text:
# Energy–momentum relation

In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating any object's rest (intrinsic) mass, total energy, and momentum:

${\displaystyle E^{2}=(pc)^{2}+\left(m_{0}c^{2}\right)^{2}\,}$

( 1 )

holds for a system, such as a particle or macroscopic body, having intrinsic rest mass m0, total energy E, and a momentum of magnitude p, where the constant c is the speed of light, assuming the special relativity case of flat spacetime.[1][2][3]

The Dirac sea model, which was used to predict the existence of antimatter, is closely related to the energy-momentum equation.

## Connection to E = mc2

The energy–momentum relation is consistent with the familiar mass–energy relation in both its interpretations: E = mc2 relates total energy E to the (total) relativistic mass m (alternatively denoted mrel or mtot ), while E0 = m0c2 relates rest energy E0 to (invariant) rest mass m0.

Unlike either of those equations, the energy–momentum equation (1) relates the total energy to the rest mass m0. All three equations hold true simultaneously.

## Special cases

1. If the body is a massless particle (m0 = 0), then (1) reduces to E = pc. For photons, this is the relation, discovered in 19th century classical electromagnetism, between radiant momentum (causing radiation pressure) and radiant energy.
2. If the body's speed v is much less than c, then (1) reduces to E = 1/2m0v2 + m0c2; that is, the body's total energy is simply its classical kinetic energy (1/2m0v2) plus its rest energy.
3. If the body is at rest (v = 0), i.e. in its center-of-momentum frame (p = 0), we have E = E0 and m = m0; thus the energy–momentum relation and both forms of the mass–energy relation (mentioned above) all become the same.

A more general form of relation (1) holds for general relativity.

The invariant mass (or rest mass) is an invariant for all frames of reference (hence the name), not just in inertial frames in flat spacetime, but also accelerated frames traveling through curved spacetime (see below). However the total energy of the particle E and its relativistic momentum p are frame-dependent; relative motion between two frames causes the observers in those frames to measure different values of the particle's energy and momentum; one frame measures E and p, while the other frame measures E and p, where EE and p′ ≠ p, unless there is no relative motion between observers, in which case each observer measures the same energy and momenta. Although we still have, in flat spacetime:

${\displaystyle {E'}^{2}-\left(p'c\right)^{2}=\left(m_{0}c^{2}\right)^{2}\,.}$

The quantities E, p, E, p are all related by a Lorentz transformation. The relation allows one to sidestep Lorentz transformations when determining only the magnitudes of the energy and momenta by equating the relations in the different frames. Again in flat spacetime, this translates to;

${\displaystyle {E}^{2}-\left(pc\right)^{2}={E'}^{2}-\left(p'c\right)^{2}=\left(m_{0}c^{2}\right)^{2}\,.}$

Since m0 does not change from frame to frame, the energy–momentum relation is used in relativistic mechanics and particle physics calculations, as energy and momentum are given in a particle's rest frame (that is, E and p as an observer moving with the particle would conclude to be) and measured in the lab frame (i.e. E and p as determined by particle physicists in a lab, and not moving with the particles).

In relativistic quantum mechanics, it is the basis for constructing relativistic wave equations, since if the relativistic wave equation describing the particle is consistent with this equation – it is consistent with relativistic mechanics, and is Lorentz invariant. In relativistic quantum field theory, it is applicable to all particles and fields.[4]

## Origins and derivation of the equation

The Energy–momentum relation was first established by <NAME> in 1928 under the form ${\displaystyle E={\sqrt {c^{2}p^{2}+(m_{0}c^{2})^{2}}}+V}$ , where V is the amount of potential energy. [5]

The equation can be derived in a number of ways, two of the simplest include:

1. From the relativistic dynamics of a massive particle,
2. By evaluating the norm of the four-momentum of the system. This method applies to both massive and masseless particles, and can be extended to multi-particle systems with relatively little effort (see § Many-particle systems below).

### Heuristic approach for massive particles

For a massive object moving at three-velocity u = (ux, uy, uz) with magnitude |u| = u in the lab frame:[1]

${\displaystyle E=\gamma _{(\mathbf {u} )}m_{0}c^{2}}$

is the total energy of the moving object in the lab frame,

${\displaystyle \mathbf {p} =\gamma _{(\mathbf {u} )}m_{0}\mathbf {u} }$

is the three dimensional relativistic momentum of the object in the lab frame with magnitude |p| = p. The relativistic energy E and momentum p include the Lorentz factor defined by:

${\displaystyle \gamma _{(\mathbf {u} )}={\frac {1}{\sqrt {1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}}={\frac {1}{\sqrt {1-\left({\frac {u}{c}}\right)^{2}}}}}$

Some authors use relativistic mass defined by:

${\displaystyle m=\gamma _{(\mathbf {u} )}m_{0}}$

although rest mass m0 has a more fundamental significance, and will be used primarily over relativistic mass m in this article.

Squaring the 3-momentum gives:

${\displaystyle p^{2}=\mathbf {p} \cdot \mathbf {p} ={\frac {m_{0}^{2}\mathbf {u} \cdot \mathbf {u} }{1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}={\frac {m_{0}^{2}u^{2}}{1-\left({\frac {u}{c}}\right)^{2}}}}$

then solving for u2 and substituting into the Lorentz factor obtains its alternative form in terms of 3-momentum and mass, rather than 3-velocity:

${\displaystyle \gamma ={\sqrt {1+\left({\frac {p}{m_{0}c}}\right)^{2}}}}$

Inserting this form of the Lorentz factor into the energy equation:

${\displaystyle E=m_{0}c^{2}{\sqrt {1+\left({\frac {p}{m_{0}c}}\right)^{2}}}}$

followed by more rearrangement yields (1). The elimination of the Lorentz factor also eliminates implicit velocity dependence of the particle in (1), as well as any inferences to the "relativistic mass" of a massive particle. This approach is not general as massless particles are not considered. Naively setting m0 = 0 would mean that E = 0 and p = 0 and no energy–momentum relation could be derived, which is not correct.

### Norm of the four-momentum

The energy and momentum of an object measured in two inertial frames in energy–momentum space – the yellow frame measures E and p while the blue frame measures E′ and p′. The green arrow is the four-momentum P of an object with length proportional to its rest mass m0. The green frame is the centre-of-momentum frame for the object with energy equal to the rest energy. The hyperbolae show the Lorentz transformation from one frame to another is a hyperbolic rotation, and ϕ and ϕ + η are the rapidities of the blue and green frames, respectively.

#### Special relativity

In Minkowski space, energy (divided by c) and momentum are two components of a Minkowski four-vector, namely the four-momentum;[6]

${\displaystyle \mathbf {P} =\left({\frac {E}{c}},\mathbf {p} \right)\,,}$

(these are the contravariant components).

The Minkowski inner product ⟨ , ⟩ of this vector with itself gives the square of the norm of this vector, it is proportional to the square of the rest mass m of the body:

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=\left(m_{0}c\right)^{2}\,,}$

a Lorentz invariant quantity, and therefore independent of the frame of reference. Using the Minkowski metric η with metric signature (− + + +), the inner product is

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=-\left(m_{0}c\right)^{2}\,,}$

and

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =P^{\alpha }\eta _{\alpha \beta }P^{\beta }={\begin{pmatrix}{\frac {E}{c}}&p_{x}&p_{y}&p_{z}\end{pmatrix}}{\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}{\begin{pmatrix}{\frac {E}{c}}\\p_{x}\\p_{y}\\p_{z}\end{pmatrix}}=-\left({\frac {E}{c}}\right)^{2}+p^{2}\,,}$

so

${\displaystyle -\left(m_{0}c\right)^{2}=-\left({\frac {E}{c}}\right)^{2}+p^{2}.}$

#### General relativity

In general relativity, the 4-momentum is a four-vector defined in a local coordinate frame, although by definition the inner product is similar to that of special relativity,

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=\left(m_{0}c\right)^{2}\,,}$

in which the Minkowski metric η is replaced by the metric tensor field g:

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=P^{\alpha }g_{\alpha \beta }P^{\beta }\,,}$

solved from the Einstein field equations. Then:[7]

${\displaystyle P^{\alpha }g_{\alpha \beta }P^{\beta }=\left(m_{0}c\right)^{2}\,.}$

Performing the summations over indices followed by collecting "time-like", "spacetime-like", and "space-like" terms gives:

${\displaystyle \underbrace {g_{00}{\left(P^{0}\right)}^{2}} _{\text{time-like}}+2\underbrace {g_{0i}P^{0}P^{i}} _{\text{spacetime-like}}+\underbrace {g_{ij}P^{i}P^{j}} _{\text{space-like}}=\left(m_{0}c\right)^{2}\,.}$

where the factor of 2 arises because the metric is a symmetric tensor, and the convention of Latin indices i, j taking space-like values 1, 2, 3 is used. As each component of the metric has space and time dependence in general; this is significantly more complicated than the formula quoted at the beginning, see metric tensor (general relativity) for more information.

