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Acceleration of gravity from earth, on objects, traveling at near the SOL.
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Non-Abelian:

--- Quote from: MIS42N on December 20, 2020, 12:33:51 am ---I have always been suspicious of so called time dilation of fast moving objects. There isn't a way to measure time, we rely on observing repetitive phenomena (like atomic vibration), and counting how many repetitions of the phenomenon occur between some other events (like the sun appearing at the same angle above the horizon).
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Sure, but that's how you measure time. And you are not counting repititions in the sense of a deterministic repetitive process so that the precision of the measurement depends on the statistics, not the time for the process itself to occur (which is probabalistic). You end up with a better clock by having large statistics on probabilistic events than trying to create a clock from a fixed repetition rate.


--- Quote ---If a spaceship were able to go to Alpha xxx our nearest stars (say 4 light years away) at 1/2 the speed of light, then return at 1/2 the speed of light, I think what we would see is the spaceship taking 12 years to get there and 4 years to get back. Any people in the spaceship would see it take 8 years to get there and 8 years to get back. Apart from the slowdown due to acceleration, the people on earth and on the spaceship will age about the same amount. It seems to me any other outcome results in a paradox. Using the spaceship as a reference or using the earth as a reference, shouldn't the result be the same? The spaceship and earth start in the same spacetime and end in the same spacetime.
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Using the speed of light in relativity to understand relativity is an archaic way of doing relativity. The reason Einstein did it that way is that he was trying to explain electrodynamics by modifying Newtonian echanics rather than the other way around, which is what everyone else was trying to do at the time. Einstein had no concept of geometry at that time and it wasn't until Minkowski pointed out in 1908 that reltivity could be understood as a 4-dimensional spacetime that it began to be understood as geometry. I don't have any idea why it is still taught this way. when a physicist says "speed of light," he/she doesn't really mean that (depending on context). It's just that other physicists know what it means in context and it's easier than saying something else.

The speed of light really has nothing to do with it except for the fact that massless particles propagate at the same velocity in every frame and the photon is known to be massless to the best limits of experiment. A theory of electromagnetism in which the photon is NOT massless and which would require Maxwell's equations to be changed was known as early as 1914 (the Proca Lagrangian). So, really, the speed of light thing properly belongs in a theory of electromagnetism.

Relativity itself simply treats space and time on equal footing. You can posit two different theories of space and time which have physical plausibility. One is Galilean relativity in which time is parameter and relativity (special or general) in which time is a coordinate and "proper time" is an invariant quantity. In regular euclidian space you measure the lengths of lines (loosely speaking) via the "metric" ds^2 = dx^2 + dy^2 + dz^2, i.e. the pythagorean theorem. The equivalent in relativity is ds^2 = dx^2 + dy^2 + dz^2 - dt^2  (where time here is measured in meters since the constant c just converts seconds to meters and really doesn't mean anything physically). Experiments show that galilean relativity is not correct, which leaves general/special relativity.


--- Quote --- But during the journey observations are distorted due to the speed of light.
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As noted, the speed of light has nothing to do with it. Relativity is almost the same as changing the origin point on a circle from (x,y) to (x',y').

x' = x cos(a) + y sin(a)
y' = y cos(a) - x sin(a)
Instead of using sines and cosines, you use sinh and cosh:

x' = x cosh (a) - t sinh (a)
t' = t cosh (a) - x sinh (a)

In this case a velocity is just a hyperbolic rotation v = tanh(a). Feel free to change the t and t' to ct and ct' if you're uncomfortable with time in meters in which case you also insert a c before the tanh for the velocity. As an exercise, show that x^2 - t^2 = x'^2 - t'^2, so that the result x^2 - t^2 is not frame dependent.


--- Quote --- Does relativity deal well with correlating observers in different reference frames?
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Sure. Since you have invariants that are frame independent, different observers relate what they observe by changing coordinates, just like you would if you and a friend were sitting in two different places and each of you wanted to use where you were sitting as the origin of your coordinates.

As an aside, this also ends up meaning that the physically meaningful quantities for capacitance and inductance are geometric. (1pf ~ 1cm if I recall). Not so useful for engineering.
Non-Abelian:

--- Quote from: nctnico on December 20, 2020, 01:30:03 am ---Unfortunately it doesn't work that way. You can only apply special relativity theory on objects which have a constant velocity compared to the reference object. As soon as there is accelleration / de accelleration involved you need general relativity theory. Examples of objects to which the special relativity applies to are satellites orbiting the earth.

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Actually, you don't need general relativity to deal with accelerations. In the case of satellites, you do need general relativity for gps satellites. The general relativistic effects are small, but enough that gps would not work without those corrections. I'm not sure if anyone bothers with it for satellites that don't need to be syncronized.
MIS42N:

--- Quote from: nctnico on December 20, 2020, 01:30:03 am ---
--- Quote from: MIS42N on December 20, 2020, 12:33:51 am ---Consider an object moving away from earth near SOL. If we are in continuous contact with the object, it's clock will slow down while it is being accelerated, then it will appear to be running slower because of redshift in the received signal. But is it actually running slower when the object reaches its desired speed and is not being accelerated? From the point of view of the object, it will see the earth recede at the same speed, and signals from earth will be redshifted.

