Is there a formula to measure time dilation?

Question

Travelling at what percent of C (speed of light) would you experience 99secs as the moving object to 100secs of a stationary object?

Answer 1

Yes, there is such a formula. It's : t_mov = sqrt [1 - (v/c)^2] t_stat, where t_stat is the elapsed time in the stationary frame, and t_mov is the corresponding elapsed time in the moving frame.

[Note that I carefully say "frame" rather than "object," because that elapsed rate is a pairwise property of all objects moving with the same motion as the object under consideration, one in each separate frame. In other words, there are two frames, a "moving" frame (subscript "_mov") and a "stationary" frame (subscript "_stat"), and these time transformations apply between any two objects, one in one of those frames, the other in the remaining frame.

Nevertheless, a word of warning : These time relationships have to be VERY carefully reconsidered, however, when it's a question of actually "seeing" what a rapidly moving clock is exhibiting, and comparing it to what your own clock is showing.

In that case, there are additional effects combined with these relativistic considerations, esentially those of the usual Newtonian Doppler shifts, or more generally, aberration due to the angle of the relative motion with respect to the angle of light (and therefore information) transmission. (### But see my footnote, below.)

Even physicists who use and write about this every day can forget this in moments of their own mental aberration. It's a case of over-familiarity breeding, if not contempt, a certain laxness in the precision with which they say or write what they really know deep down. Brian Greene, of "Elegant Universe" fame, is a case in point, but then so are many others, even some far more distinguished relativists. There's no substitute for carefully parsing statements about this, whether one is an expert or an amateur.]

Leaving that little morality tale aside, and returning to your question (which I interpret as involving the fundamentally different frame-dependent rates of time elapsing, without additional effects involved in learning about it!), one gets :

(1 - (v/c)^2) = [(t_mov)/(t_stat)]^2 = (99/100)^2 in your case.

Then (v/c)^2 = 1 - (99/100)^2 = 0.0199.

Therefore v/c = sqrt (0.0199) = 0.14106736 ... .

In other words, you would need to be travelling at slightly more than 14 % of the speed of light (14.106736 ...%, if you're a stickler for accuracy) in order for time to appear to be slowed down by 1 %, i.e. for just 99 secs to appear to go by for the moving object when 100 secs goes by for the stationary object.

Live long and prosper.

### FOOTNOTE : There is one situation where such "aberrational effects" can be avoided. That is by having the rapidly moving objects NEVER changing their distance from the stationary observer. How can this be ?! --- By having them move rapidly around the observer in a circle !

This was actually done, in effect, in a classic experiment with highly relativistic muons, at CERN in the 1960s. These experiments very accurately confirmed the correctness of the formula I gave above, to within an astonishing accuracy. The experiment was performed with naturally decaying muons moving at v ~ 09994c (as determined by a bunch of them circling around and passing a fixed detector). Their usual "at rest" decay lifetimes were lengthened (as far as the central observer could measure them) by about a factor of 30, from (roughly) 1.5 microsecs to about 44 microsecs!

These results, so close to the speed of light, where the effect increases hugely for quite small differences in v from c, were accurate to within about a percent --- truly a dramatic and heroic confirmation that cranks who oppose special relativity still choose to ignore, even if they ever bother to learn or read about it.

Answer 2

Yes

Answer 3

So, the times and distances are related as follows:
L = ct'
L2 + v2*t2 = c2*t2 (Pythagorean theorem)

Eliminate L from the equations:
c2*t'2 = c2*t2 - v2*t2
t' = t*sqrt(1-v2/c2)