for example, it is the cause of the factor of 2 difference in the gravitational lensing between Newtonian gravity and GR.
Btw, that's another thing I have been wondering for long. I have read in several places, now including also here, that also according to Newtonian mechanics light would bend when close to massive objects (but, as you say, significantly less than it does according to GR and reality).
I do not understand this. Photons are massless (which is the reason why they travel at c in the first place), so according to Newtonian mechanics the force of gravity on them would be zero, which would mean they are not affected by gravity at all, and thus should always travel straight in vacuum.
(AFAIK photons are considered to have relativistic mass, which is different from rest mass, which is still zero, according to SR/GR. However, there's no such a thing as relativistic mass in Newtonian mechanics, where velocity does not affect the mass of an object. Hence a massless particle would still be massless even if traveling at c, or at any other velocity.)
And how exactly do you know that light has zero rest mass? After all, classical electromagnetism predicts that light with energy E has momentum with magnitude p = E/c; in Newtonian physics, momentum is p = mv, which means light has momentum p = mc hence mass m = p/c = E/c^2. Having mass, light is attracted by gravitational fields, as well as generating its own field. This is consistent with what I talked about earlier about the possibility of treating gravity as curvature of space-time in Newtonian gravity -- Élie Cartan was the first to show this when he did just such a model.
When you move to relativity, the theory basically demands that light has zero rest mass; one reason being that the energy E, momentum p and rest mass m of a massive body are related by the equation E^2 = (pc)^2 + (mc^2)^2. The relationship between E and p for light (p = E/c) is still valid (being a consequence of Maxwell equations, and not of Newtonian mechanics); the two relationships are the same if the rest mass of light is zero. Hence, the rest mass of light being zero is a prediction of relativity -- one which can't really be directly measured because there is no rest frame in which to measure the rest mass of light.
Moreover, when you move to GR, light will also respond to gravitational fields (being the curvature of space-time), as well as generating its own field (because it has energy and momentum, which are the sources of gravity in GR).
Here's a simple question that confuses many people, so it's nice to talk about.
As we all know, if we drop an object at rest from a certain height (for example, the top of a tower), it'll follow a path along the vertical line (hitting the base of the tower, if it was built having an angle of 90º). However, this is only a good approximation for small heights, since the Earth is rotating around its axis, so in fact the path has a small displacement from the vertical line.
So, as a thought experiment consider that we have an observer at the Equator that sees an object at rest at a height h directly above him. Suddenly this object is released and starts to fall towards the ground. One could argue that, during the time of the fall the Earth has dragged the observer a little from West to East, so the object must land to the west of the observer.
Explain why the previous reasoning is wrong and the object in fact lands east of the observer.
So, as a thought experiment consider that we have an observer at the Equator that sees an object at rest at a height h directly above him.
At rest compared to what? Do you mean stationary with respect to the surface of the Earth?
But if it's stationary with respect to the surface of the Earth, that means that it's actually orbiting the Earth (at a speed of one revolution per day). In other words, it's moving along with the Earth. This lateral movement does not stop when the object is let go. Thus it will stay over that same point as it falls down.
(Of course the atmosphere can mess things up a lot.)
An interesting question is what happens due to frame-dragging (or whether it has any effect at all).
Yes, stationary with respect to the Earth's surface, since it's the observer at the equator that's recording its motion.
Relativity is not involved in this problem.
You got the first part right, but the conclusion is incorrect. Are you familiar with the concept of angular momentum? /spoils the problem
It has to move faster to stay exactly above the observer since being higher means you have to travel larger distances to orbit the earth in the same time. As it falls down it should retain this higher velocity. Unfortunately, I didn't manage to figure this out by myself, even though it should be a fairly easy problem, it seems counter-intuitive. (Hurray for run-on sentences) I think our flawed intuition about this might be based on conveyor belt like constructions we can observe on the earth, and we think of the earth as a conveyor belt that attracts air. Since it doesn't attract air/objects that are further away as strongly, we think they wouldn't be dragged along as much by it. But that's just totally wrong, isn't it?
But if it's stationary with respect to the surface of the Earth, that means that it's actually orbiting the Earth (at a speed of one revolution per day). In other words, it's moving along with the Earth. This lateral movement does not stop when the object is let go. Thus it will stay over that same point as it falls down.
