Markus Hanke

That depends what you mean by "experience". For example, take three test particles which are spaced radially, and let them freely fall towards a central mass, like so : As time goes by, they all approach the central body, but they also increase the distance between them. Why ? Not because there are any forces acting on them ( remember there is no acceleration, so a=0, and hence F=ma=0 on all these particles ), but because they are in a spacetime that is not flat, so as they age into the future ( they can't do anything else ! ), their spatial position will change. That's curvature right there for you - it is not just some abstract, theoretical concept, but something quite tangible, and it can be easily observed. The "engine" that drives all dynamics here is the fact that everything ages into the future, and, because time and space are intrinsically linked, spatial positions and distances change as that happens. This of course also works if you arrange the particles horizontally instead of radially : While the mathematical description might be quite involved ( I plead no contest in that regard ), the actual meaning of the equations is really quite simple, and anyone can grasp it - they are just descriptions of what happens to test particles.

The meaning of "straight" in this context is different than just being "straight in space". What it means is that no forces act on the satellite anywhere along its trajectory - you can easily see this by placing an accelerometer on board, which will read zero at all times, so there are no forces in its rest frame. In other words, the satellite is in free fall at all times. If you plot its world line in spacetime ( not just its trajectory in space ! ), you find that between any two events you choose, this world line is such that it traces out the longest proper time - this is called the "principle of extremal ageing", and in that sense the world lines of test particles in free fall constitute the most direct and straight connection between events in spacetime, and such world lines are called geodesics. This does not imply that the test particle's spatial trajectory ( = the orbit of the satellite ) is a straight line - it implies only that, given initial conditions, it is the trajectory that allows the satellite to trace out the most proper time as recorded by its own clocks. Always remember that we are in spacetime, not just in space.

I don't really understand the point in this - even the standard field equations admit dynamic ( i.e. time-dependent ) metrics as valid solutions, so you don't even need the ADM formalism for this. Why is that so special ?

That is interesting, I have not come across this statement before. Could you elaborate on this a bit ? Let's say I have a composite system of two fermions, and I create an entanglement relationship between them. I am interested only in their spin state, so either particle can be either spin-up or spin-down. Where does uncertainty come into this ?

You are right in that as of now there is no empirical evidence for the existence of wormholes and similar topological constructs; nonetheless, the fact remains that these are mathematically valid solutions to the gravitational field equations. Since those very same equations are otherwise in extremely good accord with experiment and observation, I think a case can also be made for not readily dismissing these phenomena purely on ideological grounds. There is certainly no scientific reason to a priori rule them out. In fact, as recent work on the ER=EPR conjecture has shown, such concepts may well have a role to play in the fundamental makeup of spacetime.

While I do not advocate any particular alternative of GR over another, in this context I thought it appropriate to mention that there is a class of bimetric theories of gravity which closely mimic weakly interacting dark matter effects, without requiring any new particles ( which would need to fit neatly into the Standard Model, with all associated difficulties ). One example - among others - is Sabine Hossenfelder's proposal : https://www.researchgate.net/publication/235472718_Bimetric_theory_with_exchange_symmetry This is of course all conjecture, but IMHO does warrant further study.

The fundamental principle you are referring to is the fact the laws of physics are the same in all inertial frames; the invariance of c is a consequence of this. This principle is not violated in your example, since no matter which rest frame you choose, no relative velocity will exceed the speed of light.

You seem to be asking whether or not the spin state of the particle, which is definite after the measurement, changes over time. The answer is that this depends on whether or not the particle interacts with other particles, but this information isn't given in your setup. You also need to bear in mind that after the initial measurement at t=0, the particle pair is no longer entangled, since wave function collapse always destroys entanglement. The other subtle issue I see here is the question of simultaneity, since t=0 may not necessarily mean the same thing to Alice and Bob, depending on the spatial and temporal relationships between the rooms.

Perhaps it would help to start at the most basic entity for a system in motion, being its Lagrangian. The action for a relativistically moving particle is of the form [latex]\displaystyle{S=\int \left ( -mc^2\sqrt{1-\frac{v^2}{c^2}} \right )ds}[/latex] From this, you can then derive a quantity called the 4-momentum by taking [latex]\displaystyle{p_{\mu}=-\frac{\partial S}{\partial x^{\mu}}=\left ( \frac{E}{c},-\mathbf{p} \right )}[/latex] The usual energy-momentum relation is then quite simply the norm of this 4-vector : [latex]\displaystyle{p^{\mu}p_{\mu}=\left | \mathbf{p} \right |^2-\frac{E^2}{c^2}=-m^2c^2}[/latex] So, the basic idea is that energy-momentum is a 4-vector, and the norm of that 4-vector is defined to be the proper mass of the particle. Because we are in flat Minkowski spacetime, calculating the norm of a 4-vector is equivalent to applying the Pythagorean theorem to the three elements of the equation ( proper mass, energy = time component of vector, and momentum = magnitude of spatial component of vector ). The first step ( Lagrangian to 4-momentum ) can also be made mathematically precise via the calculus of variations.

