21:04
  • $\begingroup$ Oh, if you are talking about an answer pertaining to quantum field theory, then it does not really work with the whole throwing a photon model. In the end quantum field theory is more complicated $\endgroup$ – Jimmy360 Mar 21 '15 at 21:10
  • $\begingroup$ There is a good boomerang analogy Think of the photon not as an ball, but as a boomerang. @user69715 $\endgroup$ – Jimmy360 Mar 21 '15 at 21:14
  • $\begingroup$ If you consider the fact, that the action of electromagnetism and gravitation is not instantaneous, this explanation is not sufficient. The field itself has to carry momentum. $\endgroup$ – Sebastian Riese May 10 '15 at 23:29

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