In general, strings in string theory may be open or closed.
Open strings are topologically line intervals with two endpoints; closed strings are topologically circles. The fields defined on the strings may be periodic as a function of \(\sigma\), the spatial coordinate along the string, or they may obey various other boundary conditions. Let's look at them.
Strings are parameterized by the single spatial coordinate \(\sigma\), pronounce "sigma", and as they're moving in space, they're painting a twodimensional generalization of particles' onedimensional world lines, the world sheets. Among the two dimensions of the world sheet, one dimension is spatial, it's our \(\sigma\), and one is temporal and it's usually called \(\tau\), pronounce "tau". Note that "sigma" and "tau" are Latinized as "s" and "t" and those letters stand for "space" and "time", respectively.
As long as the laws on the string are uniform as a function of \(\sigma\), it's clear that \(\sigma\) can't be either infinite or semiinfinite because the total angular momentum, energy, and other quantities carried by the string would have infinite values and the string would have an infinite entropy. So realworld lowenergy strings must definitely be encoded just by a line interval. It's a convention to rescale (or reparameterize) \(\sigma\) in such a way that it belongs to the interval\[
\sigma\in (0,\pi).
\] However, the end of the world – and end of the string – is a subtle place. The laws of physics may have trouble to determine what should happen there. I won't go into this technicality but if you will study string theory at a technical level, you may want to know it: When you derive the EulerLagrange equations by varying the world sheet action, you need to get rid of the boundary terms and it's only possible if the string obeys certain conditions at \(\sigma=0\) and \(\sigma=\pi\).
Closed strings
The simplest way to solve the problem with the "end of the string" is to have no ends of the string. We may just glue the endpoints. Any wave that is going to the left and reaching \(\sigma=0\) will simply reappear at the right side, near \(\sigma=\pi\). The string is "periodic". More precisely, the fields on the strings are periodic functions of \(\sigma\). What are the fields on the string? They're fields \(X^\mu(\sigma,\tau)\) encoding where a particular point of the string (or world sheet) is located in the target spacetime. There may be fermionic counterparts such as \(\psi^\mu(\sigma,\tau)\) or \(\theta^a(\sigma,\tau)\) and some friendly ghosts and superghosts denoted by letters such as \(b,c,\beta,\gamma\).
I will only talk about \(X^\mu(\sigma,\tau)\) in this blog entry.
So for closed strings, we simply have\[
X^\mu(\sigma=\pi,\tau) = X^\mu(\sigma=0,\tau).
\] The two end points are identified, the string becomes a circle (topologically), and the world sheet is a cylinder. All the boundary terms \(B_0^\pi\) vanish simply because the values at \(\sigma=\pi\) cancel against those at \(\sigma=0\). When you decompose \(X^\mu(\sigma,\tau)\) into Fourier modes, the most natural modes go like \(\exp(2in\sigma)\) where \(n\) is an integer. You will find out that there are leftmoving "complex" waves as well as rightmoving "complex" waves on the world sheet.
The relevant Fourier modes of \(X^\mu(\sigma,\tau)\) and its canonical momentum (essentially velocity), \(P^\mu(\sigma,\tau)\), are known as \(\alpha^\mu_n\) and \(\tilde \alpha_n^\mu\) where \(\mu\) is a spacetime Lorentz index and \(n\) is an integer. Whenever \(n\) is negative, these \(\alpha\) operators are creation/raising operators of a harmonic oscillator; for positive values, they're the annihilation/lowering operators.
The single bosonic string has a ground state, \(\ket 0\), and it may be excited by those creation operators to obtain states such as\[
(g_{\mu\nu}+B_{\mu\nu}) \alpha_{1}^\mu \tilde\alpha_{1}^\nu \ket 0
\] where the fields \(g,B\) – the symmetric metric tensor and the antisymmetric Bfield twoform that are somewhat naturally combined into a general 2index tensor – may depend on the zero modes \(X_0^\mu\), the average of \(X^\mu(\sigma,\tau)\) over \(\sigma\).
