3 ms·

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 ei

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.

tedunangst 12y ago
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.
- haberman 12y ago
  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.
  - tedunangst 12y ago
    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.)
    - Smudge 12y ago
      It depends on the location of the observer. (coming towards you = blue, moving away from you = red)
  - Nursie 12y ago
    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 http://en.m.wikipedia.org/wiki/Chimborazo
- ithkuil 12y ago
  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%.
deleted 12y ago
[deleted]
HCIdivision17 12y ago
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 http://www.wolframalpha.com/input/?i=orbital+velocity+of+ISS [1] http://en.wikipedia.org/wiki/Machine_epsilon 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.
- haberman 12y ago
  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.
  - 12y ago

tedunangst 12y ago

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.

haberman 12y ago
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.
- tedunangst 12y ago
  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.)
  - Smudge 12y ago
    It depends on the location of the observer. (coming towards you = blue, moving away from you = red)
- Nursie 12y ago
  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 http://en.m.wikipedia.org/wiki/Chimborazo
ithkuil 12y ago
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%.

deleted 12y ago

[deleted]

HCIdivision17 12y ago

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 http://www.wolframalpha.com/input/?i=orbital+velocity+of+ISS [1] http://en.wikipedia.org/wiki/Machine_epsilon 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.

haberman 12y ago
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.
- 12y ago