Expert: Philip Stahl - 8/23/2010

Question
Dear Sir,
I have read about the Time Dialation on the web. As I understood it says that if we move with the speed of light in a spaceship for one day our clock on the space ship will show only 24 hrs but in the mean time 20000 yrs will be past on the Earth.
Now I want to know that Voyeger spacecrafts are more than a light day away from Earth now & radio transmission signals take same time time by the speed of light but in this case why these signals reach earth in real time?
I hope you understood what i want to ask.
thanks & regards,  

Answer
Hello,

First, no one can move at the "speed of Light" - certainly no material object.  The relativistic relation of a moving mass, m, to its rest mass m(0) is, in fact:

m = m(o)c^2 / [1 – {v/c)^2] ^1/2

where m(o) is the rest mass. Now, if the velocity v = c, then:

m = m(o)c^2/ [1 - c^2/c^2]^1/2 = m(o)c^2/ 0 = oo

In other words, the moving mass would have to have infinite inertia, which is to say, infinite mass. (Another way of saying it is precluded is that all the mass in the existing universe would have to be converted into energy and even that would not allow a speed of c!)  So all your numbers about the clock showing 20,000 yrs. and 24 hrs. is pure fantasy.

Now, there are genuine time dilation effects which we can ascertain, and these apply to the particles known as muons.

Consider then, the case of a muon formed high in the atmosphere and travelling at 0.99c for a distance of 4.6 km before it decays: e.g.

muon -> (e-) + neutrino + (anti-neutrino)

How long does the muon survive as measured in its own rest frame, and how far does the muon travel as measured in its own frame? If time dilation applies, we expect that the time in its own rest frame will be significantly longer than for a stationary observer (e.g. on Earth's surface) observing it. We also expect its distance traveled will be much shorter!

Given a proper time, t': and t = t'/ y

where y = [1 - v^2/c^2]^1/2 then t' = ty

So the proper time t' = t, where t = x/c = (4.6 x 10^3 m/s)/ 2.98 x 10^8 m/s) = 1.55 x 10^-5 s

which is the muon lifetime relative to an observer on Earth.

and so: t' = ( 1.55 x 10^-5 s) (o,141) = 2.18 x 10^-6 s or ~ 2.2 us (micro-seconds)

which is the commonly observed lifetime for muons in their frame of reference.

The distance traveled is: L = (4.6 x 10^3 m) (0.141) = 649 m

Using only L, one might suppose the muons would never reach Earth's surface since the path length is too short. But it is precisely time dilation that accounts for the fact a large number DO reach the Earth and are detected.

The same principle as the above can be used to ascertain potential space journeys for any astronauts traveling to other planets at near light speeds. Consider, for example, two astronauts traveling to Proxima Centauri at 4.2 Ly distance. If their craft manages to travel at 0.95c nearly all the way, then:

How much time would elapse on the clocks of Earth observers?

Well, this would be: (4.2 Ly)/ 0.95c = 4.4 yrs.

However, the astronauts would disagree - and compute:

t' =(4.4 yrs.) [1 - (0.95c)^2/c^2]^1/2 = 1.37 yrs.

Re: your question to do with the Voyager spacecraft, I am unclear as to what exactly it is you are asking. Please consider the examples I've given, as with muons, then try to reframe your question in that light.