Solving Galilean Transform Homework with Gradients in (u,v)

  • Thread starter Thread starter MarkovMarakov
  • Start date Start date
  • Tags Tags
    Galilean Transform
Click For Summary
SUMMARY

The discussion focuses on transforming variables in a Galilean transformation context, specifically from \((\vec x(t), t)\) to \((\vec u = \vec x + \vec a(t), v = t + b)\). The key expression to rewrite involves the gradient \(\nabla_{\vec x} f(\vec x, t) + \frac{d^2 \vec a}{dt^2}\). The transformation of the first term is established as \(f(\vec x, t) \to f(\vec u - \vec a, v - b)\). The second term requires the application of the chain rule, leading to the formulation \(\frac{\partial}{\partial x} = \frac{\partial u}{\partial x} \frac{\partial}{\partial u}\).

PREREQUISITES
  • Understanding of Galilean transformations
  • Familiarity with gradient notation and vector calculus
  • Knowledge of the chain rule in multivariable calculus
  • Basic concepts of variable substitution in physics
NEXT STEPS
  • Study the application of the chain rule in vector calculus
  • Explore advanced topics in Galilean transformations
  • Learn about the implications of variable changes in physics
  • Investigate the use of gradients in different coordinate systems
USEFUL FOR

Students and professionals in physics, particularly those studying mechanics and transformations, as well as mathematicians focusing on calculus and vector analysis.

MarkovMarakov
Messages
32
Reaction score
1

Homework Statement


If there is a change of variables:
(\vec x(t),t)\to (\vec u=\vec x+\vec a(t),\,\,\,v=t+b) where b is a constant.

Suppose I wish to write the following expression in terms of a gradient in (\vec u, v)

\nabla_\vec x f(\vec x,t)+{d^2\vec a\over dt^2} How do I do that?

Homework Equations


Please see above.

The Attempt at a Solution


For the first term, I think
f(\vec x, t)\to f(\vec u -\vec a, v-b)
I am not sure what to do with the second term though.
 
Physics news on Phys.org
You should use chain rule I think, so

\frac{\partial }{\partial x}=\frac{\partial u}{\partial x}\frac{\partial}{\partial u}

If I understood your question, this is what you are looking for.
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
18
Views
3K