Recent content by deedsy

  1. deedsy

    Wave in inhomogeneous medium (ray equation)

    ok, without including the (ds) I was able to get the answer! ##\frac{d}{ds} (n\hat{t}) = \nabla n\hat{j}## Thanks so much for your help TSny
  2. deedsy

    Wave in inhomogeneous medium (ray equation)

    actually, I'm not sure i need the (ds) in the unit vector equation
  3. deedsy

    Wave in inhomogeneous medium (ray equation)

    ##\frac{d\hat{t}}{ds} = \frac{d\hat{t}}{d\theta} \frac{d\theta}{ds}## ##\frac{d\hat{t}}{d\theta} \frac{d\theta}{ds} = \frac{-d\hat{t}}{d\theta} \frac{\nabla n sin(\theta)}{n}## ##\frac{d\hat{t}}{ds} = -\frac{\nabla n sin(\theta)}{n} \frac{d\hat{t}}{d\theta}## so now I need my expression...
  4. deedsy

    Wave in inhomogeneous medium (ray equation)

    wow, yeah, so after the real product rule, I have ##\frac{d}{ds} (n\hat{t}) = \nabla n cos(\theta) \hat{t} + n \frac{d\hat{t}}{ds}## so I need an expression that can show how ##\hat{t}## changes with ##s##, but I'm having trouble with that. You said before you found an expression between...
  5. deedsy

    Wave in inhomogeneous medium (ray equation)

    Okay, so I'm going to reduce ##\frac{d}{ds} (n \hat{t})## down to ##\vec{\nabla}n## ##\frac{d}{ds} (n \hat{t})## ##\frac{dn}{ds} + \frac{d\hat{t}}{ds}## Using a previous result.... ##\frac{-A cos(\theta)}{sin^2(\theta)} \frac{d\theta}{ds} + \frac{d\hat{t}}{ds}## Using another previous result...
  6. deedsy

    Wave in inhomogeneous medium (ray equation)

    thanks TSny, that tip led me to the right answer! Now I'm trying to verify the ray equation ##\frac{d}{ds} (nt) = \nabla n## with that result; so far, ##\frac{d\theta}{ds} = \frac{-(dn/dy) sin(\theta)}{n}## since n is only dependent on the y direction, ##\nabla n = \frac{dn}{dy}##, so...
  7. deedsy

    Wave in inhomogeneous medium (ray equation)

    Homework Statement A wave travels in a stratified medium whose index of refraction is a function of the coordinate y. Show that the angle ##\theta## between a ray and the y-axis obeys the following law: ## \frac{d\theta}{ds} = \frac{-(dn/dy) sin(\theta)}{n} ## , where the distance s is measured...
  8. deedsy

    Work done by climbing stairs

    But like you said before, when you are walking you are propelling yourself forward; this wouldn't be possible without friction. But since your foot is not slipping while you push (no displacement), the static friction is doing no work.
  9. deedsy

    Work done by climbing stairs

    You'd also have to assume there is no air resistance
  10. deedsy

    Work done by climbing stairs

    I don't believe so, as long as you aren't skidding/shuffling while you walk (no frictional forces over a distance)
  11. deedsy

    Work done by climbing stairs

    there will be a normal force out of the stair as you walk on them. However, since your foot is situated, no frictional force is moving through any distance, so no Work is being done there.
  12. deedsy

    Work done by climbing stairs

    the fancy way to write Work is ##W=\int F \cdot d\vec{r} ## So, my understanding is, The force is all in the vertical direction due to gravity (there's no mention of any frictional effects on the stairs in the problem), so when the person is walking in the horizontal direction, the angle...
  13. deedsy

    Splitting Fractions (Integrals)

    this can be solved using a trig substitution for x. Can you think of what this substitution should be to simplify the denominator?
  14. deedsy

    Mathematics useful for Physics classes

    E&M - vector analysis is useful to understand all the derivations Quantum - all kinds of integrals, linear algebra/matrices, spherical harmonics, Dirac notation Thermo - so far, a lot of partial derivatives
  15. deedsy

    Dark energy and type 1a supernovae?

    Here's a graph of apparent (observed) magnitude and redshift. You can see that relative to a matter dominated universe (\Omega_\Lambda=0) a supernova at a given redshift appears dimmer (higher apparent magnitude) in an accelerating universe that contains dark energy (\Omega_M = 0.25...
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