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Does the spacetime curvature according to Einstein really exist?


tmdarkmatter

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1 hour ago, tmdarkmatter said:

Is light in its different frequencies able to push massy objects (some kind of light sail effect)? I suppose that yes

Yes, and the force it exerts is P/c (power divided by the speed of light) for absorption. (reflection doubles this)

So 3.33 nanonewtons per watt. This is not a large force.

3.33 micronewtons per kilowatt, meaning around 5 micronewtons per square meter from the sun, which is by far the brightest source for earth.

 

 

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1 minute ago, tmdarkmatter said:

the brightest source of visible light

If you think there are other sources that come anywhere close to this, cough up the evidence. The astronomers that look for such sources will be very interested.

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I refer to this: Moon glows brighter than sun in images from NASA's Fermi (techexplorist.com)

1 hour ago, mistermack said:

If you plot the motion of a body, against time, you get a curve. The curve is in the plotting. What exists in reality is a three dimensional motion of a body. 

So what does the bending of light passing by the sun mean for you? Is it because of spacetime curvature?

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16 minutes ago, tmdarkmatter said:

So what does the bending of light passing by the sun mean for you? Is it because of spacetime curvature?

No, for me, spacetime curvature is how it is modelled. What it means for me is that the light is spending time in inertial space that is accelerating towards the Sun. 

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2 minutes ago, mistermack said:

What it means for me is that the light is spending time in inertial space that is accelerating towards the Sun. 

So for you the sun is an object that is constantly absorbing space. That is interesting. It is like throwing a spear through a waterfall.

Please give me a -999 reputation. Lol. I don´t care. I prefer to have answers to my questions.

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10 minutes ago, tmdarkmatter said:

It is like throwing a spear through a waterfall.

I don't get the similarity. 

Inertial frames close to a massive body like the Sun accelerate towards it. If a photon passes through, it accelerates at the same rate. But only for a short time, because it's moving so fast. So it's only slightly deflected before it leaves the zone of high acceleration.  

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Just now, tmdarkmatter said:

How would you define this? Space accelerating towards the sun?

If you are free floating, in the vicinity of the Sun, you accelerate towards it, because the inertial frame you are in is accelerating towards it. 

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15 minutes ago, mistermack said:

because the inertial frame you are in is accelerating towards it.

The interesting part is that if there is a constant acceleration of the inertial frame, why is the atmosphere of earth not being compressed to a thin layer on the ground? The air is exerting force against this acceleration. How should that be possible if spacetime is bent? This rather seems to be an equilibrium of forces between the air pressure and a gravitational *force*. The lower in the atmosphere you go, the higher the pressure of air must be in order to push against the increasing gravitational force. The idea of spacetime curvature does not fit here. Is there a spacetime anticurvature of the air to cancel the spacetime curvature? This does not make sense. You might say that a tower is also exerting force against the curvature, but air particles are not a fixed structure. Why would they not follow the frame acceleration or why only to a certain point?

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4 minutes ago, tmdarkmatter said:

The interesting part is that if there is a constant acceleration of the inertial frame, why is the atmosphere of earth not being compressed to a thin layer on the ground?

It is. The earth has a radius of ~6400 km. The atmosphere is only substantial for a few km. That’s pretty thin.

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13 minutes ago, swansont said:

It is. The earth has a radius of ~6400 km. The atmosphere is only substantial for a few km. That’s pretty thin.

Just to endorse that in a visual way, here is 10,000 metres, roughly the liveable limit, shown on Google Earth. (I love google earth)

It's the yellow line starting in Manhattan going left to New Jersey. You can just make it out. And below, in proportion to more of the US.

image.png.b96715e2bd119c96a184693030ef80c0.png

image.thumb.png.e59730c98d56bb6dee08c354ad95495d.png

 

 

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3 hours ago, mistermack said:

Inertial frames close to a massive body like the Sun accelerate towards it.

How is described, in this interpretation of GR, slowing of the falling close to BH event horizon?

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6 minutes ago, Genady said:

Thanks, I'll read the paper. However, I can't watch videos

I'm sorry, I didn't notice that you said slowing of an object close to the event horizon. I've never heard of that, but I can't imagine that an object would slow. It's probably an optical illusion, as the light from the object stops coming outwards, and starts to reverse inwards. In the river picture, the light is like a salmon that can no longer make headway in the current. (I'm guessing)

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26 minutes ago, mistermack said:

I'm sorry, I didn't notice that you said slowing of an object close to the event horizon. I've never heard of that, but I can't imagine that an object would slow. It's probably an optical illusion, as the light from the object stops coming outwards, and starts to reverse inwards. In the river picture, the light is like a salmon that can no longer make headway in the current. (I'm guessing)

In the proper time of an observer far from the event horizon, a falling object slows down before crossing the horizon and never crosses it. IOW, observed or not, there is no point on the far observer's time axis / clock that would correspond to a moment of the falling object crossing the horizon.

