Special Relativity Time Dilation Santa question

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1. Nov 10, 2014

applestrudle

I am doing a presentation and want to make sure I'm not misunderstanding something very fundamental.

My argument goes like this:

t0 = ϒt

Santa is moving very fast and from his point of view he is in proper time. This means that if it takes him t0 seconds to deliver a present, the amount of time he observes passing on Earth is ϒt seconds and ϒt > t0.

This means on Earth it takes a lot longer to deliver all the presents (ϒt seconds) but for him it only takes t0 seconds. So he gets the job done in a shorter period of time.

Also, length contraction:

On Earth Santa needs to travel X m (X m is the distance between all the houses) but since he is moving so fast, the distance is reduced to

X0 = $$\frac{X}{\gamma}$$

The two effects mean he has to travel a shorter distance and more time passes on Earth compared to the 12 hours in which he delivers the presents.

So he can successfully do the task.

2. Nov 10, 2014

Staff: Mentor

Calenders work with earth time zones. I don't know where the 12 hours come from, but they certainly refer to the time as measured by all the families.
Time dilation does not help Santa, unless his shift ends after a few hours and he is not allowed to work overtime.

3. Nov 10, 2014

pervect

Staff Emeritus
I'm not sure where the 12 hours comes from either (why not 24?). But you can analyze everything from the Earth frame first, then consider Santa's frame second.

In the Earth frame, it takes (2 pi r) / c = .133 seconds to travel around the Earth at the speed of light, which is as fast as you can go. The total length of Santa's trip, which has to zig zag from house to house, must be less than around 323,000 circumnavigations around the Earth. I'm afraid I don't have an estimate of the shortest path that visits each house once, I can't convince myself for sure whether or not it's under the limit or not. though I can compute that the distance between 320,000 great circles going through the north pole would be 124 meters at the equator. But this is an inefficeint path, Santa doesn't need to drop many presents off over the ocean regions, also the paths are far a part at the equator but closely spaced at higher lattitudes.

If you give him a full day rather than 12 hours, of course, the time margin is better.

From Santa's point of view, due to time dilation on his trip, less than 12 hours will pass.

I'm not sure if we are assigning Santa some time to drop off each present, if we did, this part of the time budget would not be time dilated assuming Santa has to stop to drop off the presents. Only the travel time would be time dilated. Of course the total time (earth time) has to be divided between travel an delivery.

4. Nov 10, 2014

Staff: Mentor

This is true when you properly specify how the times are compared; but unfortunately a proper comparison hurts your case rather than helps it. The proper way to compare is this: have Santa start and end his entire journey at the same point (the North Pole). Then the time elapsed for Santa between leaving the North Pole and returning (after delivering all the presents) is t0, but the time elapsed for one of his elves who stays at the North Pole is ϒ * t0, which is greater than t0. So if 12 hours elapses for the elf, Santa has less than 12 hours to deliver all the presents (because less than 12 hours elapses for him).

This is true for a journey in a straight line. Santa's actual journey is not in a straight line, which complicates things, but I think it's still OK to say that the distance he travels, from his perspective, is shorter than the distance he has to travel from an Earth perspective.

5. Nov 10, 2014

phinds

One little known fact that is not presented here so far is that all of the extreme acceleration and deceleration (when Santa actually makes the stops) is what gave Rudolph the red nose. All the blood keep rushing to his head in the decelerations and didn't fully redistribute itself. Eventually he got the most extreme case of varicose veins you've ever seen.