A gravitating mass causes time to run slow. So if you have an object with energy E in flat space-time and you place it in the vicinity of this gravitating mass it will now have a lower energy.
But energy must be conserved, so what’s an object to do? Well if it accelerates, it gains kinetic energy and therefore energy can be conserved. So that’s what it does.
A spherical array of test masses in free fall above the surface of the earth will contract in the tangential and elongate in the perpendicular dimensions – a very real example of spacetime curvature.
[QUOTE=Mathochist]
As measured by remote observers.
[/quote[
As measured by any observer whose radial distance is greater. Time curvature is responsible for one half the deflection of a light ray (or photon) near a gravitating object.
Of course of course. (I have to remember to get rid of that sig tomorrow)
As measured by any observer whose radial distance is greater. Time curvature is responsible for one half the deflection of a light ray (or photon) near a gravitating object.
Of course of course. (I have to remember to get rid of that sig tomorrow)
The point is that time only appears to “run slow”. Time always runs fine locally. Acceleration really does occur, though, and can be measured locally by deviation from canonical “test geodesics” (as described in Misner-Thorne-Wheeler). Thus your argument doesn’t really hold water.
Since the OP has been answered, I’m tossing in a hijack question of my own.
(bolding mine)
This is my understanding as well. Assuming this interpretation is correct, what is the justification is for the existence of a graviton? (The particle aspect of gravity.) How can a curvature in spacetime – which is not a force – have a particle aspect?
Well, the rest of the “forces” aren’t really forces as we first understood them either, you know, and neither are their particles our familiar “billiard balls”.
I think the best way to think about particles in quantized field theories is with a toy problem. Consider a stretched guitar string. Classically, we know that its state of vibration is a sum of basic vibrations, each of which breaks the string into an integral number of pieces: the whole string vibrating, each half vibrating with a node in the center, each third vibrating with two nodes, and so on. In general, the nth basic vibration will have maximum displacement A[sub]n[/sub]sin(nx pi/L), where L is the length of the string. When we apply the rules of quantum physics, we find that A[sub]n[/sub] can only be a multiple of some a[sub]n[/sub]. The state of vibration can be described as the sum of k[sub]n[/sub] copies of the state with maximum displacement a[sub]n[/sub]sin(nx pi/L) for each n. We might call these fundamental solutions of the field equations “stringons”, and say that the state has k[sub]1[/sub] stringons in state 1, k[sub]2[/sub] stringons in state 2, and so on for each n. This is the particle aspect of the “field” of the string, which measures how far it’s been pulled from its rest position at any given point along its length.
Understanding how gravity as curvature can be a field (and thus amenable to this sort of technique) is a little trickier, but ultimately the curvature is determined by the
“metric tensor”. Around every point, at least, there’s a four-dimensional ball where the topology is just like that of R[sup]4[/sup]. The whole of spacetime is made by patching together a bunch of these balls with certain rules holding. At every point within each ball, we should be able to describe the geometry by a symmetric four-by-four matrix with signature (3,1). This function should be smooth as we vary from point to point and it should patch together nicely as we move from ball to ball. Within each ball, though, we can regard it as a collection of ten functions satisfying a certain differential equation, just like the stretched string was described by one function satisfying a differential equation. This is the “gravitational field”.
Lets say you’re on a space station orbiting a neutron star and one of your fellow astronauts descends to the surface (squish) and some years later returns. Do you really think you and he will have aged the same?
Light must travel at c and yet radially directed photons at the event horizon of a black hole are trapped there. How can this be unless time slows in the vicinity of a massive object?
[another hijack]
So there is the principle of least action. I get that more or less, I think. But I am pretty sure that I have read that a body will follow a geodesic that maximizes its path. That seems counter-intuitive – I would think that the path would be minimized. Is this description based on 3D space without considering time (4D)?
[/hijack]
No, because the two astronauts have followed curves of different lengths in spacetime. The situation described in the FAQ doesn’t apply to the OP because the astronaut approaching and receding from the hole is not following a geodesic: his path through spacetime is not straight, which is what the OP was asking about.
This is actually an effect of an arbitrary sign choice. Some physicists write the “spacetime interval” as ds[sup]2[/sup]=dx[sup]2[/sup]+dy[sup]2[/sup]+dz[sup]2[/sup]-c[sup]2[/sup]dt[sup]2[/sup]. With this convention, spacelike vectors have positive squared length, lightlike vectors have zero squared length, and timelike vectors have negative squared length. This sign change between spacelike and timelike tangent vectors has the effect of changing the extrema we’re trying to find.
If the spacetime interval was written ds[sup]2[/sup]=c[sup]2[/sup]dt[sup]2[/sup]-dx[sup]2[/sup]-dy[sup]2[/sup]-dz[sup]2[/sup], the two situations would be reversed.
Mind you, even before GR, the “principle of least action” had been generalized to the “principle of stable action”. You’re looking for place the second derivative (in a certain sense) vanishes, which may be a maximum or a minimum (or weirder).
Well, in general it doesn’t even apply there. You’re thinking of surfaces of constant curvature. There are 4-manifolds of constant curvature, but I don’t think that enters into this.
Mind you, the same “tidal acceleration” effect occurs in Newtonian gravitation. Try drawing a heavy point (for the center of a gravitating body), a light circle, eight heavy points equally spaced around the circle with two of them along the line between the center of the circle and the first body, and light lines from each point on the circle and the first body. Now, if we assume the test bodies have one unit of mass (and ignore their gravitational effects of each other) and the main body has exactly enough mass that the Newtonian formula reduces to
a® = 1/r^2
Now, solve this equation for each of the eight initial distances and draw all eight points along their respective lines for a few different values of elapsed time. You’ll see the circle of points elongate into an ellipse as it moves towards the gravitating point, and all without ever invoking spacetime curvature.
Now, this is not to say the two have nothing to do with each other. In the context of a theory like GR, tidal acceleration is a clearly visible effect of curvature, but it can’t be taken by itself as evidence of such a theory. To do so would beg the question.
If two parallel geodesics converge the curvature is positive, and vice versa. So it’s positive in the two-dimensional space-time defined by time and the tangential direction and negative in the space-time defined by time and perpendicular direction. But the curvature in the perpendicular direction is twice as strong.