# How much energy did it take for God to part the Red Sea?

EDIT

Why this is a world-building query. If I had framed this as, "In my world where magic requires the expenditure of energy, a powerful wizard decides to open a path through a large lake,etc. The world I am supposing is that God himself can create and destroy energy (or has an infinite supply) but still uses energy to do his miracles and acts of creation. This is different from simply magicking things in and out of existence as the Bible perhaps implies.

The Red Sea has a surface area of roughly 438,000 km2 (169,100 mi2),1 is about 2250 km (1398 mi) long and, at its widest point, 355 km (220.6 mi) wide. It has a maximum depth of 3,040 m (9,970 ft) in the central Suakin Trough,[3] and an average depth of 490 m (1,608 ft).

Assumptions for the purposes of this question

We cross where the sea is 100km wide, the depth is 500m at its maximum, the seabed is an arc of a circle.

The density of Red Sea Brine is 1100 kg/m3

The corridor for the Israelites is 10m wide. They can walk at a steady pace of 4kph regardless of slope.

Question

If we simplify by saying they all leave and arrive at the same time:

(a) How long will God have to hold the waters apart and

(b) What will be his total energy expenditure both for opening the corridor and then holding it open?

• Comments are not for extended discussion; this conversation has been moved to chat. – L.Dutch - Reinstate Monica Nov 30 '18 at 2:15
• I wish you had asked God to part a generic, idealized sea instead of the Red Sea. The thing that bugs me the most is the seafloor requirement -- most seas don't have a circular bottom and an idealized sea (as opposed to a bathtub) would have a flat bottom rather than circular. The Red Sea has a tectonic rift along its length and it's a bummer to leave out the most interesting thing about the sea. – Spencer Nov 30 '18 at 17:45
• Since this is a "Hard Science" question, I'll just leave this here... US Scientists use Computer Model to demonstrate how a strong East wind could have parted the Red Sea (pub. 2010) – Chronocidal Dec 3 '18 at 14:34

Redrawing the original diagram to make it clear what I'm assuming the parted volume looks like:

The section through the sea bed is specified to be a circular arc and so, given the constraints, we can work out the radius of the circle using Pythagoras' theorem, and the length of the arc with some simple trigonometry:

The distance walked is 100,007m, and at a constant speed of 1.111ms-1 that gives a crossing time of 9.00x104s (25 hours).

To calculate the energy required to displace this volume of water, we need a bit of calculus. Assuming that the displaced water is lifted up to the top of the sea, and assuming that the effect on the overall sea level is negligible, then we need to calculate how much energy it takes to raise every drop of displaced water to sea level (h=0).

To raise one kg of water by one meter requires 9.81 joules of energy. But the height through which we need to raise the water varies continuously with depth. We need to divide the displaced volume into (infinitely many) horizontal slices – so that each slice is at the same height – and then sum the energy required to raise each slice.

• the width of each slice is a function of height
• the area of each slice is a function of width
• the "specific mass" of each slice (its mass per vertical meter) is a function of area
• the energy required to raise each slice is a function of specific mass and of height

– therefore we can write down the required energy per slice as a function of height.

$$E=-9.81 m_s h$$ $$m_s = 22000 w$$ $$w = \sqrt{-h^2 + 5.0\times10^6 h + 2.5\times10^9}$$ $$E=-2.158\times10^5h\sqrt{-h^2 + 2.5\times10^6 h + 5.0\times10^9}$$

(I've omitted most of the algebra). The required total energy is then

$$E_T = -2.158\times10^5\int_{-500}^0{h\sqrt{-h^2 + 5.0\times10^6 h + 2.5\times10^9}}dh$$

It's not obvious whether someone who was better than me at calculus could solve that analytically; anyway, my computer comes up with the numerical solution

$$E=2.248\times10^{19}\;\mathrm{J}$$ $$=22.5\;\mathrm{exajoules}$$

For comparison, this would completely drain 2.3 million billion fully-charged iPhone X batteries.

This does not take into account the energy to keep the waters parted. As others have pointed out, this depends exactly how the miracle is done; if it was done by raising the sea floor, or by creating some kind of wall to hold the water back, those are static arrangements which don't require any energy input at all. If it was more like the Charlton Heston movie – with the water seemingly flowing as normal, but being continually driven back on either side – then that would take considerable power. The description in Exodus 14 mostly sounds like the tide going out (i.e. the sea floor being raised), but verse 29 says "the waters were a wall to them on their right hand and on their left", which does sound more like the movie version.