While \(\ket 0\) corresponds to the tachyon with \(m^2\lt 0\), the first (doubly) excited states above are massless, as expected for a graviton (and the Bfield as well). One may excite the harmonic oscillators many times to make the closed string behave as a massive particle with masses given by \(m^2=4N/ \alpha'\) where \(N\) is an integer (the sum of all the numbers \(n\) in the operators \(\alpha_{n}^\kappa\); it must be equal to the same total excitation of \(\tilde\alpha\) operators).
We may also "twist" the boundary conditions by a symmetry. For example, the coordinate \(X^3(\sigma,\tau)\) may be antiperiodic rather than periodic,\[
X^3(\sigma=\pi,\tau) = X^3(\sigma=0,\tau).
\] The minus sign is new. In this case, the function is an antiperiodic function of \(\sigma\) and the indices of the Fourier modes will be in \(n\in \ZZ+1/2\). This will still cancel the boundary terms because the action is bilinear in (derivatives of) \(X^\mu\) or, more conceptually, because the extra operation \(X^3\to X^3\) we did before we glued the endpoints is a symmetry of the world sheet action.
Because \(X^3\) doesn't want to oscillate too much (it increases the energy i.e. the mass of the particle that the string behaves as) and because it's odd, it's clear that \(X^3\) will oscillate in the vicinity \(X^3=0\) in the spacetime. So these "twisted" closed strings will correspond to particles that are stuck near the \(X^3=0\) plane in the spacetime. Twisted boundary conditions for closed strings mean that the states of the closed string we may get in this way live in the socalled "twisted sectors", new kinds of states that appear in "string theory on orbifolds". In fact, the wound strings (strings wrapped around a cylinder in spacetime \(w\) times) may be viewed as special examples of a twisted sector.
But I don't want to focus on closed strings and orbifolds. Instead, my main target were the open strings and Dbranes.
Open string boundary conditions
There exists another way to cancel the boundary terms at the end of the string: to force that the contribution to the boundary terms is zero from both \(\sigma=0\) and \(\sigma=\pi\) separately. The traditional condition for \(X^\mu\) that's been around since the early 1970s is the socalled Neumann boundary condition:\[
\pfrac{}{\sigma}X^\mu_{\sigma=0} = 0
\] and similarly for \(\sigma=\pi\). The phrase "Neumann boundary condition" simply refers to the vanishing of the \(\sigma\)derivative of the field \(X^\mu\), in this case. Note that the \(\sigma\)derivative may also be called "normal" (perpendicular to the boundary at \(\sigma=0\)) so you may also say that the normal derivative vanishes.
If one omitted the differentiation with respect to \(\sigma\), he would get the socalled Dirichlet boundary condition such as \(X^\mu=0\) at \(\sigma=0\), named after a mathematician who was born 30 years before Carl Neumann. However, \(X^\mu=0\) and the endpoint of a string means that the plane \(X^\mu=0\) is special. Such laws for open strings break the translational, rotational, and Lorentz symmetries in the spacetime which is why people wouldn't allow such boundary conditions for open strings in the 1970s and 1980s.
On the other hand, open strings with Neumann boundary conditions were studied carefully for decades – it's really these open strings that were the first ones to appear in string theory (before closed strings). While closed strings have leftmoving and rightmoving waves, \(X^\mu(\sigma)\) at open strings may be decomposed into "standing waves" Fourier modes going like \(\cos n\sigma\). Note that the positive and negative values of \(n\) give you the same cosinelike function. However, you still get oscillators \(\alpha_{n}^\mu\) for both positive and negative values of \(n\) (which are independent) because there's another doubling of the degrees of freedom, the existence of the momenta \(P^\mu(\sigma,\tau)\).