## Units of energy, mass and momentum

In natural units where c = 1, the energy–momentum equation reduces to

${\displaystyle E^{2}=p^{2}+m_{0}^{2}\,.}$

In particle physics, energy is typically given in units of electron volts (eV), momentum in units of eV·c−1, and mass in units of eV·c−2. In electromagnetism, and because of relativistic invariance, it is useful to have the electric field E and the magnetic field B in the same unit (Gauss), using the cgs (Gaussian) system of units, where energy is given in units of erg, mass in grams (g), and momentum in g·cm·s−1.

Energy may also in theory be expressed in units of grams, though in practice it requires a large amount of energy to be equivalent to masses in this range. For example, the first atomic bomb liberated about 1 gram of heat, and the largest thermonuclear bombs have generated a kilogram or more of heat. Energies of thermonuclear bombs are usually given in tens of kilotons and megatons referring to the energy liberated by exploding that amount of trinitrotoluene (TNT).

## Special cases

### Centre-of-momentum frame (one particle)

For a body in its rest frame, the momentum is zero, so the equation simplifies to

${\displaystyle E_{0}=m_{0}c^{2}\,,}$

where m0 is the rest mass of the body.

### Massless particles

If the object is massless, as is the case for a photon, then the equation reduces to

${\displaystyle E=pc\,.}$

This is a useful simplification. It can be rewritten in other ways using the de Broglie relations:

${\displaystyle E={\frac {hc}{\lambda }}=\hbar ck\,.}$

if the wavelength λ or wavenumber k are given.

### Correspondence principle

Rewriting the relation for massive particles as:

${\displaystyle E=m_{0}c^{2}{\sqrt {1+\left({\frac {p}{m_{0}c}}\right)^{2}}}\,,}$

and expanding into power series by the binomial theorem (or a Taylor series):

${\displaystyle E=m_{0}c^{2}\left[1+{\frac {1}{2}}\left({\frac {p}{m_{0}c}}\right)^{2}-{\frac {1}{8}}\left({\frac {p}{m_{0}c}}\right)^{4}+\cdots \right]\,,}$

in the limit that uc, we have γ(u) ≈ 1 so the momentum has the classical form pm0u, then to first order in (p/m0c)2
(i.e. retain the term (p/m0c)2n
for n = 1 and neglect all terms for n ≥ 2) we have

${\displaystyle E\approx m_{0}c^{2}\left[1+{\frac {1}{2}}\left({\frac {m_{0}u}{m_{0}c}}\right)^{2}\right]\,,}$

or

${\displaystyle E\approx m_{0}c^{2}+{\frac {1}{2}}m_{0}u^{2}\,,}$

where the second term is the classical kinetic energy, and the first is the rest mass of the particle. This approximation is not valid for massless particles, since the expansion required the division of momentum by mass. Incidentally, there are no massless particles in classical mechanics.

## Many-particle systems

In the case of many particles with relativistic momenta pn and energy En, where n = 1, 2, ... (up to the total number of particles) simply labels the particles, as measured in a particular frame, the four-momenta in this frame can be added;

${\displaystyle \sum _{n}\mathbf {P} _{n}=\sum _{n}\left({\frac {E_{n}}{c}},\mathbf {p} _{n}\right)=\left(\sum _{n}{\frac {E_{n}}{c}},\sum _{n}\mathbf {p} _{n}\right)\,,}$

and then take the norm; to obtain the relation for a many particle system:

${\displaystyle \left|\left(\sum _{n}\mathbf {P} _{n}\right)\right|^{2}=\left(\sum _{n}{\frac {E_{n}}{c}}\right)^{2}-\left(\sum _{n}\mathbf {p} _{n}\right)^{2}=\left(M_{0}c\right)^{2}\,,}$

where M0 is the invariant mass of the whole system, and is not equal to the sum of the rest masses of the particles unless all particles are at rest (see mass in special relativity for more detail). Substituting and rearranging gives the generalization of (1);

${\displaystyle \left(\sum _{n}E_{n}\right)^{2}=\left(\sum _{n}\mathbf {p} _{n}c\right)^{2}+\left(M_{0}c^{2}\right)^{2}}$

( 2 )

The energies and momenta in the equation are all frame-dependent, while M0 is frame-independent.

### Center-of-momentum frame

In the center-of-momentum frame (COM frame), by definition we have:

${\displaystyle \sum _{n}\mathbf {p} _{n}={\boldsymbol {0}}\,,}$

with the implication from (2) that the invariant mass is also the centre of momentum (COM) mass–energy, aside from the c2 factor:

${\displaystyle \left(\sum _{n}E_{n}\right)^{2}=\left(M_{0}c^{2}\right)^{2}\Rightarrow \sum _{n}E_{\mathrm {COM} \,n}=E_{\mathrm {COM} }=M_{0}c^{2}\,,}$

and this is true for all frames since M0 is frame-independent. The energies ECOM n are those in the COM frame, not the lab frame.

### Rest masses and the invariant mass

Either the energies or momenta of the particles, as measured in some frame, can be eliminated using the energy momentum relation for each particle:

${\displaystyle E_{n}^{2}-\left(\mathbf {p} _{n}c\right)^{2}=\left(m_{n}c^{2}\right)^{2}\,,}$

allowing M0 to be expressed in terms of the energies and rest masses, or momenta and rest masses. In a particular frame, the squares of sums can be rewritten as sums of squares (and products):

${\displaystyle \left(\sum _{n}E_{n}\right)^{2}=\left(\sum _{n}E_{n}\right)\left(\sum _{k}E_{k}\right)=\sum _{n,k}E_{n}E_{k}=2\sum _{n
${\displaystyle \left(\sum _{n}\mathbf {p} _{n}\right)^{2}=\left(\sum _{n}\mathbf {p} _{n}\right)\cdot \left(\sum _{k}\mathbf {p} _{k}\right)=\sum _{n,k}\mathbf {p} _{n}\cdot \mathbf {p} _{k}=2\sum _{n

so substituting the sums, we can introduce their rest masses mn in (2):

${\displaystyle \sum _{n}\left(m_{n}c^{2}\right)^{2}+2\sum _{n

The energies can be eliminated by:

${\displaystyle E_{n}={\sqrt {\left(\mathbf {p} _{n}c\right)^{2}+\left(m_{n}c^{2}\right)^{2}}}\,,\quad E_{k}={\sqrt {\left(\mathbf {p} _{k}c\right)^{2}+\left(m_{k}c^{2}\right)^{2}}}\,,}$

similarly the momenta can be eliminated by:

${\displaystyle \mathbf {p} _{n}\cdot \mathbf {p} _{k}=\left|\mathbf {p} _{n}\right|\left|\mathbf {p} _{k}\right|\cos \theta _{nk}\,,\quad |\mathbf {p} _{n}|={\frac {1}{c}}{\sqrt {E_{n}^{2}-\left(m_{n}c^{2}\right)^{2}}}\,,\quad |\mathbf {p} _{k}|={\frac {1}{c}}{\sqrt {E_{k}^{2}-\left(m_{k}c^{2}\right)^{2}}}\,,}$

where θnk is the angle between the momentum vectors pn and pk.

Rearranging:

${\displaystyle \left(M_{0}c^{2}\right)^{2}-\sum _{n}\left(m_{n}c^{2}\right)^{2}=2\sum _{n

Since the invariant mass of the system and the rest masses of each particle are frame-independent, the right hand side is also an invariant (even though the energies and momenta are all measured in a particular frame).

## Matter waves

Using the de Broglie relations for energy and momentum for matter waves,

${\displaystyle E=\hbar \omega \,,\quad \mathbf {p} =\hbar \mathbf {k} \,,}$

where ω is the angular frequency and k is the wavevector with magnitude |k| = k, equal to the wave number, the energy–momentum relation can be expressed in terms of wave quantities:

${\displaystyle \left(\hbar \omega \right)^{2}=\left(c\hbar k\right)^{2}+\left(m_{0}c^{2}\right)^{2}\,,}$

and tidying up by dividing by (ħc)2 throughout:

${\displaystyle \left({\frac {\omega }{c}}\right)^{2}=k^{2}+\left({\frac {m_{0}c}{\hbar }}\right)^{2}\,.}$

( 3 )

This can also be derived from the magnitude of the four-wavevector

${\displaystyle \mathbf {K} =\left({\frac {\omega }{c}},\mathbf {k} \right)\,,}$

in a similar way to the four-momentum above.

