p = m v;
vs. rest energy
momentum: μ Sinh[θ] μ β
energy: μ Cosh[θ]
Series[Cosh[β],{β,0,4}]
In[]:=
1+
2
β
2
+
4
β
24
+
5
O[β]
Out[]=
μ = m c
​
If KE is μ v^2/c^2
If energy is m c^2 ( 1 - β^2 + ...)
Series[mc^2/Sqrt[1-v^2/c^2],{v,0,6}]
In[]:=
2
c
m+
m
2
v
2
+
3m
4
v
8
2
c
+
5m
6
v
16
4
c
+
7
O[v]
Out[]=

​

E^2 - p^2
For a speed 0 edge, E(β) = μ c Cosh[β] (
Factor of c
Between two elementary time hypersurfaces, difference in the spatial position for the edge is c Δt
We are counting causal edges per unit area defined by the causal graph
Measuring a flux through timelike surfaces we have to account for the direction of the flow;
for the spacelike surface we don’t
​
​
Out[]=
vertical distance in every square is Δt; horizontal distance is c Δt
1 elementary light cone filled with stuff has spatial extent c Δt Cosh[β] / Δt
Δt Sinh[β] / Δt
Imagine the vertical flux is μ ( per elementary light cone ) [ number of causal edges inside a single elementary light cone ]
Series[Cosh[β],{β,0,2}]
In[]:=
1+
2
β
2
+O(
3
β
)
Out[]=
Series[Sinh[β],{β,0,4}]
In[]:=
β+
3
β
6
+
5
O[β]
Out[]=
E = c μ ( 1 + 1/2 v^2/c^2 + ...)
p = μ v/c
μ = m c
(mCosh[b])^2-c^2(cm)^2Sinh[b]^2//FullSimplify
In[]:=
2
m
(
2
Cosh[b]
-
4
c
2
Sinh[b]
)
Out[]=
Sinh[ArcTanh[b]]//FullSimplify
In[]:=
b
1-
2
b
Out[]=