# Is my weight during the night a little bit more than during the day?

Angelika10
During the day, I am between sun and earth. During the night, I am behind the Earth if I'm looking from the sun.
So its
Day: sun-me-earth
night: sun-earth-me

So the forces of gravity from sun and Earth add during the night, but during the day, the sun attracts me away from the earth.
Therefore, I should weight a little less during the day than during the night, isn't it? Probably I would need a high precision scale... but the effect should be measurable, isn't it?

And how would I calculate that with general relativity?

Delta2

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Newtonian mechanics will be sufficient for the biggest part.
In your reckoning why are there two tidal cycles per day (not one)?

sophiecentaur, vanhees71 and FactChecker
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During the day, I am between sun and earth. During the night, I am behind the Earth if I'm looking from the sun.
So its
Day: sun-me-earth
night: sun-earth-me

So the forces of gravity from sun and Earth add during the night, but during the day, the sun attracts me away from the earth.
Therefore, I should weight a little less during the day than during the night, isn't it? Probably I would need a high precision scale... but the effect should be measurable, isn't it?

And how would I calculate that with general relativity?
Weight is generally considered in the frame of the Earth, which is already in free fall relative to the Sun so the Sun's influence is negligible to first approximation. However, there are also the second order effects due to tidal gravity, but those are both away from the Earth at noon/midnight so they both act contrary to the Earth's own gravity.

As already stated, you do not need GR to make this computation.

Delta2, Angelika10, vanhees71 and 2 others
Angelika10
Weight is generally considered in the frame of the Earth, which is already in free fall relative to the Sun so the Sun's influence is negligible to first approximation.
Thank you for the answer! Ah, free fall condition. I understand this. But, after understanding that it's free fall, why is it only negligible "to first approximation"?
However, there are also the second order effects due to tidal gravity, but those are both away from the Earth at noon/midnight so they both act contrary to the Earth's own gravity.
Is there tidal gravity left knowing that the Earth (and me) is in free fall? What is tidal gravity? Why does is act contrary to Earth's own gravity? It makes me weighting less, always?

As already stated, you do not need GR to make this computation.
Is there a difference between Newton and GR?

Delta2
Angelika10
Newtonian mechanics will be sufficient for the biggest part.
In your reckoning why are there two tidal cycles per day (not one)?
It's actually 1 cycle. ...

Mentor
It's actually 1 cycle. ...
There are two high tides and two low tides every day. That’s two cycles.

sophiecentaur
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Thank you for the answer! Ah, free fall condition. I understand this. But, after understanding that it's free fall, why is it only negligible "to first approximation"?

Is there tidal gravity left knowing that the Earth (and me) is in free fall? What is tidal gravity? Why does is act contrary to Earth's own gravity? It makes me weighting less, always?

Is there a difference between Newton and GR?
Not really. Usually the problem is that often the principle of equivalence is not stated carefully enough. So sometimes you get the misconception that gravitational fields can be completely understood as inertial forces or that a reference frame freely falling in a gravitational field is inertial.

In the Newtonian theory it is easy to see that this holds only true for homogeneous gravitational fields. Indeed, if you have a point particle in a gravitational field ##\vec{g}=\text{const}## subject to arbitrary other forces ##\vec{F}##, the equation of motion reads
$$m \ddot{\vec{x}}=m \vec{g} + \vec{F}.$$
Then define a new "free-falling frame of reference" by ##\vec{x}=\vec{x}'-\frac{1}{2} \vec{g} t^2## you get
$$m \ddot{\vec{x}}'=\vec{F}.$$
It's also immediately clear that this cancellation doesn't work for inhomogeneous gravitational fields, i.e., for ##\vec{g}=\vec{g}(\vec{x})##. It's also clear that real gravitational fields due to some mass distribution like the Earth or the Sun are only approximately homogeneous over sufficiently small regions. Thus you always have some gravitational field also in the freely falling reference frame. Since these inhomogeneous fields make the tides due to the Sun and the Moon on Earth these are called tidal forces.

dextercioby, Delta2 and Angelika10
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Is there tidal gravity left knowing that the Earth (and me) is in free fall? What is tidal gravity? Why does is act contrary to Earth's own gravity? It makes me weighting less, always?
Tidal gravity is due to an extended system being in an inhomogeneous gravitational field. The system as a whole moves according to the mean gravitational force on it, but the gravitational acceleration is not the same everywhere. During day, you will be closer to the Sun and therefore the gravitational acceleration will be larger than the mean one, i.e., stronger attraction towards the sun, meaning a tidal acceleration towards the Sun (away from the Earth since it is day). During night, you will be further away from the Sun and therefore the gravitational acceleration will be weaker than the mean one, i.e., weaker attraction towards the sun, meaning a tidal acceleration away from the Sun (away from the Earth since it is night).