Since the reduced Planck constant ħ and the speed of light c both appear and clutter this equation, this is where natural units are especially helpful. Normalizing them so that ħ = c = 1, we have:

${\displaystyle \omega ^{2}=k^{2}+m_{0}^{2}\,.}$

## Tachyon and exotic matter

The velocity of a bradyon with the relativistic energy–momentum relation

${\displaystyle E^{2}=p^{2}c^{2}+m_{0}^{2}c^{4}\,.}$

can never exceed c. On the contrary, it is always greater than c for a tachyon whose energy–momentum equation is[8]

${\displaystyle E^{2}=p^{2}c^{2}-m_{0}^{2}c^{4}\,.}$

By contrast, the hypothetical exotic matter has a negative mass[9] and the energy–momentum equation is

${\displaystyle E^{2}=-p^{2}c^{2}+m_{0}^{2}c^{4}\,.}$

## References

1. ^ <NAME>, Daniel; <NAME> (2010) [1973]. An Introduction to Mechanics. Cambridge University Press. pp. 499–500. ISBN 978-0-521-19821-9.
2. ^ <NAME>; <NAME> (2009). Dynamics and Relativity. Wiley. pp. 149, 249. ISBN 978-0-470-01460-8.
3. ^ <NAME> (2006). Relativity. DeMystified. Mc Graw Hill (USA). p. 20. ISBN 0-07-145545-0.
4. ^ <NAME> (2008). Quantum Field Theory. DeMystified. Mc Graw Hill (USA). pp. 11, 88. ISBN 978-0-07-154382-8.
5. ^ <NAME>., <NAME>. (1985) Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. 2nd Edition, John Wiley & Sons. New York. p.132.ISBN 0-471-87373-X
6. ^ <NAME>; <NAME> (2009). Dynamics and Relativity. Wiley. pp. 258–259. ISBN 978-0-470-01460-8.
7. ^ <NAME>; <NAME>; <NAME> (1973). Gravitation. W.H. Freeman & Co. pp. 201, 649, 1188. ISBN 0-7167-0344-0.
8. ^ <NAME> (1967). "Possibility of faster-than-light particles". Physical Review. 159: 1089–1105. Bibcode:1967PhRv..159.1089F. doi:10.1103/PhysRev.159.1089.
9. ^ Z.Y.Wang (2016). "Modern Theory for Electromagnetic Metamaterials". Plasmonics. 11 (2): 503–508. doi:10.1007/s11468-015-0071-7.
• <NAME> (1988). 3000 Solved Problems in Physics, Schaum Series. McGraw-Hill. pp. 704–705. ISBN 978-0-07-025734-4.
• <NAME> (2010). The Cambridge Handbook of Physics Formulas. Cambridge University Press. p. 65. ISBN 978-0-521-57507-2.
• <NAME> (1994). McGraw-Hill Encyclopaedia of Physics (2nd ed.). McGraw-Hill. pp. 1192, 1193. ISBN 0-07-051400-3.
• <NAME>; <NAME> (1991). Encyclopaedia of Physics (2nd ed.). VHC Publishers. p. 1052. ISBN 0-89573-752-3.


text:
## Graded consequence relations and fuzzy closure operator.(English)Zbl 0872.03012

The role of a closure operator in the framework of formal (binary) logics has been recognized very early by Tarski: similarly, many papers have proved the usefulness of a fuzzy closure operator in Zadeh’s logic of approximate reasoning. It is well known that there exists a bijection between the class of consequence relations and the class of closure operators on a set of formulas. In 1988, <NAME> [Fuzzy logic in knowledge-based systems, decision and control, 247-257 (1988; Zbl 0666.03022)] extended the concept of a consequence relation to the theory of fuzzy sets by introducing a graded consequence relation.
This paper concentrates on the connections between fuzzy closure operators and graded consequence relations. A slightly modified version of the above-mentioned bijection could be established: there exists a bijection between the class of graded consequence relations and a subclass of the class of fuzzy closure operators namely those that can be represented as a chain of (crisp) closure operators.
Reviewer: E.Kerre (Gent)

### MSC:

 03B52 Fuzzy logic; logic of vagueness 68T27 Logic in artificial intelligence 03E72 Theory of fuzzy sets, etc. 03B50 Many-valued logic

### Keywords:

fuzzy closure operator; graded consequence relation

Zbl 0666.03022
Full Text:

### References:

 [1] Biacino L., J. Math. Anal. Appl 172 pp 179– (1991) · Zbl 0777.60004 [2] Biacino L., Inform. Sci. 32 pp 181– (1984) · Zbl 0562.06004 [3] Biacino L., International Journal of Intelligent Systems 7 pp 445– (1992) · Zbl 0761.68090 [4] <NAME>. and <NAME>. 1993.Closure operators for fuzzy subsets, EUFTT 93, First European Congress on Fuzzy and Intelligent Technologies, Aachen1441–1447. [5] <NAME>., Fuzzy Sets and Systems. [6] <NAME>. L., Journal of Applied non-classical Logic. [7] <NAME>., Fuzzy Logic in Knowledge-Based Systems, Decision and Control pp 247– (1988) [8] <NAME>., Journal of Applied non-classical Logic. [9] <NAME>., Artificial Intelligence 70 pp 33– (1994) · Zbl 0821.03014 [10] <NAME>., Mathematical Logic Quarterly. 40 pp 357– (1994) · Zbl 0811.03019 [11] <NAME>., Kybernetes 4 pp 169– (1975) · Zbl 0352.02044 [12] <NAME>., Math. Logik Grundlag. Math. 25 pp 45– (1979) · Zbl 0435.03020 [13] <NAME>., Multiple conclusion logic (1978) · Zbl 0381.03001 [14] <NAME>., Synthese 30 pp 407– (1975) · Zbl 0319.02016
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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# Synthetic division calculator

This calculator divides two polynomials using synthetic division. It finds the quotient and the remainder when a polynomial is divided by $x-c$ The calculator shows all the steps and a detailed explanation for each step.

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Divide $3x^3-5x+2$ by $x-4$ using synthetic division.
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Find the remainder when $5x^4-2x^3-4x^2 + 2$ is divided by $x-2$.
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Divide $-x^5-5x^3-x^2+2$ by $3x-1$.
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## Synthetic division

Synthetic division is, by far, the easiest and fastest method to divide a polynomial by $\color{blue}{x - c}$, where $\color{blue}{c}$ is a constant. This method only works when we divide by a linear factor. Let's look at two examples to learn how we can apply this method.

### Example 1 : Divide $x^2 +3x - 2$ by $x - 2$.

Step 1: Write down the coefficients of $2x^2 +3x +4$ into the division table.

$$\begin{array}{c|rrr} \color{blue}{\square} &2&3&4\\ & & & \\ \hline & & & \end{array}$$

Step 2: Change the sign of a number in the divisor and write it on the left side. In this case, the divisor is $x - 2$ so we have to change $\, -2 \,$ to $\, \color{blue}{2}$.

$$\begin{array}{c|rrr} \color{blue}{2} &2&3&4\\ & & & \\ \hline & & & \end{array}$$

Step 3: Carry down the leading coefficient

$$\begin{array}{c|rrr} 2 &\color{orangered}{2}&3&4\\ & & & \\ \hline &\color{orangered}{2}& & \end{array}$$

Step 4: Multiply carry-down by left term and put the result into the next column

$$\begin{array}{c|rrr} \color{blue}{2} &2&3&4\\ & &\color{blue}{4} & \\ \hline &\color{blue}{2}& & \end{array}$$

Step 5: Add the last column

$$\begin{array}{c|rrr} 2 &2&\color{orangered}{3}&4\\ & &\color{orangered}{4}& \\ \hline &2&\color{orangered}{7}& \end{array}$$

Step 6: Multiply previous value by left term and put the result into the next column

$$\begin{array}{c|rrr} \color{blue}{2} &2&3&4\\ & &4&\color{blue}{14} \\ \hline &2&\color{blue}{7}& \end{array}$$

Step 6: Add the last column

$$\begin{array}{c|rrr} \color{blue}{2} &2&3&\color{orangered}{4}\\ & &4&\color{orangered}{14} \\ \hline &2&7& \color{orangered}{18} \end{array}$$

Step 7: Read the result from the synthetic table.

$$\begin{array}{c|rrr} 2&2&3&4\\ & &4&14\\ \hline &\color{blue}{2}&\color{blue}{7}& \color{orangered}{18} \end{array}$$

The quotient is $\color{blue}{2x + 7}$ and the remainder is $\color{orangered}{18}$.

