Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
  1. Consider a correlation function between two points $A(x_1,t_1)$ and $B(x_2,t_2)$, we need to integrate over paths which could be infinite long. But the time length $(t_1-t_2)$ is finite, so if $A$ and $B$ are the coordinates of one single particle, then all of the paths from $A$ to $B$ should be time-like curves, the maximum length should be $c\times(t_1-t_2)$, which is not infinitely long. It seems special relativity could be violated here.

  2. Consider any loop integration of higher order correction to a Feynman diagram calculation, we have infinitely many off-shell processes and "internal virtual particles", which could be created and annihilated without taking any time. Does this violate special relativity?

Is there any better reason than simply saying these are "internal virtual processes"? They do affect our final real physical observations!

for the second case, one might argue that no particles here but only fluctuating fields? But when, where, and how are those fields created? This picture is not that physically clear to me.

Any better insight?

share|improve this question
    
Related: physics.stackexchange.com/q/18835/2451 –  Qmechanic Feb 11 '12 at 10:00

2 Answers 2

Concerning the path integral, there is no problem with integration over too long paths. Integration is not a particle motion, but is a sum like $A=a+b+c+d+...$. You may plug in the right-hand side anything resulting in zero, for example, $z-z$, but it does not give you the right to consider your $z$ as physical motion. The path integral is a sum of many things that are not related to the physical motion. The fact of using the Lagrangian does not makes each particular term (path) physical. Only sum (integral) makes sense.

Off-shell "processes" are just Fourier-presentations of functions of time and space, each Fourier variable $p_x$, $p_y$, $p_z$, and $\omega$ are dumb independent variables. They are not restricted with any relationship like $p^2-\omega ^2 =m^2$ and their denotions as physical variables are just deceptive.

share|improve this answer

from The Reference Frame:

Another wrong expectation that a beginner could have - and usually has - is that if you allow the summation over all trajectories of a particle, the typical particles will move faster than light most of the time and this will automatically result in a violation of the special theory of relativity. So an overzealous physicist-beginner could argue that the path integral needs to be "regulated" in the political sense. We must manually prevent the particle from moving faster than light, right? That's how the newbie could imagine the propagators in quantum field theory to be evaluated.

However, this expectation is completely incorrect, too. While most of the histories that contribute to the path integral contain points or features of fields that are moving superluminally most of the time, the resulting physical predictions will actually be fully compatible with relativity as long as the action you started with is relativistic.

This requires some calculations and cancellations (between particles and antiparticles, among other things) but if you do it right, you will see that I am right, too. The physically meaningful quantities will end up being consistent with relativity even thought the sum over mostly superluminal histories of point-like particles underlies all the propagators in any Feynman diagram.

No "regulation" of the violent behavior of the path integral is needed. Quite on the contrary. Any "intervention" into the path integral that would drop some histories that fail to obey certain inequalities - that you incorrectly assume must be imposed on the individual basis - will result in a violation of the consistency rules such as the conservation of probabilities.

Relativity only implies and requires that the ultimate testable predictions prohibit the genuine information from propagating faster than light. But the intermediate steps - and the individual histories included in Feynman's sum - can never be harassed by similar conditions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.