# Deriving t=T/√(1-v²/c²)

## Main Question or Discussion Point

you are given that in a certain scenario
-(cT)2 = (vt)2 - (ct)2

then it says Hence:
t=$$\frac{T}{\sqrt{1-v^{2}/c^{2}}}$$

But this doesn't rearrange to that equation, and I don't like taking formulae as fact without seeing them derived or working them out for myself.

Obviously with next to no knowledge about special relativity at the moment, I stand no chance of deriving it for myself so can anyone point me to where it is shown, or maybe even do it?

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Homework Helper
Are you sure it doesn't rearrange to this equation?

I'll show my working...

-(cT)$$^{2}$$ = (vt)$$^{2}$$ -(ct)$$^{2}$$

-(cT)$$^{2}$$ = t$$^{2}$$(v$$^{2}$$-c$$^{2}$$)

$$\frac{-(cT)^{2}}{v^{2}-c^{2}}$$ = t$$^{2}$$

∴ t = $$\sqrt{\frac{-(cT)^{2}}{v^{2}-c^{2}}}$$

I consistently get this from rearranging. Am I just being rubbish at maths?

Yep, I just did the derivation my self, it does work out. You just need to check your algebra.

I've tried and tried and I still get the same... can someone check what I've done above and tell me what's wrong with that algebra?

nicksauce
Homework Helper
Factor out T from the square root. Divide the numerator and denominator by -c^2.

Thanks for that pointer.... can I use a long weekend as an excuse for missing that? nicksauce
Homework Helper
I'm not here to judge :)

Another possible derivation:

$$-(cT)^2 = (vt)^2 - (ct)^2$$

$$(vt)^2 - (ct)^2 = -(c^2 T^2)$$

$$\frac{(vt)^2 - (ct)^2}{(c^2)} = -T^2$$

$$-\frac{t^2 v^2}{c^2} + t^2 = T^2$$

$$t \sqrt{-\frac{v^2}{c^2}+1} = T$$

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

or:

$$-(cT)^2 = (vt)^2 -(ct)^2$$

$$(cT)^2}\\ =\\ (ct)^2 -(vt)^2\\ =\\ t^2(c^2-v^2)$$

$$\frac{c^2T^2}{c^{2}-v^{2}}\\ =\\ t^{2}\\ =\\ \frac{T^2}{(1-v^2/c^2)}$$

$$t = \frac{T}{\sqrt{(1-v^2/c^2)}}$$

Fredrik
Staff Emeritus
Gold Member
...or set c=1. (This is just a choice of units, e.g. times are given in seconds and distances in light-seconds).

$$-T^2=(vt)^2-t^2=(v^2-1)t^2$$

$$t=\frac{T}{\sqrt{1-v^2}}$$

If you absolutely must have a c in the final result, it's very easy to restore the factors of c that you have omitted. T and t are both in seconds, so they don't need any factors of c. (You can of course replace t and T with cnt and cnT respectively, but then you can just divide both sides of the equation with cn to get rid of those factors again). The square root must be dimensionless, so the 1 doesn't need any factors of c, and the v must be replaced by v/c.

I know the o.p.'s specific question has been answered so I hope it isn't too inappropriate to take the thread a bit further to something I find both interesting and illustrative.
The factor from the above terms t/T is the time dilation effect that applies when an observer measures events in a different inertial frame (e.g. a moving clock). This is the Lorentz contraction factor gamma. When v -> c the denominator goes to 0 so gamma ranges from 1 to infinity. I prefer 1/gamma, for this blurb I like to use the letter G for this. G ranges from 0 to 1 which is easier to deal with, and when the above terms are rearranged we can get G^2 = 1 - (v/c)^2. This is the equation for a unit circle.
What it illustrates is the idea that two objects in the same frame coexist at a point on the unit circle, let's say at 1 on the real axis. As one object moves faster relative to this point, it moves in an arc (geometrically speaking) along the circle creating a separation angle.
The relative velocity (the only known speed between the two objects) v/c goes from 0 to near 1 as G goes down toward 0; I think that this factor G represents the 'apparent' change of the moving object's characteristics seen by the observer as G is the orthogonal component of their relative motion mapped onto the observer's axis.
This is directly related to one derivation of the contraction factor using a light clock. As an observer moves relative to the clock, the light's reflective path gets longer due to the same 'separation angle' between the objects (their velocities). It also illustrates that v/c will not reach 1; it can get really close but if it did then the light would never reflect off the light-clock mirror. (this is just an illustration, not an explanation. I'm still working on that.)
Ron