Confused about issues related to Time

kakram

A second is measured thus:

"the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium atom."

What if the cesium atom itself is moving close to the speed of light? Would that not change the time measured?

Also - at the centre of a black hole .... time stands still.. As one moves away from the black hole ... time starts moving again because less gravity .... if we keep moving away from heavy objects into an area of space where there is NO gravitational pull ... does time then run at *infinity*? Whats the upper limit of how 'fast' time flows?

I am getting a little confused when I think about there being no absolute time. Because even the duration of x periods of an atom - surely that 'duration' will change depending on if a) the atom is moving close to the speed of light, and b) what if it is at the centre of a black hole?

tiny-tim

hi kakram! welcome to pf!
it means as measured in the rest-frame of the atom
(no, time stands still at the boundary of the black hole)

gravitational time dilation = √(g₀₀(clock))/√(g₀₀(observer)),

where g₀₀ = 1 - R/r, r is distance, and R is the radius of the black hole …

so at r = ∞ ("where there is NO gravitational pull", as you put it), g₀₀ = 1 (not 0)

kakram

Thank you for the quick reply..

ok... I sort of understand what you are saying... as I studied A level Physics 25 years ago and I'm no Feynman.... any chance of an explanation in 'Laymans' terms ? :-)

BruceW

You're absolutely right. Which is why the definition of a second should actually be the duration of x periods of an atom when that atom is not moving relative to you, and is not under some gravitational influence which is different to you.

In other words, the atom serves as a good definition of one second only when it is at the same position as you and has no relative motion to you. (since if it is at the same position as you, it must be in the same gravitational field as you).

Alternatively, you could measure the duration of x periods of an atom which was moving at some known speed relative to you, then calculate the time dilation to get the definition of a second.

kakram

OK.. starting to get clearer.

That means at the boundary of a black hole ... if *I* am there... my local time is still passing....
- people outside see *me* as stuck in time...
- I see people/stars/galaxies ageing and dying and rebirthing at super (infinite) speed?

If im stuck at the boundary of a black hole - would I observe everything else (not at the boundary) passing infinitely faster? That seems to be the implication to me.

tiny-tim

hi kakram!
you can't be stuck at the boundary, it's not physically possible …

(any body at the boundary has to "fall in")

you can be stuck just above the boundary, where everything distant passes, say, a billion times faster

and if you came back to Earth, you would find everything a billion times older

BruceW

Yeah, this all seems right, as long as you replace the words "at the boundary" with "just outside the boundary"

Locally, time always seems to pass normally. (Unless you are inside a black hole, but in that case all the laws of physics break down).

Of course, in a very intense gravitational field (like that of a neutron star), the word 'local' must be defined over an extremely tiny space, so it is possible for a watch on your ankle to tick at a different rate to a watch on your wrist. (i.e. in some cases, the length scale of a person is not local enough for the curvature of spacetime to approximate 'flat').

Drakkith

Hrmm, to my knowledge an infalling observer would experience time at the normal rate that they always do for themselves, not have time stand still.

mikeph

A second as measured by a caesium atom in its own rest frame is the same no matter what gravitational field it's in. That's implicitly shown in tiny-tim's 2nd line above, the time dilation is dependant on the clock's position AND the observer's position. Hence if the clock is in the same position as the observer (relative to the gravitational field), it's measuring time in its own rest frame; the numerator and denominator cancel and there is no time dilation.


Edit - oops, I only read 2 replies before typing, maybe this comment is out of date.