Physical Formulas, Just a useful collection of terms

General (differential calculus)
√(1 + x) = 1+x/2-x²/8+x³/16-5x⁴/128+7x⁵/256-21x⁶/1024+33x⁷/2048-429x⁸/32768+715 x⁹/65536 - .... when x is small compared to 1.
General (waves)
ω is a general angular frequency (radians per second)
f is a general frequency (cycles per second) which characterizes the rate at which between periodic features in time arrive.
ω = 2πf (This follows from 1 cycle = 2π radians )

k is a general wavenumber with units of inverse length.
λ is a general wavelength describing the distance which characterizes the length between periodic features.
k = 2π/λ

A generic wave in 1 dimension can be written as:
u(x,t) = f(kx - ωt) = f(2πx/λ - 2πft)

The simplest wave is a sinusoidal wave, but since
sin(x) = (e^ix - e^-ix)/(2i)
and
cos(x) = (e^ix + e^-ix)/2
it is often convenient to work with the exponential function.

And so u(x,t) = Ae^i(kx - ωt) is a simple wave in one dimension. More complex waves can be built up by summing sinusoidal waves, and this is the Fourier transform.

The benefits of wavenumber and angular frequency are two-fold. Not only do we remove the factors of 2π which might clutter up our analysis, but plane waves in higher dimensions can be written as u(x,t) = Ae^i(k·x - ωt) where vectors (shown in bold) are multiplied with the inner product.

Elementary Quantum Relativistic Physics

p is momentum of an entity with respect to an inertial observer
h is Planck's constant
λ is the de Broglie wavelength as measured by that same inertial observer

|p| = h/λ
(This generalizes to p = (h/2π)k when p is a vector.)

E is the relativistic energy with respect to an inertial observer
f is the de Broglie frequency (cycles per second) as measured by that same inertial observer

E = hf

v is the velocity of an entity with respect to an inertial observer
c is the speed of light in vacuum

v = pc²/E

m is the invariant mass of an entity

E² = m²c⁴ + p²c²

When |pc| is very small when compared to |E|, we can approximate these relations as:
E = √(m²c⁴ + p²c²) = (mc²)√(1 + (p/(mc))²) = (mc²)( 1 + (p/(mc))²/2 - (p/(mc))⁴/8 + ...) ≈ mc² + p²/(2m)
|v| = |p|c²/E = |p|/(m + p²/(2mc²) ≈ |p|/m
or p ≈ mv
E ≈ mc² + p²/(2m) ≈ mc² + m²v²/(2m) ≈ mc² + ½mv²

So we see Newtonian kinematics are implied by relativistic kinematics as the low-momentum approximation.

The high-momentum approximation is gotten immediately by assuming m=0.
Then E = |p|c and |v| = |p|c²/(|p|c) = c

Next: Introduce the Lorentz Transforms and the full relativistic relation between speed and momentum and energy.

This post has been edited by rpenner on Feb 2 2010, 07:36 AM

Excellent thread.

Might I suggest the addition of the Euler Lagrange equation (not easy without tex, sorry about that).

Also, I vote that this thread be pinned if you can do that.

General (Monoids)
Let's say there was something. And you could either do something to it or leave it along.
Doing something to it would change it, and we could make this look mathy by describing the change as new = f(old). But there are a great many things that we could do that would also result in changes. Let us parameterize f by some set listing the things we could do. Then we have new_A = f_A (old). But we could also do nothing. Let us call f-that-means-do-nothing as f_I. Then f_I(old) = new_I = old. In fact, f_I(anything) = anything. So f_I is an identity transform.

Now we could also do multiple things multiple times. Clearly f_I(f_A(anything)) = f_A(f_I(anything)). But f_B(f_A(somthing)) might not be the same as f_A(f_B(somthing)). And there might be a way to say "do A then B" and we could call that C. In that case f_B(f_A(anything)) = f_C(anything)).

Mathematicians would call this the operation of a monoid on a set. The set would be the old's and the new's. And the monoid would be I, A, B, C, .... and the rules which let you talk about combining them. But most of the interesting math is on the side of telling how combining things makes things. The rules for a monoid are simple:

• For any two elements of a monoid, A and B, you can always get an element by combing them, which we write as A ⊗ B.
• There is an element, I, such that I ⊗ A = A ⊗ I = A for any element of the monoid, A.
• For any three (not necessarily distinct) elements of the monoid, A ⊗ ( B ⊗ C ) = ( A ⊗ B ) ⊗ C.

