Is it possible to have a system of units in which both $\frac{1}{4\pi\epsilon_0}$ and $c$ equal 1?
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1$\begingroup$ I think if you correctly define the units of $\mu_0$ and those are your only constraints then this is definitely possible. $\endgroup$– PaulCommented Aug 27, 2015 at 17:30
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$\begingroup$ How do I know if I've correctly defined $\mu_0$? My point is exactly to "eliminate" the $\frac{\mu_0}{4\pi}$ in biot savart law, which disappears in this system. $\endgroup$– TMGCommented Aug 27, 2015 at 17:56
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1$\begingroup$ Please note that while you can choose a unit system in which various constants have value 1, it does not make sense to say that an equation is written in any particular system of units. When people write e.g. Schrodinger's equation as $i \partial_t |\Psi\rangle = H |\Psi\rangle$ claiming that they've "set $\hbar = 1$", what's really going on is that they've redefined the symbol $H$ to mean "the Hamiltonian divided by $\hbar$". This common abuse of language in physics is an endless supply of confusion and I wish people would stop doing it and just say what they're really doing. $\endgroup$– DanielSankCommented Aug 27, 2015 at 18:00
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$\begingroup$ To avoid the confusion described in my previous comment, I like to define $\hat{\Omega} \equiv \hat{H}/\hbar$. Not only is this more explicit about where the $\hbar$ went (thus making it possible to understand the dimensions of the quantities in my equations), it's also more descriptive because $\hat{H} / \hbar$ is an angular frequency. $\endgroup$– DanielSankCommented Aug 27, 2015 at 18:02
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1$\begingroup$ @TMG You say "of course" but this issue confused people all the time. It comes up on this site and the chat room not infrequently. In any case, the point is that if you claim that $\hbar$ is magically equal to 1 then the dimensions of the equations make no sense. We must distinguish between units and dimensions; they are not the same thing and conflating their meaning leads to confusion. $\endgroup$– DanielSankCommented Aug 27, 2015 at 18:29
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2 Answers
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In any unit system you want to define, you can set as many constants to 1 as you want so long as all of the constants are independent of each other. That is, the units for any one normalized constant cannot be obtained by manipulating the other normalized constants. That said, you cannot set numerical constants to 1. $\pi$ is always $3.141592.....$ because it is just a number, not a physical constant.
For instance, you could set $c=1$ and the Planck length, $\mathscr{l}_p$, to 1 (which is the same as setting $\sqrt{\hbar G}$ to 1 if $c$ is already 1). However, after doing so, you could not normalize (set to one) any other physical constant with units of $[\text{Length}]^m[\text{Time}]^n$, because any such units could be derived by manipulating $c$ and $\mathscr{l}_p$.
So in practice, you could define a system where both $c=1$ and $\frac{1}{4\pi\epsilon_0}=1$. This would constrain the value of $\mu_0$, however.
N.B. While you can arbitrarily define unit systems like this, sometimes it's best to step back and ask yourself if that is at all useful. I can define a unit system where $25m/s=1$ and $80kg=1$, but there's no benefit in doing so. If you're going to define a different unit system, make sure it is actually useful first.
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$\begingroup$ What would be the consequences of defining the value of $c$ and $\frac{1}{4\pi\epsilon_0}$ and $\mu_0$ ($=4\pi$ in this system)? $\endgroup$– TMGCommented Aug 27, 2015 at 17:59
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$\begingroup$ @TMG I don't understand the question. If you define $c$ and $\frac{1}{4\pi\epsilon_0}$ to be 1, the consequence is that $\mu_0=4\pi$ is a necessary constraint. $\endgroup$– JimCommented Aug 27, 2015 at 18:03
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When talking about a unit system, one thing to consider is that when you "set a constant to $1$", what you are really doing is expressing that constant in units such that the value of the constant is 1 (of those units). So, for example, $c = 299792458 \, \text{m}/\text{s}$, but if I choose units $d$ (distance) and $t$ (time) such that $1 d/t = 299792458 \, \text{m}/\text{s}$, then I can say that $c = 1 d/t$.
Now say that I want to define the Planck length, $1.616 \cdot 10^{-35} \text{m}$, to be 1 as well. Even though I already have a distance unit $d$, I can still do this because the only constraint I have is on the ratio between the distance and time units. However, once I do that, I now have 2 constraints and 2 units between them, meaning that I cannot set the orbital period of Earth to be equal to 1 time unit because I've defined both $d$ and $d/t$, and thus I've defined $t$ as well.
The same goes for your system of units: you can define $c = 1$, thus fixing the ratio of the dimensions $$\frac{\text{distance}}{\text{time}} \, ,$$ and you can set $1 / (4\pi\epsilon_0) = 1$, thus fixing the relationship between the dimensions $$\frac{\text{time}^{4}\cdot \text{current}^{2}}{\text{distance}^3\cdot \text{mass}} \, .$$ As you can see, you still have a number of degrees of freedom left to set other constants to $1$. However, because $\mu_0$ is a quantity with units
$$\frac{\text{distance}\cdot \text{mass}}{\text{time}^{2}\cdot \text{current}^{2}} = \left(\frac{\text{time}^{4}\cdot \text{current}^{2}}{\text{distance}^{3}\cdot \text{mass}}\right)^{-1}\cdot\left(\frac{\text{distance}}{\text{time}}\right)^{-2} \, ,$$
$\mu_{0}$ is already expressible as a function of units that have been defined by the previous constraints. Therefore, you cannot set $\mu_0$ to be 1 by redefining any of the units.
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1$\begingroup$ I edited this answer to improve clarity. Please note that 1) you can use the \text{} macro to get normal text in a math environment, 2) punctuation of equations must be inside the double dollar signs in order to show up on the right line, and 3) really long sentences are hard to read. $\endgroup$ Commented Aug 28, 2015 at 7:08