Taylor expansion of a scalar potential field

In summary, the potential ##U(\phi)## is a function of the scalar field ##\phi## and the mass dimensions of the couplings are: ##[\lambda]=0##, ##[a]=1##, and ##[\epsilon]=4##. The potential has a local minimum at the point ##\phi=\phi_{-}## and is of order 4 in the fields.
  • #1
spaghetti3451
1,344
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Consider the potential ##U(\phi) = \frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)##, where ##\phi## is a scalar field and the mass dimensions of the couplings are: ##[\lambda]=0##, ##[a]=1##, and ##[\epsilon]=4##.

Expanding the field ##\phi## about the point ##\phi=\phi_{-}## (##\phi = \phi_{-}+ \varphi##) and keeping terms up to dimension four, we find

##U(\varphi)=\frac{m^{2}}{2}\varphi^{2}-\eta\varphi^{3}+\frac{\lambda}{8}\varphi^{4}##,

where ##m^{2}=\frac{\lambda}{2}(3\phi_{-}^{2}-a^{2})## and ##\eta = \frac{\lambda}{2}\lvert\phi_{-}\lvert##.

How do you derive this potential ##U(\varphi)## in explicit steps? Can you provide just the first two lines? I'll work out the rest for myself.
 
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  • #3
##\phi_{-}## is some fixed value of the potential ##\phi##.
 
  • #4
Alright, I have made some progress with the problem.

##U(\varphi) = U(\phi)+U'(\phi)(\varphi-\phi)+\frac{U''(\phi)}{2}(\varphi-\phi)^{2}+\frac{U''(\phi)}{6}(\varphi-\phi)^{3} \cdots##

##=\frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)+\frac{1}{2}\bigg(\frac{\lambda}{4}(\phi^{2}-a^{2})(2\phi)-\frac{\epsilon}{2a}\bigg)(-\phi_{-})+\frac{1}{6}\bigg(\frac{3\lambda}{2}\phi^{2}-\frac{\lambda}{2}a^{2}\bigg)(-\phi_{-})^{2}##.

Now, I'm not really sure how many terms to keep based on the comment 'keeping terms up to dimension four'.

Can you help with that?
 
  • #5
Well, the expression is of order 4 in the fields. So you just keep all terms, and I still don't get what your manipulations mean. The first line makes sense, because it's the Taylor expansion around ##\phi##, but the 2nd line looks totally unrelated to the first.

Usually you shift the field such that the linear term vanishes. Most convenient is to expand around a (local) minimum of the potential.
 
  • #6
vanhees71 said:
Well, the expression is of order 4 in the fields. So you just keep all terms,

The paper from which I took this problem mentions 'keeping terms up to dimension four' - why do they keep terms up to dimension four? I think the answer to this question lies in your third comment.

vanhees71 said:
and I still don't get what your manipulations mean. The first line makes sense, because it's the Taylor expansion around ##\phi##, but the 2nd line looks totally unrelated to the first.

Sorry, my bad!

##U(\varphi) = U(\phi)+U'(\phi)(\varphi-\phi)+\frac{U''(\phi)}{2}(\varphi-\phi)^{2}+\frac{U''(\phi)}{6}(\varphi-\phi)^{3}+\frac{U'''(\phi)}{24}(\varphi-\phi)^{4} + \cdots##

##=\frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)+\bigg(\frac{\lambda}{4}(\phi^{2}-a^{2})(2\phi)-\frac{\epsilon}{2a}\bigg)(-\phi_{-})+\frac{1}{2}\bigg(\frac{3\lambda}{2}\phi^{2}-\frac{\lambda}{2}a^{2}\bigg)(-\phi_{-})^{2}+\frac{1}{6}\bigg(3\lambda\phi\bigg)(-\phi_{-})^{3}+\frac{1}{24}\bigg(3\lambda\bigg)(-\phi_{-})^{4}##.

But, I think this is unnecessary in lieu of your next comment.

vanhees71 said:
Usually you shift the field such that the linear term vanishes. Most convenient is to expand around a (local) minimum of the potential.

##U(\varphi)=U(\phi_{-})+U'(\phi_{-})(\varphi-\phi_{-})+\frac{U''(\phi_{-})}{2}(\varphi-\phi_{-})^{2}+\frac{U''(\phi_{-})}{6}(\varphi-\phi_{-})^{3}+\frac{U'''(\phi_{-})}{24}(\varphi-\phi_{-})^{4} + \cdots##

Now, I did not mention it before, but ##\phi=\phi_{-}## is a local minimum, so that ##U'(\phi_{-})=0##.

Also, keeping terms to dimension four means that we truncate after the second power in ##(\varphi-\phi_{-})##.

Therefore, ##U(\varphi)=U(\phi_{-})+\frac{U''(\phi_{-})}{2}(\varphi-\phi_{-})^{2}##.

Am I correct so far?

Also, why do we shift the field such that the linear term vanishes?
 
Last edited:
  • #7
because it simplifies the further calculations a lot. In your case, if ##U''(\phi_-) > 0## for small deviations from ##\phi_-## you can treat the problem as a harmonic-oscillator problem, i.e., a linear differential equation.
 

Related to Taylor expansion of a scalar potential field

1. What is Taylor expansion of a scalar potential field?

Taylor expansion of a scalar potential field is a mathematical technique used to approximate a function using a series of polynomial terms. It is commonly used in physics and engineering to analyze and model scalar potential fields.

2. Why is Taylor expansion used for scalar potential fields?

Taylor expansion allows us to approximate complex functions with simpler polynomial functions, making it easier to analyze and understand scalar potential fields. It also allows us to make predictions and estimate values at different points in the field.

3. How is Taylor expansion of a scalar potential field calculated?

The Taylor expansion of a scalar potential field is calculated by taking the derivatives of the function at a specific point and evaluating them at that point. The resulting terms are then combined to form a polynomial expression which approximates the original function.

4. What are the applications of Taylor expansion in scalar potential fields?

Taylor expansion is widely used in physics and engineering to analyze and model scalar potential fields. It is used in fields such as electromagnetics, fluid dynamics, and thermodynamics to predict and understand the behavior of scalar potential fields in these systems.

5. Are there any limitations to using Taylor expansion for scalar potential fields?

Yes, there are limitations to using Taylor expansion for scalar potential fields. It is only an approximation and may not accurately represent the function at points far from the expansion point. It also assumes that the function is continuous and differentiable at the expansion point, which may not always be the case.

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