General relativity finds that gravity dilates the wavelength of light and the time period of an atomic clock. What does gravity do to the length of a (material) rod? If you know the answer, please cite me a reference or two.
You are no doubt feeling some frustration without any satisfactory response forthcoming. Has been my battle too - but have come to realize that in GR spatial distortion is a bit of a slippery eel thing. 'Go down there' to check and you shrink/bend/whatever same way as the rod, 'stay out here' and a host of issues re measurement and what it really means crop up. There is a short answer that may or may not satisfy you right here in PF: https://www.physicsforums.com/showthread.php?t=145285Special relativity is not in discussion here.
ln gereral relativity, here is how the frequencies of a vibrating particle are related:
(frequency at infinity) = (1 + k.m/r) (frequenct at r), where k is modified gravitational constant.
You can deduce dilatation of light wavelength and clock time period from the above.
Is that eq. true for a rod's length?
There is simply no "distance measure" we can use that is unaffected by gravity, so whatever effect gravity has on the rod, it will have on the distance measure too.
So can we place identical light sources on the ends of a radially placed rod in the vicinity of a mass and then observe the redshift (say) from a distant observational point...to measue distance??there is a frame-invariant physical observable we can use to measure it--the gravitational redshift/blueshift between observers "hovering" at different heights in the field
I don't understand what that means....You know that special relativity is a special case of GR I am sure....But questions of distance (length) get less precise in curved spacetime.Special relativity is not in discussion here.
How will you translate the redshift measurements into distance measurements?So can we place identical light sources on the ends of a radially placed rod in the vicinity of a mass and then observe the redshift (say) from a distant observational point...to measue distance??
Neglecting possible higher order derivative effects, what is written above is of course a logical necessity simply because, as I put it in #4 "'Go down there' to check and you shrink/bend/whatever same way as the rod," If I'm not mistaken SinghRP understands that bit, but like myself is trying to get a handle on relating, in some physically meaningful way, effect of gravity on 'length down there' as it relates to us 'out here'. Bottom line - if curvature of spacetime has any real meaning for space as well as time (the uncontroversial redshift part), there must be evident effects observable to us 'out here'. And not just bending of light. For instance, there must be some sense in which say a neutron star will appear, after all mechanical stress/strain, gravitational lensing etc, is factored in, to have a different diameter based on 'warped length scale' - a geometric effect independent of any coordinate system used. So, the statement 'neutron star xyz has a diameter of 20km' refers to what - local measure or our measure? It was argued elsewhere that while something will be there, too much inherent 'freedom of measurement choice etc' for definite, unambiguous predictions. The slippery eel.Now, having done all this, I lower the rod into the gravity well and place it radially. But GR predicts that, when I do this, I will measure *no* stress on the rod! (Strictly speaking, I will measure some stress due to the acceleration the rod has to sustain to "hover" at a constant radial coordinate, but we already talked about that above; once we correct for that, there will be no measurable stress in the rod. There may also be effects of tidal gravity, which I don't think we need to go into detail about here; they would need to be factored out the same way acceleration is.) When people in those various threads that were linked to above talk about there being no locally measurable effects of the difference in metric coefficients (radial vs. tangential, in Schwarzschild coordinates), this is the sort of thing they're talking about: whenever you look at a frame-invariant, physical observable, locally, you find that there is *no* difference radial vs. tangential.
Remember that in the passage you quoted I was measuring distance by measuring stress. Stresses are invariant physical observables. So the "shrinking/bending" of the measuring device is irrelevant. I was making an even stronger statement: in terms of stresses, there is *no* "shrinking/bending" of measuring devices "down there" relative to "up here". The only effects of the K factor "down there" are *how many measuring devices* will fit between concentric spheres, relative to the Euclidean prediction. I don't know how I can make it any clearer.Neglecting possible higher order derivative effects, what is written above is of course a logical necessity simply because, as I put it in #4 "'Go down there' to check and you shrink/bend/whatever same way as the rod,"
It has no effect on *length*, per se. It has an effect on "how much length" is present, radially, in between concentric spheres, relative to the Euclidean prediction. That *is* a physically meaningful way of describing the difference between "down there" and "up here". The K factor, as I defined it, is a physical observable. I told you how to physically measure it.If I'm not mistaken SinghRP understands that bit, but like myself is trying to get a handle on relating, in some physically meaningful way, effect of gravity on 'length down there' as it relates to us 'out here'.
