Equivalence Principle

I've two questions. The first is likely trivial, it must be something physicists know, something their education teaches them, something about saying there is no difference, when it doesn't necessarily mean there is no difference, but that there is no difference that makes any difference. The example that puzzles me most is the Einstein thought experiment that pretty baldly states that if you're in a box, like an elevator, there is no way to tell the difference between being at rest on the earth's surface and being subjected to 1G acceleration in a microgravity environment. Since I can easily think of two ways (I'm certain there are other ways), with proper instruments, to tell the difference from inside the box, I wonder what "no difference" means.
- Rick Holcomb (age 58)
Austin, TX, US

We'd be fascinated to hear what your ideas are. One can make a conventional choice between the two possibilities if one can peek outside the box. If a large planet is sitting under the box, it's convenient to call the effect gravity. If instead an active rocket is sitting under the box, it's convenient to call it acceleration. However, choices of coordinate systems obeying the laws of physics are not limited to those conventional picks.

Mike W.

(published on 07/08/2008)

I have a couple ideas: 1) Look at whether an electric charge radiates. For an electric charge sitting on the surface of the earth does not radiate energy (otherwise we could build a 'free energy' power plant around this), yet an electric charge accelerated on a rocket (or even just an electrical wire) must radiate according to Maxwell's equations. 2) Because the vaccuum state itself spontaneously broke the symmetry of the early universe, doesn't the non-zero expectation value of the Higgs in vacuum allow us to determine our acceleration with respect to the vacuum itself? The density of the higgs field will be changing in time for an accelerating observer, just like an observer would see a change in electron density if he was accelerating through a uniform gas of electrons. The difference here is that we don't need to see any 'outside bodies' (like the electron gas in that example, or say the 'cosmic microwave background'), instead the experiment would depend only on the properties of the vacuum itself! The properties of vacuum locally, IN the box. I have asked these questions to people before, and no one was sure. One prof tried to do a quick literature search for me (regarding the first one), and said there appear to be conflicting answers in publications, so he wished not to comment. And no one I asked felt comfortable commenting on the second one at all. There is no way a beginning college student is finding 'flaws' in any physics here, so I am well aware I must be making a mistake. But I can't learn if everyone is afraid to try to answer because they may be wrong. So please help!
- John Holcomb (age 21)
Austin, TX, US

John- Are you sure you don't want to transfer to the University of Illinois?
I'm passing your questions on to some colleagues and will post their answers when received.

Meanwhile, I've heard the first issue discussed occasionally, and am sure that the equivalence principle still works but forget how the argument about the integral of the radiated fields over time works.  I believe that the second question has a simpler answer, since you're really asking about the changing density due to changes in relative velocity with respect to the medium. Unlike the electron gas or the microwave background fields, however, the vacuum fields can be Lorentz invariant, i.e. look exactly the same regardless of nominal velocity.

Mike W.

Aha, a colleague confirms my answer to the second issue and writes of the first issue:

This problem is generally considered to have been solved by Boulware:
“Radiation from a uniformly accelerated charge”, Annals of Physics 124, 169-187 (1980).

His abstract says:
The electromagnetic field associated with a uniformly accelerated charge is studied in some detail.  The equivalence principle paradox that the co-accelerating observer measures no acceleration while a freely falling observer measures the standard radiation of an accelerated charge is resolved by noting that the radiation goes into a region of spacetime inaccessible to the co-accelerating observer.

However, this has been contested by Parrott:  , published in Found.Phys. 32 (2002) 407-440.

I think that the issue concerns the end effects.  "Uniformly accelerated" requires an infinitely long time interval in both the past and future.

(published on 01/25/2010)