Why does time slow down the faster you go?

[arabianights]

time doesn't slow down when you travel at faster speed, it flows at the same rate in any inertial system

time flow only changes pace when you accelerate or decelerate,

 

[PeterDonis]

I assume we can kick each whatever direction we want to escape to infinity in any desired path. In particular, there is an obvious better kick direction for C.

Yes, an object can achieve escape velocity in any direction (even *towards* the gravitating body, as long as it doesn't actually hit it--i.e., a close grazing approach), and still escape. So the obvious way to get C to escape is to boost it in the direction it's already moving because of the Earth's rotation.

Now I am a little queasy that these are not correlated to the time flow rates. Yet I find any other conclusion hard to accept.

That's why I posed the question. I felt queasy too until I thought about it some more. See below.

My immediate thought is that this consideration should have told me that someone co-rotating at the equator has a different energy to one stationary at the pole! When launching rockets into orbit, less energy is required if you start near the equator (C). If I recall, even Jules Verne took account of this in "from the Earth to the moon", and it is taken advantage of by the US launch sites, amongst others.

Yes.

The other two sites (A and B) require the same energy if they are at the same gravitational potential but are stationary w.r.t. the centre of the Earth.

But *are* they at the same "gravitational potential"? That's the question.

Moving from state A to state B requires some work against gravity. The result is time for B runs faster than A. To get from B state to C state, you have to do more work to add KE (1000 mph in polar inertial frame). However, this KE, leaving potential unchanged, but adding velocity, makes time slow down.

Note, I always insisted that the earth, including the sea, was oblate due to rotation. I said this a number of times, arguing that if you got to the equator on a frictionless surface just above the sea, time would be slower than at the pole. I may not have always been clear about the implication that this implies significant work is needed to get from the pole to the equator even on a frictionless surface.

The Earth is certainly oblate due to rotation. That's been known for a long time:

http://en.wikipedia.org/wiki/Figure_of_the_Earth

The equation for "gravitational potential", according to GR, is, approximately, this:

\frac{d \tau}{dt} = \sqrt{1 - \frac{2 G M}{c^{2} r} - \frac{v^{2}}{c^{2}}}

where I have included explicit factors of G and c, contrary to my usual practice.

I say "approximately" because there are at least two types of corrections that could be applied. One is that the Earth is not a perfect oblate spheroid, so there is a correction to the M/r term. It's been a while since I computed it, but IIRC it's small compared to the base correction expressed above. The other is that the Earth is rotating, so the spacetime metric around it is not perfectly Schwarzschild, which is what the above equation assumes; it's actually slightly Kerr, meaning there is an extra term for "frame dragging" due to the Earth's angular momentum. IIRC this term is way smaller than the others, so we can ignore it here.

Since the Earth itself is rotating, the shape of its surface (in the idealized case we've been considering) is determined by balancing v vs. r at every point to keep d \tau / dt constant. So the Earth's surface is equipotential, as Elroch said. But since "equipotential" includes the effects of v, if you are on the Earth's surface but your v differs from the local v of the surface at that point, your potential will *not* be the same as that of the Earth's surface at that point. So observer B *is* at a higher *potential* than observer A, as PAllen said; it *will* take work against gravity to move from A to B, even if the Earth's surface is frictionless.

So I agree with PAllen's answer here:

Next large kick needed for B, and most for A.

Since it takes work against gravity to move from A to B, that work can be subtracted from the rocket energy required to launch from B, so more energy is required for A than B. And since C has extra kinetic energy compared to B, but is at the same height, it takes less energy to launch C than B, as everyone seems to agree. So the full ordering is C < B < A.

Now, what about time flow rates? We have C < B < A for launch energy, but C = A < B for time flow rate. That's because the intuitive guess that potential energy, in the sense of "height", should be the same as time flow rate only holds for observers that are *static* in the field. A and B are static, but C is not. (Note that "static" means "not rotating with the Earth", because "static" is determined by the time translation symmetry of the spacetime as a whole, i.e., of the metric.) So we can directly relate the difference in time flow rate between A and B to the difference in their heights. But C is moving, so his time flow rate is a combination of two effects (as shown in the equation I gave above), his height and his velocity; and because the Earth's surface is equipotential, the slowdown due to his velocity is just right to offset the speedup due to his height, when compared with A. But compared to B, he's at the same height, so the only difference is his velocity, hence C is slower than B (and by the *same* amount as A, of course).

 

[Elroch]

Thanks for trying to summarize all this. I few little thoughts below...

