Notations for differentials

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  • #1
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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....
 

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  • #2
WWGD
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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?
 
  • #3
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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)##
 
  • #4
WWGD
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This is defined for time series ( for context) or a general definition?
 
  • #5
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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.....
 
  • #6
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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...
 

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