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Why is the speed of light $c$ used to determine the value of permittivity in free space?

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Qmechanic
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  • This has nothing to do with mathematics. Try the physics stack exchange. – copper.hat Jan 11 '24 at 03:01
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    In Gaussian units $\epsilon_0$ and $\mu_0$ don’t even exist because there is no need for them. This tells you that they are simply an unfortunate artifact of SI units and have no physical significance, despite their fancy names. – Ghoster Jan 11 '24 at 04:24
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    @Ghoster Your comment seems like an oversimplification. I sometimes work in a system of units where c=1 and G=1, so they don't exist in the sense that they never appear in any equation, but that does not mean that the speed of light and the gravitational constant have no physical significance. – David Bailey Jan 11 '24 at 06:17
  • @DavidBailey The particular values of $c$ and $G$ have no physical significance, since a choice of units can give them any value you like. Only the values of dimensionless ratios have physical significance. The only physical significance of $c$ is that it isn’t zero and it isn’t infinite. The statement “$c$ is finite” has, of course, amazing significance. – Ghoster Jan 11 '24 at 06:32
  • @DavidBailey What do you consider to be the “physical significance” of $\epsilon_0$? I consider its significance to be making SI units work, which doesn’t seem physical to me. There is no reason charge or current needs to be a base quantity along with mass, length, and time. – Ghoster Jan 11 '24 at 06:47
  • @Ghoster Yes, I certainly agree that the particular values of $\epsilon_0$ and $\mu_0$ are arbitrary since they depend on the unit system, but that is different from saying that $\epsilon_0$ and $\mu_0$ have no physical significance. My point was just that if a quantity disappears from formulae in a certain system of units, this does not mean that the quantity has no physical significance. What counts as "physical significance" is a discussion outside the scope of this question, but I expect opinions might differ. – David Bailey Jan 11 '24 at 17:13
  • @ghoster Your comment does not answer the question – Mark Viola Jan 11 '24 at 18:01
  • @MarkViola That’s why it’s a comment and not an answer. – Ghoster Jan 11 '24 at 18:01
  • @ghoster Your comment seems "non-physical." ;-)) – Mark Viola Jan 11 '24 at 18:03
  • @DavidBailey, *all* $c$ or $G$ or $\hbar$ or $\epsilon_0$ need to be are: 1. real, 2. positive, 3. finite . As long as these constants are all of real, positive, and finite, there is no physical significance what their values in whatever units are. As long as the 26 or so dimensionless constants stay unchanged, the rest of reality will scale accordingly and we will never know the difference. Light will travel 299,792,458 "meters" (as we define the meter) in the time we call a "second". – robert bristow-johnson Feb 24 '24 at 20:55

2 Answers2

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Why is the speed of light $c$ used to determine the value of permittivity in free space?

$$ \epsilon_0 = \frac{1}{\mu_0 c^2}\;. $$

This is a definition.

In SI units, Maxwell's equations can be reformulated into a wave equation (like $\frac{d^2\vec E}{dx^2}-\epsilon_0\mu_0\frac{d^2\vec E}{dt^2}=0$), where the parameter $\frac{1}{\epsilon_0 \mu_0}$ has units of speed squared. We call this speed $c$.

So, the squared speed is: $$ c^2 = \frac{1}{\epsilon_0 \mu_0} $$

hft
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The speed of light $c$ appears in this relation because of how the SI system of units is defined based on electromagnetic theory, where the permittivity and permeability of free space are:

$$\epsilon_0=\frac{e^2}{4\pi\alpha\hbar c} \qquad \mathrm{and} \qquad \mu_0=\frac{4\pi\alpha\hbar}{e^2c}$$

so

$$\epsilon_0 = \frac{1}{\mu_0 c^2}$$

This last relationship only depends on special relativity and is defined to be exact in the most recent update of the SI System of units. In the current SI system, the speed of light $c$, elementary charge $e$, and reduced Planck constant $\hbar$ all have defined values, so experimental measurements of $\epsilon_0$, $\mu_0$, or $\alpha$ are all equivalent. Direct macroscopic measurements of of $\mu_0$ or $\epsilon_0$, e.g. with a current balance or a parallel plate capacitor, are orders-of-magnitude less accurate than the best measurements of the fine structure constant $\alpha$, so the values $\epsilon_0$ and $\mu_0$ are calculated from the measured value of $\alpha$ and the defined values of $c$, $e$, and $\hbar$.

The answers to "Why does the vacuum even have permeability and permittivity?" are also relevant.

David Bailey
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    Using Planck’s constant to try to explain $\epsilon_0$ and $\mu_0$ to students is theoretically misguided, regardless of modern-day measurement considerations. Classical electromagnetism does not need quantum mechanics to make sense. – Ghoster Jan 11 '24 at 18:27
  • @Ghoster I interpret the question not as a request for a general explanation of $\epsilon_0$ and $\mu_0$, but why $c$ appears in the formula used to determine the actual value of $\epsilon_0$. If the OP just wants to know where $\epsilon_0=1/\mu_0 c^2$ comes from, then referencing classical electromagnetism is sufficient, but to explain how the actual value of $\epsilon_0$ (currently $8.8541878128(13) \times 10^{-12 }$ F/m) is determined requires the actual formula used, which does include Planck's constant. – David Bailey Jan 11 '24 at 19:17