Markus Hanke

Sure, there are many ways to do this. Momentum is defined to be the derivative of the Lagrangian with respect to velocity, i.e. it describes how kinetic energy relates to relative velocity, as a function of rest mass - this is simply a generalisation of the good old p=mv, as we all know it from our school days: \[p^{\mu } =\frac{\partial L}{\partial v^{\mu }} =mu^{\mu }\] with the Lagrangian for relativistic motion being \[L=-mc\int ds=-mc\int \sqrt{\eta _{\mu \nu } dx^{\mu } dx^{\nu }}\] Energy, on the other hand, is defined as \[E=p^{\mu } v_{\mu } -L\] which is in essence a restatement of the fact that energy is the conserved quantity that arises from time-translation invariance in spacetime (see Noether‘s theorem for formal proof). Put these together and rearrange to get \[E=\sqrt{m^{2} c^{4} +p^{2} c^{2}}\] as stated above. You could also simply look at the general form of a 4-momentum vector, and see immediately that its temporal and spatial parts are related as above. This would be a standard way to derive this, but there are many, many other ways to do this, both informally and in very formal ways. For massive particles at rest you have m<>0 and p=0 - insert into the above to get \[E=mc^2\] as requested. You can also derive all of this from first principles - the symmetries of Minkowski spacetime - in a very formal way by using Noether‘s theorem; this gives you the energy-momentum tensor as a conserved quantity, and from its vanishing divergence you can derive the above expression as well. The problem here isn‘t any of the above, because you can find all of this in pretty much any standard undergrad text on Special Relativity. This is basic stuff, and it is not in contention. The problem though is that your response to this will be along the lines of: “You don’t understand anything, you are just parroting what you have read!”. Am I right?

First of all, this particular relation is valid only for massive particles at rest (p=0); it’s the limit case of the full energy-momentum relation \[E^2=m^2c^4+c^2p^2\] This relation follows from the fact that the inner product of the 4-momentum with itself (i.e. its Minkowski norm) is an invariant: \[\eta _{\alpha \beta } p^{\alpha } p^{\beta } =m^{2} c^{2}\] So the validity of the energy-momentum relation is a direct result of the transformation properties of the 4-momentum under Lorentz transformations. Since you have already stated that Lorentz invariance is indeed a given, you can’t then go on and dispute the energy-momentum relation, especially not if you try to replace it with an expression that is dimensionally inconsistent and not itself Lorentz-invariant.

Just to add to what has already been said: there is, in principle, no law of nature that stops macroscopic systems from behaving quantum-mechanically. The problem is only that you need to prevent decoherence from occurring - meaning you need to prevent the system in question from interacting with its environment. This is relatively easy to do for very small systems, but becomes exponentially harder the larger the system in question becomes. Putting a single atom into a superposition of states isn't too difficult, given a suitable setup; doing the same with (e.g.) an elephant is - for all intents and purposes - a virtual impossibility.

I think his is a philosophy of acknowledging the overwhelming experimental and observational evidence for the validity of SR within its domain of applicability. Anyone who comes along and claims that SR isn't a valid model will thus have some serious explaining to do. Because of this sheer amount of observational evidence, claims to the contrary will generally be dismissed pretty much by default, on the simple principle that extraordinary claims require extraordinary evidence - it's thus your responsibility to show yourself right, not ours to show you wrong. And if you don't mind me saying so, for the case of SR your evidence will need to be exceptionally extraordinary indeed, given how well-tested and verified a model this is. This means you need to show that your own model can replicate all the results of SR, can function as a local approximation to GR, and can make verifiable predictions that standard theory can't. I dare say you will find this hard to do.

Against what, exactly?

Yes, this is a better way to look at it than thinking of space as a thing that expands. Yet another way to look at it is to say that distance measurements in this type of spacetime are time-dependent - so the outcome of a distance measurement between two otherwise fixed points explicitly depends on when that measurement is taken. There needn't be any reference to motion of any kind.

Fixed that for you.

