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exmarine
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What does the metric look like for constant uniform acceleration, say in the x-direction?
ds^2 = g_tt (cdt)^2 + g_xx (dx)^2
g_tt = ?
g_xx = ?
ds^2 = g_tt (cdt)^2 + g_xx (dx)^2
g_tt = ?
g_xx = ?
exmarine said:by the equivalence principle, it seems like my pocket watch should run at different rates, depending on how hard I am being accelerated, say in the x-direction?
Yes, because of their different position along the line of proper acceleration of their common rest frame. Not because they themselves might experience different proper accelerations.exmarine said:I thought the Pound-Rebka experiment showed that proper clocks run at slightly different rates at the top and bottom of the tower.
No, that would contradict the clock hypothesis:exmarine said:Then by the equivalence principle, it seems like my pocket watch should run at different rates, depending on how hard I am being accelerated, say in the x-direction?
Gravitational time dilation is not a local effect.exmarine said:that gravity and ACCELERATION were equivalent locally.
That is not a clock rate. That is the clock rate ratio compared to a clock at infinity.exmarine said:I understand that a clock rate here on the surface of the Earth is equal to the square root of one minus the ratio of r_s over r, etc.
No. Gravitational time dilation is related to gravitational potential difference, not to proper acceleration.exmarine said:Then if I am in distant space and fire the rocket motor at about 1 g acceleration, will my onboard clock run at the same rate as one here on the surface of the earth?
exmarine said:I thought the equivalence principle was that it is impossible for any experiment to tell the difference between acceleration and gravity, that gravity and ACCELERATION were equivalent locally.
A.T. said:Gravitational time dilation is not a local effect.
exmarine said:if I am in distant space and fire the rocket motor at about 1 g acceleration, will my onboard clock run at the same rate as one here on the surface of the earth?
bcrowell said:GR doesn't have anything that really acts exactly like a uniform gravitational field. See
http://physicsforums.com/showthread.php?t=377254
A.T. said:I was wondering about a non-concentric spherical cavity inside of a uniform density sphere, which does have uniform Newtonian gravity. But I guess there is no known analytical EFE solution for it.
I meant only inside the cavity:bcrowell said:I don't understand why you would say that the system you describe would have uniform Newtonian gravity. I think the field would vary in both magnitude and direction.
A.T. said:I meant only inside the cavity:
http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm
I think you meant problem 6 of chapter 5!bcrowell said:he trouble is that no such spacetime exists. See ch. 6, problem 6 of my GR book, http://www.lightandmatter.com/genrel/
Yep, thanks!Shyan said:I think you meant problem 6 of chapter 5!
The equation for calculating the metric for uniform constant acceleration is a = Δv/Δt, where a is the acceleration, Δv is the change in velocity, and Δt is the change in time.
The metric for uniform constant acceleration is a measure of the acceleration at a specific moment in time, while average acceleration is a measure of the overall change in velocity over a period of time. The metric for uniform constant acceleration only takes into account the change in velocity at one point, while average acceleration considers the change in velocity over a period of time.
The metric for uniform constant acceleration is important in physics because it is used to describe the rate at which an object's velocity changes over time. This is necessary for understanding and predicting the motion of objects and systems in the physical world.
The units of measurement for the metric for uniform constant acceleration are meters per second squared (m/s2).
Yes, the metric for uniform constant acceleration can be negative. A negative acceleration indicates that an object is slowing down, while a positive acceleration indicates that an object is speeding up.