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This is technically a homework thread, but I'm not asking for someone to do all my work for me and solve it completely, all I need is a nudge in the right direction because I'm fairly lost and can't find help anywhere else. If even this is not allowed, then I'm sorry 
The question:
+ Show Spoiler +Find the one dimensional particle motion in the Morse Potential
What I have so far:
+ Show Spoiler +I now need to change the variables to get something integrable, but I'm unsure what to use. For example: doesn't work because I need the du expression to be free of the variable x. I think I need to divide by something but I just can't work out what that is. Any tips or hints would be greatly appreciated, thanks.
Oh, and http://thornahawk.unitedti.org/equationeditor/equationeditor.php is what I use for the equations.
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is awesome32263 Posts
nvm this doens't work ;_;
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use the lagrangefunction.
L=T-U , T beeing kinetic and U potential energy.
this gives you directly the differential equation for the motion.
q beeing x in your case.
The solution of the resulting motion equation depends on your initial values ofc. depending on them you could solve at least the linear part.
You could also expand tp the 2nd order your inital potential around its minimum and approximate the tail to its right. use the solution of one as boundary condition for the other.
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On October 31 2009 09:13 aqui wrote:use the lagrangefunction. L=T-U , T beeing kinetic and U potential energy. this gives you directly the differential equation for the motion. q beeing x in your case.
Thats pretty much what I used to get to the first step I've written down, its from there that the problem I have is.
I know how to get the EoM, its just that specific integral I'm stuck on. Thanks though. 
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United States24483 Posts
I'm so glad I don't have to do this anymore lol
What course is this exactly and what program is it for? When I learned lagrangians in mechanics I don't think I ever was given this potential to use.
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On October 31 2009 09:38 micronesia wrote:
I'm so glad I don't have to do this anymore lol
What course is this exactly and what program is it for? When I learned lagrangians in mechanics I don't think I ever was given this potential to use.
Its just called advanced mechanics, its part of 2nd year in my degree. (theoretical physics) This question is part of a section on integrating the equations of motion.
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You can just use computer software to evaluate that integral, or look it up in a table of integrals. If you really have to do it by hand, I think you would have to substitute some sort of trig function.
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On October 31 2009 09:30 deconduo wrote:Show nested quote +On October 31 2009 09:13 aqui wrote:use the lagrangefunction. L=T-U , T beeing kinetic and U potential energy. this gives you directly the differential equation for the motion. q beeing x in your case. Thats pretty much what I used to get to the first step I've written down, its from there that the problem I have is. I know how to get the EoM, its just that specific integral I'm stuck on. Thanks though.
thought you just solved t=x/v. because thats what stands there^^ the integral looks like a bitch though^^ no idea how to solve it, sry:p
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I dont know how to solve that integral... but:
http://www93.wolframalpha.com/input/?i=integrate [1%2F%28%28k-%281-e^%28-ax%29%29%29%29^0.5%2C+x]%2C+a%3D{-infinite%2Cinfinite}%2Ck%3D{-infinite%2Cinfinite}
dont know if everything is set up correctly, so give a look before follow it blindly 
edit: sorry.. thats the best way to put the link.. copy all of it
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If you have some idea that 'a' is very small or that 'x' will be very small, then you can use the binomial expansion or the Taylor expansion to simplify the expression to something integrable. Without any paper immediately on hand, the obvious substitution looks like u = ax. Then you can probably do something like partial fractions.
Other than that, just try the Integrator.
http://integrals.wolfram.com/index.jsp?expr=1/sqrt(1-(1-exp(-3x))^2)&random=false
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Thanks for the integrator thing, but I have to do it by hand and show the work unfortunately. I'll see if I can work backwards from the result though.
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Now that I look at it more closely, I don't see where you got that integral at all.
If you use the Lagrangian (or in this case, just say F = -grad(u), which I think is faster), you should get the expression
mx'' = -2Aa(exp(-ax)-exp(-2ax))
From there you should be able to separate variables and integrate twice. There is a substitution needed (for the x side) and some algebraic manipulation, then you will end up with two integrals for the x side which should solvable. Then just backtrack your substitution carefully after you've solved.
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after the substitution
we have
the next substitution is
where
if E>A (infinite motion) then the equation in the denominator has 2 different real roots, if E<1 this equation has no real roots. These two cases should be considered separatly.
Let us consider the first case.
where
where
The second case is similar but there will be arctan. It looks messy, so maybe some simplifications are available.
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impressive... most impressive.
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