Starting polynomial $x^2 +3x - 2$ can be written as:

$$x^2 +3x - 2 = \color{blue}{2x + 7} + \dfrac{ \color{orangered}{18} }{ x - 2 }$$

### Example 2 : Divide $x^4 + 10x + 1$ by $x + 2$.

Step 1: Write down the coefficients of $x^4 - 10x + 1$ into the division table. (Note that this polynomial doesn't have $x^3$ and $x^2$ terms, so these coefficients must be zero)

$$\begin{array}{c|rrr} \color{blue}{\square} &1&0&0& 10&1\\ & & & & &\\ \hline & & & & & \end{array}$$

Step 2: Change the sign of a number in the divisor and write it on the left side. In this case, the divisor is $x + 3$ so we have to change $\, +3 \,$ to $\, \color{blue}{-3}$.

$$\begin{array}{c|rrr} \color{blue}{-3}&1&0&0&10&1\\ & & & & &\\ \hline & & & & & \end{array}$$

Step 3: Carry down the leading coefficient

$$\begin{array}{c|rrr} \color{blue}{-3}&\color{orangered}{1}&0&0&10&1\\ & & & & &\\ \hline &\color{orangered}{1}& & & & \end{array}$$

Multiply carry-down by left term and put the result into the next column

$$\begin{array}{c|rrr} \color{blue}{-3}&1&0&0&10&1\\ & &\color{blue}{-3}& & &\\ \hline &\color{blue}{1}& & & & \end{array}$$

$$\begin{array}{c|rrr} -3 &1&\color{orangered}{0}&0&10&1\\ & &\color{orangered}{-3}& & &\\ \hline &1&-3 & & & \end{array}$$

Multiply last value by left term and put the result into the next column

$$\begin{array}{c|rrr} \color{blue}{-3} &1&0&0&10&1\\ & &-3&\color{blue}{9}& &\\ \hline &1&\color{blue}{-3} & & & \end{array}$$

$$\begin{array}{c|rrr} -3 &1& 0&\color{orangered}{0}&10&1\\ & &-3&\color{orangered}{9}& &\\ \hline &1&-3&\color{orangered}{9}& & \end{array}$$

Multiply last value by left term and put the result into the next column

$$\begin{array}{c|rrr} \color{blue}{-3} &1& 0&0&10&1\\ & &-3&9& \color{blue}{-27}&\\ \hline &1&-3&\color{blue}{9}& & \end{array}$$

$$\begin{array}{c|rrr} -3 &1&0&0&10&\color{orangered}{1}\\ & &-3& 9 & \color{orangered}{-27}&\\ \hline &1&-3&9& \color{orangered}{-17}& \end{array}$$

Multiply last value by left term and put the result into the next column

$$\begin{array}{c|rrr} \color{blue}{-3} &1&0&0&10&1\\ & &-3& 9 &-27&\color{blue}{51}\\ \hline &1&-3&9&\color{blue}{-17}& \end{array}$$

$$\begin{array}{c|rrr} -3 &1&0&0&10&\color{orangered}{1}\\ & &-3& 9 &-27&\color{orangered}{51}\\ \hline &1&-3&9&-17&\color{orangered}{52} \end{array}$$

Step 7: Read the result from the synthetic table.

$$\begin{array}{c|rrr} -3 &1&0&0&10&\color{orangered}{1}\\ & &-3& 9 &-27&\color{orangered}{51}\\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-17}&\color{orangered}{52} \end{array}$$

The quotient is $\color{blue}{x^3 - 3x^2 + 9x - 17}$ and the remainder is $\color{orangered}{52}$.

Starting polynomial $x^4 + 10x + 1$ can be written as:

$$x^4 + 10x + 1 = \color{blue}{x^3 - 3x^2 + 9x - 17} + \dfrac{ \color{orangered}{52} }{ x + 3 }$$
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text:
Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=12$
Dividing both sides by $6$, then the solution to the given equation, $6x=72$, is \begin{array}{l} \dfrac{6x}{6}=\dfrac{72}{6} \\\\ x=12 .\end{array}


text:
# Potential Barrier

1. Feb 19, 2009

### Feldoh

1. The problem statement, all variables and given/known data
Consider the diagram described by: http://filer.case.edu/pal25/well.jpg [Broken]

If a particle with $$E = \frac{-\hbar^2 q^2}{2m}$$ comes in from negative infinity with amplitude 1, what is the wave function for negative x?

Oh and V(x) < -a and > a = 0

2. Relevant equations

3. The attempt at a solution
I think the problem needs to be broken into two cases, depending on if E > V or E < V in region 2.

If E > V then...

Region I:
$$\frac{\partial ^2 \psi}{\partial{x}^2} = -q^2 \psi$$

Which gives the solution:

$$\psi (x) = exp(iqx) + A exp(-iqx)$$

So we have the wave equation be we still need to determine the coefficient A, so we have to solve for the other two regions...

Region II:
$$\frac{\partial ^2 \psi}{\partial{x}^2} = -\frac{2m}{\hbar ^2}(E-V_0) \psi$$

Which gives the solution:

$$\psi (x) = B exp(ikx) + C exp(-ikx)$$ where $$k^2 = \frac{2m}{\hbar ^2}(E-V_0)$$

Region III:

$$\psi (x) = D exp(iqx)$$

-----------------------

So I guess I need to do continuity conditions for these equations? I have a feeling this is going to be really tedious and that there is a different way?

Last edited by a moderator: May 4, 2017


text:
# Restrict macro to an environment

I am quite a noob when it comes to customising LaTeX/XeLaTeX. I would like to know if there's a way to limit the usage of a macro to a specific environment.

Consider the following macro and environment:

\newcommand\myMacro[2]{#1 ... #2}
\newenvironment{myEnv}{...}{...}

I would like the following:

\begin{myEnv}
% The following two calls to \myMacro are valid.
\myMacro{...}{...} .... \myMacro{...}{...}
\end{myEnv}
% This one should not be valid and throw an error.
\myMacro{...}{...}

I know this might not be the best design. I just want to make sure that I can't use \myMacro outsite of myEnv.

-

All definitions inside an environment are local, so you can just define the macro at the begin of the environment. You need to double all # when doing so:

\newenvironment{myEnv}{%
\newcommand\myMacro[2]{##1 ... ##2}%
}{...}

Another alternative is to define the macro with a different name outside the environment and copy it to the official name using \let. Such internal macro normally use @ in their names so that users can't accidentally define macro with the same name.

\makeatletter
\newcommand\my@Macro[2]{#1 ... #2}
\newenvironment{myEnv}{%
\let\myMacro\my@Macro
}{...}
\makeatother

-
Thank you! I feel so stupid when I don't need some obscure TeX hack and the answer's so simple... Thank you! –  Minustar Apr 28 '11 at 7:22

You can also write:

\def\my@private@macro{code}

in the preamble, and then, when you define the environment:

\newenvironment{myenv}{\let\mymacro\my@private@macro + begin env code}{end env code}

-


text:
# Chapter 5 - Section 5.5 - Factoring Special Forms - Exercise Set: 49

$(x+2)^2$

#### Work Step by Step

RECALL: $a^2+2ab+b^2=(a+b)^2$ The given trinomial can be written as: $=x^2+2(x)(2) + 2^2$ This trinomial is in the same form as the perfect square trinomial above where $a=x$ and $b=2$. Factor the perfect square trinomial using the formula above to obtain: $=(x+2)^2$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Barnard and Child inequality exercise

Prove that, $$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)≥abc(a+b+c)^3$$ For positive $$a,b,c$$

The exercises in this book are making me crazy. Any help would be appreciated.

