Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Finding Lorentz acceleration transformation for arbitrary direction
Reply to thread
Message
[QUOTE="TSny, post: 6595232, member: 229090"] I was able to reduce the above equation to the expression that you want to get, namely $$\vec a ' = \frac{\vec a}{\gamma^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2}\right)^2} - \frac{(\gamma - 1) (\vec a \cdot \vec u) \vec u}{\gamma^3 u^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2} \right)^3} + \frac{(\vec a \cdot \vec u) \vec v}{\gamma^2 c^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2}\right)^3}.$$ So, check over your work starting from your expression for ##a'## that I quoted above. You can verify that your final answer is incorrect by considering the case where ##\vec v## and ##\vec u## are perpendicular, so that ##\vec v \cdot \vec u = 0##. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Finding Lorentz acceleration transformation for arbitrary direction
Back
Top