# The relationship between F=ma and E=mc2

1. Jan 6, 2016

### Starlover

I'm wondering about the relationship between F=ma and E=mc2. Is it simply that, at relativistic speeds, E=mc2 replaces F=ma? (much like D = v x t is replaced by the Lorentz Contraction at relativistic speeds)

2. Jan 6, 2016

### Ibix

$E=mc^2$ is a special case, only valid when an object is not moving, of $E=\gamma mc^2$. If you Taylor expand the $\gamma$, the first term is $mc^2$ and the second is Newtonian kinetic energy, $mv^2/2$.

In relativity, $F=ma$ is replaced by a rather more complex expression, which varies between $F=\gamma ma$ and $F=\gamma^3ma$ depending on the angle between the force being applied and the velocity of the object. Note that accelerations are not necessarily parallel to forces.

In other words, the two expressions you asked about don't have a lot in common except that they are both parts of kinematic theories. One describes an energy term that does not change, which is how it got overlooked in the development of Newtonian relativity. The other relates force and acceleration.

3. Jan 6, 2016

### Starlover

Thank you, Ibix! I see I was way off. I'm glad I asked! :)

4. Jan 6, 2016

### pervect

Staff Emeritus
The Newtonian equation, F=ma is replaced in relativity (relativistic dynamics) by F = dp/dt, where p is momentum. So force is the rate of change of momentum with respect to time.

F=dp/dt is valid in Newtonian mechanics as well, so it's valid in both relativistic and Newtonian mechanics.

To anyone who remembers their calculus, this should be a sufficient explanation. I suspect that many PF readers who ask this question don't remember (or haven't yet had) calculus, so the explanation doesn't always seem to "get through" unfortunately..

I'll go through the math in more detail, but understanding the technical points does require one to know/remember their calculus - at least the way I am going to present it. If we start with p = m*v, which is universally true both in relativistic and Newtonian mechanics, we next apply the chain rule for derivatives to simplify the expression. Thus we write dp/dt = (dm/dt)*v + m (dv/dt). When m is constant, dm/dt is zero, the first term disappears, and dp/dt reduces to f = m dv/dt = ma. In relativistic dynamics, p = $\gamma m v$, so $dp/dt = (d \gamma/dt) m v + \gamma (dm/dt) v + \gamma m (dv/dt)$, where $\gamma = 1/\sqrt{1-(v/c)^2)}$

5. Jan 6, 2016

### Starlover

Thank you, Pervect. It was kind of you to take all this time!

6. Jan 6, 2016

### DrStupid

F = dp/dt actually is the Newtonian equation. F = m·a comes from Euler.