Yes, one revolution per day. But one revolution per day on the surface is ~40.000km/day, while one revolution high up in the atmosphere is quite a bit faster.
When the object drops down, the horizontal velocity will be slowed by the atmosphere (assuming no wind etc), but it'll remain greater than the horizontal velocity on the ground, thus traveling further than the guy on the surface.
This is of course simplified. Just talking about the horizontal velocity when the object is let go is slighly misleading: when the earth continues to turn, the vector that was once horizontal isn't parallel to the surface near the observer any more, and the downwards momentum isn't perfectly downwards any more, so you'll need to constantly translate those vectors. (Or do the right thing and model it as angular momentum)
But this effect is only minor on reasonable altitudes and the short durations of the respective drops. "Dropping" a geostationary object from higher than ~36.000km will result in the object flying away, and I'm not sure where it'll land if you drop it from slightly below that.
Very good, you guys managed to explain it simpler than me!
It's better to describe any motion looking at the quantities that don't change, because it almost always leads to simple equations, in this case that quantity is angular momentum, which can be seen as a measure of how much "rotation" a system has (if this makes any sense). It's analogous to that common experiment where someone is spinning on a rotating chair with the arms open, and suddenly pulls the arms close to the body and as a consequence, he/she spins faster.
It happens because by pulling the arms, the person reduces the radius of rotation and thus must spin faster in order to conserve angular momentum.
As was said, being at rest with respect to the Earth means that the object is spinning with the same angular velocity that the Earth has, so initially it has a non-zero angular momentum. Gravity will always point to the Earth's center so it can't change angular momentum, the same way that you can't hope to make a wheel spin by applying a force directed at its center. When the object starts falling, its radius of rotation is reduced, so it has to spin at a greater angular velocity to conserve angular momentum. The observer's angular velocity was always identical to Earth's, while the object's velocity started at the same value but kept increasing as it fell so it must arrive East sooner than the observer.
Describing the motion at even higher altitudes is a bit more complicated, because we have to use the classical theory of gravitation. So, here comes a small wall of text:
We can consider that after the object is released it forms a two-body system with the Earth and, since the Earth's mass is much larger, we apply Kepler's first law and the path followed by the object (for reasonable heights, of course) is an ellipsis that has the Earth's center as focus. This orbit will have a maximum distance from Earth's center (the one that it was initially, before the drop) and a minimum distance (which is 180º from the starting point).
Seeing it this way, we notice that the object falling on the ground is simply a consequence that it has so little energy that in order to complete its elliptic orbit it should pass closer to the Earth's center than the Earth's radius allows. As we drop it higher we can see that eventually there'll be an altitude where the minimum distance is exactly Earth's radius. It'll make a turn of 180º and pass close to the ground at an extraordinary speed before rising again (of course in practice the atmosphere would dissipate the energy, but we're talking about idealized behavior anyway).
As we drop it from even higher altitudes the minimum distance keeps increasing and it keeps in its elliptic orbit up to a point where the object's energy is just enough to get rid of the eccentricity of the ellipsis. At this point the gravitational force equals the centripetal force needed to keep it rotating at the earth's angular velocity, the object becomes geostationary and follows a perfectly circular orbit.
With an even higher initial altitude, the orbit is again an ellipsis, with the difference that the initial height is now the minimum distance, not the maximum one as before. As altitude is increased even further, there'll be a point where we say the energy of the object reaches 0 (it was negative up to that point), gravity can no longer hold the object and it escapes in a parabolic orbit. Finally if you drop it slightly higher than that, it gets positive energy and escapes Earth's gravitational field in a hyperbolic orbit.
Yes, one revolution per day. But one revolution per day on the surface is ~40.000km/day, while one revolution high up in the atmosphere is quite a bit faster.
When the object drops down, the horizontal velocity will be slowed by the atmosphere (assuming no wind etc), but it'll remain greater than the horizontal velocity on the ground, thus traveling further than the guy on the surface.
we could think of the Earth as a big circle traversing along the time axis.
Btw, this is another thing that puzzles me. What causes everything to traverse on the time axis, and why cannot we move freely on the time axis in the same way as we can on the other spatial axes (ie. why can't be freely move faster or slower in time, or even backwards)?