The first monoid we learn about is called the natural numbers under addition. A ⊗ B is just a new name for A + B and I = 0, because 0 + A = A + 0 = A.

Another monoid is the natural numbers under multiplication. Of course, here I = 1. It is still a monoid if you remove 0, because if neither A nor B are zero then AB is not zero either. But you can't remove just 6 because 2 * 3 = 6.

(left + right + right) is a monoid on the natural numbers with I = 0.
1 ⊗ 2 = (1 + 2 + 2 ) = 5 but 2 ⊗ 1 = ( 2 + 1 + 1 ) = 4
Multiplication of quaternionic integers (a,b,c,d)⊗(A,B,C,D) = (aA - bB - cC - dD, aB + bA + cD - dC, aC - bD + cA + dB, aD + bC - cB + dA) is an example of a monoid with I = (1,0,0,0) which X⊗Y might not equal Y⊗X.
(0,1,0,1)⊗(1,0,1,0)=(0,0,0,2) but (1,0,1,0)⊗(0,1,0,1)=(0,2,0,0)
[Edited 2012-04-15]

For a final example, not all monoids have an infinite number of elements. Boolean AND works on a universe of just TRUE and FALSE.

General (Groups)
A group is a monoid where every element has an inverse element.
Often we can glue on inverses where none exist.
The natural numbers under addition + inverse = the integers under addition. I = 0 is its own inverse.
The natural numbers (excluding zero) under multiplication = the positive rational numbers under multiplication. Not only do we have to glue on 1/2 to be the inverse of 2, but 3 times 1/2 also needs to be included, etc.

Back to our original example. Inverses let us undo what has been done.

f_1/2(f_2(anything)) = f_1(anything) = anything.

So a group is a monoid that lets us describe undoing.

General (Lie Group)
(Pronounce Lie as if it was spelled "Lee")

What if we had a group were no matter what we wanted to do, we could always break it up into two equal steps from the identity. If A was in the group, then so was B, such that B ⊗ B = A and C was in the group such that C ⊗ C ⊗ C ⊗ C = A, etc. Then would we be well set? Not quite since we would like at least one more property. We would like I < C < B < A in some sense. Generalizing, we would always like to write A = scale_up(r,Ã) where à is a direction and r is a distance. Then B = scale_up(r/2,Ã) and C = scale_up(r/4,Ã) and we would like scale_up(0,Ã) = I, and a concept of smoothness. This is called a Lie group, and it introduces a concept of continuity which allows us to bring in concepts from differential calculus.

General (Special Orthogonal Group)

The special orthogonal group in one dimension, SO(2), is like complex multiplication of the numbers with absolute value 1. We can write these numbers as e^iθ = cos θ + i sin θ = scale_up(θ,e^i). Each multiplication by these numbers is like a rotation about the 0. So a smaller θ implies a smaller rotation, at least when θ is small.

Clearly scale_up(0,e^i) = e^0 = 1, which makes sense. And θ makes a lot of sense about out idea of breaking things up.

But we can also define (because of smoothness) lim x->0 (scale_up(x,Ã) - I)/x = d(e^ix -1)/dx = i. So that for very small θ, scale_up(θ,Ã) ≈ 1 + iθ.

So not only do small values of r (or θ) add like real numbers, but in a certain sense there is a well-defined direction near I (or near r=0). That's what we meant by smoothness.

The special orthogonal group in three dimensions SO(3) can be thought as rotating three dimensional objects every which way. Here instead of using complex numbers, we can use multiplication of 3x3 real matrices. Instead of only having one possible direction, we can have three independent directions.

But given a 3x3 rotation matrix A, we can parameterize A = scale_up(θ,Ã) = P + (1 - P) cos θ + Q sin θ. Where P and Q are special matrices which can be related to three normalized numbers which indicate a direction.

Yadda, yadda, yadda. The important thing is that the universe of rotation matrices form a Lie group, and every Lie group is a group, which means every rotation can be undone.

The other thing, is that the rotation in 2- or 3- dimensions preserves the length of all line segments, the areas of all triangles and the measure of all angles. All rotations are rigid motions in the Euclidean sense.