My attempts:

I opened the LHS brackets and used the cube identity on RHS I get expressions which are somewhat similar but no idea how to proceed furthermore: $$3\sum_{cyc}(a^3b^3+a^4bc+a^2b^2c^2)\geq abc\sum_{cyc}(a^3+3a^2b+3a^2c+2abc).$$

• I solved your problem. If you want to see my solution, show please your attempts. – <NAME> Feb 19 at 4:11
• I opened the LHS brackets and used the cube identity on RHS, – <NAME> Feb 19 at 4:15

By Holder $$3\prod_{cyc}a^2b\prod_{cyc}a^2c=\prod_{cyc}1\prod_{cyc}a^2b\prod_{cyc}a^2c\geq$$ $$\geq\left(\sum_{cyc}\sqrt[3]{1\cdot a^2b\cdot a^2c}\right)^3= \left(\sum_{cyc}\sqrt[3]{a^4bc}\right)^3=abc(a+b+c)^3.$$
• Shouldn’t all the $\prod$ be replaced with $\sum$? – Mindlack Feb 19 at 14:52
By your work, AM-GM and Muirhead we obtain: $$3\prod_{cyc}a^2b\prod_{cyc}a^2c-abc(a+b+c)^3=\frac{1}{2}\sum_{sym}(3a^3b^3+2a^4bc+a^2b^2c^2-6a^3b^2c)\geq$$ $$\geq \frac{1}{2}\sum_{sym}\left(6\sqrt[6]{(a^3b^3)^3(a^4bc)^2a^2b^2c^2}-6a^3b^2c\right)=3\sum_{sym}\left(a^{\frac{19}{6}}b^{\frac{13}{6}}c^{\frac{2}{3}}-a^3b^2c\right)\ge0.$$ The last inequality is true because $$\left(\frac{19}{6},\frac{13}{6},\frac{2}{3}\right)\succ(3,2,1).$$
• What does that modified $>$ symbol mean? – preferred_anon Feb 19 at 12:30
• $a\succ b$ says $a$ is a majoranta of $b$. Try to check it in google. I can not explain more because I am driving now. In evening, OK? – <NAME> Feb 19 at 13:05


text:
# 异常退出

 I know this question is a bit old, but for future references and googlers, here is a complete answer from another, hard to find, stackoverflow question: Android App Restarts upon Crash/force close create a class used to handle unCaughtException public class MyExceptionHandler implements java.lang.Thread.UncaughtExceptionHandler { private final Context myContext; private final Class myActivityClass; public MyExceptionHandler(Context context, Class c) { myContext = context; myActivityClass = c; } public void uncaughtException(Thread thread, Throwable exception) { StringWriter stackTrace = new StringWriter(); exception.printStackTrace(new PrintWriter(stackTrace)); System.err.println(stackTrace);// You can use LogCat too Intent intent = new Intent(myContext, myActivityClass); String s = stackTrace.toString(); //you can use this String to know what caused the exception and in which Activity intent.putExtra("uncaughtException", "Exception is: " + stackTrace.toString()); intent.putExtra("stacktrace", s); myContext.startActivity(intent); //for restarting the Activity Process.killProcess(Process.myPid()); System.exit(0); } } Then in every thread (usually you have just one, unless you start a new thread (Async or ... , ofcourse you know about threads if you are able to make new ;) ), set this class as the DefaultUncaughtExceptionHandler Thread.setDefaultUncaughtExceptionHandler(new MyExceptionHandler(this, YourCurrentActivity.class)); Remember though!Do it in the very last step of developing your application, you HAVE to at least try handle all of your exceptions one by one before leaving them for DefaultUncaughtExceptionHandler


text:
## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 4 years ago What is the area of a sector with radius 6" and measure of arc equal to 120°?

• This Question is Closed
1. Hero

$\frac{120}{360} = \frac{x}{ \pi (6)^2}$

2. Hero

Solve for x

3. Hero

In general use this formula: $\frac{\theta}{360} = \frac{x}{2 \pi r}$ where x = area of sector

4. anonymous

Area would be $(\pi*r^{2}) * 120/360$ = $(\pi*6^2)/3 \approx 37.7 units$

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text:
# Write in Standard Form n^3+4n^5+n-n^3

To write a polynomial in standard form, simplify and then arrange the terms in descending order.
Combine the opposite terms in .
Subtract from .


text:
## anonymous 4 years ago What is the equatin that would be used for a half-life problem?

1. anonymous

I use $P(t) = P_0(\frac{1}{2})^{\frac{t}{t_H}}$where P0 is the initial population and t_H is the half-life

2. anonymous

Thank you, I am given something with the half life of 29 years and a time span of 30 years, and asked to find the fraction of products in that time span How would I approach that problem?

3. anonymous


text:
# Can a passenger jet be towed to safety if it breaks down in midair?

A car can be towed by a tow truck to the repair shop when it breaks down on the highway. Is it possible for a passenger plane to be towed by an emergency tow plane along the air corridor to the next nearest airport assuming it cannot reach there on its own power (example if it runs out of fuel or both pilots are incapacitated)?

Update: This question actually consist of two parts. Whether it is feasible and whether this is practical. I have seen a small tugboat towing a barge; a man pulling a firetruck. Before writing off this idea completely, can someone address what's stopping a fighter plane from towing a passenger jet if a proper point of harness is installed?

• The plane can't very much wait for the tow like a car. In actual fact, it can still fly/glide for some time and that's how you have a chance to bring to safety. Either that or it's going to crash quickly, making the question moot. If all passengers are incapacitated, other planes can only watch it crash or, conceivably, shoot it down. Oct 6 '14 at 7:53
• Note the difficulty of trying to tow a car before it stops. Oct 6 '14 at 8:38
• If both pilots are incapacitated who is gonna call in for the Tow??? for the moment if a plane runs out of fuel or both pilots are incapacitated in midair there's one thing you can do. Pray Oct 6 '14 at 14:34
• @Tasos: Surely if you can create a means to actually tow a passenger jet, you can create a means to determine if it's needed. Perhaps a deadman switch, a cockpit video stream to ATC, or it just happens automatically if the plane strays too far from its flight plan and nobody responds.
– Gabe
Oct 6 '14 at 17:29
• This isn't very close, but still ...en.wikipedia.org/wiki/Pardo's_Push Oct 7 '14 at 3:56

Towing is quite feasible, and even landing while in tow is possible - I've done it myself in gliders. The hard part is catching up with the stricken plane.

To stay with Keegan's Boeing 777: It most likely will travel at Mach 0.8 and somewhere between 30.000 and 39.000 ft when disaster strikes. Let's further assume that power fails instantly, so the aircraft, traveling at 240 m/s, will develop a sink rate of 15 m/s just to continue flying. When it sinks, the increasing density allows it to slow down, so at sea level it will fly at 130 m/s, sinking with 8.125 m/s. For simplicity, assume a linearly decreasing sink rate, and the remaining time aloft will be less than 1000 sec or around a quarter of an hour.

Now the towplane has 15 min to scramble the crew, run up the engines, get take-off clearance and to actually fly an intercept to the stricken airliner. How realistic is that? How many airports are needed to have 777-sized towplanes ready, and how many otherwise unproductive 777s would this scheme need?

A car can stop and wait a couple of hours for the tow to reach it. An aircraft can not.

• Finally a sensible answer. Oct 6 '14 at 7:31
• 15 minutes is possibly slightly pessimistic: Air Transat 236 was aloft for 19 minutes after the second engine flamed out and could have gone a little longer if they hadn't been deliberately losing altitude to land at Lajes. But this makes no practical difference: even 20-25 minutes wouldn't give enough time for a rescue. Oct 6 '14 at 8:56
• @DavidRicherby: My calculation assumed MTOW, and a lighter aircraft will take longer to glide. Your 20 min are a better estimate, but as you said, this does not make a practical difference. Oct 6 '14 at 15:11
• @DJClayworth Your other objections are spot-on but objecting that there are currently no "tow-truck" planes doesn't make sense. Obviously, if such a system were invented, you'd need to attach some kind of device to planes to allow them to be towed and also equip or build a number of "tow-truck" planes. The mere fact that there are no such planes now (because the system hasn't been invented yet!) is no barrier per se to them existing in the future. Oct 6 '14 at 15:20
• I think this answer, like many others, assume a scenario where the plane fails, then a tow plan is come up with. Maybe a more realistic situation where this might work is one where we know the plane is going to fail in the future. For example, if there is a fuel leak, and the plane is flying over a rainforest and so unable to land safely. With still potentially hours to go until failure, maybe a rendezvous can be arranged? The towplane doesn't have to land with the jet, only extend its glide long enough to reach safety, then detach. Oct 7 '14 at 15:07

I am picturing something like a 777 being towed due to a complete engine failure. Please correct me if you had something else in mind.

This is unheard of, but it would be possible, and is a very interesting thought.

The first thing that comes to mind is something like a C-17 being refueled by a KC-135

Of course, a loaded 777 will weigh more carrying passengers than a C-17 loaded with cargo, but these mid-air refueling operations, logistically, are not much different that what would be needed to tow an airborne plane.

Here are the differences I can think of - whether or not they are feasible, I'm not sure.

