/sci/ - Science & Math

[Anonymous No.5689495]

<div class="math">\vec{F}-m\vec{a}=0</div>

[Anonymous No.5689500]

<div class="math">\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{q}_{i}}-\frac{\partial L}{\partial q_{i}}=0</div>

[Anonymous No.5689501]

y - mx - b = 0

F - ma = y - mx - b

b = c^2 - a^2

F - ma = y - mx - c^2 + a^2

a = d/dx v

F - ma = y - mx - c^2 + d/dxv

[Anonymous No.5689507]

<span class="math">\dot\rho=-{\frac{i}{\hbar}}[H,\rho]+\sum_{n,m = 1}^{N^2-1} h_{n,m}(L_n\rho L_m^\dagger-\frac{1}{2}\left(\rho L_m^\dagger L_n + L_m^\dagger L_n\rho\right))<span class="math">[/spoiler][/spoiler]

[Anonymous No.5689510]

If v<<u and u<<m then v<<m and
dv/dm = dv/du * du/dm m-almost everywhere

[Anonymous No.5689541]

<div class="math">\vec{E}+\nabla \Phi=0</div>

[Anonymous No.5689602]

Take our action <span class="math"> S\left[ \vec{r} \right] = \int_0^t \left[ \frac{m \vec{v}^2}{2} - V(\vec{r}) \right] \mathrm{d}t [/spoiler] and calculate the first variation, considering a small deviation, <span class="math"> \delta \vec{r} [/spoiler] from our classical path <span class="math"> \vec{r}_\mathrm{cl} [/spoiler]

<div class="math"> S \left[ \vec{r}_{\mathrm{cl}} (t) + \delta(\vec{r} (t)) \right] = \int_0^t \left{ \frac{m}{2} \left[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \vec{r}_{\mathrm{cl}} + \delta \vec{r} \right) \right]^2 - V \left( \vec{r}_{\mathrm{cl}} + \delta \vec{r} \right) \right} \mathrm{d} t </div>
<div class="math"> \delta r \ \mathrm{small} \ \rightarrow \ \mathrm{disregard} \ \mathcal{O}(\delta r^2) </div>
<div class="math"> = \int_0^t \frac{m}{2} \dot{\vec{r}}_{\mathrm{cl}}^2 + m \dot{\vec{r}}_{\mathrm{cl}} \delta \dot{\vec{r}} - V(\vec{r}_{\mathrm{cl}} ) - \frac{ \partial V}{\partial \vec{r}} - \delta \vec{r} \mathrm{d}t </div>
<div class="math"> = \int_0^t \frac{m}{2} \dot{\vec{r}}_{\mathrm{cl}}^2 - V(\vec{r}_{\mathrm{cl}}) \mathrm{d}t - \int_0^t \left{ m \ddot{\vec{r}}_{\mathrm{cl}} + \frac{\partial V}{\partial \vec{r}} \right} \cdot \delta \vec{r} \mathrm{d}t = \ \mathrm{classical action} - \ \int_0^t \delta S \cdot \delta \vec{r} \mathrm{d}t</div>

every possible fluctuation to vanish, i.e for the first variation to equal zero, we get
<div class="math"> m \ddot{\vec{r}} = -\frac{\partial V}{\partial \vec{r}} </div>
<div class="math"> \vec{F} =m \vec{a} </div>

[Anonymous No.5689608]

Take our action <span class="math"> S\left[ \vec{r} \right] = \int_0^t \left[ \frac{m \vec{v}^2}{2} - V(\vec{r}) \right] \mathrm{d}t [/spoiler] and calculate the first variation, considering a small deviation, <span class="math"> \delta \vec{r} [/spoiler] from our classical path <span class="math"> \vec{r}_\mathrm{cl} [/spoiler]

<div class="math"> S \left[ \vec{r}_{\mathrm{cl}} (t) + \delta(\vec{r} (t)) \right] = \int_0^t \left\{ \frac{m}{2} \left[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \vec{r}_{\mathrm{cl}} + \delta \vec{r} \right) \right]^2 - V \left( \vec{r}_{\mathrm{cl}} + \delta \vec{r} \right) \right\} \mathrm{d} t </div>
<div class="math"> \delta r \ \mathrm{small} \ \rightarrow \ \mathrm{disregard} \ \mathcal{O}(\delta r^2) </div>
<div class="math"> = \int_0^t \frac{m}{2} \dot{\vec{r}}_{\mathrm{cl}}^2 + m \dot{\vec{r}}_{\mathrm{cl}} \delta \dot{\vec{r}} - V(\vec{r}_{\mathrm{cl}} ) - \frac{ \partial V}{\partial \vec{r}} - \delta \vec{r} \mathrm{d}t </div>
<div class="math"> = \int_0^t \frac{m}{2} \dot{\vec{r}}_{\mathrm{cl}}^2 - V(\vec{r}_{\mathrm{cl}}) \mathrm{d}t - \int_0^t \left\{ m \ddot{\vec{r}}_{\mathrm{cl}} + \frac{\partial V}{\partial \vec{r}} \right\} \cdot \delta \vec{r} \mathrm{d}t = \ \mathrm{classical action} - \ \int_0^t \delta S \cdot \delta \vec{r} \mathrm{d}t</div>

every possible fluctuation to vanish, i.e for the first variation to equal zero, we get
<div class="math"> m \ddot{\vec{r}} = -\frac{\partial V}{\partial \vec{r}} </div>
<div class="math"> \vec{F} =m \vec{a} </div>

So the path integral approach approximates to the classical Newtonian view.

[Anonymous No.5689615]

>>5689541

[metric !1tensorX8Q No.5689635]

>>5689608
It doesn't approximate it, it exactly matches it. The path that minimizes the action is exactly the path where F=ma is satisfied.

[metric !1tensorX8Q No.5689644]

>>5689608
>>5689635
Wait, now I'm confused. You said "path integral" at the end, though the rest of your post didn't even mention the path integral. But yes, the Newtonian path where F=ma is satisfied approximates the path integral approach when S>>h.

[Anonymous No.5689701]

>>5689501

F=ma

a^2+b^2=c^2

E=mc^2

y=mx+b

F^2=m^2 (E/m-y^2-(mx)^2+2ymx)

F^2=Em+2yxm^3-y^2m^2-m^4x^2

[Anonymous No.5689735]

1 = x^1 * x^-1

I = AA^-1

EX NIHILO

[Anonymous No.5689736]

>>5689701
<span class="math"> F>E [/spoiler] Due to the alphabet

<span class="math">F=ma[/spoiler]
<span class="math">E=mc^2[/spoiler]
<span class="math">ma>mc^2[/spoiler]
<span class="math">a>c^2[/spoiler]

[Anonymous No.5689741]

>>5689701
Why is this so funny

[Anonymous No.5690859]

[Anonymous No.5690876]

>>5689701
someone call the cops

[Anonymous No.5690887]

>>5689701
hilarious

[Anonymous No.5690907]

>>5689501
>>5689701
did you guys just prove satire is possible to convey through entirely nonverbal language

[Anonymous No.5691784]

What does that mean?