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When traveling at the speed of light (v=c), left under the radical you would have 0. This answer would be undefined or infinity if you will (let's go with infinity). The reference time (T0) divided by zero would be infinity; therefore, you could infer that time is 'frozen' to an object traveling at the speed of light.

However, doesn't the same thing happen on our frame here on earth if light travels at c then the gamma factor is gonna be 0, so because time is frozen in our frame ,why do we see photons reaching us ?

user198045
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2 Answers2

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In special relativity, there are no valid reference frames from the "point of view" of a photon. It's not really physical to talk about such a reference frame.

Also, the gamma factor is defined as:

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

For all inertial frames $F^{\prime}$, travelling at speed $v$ with respect to frame $F$ the necessary condition is imposed that:

$v < c$

Note this is a strict inequality (i.e. not $\leq$), as you cannot mathematically or physically have a Lorentz frame with $v = c$, there is no "photon frame of reference".

Thus as $v \rightarrow c$, you have $\gamma \rightarrow \infty$ (note: not $0$), and also it is impossible to have a gamma factor with a value of $0$ the range of $\gamma$ being:

$1 \leq \gamma $

Thus in any real system you get no "freezing of time" or dividing by $0$ in equations.

I hope this clears some stuff up.

LUPHYS
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That is not how it works. To us, time appears to pass much more slowly in a system that is moving rapidly with respect to us. If * you * were part of that system, you would observe the opposite: that we were moving rapidly and that our clocks appear to run more slowly than yours. Time is not frozen in either frame, regardless of their relative speed.

S. McGrew
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