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My question came from the talk of how gravitational wave stretches and compresses space time.

Say there are two protons that are 1 centimeters apart, as a G-wave passes through them, would the electrostatic force experienced by the protons change?

What about Plank's constant? If two particles are x number of Plank distances apart, is the new x smaller than the old x?

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Suppose we choose our coordinates so both protons are on the $x$ axis, at $x = -0.5$cm and $x = +0.5$cm:

Protons

The distance $d$ between the protons is obviously 1cm - well, that may seem obvious but actually it's only true in flat spacetime. More generally the geometry of spacetime is described by a quantity called the metric tensor, $g_{\alpha\beta}$, and the proper distance between the two protons is given by:

$$ d = \int_\text{x=-0.5cm}^\text{x=0.5cm} \sqrt{g_\text{xx}}\,dx $$

In ordinary flat spacetime the value of $g_\text{xx}$ is constant at one, and the integral turns into:

$$ d = \int_\text{x=-0.5cm}^\text{x=0.5cm} dx = 1 \,\text{cm} $$

as we expect. Now suppose we have a gravitational wave coming out of the screen towards you. This causes an oscillating change in the spacetime geometry that looks like this (picture from Wikipedia):

Gravitational wave

Whatb this is supposed to illustrate is that the value of $g_\text{xx}$ (and $g_\text{yy}$) oscillates with time, so it alternately becomes greater than one and less that one. In that case our integral becomes:

$$ d(t) = \int_\text{x=-0.5cm}^\text{x=0.5cm} \sqrt{g_\text{xx}(t)}\,dx $$

and our distance $d(t)$ is no longer a constant but oscillates above and below $d= 1\text{cm}$ as the gravitational wave passes through.

This isn't some mathematical trick, the gravitational wave really does cause the distance $d$ to change with time. If you shone a light ray between the protons and timed how long it took you'd find that time oscillated as well. Assuming the oscillation is slow compared to the time the light rays takes, the time the light ray would take to travel between the protons, $T$ would be:

$$ T(t) = \frac{d(t)}{c} $$

The electrostatic force between the protons would also change with time:

$$ F(t) = \frac{ke^2}{d^2(t)} $$

However Planck's constant is just a constant and this wouldn't be changed by the gravitational wave.

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  • $\begingroup$ Thanks for a sophisticated and clearly explained answer. I just want to follow up by asking what Plank's distance really means. It appears to me as though any length of space can be divided into integer number of Plank distances, like the "pixels" of space. G-W's compression of space time seems to defeat the model because it would either mean that the number of Plank distances between two points decreased or that Plank constant itself decreased, neither or which seem satisfactory. So I was wondering if there is some explanation to the idea of Plank distance as indivisible parts of space. $\endgroup$ – user289661 Feb 21 '16 at 20:17
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    $\begingroup$ @user289661: The Planck distance is not an indivisible part of space. Space is not made up from atoms of Planck lengths. In fact as far as we know space is not discrete at all. The Planck length is just the shortest distance that can be experimentally probed. So there isn't a conserved number of Planck lengths between your two protons that is being compressed or stretched. As the gravity wave passes through the number of Planck lengths between the protons changes. $\endgroup$ – John Rennie Feb 22 '16 at 6:19
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As all your measurement equipment (and even time) also stretches/compresses with the space, no change should be noticed in plank length, and in the electrostatic force.

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  • $\begingroup$ Do you suggest that light will then be able to travel more distance in the same amount of time? Since light traveled for a different amount of time over the same distance. $\endgroup$ – user289661 Feb 21 '16 at 7:59
  • $\begingroup$ The speed and wavelength of light will be adjusted accordingly as well. $\endgroup$ – kpv Feb 21 '16 at 8:05
  • $\begingroup$ Nothing adjusts the speed of light. $\endgroup$ – Blackbody Blacklight Feb 23 '16 at 13:31
  • $\begingroup$ Adjusts means it remains c in spite of stretch or compress of space time. $\endgroup$ – kpv Feb 23 '16 at 15:52

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