• Logistically: the air force has their rendezvous planned down to a couple seconds before the plane that needs fueling is even airborne. In an emergency situation everything from planning to execution would have to happen in a VERY small amount of time.
• When refueling, no major tension is put on the refueling hose. If a 777 were being towed the tension would be AT LEAST the amount of air resistance acting on the 777.
• Designing a hook that can bear the huge forces and moments involved would take a lot of work.

I did a little math to calculate the forward force that would be required to pull a (fully loaded) 777 to keep it above stall speed.

You would need about 43,000lbs - which is not too much in the world of airliners. For example, the 777 has a take off weight of 660,000lbs - and each engine produces over 100,000lbs of thrust. So it would be very easy to have a well equipped aircraft or two to tow a 777 - once connected.

The total force in the rope, assuming the towing airplane is slightly above the airplane being towed would simply be 42,500lb/cos(angle) so if the length of the rope is a generous 300ft and the towing airplane is 50ft above the towed aircraft, then the total force would be 43,100lb.

Just in case people don't trust my numbers, here's a screenshot of the math. I had a 777 model on my computer that I used to get the data from:

• Quite an interesting discussion has emerged surrounding this answer, but it's a bit much to keep in comments. I've migrated that discussion to chat. Oct 7 '14 at 15:43
• Thanks for doing the maths. I believe that you have forgotten to plug in the drag coefficient Cd when calculating the drag force. Also, a Cd of 0.2 is quite high for a streamline aircraft. I suspect the towing force to be much lower than what you have calculated. Oct 8 '14 at 3:30
• A Cd of 0.02 is more likely. That gives $0.02*0.5*0.777*70.5^2*100 = 3861.88$
– Ian
Oct 8 '14 at 12:07
• Wouldn't the force required to tow a 777 be exactly equal to the force that it's engines would normally produce in level flight? Nov 12 '14 at 22:44
• @reirab: That would be true only if you towed the 777 at its usual cruising speed. Towing it more slowly would reduce the force required, unless the speed dropped low enough to put the 777 "behind the power curve". Nov 12 '14 at 23:00

If both pilots are incapacitated, towing the plane is moot because it won't be able to land.

In the case of fuel starvation, an airliner that loses its engines at cruise altitude can glide for maybe 20–25 minutes before it descends to sea level. If there was an airport close enough to scramble a rescue plane from, then the broken plane would have a decent chance of just gliding to a landing at that airport.

• +1 because this is a very valid point. If only the airplane is disabled, it would be much quicker/simpler/safer to land it than to scramble something up to pull it around; and if the crew is incapacitated - unfortunately even if you could scramble something up that quickly - it would not be of much use. Oct 6 '14 at 9:07
• There have actually been several incidents of pilots becoming incapacitated and, indeed, fighters can intercept them pretty quickly. For a recent example, there was a TBM 900 that was flying from Rochester, NY to Naples, FL where the pilots became incapacitated. They were intercepted by fighters and escorted all the way to Jamaica before they presumably ran out of fuel and crashed. In this particular incident, if the interceptors had been able to pull the TBM down to a lower altitude, the pilot might have regained consciousness, but that would be really hard to do without his help. Nov 12 '14 at 22:55
• @reirab Scrambling a fighter to attend an unresponsive light aircraft is hardly the same thing as scrambling something capable of towing an airliner. Nov 12 '14 at 23:37
• Very true. I mentioned as much in one of my comments on another answer. The only I way I could think of to do it with something the size of an airliner would be with multiple fighters, preferably attaching somewhere around the engine pylons (since that's where the aircraft is designed to be pulled from anyway.) That would be really hard to pull off, though. Having said that, the last sentence of your answer is not quite true. Even if the interceptor only flew at the same speed as the airliner, they could meet halfway from the field, doubling the potential range. Nov 12 '14 at 23:51

Its not unheard of - gliders are often carried up by tow planes. However, hitching up a plane on the ground and towing it up is completely different than trying to do the following:

1. Catching up to a plane flying at great speed.
2. Positioning yourself in front of it.
3. Hooking up a tow hook.
4. Pulling it fast enough to maintain flight.

Suppose you were somehow able to do the first 4, you are still left with the small point of having to land two aircraft that are tied to each other.

So its is nearly impossible to do what you are suggesting.

An airplane will maintain flight as long as there is sufficient lift on the wings.

To tow a disabled plane is really not required, because even if all engines fail (a highly unlikely event), there are contingencies in place to power the control surfaces and communications so that the pilot can glide the plane to a safe landing. See this question for examples of incidents where all power was lost.

Lets assume that everyone on board becomes incapacitated. In this scenario (which has happened before) the plane will continue on its programmed flight path until it runs out of fuel, at which point it will drop in altitude eventually until it crashes.

In April 2012 the pilot of a Cessna 421 was incapacitated (possibly due to hypoxia - lack of oxygen) - the aircraft continued on its programmed course until it eventually ran out of fuel (flightaware flight track):

September 10 had another similar incident.

• Good answer, but I think that most of the items in your list are actually very feasible. Oct 6 '14 at 5:00
• I highly doubt it - just thinking off the top of my head what the power requirements would be to do such a thing - I don't know of any aircraft that has the capability. Oct 6 '14 at 5:01
• @BurhanKhalid not an aircraft of comparable size, but you could have something like a C-130 tow a C172. Probably not what OP has in mind as most such questions come from people who think of airliners when thinking aircraft and little else. Oct 6 '14 at 7:46
• (+1) The most famous crew-incapacitated/flight-runs-its-course accident must be Helios Airways Flight 522. Fighter jets were able to reach the plane and escort it for over half an hour so asking whether something else could be done in this situation is not an entirely absurd question. Oct 6 '14 at 7:52
• You wouldn't have to land the two planes tied together. The tow could be released shortly before the landing airport to allow the disabled plane to glide in. Of course, they'd have to land it first time but I doubt you'd be able to do a go-around while towing a large jet anyway. Oct 6 '14 at 8:52

The simple answer is that it's presently not possible, certainly not for large passenger jets.

That said, it's possible to do rather impressive things like towing a glider or setting up a rendez-vous in mid-air for refueling. Towing would therefore not seem completely out of the realms of possibilities and Keegan's calculations suggest it would even theoretically be plausible (I didn't try to verify the calculations and I am probably not knowledgeable enough to do so).

One problem is that if all engines fail, staying put and waiting for a tow is not an option. Fortunately, planes do not “fall off the sky” either, they can glide in a controlled fashion and that's the actual contingency plan. It has happened before, perhaps most famously to the Gimli glider.

As <NAME> explained, an additional problem in this situation is that, while it's possible to fly the plane to safety, there is not much time available. Imagining a tow plane is one thing, but having one at the ready to reach a specific point anywhere in the world in 10-20 min is something else entirely and would be prohibitively expensive.

Now, if the pilots are incapacitated, things are quite different. Modern planes can maintain speed and altitude and basically continue to fly. This has also happened before, e.g. to Helios Airways Flight 522 and in that particular case there was more than enough time for fighter jets to reach the distressed airliner and escort it for over half an hour. Ultimately, however, they couldn't do anything else than watch it run out of fuel and crash.

The problem is that “towing” in some way or other is neither needed nor helpful. After all the plane is still entirely capable of flying on her own. What you need is taking control and that's presently not possible either. To my (non-expert) eyes, adding some remote control functionalities seems perfectly doable with current technology (cf. drones) but that would mean adding additional complexity, potential for failure, etc. I assume the cost/benefit analysis does not justify it and, as far as I know, there is no commercial passenger plane with anything like that.

• Imagine what the "bad guys" would be able to use such a remote control setup for. Who gets to override the override?
– user
Oct 8 '14 at 10:53
• @MichaelKjörling Yes, exactly, that would be one of the potential failures of the system. Thanks for the link! Oct 8 '14 at 10:54
• @MichaelKjorling presumably the pilot can override the override, if he or she is not incapacitated. Apr 10 '15 at 11:19

Cars and passenger aircraft are designed and maintained in very different ways.

## Redundancy

Cars only have one engine, unlike almost the aircraft that carry most passengers. Cars cannot continue their journey if their only engine breaks down or if an axle breaks. Aircraft have multiple independent sets of many of the most critical parts like engines and flight control systems, they are specifically designed to be able to continue when many things break down. An aircraft can continue when one engine breaks down, it can still be flown and make a landing when all engines have broken down. Aircraft have redundant systems to allow for continued operation after multiple failures.

## Maintenance and monitoring

Cars might only be looked at by a qualified mechanic once a year or once every 18 months. Even at this time, it is likely that they mechanic just changes the oil in the engine, checks the fluid levels and brake wear and does a few other basic checks specified by the manufacturer. They don't check everything in the car.

