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HankSolo
2002-Dec-02, 09:02 PM
I know we've previously discussed whether gravity is a force or the effect of the curvature of space-time, and the characteristics of light. But there's a couple of questions still bugging me.

If the pull of gravity is due to the curvature of space-time, then there should be no such thing as escape velocity. An object travelling near a star would follow the same path (curve at the same angle) regardless of its velocity, because it is travelling through the same curved space and thinks it's still going straight no matter what the velocity. I know that this isn't what actually happens, and that a slower moving object gets pulled in more than a fast moving object, and an escape velocity is necessary to break free of this pull. But wouldn't that suggest that gravity is a force being exerted on the object, and not the effect of curved space-time?

Second question: We can see a star thousands of light-years away, because photons generated from that star are reaching our eyes. If we move over an inch, we can still see the star, meaning that even at this distance, photons from that star bombard every single spot around us at all times, allowing us to see the star at all times no matter where we are, even though it is thousands of light-years away. The density at the source is unimagineable. Is it possible to be far enough away that there is any sort of measureable gap between photons, and objects fade in and out of view? (without the distorting effects of atmostphere or gravity)

Thanks in advance for any insight.

Silas
2002-Dec-02, 09:32 PM
On 2002-12-02 16:02, HankSolo wrote:
I know we've previously discussed whether gravity is a force or the effect of the curvature of space-time, and the characteristics of light. But there's a couple of questions still bugging me.

If the pull of gravity is due to the curvature of space-time, then there should be no such thing as escape velocity. An object travelling near a star would follow the same path (curve at the same angle) regardless of its velocity, because it is travelling through the same curved space and thinks it's still going straight no matter what the velocity. I know that this isn't what actually happens, and that a slower moving object gets pulled in more than a fast moving object, and an escape velocity is necessary to break free of this pull. But wouldn't that suggest that gravity is a force being exerted on the object, and not the effect of curved space-time?

This one, I can answer... "Escape velocity" is a measure of the energy required to move from one area of "curved space" to a nearby area where space is less curved.

Here, on the surface of the earth, space is "curved" to the tune of 9.8 meters per second per second. Way up in distant space, space is much flatter. An object, fired straight up at 11 km/s will gradually lose kinetic energy as it moves through these regions of lesser and lesser curvature. It's "paying its bills," so to speak, by trading kinetic energy for potential energy...and that potential energy is partly caused by the curvature of space.

Silas

Karl
2002-Dec-02, 09:44 PM
On 2002-12-02 16:02, HankSolo wrote:

Second question: We can see a star thousands of light-years away, because photons generated from that star are reaching our eyes. If we move over an inch, we can still see the star, meaning that even at this distance, photons from that star bombard every single spot around us at all times, allowing us to see the star at all times no matter where we are, even though it is thousands of light-years away. The density at the source is unimagineable. Is it possible to be far enough away that there is any sort of measureable gap between photons, and objects fade in and out of view?

Why do you say this is unimaginable? Why don't you just calculate what it is?

The answer is yes, most faint objects are detected by integrating single photons using very cold detectors to keep the thermal noise down. The objects do not fade in and out, you need to hold on the object long enough for the number of photons to integrate up above the noise level.

GrapesOfWrath
2002-Dec-02, 11:41 PM
On 2002-12-02 16:02, HankSolo wrote:
If the pull of gravity is due to the curvature of space-time, then there should be no such thing as escape velocity. An object travelling near a star would follow the same path (curve at the same angle) regardless of its velocity, because it is travelling through the same curved space and thinks it's still going straight no matter what the velocity. I know that this isn't what actually happens, and that a slower moving object gets pulled in more than a fast moving object, and an escape velocity is necessary to break free of this pull. But wouldn't that suggest that gravity is a force being exerted on the object, and not the effect of curved space-time?
You are right that that is not what actually happens. The disconnect is that you are thinking of curved space--but the appropriate concept is curved spacetime. In spacetime, the parabolic curvatures of a lazily tossed beachball and a supersonic bullet are virtually the same. The long flight time of the beachball is multiplied by c and figures in as an additional term, stretching the path out incredibly.