As far as the mass of the particle that the string behaves as is concerned, the open string oscillators \(\alpha^\mu_n\) behave almost identically as their namesakes from the closed string case. However, for open strings, there is no \(\tilde\alpha_n^\nu\); we only have one set of oscillators. In this sense, the Hilbert space of a closed string is "close" to the tensor square of the open string Hilbert space; in other words, the open string is a "square root" of a closed string.
Note that in the case of the closed string, we needed at least one leftmoving excitation \(\alpha\) and one rightmoving excitation \(\tilde\alpha\) to get away from the tachyonic ground state \(\ket 0\). That's why the field associated with these particles – excited strings – had two Lorentz indices. That's why gravity inevitably appeared at the massless level of the closed strings. Here, for open strings, we only need one oscillator to get from the open string tachyon ground state \(\ket 0\) – the main hero of an article about Sen's tachyon condensation – to the massless level. The first excited state is\[
A_\mu \alpha_{1}^\mu \ket 0
\] which is multiplied by a field with one Lorentz index. I called it \(A^\mu\) for a good reason: when you derive what laws are imposed on this field by the maths of string theory, you will find out that it really does behave exactly as a gauge field, with a gauge symmetry and the usual terms in the effective Lagrangian.
Just like the massless level of a closed string produces gravity (the metric tensor) and the Bfield (a twoindex potential under which the strings themselves are "charged"), aside from the (scalar) dilaton I neglected above (although it wouldn't have been too hard to add an \(\exp(\phi)\)like factor in front of the \((g+B)\)), the open string's massless level produces gauge fields. If all the boundary conditions are Neumann, these open strings – and therefore the gauge field – is defined everywhere in the spacetime. Nevertheless, it's better to call their locus "a spacetimefilling Dbrane" rather than the "spacetime" for reasons that could become clearer later.
Tduality
Let's return to closed strings for a little while. We have discussed the \(n\)th Fourier modes (over the circle that is the closed string) for nonzero values of \(n\). However, we haven't spent much time with the \(n=0\) case, the "zero mode". Well, you may integrate \(X^\mu\) over \(\sigma\) (and divide the result by \(\pi\), the length of the \(\sigma\)interval) to get the "center of mass" position of the string if you wish; and you may also integrate the momentum \(P^\mu\) over \(\sigma\) to find out the total momentum carried by the string.
A funny thing is that you may also integrate the \(\sigma\)derivative of the \(X^\mu\) field\[
\Delta X^\mu = \int_0^\pi \dd\sigma \pfrac{}{\sigma} X^\mu.
\] I denoted the result as \(\Delta X^\mu\) because it's nothing else than the "jump" of the coordinate induced by the trip around the closed string. If the strings are closed and living in an infinite space, it must be zero. However, if you make the coordinate \(X^\mu\) for a particular value of \(\mu\) periodic with a period of \(2\pi R\), so that the dimension may be imagined as a circle of radius \(R\), you must also allow closed strings that obey\[
X^\mu(\sigma=\pi,0) = X^\mu(\sigma=0,\tau) + 2\pi R w, \quad w\in\ZZ.
\] I have already mentioned that these \(w\) times wound strings may be viewed as generalized "twisted sectors". At any rate, the "endpoint" of the closed string may be located at a point shifted relatively to the "beginning point" of the closed string by \(2\pi R w\) because such a shifted point corresponds to the same location of the spacetime: we made the spacetime periodic in \(X^\mu\).
These wound strings are completely new features of string theory that don't exist in pointlike particle quantum field theories: points can't be "wound" around a circle.
The winding number \(w\) contributes to the mass of the string. The formula for the squared mass is actually:\[
m^2 = \frac{4(N1)}{\alpha'} + \frac{n^2}{R^2} + w^2 (2\pi R T)^2.
\] The first term quantifies the total excitation of the nonzero modes along the string; there is \(N1\) instead of \(N\) because the ground state is a tachyon. The second term adds the usual energy from the quantized momentum; note that even in pointlike particle field theories, the momentum of a particle along a circle of radius \(R\) has to be of the form \(n/R\); the last term is the winding and the coefficient depends on the string tension \(T=1/2\pi \alpha'\).
