How is rest mass $m_0$ in $E=m_0c^2$ related to mass $m$ in $F=ma$?

A particle at rest has energy given by $$E=m_0c^2$$ where $$m_0$$ is defined as its rest mass. So rest mass is essentially a measurement of a particle's intrinsic amount of potential energy.

How is this rest mass $$m_0$$ related to the mass $$m$$ we often use in equations like $$F=ma$$ and $$W=mg$$ where $$F$$ is force and $$W$$ is weight?

It does not seem to me that they should be the same. Since $$m_0$$ is a measurement of intrinsic potential energy of a particle while $$m$$ is a measurement of how much a particle accelerates given a force $$F$$.

To be frustratingly concise, they are the same quantity. In fact, that was the whole genius of Einstein to realize that the inertia of a particle (the "measurement of how much a particle accelerates given a force") is really a measure of its energy. Thus, the title of his $$1905$$ paper: "Does the inertia of a body depend on its energy content?".

Now, the playground for this story is a bit muddled due to one of the most unfortunate notions in the history of modern physics: the relativistic mass. This is the reason you misspelled the Einstein mass-energy relation as $$E=m_0 c^2$$ instead of correctly spelling it as $$E_0=mc^2$$. But, I will come to this point later, for now, I will try to bypass this issue. Anyway, let's dive in.

So, in special relativity, the total energy of a particle (or, simply, the energy of the particle) and the momentum of a particle are given by $$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=mc^2\Big[1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\cdot\cdot\cdot\Big]$$$$\vec{p}=\frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}=m\vec{v}\Big[1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\cdot\cdot\cdot\Big]$$Here, the $$m$$ that I have used is the mass of the particle, or, as people liked to call it in the early days of relativity, the rest mass of the particle. Now, the key feature to notice here is that it is the same $$m$$ in the expression for the energy as it is in the expression for the momentum. Now, when you take the limit in which $$\frac{v}{c}$$ is small and retain the first non-trivial contribution from $$v$$, you get the usual Newtonian relations $$E=mc^2+\frac{1}{2}mv^2$$ $$\vec{p}=m\vec{v}$$Now, at this moment, you can easily see that it is the same $$m$$ in the formula for the Newtonian $$\vec{p}$$ that would appear in $$\vec{F}=m\vec{a}$$. Moreover, you also notice that it is the same $$m$$ in all the four expressions that I have written. In particular, it is the same $$m$$ in $$E=mc^2+\frac{1}{2}mv^2$$ (which contains the famous $$E_0=mc^2$$ where $$E_0$$ means the energy of the particle when $$v=0$$, or, in other words, its rest energy). In Newtonian mechanics, we don't carry along the $$mc^2$$ term in the expression for $$E$$ because, essentially, we assume that $$m$$ doesn't change and thus, $$\frac{1}{2}mv^2$$ alone is a good enough (conserved) quantity which is worthy of being named energy. But, the main point is that it is the same $$m$$ that dictates the inertia (via entering the formula for the momentum) of the particle that dictates its rest-energy (via entering the formula for, well, the rest-energy).

So, in order to preserve the relation that momentum is mass times velocity, people invented the term relativistic mass $$M$$ defined as $$\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}$$ and happily wrote $$\vec{p}=M\vec{v}$$ while being relativistically correct. But, apart from being a linguistic nightmare where you necessarily needed to invent the term rest mass to distinguish the relativistically invariant mass from the made-up relativistic mass, it was also a physically problematic scheme as best explained by Lev Okun in this article. So, now, we only have one mass, the mass, or if you really wish, the rest mass. But, the total energy is not equal to $$mc^2$$ it is only the rest energy which is equal to $$mc^2$$. So, one should only write $$E_0=mc^2$$. With the forbidden relativistic mass, you could simply write $$E=Mc^2$$ where $$E$$ would have the right to be called the total energy and not just the rest energy, but, it was just not worth it!

$$F=ma$$ is a formula in classical mechanics. $$E=m_0c^2$$ is a formula from special relativity and used extensively in quantum mechanics.

In classical mechanics mass is a conserved quantity, that is how the Archimedes principle works. In special relativity $$m_0$$ is the length of the energy momentum four vector for each particle, and what is conserved is energy and momentum. A many particles system has additive four vectors and an invariant/rest mass which will be larger than the addition of the sum of the invariant/rest masses of the constituent particles.

At the limit for c approachng zero the two definitions coincide , i.e. "conservation of mass" effectively holds for the summing of invariant/rest masses.

So, in units, mass is always a mass .

The term "rest mass" and its notation as $$m_0$$ have been extinct among relativists for 50 years. The standard modern notation is that a particle has a mass $$m$$, which is invariant. The quantity $$m\gamma$$, formerly known as relativistic mass, can be called simply $$m\gamma$$, or, in units with $$c=1$$, it can be interpreted as the mass-energy $$E=m\gamma c^2$$. This notation and terminology has been gradually filtering down into textbooks, but we're still waiting for that to happen with popularizations. So in modern terminology, your question would be:

How is mass $$m$$ expressed in a relativistic version of $$F=ma$$?

Force is a concept that is seldom needed in relativity. If we do want to use it, then by analogy with Newtonian mechanics, we can define a relativistic force vector $$\textbf{F} = m\textbf{a},$$ where $$\textbf{a}$$ is the acceleration four-vector and $$m$$ is the mass of a particle that has that acceleration as a result of the force $$\textbf{F}$$. This is equivalent to $$\textbf{F} = \frac{ d \textbf{p}}{ d \tau},$$ where $$\textbf{p}$$ is the mass of the particle and $$\tau$$ its proper time. Since the timelike part of $$\textbf{p}$$ is the particle's mass-energy, the timelike component of the force is related to the power expended by the force. These definitions only work for massive particles, since for a massless particle we can't define $$\textbf{a}$$ or $$\tau$$. $$\textbf{F}$$ has been defined in terms of Lorentz invariants and four-vectors, and therefore it transforms as a god-fearing four-vector itself.

The trouble with all this is that $$\textbf{F}$$ isn't what we actually measure when we measure a force, except if we happen to be in a frame of reference that momentarily coincides with the rest frame of the particle. As with velocity and acceleration, we have a four-vector that has simple, standard transformation properties, but a different $$\textbf{F}_\textbf{o}$$, which is what is actually measured by the observer with four-velocity $$\textbf{o}$$. It's defined as $$\textbf{F}_\textbf{o} = \frac{ d \textbf{p}}{ d t},$$ with a $$d t$$ in the denominator rather than a $$d \tau$$. In other words, it measures the rate of transfer of momentum according to the observer, whose time coordinate is $$t$$, not $$\tau$$ --- unless the observer happens to be moving along with the particle. Unlike the three-vectors $$\textbf{v}_\textbf{o}$$ and $$\textbf{a}_\textbf{o}$$, whose timelike components are zero by definition according to observer $$\textbf{o}$$, $$\textbf{F}_\textbf{o}$$ usually has a nonvanishing timelike component, which is the rate of change of the particle's mass-energy, i.e., the power. We can refer to the spacelike part of $$\textbf{F}_\textbf{o}$$ as the three-force.

One can show that an object moving at relativistic speeds has less inertia in the transverse direction than in the longitudinal one. A corollary is that the three-acceleration need not be parallel to the three-force.

So in summary, ---

• In terms of four vectors, Newton's second law uses mass $$m$$.

• In terms of the three-vectors measured by an observer, Newton's second law cannot be expressed in the Newtonian form using either the mass $$m$$ or the mass-energy $$m\gamma$$.

There is a more detailed treatment of this topic, with examples, in section 4.5 of my online SR book.