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Candy Theory of Relativity

Categories: Candy,Candy Recipes,Chocolate Candy

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Sugar molecule

(Image Courtesy of The Exploratorium)

Many people are probably aware of Einstein’s Theory of General Relativity, but few are likely aware of its relation to candy. When this scientific tenet is applied to candy, the theory states that a candy’s taste is relative to its surroundings.

Okay, so maybe that’s a bit of a stretch, but it is important to note that candy is affected by its surroundings. One important aspect people often ignore is temperature. Commonly, most people eat candy in this manner: you’re driving and suddenly get a craving, park at the nearest 7-11, buy that Twix staring at you from the candy shelf, and have consumed half the bar by the time you’re back in car.

Instances like this are examples of candy being consumed at room temperature. There’s nothing wrong with this way of eating candy, and in fact, candy like chocolate is often considered to taste best at room temperature. But have you ever tried a Twix frozen? I have, and I must confess it’s a great way to enliven your favorite confections that might have lost their zest over the years.

And frozen isn’t just the only temperature to explore. Just cooling your favorite candy by sticking it in a fridge can add unexpected delights. Chocolate has a great feeling in your mouth when it’s served a few degrees below room temperature. Even gummy candies are transformed into chewier eats the more they’re chilled.

Don’t stop with cold. Explore the other spectrum of heat! Big Hunk advertises on its wrapper that you should try heating them for a little bit in the microwave to get a great melty, gooey surprise. You can even venture to the complete extreme and melt down your favorite chocolate bar as a sauce to pour on top of your favorite ice cream.

Whether you’re heating up a Peep or cooling a Hershey’s Kiss, candy is a dynamic food that needs to be tasted in all its states. Sometimes, room temperature is the best fit because not all candies work in all temperatures (never freeze a 3 Musketeers!), but you’ll never know without a little experimentation. So the next time you reach for your favorite candy, give it a try and think outside of the box (or wrapper as is usually the case)!



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8 Responses to “Candy Theory of Relativity”

  1. 1
    Rosa says:

    Frozen Charleston Chews are a revelation!

  2. 2
    Brian says:

    On Robby’s advice, I tried freezing Twix Java bars and they are great that way! I would have never tried that without Robby’s advice!

  3. 3
    Sania says:

    Why shouldn’t you freeze a 3 Musketeers?

  4. 4
    Wolfidy says:

    I have frozen a 3 Musketeers. It was really hard to eat. I think ifI wanted to try it again, I would do those little bite size square ones instead of the whole full-size bar.

  5. 5
    Robby says:

    The problem with a 3 Musketeers is, in my opinion, the nougat. I don’t like nougat chilled because it becomes incredibly hard and tastes weird. It’s almost like a more extreme version of the frozen Charleston Chew. I say try it if you want to know what I am talking about, but I personally avoid freezing any candy with nougat or marshmallow. I suggest basic chocolates, liquid centers, caramel-filling, or peanut butter to work best.

  6. 6
    Jeni says:

    I love frozen Reese’s! Also frozen cherry cordial Kisses. Yum :)

  7. 7
    robert says:

    Freeze the 3 musketeers bar, before you unwrap it hit it with a hammer and shatter it. Then you have delectable little melt-in-your-mouth pieces! Try this method on milky way, too.

  8. 8
    joe nahhas says:

    Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution
    The problem that the 100,000 PHD Physicists could not solve

    This is the solution to the “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney
    Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics
    For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

    Universal- Mechanics

    All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
    r = r (x, y, z). The state of any object in the Universe can be expressed as the product

    S = m r; State = mass x location

    P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

    = change of location + change of mass

    = m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

    F = d P/d t = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r

    = m γ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

    In polar coordinates system

    r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

    F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m’[r'r(1) + rθ'θ(1)] + (m”r) r(1)

    F = [d²(m r)/dt² - (m r)θ'²]r(1) + (1/mr)[d(m²r²θ')/d t]θ(1) = [-GmM/r²]r(1)

    d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

    Let m =constant: M=constant

    d²r/dt² – r θ’²=-GM/r² —— I

    d(r²θ’)/d t = 0 —————–II

    r²θ’=h = constant ————– II
    r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ’(d u/d θ) = -h (d u/d θ)
    d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ’(d²u/d θ²) = – (h²/r²)(d²u/dθ²)
    [- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²
    2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²
    d²u/dθ² + u = GM/h²
    r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)
    r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]
    r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

    r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

    r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution

    If λ(r) ≈ 0; then:

    r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

    θ’(r, t) = θ’[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}

    h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
    h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

    θ’ (0,0) = h/r²(0,0) = 2Ï€[√(1-ε²)]/T(1-ε)²
    θ’ (0,t) = θ’(0,0)Exp(-2ỉwt)={2Ï€[√(1-ε²)]/T(1-ε)²} Exp (-2iwt)

    θ’(0,t) = θ’(0,0) [cosine 2(wt) - ỉ sine 2(wt)] = θ’(0,0) [1- 2sine² (wt) - ỉ sin 2(wt)]
    θ’(0,t) = θ’(0,t)(x) + θ’(0,t)(y); θ’(0,t)(x) = θ’(0,0)[ 1- 2sine² (wt)]
    θ’(0,t)(x) – θ’(0,0) = – 2θ’(0,0)sine²(wt) = – 2θ’(0,0)(v/c)² v/c=sine wt; c=light speed

    Δ θ’ = [θ'(0, t) - θ'(0, 0)] = -4Ï€ {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second
    {(180/Ï€=degrees) x (36526=century)

    Δ θ’ = [-720x36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

    This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

    The circumference of an ellipse: 2Ï€a (1 – ε²/4 + 3/16(ε²)²—) ≈ 2Ï€a (1-ε²/4); R =a (1-ε²/4)
    v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

    v = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her
    Let m = mass of primary; M = mass of secondary

    v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]
    v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com

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