For pointlike particles, there would only be one "squared" term, the momentum term going like \(n^2\). For closed strings, we have two of them. You immediately see that the momentum and winding contributions are very analogous to each other. You may exchange \(w\) with \(n\) and if you also adjust the coefficients appropriately, which may be done by \(R\leftrightarrow \alpha' / R\), the second and third term will get exactly interchanged! The spectrum is symmetric under the operation; the interactions are actually also symmetric. The symmetry is known as Tduality.
The Tduality must also act on the detailed fields \(X^\mu(\sigma,\tau)\); we only saw its approximate action on the Fourier modes. If you think about it, the action of the Tduality is essentially to interchange \(\partial_\sigma X^\mu\) with \(\partial_\tau X^\mu\) (only for a particular value of \(\mu\): we are Tdualizing one spacetime dimension only which must be circular!). That's because the momentum is the integral of \(\partial_\tau X^\mu\) while the winding is the integral of \(\partial_\sigma X^\mu\) and we wanted to interchange these two integers.
You may also rephrase the previous sentences by saying that Tduality keeps \((\partial_\sigma + \partial_\tau) X^\mu\) constant but the difference of these two derivatives changes the sign under Tduality,\[
(\partial_\sigma  \partial_\tau) X^\mu \to  (\partial_\sigma  \partial_\tau) X^\mu.
\] So Tduality acts like the reflection \(X^\mu\to X^\mu\) except that the reflection only applies to the "right moving" degrees of freedom of the string, not the leftmoving ones. (Or vice versa: these two operations of Tduality only differ by the full reflection of the spacetime coordinate which is a simple symmetry of the bosonic string theory.)
Tduality for open strings
There's a lot of fun things to say about Tduality and closed strings but let's switch to open strings. What happens with an open string under Tduality? We said that the Tduality interchanged the \(\sigma\)derivative and the \(\tau\)derivative of \(X^\mu\). However, this affects a basic equation for open strings we have already mentioned, the boundary conditions, too!
We said that the "normal" open strings studied in the 1970s and 1980s had the Neumann boundary conditions\[
\pfrac{}{\sigma}X^\mu_{\sigma=0} = 0.
\] Now, because the two derivatives were interchanged, the new boundary condition (only for one value of \(\mu\), the circular direction we are Tdualizing) should be\[
\pfrac{}{\tau}X^\mu_{\sigma=0} = 0.
\] The derivative along the boundary vanishes. This really means that the whole boundary at \(\sigma=0\) must sit at the same value of \(X^\mu\). If you choose the value once, i.e. \(X^\mu=0\), it will be valid for all values of \(\tau\). In other words, we derived that the Tduality operation forces us to replace open strings with Neumann boundary conditions for everything by open strings with Neumann boundary conditions for "almost everything" except for \(X^\mu\), the Tdualized direction itself, which must now have the Dirichlet boundary condition:\[
X^\mu_{\sigma=0} = 0.
\] This makes a difference! The endpoints of such an open string are stuck at \(X^\mu=0\), a codimension one surface in the spacetime. Again, because the string can't stretch too far without dramatically increasing the mass of the resulting particle, you may say that all the lowenergy strings live at/near \(X^\mu=0\). It's the D24brane. How did I get D24? Well, the D25brane is the brane filling the whole 25+1dimensional spacetime; the number 25, by conventions, only counts the number of spatial coordinates. And I subtracted 1 from 25 because the Dbrane got localized in one direction.
(Open strings have two endpoints and both of them may have boundary conditions independent from one another; one endpoint may be stuck to one Dbrane and the other endpoint may terminate at the same or a different Dbrane. I won't go into these matters here.)
With the Dirichlet boundary conditions, the right functions to Fourierexpand into will be waves such as \(\sin n\sigma\); instead of cosines, we simply have the sines. It's almost the same thing (they're still dependent on the functions with the opposite value of \(n\), the sign doesn't influence much) except that we get zero for \(n=0\) while the cosines did give us a zero mode, too. This is a reflection of the fact that we no longer have the freedom to change the average \(X^\mu\) without changing the energy: the string is stuck at \(X^\mu=0\).