The result is manifestly wrong, as shown by the maths above, irrespective of how you might have arrived at it. Again, see the maths above. The inner product of 4-velocities cannot logically exceed c^2. Again, the inner product is an invariant, so it cannot depend on \(\gamma\). It is trivially easy to formally prove that. Also, if that weren’t so, Lorentz invariance could not hold in the real world - but of course it does, as shown by experiment and observation. I don’t know what you are actually trying to show here, but whatever it is, it is physically and mathematically meaningless. I think it is fairly obvious by now that you aren’t actually prepared to listen to any of the feedback given to you here. You are just repeating the same things again and again, and ignore the fact that you have already been shown wrong. If you were actually interested in learning something, you would listen to what has been said, and use it to improve your own understanding of the subject matter. It is not enough to just blindly do algebraic manipulations - you need to also understand the geometric meaning of these objects, and how they are related to one another, and how that fits into overall physics.

You haven‘t even addressed them yet, let alone refuted anything. I have already shown you above that the absolute norm of a 4-velocity is always exactly equal to c. It can‘t be any different of course. The mistake you made is assuming that 4-velocities can be any arbitrary 4-vector, but that isn‘t true - all 4-velocities are 4-vectors, but not all 4-vectors are automatically 4-velocities. The spatial and temporal part of a 4-vector need to be related in a specific way (as I showed above), or else it isn‘t a 4-velocity. When you take this into account, you end up with an equality, not “>=“. The same is of course true for 4-momenta, since \(p^{\mu}=mu^{\mu}\). The algebraic expressions you derived on the RHS of (1) and (2) are correct, but the “>=“ inequality is not. Point (3) is trivially wrong, since the norm of a 4-vector is an invariant, and hence cannot functionally depend \(\gamma\). You are just copy-pasting the same stuff over and over again, without addressing anything that is being said to you.

I‘m sorry, but that doesn‘t make any sense at all. The tensor transformation law applies irrespective of the value of individual components in a particular coordinate basis, so saying that all tensors should be null tensors on account of their transformation law is an entirely meaningless statement. To pick just one random example, the Minkowski tensor evidently transforms as a rank-2 tensor, but it most certainly isn‘t null, regardless of what coordinate basis you choose.

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You haven’t. You are simply repeating the same statements again and again. You are right, this was my mistake, I actually thought of the photon’s 3-momentum (which is not zero!) when I wrote my post, whereas you were referring to the 4-momentum.

I will not be opening files from untrustworthy sources, and you have failed to presented it here, so I haven’t seen what you did. However, the details are irrelevant, because the very fact that you arrived at a contradiction is the error. When you start with a 4-vector - which is a rank-1 tensor -, and take the covariant derivative, the result is a rank-2 tensor. Since both of these objects are tensors, they are subject to the same general transformation law - so there cannot be any contradiction. You’re simply mapping a rank-n tensor into a rank-(n+1) tensor. If you arrived at anything else, then the error is logically yours; there is nothing to be refuted here.

What characteristic of these people would you be studying, and what would your methodology be? And how do characteristics of people relate to the ontological existence or non-existence of a deity, or its characteristics? I wasn’t referring to the text itself, I was referring to the way people interpret it. Why are you engaging in it, then? If you are fully convinced of the veracity of your own beliefs, then it should make no difference whatsoever what anyone else thinks, and there is no need for any debates. Yet you are here trying to argue your points in front of an audience that doesn’t share your worldview, and never will. The Christian God is a learned and acquired concept - you learn of it and about it from other people, or from written texts. It’s external information, not intrinsic experience. Had you grown up in an environment where that external information was absent, the concept you now believe in so strongly would never even have occurred to you. God, as the concept is understood in Christianity, is a mental, social, historic, and cultural construct.

I don’t know what you think the Bible might suggest, since there are about as many interpretations of the text as there are readers, and they are all quite different. This is true for most religions. So if the Christian God is part of the material universe, as you seem to say here, can you then suggest a scientific experiment that might show his existence and characteristics, in a way that is repeatable and independently verifiable?

There is no “discrepancy”, so the point is moot. Non-sequitur. If you arrive at some kind of “discrepancy”, then that means you did something wrong, plain and simple. Tensor calculus is not a “theory”, it’s a mathematical framework that has been extensively developed, and is fully self-consistent.

This is not what I said - you need to go back and look at what I actually said, and you will also find the maths there. Photons have no rest mass, so according to your expression, the inner product of photon momentum with itself is zero. Since the inner product can vanish only if the two vectors are either perpendicular, or one of the vectors has zero magnitude, that means that according to you the photon has no net momentum. Just repeating the same thing again does not make it any less wrong.

As joigus said, there are no conflicts or discrepancies at all. As you correctly say, the covariant derivative yields a tensor, so it automatically has the correct transformation properties. This is so pretty much by definition, because otherwise it wouldn’t be a covariant derivative at all!

\[x^2\]

This is meaningless. If you are projecting a shorter 4-vector onto a longer one, the result can never exceed the length of the longer vector, that’s just basic geometry. In other words, the longest a projection can ever be is that of a 4-vector onto itself, which, in Minkowski spacetime, is thus -1. Since the inner product is invariant, this is true for all 4-vectors in all frames. Consider a general 4-velocity of the type \[u^{\mu } =( \gamma ,\gamma v_{x} ,\gamma v_{y} ,\gamma v_{z})\] The inner product with itself is \[u^{\mu } u_{\mu } =-\gamma ^{2}\left( 1-v^{2}\right) =-\gamma ^{2} /\gamma ^{2} =-1\] as expected. So your claim is wrong. If this were true, the momentum of a photon would be zero. This is evidently false, as we know already from experiment and observation. Mathematically, you can show this in a similar manner as above. So again, you are wrong. I’ve already shown above that the inner product of a 4-velocity with itself is -1, so the magnitude of a 4-velocity in spacetime is always exactly c. Since the inner product is invariant, this is true in all frames, so it can’t be a function of the gamma factor. Also, if you look at this expression, you should notice immediately that the resultant magnitude of v does not correspond to the gamma factor you are inputting, so the expression is meaningless. You are wrong on this one, too. Yes, because all three points you have presented contradict both basic maths, as well as observation in the real world. It’s simply wrong.

Peer review is not meant to be “fair” (what does that even mean?) - on the contrary, it is designed to be as critical as possible, so as to really put the ideas within the publication to a rigorous test. It is the most effective way to tease out any problems; remember, you want to end up with something that actually works, in the sense of the scientific method. You are missing the salient point. Science has nothing at all to say as to the existence or supposed characteristics of anything that isn’t part of nature, including any and all notions of deities. This is quite simply outside the domain of applicability of science, because the very notion of “God” is not amenable to the scientific method. So science neither rejects nor endorses the Christian faith, because it deals with a different domain of enquiry. However - and this is the important point - if someone proclaims an element of their faith as being objective truth, then this claim will of course be challenged by science. Some such claims may turn out to be compatible with scientific evidence, so they are fine; others may not be, and those will be rejected. To give a simple example - if someone claims, based on certain readings of the Bible, that the Earth is ~6000 years old, then science will certainly reject this, because that claim is evidently false based on all available scientific data. So the issue isn’t faith and belief - the issue is only when people try to misrepresent their beliefs as objective, scientific facts. That’s what’s called a category mistake, and it will always be challenged.

No, a flat manifold without boundary is simply one of the topologies that naturally arise from the mathematics of the Lambda-CDM model, when certain observational parameters take on specific values. It’s just one among several possible options. It can be closed on itself - so there can be a largest possible separation between two suitably chosen points, without there being any kind of boundary. Much like (to pick a lower-dimensional analogy) a spherical surface. A manifold being finite in extent does not imply the presence of a boundary.

The mechanism is one of spontaneous symmetry breaking, so it would depend on whatever parameters appear in the Lagrangian. In this specific case it would be the speed of light, Planck’s constant, as well as the coupling constants between the various fields. The Higgs mechanism has no effect on the value of fundamental constants (other than ones related to the masses of particles of cause, which aren’t really ‘fundamental’). It only gives rise to some new coupling constants within the Lagrangian, since after the event you end up with more fields than you started off with.

This is evidently not true, since the very mechanism that creates the property of mass (the Higgs mechanism and spontaneous symmetry breaking) already presupposes the existence of at least a Minkowski background spacetime prior to said process; without this, there would be no universe as we know it today.

We are not really in a position to give any kind of medical advice or diagnosis here, especially not based on very non-specific symptoms such as the ones you describe; if you are concerned about your friend, you need to urge them to go and see a medical professional.