The energy to continually push the water back would depend on the rate at which it flowed into the evacuated volume, and on the height through which each drop of that water had to be raised to remove it again. A "realistic" model of the fluid dynamics would be beyond me, and probably very complicated, but if we imagine each wall as the opening of a giant pipe, where:

• cross-sectional area is 4.7x106m2
• mean depth below sea level is 200m
• mean water pressure is 2x106Pa

– then (adapting the math for volume flow on this page) we can estimate that each wall is pumping out 9.4x109kg of water per second. It takes 1962J to raise 1kg of water by 200m; multiplying these numbers, we get a power requirement of 1.84x1013W (about 18 terawatts), or 36TW for both sides. That figure seems quite high, but I don't know if it's an error on my part or not; it's certainly true that this would take a formidable amount of power. Anyway, over 25 hours that comes to 1.66 exajoules, or 0.3% of the world's current annual energy consumption.

# Theophysics

It might seem like this sort of discussion is moot, or a joke, because divine intervention invalidates the laws of physics anyway. But that's an artifact of the very modern, deeply flawed idea that science and theology are somehow opposites. Enlightenment scientists like Isaac Newton saw them as basically the same thing, and they didn't have a problem with the idea that literal miracles could be reconciled with physical laws. It's actually quite a broad and interesting philosophical area.

One common approach to the parting of the Red Sea, for example, would be what we might now call a "Maxwell's Demon" model. If the universe is completely deterministic (or if, wherever quantum dice rolls are involved, God has manipulated the outcome), then every atom has been ordered, from the beginning of time, such that at the appointed time, the momenta of all the molecules in the Red Sea "just so happen" to be arranged in a highly improbable way that causes the waters to part.

The important thing to note is that this doesn't violate the presumed rules of physics; it's just that God has played an unfathomably crafty game within the rules. Conservation of energy is not violated (implicitly, the energy "comes from" the thermal energy of the surrounding environment). The laws of thermodynamics are not violated because, even though it seems like a huge amount of entropy has vanished from the system, what's actually happened is that there was an entropy deficit all along, but it was hidden – essentially, encrypted – in the state vectors of every particle in the universe from the start of time. This does have thermodynamic implications for whatever was going on prior to the book of Genesis, but that is beyond the scope of the conversation.

There are many other approaches to this sort of question (and to related questions concerning free will), but my point is that if you assume stuff like this happened at all, that doesn't mean you have to throw science out the window.

• In your calculus, you omitted a factor of "h" in your energy calculation before integrating. When E=g * ms * h you substitute for g and ms (which depends on "h"), but then you dropped the factor of h, which drastically changes the magnitude of your integration. – Mathaddict Nov 29 '18 at 22:42
• Amazing diagram - what software did you use? It's at this point that I start to doubt my wisdom in choosing the Hard-Science tag. I might have known that the deeper the analysis, the less I would understand it and so be able to judge how good an answer is. – chasly - supports Monica Nov 29 '18 at 22:46
• @Mathaddict you're right, and I was also missing a minus sign and a number 2 elsewhere (TBH I'd be impressed if those are the only mistakes I made...). My answer is still very different to yours, though it does seem like you're taking the depth to be 50m instead of 500m (or 50km instead of 0.5km?) – bobtato Nov 30 '18 at 1:19
• @chasly I used Adobe Illustrator (CS3!). As per Mathaddict's answer, you don't actually need to use calculus – I started doing it that way because I was originally going to do something a bit more ambitious. But if our two results eventually end up the same, it'll be a good cross-check! – bobtato Nov 30 '18 at 1:26
• @bobtato I used 50 km in my calculations because I was looking at half the width of 100km to find the angle of the right triangle in my circle construction. This is why I double the tangent in my cross sectional area calculation. I'm doing all my calculations in km instead of m so I'm actually using 0.5 as my depth. – Mathaddict Nov 30 '18 at 17:26