If a spaceship were able to go to Alpha xxx our nearest stars (say 4 light years away) at 1/2 the speed of light, then return at 1/2 the speed of light, I think what we would see is the spaceship taking 12 years to get there and 4 years to get back. Any people in the spaceship would see it take 8 years to get there and 8 years to get back. Apart from the slowdown due to acceleration, the people on earth and on the spaceship will age about the same amount. It seems to me any other outcome results in a paradox. Using the spaceship as a reference or using the earth as a reference, shouldn't the result be the same? The spaceship and earth start in the same spacetime and end in the same spacetime. But during the journey observations are distorted due to the speed of light. Does relativity deal well with correlating observers in different reference frames?

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Unfortunately it doesn't work that way. You can only apply special relativity theory on objects which have a constant velocity compared to the reference object. As soon as there is accelleration / de accelleration involved you need general relativity theory. Examples of objects to which the special relativity applies to are satellites orbiting the earth.

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I would say satellites orbiting earth need general relativity because they are in a gravitational field and therefore being accelerated toward earth. They negate that by travelling in a straight line in a curved spacetime. And once the spaceship gets up to speed, it DOES have constant velocity relative to earth. The observed velocity should be constant both from the spaceship and earth, just they are different. The measured elapsed time on the spaceship will be slightly different to that on earth because we are in both the earth's and the sun's gravitational well and the spaceship isn't. But I don't think it makes the sort of difference people call time dilation.
Nominal Animal:

--- Quote from: Non-Abelian on December 20, 2020, 12:38:25 am ---
--- Quote from: Nominal Animal on December 19, 2020, 10:06:09 pm ---
--- Quote from: Non-Abelian on December 19, 2020, 09:40:51 pm ---Second, the curvature is in spacetime, not space.

--- End quote ---
You managed to pick the one point where I miswrote spacetime, and made that a point.

Perhaps you should read my first post (fourth in this thread), and point out why you believe Dr. Bairds post invalidates my point.

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His point was the same as the one you seem to be trying to make - that the invariant mass is the physically relevant quantity, which is correct. You can only disagree with that by trying to say physical quantities depend on what coordinates you choose, which would be ridiculous.

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I'm trying to explain how spacetime curvature depends only on energy (which, if we are precise, is not just invariant mass, but also e.g. the electric fields) and momentum.

Consider his title: "Can you go fast enough to get enough mass to become a black hole".  Yes, you can interpret his text to make a point about invariant mass; or you can read his text at the face of it, and see that he completely ignores the effect of momentum on the curvature of space.  (I believe the title is technically misleading: it's like posing a question like "Will I live forever, if I eat a young, healthy, unrestrained, Siberian Tiger every day?", then completely ignoring the context of "tiger blood" and traditional remedies, and self-answering the question by stating "No, because your stomach isn't big enough."  See?  You can argue it is technically correct, but in the context of claiming to explain things and answering the stated question, it is horribly incorrect: it avoids the core answer, which is that there is nothing in consuming Siberian Tigers that affects longevity differently than consuming basically any other mammal.)

Similarly, you do not need invariant mass to become a black hole.  Momentum does suffice.

From the point of view of an external observer, as long as they cannot precisely measure the velocity in barycentric coordinates, they cannot really tell whether the approaching object curves spacetime because of their mass (energy), or because of their momentum.

Part of the difficulty I'm facing here is the lack of valid examples geared toward a layman.  Any event that could impart sufficient momentum to a single point-like mass to become a black hole is necessarily energetic enough to make the entire point moot.  (Based on known ways of transferring momentum, we can immediately say it is only possible in a system that is already a black hole.)  It is completely unrealistic.   So, whenever momentum is discussed wrt. black holes, it tends to be about angular momentum, which causes interesting measurable properties on black holes even when so small its effect on spacetime curvature is neglible.

Yet, contrary to what one might think by reading Dr. Baird, how much curvature a fixed invariant mass spaceship causes to spacetime, most definitely depends on its velocity.  Sourcecharge used his text to refute my points.  I do not make much of a difference between "typical reader gets the wrong impression and understanding" and "writer being wrong", because the effect is the same, and physics is about describing reality, not intent.
Non-Abelian:

--- Quote ---Yet, contrary to what one might think by reading Dr. Baird, how much curvature a fixed invariant mass spaceship causes to spacetime, most definitely depends on its velocity.
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The spaceship is at rest in its own frame, so its velocity is zero. It makes no sense to refer t the velocity of something without referring to what that velocity is relative to, in which case the situation is recipricol and you can take either to be at rest or neither to be at rest relative to something else.


--- Quote ---Part of the difficulty I'm facing here is the lack of valid examples geared toward a layman.
--- End quote ---
I'm not a layman. I have a PhD in nuclear physics. Feel free to do this using differential geometry.
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