Harry Potter ? He sure knew his teleportation techniques

Based on what you have written here, what I think is that you know very little about quantum mechanics; as such you quite simply are not in a position to advance any meaningful "theories". What I suggest you should do is engage in a serious and very detailed study of existing quantum theory - this will put you in a position of knowledge and understanding, and that is the only position from which it is possible to make meaningful contributions to physics. Having imagination is great, and it is important, but in the context of physics it is useless without knowledge.

Xinhang, in post #47 you say : which is evidently in direct contradiction to empirical data, and hence nonsense. In post #51 then you suddenly say : which seems to imply that you think time dilation is after all a real phenomenon. You are contradicting yourself here, you can't have it both ways. So does - according to you - time dilation exist, or not ? If you decide to respond to this, then bear in mind that in physics "time" is defined as being what clocks measure; you cannot decouple this concept from its physical measurement, or else you are no longer doing physics.

@geordief, you might like this interactive Minkowski diagram generator - it's great for playing around with various parameters in order to get a "feel" for what is going on : http://www.trell.org/div/minkowski.html

In post #7, under the metric tensor paragraph, it should read [math]g_{\mu \nu}[/math], and not [math]G_{\mu \nu}[/math], to avoid confusion with the Einstein tensor. Just a small thing though

Thank you

In my humble opinion, which may or may not turn out to be helpful to you, it is best to find a perspective on relativity which is both geometric and physical. Consider the following simple scenario : let's say you take into your hand a bunch of very fine coffee grounds, and form that into a perfectly spherical ball. Let us further assume that the particles that ball is composed of do not interact with each other in any way ( that is of course an idealisation, but you probably get the idea that this is merely an analogy to demonstrate a principle ). You now release that ball of test particles ( coffee grounds ), and observe how it behaves under a number of conditions. In the most trivial case, you perform this experiment very far away from any source of gravity. What happens ? The answer is that nothing happens - the ball remains stationary with respect to any ( close by ) reference point you choose, and neither its volume nor its shape change in any way. This corresponds to the case of a spacetime that is approximately flat. In the next scenario, you release the ball of test particles in the exterior vacuum somewhere in the vicinity of a source of gravity, such as a planet or a star, again assuming that no influence other than gravity is present here. What happens now ? As time goes by, you will notice the ball starting to move towards the centre of the source of gravity; as the ball approaches the central mass, you will also notice that its shape begins to change - it becomes elongated along the direction of motion, and "squeezed" perpendicularly to it. However, if you perform a measurement of its volume ( by whichever means you choose ), you will - perhaps counterintuitively so - find that the volume of the ball does not change, meaning the deformation of its shape is not arbitrary, but is such that its volume remains unaffected. That is the situation in vacuum. In the last scenario, you place the ball in the interior of a source of gravity. Strictly speaking we would need to find a way for the ball not to be affected by the mass/energy itself, or else we would no longer be dealing with free fall. You can create such an approximate situation by - for example - drilling a very narrow shaft through your planet. What happens now ? Once again, the ball of test particles will move towards the centre of the source of gravity, and once again its shape will deform. However, this time you would find that both shape and volume of your ball of test particles changes as it freely falls. This is the situation in the interior of mass-energy distributions ( as opposed to the exterior vacuum ). All of these are empirical findings about how test particles ( specifically a spherical collection of them ) will behave under the influence of gravity. To understand how this is connected to General Relativity, just put some simple mathematics around it. Specifically, let us look at the volume of our ball of test particles. In scenario (1) there is no gravity, so nothing at all happens. This is trivial, so will ignore it. In scenario (2) - the vacuum case outside a source of gravity - we find that the volume of the ball remains constant; we slightly rephrase this and state that the rate at which the volume of a ball of test particles begins to change as we release it, is zero. In mathematical notation : [latex]\displaystyle{\left ( \frac{\ddot{V}}{V} \right )_{t=0}=0}[/latex] General Relativity uses the languages of tensors to express this very same idea - the mathematical object which describes the rate at which a small ball of test particles begins to change its volume is called the Ricci tensor; using this, we can write the above simply as [latex]\displaystyle{R_{\mu \nu}=0}[/latex] These are precisely the Einstein field equations in vacuum, and their meaning is very simple - if you take a ball of test particles and allow it to fall freely in vacuum, the rate of change of its volume is zero, i.e. its volume remains constant. Note that this says nothing about the shape of the ball, only about its volume, so this equation doesn't mean we are in a flat spacetime - quite the contrary, as the ball will be deformed as it falls. In scenario (3), the ball is in the interior of an energy-momentum distribution. In this case, both shape and volume of the ball change over time, and hence we find that the rate at which the volume begins to change is no longer zero, but rather [latex]\displaystyle{\left ( \frac{\ddot{V}}{V} \right )_{t=0}=-\frac{1}{2}\left ( \rho + P_x+P_y+P_z \right )}[/latex] This means that the rate at which the ball changes its volume is proportional to the energy density at its centre, plus the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction. Expressed in the language of tensors, this reads as [latex]\displaystyle{R_{\mu \nu}=\kappa\left ( T_{\mu \nu}-\frac{1}{2}Tg_{\mu \nu} \right )}[/latex] These are precisely the full Einstein field equations, and their meaning - in geometric and physical terms - is as given above. The form I am giving the equations in looks different than what most textbooks will write, but they are exactly equivalent ( my form follows from the usual form via an operation called trace-reversal ). So, if you wish to understand the meaning of the Einstein equations, just think about small balls of test particles in free fall - and that is all there really is to it. Of course, while this is easy to grasp conceptually, performing any actual calculations with this is fiendishly difficult, since the various tensors used in the equation follow from the metric in non-linear and non-trivial ways. Perhaps the above may be helpful to you.