However, it's still possible to excite the open string tachyon ground state \(\ket 0\) by the oscillators \(\alpha_{1}^\kappa\) both for \(\kappa\neq \mu\) which behave just like before and \(\kappa=\mu\) which are new because of the Dirichlet boundary conditions for \(X^\mu\). We get the same number of polarizations – massless bosonic fields in the spacetime – and most of them are components of a gauge field again.
Nevertheless, one subtlety is new: the field creating the particle identical to the string in the \(\alpha_{1}^\mu\ket 0\) state, for the particular value of \(\mu\) that we Tdualized, will no longer be a component of the gauge field \(A_\mu\). Instead, it will be a scalar field \(X^\mu\) that is physically interpreted as the transverse location of the brane in the direction \(\mu\).
It's kind of cool. For the fullNeumann boundary conditions, we had a spacetimefilling D25brane and a massless gauge field \(A_\mu\) that is defined everywhere in the spacetime (well, on the spacetimefilling Dbrane). Now, the gauge field is only defined on a 24+1dimensional hypersurface, the D24brane. The gauge fields in 24+1 dimensions have one fewer polarization. But it's compensated by the fact that there's one scalar field that remembers the transverse location \(X^\mu\) of a point on the D24brane as a function of the remaining 24+1 dimensions! You can actually make the D24brane wiggle inside the spacetime if you "pour" a coherent state of open strings in the \(\alpha_{1}^\mu\ket 0\) state on the Dbrane.
Indeed, if you derive the effective action for all these open string modes from string theory, and it's a general fact that you may derive *everything* from string theory, you will find out that there are the usual YangMills terms for the gauge field component; and there are new terms that give the D24brane its tension. The D24brane may wiggle, stretch, and try to return to a placid state just like the strings themselves. A difference is that for the fundamental strings, we inserted the equations doing these wiggly things as defining equations of the theory; for the Dbranes, their wiggling (much like the dynamics of the gauge fields inside them) is derived from the open strings that can be attached to them!
At the beginning, I mentioned the closed string state that is responsible for the spacetime metric; "pouring" closed strings in this state into the spacetime is physically *identical* to deforming the shape of the spacetime. The closed strings in this graviton states are not just "similar" to the variations of the spacetime shape and curvature; the variations of the spacetime shape and curvature *are* the closed strings in this vibration mode. There's no other way to curve the spacetime differently than to pour closed strings. All the dynamics behaves in agreement with all the principles of general relativity.
Equivalently, the D24brane doesn't have "completely new, nonstringy" degrees of freedom that would define its shape within the spacetime or the values of the gauge fields that only exist inside this D24brane. Instead, all these degrees of freedom are among the "standard" stringy degrees of freedom that always existed, assuming that you also allow strings with the boundary conditions corresponding to a given Dbrane. Again, the open strings are not just "analogous" to wiggles on the D24brane; the wiggles on the Dbrane are physically *identical* to some particular vibrating open strings. They're the same thing. Aside from the open strings, there are no other wiggles of the Dbranes.
In perturbative string theory, everything – including all the spacetime fields and shapes and particles as well as Dbranes' properties such as their shape and electromagnetic fields – is made out of strings. When you go to strong coupling, the compositeness becomes "less useful" for accurate calculations – you need to calculate lots of corrections etc. – and the fundamental strings become less fundamental. At a generic coupling, you can't unambiguously say which objects are fundamental and which objects are made out of them. There is actually often a symmetry, Sduality, that exchanges them.
However, it's still important to learn how string theory works at a very weak coupling and indeed, everything is made out of fundamental strings and it makes a perfect sense.
