Observers in Special Relativity

Question

Suppose I am an observer in Minkowski space, observing a particle traveling with a constant velocity $v$. If I want to calculate the particle's Lorentz factor, given that the particle is traveling at a constant velocity, I can say $\gamma = \frac{\Delta (c\cdot t)}{\Delta \tau}$. My question is, to get a value for $\Delta t$, do I look at my clock, or do I look at the particle's clock?

Answer 1

For any observer observing two events, $E_1$ and $E_2$, the Lorentz factor is:

$$ \gamma = \frac{\Delta t}{\Delta \tau} \tag{1}$$

where $\Delta t$ is the time interval measured by the observer, i.e. $t_2 - t_1$, and $\Delta \tau$ is the proper time difference calculated by the observer. To see this consider the example you give of our observer watching a particle moving with velocity $v$. For convenience we take $E_1$ to be $(0, 0)$ so $E_2$ is $(t, vt)$ where $t$ is the time measured by the observer. The proper time is given by:

$$ c^2\tau^2 = c^2 t^2 - v^2t^2 $$

so substituting into equation (1) gives:

$$ \gamma = \frac{t}{\sqrt{t^2 - \tfrac{v^2t^2}{c^2}}} $$

which simplifies to:

$$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $$

as expected.