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5 minutes ago, Genady said:

In the proper time of an observer far from the event horizon, a falling object slows down before crossing the horizon and never crosses it. IOW, observed or not, there is no point on the far observer's time axis / clock that would correspond to a moment of the falling object crossing the horizon.

I can't argue the case, I don't know the theory. What I would guess though, is that the ordinary conditions no longer apply at that instant. After all, at the even horizon, the object would be travelling at or above the speed of light, relative to a distant observer, which would probably have some drastic effect on the calculations. 

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6 hours ago, Genady said:

In the proper time of an observer far from the event horizon, a falling object slows down before crossing the horizon and never crosses it. IOW, observed or not, there is no point on the far observer's time axis / clock that would correspond to a moment of the falling object crossing the horizon

This can be explained because time is passing slower and slower when getting closer to the event horizon, and when finally arriving at the event horizon, the time will pass infinitely slowly. Therefore objects never cross the horizon. Once you get there, you will see galaxies around you moving at fast speed until everything (the entire universe) falls into the black hole, just when you are crossing the event horizon.

11 hours ago, mistermack said:

Gas is an elastic medium, because of the motion and collisions of the molecules. 

It takes a force to compress air, so gravity must also be a force, not a mathematical wonder. Nature does not follow geodesics, the geodesics are a way we are currently using to interpret nature.

11 hours ago, swansont said:

It is. The earth has a radius of ~6400 km. The atmosphere is only substantial for a few km. That’s pretty thin.

Several km is pretty thin, but still very far away from a thin line.

You might think how I would incorporate time in my idea. Well, I think that time should somehow have to be related to the amount of radiation coming from all directions. So when there is a gravitational shadow or when the object is moving at fast speeds, it is being hit by less radiation or radiation only coming from a certain angle. Therefore time passes slower. Close to a black hole, the gravitational shadow and deviation of light should be so big, that an object would only be hit by a small ray of radiation coming just perpendicular to the surface of the black hole (from behind). This highly concentrated radiation pushing only into one direction should it make very hard for time to pass. Maybe because quarks, electrons, neutrons etc. have to struggle to move around their center, because when coming back against this high intensity radiation they struggle so hard that time must be slowed down in a certain way to make it possible for these particles to still move at the speed of light, not losing their properties. So when we are there, our atoms are struggling for their existance in a completely asymmetric environment and this struggling slows down time. Its like a tree in a storm that does not want to lose its leafs. When the tree is in interstellar space, there is almost no wind, when the tree is on earth, there is a slight breeze, and when the tree is close to a black hole, there is a heavy storm.

Just imagine an electron cloud being hit by radiation coming only from one direction. The cloud would be displaced so the electron needs more time at an average to complete a full lap around the nucleus. The pinhead instead of being in the middle of the soccer field is now displaced close to one of the borders and the soccer field is increased in size at the other border. But the electron still has to find its way around the pinhead within this altered soccer field to comply with the laws of physics.

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48 minutes ago, tmdarkmatter said:

Therefore objects never cross the horizon. ... when you are crossing the event horizon.

So, do they or don't they cross horizon in this explanation?

 

49 minutes ago, tmdarkmatter said:

Once you get there, you will see galaxies around you moving at fast speed

No, you will not. Once you get there, nothing extraordinary happens.

 

50 minutes ago, tmdarkmatter said:

everything (the entire universe) falls into the black hole

This is not what would happen. Your description is incorrect.

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Of course I am supposing that time would pass at an incredibly slow pace close to zero. If time does not pass, the galaxies around you and the entire universe can dance around for the time they want to. Therefore, once you finally cross the event horizon, this should be the end of the universe where you were and the beginning of a new universe in this black hole. In an instant, you would see the entire future of the universe you were in, because you reached a point where time does no longer pass for you within that universe. I think the best way to desribe this would be a huge white lightning, because you would receive an almost infinite amount of radiation coming from these galaxies in the instant of time when you cross the horizon. And according to how the universe will end, you will either see all the galaxies approaching you and combining into one black hole or you would be seeing them all moving away (if the idea of big bang and expansion is right).

But of course I am only supposing and guessing, so this topic will be closed soon anyway.

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8 hours ago, mistermack said:

I can't argue the case, I don't know the theory. What I would guess though, is that the ordinary conditions no longer apply at that instant. After all, at the even horizon, the object would be travelling at or above the speed of light, relative to a distant observer, which would probably have some drastic effect on the calculations. 

The chart below illustrates what happens in two inertial frames as one falls and the other remains far away. Time, in units of the BH mass M, shown on the horizontal axis. Distance from the BH singularity, also in units of the BH mass, shown on the vertical axis. 

r = 2M (r/M = 2) is the event horizon, shown as a dashed line. 

The falling observer starts at the distance r = 6M, accelerates and falls through the event horizon and into the singularity in about t = 15.3M, in his own frame.

In the stationary observer's frame, OTOH, the falling one accelerates, albeit more slowly, until about t = 20M, then decelerates, slows down to zero asymptotically, and never crosses the horizon.

image.thumb.jpeg.d265a02ae6f0bd1296baaed330f0fac1.jpeg

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