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So in my QFT course, my professor said that you can set $c$ and $\hbar$ to 1. And he gave us an example:

$$E = mc^{2}$$

And then set $c = 1$:

$$E = m$$

This seems completely ludicrous to me to do. Doesn't it change the result? Why can this be done and why isn't it wrong?

I mean, $E = mc^{2}$ gives you one answer and $E = m$ gives you another, completely different answer!

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    $\begingroup$ Does this answer your question? What exactly are we doing when we set $c=1$? $\endgroup$
    – jacob1729
    Commented Oct 18, 2020 at 13:05
  • $\begingroup$ Rhetorical question: Why can we set $k$ in $F=kma$ to one when it changes the result? Newton's second law does not say that $F=ma$, or even $F=dp/dt$. It says that (emphasis mine) "the change of momentum of a body is proportional to the impulse impressed on the body ..." In other words, it says $F=k\,dp/dt$, where $k$ is a constant. The modern view is that any system of units in which $k$ is not one and unitless is an inconsistent set of units. An even more modern view is than any system of units in which $c$ and $\hbar$ are not one and unitless is an inconsistent set of units. $\endgroup$ Commented Oct 18, 2020 at 17:38

2 Answers 2

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It's easiest to think about this as a matter of units. What is the value of $c$? $3 \times 10^8 \mathrm{m/s}$? $6.7 \times 10^8$ miles/hour? Suppose I invented two new units, the florp (for space) and the zoob (for time). It just so happens that the speed of light in my system is exactly one florp per zoob, so you don't have to keep track of its numerical value. In my system it's fine to write $E = m$, since the $c^2$ never changes the actual number we get at the end.

Now, the important thing to remember here is that at the end you want to make sure that your answer is in the right units, which may require adding $c^2$ or $\hbar$ back in. That doesn't really matter until you're getting to the point of calculating actual physical numbers which -- as you'll see in this QFT class -- is the last part of a long problem. Easier to do all the integrals and algebras without $c$ and then put it back by dimensional analysis later.

Also note that you can only do this to a few constants at a time -- you can't have both, say, the mass of the electron AND the mass of the proton be set to 1.

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If you ask two doctors how much you weigh and one doctor says “you weigh $100\text{ kg}$” and the other doctor says “you weigh $220\text{ lbs}$”, would you claim that they have given you completely different answers? No. They gave you the same answer using different units.

Setting $c$ or $\hbar$ to 1 is simply choosing to use units where those quantities are 1. For example units of years for time and units of light years for distance in the case of setting $c=1$

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