# How is the vacuum energy affected by nearby particles?

1. Feb 28, 2015

### friend

When calculating the dynamics of a particle, we can use Feynman's path integral to determine the observables involved. This integrates over every possible path that the particle can take. We can even use it to calculate the vacuum energy of empty space. Now what happens to that vacuum energy calculation at a particular point if there is a nearby particle. Since the path integral calculates every possible path, some of those paths will be through the nearby particle. So I wonder what happens to the vacuum energy calculation if a nearby particle is introduced? Does it take longer for process than compared to no nearby particle? Does it take less space to get the same effect?

As I understand it, a particle can influence the surrounding vacuum by polarizing the virtual particles and producing a screening effect so that other particles don't feel the full strength of a charge, for example. So I wonder if the reverse is true. Does a nearby particle influence calculations of the vacuum?

2. Mar 1, 2015

### Staff: Mentor

Particles do not have a size, you cannot go "through" a particle.

You can have interactions with this particle, sure, but those are completely different Feynman graphs.

3. Mar 1, 2015

### friend

Yes, I suppose it contributes an impulse at one point, so that I would expect that it does contribute something. I ask because it sounds like it might be a way to explain gravity. Perhaps calculations with nearby particles for any process occurs in a shorter distance and in a longer time than if there were no nearby particles.

4. Mar 2, 2015

### friend

This seems well motivated to me because contraction of space and dilation of time associated with gravity would appear from a different frame to be equal to the same process occurring in a shorter space in a longer time. And Feynman's path integral takes into account all of space and everything round a test point or test particle. Thus the question does the Feynman path integral take into account other particles in its vicinity? I suppose the walls that form the two slits in the double slit experiment might be considered "other particles" being taken into consideration. That would be an extreme example. Can this be broken down to the individual particles of the wall?

5. Mar 2, 2015

### Staff: Mentor

If you want to ask whether that is possible or not, the answer is no.
Then do the calculations, write a paper about it an publish it. This forum is not the right place for wild speculations.

6. Mar 2, 2015

### friend

Yes, I may have to do everything all myself. Not a pleasant prospect. But first I wanted to see if there was any effort by anyone else that has been done. Or more importantly, if there is any no-go theorems preventing such calculations. And since I'm only bringing up questions, I'd appreciate a little more patience. Thank you.

7. Mar 2, 2015

### Staff: Mentor

All calculations of interactions always take self-interactions ("everything vacuum-related") into account. There is nothing new. And certainly nothing that will lead to gravity-like effects.

8. Mar 2, 2015

### friend

As I recall, the strength of gravity is 10-33 times the strength of electromagnetism. And we've only calculated the accuracy of quantum effects within 12 orders of magnitude. I don't remember the exact numbers. I think your dismissal may be premature.

Also, I don't think QFT would even be relevant. As I recall, QFT integrates wrt fields and does not take into account where the particles are, only whether they were created or annihilated. But perhaps we would want to integrate wrt spacetime, since we would be interested in effects at various distances, right? So I guess I'm asking with what accuracy have we calculated QM effects (not QFT)?

9. Mar 3, 2015

### Staff: Mentor

It does not matter how weak gravity is for particles of specific mass if you do not get any mass-dependent, charge-independent effect.

Your whole question only makes sense in the context of QFT. Otherwise there are no path integrals and no virtual particles.
The position does matter if you calculate the interaction between particles.

10. Mar 3, 2015

### Staff: Mentor

I don't see how to make sense out of that question. QFT is a relativistic formulation of quantum mechanics, so the set of correct predictions it makes has to be a superset of the set of correct predictions that we get from non-relativistic quantum mechanics.

This discussion is in some danger of shifting from an explanation of how quantum mechanics is understood to work today to an argument over whether that explanation is correct... if it turns into a discussion of personal theories the thread may be closed.

Last edited: Mar 3, 2015
11. Mar 3, 2015

### friend

I may have forgotten that part of my QFT. Is it possible to summarize how QFT accounts for the position of particles? I thought QFT was assuming particles come in from infinity and leave to infinity. I don't see how that account for where particles are located.

12. Mar 3, 2015

### Staff: Mentor

That still means they can be close together in between for a finite time. And there are ways to handle bound systems like hadrons or atoms as well where the particles do not leave to infinity. It does not make sense to ask for the exact position of a particle at a specific point in time, of course - it is still quantum mechanics.

13. Mar 4, 2015

### friend

I recently watched a video where Leonard Susskind was explaining how gravity might be related to entanglement. So I might be thinking along the same lines when I ask how the wave function of a particle (or any quantum process) might be affected by nearby particles.

This brings me to the question, is quantum entanglement described in the second quantization procedure of QFT? Or is entanglement only part of the first quantization procedure of quantum mechanics?

14. Mar 4, 2015

### Staff: Mentor

Never extrapolate based on attempts to explain science to the general public. That rarely works because all those descriptions are way too simple to be used for anything.

QFT reproduces everything nonrelativistic quantum mechanics can describe. Sure.