Probably not. You're looking at a height difference of about 9 km [1], at a radius of 6380 km from earth's core [0] versus about an equatorial 6371 km. Rotational velocity at the equator is about 1040 mph [5], while Everest is looking at about 6380 km / 6371 km * 1040 mph = 1041.5 mph, assuming we moved Everest to the equator and kept everything else the same. One or two mph is really small compared to c, so much so WA just sorta says gamma = 1 [4].

And so is a mere 0.15% increase in height. But compare the gravitational acceleration at these two heights:

[2] Everest = 9.76322 m/s^2

[3] Sea Level = 9.831 m/s^2

Sea level is 0.69% stronger. That's somewhat more significant, and given that gravitational energy follows the inverse-square law, it makes some sense that it would amplify differences more than the linear effects of speed.

[0] https://www.google.com/search?q=radius+of+earth

[1] https://www.google.com/search?q=height+of+everest

[2] http://www.wolframalpha.com/input/?i=gravitational+accelerat...

[3] http://www.wolframalpha.com/input/?i=gravitational+accelerat...

[4] http://www.wolframalpha.com/input/?i=lorentz+factor+at+1.5+m...

[5] http://www.wolframalpha.com/input/?i=rotational+velocity+of+...

EDIT: forgot the reference to rotational velocity (and spelling). Note that the point is scale here. Gravity is n^2 versus velocity's n. It'll have a stronger effect in most cases (and any where speed overcomes, well, we call them relativistic speeds, and they'll often be some appreciable fraction of c, like 1% or more).

DOUBLE EDIT: 'AnimalMuppet makes a good point that the entire estimate for Everest's rotational velocity is flat out wrong. So I've changed my premise to match my initial incredibly incorrect assumption.

If I were to attempt to be correct about the problem, we'd see that Everest is cos(27.9881 degrees [10]) = 0.8837 [11] as fast as the equator, which sorta blasts out any height changes by long shot. (Note I'm assuming the earth to be a perfect, frictionless sphere here, and definitely not a oblate spheroid.)

[10] https://www.google.com/search?q=latitude+of+everest

[11] https://www.google.com/search?q=cos(27.9+degrees)

You're forgetting that Everest isn't on the equator. Distance from the center of the earth doesn't matter; distance from the axis of rotation matters. So you need to multiply Everest's (height + radius) by the cosine of its latitude.

You're entirely correct, and I'm going to modify my answer to a convenient lie (in the true style of a physicist).

The point stands, but now I'm interested in what kind of cones we could make where relative heights and latitudes yield identical velocities. Prolly as useful a question as most XKCD What If? notes :)

It's gravity. The mountaintop doesn't travel faster through space-time. There's no such thing as absolute speed, there's only the velocity compared to something else.

Special relativity covers time dilation due to relative velocity, general relativity covers time dilation due to gravity.

Even in relative terms, the top of Mt. Everest is traveling faster than objects at sea level, no?

Wikipedia says: "Time dilation is caused by differences in either gravity or relative velocity. Both factors are at play in the case of ISS astronauts (and are actually opposing one another)."

Why wouldn't the same apply to the top of Mt Everest. Sure, unlike the ISS, Everest is actually attached to the Earth, but still it is traveling a longer distance than the oceans over the same period, and therefore would be moving faster in relative terms AFAICS.

You're correct, and my napkin guess is that it's going about 1.5mph faster than sea level. But the ISS is going a whopping 16100 mph faster than sea level [0].

Popping that into MS Mathematics using the Lorentz equation, we see that gamma now is 0.2 billionths above unity. Which is super small, but definitely detectable. In contrast, 1.5mph doesn't yield anything - I'm guessing (1.5/670616629)^2 is nearing the floating point epsilon (I guess MS Math is using floats?) [1].

[0] http://www.wolframalpha.com/input/?i=orbital+velocity+of+ISS

[1] http://en.wikipedia.org/wiki/Machine_epsilon

EDIT: Right, as noted elsewhere, this assumes Everest is at the equator. Which Wikipedia tells me is not a great approximation for where Nepal is. So assume that Nepal ended up in the wrong place for a little while, and the arguement holds. Otherwise, we've got a bit more calculating to do.

Thanks for the info!

What's the right answer if someone makes this counterargument: in the inertial reference frame of the Earth the Earth is not rotating, so neither the sea nor Mt. Everest are moving at all, so there is no relative velocity difference and no time dilation from velocity difference.

Btw, I'm not sure about the phrase "I'm guessing (1.5/670616629)^2 is nearing the floating point epsilon" (unlike physics, I do have a fair amount of expertise in floating point formats). The way you put it suggests that the "machine epsilon" represents the smallest increments that floating point can represent, and therefore suggests that (1.5/670616629)^2 represented as float will underflow to zero or be so inaccurate as to be meaningless (sorry if I'm misreading you).

I find the definition of "machine epsilon" given on that page somewhat confusing. I think it's more intuitive to think of floating point error in terms of percentages. That page says the "machine epsilon" for float is ~1e-7; an equivalent and IMO more intuitive way to say it is that float is accurate to ~0.00001%.

The range of float goes far smaller than 1e-7; FLT_MIN is ~1e-38, and that's not even considering subnormal numbers. So float can very easily represent the results of (1.5/670616629)^2 ≈ 5e-18, and like any other representable float this is accurate to no worse than ~0.00001%, which is pretty decent.

