# B Einstein elevator

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1. Jun 18, 2016

### Stephanus

Dear PF Forum,
I've watched some youtube video or read about it, I forgot. That our head is wiser than our foot, because time slow for our foot.
Okay..., this could have been something in GR formula. My feet is at 6200 km from the earth center and my head is at 6200 + 1.7m. Even in Newton gravity formula my head wouldn't feel 9.8m/s2. Perhaps 9.7999999 something m/s2?
Now in Einstein elevator. Would our head be older than our feet? Since there's no gravity there.
And our head would be accelerated 9.8 m/s2 just as our feet?
Thank you very much.

2. Jun 18, 2016

### Stephanus

I forgot to add another paragraph.
I ask this because I remember that in Einstein elevator, if someone one the top floor shine a light bulb, the receiver from the bottom receives slightly more signal than a receiver from the top.
Would the effect happens EXACTLY as on earth?
Thank you very much.
This is the question that I'd like to know actually.

3. Jun 18, 2016

### Staff: Mentor

No, not quite exactly. And the reason is what you identified in your first thread, that the gravitational force at you feet is very slightly greater than at your head. That is a tidal effect.

The equivalence principle that says being in an accelerating elevator is like being in a gravitational field only applies when tidal effects are small enough to ignore.

4. Jun 18, 2016

### Staff: Mentor

Yes, and it's worth taking a moment to calculate roughly what the difference might be.

5. Jun 18, 2016

### Janus

Staff Emeritus
The thing to keep in mind is that the difference between time rate of your head and your feet is not related to the difference in g force they experience, but to the difference in potential, or how much work it would take to move a test mass from one point to another. Since g force falls off with height on the Earth, you would have to figure that in when determining the difference in potential and the time difference for the person standing on Earth. On the other hand, for the person in the elevator, g-force does not fall off between feet. In other words, it would take more energy to lift a test mass from floor to head height in the Elevator than it would on the Earth, and thus the time rate difference in the elevator would be the greater of the two.

6. Jun 18, 2016

### Stephanus

Thank you very much
Okay...
The earth radius from equator is 6378.1 km, let's call it r
https://en.wikipedia.org/wiki/Earth
or 6378.100m

This is what makes me irritated. I'm calculating 1.7 m against a 0.1 km rounding. But, I'll do it anyway
I live in Jakarta 6° 8'59.61"S 106°42'9.06"E (Google Earth)
Frankly I can't calculate how far I'm from the center of the earth, considering I'm not exactly in the equator, but closely enough. More over, I don't know how high my house from the sea level. Heck, even my house is some 50cm higher than the road in front of it. But, I'll do it anyway.

First of all, the gravitational constant:
https://en.wikipedia.org/wiki/Gravitational_constant
6.674 * 10-11 * N * m2/kg2 let's call it G

The mass of the earth: 5.97237×1024 kg, let's call it M
https://en.wikipedia.org/wiki/Earth

So the force of gravity, let's call it F for me (my feet) and earth from 6378.km is...
$F = G \frac{M * \text{my weight} * kg }{r^2}$ not that I don't want to disclose my weight . But surely the staffs/mentors who read this will understand why M2 is not important
$F = 6.674 * 10^{-11} N * \frac{m^{2}}{kg^2} \frac{5.97237 * 10 ^ {24}\text{kg} * \text{my weight} * kg}{6378100^2 m^2}$
$F = 9.80627 * N * \frac{m^{2}}{kg^2} * \frac{1 \text{ kg} * \text{my weight} kg}{m^2}$
$F = 9.80627 * N * \text{my weight}$
So, this is the force that is needed to keep me standing on the ground.
How much acceleration is experienced for an object as heavy as me if applied this number of force?
Because force is mass * acceleration, and newton is in Kg * m / t2, or F = m * a, a =F/m, so...
$a = \frac{9.80627 * kg * \frac{m}{t^2} * \text{my weight}}{\text{my weight } kg}$
$a = 9.80627 * \frac{m}{t^2}$
So, this is the acceleration that is felt by my feet.
What is the acceleration felt by my head?
I don't have to go through all the calculation above, just this one
$F = ... \frac{...}{(r+1.7m)^2}$
$a_{head} = 9.80627 * \frac{6378100^2}{6378101.7^2}$
$a_{head} = 9.806264773$
$a_{feet} = 9.80627$
I don't know if my calculation is correct.
Thanks for the attentions.

7. Jun 18, 2016

### Janus

Staff Emeritus
8. Jun 18, 2016

### Stephanus

So..., the time rate differs depending on acceleration. No gravity involved?
Is this GR or SR concept?

Just for an instant I thought Einstein elevator has anything to do with Bell's Paradox. But it's not, right. Because in Bell's Paradox, the distance wrt a rest observer stays the same, though there's acceleration involved.

9. Jun 18, 2016

### Stephanus

Ok, thanks. Then we only need to calculate distance only.

10. Jun 18, 2016

### Janus

Staff Emeritus
The time rate differs by the difference in potential. On the Earth that difference is due to gravity, in the elevator it is due to acceleration.