Needless to say, there are Dbranes of all dimensionalities between 25 and –1. Yes, it's minus one. You may Tdualize several circular spacetime dimensions to get some intuition for what happens and how the degrees of freedom are related to the original spacetimefilling Dbranes. The generalization is straightforward.
All these things may also be extended to the truly physically relevant theory, the 9+1dimensional superstring. In that theory, we also need to deal with the world sheet fermionic fields and their boundary conditions and we find out that the evendimensional and odddimensional branes are a bit different. Only one of the two groups may be supersymmetric and the others can't. (In type IIA and type IIB, the role of "even" and "odd" get interchanged.)
Moreover, superstring theory adds one more cool thing that isn't present in bosonic string theory: the supersymmetric Dbranes (Devenbranes in type IIA and Doddbranes in type IIB string theory) are charged under new antisymmetric tensor fields, the RamondRamond \(p\)form fields that arise at the massless level of the closed superstring (in the sector whose ground state looks like a spacetime spinor \(\otimes\) spacetime spinor). Prior to the discovery of Dbranes, it looked like those fields were somewhat "useless" because nothing was charged under this particular generalization of electromagnetism. As Joe Polchinski showed, Dbranes are charged under them: everything makes some sense in string theory, everything plays some role, and all these roles fit together. String theory never offers you an incomplete story, a Hollywood movie that wasn't finished because they found out there was no sensible way to explain what they already shot. String theory is a fully consistent theory that makes everything meaningful.
Everyone who wants to feel some familiarity with basic string theory should spend some time by thinking about the different boundary conditions and what they mean because the differences between things that were thought to be qualitatively different – like branes of different dimensions, with or without scalars, and so on – boil down to a very modest "technical difference" at the level of string theory. The diversity of physical phenomena one may derive from string theory by looking at objects with slightly different boundary conditions etc. is of course a part of the ability of string theory to unify everything.
And that's the memo.
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2 hours ago
snail feedback (14) :
Aah, just this morning I thought I'd like to see a new serious but pedagogical TRF article (like this one) not only because I like them but because one could probably interpret them as a sign that Lumo feels a it better too :) ?
I always imagine that not only do I have to be concentrated and attentive while reading, if I want to get it, but Lumo has to feel good enough to be able write and put up such nice serious articles.
So I'm double happy about this one ;) :)
(Will have to read it later, since I'm at work now...)
Cheers
Darn, I've forgotten (not sure if I've already seen it somewhere) why m2=4N/α′ for particles, corresponding to excitations of closed strings :/ ?
Strings may only have zero or two endpoints? Why do you exclude strings with an odd number of endpoints? For example, a circle with a short string dangling from it. It requires a 3pronged vertex, but is there a deep reason to forbid them?
Lubos, thanks for this very nice high level summary of bosonic string theory and Dbranes. It is great to have this all crisply summarized in one place  I will use it as a user's guide to big chunks of Zweibach's String Theory for Undergraduates.
Great question. The local novelty is of course the vertex. I've been thinking about whether it may be consistent or not, too. Try to define the boundary conditions for X(sigma,tau) at the vertex. It surely isn't straightforward. You must cancel the boundary terms etc.
Then there is a worry that these vertices are indestructible, and nothing should be indestructible in a complete theory. Try to design interactions that could pairannihilate such vertices. You will probably fail, too.
At any rate, world sheets with such vertices don't look like manifolds; they're not locally sets of R^2, neither open nor closed sets, and this would probably mess the nice UV properties of string theory.
Also, there wouldn't be a stateoperator correspondence for these Mercedesshaped strings.
Dear Dilaton, the string is an infinitedimensional harmonic oscillator. Each Fourier mode  the coefficient of exp(2n.i.sigma) in the Fourier expansion of X(sigma)  behaves as a harmonic oscillator. Well, you have 24 times more those oscillators.
Each oscillator has some frequency and the excitation level of this oscillator is integer, just like for any other harmonic oscillator in quantum mechanics.