On the topic of the floating point mechanics, I'll readily defer to you. I'll be honest: I was grasping for an explanation of why the result was simply a solitary "1", and there's likely a much smarter explanation (like "After a number of decimal places, nobody gives a damn." - false, but useful typically).

As for the counterargument: I don't think that works. And that's taking into account the harrowing liberties I'm willing to assume for the sake of a physics argument. In the inertial reference frame of the Earth, the Earth is rotating - as in, there's rotational inertia in that frame. This has measureable effects on stuff, from time dilation from velocity (very small) to time dilation from gravity warping spacetime (noticeably larger, but still small).

Put another way, here's a thought experiment. The mantel of the planet is molten rock, and can be treated as a viscous fluid. If the planet were spinning, the fluid would bulge out at the latitudes where the planet is spinning fastest (centrifugal force stuff). Otherwise the planet would be a sphere. This is directly measurable, and - in fact - the planet's a sphere. Mostly. It bulges out a bit at the equator [0].

So if you're careful, you'll note it's never fair to say you're in a non-rotating inertial frame on earth. But in practice it almost never matters. Unless you're doing something crazy like measuring femtillionths of a second with one of the most sensitive devices we can build - then we start being a bit wrong. Or you're just trying to be accurate with your GPS satellites (which are way higher and faster than a mountain top).

I rant a bit about this, as this sort of counterargument comes up a lot. I think it's due to the completely unreasonable mismatch of scales people are used to. Feynman ranted a bit on it about QM, and we're running into it here with relativity [1]. Here we're trying to talk about something reasonably, and the levels of precision are completely unreasonable: we're talking about a couple mile's difference over the span of thousands of miles to have an effect on the order of a few (bi|tri)llionths of a second; how do you keep such a scale in mind? It's like the silly analogies of hitting a baseball in NYC and nailing a bumble bee in San Fransico for precision. By the same token, we're hurtling through space, whipping around the sun and being wobbled so hard by the moon the ocean sloshes over our beachfronts. And that seems perfectly normal, even though it directly implies enormous forces at work.

/rant (and sorry for that - I find this sort of thing facinating)

TL;DR: I think that counterargument is simply wrong. But completely understandably so.

[0] http://image.gsfc.nasa.gov/poetry/ask/a11818.html

[1] http://bouman.chem.georgetown.edu/general/feynman.html

Your counter argument is correct, in my opinion.

Is the top of Mt. Everest traveling faster? The distance between here and there is a constant, no? From my frame of reference it's not moving at all.

your scenario to the extreme: a spacecraft is traveling at 0.5c in an circular path around earth at 1/pi light days of distance. From your (non inertial) frame of reference it's not moving at all. Yet the spacecraft is experiencing time dilation of about 15%.

Mt. Everest is higher. Because of the Earth's rotation, every point on earth traces 360 degrees every day. However the radius of these arcs depends on elevation. Ergo the top of Mt. Everest travels farther every day due to rotation that any other place on earth. That is the same as saying it is moving faster.

EDIT: I guess compared to a rotating reference frame this isn't true? Clearly this isn't my area of expertise.

Not that I disagree with your point in any way, but I thought you might like to know - the top of Everest is not the furthest point from the centre of the earth. It is the highest point above sea level, but given the slightly ellipsoid nature of the planet, the spot with the largest radius value is actually in South America - http://en.m.wikipedia.org/wiki/Chimborazo

I think a way to rephrase the question is to ask, is the light coming from the top of mt. everest red shifted or blue shifted or neither? (I'm not certain of the answer, either. Physics was a long time ago.)

It depends on the location of the observer. (coming towards you = blue, moving away from you = red)

There is no absolute velocity, but there is absolute acceleration. And the mountaintop is accelerating more because it's in circular motion with a larger radius.

However, the effect of gravity is much more significant (if it weren't, the mountaintop would be flying out into space due to centripetal force).

But if I'm standing on the Earth's surface, it's not accelerating relative to me. It's always at the same distance and direction.

Since acceleration is change in velocity, it's impossible to have absolute acceleration without absolute velocity. You can only say that it's accelerating compared to X.

It actually is accelerating relative to you, in exactly the same way that a kid on the outside of a spinny-go-round is accelerating harder than a kid on the inside. Of course, there's not a huge difference, as a couple km is spare change to the radius of the planet, but it's there.

There's definitely a constant vector difference between you and Everest's peak, and that's likely what you're thinking of. But Everest itself is under a greater strain to maintain that vector compared to you. And that's actually a measurable difference (where measurable is on scales of stupefying precision not normally used in everyday life).

And as another noted, it really should maintain that relative vector - if it didn't, the earth wouldn't be in steady-state and you'd have a changing position vector between you and it, and that'd imply one of you was moving... hopefully it's you.

(Let's ignore all the techtonic complications, as that really ruins the simplicity of the argument :)

EDIT: As noted elsewhere in the thread, Everest isn't on the equator, and - if I may be bold enough to presume - you likely aren't either. So it's not clear if you or Everest is on the outside of the spinny-go-round in the analogy.

For the purposes of wondering if velocity or gravitational effects on spacetime are the dominant factors, this has no effect. If you're actually interested in the effective relative centrifugal forces between you and Everest, then it's damn near everything.

> Special relativity covers time dilation due to relative velocity, general relativity covers time dilation due to gravity.

It's actually due to acceleration, whether from gravity or from motion. And, given earth's rotation, points on the surface accelerate (except for the poles).

But, as I pointed out elsewhere in this discussion, it's distance from the axis that determines acceleration from rotation, not distance from the center of the earth. (Distance from the center does determine gravity, though).