f

content/physics/2023-10-29-physics.md

@@ -2,7 +2,7 @@
 title = "Physics"
 author = ["littlehome"]
 date = 2023-10-29
-lastmod = 2023-10-30T17:22:49+09:00
+lastmod = 2023-10-30T17:55:43+09:00
 draft = false
 +++
 
@@ -19,6 +19,7 @@ Summary of lecture 1 - 10.
     Whereas aristoteles would have said, you need a force to move an object.
 2.  \\( F = ma \\)
     This doesn't survive QM.
+    You need an inertial observer. (ie, one who has no accerelation)
 3.  \\( F\_{12} = - F\_{21}\\)
 
 
@@ -29,7 +30,7 @@ Summary of lecture 1 - 10.
 &  v \frac{dv}{dt} = a v \\\\
 & \frac{d}{d t}\left(\frac{v^2}{2}\right)=a \frac{d x}{d t} \\\\
 & d\left(\frac{v^2}{2}\right)=a d x \\\\
-& \int a\left(\frac{v^2}{2}\right)=\int a d x \\\\
+& \int d \left(\frac{v^2}{2}\right)=\int a d x \\\\
 & \frac{v^2}{2}-\frac{v\_0^2}{2}=a\left(x-x\_0\right) \\\\
 \end{align\*}
 
@@ -57,6 +58,8 @@ We are considering a motion where rotation center is fixed. Then, analogue of di
 p = m v && L = I w \\\\
 \end{alignat\*}
 
+{{< figure src="/ox-hugo/circle.png" width="150px" height="150px" >}}
+
 \begin{align\*}
   & \vec{R}=R(\vec{i} \cos \omega t+\vec{\jmath} \sin \omega t) \\\\
   & \frac{d \vec{R}}{d t}=R(-\omega \vec{i} \sin \omega t+ \omega \vec{\jmath} \cos \omega t) \\\\
@@ -77,7 +80,7 @@ We have an analogue of \\( F = m a\\) in a circular motion.
 \end{align\*}
 
 
-## conservation of angular momentum {#conservation-of-angular-momentum}
+## Conservation of angular momentum {#conservation-of-angular-momentum}
 
 Without external torque, angular momentum is preserved.
 
@@ -90,7 +93,7 @@ m v\_{1} = m v\_{2} && I w\_{1} = I w\_{2} \\\\
 
 ## Question {#question}
 
-Newton's 2nd law is a conjecture. and Testing it needs to use the conjecture as well.
+Newton's 2nd law is a conjecture. and testing it needs to use the conjecture as well.
 If we use spring as discussed in the lecture to get the \\(F = -kx\\), deriving it (by experimentation) would also have used the law.
 
 So using the two equation as in the lecture beats the purpose of showing \\( F = ma \\), in a sense that we don't know what the true \\( F \\) is for a string.
@@ -120,7 +123,7 @@ What happens to internal forces and their effect? they all cancel out.
 
 conservation of linear momentum says \\( mv \\) is constant
 conservation of angular momentum says \\( Iw \\) is constant
-But \\( Iw = m r^2 \* frac{v}{r} = m r v\\), so it says a different quantity is constant.
+But \\( Iw = m r^2 \* \frac{v}{r} = m r v\\), so it says a different quantity is constant.
 
 Now if i were, given an instantaneous information of m and v, I can tell its linear momentum,
 but I can't tell the angular momentum unless I need a frame of reference (whree the rotation axis is).
@@ -131,6 +134,8 @@ It turns out, you need also need a frame of reference for the linear momentum.
 You need an inertial frame of reference for linear momentum.
 You need an rotaional(?) frame of reference for angular momentum.
 
+Also, as pointed out later, \\( m r v \sin\theta \\) is conserved.
+
 And so it happens there are deeper explanation which I don't yet understand:  Noether's theorem. The conservation of linear momentum corresponds to the translational symmetry of space, while the conservation of angular momentum corresponds to its rotational symmetry
 
 
@@ -172,13 +177,15 @@ No, if there's no axis of rotation, we can't talk about rotation, so torque is n
 
 1.  if there's no rotation, you can take any point as axis of rotation.
 
-    \\[ \sum F\_{i} (x\_{i} +a ) = \sum {F\_{i} X\_{i}} + a \sum F\_{i} \\]
+    \\[ \sum F\_{i} (x\_{i} +a ) = \sum {F\_{i} x\_{i}} + a \sum F\_{i} \\]
      i.e. you can translate by any a, when \\( \sum F\_{i} = 0 \\)
 
 2.  angular momentum can be defined when there's no rotation.
+
     \begin{align\*}
-    &amp;&amp; \vec{L} = \vec{R} \* \vec{P} <br />
+    & \vec{L} = \vec{r} \times \vec{p} \\\\
+    & \frac{d \vec{L}}{d t} = \frac{d \vec{r}}{d t} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{d t} \\\\
+    & L = | \vec{r} \* \vec{p} |  = r p \sin \theta
+    \end{align\*}
 
-&amp;&amp; \frac{d \vec{L}}{d t} = \frac{d \vec{R}}{d t} \* \vec{P} + \vec{R} \* \frac{d \vec{P}}{d t} <br />
-\end{align\*}
-![](/ox-hugo/angular.png)
+{{< figure src="/ox-hugo/angular.png" caption="<span class=\"figure-number\">Figure 1: </span>Angular momentum is constant if object is moving linearly" width="150px" height="150px" >}}

hugo.toml

@@ -35,6 +35,10 @@ theme = 'hugo-book'
   name = "accounting"
   url = "/accounting"
   weight = 7
+[[menu.after]]
+  name = "physics"
+  url = "/physics"
+  weight = 7
 
 [params]