# Time it takes the Hebrews to cross

The Hebrews walk half of the circumference of an ellipse with the dimensions that you specify. The circumference of an ellipse is

$$C = 4a\int_0^{\pi/2}\sqrt{1-e^2\sin^2\theta}d\theta,$$

where $$a$$ is the semi-major axis and $$e$$ is the eccentricity. The semi-major axis is half of the 'long way' across the ellipse, which is 50,000 meters. The eccentricity is

$$e = \sqrt{1-b^2/a^2},$$ and $$b$$ is the semi-minor axis, or 500 meters. Thus $$e = 0.9999500$$. Lets just round that to 1. If we do so, everything becomes a lot easier. The integral becomes

$$\int_0^{\pi/2}\cos\theta d\theta,$$ which evaluates to 1, so half of the circumference of the ellipse is just $$2a$$ or 100 km. At 4 kilometers per hour, this takes 25 hours to cross.

# Now, assume God is omnipotent

I'm Catholic, so I think this is a good assumption. God created all things, and he can presumably un-create all things. If God simply wants to un-create the waters in the Red Sea, he can do so, for a net energy gain of $$E = mc^2$$, per mass unit. But, he can then create those same waters at some distance away and let them simply flow back towards the void that he is un-creating. The cost of creating those same waters is also $$E = mc^2$$.

Normally, I would say that expecting a process to be 100% thermodynamically efficient is silly, but we are talking about God here. So assuming that God is omnipotent, and that he can create and annihilate matter with perfect thermodynamic efficiency, then the net energy cost to God is 0, to hold the waters open for any amount of time.

Sorry, but if you want me to assume God is not omnipotent, you will have to put it in your question and I will edit this answer appropriately.

• +1 for requiring God to follow the physics laws that hard-science answers demand, but not limiting yourself to the obvious ways to make it happen! – Cort Ammon Nov 29 '18 at 3:39
• since water going to constantly try to fall back in the gap, you can probably rough it at 10 joules of energy per kg, per meter of displacement, per second, which is how much energy it would take to lift that mass that distance. – John Nov 29 '18 at 3:55
• @John Why does God have to lift anything? What are space and time to the Creator of both? – kingledion Nov 29 '18 at 4:01
• @kingledion that's assuming they are the creator of both, and not just another random god showing off 😉 – John Nov 29 '18 at 4:10
• Why are you assuming an ellipse? Why are you assuming the Catholic god? What happens when the suddenly created cubic meters of vacuum suddenly collapse? – L.Dutch - Reinstate Monica Nov 29 '18 at 6:08

How long does it take?

Call R the radius of the circle that the seabed follows: $$R^2=(\frac{w}{2})^2+(R-d)^2=2500+R^2-R+0.25$$. Solving for R gives $$R=2,500.25 \text{ km}$$. Then the arclength of the circle on the seafloor is $$D=2R \times tan(\frac{50}{R})= 100.0333428 \text{ km}$$ to be traveled at 4 km/hr gives

a) 25 hrs and 30 seconds. (with 0.008522 fractions of a second remaining)

*Remember the assumption in the question is that they all leave and arrive at the same time.

How much energy does it take?

Instead of assuming that the water just winks out of existence and is actually moved into the surrounding red sea, then we can solve this using a potential energy calculation. First the Volume of water displaced will be $$V=A*10 \text{ m}$$. And $$A=2 \times tan(\frac{50}{R}) \times pi \times R^2 - (R-d) \times \frac{w}{2}= 660,593.9 \text{ km}^2$$, so

$$V=6,605.939 \text{ km}^3$$, and at a density of 1100 kg/m^3. the mass, $$m = 7.2665329 \times 10^{15} kg$$ of water.

Regardless of how this water gets there, it will, from an energy perspective, need to be lifted to the surface of the sea, so we'll need the centroid of the water being moved to see how far it needs to be lifted. The formula for the distance from the bottom of the sea to the centroid is $$x=R-4/3 \times \frac{R*sin^3(a)}{2a-sin(2a)}$$, where $$a$$ is the angle $$a=tan(\frac{50}{R})$$. Doing the calculations, $$x=0.3 \text{ km}$$. So the height that the water will be lifted is

$$h=d-x=0.2 \text{ km}$$ (we could also use calculus to do this, but we don't need it)

The gravitational potential energy required is $$E=mgh$$, where g=9.807 {m/s$$^2$$ so...

b) $$E=7.2665329 \times 10^{15} \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.3 \text{ km} = 2.1378866 8 10^{16} \text{ kJ}$$

It doesn't take any additional energy to hold the water in place, energy is measured as a force over a distance, and since there is no movement when it is holding still, there isn't any additional energy needed to hold it in place.