An aircraft is more likely to be monitored continuously. It is not unusual for the engines to be sending data to the engine-makers in flight so that the engine makers can measure the health of the engine and can predict when the engine needs maintenance before any problem arises for the aircrew. After every flight, the crew of an aircraft report any unusual behaviour of the aircraft to the maintenance crew so that incipient problems can be dealt with before there is any danger of breakdown.

Car owners are not this diligent

## Towing

Cars are expected to break down and are designed to be towed. Aircraft are maintained so that they don't need this and the airframes are not designed to be towed in the air, there is no towing point with enough strength other than that designed into the landing-gear for gentle low-speed towing on the ground.

You cannot attach a towline to an aircraft that is not under good control, if the pilots are incapacitated or if the aircraft is damaged such that steady flight is impossible (e.g. phugoid cycles etc). If the aircraft is under good control it is probably able to fly or glide to an alternate airfield or attempt an emergency landing.

## Time is of the essence

Say a 747 lost all four engines, there would probably not be enough time to launch a rescue tow-aircraft from a nearby airport and reach the plane in time to do anything useful in mid air. Particularly if the 747 is in mid-Pacific or flying through a volcanic ash-cloud. It is better to design the aircraft to cope as well as we can make it.

A car can wait an hour or two at the side of the road. An aircraft cannot wait.

• Unless the pilots are incapacitated the plane could be turned to glide towards the tug. That would save some time at least.
– Ant
Oct 17 '14 at 16:19

As others have pointed out, the plane that would benefit from towing would go down before it could be reached by a tow plane.

Even if the tow plane could catch up to the glider it would be difficult to do the hookup quickly. You would need some kind of special apparatus to capture the tow hook and attach it. Doing this would be very difficult because the aircraft would be in different configurations.

For example, lets say it is two 737, one with engines out. The stall speed for a 737 is 150 knots, so it might be going, say 180 knots with a glide ratio of 15:1. At this rate it will be descending at 1200 ft/min. That is a FAST descent rate, almost crashing/out of control. For example, if you descend faster than 1000 ft/min in a flight test you will be automatically failed.

The cruising speed for the tow plane is 450 knots, so he would have to somehow slow down to 180 knots AND be descending at 1200 ft/min AND stay directly in front of the glide plane while the hookup occurred. Doing all three of these things would be very, very difficult.

• That's a good point. Oct 11 '14 at 2:38

However, we could widen the discussion by fitting the stricken aircraft with a docking mechanism on the underside. Then the rescue aircraft (with a matching docking unit on the top side) could maneuver beneath it and lock on. We would then have a composite structure like the 747-space shuttle transporter configuration. A sufficiently powerful rescue ship could then fly the pair down to a safe landing.

Further enhancements occur: (a) The docking port could have a manhole through which the PAX could be evacuated into the rescue ship. (b) A command-interface could enable the rescue pilot to control the flight surfaces on the stricken plane (assuming they were still functional), thus making it easier to fly.

• Bear in mind that an empty Space Shuttle weighs 78 tonnes, whereas even a Boeing 757 full of fuel and passengers can weigh up to 125 tonnes and a 777 can be up to 350t. (And let's not get into the 590-tonne Airbus A380.) Oct 10 '14 at 22:34
• And even then, carrying the Space Shuttle reduced the aircraft's range from 5,500 miles to 1,000 miles. Still a considerable distance, granted, but goes to show just how much that drag does! (The Shuttle Orbiter wasn't that non-aerodynamic from the front.)
– user
Jun 4 '16 at 14:48

There are several reasons that this idea would not be practical.

An gliding aircraft flying at around 30,000ft would have, say, 20 mins before it hits the ground. If there is a suitable airfield within that range it stands a decent chance of landing safely.

The stricken aircraft would be travelling at, say, Mach 0.8 so only a fast fighter jet would be capable of chasing and catching it from behind in the available time. Any other rescue aircraft would have to approach from the front, circle and manoeuvre into position.

The time constraint makes this almost impossible. Assuming the rescue aircraft took 10 mins to prepare and take off, then the manoeuvring and attachment of the tow could add another 5 mins. That would allow only a few minutes to actually catch the stricken aircraft.

Gliders are designed to be very light aircraft and get towed up by relatively powerful tugs. They then release the tow - it is not re-attached! The airframe of a glider is designed to take the forces involved in towing; that of a passenger aircraft is not. Would it be possible to re-design all passenger planes to cope with this very unlikely scenario of total engine failure without adding weight and reducing efficiency? I doubt it.

As mentioned by <NAME>, air-to-air refuelling is currently a very expert skill practised by a handful of the best Air Force pilots. Some cunning new technology would be needed before the principle could be used on commercial aircraft.

Rather than this idea, a more useful and practical suggestion might be to add some sort of remote control function to be used in the case of the pilots becoming incapacitated or terrorists taking over the aircraft. Such technology would need to be robustly fail-safe and many crashes have shown that that isn't easy to achieve.

• Who gets to override the override?
– user
Oct 8 '14 at 10:59
• And you don't need to apply this new technology just to any commercial aircraft -- you need it to work on commercial aircraft whose engines have failed! Oct 10 '14 at 22:28

There has to be consideration for whether the towing process would remain at a stable altitude, or just provide assist to the nearest airport, calling into question the amount of power that would be required to assist. Given the amount of turbulence that would be generated from the towing craft (and the bigger it is, the worse it would be) it seems unlikely the disabled aircraft would be able sustain stable flight and structural integrity.

You also need to consider the relatively small number of aircraft that become completely "disabled" without being impacted by some related catastrophic event that otherwise compromises it's flight worthiness. There are very few instances of US Air 1549 and most would be too close to the ground to allow any time to react without the famed Star Trek Tractor Beam

Perhaps with a highly specialized design process some of the challenges could be overcome, but the practicality of being able to actually have the proper assistance available would negate the feasibility.

• Air-to-air refuelling shows that it's perfectly possible to have two aircraft flying in the sort of close formation that would be required for towing. On the other hand, air-to-air refuelling is a difficult manoeuvre to carry out in a fully operational plane even when the pilots do it every day; for a pilot to do it once in their career, in a plane with no engines, sounds like a very big ask. Oct 7 '14 at 8:16
• @DavidRicherby Air to Air refueling, correct me if I'm wrong, is a well planned operation handled by experts in the field, and even then it does not involve the stresses of towing an unpowered aircraft. Oct 7 '14 at 10:46
• @NPSF3000 You are correct. But it shows that it is possible to have two large planes flying in the same sort of close formation you would need for towing, which contradicts the claim in this answer that the towing aircraft would generate so much turbulence that the towed craft would be unable to fly. Oct 7 '14 at 11:09

With enough engines and all the right hardware...sure, it's possible. See this report on tow launching an F-106: http://www.nasa.gov/centers/dryden/history/pastprojects/Eclipse/index.html

The problem is that it's just not practical. Nevermind the actual execution of a pickup maneuver. As mentioned in previous answers the costs of a standby fleet required to execute reliable pickups are very high. There would be zero utilization of such a fleet.

Your edit asks about a fighter jet serving as the tow plane. Yeah, sure, it's technically possible to build a trap of some kind to be a tow rig on an airliner. If afterburner is required to keep both planes aloft tow time would be extremely limited though.

This has happened under wartime conditions.

Pardo's Push Captain <NAME> pushed another F4 Phantom that had lost its engines into Laotian airspace after a raid on N Vietnam, to where the pilots could eject into less than hostile ground. All while Pardo's plane lost one engine due to battle damage and a fire.

<NAME> pushed an F86 Sabre that had run out of fuel, out of N Korean airspace and over the ocean. Unfortunately, the pilot drowned when he ejected, landed in the ocean, and became entangled in his chute lines.

However... If you undertake a study of high profile commercial aviation incidents, you will find that almost none of them could have been mitigated with this scheme. Here's a quick spot check...

AF447 High altitude stall in the middle of the Atlantic. Went down in less than ten minutes.

JAL 123 loss of horizontal stabilizer... aircraft uncontrollable. Would have just taken the tow plane down.

USAir 427 malfunctioning rudder control unit on landing approach - no time, aircraft uncontrollable.

ValuJet 592 Fire in the cargo hold. Burned through control lines and probably incapacitated the crew.

BA 9 Lost engines due to volcanic ash. Three engines restarted. "I trust you are not in too much distress"

USAir 1549 Lost engines due to bird collision shortly after takeoff. Not enough time to do anything other than ditch.

Alaska Air 261 Vertical stabilizer jackscrew malfunctioned due to poor maintenance. Aircraft uncontrollable.