However, at dusk and dawn, the tidal acceleration is towards the Earth.

This is why there are two tidal wave cycles per day (although this is governed mainly by the tidal influence of the Moon - although the Sun can play a significant role in making the tides weaker/stronger).

Angelika10
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What is tidal gravity? Why does is act contrary to Earth's own gravity?
You are moving in a circle around the sun, with speed equal to the speed of the center of mass of the Earth in its freefalling orbit. But you are a different distance from the sun than is the earth’s center of mass, so if you are moving at that speed you are not in a freefalling orbit.

It may be easiest to see what is going if you compare the centrifugal force from your circular motion about the sun ##mv^2/r## with the sun’s gravitational force ##mM_S/r^2## when the sun is directly overhead and directly underfoot.

vanhees71 and Angelika10
Angelika10
There are two high tides and two low tides every day. That’s two cycles.
Ah, ok, that you mean: the real tides (of the oceans). But those are triggered by the moon moving around the Earth and the moon takes 12 hours for one cycle.

vanhees71
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Ah, ok, that you mean: the real tides (of the oceans). But those are triggered by the moon moving around the Earth and the moon takes 12 hours for one cycle.
The tidal lunar day is 24 hours and 50 minutes (the time between consecutive zeniths of the Moon), which is mainly due to the rotation of the Earth, but also has a contribution from the Moon's orbit.

sophiecentaur, vanhees71 and hutchphd
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But those are triggered by the moon moving around the earth
The position for the Moon relative to the Earth changes by only a little during a day. Both Sun and Moon are pretty much in their same places over any 24 hour period. The difference in positions accounts for the Spring / Neap / Spring / Neap cycles of max and min tidal heights over the month.
The two cycles a day is due to the Earth moving around the Barycentre too.

vanhees71
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The two cycles a day is due to the Earth moving around the Barycentre too.
That's not right, is it? There'd be two high tides a day even if the Earth and Moon were nailed down (don't ask how...) as long as the Earth rotated. The "extra" bit of the actual tidal sequence of two and a bit tides a day is due to orbital motion changing the position of the Moon relative to the Sun. It would be just less than two tides per day if the Moon orbited retrograde.

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There'd be two high tides a day even if the Earth and Moon were nailed down (don't ask how...) as long as the Earth rotated.
Perhaps, at first sight, that sounds reasonable but you have to ask yourself what would cause a bulge on the side opposite the Moon. There is no 'gravity shadow', on the far side. There has to be some movement involved, to 'throw the water' out away from the Moon's direction.
If the two were "nailed down", there would be a single bulge in the side facing towards the Moon. If the Earth were fixed (stationary) on a pole and the Moon went round it, that bulge would appear to go round once a month.As the Earth rotated, it would only pass a single bulge each day.
There has to be some acceleration about a point that's not the centre of the Earth and that point is the barycentre.

vanhees71
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Perhaps, at first sight, that sounds reasonable but you have to ask yourself what would cause a bulge on the side opposite the Moon. There is no 'gravity shadow', on the far side. There has to be some movement involved, to 'throw the water' out away from the Moon's direction.
That's just normal tidal acceleration. The tidal acceleration on the far side is away from the Earth.

If the two were "nailed down", there would be a single bulge in the side facing towards the Moon. If the Earth were fixed (stationary) on a pole and the Moon went round it, that bulge would appear to go round once a month.As the Earth rotated, it would only pass a single bulge each day.
No, this is incorrect. The tidal acceleration stretches along the direction of separation and squeezes in the transversal direction.

vanhees71
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That's just normal tidal acceleration.
I don't know what that expression means, I'm afraid. How would there be any 'acceleration' in a stationary system?

