Markus Hanke

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.

They do have inertia, just like all bodies with non-vanishing rest mass. It is just that their gravitational interaction is extremely small. They are effected by Newton's laws. They are not different, both domains obey the same set of laws of mechanics, being quantum mechanics. It is just that for macroscopic objects, quantum effects become so vanishingly small as to be entirely negligible, thereby reducing the observable laws to standard Newtonian mechanics. In the microscopic domain that is not the case, so the laws "look" different at first glance. In other words - Newtonian mechanics is just a subset of quantum mechanics for systems where quantum effects are negligible. Difficulties arise once we leave the domain of applicability of Newtonian mechanics, namely once we get into the domain of General Relativity; this is because GR is most definitely not a subset of quantum mechanics, so there appears to be a contradiction here. For that reason, finding a self-consistent model of quantum gravity is one of the core areas of research in theoretical physics at present.

Why is this nonsense still in the science section ? The link provided by the OP is a crackpot site, and the idea of "gravity is electromagnetism" is a well known crackpot concept which has been done to death countless times on countless crank forums. This really shouldn't be in the main sections of the forum - at best this belongs in "Speculations".

I don't really follow you, to be honest...what exactly is the scenario you are considering, and what are the conclusions you draw ?

You are contradicting yourself here - first you say that it is not the acceleration, then you say that you need to match the accelerations to get the same result. So which one is it ? Regardless of the exact circumstances though, the difference - if there is one - in proper times arises because one of the frames is inertial, whereas the other one is not. If both frames where inertial, you could never get differences in proper times between the same two events; if one or both are non-inertial, then the result depends on the space-time geometry within those frames, in other words, the metric tensor. This is what I tried to demonstrate with the maths you quoted me on earlier.

I dream and think in English 95% of the time, even though that is not my first language. Mind you, it took a few years after I started to be exposed to English for that to happen. Good luck

In order to get the same result, you would need to match the net effects of the accelerations in both frames. Their proper times will then agree. But again, the "twin paradox" is a well defined scenario, where you compare an accelerated with an unaccelerated frame. If you alter that setup you can no longer call it "twin paradox", but some other experimental setup of your own making.

I'm fluent in both English and German, and have elementary knowledge of French, Spanish, Samoan, Mandarin ( written & spoken ), and conversational mathematics. Yes indeed - I consider mathematics a language in its own right.

You can do that of course ( "back and forth" equates to acceleration ), but then you have officially left the confines of the "twin paradox" scenario.

I'm in agreement with you, elfmotat. That is pretty much what I was saying anyway; it boils down to the fact that there is a difference in metric tensors between the observer at infinity and the observer inside the cavity, even though the actual values are not fixed and simply are a result of the boundary conditions imposed. Hence the time dilation. So I think we are all good

That's true, but you need to consider the boundary conditions of this problem as well. The metric inside the cavity needs to connect smoothly to the interior metric of the shell itself to ensure global differentiability, which then again needs to connect smoothly to the exterior Schwarzschild metric. It would be interesting to do the numbers here, but just by looking at it I don't think it will work if the metric in the cavity is Minkowskian. If that is the case we will get either a discontinuity at the boundary, or a hypersurface where the metric is smooth but not differentiable, both of which is unacceptable. I don't believe it is possible for the elements of the metric tensor inside the shell itself to be exactly +/-1 at any point including the boundary, since this would imply a Minkowskian vacuum with vanishing energy-momentum tensor; so, in order to maintain smoothness and differentiability at the boundary, the vacuum metric in the interior of the cavity cannot be +/-1 anywhere either, or else we have a boundary problem. But then again, that's just my two cents' worth from the point of view of differentiable manifolds, I might well be wrong. Has anyone got any references to a fully worked calculation for just such a case ? I couldn't find anything. I think here's the solution : locally the metric inside the cavity is Minkowskian; however, if we want a global coordinate system which asymptotically approaches Minkowski at infinity, then the cavity will be flat, but not Minkowskian. This takes care of the time dilation issue, and tallies nicely with Newton.

Having thought about it further, it would appear I made a serious mistake here, so I have to retract the above statement. We will indeed find a region of completely flat space-time, but not the Minkowski metric; instead, we will have a metric tensor the elements of which are all constants, but not equal to +/- 1. These constants will be some function of the shell mass and the cavity radius. Physically this means there is no gravitational field in the cavity, but a clock located there is still dilated compared to a clock located at rest infinitely far away. This tallies nicely with the Newtonian shell theorem, and in terms of classical mechanics, can be thought of as a non-vanishing gravitational potential inside the cavity. It should be noted that there are no potentials in GR, it is the metric tensor which is the source term for gravitational time dilation.

Take the first sentence for example : This is without any meaning.

I am afraid you are using too much non-standard terminology for me to give any meaningful reply; I quite simply have no idea what you are trying to say.