Does Earth's Rotation Affect Time Dilation Between Day and Night?

[daytripper]

hello everyone, to start off, I'm in no way a physics expert but I think I understand time dialation. for all of this, use the sun as a reference frame. I was reading a thread on here and I thought of something. I'm pretty proud of myself for thinking of this but I would like someone with more experience to verify it for me. The night side of the Earth is moving more quickly than the day side of the earth. The Earth rotates at a speed of about .47 km/s and the Earth revolves at a speed of about 29.8 km/s. This means that the night half of the Earth (for this, let's just ignore the 23.3 degree tilt of the Earth's axis and assume that it's verticle, relative the Earth's orbit) is moving at an average rate of 29.8 + .47 km/s (the .47 is an average rate, it actually only .47 at the furthest point from the sun). The day side, on the other hand, is moving at a rate of 29.8 - .47 km/s (again, average). Does this mean that, since the day side is moving slower than the night side, that clocks, according to time dialation, move slower during night than during the day? I've done some rough calculations with this and if all I've said is true, here's a fun bit of trivia: you age 1.0000000003116756 times slower and weigh 1.0000000003116756 times less during the night than you do during the day. haha. Thank you for your time and replies
edit: I just realized that the fact that the day and night halves of the Earth move at an equal rate relative to each other, my idea might be incorrect. I don't know if this would disprove the idea though, since the night side IS moving more quickly than the day side relative to the sun.

 

[daytripper]

This would be true for an obsever at the barycenter of the solar system, but it would not be true for an obsever at the center of the Earth.

i don't quite understand your response but then again, my original statement wasn't exactly lucid. let's take 2 points on the equator of the Earth (given that the Earth has a verticle axis). A will be the point closest to the future positiotn of Earth and B will be the point closest to the previous position of earth. There's two observers, one on A, one on B. They sync their clocks. Then A moves to B and B moves to A. Will the observer that moved to point A (let's call him bob)'s clock be running just <i>slightly</i> behind the observer that moved to point B(lets call her sue)'s clock? because bob moved faster relative to the sun than sue did. because they're both moving at different rates relative to barycenter of the solar system. or does it not work that way? I read something about the relative addition rule or something like that, where A moving relative to B and B moving relative to C needs a special formula to find the speed of A relative to C so that you can't have anyone moving faster than c by adding velocities.

 

[pervect]

That's OK, I answered too quickly, and have since deleted that response. I'm going to double-check a few things and get back to you.

 

[pervect]

Basically, you need to include the gravitational dilation effect of the sun to answer this question. When you do, you find that all clocks on the Earth's geoid tick at *almost* the same rate. There is a very small effect due to the suns tidal force, though.

Here's a reference

http://www.cosmic.ucar.edu/related_papers/1997_ashby_relativity.pdf

Around pg 19 (on the pdf document, the printed label says pg 10) reference is made to the fact that all clocks "on the geoid" click at approximately the same rate.

Thus the very useful result has emereged, that ideal clocks at rest on the geoid of the rotating Earth all tick at the same rate.

Around pg 2 of the pdf document, it is pointed out that there is a very small effect due to the sun, on the order of 10^-16, due to the tidal forces of the sun. These are a million times smaller than you've calculated without including the counteracting effect of gravitational time dilation due to the sun's gravity.

So you are right that there is an effect, but it is *very* small, and you have to include both gravitational time dilation as well as the velocities to compute it correctly.

Also, I do not believe it will be a "day-night" effect as much as a "high-tide, low-tide" effect. (The period of the effect will not be 24 hours, but 12).

 

[daytripper]

Basically, you need to include the gravitational dilation effect of the sun to answer this question. When you do, you find that all clocks on the Earth's geoid tick at *almost* the same rate. There is a very small effect due to the suns tidal force, though.

Here's a reference

http://www.cosmic.ucar.edu/related_papers/1997_ashby_relativity.pdf

Around pg 19 (on the pdf document, the printed label says pg 10) reference is made to the fact that all clocks "on the geoid" click at approximately the same rate.



Around pg 2 of the pdf document, it is pointed out that there is a very small effect due to the sun, on the order of 10^-16, due to the tidal forces of the sun. These are a million times smaller than you've calculated without including the counteracting effect of gravitational time dilation due to the sun's gravity.

So you are right that there is an effect, but it is *very* small, and you have to include both gravitational time dilation as well as the velocities to compute it correctly.

Also, I do not believe it will be a "day-night" effect as much as a "high-tide, low-tide" effect. (The period of the effect will not be 24 hours, but 12).

Allright, I understand. I guess if you wanted to get an EXACT answer, you'd have to factor in the gravitational time dilation of the moon as well. One last thing though, you said that there's a slight difference due to the sun's gravitational time dialation. Just to be sure, would the difference in velocities between the night half of the Earth and the day half of the Earth (wrt the sun) also make a very small affect or is that irrelevant since their both traveling at the same rate relative to one another? After thinking about it, I think it would be irrelevant, but I'm not sure.

 

[pervect]

Clocks deeper in a gravity well run slower than clocks higher up in a gravity well - this is called gravitational time dilation. I can give you a reference for more reading if you like, the basic formula for the amount of slowing is

$$T = \frac{1}{\sqrt{1-\frac{2 G M}{c^2 r}}}$$

where r is your distance from the sun.

During nightime, you are moving faster, but you are also further away from the sun (higher up in a gravity well). So your clocks tick more slowly because you are moving faster, but more quickly because you are higher up in a gravity well. The two effects essentially almost cancel out.

The actual analysis is easier if one takes advantage of the properties of GR and does it in the coordinate system of the Earth rather than from the coordinate system of an observer at the barycenter of the solar system.

Basically, if the Earth were an idealized fluid in equilibirum, the equations of relativity say that the shape of the Earth would adjust itself automatically so that all clocks on the surface of the fluid ran at exactly the same rate. This happens because the equations of motion for the fluid make it a necessary property for equilibrium of the fluid. You can find this statement online in several places, and in textooks like MTW, but a detailed explanation gets a bit long and technical.

The sea isn't a totally ideal fluid, and it's not perfectly in equilibrium either, but the tides move slowly enough that essentally the surface at which clocks all tick exactly at the same rate is fairly close to the actual surface of the water. And the water does not move that far relative to the surface of the land, so you can already see that the effect is very tiny - the effect will be of the same order as the gravitational effect of tens of feet of elevation of the clock. (See the formula for gravitational time dilation).

It turns out that the biggest difference in clock rates occurs not between noon and midnight (when the sun is directly overhead and on the other side of the Earth), but between noon and 6 pm.

 

[daytripper]

Thank you very much for your help, this cleared it up rather nicely.
-Daytripper