Christoffel symbols knowing Line Element (check my result)

In summary, the line element given is ds^2=2x^2dx^2+y^4dy^2+2xy^2dxdy and the only non-zero component of the Christoffel symbols is \Gamma^x_{xx}=\frac{1}{x}. The conversation also includes a discussion about the correct solution and a suggestion to use a coordinate change to simplify the calculation. The solution obtained by the expert includes a second non-zero Christoffel symbol, \Gamma^y_{yy}=\frac{2}{y}, and the expert recommends trying the coordinate change \chi = x^2, \eta = y^3 to see its effect on the metric.
  • #1
Hi! I'm asked to find all the non-zero Christoffel symbols given the following line element:

[tex]ds^2=2x^2dx^2+y^4dy^2+2xy^2dxdy[/tex]

The result I have obtained is that the only non-zero component of the Christoffel symbols is:

[tex]\Gamma^x_{xx}=\frac{1}{x}[/tex]

Is this correct?

MY PROCEDURE HAS BEEN:

the Lagrangian squared is given by:
[tex]\mathcal{L}^2=2x^2\dot{x}^2+y^4\dot{y}^2+2xy^2\dot{x}\dot{y}[/tex]

where [tex]\dot{\quad}\equiv d/d\tau[/tex], with [tex]\tau[/tex] being the proper time.

The Euler-Lagrange equations for this Lagrangian are:
[tex]
\begin{split}
\frac{d}{d\tau}\Bigg(\frac{\partial\mathcal{L}^2}{\partial{\dot{x}}}\Bigg)-\frac{\partial\mathcal{L}^2}{\partial x}=0\quad\longrightarrow\quad&\frac{d}{d\tau}\Big(4x^2\dot{x}+2xy^2\dot{y}\Big)-4x\dot{x}^2-2y^2\dot{x}\dot{y}=0\\
&8x\dot{x}^2+4x^2\ddot{x}+2y^2\dot{x}\dot{y}+4xy\dot{y}^2+2xy^2\ddot{y}-4x\dot{x}^2-2y^2\dot{x}\dot{y}=0\\
\text{(1)}\quad&2x^2\ddot{x}+xy^2\ddot{y}+2x\dot{x}^2+2xy\dot{y}^2=0\\
\frac{d}{d\tau}\Bigg(\frac{\partial\mathcal{L}^2}{\partial{\dot{y}}}\Bigg)-\frac{\partial\mathcal{L}^2}{\partial y}=0\quad\longrightarrow\quad&\frac{d}{d\tau}\Big(2y^4\dot{y}+2xy^2\dot{x}\Big)-4y^3\dot{y}^2-4xy\dot{x}\dot{y}=0\\
&8y^3\dot{y}^2+2y^4\ddot{y}+2y^2\dot{x}^2+4xy\dot{x}\dot{y}+2xy^2\ddot{x}-4y^3\dot{y}^2-4xy\dot{x}\dot{y}=0\\
\text{(2)}\quad&xy^2\ddot{x}+y^4\ddot{y}+y^2\dot{x}^2+2y^3\dot{y}^2=0
\end{split}
[/tex]

Multiplying equation (1) by [tex]y^2/x[/tex] we obtain: [tex]2xy^2\ddot{x}+y^4\ddot{y}+2y^2\dot{x}^2+2y^3\dot{y}^2=0[/tex]. Subtracting equation (2) from this result we obtain:
[tex]xy^2\ddot{x}+y^2\dot{x}^2=0\quad\longrightarrow\quad \ddot{x}+\frac{1}{x}\dot{x}^2=0[/tex]

Multiplying equation (2) by [tex]2x/y^2[/tex] we obtain: [tex]2x^2\ddot{x}+2xy^2\ddot{y}+2x\dot{x}^2+2xy\dot{y}^2=0[/tex].Subtracting equation (1) from this result we obtain:
[tex]xy^2\ddot{y}=0\quad\longrightarrow\quad \ddot{y}=0[/tex]

Knowing that the general expression for the geodesic equation is:
[tex]\ddot{x}^\alpha+\Gamma^\alpha_{\beta\gamma}\dot{x}^\beta \dot{x}^\gamma=0[/tex]

[tex]\boxed{\text{the only non-zero Christoffel symbol is therefore: } \Gamma^x_{xx}=\frac{1}{x}}[/tex]
 
Last edited:
Physics news on Phys.org
  • #2
Please show your work on homework-like questions. Otherwise it is impossible to know where you go wrong (if you do) or confirm that you did things correctly (if you did) and did not come to the correct answer by chance.
 