There are; the K factor as I defined it is just as observable "out here" as "down there". The non-Euclideanness of packing of little objects can be observed from anywhere.Bottom line - if curvature of spacetime has any real meaning for space as well as time (the uncontroversial redshift part), there must be evident effects observable to us 'out here'.
You would have to quote a specific instance of such a statement to answer this, and the answer would depend on the specific instance. As "diameters" are usually quoted, I suspect they are actually in "radial coordinate units", meaning they are really the square roots of areas divided by 4 pi, or they are circumferences divided by 2 pi (well, for "radius", not "diameter", but it's easy to relate the two). If actual "physical diameter" is meant, then yes, you would have to evaluate the K factor throughout the interior of the star to see how much actual, physical distance there was inside a sphere with the area of the star's surface, given the K factor.And not just bending of light. For instance, there must be some sense in which say a neutron star will appear, after all mechanical stress/strain, gravitational lensing etc, is factored in, to have a different diameter based on 'warped length scale' - a geometric effect independent of any coordinate system used. So, the statement 'neutron star xyz has a diameter of 20km' refers to what - local measure or our measure?
I think this formula, as you give it, is approximate; it assumes that km << r. But that's only a technical point, and doesn't affect the argument; I just mention it in passing.I took Einstein’s equation about the frequencies (ν) of a vibrating particle from his book “ The Meaning of Relativity”: ν∞= (1 + κm/r) νr , where m is mass under whose gravity the vibrating particle is, r is the distance between the two, and κ = G/c2, where G is the gravitational constant.
How is x measured? Remember all the difficulties I raised in my last post about that.Then I imagined a very thin wire, whose “particles” (charge e, mass δm << m) are separated by x.
How are you deriving this from Heisenberg's uncertainty principle? First of all, I don't see Planck's constant anywhere. Second, oscillations due to the uncertainty principle are not the same as oscillations due to Coulomb force.Under the electrostatic forces of its neighboring particles, each particle oscillates with a frequency. (Heisenberg’s uncertainty principle forces the oscillations.) The frequency of this oscillation can be derived elementarily to be: f = z/xq. Here z and q are parameters of the rod. For instance, for a one-dimensional rod dimensional rod, z = (1/π) (Q e2 /δm)1/2 and q = 3/2. Here Q is Coulomb’s constant.
No, it's not telling you that. It's only telling you that, if the oscillations in the rod generate some sort of signal (for example, oscillations due to electrical forces inside the rod might generate a radio signal--the rod might be a radio antenna), the frequency of that signal if it is observed far away may be different than the frequency the signal had when it was emitted from the source. In order to derive information from the signal about the inter-particle distances x at the source, you would first have to correct for what happened to the signal during its travel, which means you would have to *undo* the effect predicted by Einstein's equation on the signal.Substituting the particle’s frequency into the Einstein’s equation, we have: xr = (1 + κ m/r)1/q x∞. This tells me that: the rod is longer closer to the mass; and the rod is flattened and disintegrated near a black hole (m/r → ∞). If the gravitational field is not uniform over the rod, the rod is deformed.
It's also worth nothing that nothing about what you have said above involves any non-uniformity of the field *over the rod*. It only involves non-uniformity of the field between the rod and the point where signals coming from the rod are being observed. Non-uniformity of the field over the rod itself is tidal gravity, and I understood that we were supposed to be leaving that out.If the gravitational field is not uniform over the rod, the rod is deformed.
