J.C.MacSwell

Yes. If our sun were to spontaneously accelerate' date=' we wouldn't know about ti for 8 minutes (you have to ignore the effect of whatever caused the sun to accelerate, and perhaps a few other laws of physics, which limits the usefulness if scenarios like this)

 

By whatever means, the sun "communicates" its presence to space at the speed of light, and causes the curvature. Think of the curvature at some point as a rod extending out into space - that rod moves with the sun according to any outside observer, so the curvature is fixed at that point, relative to the sun. The curvature is there when the earth passes that point, and stays there afterwards. If the sun were to disappear, or to accellerate, the information about that (the disturbance in the curvature) will travel at c.[/quote']

 

So a body in motion in a given inertial frame throws "curveballs" to mediate or communicate the gravitational force (cause the curvature)? They are only straight in the rest frame of the body?

 

In your analogy your rod is always straight, but the path of the components, travelling along the rod at c, is itself curved in all inertial frames except the rest frame?

 

Edit: I hope I'm not causing confusion with respect to the 2 curvatures, the curvature over time of the vector path and the space curvature it produces.

As long as the star is not accelerating, the curvature does not vary in time. In the star's frame it's not moving[/b']. In an outside inertial frame, the curvature moves with the star.

The part I bolded seems right, but that would remain the same (no different) from the vector directed at the "old" position (since they are the same).

Just trying to gain incite , I know I haven't refuted anything you've said.

Now the unbolded part: It's hard to picture how that would work although as long as the star is affected only by gravity then this seems plausible (the field could "anticipate" the future position of the star)

For the outside inertial frame/s:

So if our Sun was "blindsided" by a high speed collision with a massive object

that changed it's position the gravitational vector (sun component) for our Earth would be directed at an imaginary path the Sun would have taken if not for the collision for the next 8 minutes?

Time dilation will redshift the spectrum of an radiating object moving relative to the observer.

Also the motion of the object will redshift (if the object is moving away) or blueshift (if the object is moving closer) it spectrum because of the Doppler effect.

 

For an object moving toward the observer' date=' is there a speed where the redshift cause by time dilatation will be compensated exactly by the Doppler blue shift ?[/quote']

 

I like this question and noone has answered so I will blurt out a guess and hope someone will correct or add incite:

 

My thought is no because if any speed matched/compensated then all speeds would.

 

Having said that there must be some angle (perpendicular component) such that it would work out where the blueshift is exactly compensated for by the total speed.

 

What is that angle? I think it would be somewhat profound if it was 45 degrees so that is my guess.

I don't think so. In GR, space is curved, and the curving dictates the motion through space. That curve is constant in time for unaccelerated motion, so the gradient would be directed toward the star.

And not the position of the star when it affected that curvature and gradient? In some frames what you are saying seems obvious or straight forward while in others it seems wrong. I have to give this some thought.

Is there a perihelion shift that is not explained by GR?

None I am aware of. But I thought the gravitational vectors being directed at the "old" position of the bodies was the major part of the GR explanation.

Swansont' date=' what is a perihelion shift exactly?

 

Regards[/quote']

 

The perhelion is the closest point an orbiting body gets to the orbited body in an elliptical orbit. Newtonian physics predicts that orientation (angle relative to "fixed"stars) would stay constant. GR predicts a consistent shift with each revolution.

The light we see points toward where the sun was, 8 minutes ago, i.e. there is aberration. The gravitational vector points to where the sun is now.[/b'] (This fact is often mistaken to imply that gravity is instantaneous.) How would you reconcile this?

Then why is there a perhelion shift for orbits?

So if I were to enclose myself in a sufficient amount of lead, I would become weightless?

You can run (orbit) but you cannot hide!

Is it applying any force to you? If not' date=' then none. If so, then whatever the relative force then applies is the drag your "high speed space travel" device experiences.

 

Is your question then "does CMB apply force to anything?"

 

 

 

[b']FYI, you cannot go past the speed of light in any frame[/b], according to the current model. Even if two objects are moving .99c away from each other, they will never go past light speed (or reach it if they are not pure energy). I never learned the math, but it has been mentioned on the forum within the past month if you care to look for it.

Inertial frame locally not any frame.

Thats a bit over my head. Ask Sayo

Thank-you anyway and happy birthday (slightly belated).