A few caveats. First, even in a strictly conservative field, work to go from A to B is path independent only if you apply minimum force at all times. For arbitrary motion, where you are free to add and subtract KE at will, you can do any amount of work above the minimum to get from A to B. More substantively, not all work (even assuming paths that use minimum work) has the same impact on time dilation. This is shown clearly using Peter's nomenclature for B versus C. Moving from state A to state B requires some work against gravity. The result is time for B runs faster than A. To get from B state to C state, you have to do more work to add KE (1000 mph in polar inertial frame). However, this KE, leaving potential unchanged, but adding velocity, makes time slow down. This disproves your simple model.

Of course I meant the minimum work assuming no energy being lost (I had assumed zero friction simply as a way to do the energy calculation). The sign of the input of energy can vary over the path and the reverse path gives the same number multiplied by -1 (just run time backwards and look at the scalar product of the force and the motion). Otherwise energy would not be conserved.

I don't think I (or Peter) ever suggested this frictionless surface modeled anything (even ideally) about the real earth. I introduced this to clarify that the real ocean is inherently doing work via friction on a boat moving from the pole to the equator.

I infered this was the assumption because a previous post suggested a boat nudged at the pole with a frictionless sea would reach the equator moving very slowly in a stationary frame (specifically "1000mph Westward", cancelling out the rotation), exactly as if the Earth was stationary and spherical. If the surface is frictionless, the rotation of the Earth is irrelevant. We are not going to confuse this further by incorporating frame dragging!

Actually, the point of the frictionless surface was to show the rotational KE of a comoving equatorial boat had to come from somewhere. There may have been a few slopply implications that A to B was zero work, but that wasn't germane to the main discussion. Peter's latest scenario now makes very clear (see my answer to it) that A to B involves substantial work. Then B to C involves more work. These two components of work happen to cancel as to time flow.

I accept that you can't bundle kinetic energy into a souped-up potential like I attempted. What sounds more plausible is to incorporate the related "centrifugal force" as a sort of apparent antigravity. This may be possible in a weak field approximation?

Do bear in mind that all friction is utterly irrelevant to the actual issue under discussion, which was my conservation of energy argument to compare clocks. We are perfectly free to replace the Earth's surface with a frictionless one to compare energies.

See above. Note, I always insisted that the earth, including the sea, was oblate due to rotation. I said this a number of times, arguing that if you got to the equator on a frictionless surface just above the sea, time would be slower than at the pole. I may not have always been clear about the implication that this implies significant work is needed to get from the pole to the equator even on a frictionless surface.

The higher gravitational potential due to oblateness (i.e. nearer zero) speeds up time at the equator, rather than slowing it. Doesn't matter how you get there.

I was certainly misled by the claim (I think by PeterDonis?) that if you nudged a boat towards the equator from the pole with no friction it would reach the equator heading at "1000mph" West against the rotating Earth (implying to me that it was essentially stationary compared to a non-rotating Earth). Clearly if you are putting in or extracting energy, you can make it any velocity you like, so without other explicit assumptions about the forces, so this would make no sense.

 

[PAllen]

Of course I meant the minimum work assuming no energy being lost (I had assumed zero friction simply as a way to do the energy calculation). The sign of the input of energy can vary over the path and the reverse path gives the same number multiplied by -1 (just run time backwards and look at the scalar product of the force and the motion). Otherwise energy would not be conserved.

I figured this was understood, should not have bothered with the first caveat. The second caveat remains.

I infered this was the assumption because a previous post suggested a boat nudged at the pole with a frictionless sea would reach the equator moving very slowly in a stationary frame (specifically "1000mph Westward", cancelling out the rotation), exactly as if the Earth was stationary and spherical. If the surface is frictionless, the rotation of the Earth is irrelevant. We are not going to confuse this further by incorporating frame dragging!

This is where sloppy wording came in. It is correct, but the nudge may need to be fairly large. But the intent was always that this hypothetical frictionless surface followed the oblateness of the Earth cuased by its rotation.

I accept that you can't bundle kinetic energy into a souped-up potential like I attempted. What sounds more plausible is to incorporate the related "centrifugal force" as a sort of apparent antigravity. This may be possible in a weak field approximation?

Do bear in mind that all friction is utterly irrelevant to the actual issue under discussion, which was my conservation of energy argument to compare clocks. We are perfectly free to replace the Earth's surface with a frictionless one to compare energies.

You can make a combined pseudo-potential like that. However, it doesn't change anything about work needed to get from the pole to co-rotating at the equator. You can make the rigid skin at this potential co-rotating. Still, if you have a frictionless puck at the pole, you need to do work on it to get up to the equator's elevation and more work to get it to co-rotate. I demonstrated this by arguing from an inertial frame. Peter showed the equivalent result follows in a properly used rotating frame - you need to work against the coriolis force. In no way do you get from the pole to co-rotating at the equator without work being done.