• Good answer, I was developing my too but my basic math knowledge was killing me due to its inaccuracy. – Ender Look Nov 30 '18 at 0:17
• Can you explain why it doesn't take energy to prevent the rest of the Red Sea from flooding the open channel? My thought experiment goes like this, if I'm sitting in a small pool holding two 'walls' while my child crawls through the 'dry' channel, my arms are going to get tired from the energy being expended holding the walls in place. There is a fairly simple formula for how much energy a dam needs to withstand given the height, width, and density of the water. Wouldn't that energy need to be summed over the 25 hours? – Tracy Cramer Nov 30 '18 at 0:32
• It isn't energy that the dam needs to hold the water back, it's Force. Energy is force over a distance, and the dam is standing still. Using your other analogy, if instead you holding the two walls, you put a broom handle between them, the broom doesn't expend any energy while holding the walls apart. – Mathaddict Nov 30 '18 at 17:22
• @Mathaddict, Sorry, I'm still missing something. Let's try another example. If I take a 10kg dumbbell and raise it over my head, I've expended x amount of energy to get it there. If there is no more energy required to keep that weight above my head, why does my arm get tired? – Tracy Cramer Nov 30 '18 at 17:44
• The answer is that your muscles tire both from exerting a force or from producing energy because of the way that they work. But since there is a way to hold up a weight like that using no energy (like putting it up on a shelf, which has no electrical power) you can't say that keeping it at the same elevation uses energy. The main point is that energy does not equal force, even if force sometimes tires muscles too. – Mathaddict Nov 30 '18 at 17:48

(a) is a deceptively complex question. This is basically a fluid viscosity problem comprised of 6,000,000 units (600,000 men + wives + children + "mixed multitudes" + flocks + herds). With the mix of variables between children, elderly, chaotic animals, bathroom/rest breaks, coral, mud, hectic terrain, etc you are going to see huge swings in crowd viscosity. I would think that traffic flow would essentially be stopped (at least in part) for much of the trek through the corridor.

Realistically I would estimate that at the far end of a 100km x 10m corridor the exit rate would be 10,000 units per hour. In total it would take about a month to get everyone across.

With perfect coordination and flow, no accidents, breaks or interruptions, and every man, woman, child, and beast willing to walk day and night it would take a little over 4 days for everyone to cross.

Rows of 20 units wide

6,000,000 units/row (20) = 300,000 rows

300,000 rows * 1 meter of separation = 300 kilometer caravan

300 kilometer caravan + 100 kilometer sea = 400 kilometers of travel

400/4kph = 100 hours

100/24 = 4.1 days

With densely packed units (only 0.5 square meters per unit which includes their food, water, and possessions) perfect coordination and flow, no accidents, breaks or interruptions, and every man, woman, child, and beast willing to walk day and night it would take a little over 4 days for everyone to cross.

(b) Whatever energy it takes to create or destroy energy can be created to supplement the difference. The net result is zero.

• "Whatever energy it takes to create or destroy energy", If it takes energy to create energy then you haven't really created it. You've just transformed from it from one type of energy to another. – chasly - supports Monica Nov 29 '18 at 18:49
• @chaslyfromUK That is true. God would be forever making up the ever shrinking difference. But if we want to use calculus we could prove that eventually there would be a convergence. – Skek Tek Nov 29 '18 at 19:08

kingledion pointed to an intriguing possibility within their answer: The stupendous amount of energy could be supplied with help of Maxwells demon, who somehow sorts the water molecules acording to speed etc. to generate useful work.

Out of sheer fun, I did the math by how much we would need to cool the water to gain the energy to lift it. Using the numbers from Mathadicts answer, with $$1.62*10^{15}$$ kg of water with a specific heat capacity of $$4.182 \frac{kJ}{kg K}$$, to get $$2.13*10^{16} kJ$$ we need to cool the water by ... 0.77K.

The relationship between potential (=height) and thermal energ in water is fun to ponder.