The word "non-uniform" is crucial here; for the spacing between atoms in the crystal to be affected, the field needs to be non-uniform *across the crystal*.The scenario in my statement is quite simple. The separations between the atoms in a crystal are affected differently in a non-uniform gravitational field, so the crystal will be deformed.
No. It will not always look contracted. It might even look elongated if it moves towards youSR is an apparent, not real, effect:
1. If we observe a rod moving past us with a uniform velocity v, it will look contracted in the direction of its motion by a factor (1 – v2/c2)1/2.
This is our best current belief, yes, based on the belief that there is a consistent theory of quantum gravity (we just haven't found it yet). But if it's true, then this...Gravity a real force – the fourth fundamental force, assuming there are no others. In GR, gravitation is due to the curvature which matter creates in the field of space-time geometry. The field of space-time geometry is the gravitational field. At the microscopic level, gravitons would be the quanta of gravitational field.
...is *not* correct. The graviton mediates the gravitational interaction in the same way that the gluons, weak bosons, and photon mediate the strong, weak, and electromagnetic interactions.The strong, the weak, and electromagnetic fundamental interactions are mediated by the color, the weak, and electromagnetic fields associated with the color, the weak, and electrical charges of matter or antimatter. There is no such analogy associated with gravitational interaction.
SinghRP: Your #1 entry suggested, by the use of analogy with differential aging wrt height in a gravitational potential, length also a relational measure (inferred in that one could not notice anything by travelliong from one location to another). It was on that basis I referred you to Bowler's book. Not clear now what your conception of length change relates to - but seems increasingly you mean a locally measured length change. It is further not clear whether that locally measured length change is owing to graivitational potential, or tidal forces. If the former, you will have to explain how it could be locally detected - after all any 'ruler' is subject to change just as the crystal of NaCl or whatever one is using as 'detector'. Do you imply then that local measurement is possible owing to material dependence - 'ruler' made of steel changes differently to crystal made of NaCl? If on the other hand you just mean the effect of tidal forces - that has no direct linkage to redshift. A compact gravitating mass will generate far greater tidal forces on your suspended rod than for a less dense body - given radial separations achieving the same relative redshift factor. One is a function of potential the other is a function of second order derivatives of the same. Please then clarify your conception of exactly what length change means, and what generates it.As discussed before, gravity does affect vibrators and what they emit or absorb. I am sure we can agree on those for the time being. All I wanted to have at this point is a relationship between the frequency of vibration of an atom and separation distance between atoms in a crystal. (Separations are infinitesimal and local.) So, that substitution is justified. An experiment similar to the Pound-Rebka’s may be conducted, where changes in Bragg’s reflection/diffraction patterns from a crystal (such as NaCl) may show whether spacings between the atoms are changed by gravity. Thermal and other non-gravitational effects must be sorted out. I am very much confident – just being a human here – that atoms in a crystal have larger separations in a stronger gravitational field or closer to a mass. On the other hand, I could be wrong.
It’s also worth … leaving that out.
The scenario in my statement is quite simple. The separations between the atoms in a crystal are affected differently in a non-uniform gravitational field, so the crystal will be deformed.
pervect: Several times now I have given the example of sending down a clock to the surface of a planet, leaving it a set coordinate time, and retrieving. There will be a noticed difference in elapsed time. Sure the sending down and retrieving parts create some difficulty if a one-shot procedure is employed - just separating SR from 'pure' GR effects is one. All that's required to eliminate such difficulties is to repeat the procedure a second time, leaving the clock a different period of elapsed coordinate time. Provided sending and retrieving are carried out just as before, we can easily eliminate everything but the true difference in clock rates. How can it be said there is no precise definition of 'gravity slowing down time' - and of course it goes without saying that has to be on a relational - 'here' vs 'there' basis.I think the best answer is that it is not really a good idea to think of gravity as "slowing down time" in any physical sense.