  • Like
Likes Confused Physicist
  • #3
Your approach is fine. However, your final equation for ##\ddot y## is not. Also, in this particular case, there is a shortcut by a simple change of variables that turns your line element to one with constant coefficients.
 
  • Like
Likes Confused Physicist
  • #4
Orodruin said:
Your approach is fine. However, your final equation for ##\ddot y## is not. Also, in this particular case, there is a shortcut by a simple change of variables that turns your line element to one with constant coefficients.

Oh, why is my final equation for ##\ddot{y}## not correct?

And what is that change of variables that simplifies the calculation?

Is the solution I have obtained incorrect then?
 
Last edited:
  • #5
You have an arithmetic error in your multiplication of (2).

The change of variables should be quite evident from the form of the line element. Hint: You want to make a change to variables ##\chi(x)## and ##\eta(y)## so that the line element takes the form ##ds^2 = A d\chi^2 + B d\chi\, d\eta + C d\eta^2##, where A B and C are constant.
 
  • Like
Likes Confused Physicist
  • #6
Orodruin said:
You have an arithmetic error in your multiplication of (2).

You mean when I multiply (2) by ##2x/y^2##? I'm repeating the process again, but I can't find the mistake...

Ohhhhh okay I've seen the mistake now. The mistake is on equation (2). The ##2y^3\dot{y}^2## should be ##4y^3\dot{y}^2##.

Oh no. That wasn't the mistake. I still can't see what you're pointing out.

FINAL EDIT: Okay, I've seen the mistake. The equation for ##\ddot{y}## should be: ##xy^2\ddot{y}+2xy\dot{y}^2=0##, which gives another non-zero Christoffel symbol: ##\Gamma^y_{yy}=\frac{2}{y}##.
 
Last edited:
  • #7
Right. Now try out the coordinate change ##\chi = x^2##, ##\eta = y^3## to see what it does to the metric.
 
  • Like
Likes Confused Physicist

1. What are Christoffel symbols?

Christoffel symbols are mathematical quantities used in differential geometry to describe the properties of a manifold. They represent the connection between the coordinate system and the geometry of the manifold.

2. How are Christoffel symbols related to the line element?

The Christoffel symbols are used to calculate the line element, which is a measure of distance on a curved surface. They provide information about how the coordinates of a point on the surface change as we move along a particular direction.

3. Why are Christoffel symbols important in physics?

Christoffel symbols are important in physics because they are used in the study of general relativity, which describes the curvature of spacetime due to the presence of mass and energy. They are also used in other areas of physics such as fluid mechanics and electromagnetism.

4. How do you calculate Christoffel symbols?

Christoffel symbols can be calculated using the metric tensor, which describes the local properties of a manifold. They can also be calculated using the derivative of the metric tensor with respect to the coordinates of the manifold.

5. What is the significance of knowing the Christoffel symbols?

Knowing the Christoffel symbols allows us to understand the geometry of a manifold and how it is influenced by the choice of coordinates. This is important in many areas of mathematics and physics, including the study of curved spaces and the behavior of particles in gravitational fields.

Suggested for: Christoffel symbols knowing Line Element (check my result)

Replies
7
Views
4K
Replies
9
Views
3K
Replies
15
Views
4K
Replies
5
Views
2K
Replies
10
Views
3K
Replies
1
Views
2K
Replies
2
Views
5K
Replies
6
Views
3K
Back
Top