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Let's say we want to write down the metric in a uniform field. I see two ways of going about this.
Method 1: Straightforward arguments using the equivalence principle and photons in elevators show that if a photon with initial energy E rises or falls by dy, then its energy shift is given (ignoring signs) by [itex]dE/E=g dy[/itex]. Integration shows that the time dilation factor between two different heights is [itex]\exp(\Delta \Phi)[/itex], where [itex]\Phi[/itex] is the gravitational potential. If two clocks have parallel world-lines at two different heights, then the ratio of the proper times is the square root of the ratio of the time-time elements of the metric, so we have
[tex]ds^2=e^{2\Delta\Phi}dt^2-dy^2 \qquad [1][/tex]
Method 2: Start in a frame where the metric is Minkowski. Find the motion of an observer who experiences constant proper acceleration. Transform into this observer's frame by using the tensor transformation law on the metric. The result is
[tex]ds^2=(1+ay)^2dt^2-dy^2 \qquad [2][/tex]
This is given in Semay, http://arxiv.org/abs/physics/0601179 . By the equivalence principle, it can also be interpreted as the metric experienced by an observer in a gravitational field with g=a.
If we set [itex]\Phi=gy[/itex], then these two forms are equivalent to the first non-constant order:
[tex] g_{tt} = 1+2gy+\ldots \qquad [/tex] ,
but they disagree in their higher-order terms.
What is the reason for this discrepancy?
Rindler's Essential Relativity (2nd ed., 120) suggests using the gravitational redshift to define the gravitational potential. I'm not sure if this is meant to suggest that the potential in a uniform field is not necessarily exactly [itex]gy[/itex], or if it's meant to allow the generalization to nonuniform fields (which he carries out a few pages later).
Do the two forms differ because there's an implied choice of coordinates that is different in the two cases? Is this perhaps related to Bell's spaceship paradox, i.e., to issues in defining the notion that two different objects both experience the same proper acceleration, due to the relativity of simultaneity?
Form [1] (with [itex]\Phi=gy[/itex]) has a property that I would consider indispensable for a uniform field, which is that I can't determine my y by local measurements. With form [2], I can take a vertical measuring rod with clocks at each end, and depending on what y I'm at, the ratio of the clocks' rates will be different. This seems physically wrong to me, even in the case where you interpret it as an acceleration rather than a gravitational field. An observer inside an accelerating rocket should not be able to determine what point in the motion he's presently at, using local measurements -- should he?
This is probably irrelevant, but there are also the x and z coordinates. I've seen a nice elementary argument by Born that because a measuring rod in a rotating frame of reference experiences a Lorentz contraction when oriented in the transverse direction, by the equivalence principle a rod's length varies with height in a gravitational field. This would presumably apply to x and z, not y. (This argument is apparently made in Born, 1920, Einstein's Theory of Relativity, which was an early popularization of GR. i don't have the book yet, so I'm just doing this from a summary of Born's argument.)
Method 1: Straightforward arguments using the equivalence principle and photons in elevators show that if a photon with initial energy E rises or falls by dy, then its energy shift is given (ignoring signs) by [itex]dE/E=g dy[/itex]. Integration shows that the time dilation factor between two different heights is [itex]\exp(\Delta \Phi)[/itex], where [itex]\Phi[/itex] is the gravitational potential. If two clocks have parallel world-lines at two different heights, then the ratio of the proper times is the square root of the ratio of the time-time elements of the metric, so we have
[tex]ds^2=e^{2\Delta\Phi}dt^2-dy^2 \qquad [1][/tex]
Method 2: Start in a frame where the metric is Minkowski. Find the motion of an observer who experiences constant proper acceleration. Transform into this observer's frame by using the tensor transformation law on the metric. The result is
[tex]ds^2=(1+ay)^2dt^2-dy^2 \qquad [2][/tex]
This is given in Semay, http://arxiv.org/abs/physics/0601179 . By the equivalence principle, it can also be interpreted as the metric experienced by an observer in a gravitational field with g=a.
If we set [itex]\Phi=gy[/itex], then these two forms are equivalent to the first non-constant order:
[tex] g_{tt} = 1+2gy+\ldots \qquad [/tex] ,
but they disagree in their higher-order terms.
What is the reason for this discrepancy?
Rindler's Essential Relativity (2nd ed., 120) suggests using the gravitational redshift to define the gravitational potential. I'm not sure if this is meant to suggest that the potential in a uniform field is not necessarily exactly [itex]gy[/itex], or if it's meant to allow the generalization to nonuniform fields (which he carries out a few pages later).
Do the two forms differ because there's an implied choice of coordinates that is different in the two cases? Is this perhaps related to Bell's spaceship paradox, i.e., to issues in defining the notion that two different objects both experience the same proper acceleration, due to the relativity of simultaneity?
Form [1] (with [itex]\Phi=gy[/itex]) has a property that I would consider indispensable for a uniform field, which is that I can't determine my y by local measurements. With form [2], I can take a vertical measuring rod with clocks at each end, and depending on what y I'm at, the ratio of the clocks' rates will be different. This seems physically wrong to me, even in the case where you interpret it as an acceleration rather than a gravitational field. An observer inside an accelerating rocket should not be able to determine what point in the motion he's presently at, using local measurements -- should he?
This is probably irrelevant, but there are also the x and z coordinates. I've seen a nice elementary argument by Born that because a measuring rod in a rotating frame of reference experiences a Lorentz contraction when oriented in the transverse direction, by the equivalence principle a rod's length varies with height in a gravitational field. This would presumably apply to x and z, not y. (This argument is apparently made in Born, 1920, Einstein's Theory of Relativity, which was an early popularization of GR. i don't have the book yet, so I'm just doing this from a summary of Born's argument.)