I cannot really tell you why the time dimension seems to work differently than the room dimensions. I could point out some formulas where time is treated differently, but that wouldn't really answer the question.
However, I can add a few comments:
* Movement speed is distance over time. If you're trying to talk about the speed of moving through time, you get time over time, which is always 1. Not a useful thing to talk about, unfortunately.
* You cannot even measure how "fast" you're moving through time. Let's say you create a sci-fi-bubble in your living room in which time passes twice as fast as outside the bubble. While inside, you wouldn't notice anything different, because both your perception and any clocks you may carry will proceed twice as fast, too.
The only way to know that something is different about the bubble is to compare a clock inside the bubble with a clock outside. And even then you wouldn't know if the bubble sped up time inside, or slowed down time outside of it.
* You have heard about synchronized clocks ending up with different times in several relativity examples. Would that satisfy your notion of "moving faster through time"?
About traveling backwards through time: most laws of nature don't differentiate between time directions. If we reversed time right now, without changing physics, the universe would remain almost the same.
The reason we perceive a "past" and a "future" is purely a statistical thing, second law of thermodynamics. It suggests that there have been some initial conditions at some point in time, and the universe as we know has been derived from those. It makes sense to call these initial conditions "past" and everything that's derived from our current state as "future", but I think that's mostly a perception thing.
Past and future are not a property of the time dimension. Time is mostly symmetrical (not entirely, but mostly). Past is simply the direction in time when the initial conditions were created.
Like any person with a vivid imagination, I sometimes like to come up with wild hypotheses to explain things. Since I'm not a physicist and I only have a very cursory knowledge of general relativity and QM, this wild hypothesis may very well be waaay of the mark, completely nonsensical from a physical point of view, and basically ridiculous, but nevertheless, it entices my imagination.
It could be that our universe resides inside some kind of "metaverse", which properties are quite different (or possibly not even so much different) than the properties inside this "bubble" that's our universe. It could also be that what we perceive as movement in time is actually our universe moving inside this metaverse. The "time axis" is in actuality the movement axis that our universe traverses along in the metaverse.
Masses "drag behind" in this time axis, bending it. They don't drag in the spatial axes, only in the "time axis" of the metaverse. This is the reason why time passes at a different speed close to massive objects. (Singularities are, in theory, infinitely dense, which means that they have "infinite drag" and thus they don't move at all in the time axis of the metaverse, which is why time is completely still at a singularity.)
This would be what makes the time axis different from the spatial axes, and why everything moves on the time axis (because our universe is moving in the "metaverse").
It could also be that what we perceive as movement in time is actually our universe moving inside this metaverse. The "time axis" is in actuality the movement axis that our universe traverses along in the metaverse.
Again, movement as we know it requires both space and time. If our universe is to "move" across the metaverse, then the metaverse needs a time dimension.
So how would you then explain the time dimension in the metaverse? Moving through a meta-metaverse? How many turtles down, exactly?
Again, movement as we know it requires both space and time. If our universe is to "move" across the metaverse, then the metaverse needs a time dimension.
So how would you then explain the time dimension in the metaverse?
The metaverse doesn't need to necessarily obey the same laws of physics as our universe. I'm not versed enough in physics to give a possible physical description of this hypothetical phenomenon.
The existence of movement presupposes two dimensions: an independent dimension (normally assumed to be time) and a dependent dimension (the dimension moved along). You have to be able to say "when position on dimension A is X0, position on dimension B is Y0; when position on dimension A is X1, position on dimension B is Y1."
It's not a question of laws of physics; it's a question of definitions of words.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
It's not a question of laws of physics; it's a question of definitions of words.
If I have learned anything from quantum mechanics it's that the whole concept of time is... complicated.
The so-called delayed choice quantum eraser experiment is a perfect example of this.
In the double-slit experiment, measuring which slit the particle went through removes the interference pattern. Canceling that measurement after the particle passes the slits restores the interference pattern. Thus you can affect whether the interference pattern appears or not by choosing whether to measure or cancel the measurement.
