# B Special Relativity & Relativistic Mass

1. Nov 24, 2017

### Mister T

In a particle accelerator energy is being transferred to the particle, and it's accounted for in the kinetic energy of the particle. That is, you transfer energy $E$ to the particle and the particle's kinetic energy increases by $E$.

The only issue you're having is that the relationship between speed and kinetic energy is not what you think it is. You think that as the kinetic energy increases so does the speed. And you're absolutely correct about that. It's just that it doesn't increase in the way that doesn't seem to make sense to you.

$\gamma$ is simply a mathematical function of the speed $v$. The only reason it has a physical implication is that it's used in the formula that relates the kinetic energy to $v$.

2. Nov 24, 2017

### SiennaTheGr8

Momentum: $\mathbf{p} = \gamma m \mathbf{v}$
Energy: $E = \gamma m c^2$

It is true that if something is moving near light-speed and you transfer momentum to it in the direction it's already going, the increase in its magnitude-of-momentum that you measure will be almost entirely in its $\gamma$, and hardly at all in its $v$ (I'm assuming its mass stays constant). At very low speeds, on the other hand, the "roles" of $\gamma$ and $v$ here would be reversed: the increase in $\gamma$ would be negligible, while the increase in $p$ would be almost entirely accounted for by the increase in $v$.

You'd also measure a corresponding increase in the object's energy, and that increase would be entirely accounted for by the increase in $\gamma$, regardless of the object's initial speed.

3. Nov 24, 2017

### jbriggs444

The rate of kinetic energy gain by a car powered by a 100 horse power motor can greatly exceed 100 horse power if the highway is moving at multiple miles per second. This does not violate energy conservation once you consider the force the tires apply to the highway.

4. Nov 24, 2017

### PeroK

You're getting a bit confused by the mathematics. There is only a change in kinetic energy here. There is no gamma store. In fact if you expand the expression for gamma, you get:

$\gamma mc^2 = mc^2 + \frac{1}{2}mv^2 + \dots$

Where the remaining terms in the expansion have increasing orders of $1/c^2$.

From this you can see that the relativistic expression for kinetic energy reduces to the classical expression where $v$ is small compared to $c$.

In any case, there is no need to see relativistic kinetic energy as fundamentally different from classical kinetic energy. They just have different formulas, where one is an approximation of the other.

Last edited: Nov 25, 2017
5. Nov 25, 2017

### nitsuj

Are you able to reword that? I don't understand what you're saying.

6. Nov 25, 2017

### Ibix

Working entirely with Newton - if a car is doing v and increases to v+1 then the kinetic energy increase is m(2v+1)/2. This energy increase can be tiny in a frame where v was initially small or enormous where v was initially large. But the engine has the same power output and the time taken is the same in either frame (Newton, remember). This is not paradoxical, however, as long as you take into account what happened to the road. It also changes speed slightly, and if you include its kinetic energy change then the energy difference before and after the acceleration will work out to 100 hp times the time.

If you neglect the road then energy is not conserved. But that's because you are ignoring part of the scenario, not a fundamental problem with physics.