Another way of looking at this is:

How much drag would the CMB cause to high speed space travel? "High speed" of course being relative to the BBT/CMBI frame (does anybody know any other names for it?).

Well' date=' that depends on how much of the universe you look at. Pick up a penny. It does not move in relation to your hand. You do not move in relation to the earth. (At the moment) The earth does move in relation to the solar system, but in an oval, not in a line. The milkey way moves only a little in relation to the cluster of galaxies it belongs to, and that is only aimless wandering due to gravity,[b']However[/b], the cluster of galaxies does move in relation to the other clusters. And the maga clusters move in relation to the other clusters. And the even larger clusters that contain the maga clusters move in relation to the other giga-clusters.

ect.[math]

\infty

[/math] So the giga-clusters move, therefore the maga clusters move, therefore the clusters move, therefore the galaxies move. therefore our solar system moves, therefor the earth moves, therefore you move, therefore the penny moves. All of this movement is in the general direction of away from the site of the big bang.So, small picture, no.

Big picture, yes.

I think (if I'm interpreting you properly) that this is (on average) the "big bang track" or "CMB isotropy frame". Anything on it is at rest in that frame (such as it is; distant objects may be "Hubbling along" at greater than the speed of light in each others "inertial frame" but they are both at rest in the BBT/CMBI frame).

Anything "off track" should be somewhat compelled by the anisotropy of the CMB to "tend" toward "getting back on track" or toward rest in the BBT/CMBI frame. Correct?

In 1976 or so Bill Unruh (British Columbia----a physicist at UBC)

associated a temperature called "Unruh temperature" to any given acceleration

 

if you use natural units' date=' planck units, the Unruh temp is about a/2pi

 

I forget exactly. But the temp is the same order of magnitude as the acceleration when both are expressed in natural units.

 

this means that Planck acceleration (increase speed by c in one Planck time unit) would have Unruh temperature equal to about Planck temp.

 

I think that Planck acceleration is a natural maximum on acceleration because it would be so weird (its Unruh radiation would be forming black holes)

 

So I will propose Planck acceleration to you as a possible universal maximum. IIRC it is about E51 gee. You know what a gee is, of acceleration, the planck accel is roughly E51 times that. I forget exactly.[/quote']

 

Couple of thoughts (read "blurt"):

 

If you had a maximum acceleration that would "cap" the high end of frequency for the resulting radiation (an ultraviolet uncatastrophe )...

 

A black body so "capped" would not exhibit the black body radiation "signature"...

 

A black body so "capped" would still exhibit the black body radiation "signature" due to relativity...

 

How does relativity affect temperature?

This reminds me of a question I've had that is somewhat related, to both this and perhaps the speed of light:

Is their a maximum acceleration?

Is there a natural "tendency" for objects to "tend" towards rest in the "big bang track" or "the CMB isotropy frame" if undisturbed otherwise? (other than by the CMB)

Whether a 2d world is impossible or not' date=' I want to understand how it could function if it was possible. The article I'm reading about the 4th dimension requires me to understand how I would teach a 2D being the meaning of a 3D world. How I would get him or her to comprehend it even though he cannot see or understand the concept. I am compressing the 3D world into his. Once I can do that, I can apply it to try to understand the 4th dimension concept.

 

It is a kool experiment.

 

Bettina[/quote']

 

It would be much harder teaching someone from a 1D world what a 2D world was than teaching someone from a 2D world what a 3D world was.

I have no idea what this is suppposed to mean. Care to try again?

 

10% of the radiation incident on the earth is from the sun? (I wouldn't think this would be accurate)

How about 10% from "other than the sun"? Or is that still way too high?

 

 

If you insist on using a title' date=' please us the correct one, (Dr) though addressing me by my user name is perfectly fine.[/quote']

 

Can we call you "Doc"?

So no physical object in the universe is moving away from any other physical object at c? Now from a third party, can those two physical objects be seen to be moving away from one another at c[/b']?

Yes, up to but not including, 2c. (locally)

(first question is no anyway due to Hubble but yes locally in either objects rest frame)

Exactly. I know its short (certainly way shorter than the grass where I spend most of my round.)' date=' but it still has mobility. The individual blades can be pushed closer together by the weight of the ball. That could be just enough to for the effect we are discussing.

