0
$\begingroup$

You say virtual photon caries only momentum and changes only momentum of the particle. Momentum $p$ of electron is $v*m$. Energy of moving electron is $E=\dfrac{mc^2}{\sqrt{1-v^2/c^2}}$, which means, changing velocity = changing energy. So, if You say that virtual particles do not change energy, but change momentum, You mean that it change the mass?

I didn't notice but energy depends on mass too, I forget this. Then virtual photon actually changes energy(and hence constant magnetic field changes energy)?

$\endgroup$
  • 2
    $\begingroup$ Virtual particles don't change total energy, which is conserved, but a charged particles in an electric field can certainly gain kinetic energy. Side note: If you're going to use the relativistic formula for energy, it probably also makes sense to use the relativistic formula for momentum as well. $\endgroup$ – Ricky Tensor Sep 7 '18 at 19:47
  • $\begingroup$ @RickyTensor, so virtual photons change energy of particle? $\endgroup$ – user205695 Sep 8 '18 at 8:59
  • $\begingroup$ Virtual particles don't do anything physically. They're just tools to handle higher order perturbation theory. Just conserve 4 momentum at every vertex and impose the condition $p_{total \ in}^\mu = p_{total \ out}^\mu$ to know what happens to the scattered particles. $\endgroup$ – Avantgarde Sep 9 '18 at 12:36
  • $\begingroup$ @Avantgarde, this is actually what am I confused of. How does fixed charge can always create forces around it? It should always loose energy, no? $\endgroup$ – user205695 Sep 9 '18 at 13:12
0
$\begingroup$

Changing momentum does not necessarily mean changing energy. It is possible for a virtual particle to change a particles momentum without changing its energy: Momentum is mass times velocity, but velocity is direction as well as speed. If momentum changes then by definition there has to be a force: F=dp/dt. So the virtual particle has then to exert a force. However, provided that force acts to change momentum only by changing the direction of motion of the particle; neither speeding it up or slowing it down, then no work is done on the particle; therefore no energy is transferred to or from it.

$\endgroup$
  • $\begingroup$ First of all, about "change only direction". I was wondering about a good example - circular motion. As far, as I understood correct(that probably wrong) there is only accelerated circular motion, I mean if something moves in a circle, it always accelerates, but in the same time the motion, is constant. Am I right? $\endgroup$ – user205695 Sep 8 '18 at 19:41
  • $\begingroup$ Circular motion is a good example and you are correct it implies acceleration; F=dp/dt=ma; if momentum changes it must be accelerating. But a=dv/dt so velocity cannot be constant when something is accelerating but can be changing due to a change in direction only, which is the case in uniform circular motion. $\endgroup$ – Jeff Storry Sep 8 '18 at 20:11
  • $\begingroup$ Hmm, wait a moment, v is dx,y,z\dt. If we considering one-dimensional case, what would happen? $\endgroup$ – user205695 Sep 8 '18 at 20:17
  • $\begingroup$ If the force acting and the velocity of the particle were both in same dimension, the particle would now have to change momentum and energy by changing speed and therefore kinetic energy. It was only stated that changing momentum does not necessarily ensure changing energy and this applies when velocity is changed by changing direction only. It can be added this requires the force to act perpendicular to the velocity direction, as is the case in uniform circular motion. $\endgroup$ – Jeff Storry Sep 8 '18 at 20:50
  • $\begingroup$ Well, okay, then what about Coulomb field?1) I didn't understand, virtual photon can change momentum, or only direction of speed? 2) Perpendicular classic force(like magnetic) = virtual photon with momentum perpendicular relative to particle momentum? 3) If yes, what if photon will be real and with perpendicular $p$, will it change energy and accelerate? $\endgroup$ – user205695 Sep 9 '18 at 6:48
1
$\begingroup$

Let us clarify the terminology, keeping in mind that we are discussing particle physics which is modeled quantum mechanically and obeys special relativity rules.

Particles are modeled by their four vectors, and the "length" of the four vector is the invariant mass characteristic of the particle. What is conserved are the energy and momentum between the incoming particles and the outgoing ones in an interaction.

virt

Virtual particles exist only within the Feynman diagram representation of the interaction of elementary particles, based on quantum field theory. In the above diagram the particle labeled "virtual" is an internal line with a variable four vector as constrained by the limits in the integral which the diagram summarizes. It is the incoming and outgoing momentum and energy that are conserved. The virtual "particle" is off mass shell, and it is called a "particle" because it carries the quantum numbers of the name, but not the characteristic mass.

( If you want to see how a crossection is calculated this way see here, example 6)

So it has no meaning to ask about an instantaneous value of the energy momentum of the virtual particle, as it scans the available phase space under the integral.

BTW the relativistic mass is not used when discussing particle physics seriously. It is only useful for thought experiments with space ships :). In particle physics, where virtual exchanges exist in the models, one uses four vectors and their invariant masses.

$\endgroup$
  • $\begingroup$ I little bit changed the title, maybe You didn't understand what did I mean(or I didn't understand what did You mean). However, Jeff almost answered but I have a couple of question, that He didn't answer yet. If You can, answer them. $\endgroup$ – user205695 Sep 9 '18 at 12:16
  • $\begingroup$ 1) I didn't understand, virtual photon can change momentum, or only direction of speed? 2) Perpendicular classic force(like magnetic) = virtual photon with momentum perpendicular relative to particle momentum? 3) If yes, what if photon will be real and with perpendicular p p , will it change energy and accelerate? $\endgroup$ – user205695 Sep 9 '18 at 12:16
  • $\begingroup$ About first one - considering the link, You've gave with 4-vector, I assume, when we say that only direction of momentum changes, we mean that, for example $p_x$ was 10, $p_y$ was 0, and now $p_x=0,p_y=10$ - summary, momentum didn't change. This means changing only direction? $\endgroup$ – user205695 Sep 9 '18 at 12:21
  • $\begingroup$ I am saying that virtual particle's four vector changes are infinite , under the integration which is necessary to define a virtual particle four vector. It is a continuous function . It has no meaning to ask of change $\endgroup$ – anna v Sep 9 '18 at 14:25
  • $\begingroup$ Okay, rather I do not understand You. "It is a continuous function", why vector becomes a function? $\endgroup$ – user205695 Sep 9 '18 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy