Luboš Motl

Czech Republic

motls.blogspot.com

Age: 41

Hi, I am a string theorist and a publicist.

6h
comment Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?
Dear @Stan, I am confident that most experts with 5,000+ cits on quantum gravity papers will agree with me that magnetic monopoles are unavoidable in quantum gravity. It's really because you may create black holes with a confined magnetic flux - imagine two end points of a big bar magnet separately collapse into black holes. So these objects exist in the form of black holes and even if and when the black hole evaporates all the particles with $Q_m=0$, something with $Q_m\neq 0$ has to be left, anyway. The magnetic monopole elementary particles may be identified with the lightest BH microstates
6h
comment Why does $1+2+3+\cdots = -\frac{1}{12}$?
I don't think so - my identity for $(n)+ (n+1)+\dots$ works for an arbitrary fractional $n\in R$, too. The value for $n=1/2$ is as important in superstring theory as the value for $n=0$.
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Feb
24
awarded Enlightened
Feb
20
comment Is the wave-particle duality a real duality?
@StanShunpike - the wave-particle duality has never been a "problem", it's been a defining feature and virtue of quantum mechanics (or "quantum theory") since the beginning. Even non-relativistic quantum mechanics describes the objects so that they have both wave-like and particle-like properties. Quantum field theory does the same - while it's also easier to describe all the collections of particles as a "quantization of a classical field" (or "of classical waves"). But when it comes to basic conceptual properties like the co-existence of wave-like and par-like properties, QFT changes nothing
Feb
19
comment Why does $1+2+3+\cdots = -\frac{1}{12}$?
What is wrong about your calculation is that you are assigning an incorrect value to this $1+1+1+\dots$. In that sum, one must really keep track from which value of $n$ each term $1$ comes from. So the sum $1+1+1+$ starting at $n=1$ is $-1/2$ but if it starts at $n=0$, the sum is $+1/2$, for example. However, no such ambiguity exists for the analogous value of $\zeta(-1)$. Incidentally, the general sum $(n)+(n+1)+(n+2)+\dots$ is equal to $(n-n^2/2) -1/12$. You may check that it is equal to $-1/12$ both for $n=0$ and $n=1$ and it obeys the consistency checks when removing $k$ initial terms, too
Feb
19
comment Why does $1+2+3+\cdots = -\frac{1}{12}$?
Dear @MarioCarneiro, some sums may be hard or even ambiguous but I assure you that both $0+1+2+3+\dots $ and $1+2+3+4+\dots$ are equal to $-1/12$. The sum honors everything that needs to be honored to be certain that $-1/12$ is the only right finite value that may be attributed to it.
Feb
19
comment How to get Planck length
No, it doesn't sound like astrology (it sounds like fundamental and demonstrable science) and your question should have contained "doesn't it" rather than "isn't it".
Feb
14
comment Why quantum electrodynamics?
Something is traveling in between the source and the other charge that reacts. It's the perturbations of the fields. If the charges are static (time-independent), then the fields inside are also static, so they appear not moving. But if the charged source starts to wiggle, there are literally electromagnetic waves propagating in between, by the speed of light. In QED, electromagnetic waves are reinterpreted as a collection of an integer number of photons, they also move by the speed of light.
Feb
14
comment Why quantum electrodynamics?
Dear @Primeczar, good if something has been englightening. The time delay - already existing in classical electrodynamics - is the time needed for the wave to get from one place to another. If you say something, the sound also spreads by a finite speed so there is a time delay. If the charged source of the field is static, you can't say at which moment the charge created the fields we feel now. But when things are moving, the answer is just one - it was after a delay $\Delta t = \Delta x/c$.
Feb
14
comment Why quantum electrodynamics?
E.g. you may define the field $H(x,y,z)$ which is the average of $|\vec E|^2$ in a ball of radius 1 meter around $x,y,z$. The equations describing the evolution of $H$ may look nonlocal and allow superluminal propagation of the signal but it's because $H$ is "complicated" in terms of the more elementary fields - and in terms of the elementary ones, the interactions may be seen to be local (the action is never faster than the speed of light).
Feb
14
comment Why quantum electrodynamics?
An unphysical degree of freedom, like the electrostatic potential, is one that is used in the equations we use to describe physics but that can't be measured. For example, the potential and the whole vector $A_\mu$ is unphysical because of gauge invariance, and only some derivatives like $F_{\mu\nu}$ are physical. A nonlocal variable is a functional of fields that depends on values of fields at many points in the space or spacetime, whose distance is strictly positive.
Feb
14
comment Locally every force admits a potential?
Yes, in actual Nature, friction (to the leading approximation) may be reduced to statistics of lots of interactions that are ultimately electrostatic i.e. conservative. But the more general claim isn't right. The expression for the total energy may be rather complicated and not just $p^2/2m+V$ where $V$ is a potential energy. There may be (and in general, there are, even in Nature) various more complicated functions and products of coordinates and momenta (or coordinates' derivatives, velocities etc.) and any sufficiently general form of the energy means that the forces are non-conservative.
Feb
14
comment Why quantum electrodynamics?
Electrodynamics (just like general relativity) is compatible with special relativity so no information or impulse or true matter may propagate faster than light or instantaneously, and quantum electrodynamics doesn't change anything about this fact. There are various descriptions in classical and quantum electrodynamics where it may "look" that something is instantaneous but this is always a result of using nonlocal or unphysical degrees of freedom. Elmg and gravitational waves/impulses are propagating exactly at the speed of light in the vacuum, as the equations show (both in class+quantum).
Feb
13
comment Why quantum electrodynamics?
Did you mean interaction between charged particles? Even in classical electrodynamics, CED, the interactions are due to the field in between. This force in may be described in QED using virtual particles, and different gauges, etc., this makes no sense to explain this aspect of QED in isolation from a comprehensive introduction to QED as a whole, sorry.
Feb
13
answered Why quantum electrodynamics?
Feb
8
comment Why are atoms of the same element exactly the same?
If the atoms are in the same excitation state, e.g. ground states, the wave function is exactly symmetric or antisymmetric (depending on whether the number of elementary fermion in each of them is even or odd). If there are two atoms that are in different excitation states, they effectively behave as distinguishable particles, so their wave function is neither symmetric nor antisymmetric in general. However, one may always describe the different levels as a multiplet of one particle, and then the full wave function is still a symmetric/antisymmetric wave function, now a matrix.
Feb
8
comment Why are atoms of the same element exactly the same?
Dear @BjornW, if you have two magnesium atoms, each of them has 12 electrons and the accurate description of the situation is in terms of a wave function of positions of 24 electrons (plus two nuclei) and this wave function is perfectly antisymmetric under the $S_{24}$ permutation group of the 24 electrons. Alternatively, you may overlook the exact internal structure of the atoms. Then you have a wave function that is a function of locations of 2 atoms.
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