I think even a thread in "Speculations" ought to have at least some connection to real physics. This one clearly does not.

GR basically states that the presence of energy and matter is equivalent to changes in the geometry of space-time, in that this geometry is no longer flat but possesses intrinsic curvature. This curvature manifests as gravity, because two initially parallel-moving test particles, when crossing through a region of space-time with curvature, will start to approach each other ( this is called geodesic deviation, and forms the basis of GR's maths and physics ). So in short - in GR, gravitation is a manifestation of the curvature of space-time itself.

It models gravitational interactions between bodies.

What we find at the Schwarzschild radius is a coordinate singularity in Schwarzschild coordinates; it is physically meaningless, since it can be eliminated simply by choosing a different coordinate system. An observer falling into a black hole would notice nothing special as he crosses the event horizon.

As measured by whose clock and whose ruler ?

Particles travelling at the speed of light have always done so - massless particles cannot accelerate or decelerate, therefore no "minimum" energy is required. On the other hand a particle with zero rest mass and zero total energy would not be physically meaningful, since such an entity could not interact with its surroundings; for all intents and purposes it wouldn't physically exist at all. What "gives" speed is technically acceleration, which boils down to the expenditure of energy. The difference between massive and massless particles in that regard is that the former can experience acceleration, whereas the latter can't. Ultimately the reason for this is the geometry of space-time itself; massless particles always move along so-called null geodesics in space-time, which implies ( maths omitted here ) that they cannot experience acceleration and thus always propagate at exactly the speed of light.

Of course it does, but that effect is far too small to cause any measurable time dilation effects - unless of course you can show us some numbers proving otherwise, but then again, you appear to reject any maths, so obviously you are unable to. I should also remind you that the travelling twin's proper time dilation explicitly depends on the exact path it takes before returning to earth, including all maneuvers undertaken while far away from the stationary twin, i.e. far outside the range of any measurable gravitational interaction. Likewise, it is easy to show that the total proper time dilation experienced by the observer does not depend on their masses; if you were to substitute the twins with, say, muons, you would find that the particles' lifetime is dilated by the exact same amount as the much more massive spaceship would have been. Therefore it is obvious that the time dilation in this experiment is not the result of gravitational interactions between the observers.

News to me. Can you specify exactly what those "forces acting on him from escaping twin" would be, both qualitatively and quantitively. I was once present at the launch of a rocket carrying a satellite - what forces from the escaping rocket was I supposed to be feeling ? Please be very exact here, and provide a textbook reference to the force you are proposing.

It changes the arc length of the worldline connecting the same two events in space-time. In other words - the proper times as measured by the two twins will no longer agree, which is precisely what we wanted to explain in the first place. Too imprecise. The basic proposition is that both twins start off at rest in the same frame of reference, with synchronized clocks; one twin remains at rest, whereas the other twin makes a round trip of arbitrary distance and duration, and then returns to the stationary twin. In the end both twins are once again together at rest within the same frame. They then compare their clocks which each of them has been carrying, and find that the readings don't agree. There is no contradiction. The stationary twin feels no forces acting on him, so will always consider himself at rest. The travelling twin feels the forces of his own acceleration and deceleration during his round-trip journey, so he will always consider himself moving. These two frames are physically distinct, and not interchangeable, hence there is no contradiction, and the fact that their clocks don't agree at the end of the experiment is not a paradox because they know precisely who was moving and who was stationary at the end of the experiment. Mathematics is just simply a language to express physical ideas in concise form. You learn maths just as you learn any other foreign language. Dealing with physics and refusing to use maths is like living in China and refusing to use Mandarin; not only are you going to make your own life extremely difficult ( trust me, I've been there ), but you will also forego any chance of ever really understanding or appreciating what it really is you are dealing with. That is a simple truth of life. In this example, everything that has been so verbously elaborated on over the course of 18 pages can be mathematically expressed in just a handful of lines - if everyone here spoke the language of maths to the same standard, we would not have had to engage in this lengthy discussion. It is very easy to show mathematically that the proper times of the two twins cannot agree in this scenario, and more crucially also precisely why that is. Doing the same verbally on the other hand takes literally a wall of text to do it right; everyone needs to decide for him/herself which one they prefer.