# A little help on Gravitational time dilation

Could somebody explain Gravitational time dilation in layman's term for me please!?

And in the formula of Definition, it mentioned a capital H, but it didn't explain what it is! And h' is h derivative or h prime!?

If it's h derivative, then how does that formula works!? Taking a integral through the derivative of h!? What the!? I'm sorry to say that it's definitely exceed my ability of understanding!

And if it's h prime, then what is it!?

And g(h) is the dependence of g-force on "height", so is it GMm/r^2!? So the whole formula basically is the integral of the gravitation equation (With a derivative or prime! Which I don't understand how neither of them works![2])!?

The article is so ambiguous that I think I even have lesser understanding of Gravitational time dilation after reading it!!!

[2]: I do know how derivative works, I just don't know how it worked in this particular formula!

In layman's terms, gravitational time dilation means that a clock runs more slowly in the presence of a large mass compared to a clock that is farther away. As the Wikipedia - Gravitational time dilation article says:

... proximity to Earth's gravitational well will cause a clock on the planet's surface to accumulate around 0.0219 fewer seconds over a period of one year than would a distant observer's clock. In comparison, a clock on the surface of the sun will accumulate around 66.4 fewer seconds in one year.

As it is explained there, the relationship between time dilation $T_d(h)$ as a function of distance (i.e. height) $h$ from the mass, the formula $T_d(h) = e^{\frac1{c^2}\int_0^h{g(h')dh'}}$ gives the integral for $T_d(h)$ in relation to local gravity $g(h)$.

• And that formula was what confused me!!! PLEASE read my question more thoroughly! Appreciated!!! – Pikachu620 Jul 16 '18 at 10:07
• You get that $\int g(h')dh' = g(h)$, right? – Mick Jul 16 '18 at 10:11
• I guess you could replace h' as x, then the whole thing become ∫g(x)dx, and it does NOT equal to g(∫xdx)! Does it!? And if it DOES equal to it, why didn't the formula just say g(h)-g(0)!? Why it have to use all those fancy integrals to confuse people!? – Pikachu620 Jul 16 '18 at 10:18
• Special Relativity explains the effect of velocity, as you travel faster, time dilation makes clocks seem to run slower than clocks in a stationary frame, and General Relativity explains how mass warps space-time. An understanding of integrals, differentials and limits is a starting point for understanding the mathematical expression of the concepts... – Mick Jul 16 '18 at 10:30
• I do understand the basic of calculus! Otherwise I wouldn't know that ∫g'(x)dx from 0 to h is g(h)-g(0)! And just as I commented, if ∫g(h′)dh′=g(h) , why don't they just say g(h)-g(0)!? Could you explain that to me!? – Pikachu620 Jul 16 '18 at 10:41

Mine is one answer I have not read or heard anyone else offer. I believe time is a property only of discrete matter objects and closed matter systems, similar to the way the gravitational force is a property of massive objects. Both exist on mass and matter objects, perhaps the same as quarks exist on particles.

Therefore, I propose too that time is a fifth fundamental force with the inherent power to age all matter toward a state of total equilibrium, or complete structural disorder to the point where its discreteness as matter no longer exists.

If I am correct in this, it means that gravity does not directly impose its power of attraction to matter, yet by increasing the speeds of matter objects, as well as all other masses that enter a more massive object's gravity field. it helps their time rates to pass slower than if gravity was not present.

I am not sure this will serve as an answer to your question, but I hope others will agree it is not too far-fetched to be helpful to you.

• Please try to use standard physics when answering questions here. This is not a forum for promoting your personal non-standard theory of time. – PM 2Ring Aug 10 '18 at 16:40