# B LIGO and space time changes

#### asca

Just for the sake of completness, I found this article surfing the net :
However I'd just like to share the following: what happens to the clocks in the "x" arm of the example? Isn't their pace slower than the pace of clocks placed at the beam splitter? so for any observer in the x arm the time it takes for a crest to "reach the subsequent crest" (I hope you see what I mean by that) is shorter than the time measured by a clock placed at the beam splitter. But light speed is the same, so for any observer in the x arm the "distance" between two crests is shorter than the same distance measured by an observer placed at the beam splitter, in other word his ruler is longer than an identical ruler placed at the beam splitter. The solution to my puzzle maybe lies in the assumption the GW effect was and is different, actually opposite, in the two arms, and in the assumption that the observation point is somehow not (or less or more) affected by the GW passing by. So any obesrver in the x arm observing juts what happens in the x direction would not realize a GW is passing by, the same for any observer in the y arm, only the comparison by a third observer of the X and Y observations can deduct that a GW has passed by. That is probably the key.

#### asca

Sorry, I'm not so sure about what I just wrote: my question is what happens to the clocks in the X arm and to the clocks in the Y arm? I'm not so sure about the aswer I gave before, because the clock should not be affected by the direction of observation.

#### Ibix

That paper has specifically chosen a coordinate system where the passage of time is unaffected by the gravitational wave. Free floating clocks will remain synchronised throughout.

#### asca

Meanwhile I found an even clearer (at least to me) answer to my initial question.

Still I remain puzzled by the new question: what happens in general to the clocks placed in the X and Y arms? Why does Ibix say that their pace does not change?

#### Ibix

Why does Ibix say that their pace does not change?
Because that's what the maths says. For a free floating clock at rest in the coordinate system in use, $dx=dy=dz=0$, and the metric is diagonal with $g_{tt}=1$. Why do you think they should change?

#### PeterDonis

Mentor
any obesrver in the x arm observing juts what happens in the x direction would not realize a GW is passing by, the same for any observer in the y arm, only the comparison by a third observer of the X and Y observations can deduct that a GW has passed by
That's not correct; observing just one arm can still tell you that a GW passed, because the round-trip travel time of light in the arm changes. However, it's much harder to measure that change in round-trip travel time in a single arm, then it is to measure the interference between the light in the two perpendicular arms. So an interferometer detector like LIGO is more sensitive than a single-arm round-trip travel time would be.

#### PeterDonis

Mentor
what happens to the clocks in the "x" arm of the example?
Nothing. By the analogy the paper you linked to makes with cosmological models: as the universe expands, clocks are not affected, only distances are. Similarly, as the GW passes, clocks are not affected, only the lengths of the arms are.

#### pervect

Staff Emeritus
Still I remain puzzled by the new question: what happens in general to the clocks placed in the X and Y arms? Why does Ibix say that their pace does not change?
Because that's an English language statement of the metric given in 2.1 in the paper you cite:

$$ds^2 = -c^2\,dt^2 + [1+h(t) ]\,dx^2 + [1-h(t)]\,dy^2 \quad [2.1]$$

In this equation, h(t) - the gravitational wave - doesn't modify the relationship between dt (coordinate time) and ds (which represents either distance or time intervals, which are unified in special relativity). So h(t) doesn't have any effect on time.

h(t) does modify the relationship between dx and dy (spatial coordinates) and ds, so the gravitatioanl wave does have an effect on space.

I'm not sure what pre-requisite knowledge you have, pessimestically I tend to assume you have little :(. Still, I'll take the risk of talking in ways that may go "over your head", as you seem to have some interest in the topic.

So, trying to thing about what you might need to know - you might start looking into what a metric is, as that's the mathematical key to answre your quesitons.

A good approach might be to start with understanding the pythagorean theoerm - the square of the hypotenuse is the sum of the square of the other two sides, then building upon this knowledge to understand what a purely spatial metric is. The translation of the pythagorean theorem in the language of metrics would be ds^2 = dx^2 + dy^2. s represents distance, x and y are coordinates, and d represents "a change in" or "differential". This would be hopefully familiar from calculus, if you have it. If you don't have calculus yet, that's another thing to add to your list of things you'd need to find out about.

At this point, you'd need to know what coordinates are. It's easy enough to state that coordinates are just lablels, but I've noticed in the past that sometimes this doesn't seem to be accepted by some PF posters, I'm not quiite sure what the difficult is.

Given that you know what coordinates are, and have some notion of what distances are, and the funamentals of calculus, then the metric allows you to take these coordinate labels (or rather their differentials), and computes differnetial distances from them.

After that, all you need to do is learn special relativity in general, and the metric approach to special relativity (the Lorentz interval), specifically.

After that, you'll have the needed bases to tackle this issue again with a firmer foundation. I've probably skipped a few steps of things you'd need to know along the way, but that's the short outline as I see it.

#### asca

Everything is clear now folks. When I wrote that stuff about the clocks I was momentarily carried away by the thought of a static gravitational field, but this is not the case when a wave is passing by, so I was really "off tune", sorry.

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