The relevance to the original argument is that you can't ignore this work requirement in an energy conservation argument. Whether the surface is frictionless, and you have to explicitly push the object, or there is friction and the surface does work on the object by friction, work is done to get the object from the pole to co-rotating at the equator, and this absolutely must be accounted for in conservation of energy argument.

The higher gravitational potential due to oblateness (i.e. nearer zero) speeds up time at the equator, rather than slowing it. Doesn't matter how you get there.

Sorry, typo. Said it correct numerous other places.

 

[PeterDonis]

I was certainly misled by the claim (I think by PeterDonis?) that if you nudged a boat towards the equator from the pole with no friction it would reach the equator heading at "1000mph" West against the rotating Earth (implying to me that it was essentially stationary compared to a non-rotating Earth).

I think it was me, but I didn't intend to make that claim; I was only trying to explore how Coriolis force would affect the motion, and I didn't adequately consider the potential energy aspect when I made that previous post. Sorry for any confusion.

Edit: Modified further, see later post!

 

[Noja888]

Or is this only true when I am viewing someone else's clock?

Why does time slow down the faster something travels?

How could this apply to the Earth? How would our time on Earth slow down? Would the Earth have to be rotating faster and revolving faster around the Sun?

This is not an attempt to explain the math - just a visualization.

One way to think of it...Standing under a waterfall. The movement of the water washing over you is time acting upon you. Consider this 'washing over' the a reference point for time's flow. Now let's swim with the waterfall (moving at the speed of light). The waterfall does not wash over you anymore you caught up to it, time is not acting upon you, but for everything else it is, at least in your frame of reference. Everything else's clock is the waterfall's flow speeds into the future from your point of view. But to your view time moves at the "normal" pace. Its not an complete accurate description, but may help out visualizing this concept.

Space requires Time to observe it. Time requires Space to observe it.

 

[PeterDonis]

I think it was me, but I didn't intend to make that claim; I was only trying to explore how Coriolis force would affect the motion, and I didn't adequately consider the potential energy aspect when I made that previous post.

Oops, should have clarified further here. What I meant was that in my initial post on the Coriolis force, I didn't intend to imply that it would take zero energy to move an object from the pole to the equator, even on a frictionless Earth with no energy expended to offset the Coriolis force. What I said about the Coriolis force was true (*if* one moved an object from the pole to the equator, the Coriolis force would give it a westward velocity that, by the time you reached the equator, would be 1000 mph), but I should have clarified that the only way to actually realize the scenario I was describing would be to exert a constant force on the object due South (or North, if coming from the South pole). The force being perpendicular to the Coriolis force means its presence would not change the part about the westward velocity, but the Southward (or Northward) force still needs to be there.

 

[Elroch]

Oops, should have clarified further here. What I meant was that in my initial post on the Coriolis force, I didn't intend to imply that it would take zero energy to move an object from the pole to the equator, even on a frictionless Earth with no energy expended to offset the Coriolis force. What I said about the Coriolis force was true (*if* one moved an object from the pole to the equator, the Coriolis force would give it a westward velocity that, by the time you reached the equator, would be 1000 mph), but I should have clarified that the only way to actually realize the scenario I was describing would be to exert a constant force on the object due South (or North, if coming from the South pole). The force being perpendicular to the Coriolis force means its presence would not change the part about the westward velocity, but the Southward (or Northward) force still needs to be there.

The way I look at this, I get what I believe is the same answer in a trivial way. A very, very slippery rotating globe is exactly the same as the same globe when stationary, to an object moving on its surface. What happens is you push the object south, it goes straight south, but the globe rotates underneath it. No need to worry about Corioli's force at all.

[We really don't want to make this false by including frame dragging ... ]

 

[PeterDonis]

The way I look at this, I get what I believe is the same answer in a trivial way. A very, very slippery rotating globe is exactly the same as the same globe when stationary, to an object moving on its surface. What happens is you push the object south, it goes straight south, but the globe rotates underneath it. No need to worry about Coriolis force at all.

If you are working in a non-rotating frame, there is no Coriolis force, and this is basically what you are doing here; the lack of friction allows the object to move directly south as viewed in the non-rotating frame, without the Earth's rotation affecting its motion.

The Coriolis force only arises when you are looking at things from the point of view of the rotating frame (i.e., observers rotating with the Earth). From that point of view, the observer you describe acquires westward velocity as he moves south, so the "fictitious" Coriolis force is invoked to explain why. The fact that you can make the Coriolis force disappear by a change of frame is why it (and other forces like it, such as "centrifugal force") are sometimes called "fictitious". (My high school physics teacher used to joke that they were due to "gremlins".)