TACA Air 110 Lost engines due to storm. Landed on levee without damage by unbelievably talented pilot.

Lauda Air 004 Thrust reverser deployed inflight. Aircraft out of control. We learned that the certification that the aircraft could fly with a thrust reverser deployed was inadequate from this incident.

In all of these, only TACA 110 or BA 9 might have benefited from a tow plane, and even so, there probably wasn't enough time, as both planes could remain airborne for maybe 20-30 minutes.

So it's an interesting idea, but as of today, an airliner losing all engines is extremely rare, as contemporary gas turbines are very reliable - enough to certify crossing major oceans with only two engines. The funds for modifications for a tow hookup, the training in how to do it, and the maintenance of dedicated tow aircraft, would be better spent addressing the issues that are causing most accidents.

• JL123 lost its vertical stabiliser, and AS261 lost control of its horizontal stabiliser, not vice versa. Oct 4 '19 at 1:08


text:
# An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. …

Q: An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is Po. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

(a) $\displaystyle \frac{1}{2\pi} \frac{A \gamma P_0}{ V_0 M}$

(b) $\displaystyle \frac{1}{2\pi} \frac{V_0 M P_0}{ A^2 \gamma}$

(c) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{A ^2 \gamma P_0}{M V_0 }}$

(d) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{M V_0}{ A \gamma P_0 }}$

Ans: (c)

Sol: For sudden change in pressure and volume of the gas ;

$\displaystyle PV^\gamma = constant$

Differentiating ,

$\displaystyle P \gamma V^{\gamma -1}dV + V^\gamma dP = 0$

$\displaystyle dP = -\frac{\gamma P}{V}dV$

Net force on piston is

$\displaystyle F = dP \times A$

$\displaystyle F = -\frac{\gamma P_0}{V_0}dV A$

Let the piston be moved inwards through a small distance x, then decreases in the volume of the gas,
dV = Ax

$\displaystyle F = -\frac{\gamma P_0}{V_0} \times Ax \times A$

$\displaystyle F = -\frac{\gamma P_0 A^2}{V_0} x$

$\displaystyle M \omega^2 = \frac{\gamma P_0 A^2}{V_0}$ (Where M = mass of piston)

$\displaystyle \omega^2 = \frac{\gamma P_0 A^2}{M V_0}$

$\displaystyle \omega = \sqrt{\frac{\gamma P_0 A^2}{M V_0}}$

Frequency $\displaystyle \nu = \frac{2\pi}{\omega}$


text:
Permeance

Permeance is how well matter or energy flows through something. In science, permeance is usually shown as: ${\displaystyle {\mathcal {P}}}$.


text:
# Does $f(x) = f(2x)$ for all real $x$, imply that $f(x)$ is a constant function?

If a Continuous function $f(x)$ satisfies $f(x) = f(2x)$, for all real $x$, then does $f(x)$ necessarily have to be constant function? If so, how do you prove it? If not any counter examples?

-

For all $x\in\mathbb{R}$ and $n\in\mathbb{Z}$, $f(x)=f(x/2^n)$, so $f(x)=\lim_{n\to\infty}f(x/2^n)=f(0)$ by continuity.

-

$f(x)=\sin(\log_a(x))$, select basis a so as $\log_a(2)=2\pi$, that is, $a ^{2\pi}=2$.

-
It's not continuous at zero –  Quark May 24 '13 at 17:14
It is continuous everywhere else, which I guess is the best you can do, given the argument above. –  Nishant Jul 14 at 21:09


text:
# arrc

I wanted to see just how small I could make a self-hosting compiler and arrc was the result. I wrote a 100-line bootstrap compiler in C (targeting x86 assembly), and then a self-hosting compiler in arr itself. The self hosted version is a bit longer, and totally unreadable, but it was a lot of fun to make.

The language is stack based and has some limitations that make it fun (or not, depending on your attitude) to use. For instance, you can only use 26 variable a-z. This was, of course, to make the compiler easier to self-host, but it also has the nice benefit of making anything written in the language totally unreadable.

Here’s some example code from the readme:

. # . pops from the front of the input array
, # , writes the last thing to the front of the output array

.A # a capital letter stores the last thing in that variable
a # a lowercase letter gets the value from that variable

:B # makes a label called B

@D # @ calls D

\A # \ makes a literal char
? # = compares the top two things on the stack
=F # jumps to F if the comparison is true

123 # numbers are literals
A # store 123 in a
2 1 - # == 1

:D # d label
; # ; returns

2 & + # & duplicates the top item on the stack

\A \B \C \D 1 ! , # prints C
# !:
# let n = the top item on the stack
# duplicate the n-th value on the stack (starting at 0)
# and put it on the top of the stack

The repo is available here.


text:
PDF e-Pub

## Section: New Results

### Exact biconvex reformulation of the ${\ell }_{2}-{\ell }_{0}$ minimization problem

Participants : <NAME>, <NAME>, <NAME>.

We focus on the problem of minimizing the least-squares loss function under the constraint that the reconstructed signal is at maximum k-sparse. This is called the ${\ell }_{2}$-${\ell }_{0}$ constrained problem. The ${\ell }_{0}$ pseudo-norm counts the number of non-zero elements in a vector. The minimization problem is of interest in signal processing, with a wide range of applications as compressed sensing, source separation, and super-resolution imaging, for example.

Based on the results of [31], we reformulate the ${\ell }_{0}$ pseudo-norm exactly as a convex minimization problem by introducing an auxiliary variable. We then propose an exact biconvex reformulation of the ${\ell }_{2}-{\ell }_{0}$ constrained and penalized problems. We give correspondence results between minimizer of the initial function and the reformulated ones. The reformulation is biconvex. This property is used to derive two minimization algorithm, CoBic (Constrained Biconvex) and PeBic (Penalized Biconvex).

We apply the algorithms to the problem of Single-Molecule Localization Microscopy and compare the results with the well-known IHT algorithm [22]. Both visually and numerically the biconvex reformulations perform better. Furthermore, the algorithm has been compared to the IRL1-CEL0 [23] and Deep-STORM [25]. The IRL1-CEL0 minimizes an exact relaxation [29] of the ${\ell }_{2}-{\ell }_{0}$ penalized form and Deep-STORM is an algorithm that uses deep-learning and convolutional network to localize the molecules. This work has been presented at the ISBI 2019 conference [6], as well as a more mathematical article was presented as a poster at GRETSI 2019 [12]. A full journal article has been submitted to the Biomedical Optics Express for a feature issue: Superresolution Microscopy on the 25th Anniversary of STED Microscopy and the 20th Anniversary of SIM.

Figure 1. Reconstructed images from the simulated ISBI dataset [28], 99 non-zero pixels on average. Top: Sum of the acquisitions. Middle: From left to right: CoBic, Constrained IHT and Deep-STROM. Bottom: From left to right: PeBic, Penalized IHT and IRL1-CEL0.


text:
• Diffusion of gases into the lung: How physics can help to understand physiology

• Fulltext

https://www.ias.ac.in/article/fulltext/pram/071/02/0245-0251

• Keywords

Lung; diffusion; screening; efficiency; edema.

• Abstract

In the human lung, the gas transfer between air and blood is achieved in terminal units that are called acini'. Whereas convection is still the predominant transport phenomenon at the acinus entrance, most of the acinar surface is in fact accessed by diffusion. The transition between convection and diffusion, and thus the size of the diffusion unit, depends on the air velocity at the acinus entrance. In this paper, we present a gas transport model which takes into account both the diffusion into the acinus and the diffusion across the alveolar membrane. It is shown that the physiological sizes of the diffusion unit in the lung, at rest or at exercise, can be explained by physical arguments. In that sense, diffusion is the dimensioning criterion' of the lung at the acinar level. This approach shows that, due to diffusional screening at inspiration and at rest, there exists a permanent spatial inhomogeneity of oxygen and carbon dioxide partial pressure which reduces the effective surface efficiency of the human acinus to a value of only 30 to 40%. This model casts a new light on the properties of this physiological transport system. It permits in particular to understand how several diseases among which pulmonary edema may remain asymptomatic in their early stages.

• Author Affiliations

1. Physique <NAME>, Ecole Polytechnique, CNRS, 91128 Palaiseau, France
2. CMLA, ENS Cachan, CNRS, UniverSud, 61 Avenue du President Wilson, F-94230 Cachan, France

• Pramana – Journal of Physics

Volume 94, 2020
All articles
Continuous Article Publishing mode

• Editorial Note on Continuous Article Publication

Posted on July 25, 2019