More oddly, this is so even if the choice whether to measure it or not is made well after the particle hits the detector. It's as if the choice affects the behavior of the particle in the past. Even though, classically speaking, at the moment the particle hits the detector the choice has not yet been made, it still makes or doesn't make the interference pattern depending on the choice made in the future.
It's as if the particle were at all points of its path at the same time, because even a measurement choice made later in the path affects its behavior earlier in the path. Regardless, you can still measure the time that the particle takes from the emitter to the detector, and this is a finite time that depends on the speed of the particle.
It gets really weird.
Btw, this is another thing that puzzles me. What causes everything to traverse on the time axis, and why cannot we move freely on the time axis in the same way as we can on the other spatial axes (ie. why can't be freely move faster or slower in time, or even backwards)?
I will base the answer in Relativity; it is the only theory that has anything to say in the matter thus far.
As I mentioned earlier, the universal limiting speed (the speed of light) changes the structure of space-time. This change is in both geometry and topology; but the bit that is relevant is that it allows the definition of a space-time "distance" function that is the same for all observers (whether inertial or not).
This "distance" function (called "interval" in relativity) splits geodesics (and "forced" geodesics too) in 3 classes -- timelike, spacelike and null (or lightlike).
A geodesic [its velocity vector] can never change its class: if a geodesic [velocity] is timelike, it will always be timelike; likewise if it is spacelike or null. This is a theorem of differential geometry, being a purely geometric effect.
Moreover, any spacelike vector can be turned into any other spacelike vector by a rotation (in space-time) and a change in scale; but for timelike vectors, this is only true if both vectors point to the same general direction ("future" or "past"). This is a consequence of the above: the timelike vector would need to become spacelike or null (or both) along the way to change from pointing to the future to pointing to the past, which is impossible.
In relativity, massive particles are assumed to move in timelike geodesics, while light moves in lightlike geodesics. This assigns a nonzero timelike speed vector to all massive particles, which forces them to move in time.
In relativity, massive particles are assumed to move in timelike geodesics, while light moves in lightlike geodesics. This assigns a nonzero timelike speed vector to all massive particles, which forces them to move in time.
I'm not completely sure I fully understood the explanation (since the math involved in GR goes well beyond my understanding), but it made me wonder if this is just a (mathematical) description of how it works, rather than an explanation of why it works like that.
As for the former (ie. the description of how it works), could the movement in time be a consequence of something else, perhaps the constness of c for all observers? (I'm just throwing ideas here to try to better understand the mathematical description in more intuitive ways, without needing to dive deep into the GR equations and tensor metrics, which are very difficult to grasp...)
The real reason for selecting timelike geodesics as the trajectories of massive particles in relativity is that they are the ones that ensure that such particles move at subluminal speeds in space. From this choice, you get for "free" that massive particles cannot ever either accelerate to luminal or superluminal speeds (in space), nor double back in time.
Again, movement as we know it requires both space and time. If our universe is to "move" across the metaverse, then the metaverse needs a time dimension.
So how would you then explain the time dimension in the metaverse? Moving through a meta-metaverse? How many turtles down, exactly?
is in contradiction with this:
Warp wrote:
[...] we could think of the Earth as a big circle traversing along the time axis [...]
marzojr wrote:
The visualization is correct, with the caveat that it is space-time, not space, that curves.
Either one of these statements must be incorrect (or at the very least lacking). Which one?
If movement along an axis implies a separate time axis, how can be move along the time axis?
A point in the observable universe is a point in space-time, not simply a point in space. Since as marzojr said, the observable universe of A is defined as being every point which could have affected A's current state, then a point in A's universe is a point in A's past. As a result, A cannot transmit something to the star in A's observable universe, because this would involve sending the signal backwards in time!
A' B'
\uAB /
\ /
uA \/ uB
/\
/ \
/ u \
A B
In the diagram above, A is A at some point in the past. B is B at some point in the past. A' is A some time later; B' is B some time later. uA is the universe that A can affect. uB is the universe that B can affect. uAB is the overlap. u is the universe that neither can affect.
A can affect B' but not B.
The same diagram can demonstrate observable universes: uA and u are the observable universe of A'; uB and u are the observable universe of B'.
