Converting effective potential to dimensionless function

In summary: This may simplify your calculations.In summary, the task is to convert the effective potential to a dimensionless function by substituting the expressions for r0 and U0 into the given equations. This simplifies the problem and reduces the number of parameters involved.
  • #1
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Homework Statement



Convert the effective potential to a dimensionless function by scaling to the radius of a circular orbit.

Homework Equations



[itex]U_{eff}(r) = - \frac{k}{r}+ \frac{L^2}{2mr^2}[/itex]
[itex]\frac{U_{eff}}{U_0} = -\frac{2}{r / r_0} + \frac {1}{(r/r_0)^2}[/itex]
[itex]r_0= \frac{L^2}{mk}[/itex], [itex]U_0= \frac{k}{2r_0}[/itex]

The Attempt at a Solution



I would love to try solving this, but unfortunately I can't decipher what the question actually is. What does "scaling to the radius of a circular orbit" mean, in this case?
 
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  • #2
Emspak said:

Homework Statement



Convert the effective potential to a dimensionless function by scaling to the radius of a circular orbit.

Homework Equations



[itex]U_{eff}(r) = - \frac{k}{r}+ \frac{L^2}{2mr^2}[/itex]
[itex]\frac{U_{eff}}{U_0} = -\frac{2}{r / r_0} + \frac {1}{(r/r_0)^2}[/itex]
[itex]r_0= \frac{L^2}{mk}[/itex], [itex]U_0= \frac{k}{2r_0}[/itex]

The Attempt at a Solution



I would love to try solving this, but unfortunately I can't decipher what the question actually is. What does "scaling to the radius of a circular orbit" mean, in this case?
In the equation for the potential, one of the terms represents repulsion and the other term represents attraction. Scaling this problem like this reduces the number of parameters involved from 5 to 2. This boils the equation down to its bare essence. Solving problems in terms of dimensionless variables often simplifies the compilation of the results, and often also reduces the algebraic complexity and the potential for making typos.
 
  • #3
OK, so I'm supposed to do what exactly? Again, I can't figure out what the prof wants me to actually DO to it.

I'm sorry. Pretend I am a retarded marmoset taking physics. :-)

EDIT: I'm not asking for anyone to solve this for me-- I just want to understand what action I am supposed to take.
 
  • #4
Emspak said:
OK, so I'm supposed to do what exactly? Again, I can't figure out what the prof wants me to actually DO to it.

I'm sorry. Pretend I am a retarded marmoset taking physics. :-)

EDIT: I'm not asking for anyone to solve this for me-- I just want to understand what action I am supposed to take.

You already did what you are supposed to do.

Chet
 
  • #5
Dude, not meaning to be a jerk or anything, but the Zen koan doesn't help me see what's happening here. It's a bit frustrating. He wants me to make it dimensionless, and maybe my being up all night is clouding my reasoning. Or maybe the whole thing is a lot simpler than it looks.
 
  • #6
Emspak said:
Dude, not meaning to be a jerk or anything, but the Zen koan doesn't help me see what's happening here. It's a bit frustrating. He wants me to make it dimensionless, and maybe my being up all night is clouding my reasoning. Or maybe the whole thing is a lot simpler than it looks.
I think he might have wanted you to substitute your expression for r0 into your equation for U0:
[tex]U_0=\frac{m}{2}\left(\frac{k}{L}\right)^2[/tex]

This eliminates U0 so that
[tex]\frac{u}{U_0}=\frac{2U}{m}\left(\frac{L}{k}\right)^2[/tex]
 
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What does it mean to convert effective potential to dimensionless function?

Converting effective potential to dimensionless function means taking a physical system's effective potential, which is a function of position and energy, and transforming it into a dimensionless function by removing the physical units and making it a pure mathematical representation.

Why is converting effective potential to dimensionless function important?

Converting effective potential to dimensionless function is important because it allows for a more general and universal understanding of a physical system. By removing the physical units, the dimensionless function can be applied to different systems and scenarios, making it a useful tool for scientists.

What is the process of converting effective potential to dimensionless function?

The process of converting effective potential to dimensionless function involves removing the physical units from the effective potential and then dividing by a characteristic length scale and energy scale. This results in a dimensionless function that describes the same physical system, but in a more general and abstract way.

What are the benefits of using dimensionless function instead of effective potential?

Using dimensionless function instead of effective potential allows for a more universal understanding and comparison of different physical systems. It also simplifies calculations and reduces the number of variables needed to describe a system, making it more efficient and easier to work with.

Are there any limitations to converting effective potential to dimensionless function?

Yes, there are some limitations to converting effective potential to dimensionless function. It may not be applicable to all physical systems, and it may not accurately capture all aspects of a system's behavior. Additionally, the choice of characteristic length and energy scales can affect the resulting dimensionless function and may not always be clear or easy to determine.

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