On practice putting surfaces I have observed the blades spring back after the ball has 'fallen' into the hole. Now, you make a very good point that the greens at Augusta are cut very close - closer for sure than a practice green - but I still think its a plausible explanation.[/quote']

 

I agree with your assessment. The grass can "buckle" and then recover, partially or otherwise and always with some energy loss. The ball looked to be still poised to go downhill when it "stopped" but with the grass underneath straining/slowly collapsing under the weight. Crowd noise "plausibly" or butterfly wings flapping in Australia a week earlier "possibly" could have "tipped the scale".

The simplest way to beat gravity is the "bootstrap" method. You need good balance and very strong arms and laces (regular shoe laces will probably break). You lean over so that your center of gravity is directly over your laces and pull straight up. If you pull hard enough (the force must exceed your weight plus the tension on your laces) you will rise and "hover" for a few seconds before falling over. Good balance is the key. Once you leave the ground it is like balancing a pencil on it's point so a couple of seconds is extremely good.

Definition: Let (x1' date='y1,z1), (x2,y2,z2) denote the coordinates of a point in some frame. Let D be defined as follows:

 

[math'] D = \sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2} [/math]

 

The axes are rigid if and only if d(D) = 0

 

Ok, I'm not going to let JC or Swansont confuse me. Here is the deal.

 

I have an actual meterstick. This one single meterstick is used to define distances throughout the entire universe.

 

So if I want to know the distance from my current location to the center of this universe, what I want to know is how many metersticks I would have to place in a striaght line, to reach from here to there.

 

If I want to know the distance from here to the sun, again, I want to know how many metersticks I need.

 

If I want to know the distance from the sun to some other star, again, I want to know how many metersticks I need.

 

So the Pythagorean theorem is irrelevent.

 

Again, if I want to know the distance from my current location to the sun, I want to know how many meter sticks i need, to rightnow simultaneously place in a straight line pointing right at the sun. In other words I want to know how many metersticks I can fit in between my current location, and the sun.

 

So for example, here is the current approximate distance from the earth to the sun:

 

 

Astronomical Unit

 

 

So there it is.

 

150 million kilometers

 

And since 1000 meters = 1 kilometer

 

The approximate distance from the earth to the sun is

 

[math] 1.5 \times 10^8 Km \frac{1000 m}{Km} = 1.5 \times 10^{11} m [/math]

 

So 150 billion metersticks would be needed. And this gives the mind a sense of the distance.

 

So this is why we need rigid axes.

 

Real rulers must be substitutable for the axes.

 

But not rulers that can expand or contract though.

 

When we were interested in the distance from here to the sun, we used rigid rulers. And the length of a real ruler doesn't vary much over large amounts of time. And any change in its length is certainly imperceptible.

 

So this is the empirical thought behind rigid axes.

 

In order to say this logically, it suffices to say that the distance between any two points on an axis is constant in time.

 

So given any frame, we are given three coordinate axes.

 

The distance between any two points on an axis is found by subtracting the lesser coordinate from the greater.

 

So the distance between the point 4 on the x axis, and the point 7 on the x axis is simply 7-4=3. And this distance never changes. If this distance could change, then the axis wouldn't be rigid.

 

The key thing here, is temporal constant.

 

This is the concept to be developed.

You are doing quite a job explaining Euclidean space. If you try real hard you can explain Euclidean space in a number of ways. You can come up with many more examples of Euclidean space. But it will still be Euclidean space. It will be a very good model of reality at low speeds and a very poor one when higher speeds are involved.

What do you mean it's unlikely to be Euclidean, or Lorentzian?

I think that if it was obvious we would already know what it was.

Why do you have to go and complicate things.

 

I don't want the axes to be able to expand and contract though' date=' because there was a unit of distance chosen. I want the axes rigid.

 

Let matter expand and contract, not the axes.[/quote']

 

 

If you find an absolute reference frame (if such a thing exists) it is unlikely to be Euclidean (or even simply Lorentzian).

The thing about frames' date=' is that [b']they can translate, and rotate[/b]. That means that the truth value of certain statements can vary in time. So...

 

.

...and expand and/or contract, constantly or at a varying rate, with respect to another frame.

First swansont, and now you. What is an absolute frame[/i']?

You will be telling us at your Nobel prize acceptance speech!

(I do not know but I'm sure it's not Euclidean)