This post has been edited by rpenner on Apr 15 2012, 11:22 PM

General (Euclidean geometry and trigonometry)

All right triangles can be scaled up or down and repositioned such that the hypotenuse is a radius of the unit circle and one leg is co-linear with the x-axis. That is to say, all right triangles are similar to such a right triangle with hypotenuse 1.

Pythagoras' Theorem states that for any right triangle the squares of the lengths of of the two legs, A and B, is equal to the square of the length of the hypotenuse, C.
Or A² + B² = C²
Or for the similar triangle with a hypotenuse as a radius to the unit circle:
(A/C)² + (B/C)² = 1

In analytic geometry, if A/C is the length of the leg co-linear with the x-axis, then the coordinates of the vertexes of the triangle are: (0,0), (A/C, 0) and (A/C, B/C), where the last is a point on the unit circle. All such points on the unit circle can be written as (cos θ, sin θ), where θ is the angle between side A and side C and also on the corresponding sides on the similar triangle in the unit circle. That's because similar triangles have all the same corresponding angles.

So these trigonometric functions are universal for right triangles.

Depending on where we chose to measure θ in the right triangle, one leg is adjacent to the measured angle, and the other is completely on the other side of the triangle, which we will call the opposite leg. Then the six common trigonometric functions are simple ratios of side lengths.

sin θ = opposite/hypotenuse
cos θ = adjacent/hypotenuse
tan θ = opposite/adjacent = sin θ/cos θ
csc θ = hypotenuse/opposite = 1/sin θ
sec θ = hypotenuse/adjacent = 1/cos θ
cot θ = adjacent/opposite = 1/tan θ = cos θ/sin θ

And the most famous of all trigonometric identities:
(sin θ)² + (cos θ)² = 1
is nothing but Pythagoras' Theorem for all right triangle.

Elementary Relativistic Physics
So since E² = m²c⁴ + p²c², this reminds of a right triangle with legs of pc and mc², especially when m > 0. If we chose adjacent = mc² we get the following relations:
sin θ = pc/E
cos θ = mc²/E

But sin θ = pc/E = pc²/(Ec) = v/c
which means (mc²/E)² = (cos θ)² = 1 - (sin θ)² = 1 - (v/c)²
And it follows that when m > 0, E = (mc²)/√(1 - v²/c²) and p = (mv)/√(1 - v²/c²)

This lets us rewrite m²c⁴ = E² - p²c² = [(mc²)/√(1 - v²/c²)]² - [(mc²v/c)/√(1 - v²/c²)]² = (mc²)² ( 1 - (v/c)² ) / (1 - v²/c²) = m²c⁴ .

So when m > 0, E/m and p/m are somewhat curious functions of v. But if we define a type of particle by what we observe E² - p²c² to be, then we seem to be able to ignore v when we speak of it having a well-defined mass.

But when E² - p²c² = 0, m = 0, v = c and E = pc.

This post has been edited by rpenner on Feb 11 2010, 11:30 PM

General (Trigonometric and related functions)

d e^(ax) /dx = a e^(ax)

e^x = cosh x + sinh x
e^(iθ) = cos θ + i sin θ = cosh(iθ) + sinh(iθ)

sinh ζ = (e^ζ - e^-ζ) / (2) = -sinh -ζ
cosh ζ = (e^ζ + e^-ζ) / (2) = cosh -ζ
tanh ζ = sinh ζ/cosh ζ = (e^ζ - e^-ζ)/(e^ζ + e^-ζ) = (e^(2ζ) - 1)/(e^(2ζ) + 1) = - tanh -ζ
sin θ = (e^(iθ) - e^(-iθ)) / (2i) = - sin -θ = -i sinh(iθ)
cos θ = (e^(iθ) + e^(-iθ)) / (2) = cos -θ = cosh(iθ)
tan θ = sin θ /cos θ = -i (e^(iθ) - e^(-iθ))/(e^(iθ) + e^(-iθ)) = -i (e^(2iθ) - 1)/(e^(2iθ) + 1) = - tan -θ = -i tanh(iθ)

