Gravity makes clocks run more slowly. But say you placed a clock at the point around which two suns of equal size revolved. The suns are diametrically opposed to each other, orbiting around this central point. Gravity would be cancelled out, so the clock would be drawn to neither sun, but would the clock run more slowly because of the mass of the suns? Question clarified. See post #8
Since the net g at that point is zero, I suppose the clock would carry on ticking as per normal. Though I know nuts about GR.
My know nothing answer is that if you moved towards either star then the clocks would run slower because the gravitational field would be higher. A real physicist should wander along before too long and answer the question for real...
More slowly than what? You see, you created a valid scenario but you are unable to ask a correctly formed question.
As pointed out elsewhere, what is the reference to 'running more slowly'? Probably wrt some distant (aka co-ordinate) observer. In any case as long as such observer is considered stationary wrt system center the governing factor is gravitational potential - not 'field' or 'force'. Just apply the gravitational redshift formula as per Wiki article: https://en.wikipedia.org/wiki/Redshift#Redshift_formulae which is good for an observer at any relative potential. If 'at infinity' the numerator is set to unity. Strictly speaking the fact the two suns are co-rotating will doubtless have a very slight perturbative effect. [one must invert the redshift formula above linked to get relative clock rate]
True. I'll try again. Let's say that you have a binary star system. Def. A binary system is a system of two objects in space (usually stars, but also brown dwarfs, planets, galaxies, or asteroids) which are so close that their gravitational interaction causes them to orbit about a common center of mass. A clock is placed at the common centre of mass. Will it run more slowly than a similar clock which is outside the two suns' gravitational influence entirely? I hope I've explained the scenario better this time.
Yes, this is better. The answer is that the gravitational potential is the same in both cases (zero), so the two clocks will run at the same rate.
Not true. Ignoring any slight effects of stars motion per se, gravitational potential at center location is depressed wrt 'infinity' by just twice the value for the case of clock stationary wrt single star (clock at same distance from star as before, and assuming both stars have equal mass). Hence redshift formula (inverted) as per link I gave earlier applies - clock runs slow.
This is incorrect, potential is additive, so the potential at dead center between the stars is: \(\phi_1=+\frac{GM}{r}-\frac{GM}{r}=0\) The potential at infinity is \(\phi_2=lim_{r->\infty}\frac{GM}{r}=0\) So, \(\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-\phi_2}{1-\phi_1}}=1\) No redshift.
How do you figure that one potential 'cancels' the other? Which is taken as +ve?! Think about this. Suppose one drops a stone from afar such that it passes exactly between the two stars (consider then 'frozen' in place for this exercise). If what you say were true, the velocity should be zero, just as it was at the start of fall. I think not. There will be a net conversion of potential to K E and that stone will be flying through the middle at high speed. Both potentials are negative and add as such (proviso we ignore nonlinearity of GR for now, which is reasonable here).
Dang, you are right, the potential is not null at the midpoint, it is double the potential: \(\phi_1=-\frac{GM}{r}-\frac{GM}{r}=-2\frac{GM}{r}\) The potential at infinity is \(\phi_2=lim_{r->\infty}\frac{GM}{r}=0\) So, \(\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-\phi_2}{1-\phi_1}}=\sqrt{1+\frac{2GM}{r}}\)
Yep, we agree. Might have been a case of too much gravitational potential re thought processes for a moment or two? Please Register or Log in to view the hidden image!
So time runs more slowly. Can we also infer that time runs more slowly near the centre of a galaxy than at the periphery? Ignore the black holes, which would eat the clock.