Sorry to return to this (and sorry for the long quote, but I wanted the whole context here), but I really want to understand this. Please help me understand the answers to these questions:
Assume that we have two planets A and B, and that the universe is expanding at such a rate that A and B are always receding from each other slightly faster than c. Also assume that there's a star between them that both can observe.
1) Can A send a probe to the star? If the answer is no, then why not?
2) If the answer is yes, then can the probe then proceed to B? Why?
Either one of these statements must be incorrect (or at the very least lacking). Which one?
The incorrect word is "circle". Let's say I have plotted a series of points on a graph such that they are on a straight line. Now I rotate the graph so the line coincides with the x axis. Then I could say, "these points go along the x axis". Although I used the word "go", of course it doesn't imply that this point here takes a walk along the x axis, as does this other point, and this one, and so on. Likewise, the earth does not actaully "move" through time. It's a 4-cylinder that has a beginning and an end.
Assume that we have two planets A and B, and that the universe is expanding at such a rate that A and B are always receding from each other slightly faster than c. Also assume that there's a star between them that both can observe.
You're now including cosmic expansion, a concept outside of special relativity, and you're trying to answer a question that cannot be answered strictly within relativity.
The answer depends not only on the position and relative motion of the star in the middle and the exact speeds of A and B, but mostly on the actual expansion of the universe.
Try starting here, for example:
http://www.astro.ucla.edu/~wright/cosmology_faq.html#FTL
Assume that we have two planets A and B, and that the universe is expanding at such a rate that A and B are always receding from each other slightly faster than c. Also assume that there's a star between them that both can observe.
You're now including cosmic expansion
My original question was about the expanding universe and overlapping observable universes. I haven't added anything. I just couldn't understand the original answers given to the question, and I would really like to understand.
you're trying to answer a question that cannot be answered strictly within relativity.
Sorry to return to this (and sorry for the long quote, but I wanted the whole context here), but I really want to understand this. Please help me understand the answers to these questions:
Assume that we have two planets A and B, and that the universe is expanding at such a rate that A and B are always receding from each other slightly faster than c. Also assume that there's a star between them that both can observe.
1) Can A send a probe to the star? If the answer is no, then why not?
2) If the answer is yes, then can the probe then proceed to B? Why?
AFAIK you don't need to put the fact that the universe is expanding to understand that. The drawing represents the spacetime diagram for a particle where one considers only one dimensional velocities (a universe of only one space dimension, where particles are constrained to move on a line).
The spacetime diagram plots the space dimension x measured by a given frame at the horizontal axis and the time t measured by its clock at the vertical axis. We have to consider time as a variable because unlike in classical mechanics not all observers agree on the time t elapsed. (Because of this, when considering 3-dimensional motion, the spacetime diagram would require 4 dimensions, which is impossible to draw, so we analyse motion in only one coordinate so that it can fit in a plane).
Now, consider a point at the origin. That corresponds to a particle at the coordinate x=0 at a time t=0. If that particle is moving at a constant speed v, then x = v t, this is a straight line in spacetime, its slope is related to the particle's speed. If v = 0, then such line is x = 0, which is a vertical line along the time axis. Recall that there's a universal limiting speed c. So, we normally scale the time coordinate so that the slope of the line x = c t has an angle of 45º.
So, if you plot the lines where the particle moves at the speed of light, you get two lines, one inclined 45º to the right (x=ct) and the other 45º to the left (x=-ct). If the particle is at rest, the line that describes its motion is vertical. As its speed increases, that line will gradually tend to the inclination of 45º. Because of this, an arbitrary curve that describes the particle's motion can't hope to escape the region limited by the two lines with slope of 45º. Every event outside this region can't be affected by the particle.
The concept of observable universe comes to solve a paradox of causality. Our intuition says that, if an event A happened at a time before another event B, then A could have been the cause of B. In SR, however, a frame can measure that A happened before B while another can measure that B happened before A, causing a paradox. However, if you remember, our previous discussion, a particle's motion can only affect what's inside the region delimited by those two inclined lines. So, if you have two events A and B, A could only have caused B if B is inside A's causal future (that region). Using this definition of causality, the math of SR will guarantee that if you take a frame, draw the two 45º lines from A and conclude that B is inside that region, then any other frame will get to the same conclusion so we have consistency.