Second question: We can see a star thousands of light-years away, because photons generated from that star are reaching our eyes. If we move over an inch, we can still see the star, meaning that even at this distance, photons from that star bombard every single spot around us at all times, allowing us to see the star at all times no matter where we are, even though it is thousands of light-years away. The density at the source is unimagineable. Is it possible to be far enough away that there is any sort of measureable gap between photons, and objects fade in and out of view?
A single candle puts out an immense number of photons. When you move away from it to such a distance as they are no longer dense enough to impact your eye--you lose sight of the candle. Same for the stars.

Wiley
2002-Dec-03, 01:46 AM
On 2002-12-02 16:02, HankSolo wrote:
If the pull of gravity is due to the curvature of space-time, then there should be no such thing as escape velocity. An object travelling near a star would follow the same path (curve at the same angle) regardless of its velocity, because it is travelling through the same curved space and thinks it's still going straight no matter what the velocity. I know that this isn't what actually happens, and that a slower moving object gets pulled in more than a fast moving object, and an escape velocity is necessary to break free of this pull. But wouldn't that suggest that gravity is a force being exerted on the object, and not the effect of curved space-time?

Grapes is exactly right. If Tom Glavin and I each threw a baseball straight up, the spacetime path of each baseball would be nearly identical. However, if baseball was shot out of a cannon with a muzzle velocity equal to the escape velocity (and neglecting air resistance, for all you nit-pickers), the spacetime path is very different. In each case, the path is nearly straight since c*t is much greater than the distance covered. Changing the initial velocity changes the spacetime angle of release, i.e. it changes the angles between the space and time dimensions.

Second question: We can see a star thousands of light-years away, because photons generated from that star are reaching our eyes. If we move over an inch, we can still see the star, meaning that even at this distance, photons from that star bombard every single spot around us at all times, allowing us to see the star at all times no matter where we are, even though it is thousands of light-years away. The density at the source is unimagineable. Is it possible to be far enough away that there is any sort of measureable gap between photons, and objects fade in and out of view? (without the distorting effects of atmostphere or gravity)

In addition to what Grapes and others have said, remember that we always view things through a finite sized aperture. Whether using the naked eye or a telescope, the number of photons is not as important as the number of photons per unit area. Ideally the number of photons per unit area is constant, but since we average the number of photons over the entire area, small deviations are not noticed.

DStahl
2002-Dec-03, 07:32 AM
And of course, thinking of gravity as spacetime curvature is mathematically accurate but not necessarily the "ultimate truth" of the situation. From Kip Thorne's Black Holes and Time Warps, page 400 (the chapter entitled "What is reality?"):

First, Dr. Thorne explains that one can view gravity as a field in flat spacetime, one which distorts the size of objects and the speed of clocks. He continues:

"What is the real, genuine truth? Is spacetime really flat, as the above paragraphs suggest, or is it really curved? To a physicist like me this is an uninteresting question because it has no physical consequences. Both viewpoints, curved spacetime and flat, give precisely the same predictions for any measurements performed with perfect rulers and clocks, and also (it turns out) the same predictions for any measurements performed with any kind of kind of physical apparatus whatsoever....Since the two viewpoints agree on the results of all experiments, they are physically equivalent. Which viewpoint tells the 'real truth' is irrelevant for experiments; it is a matter for philosophers to debate, not physicists. Moreover, physicists can and do use the two viewpoints interchangeably when trying to deduce the predictions of general relativity."

So a layman like me should be very cautious about assuming that one method of visualizing gravity is the absolute truth. The truth resides in the mathematics, if anywhere, I expect; and as Thorne writes, the mathematics for a gravitational field in flat spacetime and gravity as pure spacetime curvature are precisely equivalent.