(cos θ)² + (sin θ)² = 1
1 + (tan θ)² = (sec θ)² = 1/(cos θ)²
(cosh θ)² - (sinh θ)² = 1
1 - (tanh ζ)² = (sech ζ)² = 1/(cosh ζ)²


arcsinh x = ln(x + √(x² + 1))
arccosh x = ln(x + √(x² - 1)) when x >= 1
arctanh x = ln √((1+x)/(1-x)) when | x | < 1
arcsin x = -i arcsinh (ix) = -i ln(ix + √(1 - x²)) when | x | < 1
arccos x = -i arcosh (x) = -i ln(x + √(x² - 1)) when | x | < 1
arctan x = -i arctanh(ix) = (i/2) ln ((1-ix)/(1+ix)) when | x | < 1

sinh(a±b) = sinh(a)cosh(b)±sinh(b)cosh(a)
cosh(a±b) = cosh(a)cosh(b)±sinh(a)sinh(b)
tanh(a±b) = ( tanh(a) ± tanh(b) ) / ( 1 ± tanh(a)tanh(b) )
arcsinh(a)±arcsinh(b) = arcsinh(a√(1 + b²) ± b√(1 + a²))
arccosh(a)±arccosh(b) = arccosh(ab ± √(a² - 1)√(b² - 1))
arctanh(a)±arctanh(b) = arctanh(( a ± b ) / ( 1 ± ab ))


sin(a±b) = sin(a)cos(b)±sin(b)cos(a)
cos(a±b) = cos(a)cos(b)∓sin(a)sin(b)
tan(a±b) = ( tan(a) ± tan(b) ) / ( 1 ∓ tan(a)tan(b) )
arcsin(a)±arcsin(b) = arcsin(a√(1 - b²) ± b√(1 - a²))
arccos(a)±arccos(b) = arccos(ab ∓ √(1 - a²)√(1 - b²))
arctan(a)±arctan(b) = arctan(( a ± b ) / ( 1 ∓ ab ))


d sin θ / dθ = cos θ
d cos θ / dθ = - sin θ
d tan θ / dθ = 1/(cos θ)² = 1 + (tan θ)²
d arcsin x / dx = 1/√(1-x²)
d arccos x / dx = -1/√(1-x²)
d arctan x / dx = 1/(1+x²)

d sinh θ / dθ = cosh θ
d cosh θ / dθ = sinh θ
d tanh θ / dθ = 1/(cosh θ)² = 1 - (tanh θ)²
d arcsinh x / dx = 1/√(1+x²)
d arccosh x / dx = 1/√(x²-1)
d arctanh x / dx = 1/(1-x²)

Elementary Relativistic Physics (cont.)
So starting with v = pc²/E and E² = m²c⁴ + p²c², we used geometric reasoning to show that a right triangle with sides of length mc², pc and E was similar to a right triangle with sides of √(1 - v²/c²), v/c and 1, respectively. While it works, rescaling all E's to be a radius of a unit circle, it is not always the most convenient geometric metaphor, since what distinguishes particles is frequently their masses, not their energies. The mass, so our experience has taught us, is an intrinsic property of a particle, and we have well-defined masses for many particles, atoms and molecules. Study of masses has even taught us that many pure elements are composed of mixtures of different isotopes, which turn out to have a variety of unique properties correlated with their mass alone, even when the chemical properties are nearly identical.

So when m>0, we can study another similar right triangle with sides of A=1, B=(v/c)/√(1 - v²/c²) and C=1/√(1 - v²/c²), respectively. As v changes, this triangle changes shape, but the relation A² = 1 = C² - B² always holds. This is not the formula for a unit circle, but for a unit hyperbola. (A general hyperbola has the formula (x/a)² - (y/b)² = ±1, with the change in sign corresponding to a change of orientation.)

Just like points on a unit circle, x² + y² = 1, might by parametrized by any of
(x,y) = (±cos θ, ±sin θ) or (x,y) = (±sin φ, ±cos φ) or (x,y) = (u, ±√(1 - u²)) or (x,y) = (±√(1 - v²), v)
points on the unit hyperbola, x² + y² = 1, might by parametrized by any of
(x,y) = (±cosh ζ, ±sinh ζ) or (x,y) = (±sec ξ, ±tan ξ) or (x,y) = ((s²+1)/(2s),(s²-1)/(2s)) or (x,y) = (±(w²+1)/(1-w²),(2w)/(1-w²))
but none of these parametrization change the geometry.
If x = E/(mc²) = 1/√(1 - v²/c²) and y = p/(mc) = (v/c)/√(1 - v²/c²) and y/x = v/c, then these four parametrizations naturally suggest (up to a change of signs):
v/c = sinh ζ/cosh ζ = tanh ζ or ζ = arctanh(v/c) = ln √((1+v/c)/(1-v/c))
v/c = tan ξ/sec ξ = sin ξ or ξ = arcsin(v/c) = -i ln(iv/c+√(1 - v²/c²))
v/c = (s²-1)/(s²+1) or s = √((1+v/c)/(1-v/c)) = e^arctanh(v/c)
v/c = 2w/(w²+1) or w = c(1+√(1-v²/c²))/v= e^(-2 arccoth√((1+v/c)/(1-v/c)))

