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How do we know values for a true vacuum?

  1. Nov 18, 2005 #1
    This isn't a homework question or anything, it's just been bothering me as to how we know values for a true vacuum. From my understanding, free space is another term for a true vacuum. If this is correct, how were we able to attain values for the permittivity and permeability of free space if it is unattainable in a laboratory?
  2. jcsd
  3. Nov 19, 2005 #2


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    to start with, at the present time, in SI, the permeability of free space, [itex] \mu_0 [/itex], and the speed of E&M propagation (otherwise known as "the speed of light"), [itex] c [/itex] are defined. consequently the permittivity of free space [itex] \epsilon_0 [/itex] and characteristic impedance of free space [itex] Z_0 [/itex] (that depend solely on [itex] \mu_0 [/itex] and [itex] c [/itex]) are defined. while it hasn't always been the case that the speed of light has been defined (the meter and second were defined independently of each other then) now the meter is defined to be the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second. (doesn't get to your vacuum question, hold on a bit.)

    however the permeability of free space has always been defined to be [itex] \mu_0 = 4 \pi \times 10^{-7}[/itex] because the unit of electical current, the ampere was defined to be such a current that, "if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to [itex] 2 \times 10^{-7}[/itex] newton per meter of length." because of that definition of the ampere, and from the pre-existing definition of the second, the unit of charge, the ampere-second or coulomb was defined. that unit of current or of charge happens to be defined just so that the permeability is [itex] \mu_0 = 4 \pi \times 10^{-7}[/itex].

    you might want to check out the definitions, "right from the horse's mouth": http://physics.nist.gov/cuu/ there's some nice history of the SI units there, too.

    the relationship of these four constants are:

    [tex] c^2 = \frac{1}{\mu_0 \epsilon_0} [/tex]


    [tex] Z_0^2 = \frac{\mu_0}{ \epsilon_0} [/tex]

    so you can see once [itex] \mu_0 [/itex] and [itex] c [/itex] are defined, so also is [itex] \epsilon_0 [/itex] and [itex] Z_0 [/itex].

    now at one time (before the 60s) the meter was defined independently from the second and from [itex] c [/itex]. in fact, they didn't know [itex] c [/itex] precisely and had to measure it, with their predefined meters and seconds. Albert Michaelson did a famous experiment (about a century ago) with a rotating mirror system and a mirror mounted on a mountain 22 miles away to do that measurement. but that was [itex] c [/itex] in air, not a vacuum.

    but you can set up an interference experiment with light of a known color (or wavelength) where light in air is interfering with light passing through a vacuum chamber. because even though the distance traveled by both rays of light is the same, the ray through a vacuum arrives before the ray in air, and there is some wave mismatch and light and dark interference patterns show up because of "destructive" and "constructive" interference. as air is slowly let into the vacuum chamber, the interference pattern moves and one can count how many ridges of intereference pass by and from that, they can measure the number of wavelengths the "fast" light was ahead of the "slow" light and from that determine the ratio of the two speeds of light and from the speed of light in air (Michaelson) you can compute a speed of light in a vacuum. i dunno how they did it on the earth otherwise, but in space, they could have measured it directly.

    however, eventually they redefined the meter so that [itex] c [/itex] must always be 299792458 m/s.

    check out http://en.wikipedia.org/wiki/Speed_of_light#Measurement_of_the_speed_of_light . wikipedia and google are your friends.
  4. Nov 19, 2005 #3
    Thanks for that lengthy and informational post! So, is the speed of light given now that of a TRUE vacuum, or one of a near vacuum? I'm thinking true vacuum.
  5. Nov 19, 2005 #4
    Your question might have had an answer in the past but if quantum physics is correct (most believe it to be), then a `true vacuum' cannot exist. You may want to familiarize yourself with zero point energy (ZPE), the Casimer Effect, and the accelerating expansion of the universe.
  6. Nov 19, 2005 #5
    Oh yes, I am aware that a vacuum cannot be truly empty due to the presence of ZPE (virtual particles and such). I was just wondering, putting aside effects like that, would we have found values for a true vacuum state?
  7. Nov 20, 2005 #6
    I understand your question and will try to respond intelligently with the caveat that I enjoy reading about physics but I have very limited knowledge.

    We will need to consider the propagation velocity of an individual photon to avoid getting into a phase and group velocity discussion. We will need to imagine a region in space wherein gravity fields are somehow nulled out or small enough to be ignored. We will have to assume certain quantum effects do not occur during our measurement period; for instance the photon will not propagate as a positron-electron pair and thus travel briefly below C. We will need our region of space to have less density than `normal’ space to reduce the probability of interaction with virtual particles.

    The last requirement I think can be obtained by having the proton propagate between two conductive planes i.e., use the Casimer effect to decrease the energy density or increase negative energy. To obtain a true vacuum, the distance between the plates must be reduced to the Plank limit wherein our `laws’ no longer apply. To avoid that conflict, we will keep our gap at a femtometer or so.

    In that environment, I think the photon will have a higher velocity than it will in the vacuum of normal space.

    I think above violates SR as it establishes an impossible reference plane. I think your question cannot be answered and I have only demonstrated my lack of knowledge by trying.
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