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I don't know what that expression means, I'm afraid. How would there be any 'acceleration' in a stationary system?
Yes. The point is that the gravitational acceleration on the far side is smaller than the mean acceleration. Therefore, relative to the Earth centered system, the far side experiences an acceleration away from the Moon.

vanhees71
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Yes. The point is that the gravitational acceleration on the far side is smaller than the mean acceleration. Therefore, relative to the Earth centered system, the far side experiences an acceleration away from the Moon.
I think you are putting the cart before the horse here. One step at a time:
If both are 'nailed down' with no movement at all there would be only one bulge. If the Earth were a smooth sphere, it would revolve under that bulge once a day.
If just the Earth were nailed down, the bulge would follow the Moon (still no second bulge / tide because only the Moons gravity would affect the water).
If the Earth can move (intergalactic space with no Sun) then it would orbit round the Moon Earth barycentre. Take the near and far sides of the oceans and the centre of the Earth. They are all at different distances from the Barycentre (in free fall) so there will be two bulges (three distances involved). That effect has caused the two bulges. Because the three orbital periods will be different, the near side bulge will arrive a bit early and the far side bulge will arrive a bit later (same as the microgravity effects in orbiting spacecraft ).
Edit: The water is attracted to the Barycentre - not just the Moon.

hutchphd
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The tidal acceleration stretches along the direction of separation and squeezes in the transversal direction.
I just re-read this. In a stationary situation, what force / acceleration could there be in a direction 'away from' the Moon on the far side of Earth. The force on virtually every drop of water from the Moon is identical. The force on virtually every drop of water from the Earth is very similar ( a few metres in a few thousand).
I'm hoping we're discussing different situations here and that's why I have tried to build the model up from scratch.

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I just re-read this. In a stationary situation, what force / acceleration could there be in a direction 'away from' the Moon on the far side of Earth. The force on virtually every drop of water from the Moon is identical. The force on virtually every drop of water from the Earth is very similar ( a few metres in a few thousand).
I'm hoping we're discussing different situations here and that's why I have tried to build the model up from scratch.
The point with tides is that it is not a stationary situation. The entire point with tidal forces is that it is a difference of accelerations and if you have acceleration things are not stationary. I think you are just complicating things by introducing a stationary situation and somehow trying to reinvent the wheel. If you nail the Earth down you have a completely different situation and all of a sudden the main gravitational force which pulls everything in the same direction becomes the main notable effect instead of the actual differences in the gravitational field, which lead to the tidal effects. You are therefore discussing an entirely different phenomenon when you nail the Earth down.

Motore and vanhees71
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The point with tides is that it is not a stationary situation.
To be fair, the thread ended up here because @sophiecentaur said that the non-stationary situation is necessary to produce the tides and someone else disagreed.

DrStupid
The entire point with tidal forces is that it is a difference of accelerations and if you have acceleration things are not stationary.
Unless 'acceleration' means proper acceleration.

If you nail the Earth down you have a completely different situation
Unless it is 'nailed down' with a homogeneous field.

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Unless it is 'nailed down' with a homogeneous field.
I think we both meant held stationary on two massless fixtures. I was building up from that situation - adding more and more possible motion. When I introduced the Barycentre, it seemed to bring on the difficulties. But how can you do without one of those>

The two cycles a day is due to the Earth moving around the Barycentre too.
Not really. Even if Moon and Earth were falling straight towards each other along a line (no motion around the common Barycentre), the Earth would still be stretched along that line (due to the Moon's gravity gradient), and have two bulges, and two tidal circles per one rotation around its own axis.

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Is that not just re-stating the way the objects are attracted to the barycentre?

russ_watters
Is that not just re-stating the way the objects are attracted to the barycentre?
The barycentre and the motion around it are not relevant for the second bulge. The Moon's gravity gradient explains both bulges.

Richard R Richard
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No problem with that but it only works at all because there's rotation - otherwise it's a one-off extended collision and the 'periods' involved aren't there.

In other orbital discussions, would you ignore the barycentre? Are you actually saying that the two tidal bulges are not in orbit around somewhere (like the barycentre)? I appreciate that there are a number of ways of describing what goes on but the Moon's gravity gradient is not the only force on the water. The Earth keeps it in place .
I still wonder about the effect of that gravity gradient. It should mean that the bulges wouldn't be on the line of centres.