At the end, the total excitation  counting the nth oscillator as "n" units  is linked to the squared mass of the resulting particle because the "total world sheet energy" is
L0 = p_mu p^mu  N
and has to vanish, in the right units. The p^2 term comes from the zero modes and N comes from the nonzero modes, the oscillators. It has to vanish because L0 is a generator of coordinate transformations and all of those must annihilate physical states because the world sheet theory is a theory of gravity, too. From this condition, you see that p^2 = N. Add the coefficient and the shift...
Wow Lumo,
thanks for these additional helpful explanations :); did you see the "first version" of my comment (thought you were asleep then :P)... ?
Lumo seeing everything like this would be awesome, LOL :D
Cool article. Only today have I found the time to print it out and read it. A lot of deep insight summarized in one neat text. And I can even understand half of it (at least on some elementary level). I still do not understand where the CalabiYau manifolds fit in in this string theoretical description of universe. Do the strings and branes live inside the CYmanifolds, the open ends of strings begin and end on the boundaries created by CYmanifolds? How did the string theoreticians even come to CYmanifolds and found out that they are needed for the theory?
"there is N−1 instead of N because the ground state is a tachyon"
This is a helpful note, since it immediately ties into an interesting property of conservative systems (which I will define hear as a system of variables that sum to zero). This ties closely to the degrees of freedom for residuals in linear regression (http://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29#Linear_regression)
The N1 property is extremely important and tying that property to tachyons provides insight into what tachyons are. The N1 relationship allows for immediate knowledge of one degree of freedom based on knowledge of the other degrees of freedom (in the statistical context, one could also understand this as dependent and independent variables).
Dear Lubos,
The
difference between the closed and open ring is very interesting to speculate,
because the closed ring seems to have less possibilities to vibrate and has a
special position like the Higgs.
If the closed ring represents ONLY the Higgs particle
and the open ring is able to vibrate and rotate internally over only three
hinges to represent all other singular (noncompound) particles, like Photons,
Gluons, Neutrinos, the Electron and Positron...
See:
3Dimensional String based alternative particle model.
http://vixra.org/pdf/1103.0002v4.pdf
Hi again hope youre feeling better
Fascinating stuff but difficult to get my head round  some time ago I was at a lecture by David Berman from Queen Mary College  he explaind how string oscillations reflect in the mass of the particle  i.e. a particle with no mass like a photon will be a stationary string  I asked how the photons frequency is then encoded in such a motionless string  but completely lost him, maybe you could help with a simplified answer?
cheers
zbynek
Dear Zbyněk, I feel far from healthy but thanks for your hopes.
There are two problems with the apparent paradox you're proposing.
First, no string is quite motionless  the uncertainty principle prohibits motionless strings. A quantum string always carries some potential energy and some kinetic energy. A string is filled with many copies of the zeropoint energy hf/2 that you may know from the harmonic oscillator, and perhaps some multiples of these energies if the string is excited. All these frequencies "f" are comparable to the string mass or the "string frequency".
Second, the frequency of a photon (in the usual sense) has nothing whatsoever to do with the internal oscillations of a string. The frequency/wavelength of a photon is encoded in the wave such as exp(i.p.xi.omega.t) but this wave depends on x,t which are centerofmass coordinates of the string so the wave function doesn't describe a changing shape of the string at all: it describes a changing probability amplitude that the string is located at one point of space (or time) or another. The more quickly this phase is changing, the higher the frequency of the photon is. But the variability of this frequency in no way depends on the internal oscillations of a string  it's only about the overall motion of the whole object, the string, which is also why even pointlike photons can have any frequency.
What the internal oscillations of a string explain is the rest mass of the resulting particle. Just like the hydrogen atom may have one energy or an energy higher by 3/4 times 13.6 eV, strings also have many levels that correspond to particles of different rest masses.
Many thanks for the link to whystringtheory.com! It covers many elementary questions that I've been too embarrassed to ask :)
Thanks more understandable, especially since I can go through it several times (slowly)
sorry to hear youre still not right  wish you all the best
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