Opinions may vary as to which of these expressions is most beautiful to the eye. But the expression in hyperbolic trigonometric functions has a certain advantages.

ζ can take on any real value, and when v/c is small, ζ is just about equal to v/c. One more property we would like is that ζ makes math as simple as math with the reals, just like how θ makes math on the unit circle easy. Stay tuned for further development of this idea.

So we have now, for m > 0, a triangle with sides mc², mc² sinh ζ and mc² cosh ζ, when v/c = tanh ζ

This post has been edited by rpenner on Feb 12 2010, 11:31 PM

This is, without doubt, the best thread on the forum. I hope all members will make the effort to learn from it.

Kudos, rpenner. I realise that these type of endeavours initially feel like a good idea, but then seem too much like hard work. Good on you for seeing it though.

It's so hard to type up these trig identities and double-check them all :-( Plus, I know I'm taking liberties with branch cuts.

Unfortunately, I think you and Euler are wrong (not incorrect) but very few can get anything from your burden. It is true, that Universities use different text books from the same courses time to time.

I would more like "Book recommendations" anew.

That would be a better sticky IMO.

This post has been edited by Beer w/Straw on Feb 13 2010, 12:56 AM

arccosh x = ln(x + √(x - 1)√(x + 1)) is Mathematica's prefered definition.

arctanh x = (1/2) (ln(1+x) - ln(1-x)) is Mathematica's prefered definition.

likewise:
arccos x = (√(1 - x)/√(x - 1)) arccosh(x)
arccos x = i ln(-i) + i ln(ix + √(1 - x²)) and
arctan x = (i/2)(ln(1-ix)-ln(1+ix)) are preferred by Mathematica. Branch cuts are subtle.

Or as Mathematica would have it:
arccosh(a)±arccosh(b) = arccosh(ab±√(a-1)√(a+1)√(b-1)√(b+1))

This post has been edited by rpenner on Feb 13 2010, 04:33 AM

Relativistic 2-body collisions
A large body of physics is set up around analysis of two bodies which start far apart, and move freely, come close together, and then from the point where they were close together, two bodies fly apart to such a distance that they can be said to move freely. Such a close encounter is called a collision, in analogy with traffic accidents, etc. For all collisions, the total energy is conserved, and the total momentum is conserved. For elastic collisions, the masses of the particles are unchanged.

So for a completely general collision in the lab frame, we have:

Sum of initial energy-momentum = Sum of final energy momentum

Any way you slice it, you have 12 more unknowns than equations. Often, when we want to study the collision itself, we don't care about the total velocity of the system in the lab frame or its orientation to lab-chosen coordinates. But some of these we can make go away by working in the center-of-mass-energy frame, because while in the lab frame, we have (v_x, v_y, v_z) = (p_1x + p_2x, p_1y + p_2y, p_1z + p_2z) c²/(E_1 + E_2), in the center-of-mass-energy frame for the same collision, this is zero both before and after. This is not changing the collision, just our description of it in our coordinate basis.

