- #1
don_anon25
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Question answered! Thanks for the input!
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Newton's Second Law Explained - Learn the Basics
- Thread starter don_anon25
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- #2
Doc Al
Mentor
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That resistive force is not the only force acting on the object (otherwise it'll just sit there). Don't forget gravity.
- #3
don_anon25
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It moves along the x-axis...I forgot to mention that. So gravity is not taken into consideration?
- #4
Physics Monkey
Homework Helper
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If the object is initially at rest in a frame where the only force acting is proportional to velocity squared, then yes, it is obvious that the particle won't do anything. Somehow, I suspect that the question wan't meant to be that trivial.
- #5
don_anon25
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I think that velocity to be v0, not zero. Otherwise, the problem is trivial. I just wanted to check and make sure I wasn't missing anything 
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(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.)
I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
So that they become:
##
g^{\sigma \rho} \nabla_\sigma \nabla_\rho R ~g_{\mu\nu} + R ~R_{\mu\nu} - \nabla_\mu \nabla_\nu R
##
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