No problem with that but it only works at all because there's rotation - otherwise it's a one-off extended collision and the 'periods' involved aren't there.
What do you mean by "rotation"? Rotation of the Earth around it's axis is needed for the periods. Motion around the barycenter is not needed for the periods.

vanhees71
What A.T. is saying, I think, is that you don't need any periodic motion (or angular velocity) for the bulges to develop. As long as the body is in free fall in a central gravitational field, there will be two bulges. For what it's worth, any such motion will be technically speaking an orbit - even falling along a straight line on a collision course towards the barycentre is a degenerate orbit. Or being launched straight up to infinity, with escape velocity. But the reason to avoid thinking about orbits here is that it tends to muddle the water with intrusive intuitions, like assuming it has something to do with angular motion, centrifugal forces, periodicity etc. All you need is free fall and a gravitational gradient.
As they say, a picture with eighty-ish words is worth approx. 1.08 pictures. Or somesuch.

It doesn't matter how exactly the body is moving in the gravitational field, as long as it is free falling.

Now, if we literally nailed down the body (to the fabric of space-time? aether? firmament?) through its centre of mass, we'd still have an inertial frame as in #3, only with reaction force at F0. The proximal side would be under tension, the distal side under compression (i.e. one bulge only). Which is why letting oneself fall into a black hole leads to spaghettification, but trying to hover above the horizon leads to pancaking.
In yet another words, a slinky toy held by its centre of mass extends from one side only, but if we let it go (in a vacuum, with sufficient gradient to notice), it extends from both sides of its CoM.

If, however, we nailed down the body with a homogeneous gravitational field, as DrStupid suggested earlier, that magically works on the body only, and not on the source of tidal gravity, we would then be able to do the same subtraction as in #5. Since it would act with the same force on every point of the body. The gradient of the central field would then produce two bulges again.

Incidentally, the bulges in general needn't be aligned with the direction towards the centre of the field, but that is tangential to the topic at hand, and has to do with rotation under the bulges and the delayed response of the material to deformation. We can always consider a case with no relative rotation to simplify.

russ_watters, vanhees71, Orodruin and 1 other person
DrStupid
No problem with that but it only works at all because there's rotation - otherwise it's a one-off extended collision and the 'periods' involved aren't there.
In order to avoid the collision you could attach thrusters on the Moon and use it as gravity tractor for Earth. There would still be two tidal bulges.

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Thanks for that post, @Bandersnatch. It neatly lays out where I went wrong in my last post. There are two bulges if the Earth is in freefall, one if it is not.

Incidentally, the physical model I had in mind was something like an orrery, although powerful rockets mounted on the Earth to keep the center moving in a straight line while the moon circles it would be an alternative. I don't think either is plausible in GR, but in Newtonian physics the support systems can be made arbitrarily light and powerful.

russ_watters
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In order to avoid the collision you could attach thrusters on the Moon and use it as gravity tractor for Earth. There would still be two tidal bulges.
... and as long as the Earth is spinning, this will result in having two tides per rotational period of the Earth. (Just to make that clearer)

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There would still be two tidal bulges.
. . . . and if Earth were just a lump of goo, it would stretch out on the way to the collision. But is that getting us to our Moon's orbit and the tides?

I mentioned near the top that considering it as a system in free fall, the two bulges will separate out and remain separated out because of the orbit. With no orbit, you would just get a linear, one - off free fall then a crash. The period of the orbit is what defines the actual amount of the 'stretching', which doesn't change (first order situation). The period depends on all the masses involved and the separation. So I really can't see how it's "just" the effect of the varying field of the Moon that makes it all happen the way it does. Why isn't it just the order that the situation is described?

DrStupid
With no orbit, you would just get a linear, one - off free fall then a crash. The period of the orbit is what defines the actual amount of the 'stretching', which doesn't change (first order situation).
In my example above (Moon powered with rocket engines to the same acceleration as Earth) there is no crash and no orbit but the same stretching as in orbit. All that matters is the gravitational field of the Moon and the relative position of Earth in this field. Orbit or not makes no difference.

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