We can also do a rotation so that p''_1y and p''_1z are zero, and so the collision happens dead-on the x axis. We can do a third rotation to ensure p''_3z is zero. And so, in the center-of-mass frame which has been rotated post hoc to align with our collision, we have 4 less net unknowns. And if we parameterize the center-of-mass-energy angle between (p_1x, 0) and (p_3x, p_3y) as θ, (which is independent of our choice of axes) then we can write:

E_1 + E_2 = E_3 + E_4
p_1x + p_2x = 0 = p_3x + p_4x
p_1y = p_2y = 0 = p_3y + p_4y
p_1z = p_2z = 0 = p_3z = p_4z
p_3x = p_3 cos θ
p_3y = p_3 sin θ

Substituting in our relativistic relation between E, p and m, we get
E_1 = √(m²_1 c⁴ + p²_1x c²)
E_2 = √(m²_2 c⁴ + p²_1x c²)
E_3 = √(m²_3 c⁴ + p²_3 c²)
E_4 = √(m²_4 c⁴ + p²_3 c²)

For a total of 6 more unknowns than equations. With the exception of the angle θ, the 2-body collision in the center-of-mass-energy frame greatly resembles collisions in 1 dimension. Things 1 and 2 come together, then things 3 and 4 fly apart.

Mandelstam variables in the center-of-mass frame
s is the center-of-mass energy, squared.
t describes the energy and momentum transfer.
u describes the crossed-over energy and momentum transfer

Often you will find collider experiments described in terms of √s -- such as proton-antiproton collisions at the Tevatron at √s = 1.96 TeV . That is the same as saying antiprotons of 0.98 TeV hit head-on protons of 0.98 TeV, when the lab frame is the same as the center-of-mass-energy frame. What's nice about s, t and u is that they are Lorentz scalars -- unchanged by coordination changes like rotations and choice of inertial reference frame.

General Relations for 2 body collisions
E_1 + E_2 = E_3 + E_4
p_1x + p_2x = p_3x + p_4x
p_1y + p_2y = p_3y + p_4y
p_1z + p_2z = p_3z + p_4z
E_1 = √(m²_1 c⁴ + p²_1x c² + p²_1y c² + p²_1z c²)
E_2 = √(m²_2 c⁴ + p²_2x c² + p²_2y c² + p²_2z c²)
E_3 = √(m²_3 c⁴ + p²_3x c² + p²_3y c² + p²_3z c²)
E_4 = √(m²_4 c⁴ + p²_4x c² + p²_4y c² + p²_4z c²)
s = (E_1 + E_2)² - (p_1x + p_2x)²c² - (p_1y + p_2y)²c² - (p_1z + p_2z)²c²
t = (E_1 - E_3)² - (p_1x - p_3x)²c² - (p_1y - p_3y)²c² - (p_1z - p_3z)²c²
u = (E_1 - E_4)² - (p_1x - p_4x)²c² - (p_1y - p_4y)²c² - (p_1z - p_4z)²c²
s + t + u = m²_1 c⁴ + m²_2 c⁴ + m²_3 c⁴ + m²_4 c⁴

motion of center of mass-energy : (v_x, v_y, v_z) = (p_1x + p_2x, p_1y + p_2y, p_1z + p_2z) c² / ( E_1 + E_2 )

Simplification -- working in the center of mass frame with x aligned with the initial direction of collision, and scattering in the x-y plane
p_2x = - p_1x
p_4x = - p_3x
p_4y = - p_3y
p_1y = p_2y = p_1z = p_2z = p_3z = p_4z = 0
p_3x = p_3 cos θ
p_3y = p_3 sin θ

E_1 + E_2 = E_3 + E_4
E_1 = √(m²_1 c⁴ + p²_1x c²)
E_2 = √(m²_2 c⁴ + p²_1x c²)
E_3 = √(m²_3 c⁴ + p²_3 c²)
E_4 = √(m²_4 c⁴ + p²_3 c²)
s = (E_1 + E_2)²
t = (E_1 - E_3)² - (p_1x - p_3 cos θ)²c² - (p_3 sin θ)²c²
u = (E_1 - E_4)² - (p_1x + p_3 cos θ)²c² - (p_3 sin θ)²c²

Simplification -- Totally inelastic collision in the center-of-mass-energy frame

Imagine a collision so inelastic, the bodies merge. Alternately, imagine Body 4 is wholly negligible in mass, energy and momentum. As a consequence, θ is undefined.
What's left:

m_4 = 0
E_4 = 0
p_3 = 0
p_2x = - p_1x
p_3x = p_4x = p_1y = p_2y = p_3y = p_4y = p_1z = p_2z = p_3z = p_4z = 0
E_3 = E_1 + E_2 = m_3 c²
E_1 = √(m²_1 c⁴ + p²_1x c²)
E_2 = √(m²_2 c⁴ + p²_1x c²)
s = (E_1 + E_2)² = E²_3 = m²_3 c⁴
t = m²_2 c⁴
u = m²_1 c⁴

If we know 4 of E_1 (or E_2), p_1x (or p_2x), s (or m_3 or E_3), t (or m_2), or u (or m_1) then we know everything about the collision.

