Undergrad Deriving Notations for Differentials in Time-Evolving Systems

  • Thread starter Thread starter Apashanka
  • Start date Start date
  • Tags Tags
    Differentials
Click For Summary
The discussion focuses on deriving notations for differentials in time-evolving systems, specifically examining the relationship between evolving coordinates (x) and constant coordinates (X). A key equation presented is the time derivative of the Jacobian, resulting in the expression for the change in J, which is consistent with the divergence of velocity. The participants explore the definition and application of the operator d/dx in the context of differential forms, questioning its validity without a special definition. They also discuss the total derivative in relation to time series, emphasizing the dependency of velocity components on spatial and temporal variables. The conversation concludes with attempts to equate differentials and derivatives, highlighting the complexity of the operations involved.
Apashanka
Messages
427
Reaction score
15
if ##d^3x=Jd^3X...(1) ## where ##x's## evolves with time and ##X's## are constt. and ##x_i=f(X_i)##(for ##i^{th}## coordinate) where the functional form of ##f(X_i)## changes with the time evolution of ##x_i##.
Now taking time derivative of (1) and dividing throughout by (1) it is coming ##\dot J=J(\nabla•v)##(##x## and ##X## are coordinates) which is consistent.
But another thing is approximated while doing this ##\frac{d}{dt}(dx_i)=d(\frac{dx_i}{dt})=dv_i## actually I try to prove it by hand but can't...
Will anyone provide me any hints...
 
Physics news on Phys.org
D , as usually defined acts on differential forms which includes functions, but the expression d/dx is neither unless you have a special definition. How do you define this operation?
 
WWGD said:
D , as usually defined acts on differential forms which includes functions, but the expression d/dx is neither unless you have a special definition. How do you define this operation?
##\frac{d}{dt}## is defined as ##\frac{\partial }{\partial t}+(v•\nabla)##
 
This is defined for time series ( for context) or a general definition?
 
WWGD said:
This is defined for time series ( for context) or a general definition?
Here is the case where ##i^{th}## component of velocity depends on ##v_i(x,y,z,t)## ,now ##\frac{d v_i}{dt}=\frac{\partial v_i}{\partial t}+(v•\nabla)v_i## for which ##\frac{d}{dt}=\frac{\partial }{\partial t}+v•\nabla## using this I tried to find ##d(\frac{dx}{dt})=dv_x## and trying to equate it to ##\frac{d}{dt}(dx)##...but can't...
 
Okk I have done it like this ,is it the correct procedure??
##\frac{d}{dt}(dx)=\frac{d}{dt}(v_x(x,y,z,t)dt)=\frac{dv_x(x,y,z,t)}{dt}dt=dt(\frac{\partial v_x}{\partial t}+(v•\nabla)v_x)=dt[\frac{\partial v_x}{dt}+(v_x\frac{\partial }{dx}+v_y\frac{\partial }{dy}+v_z\frac{\partial }{dz})v_x]##

And taking ##d(\frac{dx}{dt})=dv_x(x,y,z,t)=\frac{\partial v_x}{\partial x}dx+\frac{\partial v_x}{\partial y}dy+\frac{\partial v_x}{\partial z}dz+\frac{\partial v_x}{\partial t}dt## which are same...
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K