Simplification -- Elastic collision in the center-of-mass-energy frame
We identify objects 3 and 4 as "the same" as objects 1 and 2, respectively. m_1 = m_3, m_2 = m_4

m_3 = m_1
m_4 = m_2
p_3 = p_1x
E_3 = E_1
E_4 = E_2
p_2x = - p_1x
p_4x = - p_3x
p_4y = - p_3y
p_1y = p_2y = p_1z = p_2z = p_3z = p_4z = 0
p_3x = p_1x cos θ
p_3y = p_1x sin θ
E_1 = √(m²_1 c⁴ + p²_1x c²)
E_2 = √(m²_2 c⁴ + p²_1x c²)
s = (E_1 + E_2)²
t = - 2 p²_1x(1 - cos θ)c²
u = (E_1 - E_2)² - 2 p²_1x(1 + cos θ)c²

Simplification -- Elastic collision of same-mass objects in the center-of-mass-energy frame

m_1 = m_2 = m_3 = m_4
p_3 = p_1x
E_1 = E_2 = E_3 = E_4
p_2x = - p_1x
p_4x = - p_3x
p_4y = - p_3y
p_1y = p_2y = p_1z = p_2z = p_3z = p_4z = 0
p_3x = p_1x cos θ
p_3y = p_1x sin θ
E_1 = √(m²_1 c⁴ + p²_1x c²)
s = 4 E²_1
t = - 2 p²_1x(1 - cos θ)c²
u = - 2 p²_1x(1 + cos θ)c²

Simplification, Working in one dimension, Elastic collision
p_1y = p_2y = p_3y = p_4y = 0
p_1z = p_2z = p_3z = p_4z = 0
m_1 = m_3
m_2 = m_4

V = (p_1x + p_2x)c²/(E_1 + E_2) = (p_3x + p_4x)c²/(E_3 + E_4)

E_1 + E_2 = E_3 + E_4
p_1x + p_2x = p_3x + p_4x
E_1 = √(m²_1 c⁴ + p²_1x c² )
E_2 = √(m²_2 c⁴ + p²_2x c² )
E_3 = √(m²_1 c⁴ + p²_3x c² )
E_4 = √(m²_2 c⁴ + p²_4x c² )
s = (E_1 + E_2)² - (p_1x + p_2x)²c²
t = (E_1 - E_3)² - (p_1x - p_3x)²c²
u = (E_1 - E_4)² - (p_1x - p_4x)²c²
s + t + u = 2m²_1 c⁴ + 2m²_2 c⁴

Neglecting the trivial solution of no interaction, if we are given m_1, m_2, p_1x and p_2x, we have:

p_3x = ( 2 E_1 E_2 (m²_2 - m²_1) (p_1x + p_2x) + (m⁴_1 - m⁴_2) c⁴ p_1x - 2 m²_1 m²_2 c⁴ p_2x - 2 (m²_1 (p_1x + 2 p_2x) + m²_2 p_1x ) (p_1x + p_2x) p_2x c² ) / ( (m²_1 - m²_2)² c⁴ - 4 (p_1x + p_2x)( m²_1 p_2x + m²_2 p_1x) c² )

p_4x = p_1x + p_2x - p_3x

(Maybe)

Other sites have information like this pinned. Can you do this?

Building on this, we can construct a number of cases where the math of Special Relativity is not scary for the pen-and-paper set.

1) Pick two positive integers, d and e, and their sum f. Then 0 < d < f.
2) Compute n = f² − d², o = 2df, p = f² + d². Then n² + o² = (f² − d²)² + (2df)² = f⁴ − 2d²f² + d⁴ + 4d²f² = (f² + d²)² = p², and n,o,p form the sides of a right triangle.
3) Then, β = v/c = n/p = (f² − d²)/(f² + d²) gives γ = 1/√(1 - v²/c²) = 1/√(1 - (f² − d²)²/(f² + d²)²) = (f² + d²)/(2df) = p/o. Likewise v/c = o/p gives γ = p/n.

http://web.mit.edu/course/3/3.091/www3/constants.html