Register to reply 
Variational method for geodesics  I'm stuck! 
Share this thread: 
#1
Oct1308, 08:50 AM

P: 5

1. The problem statement, all variables and given/known data
Hi, I am reading Ray d'Inverno's book, 'Introducing Einstein's Relativity' and there is a particular derivation of the geodesic equation that I get stumped on (chapter 7). It is a variational method and the final equation is df/dx_alphad/du{df/dx_alpha_dot}=0 where f is the Lagrangian, x_alpha, beta etc are the coordinates and _dot denotes differenation with respect to an affine parameter. Now you can carry this computation and derive the equivalent equation but using christoffel symbols. But in this derivation, one line is f=g[beta,gamma]*dx_beta/du*dx_gamma/du This equation is a sum over all indices (alpha, beta etc). df/dx_alpha=dg[beta,gamma]/dx_xlpha*dx_beta/du*dx_gamma/du My question is shouldn't this last term be differentiated with the product rule, so that df/dx_alpha=dg[beta,gamma]/dx_xlpha*dx_beta/du*dx_gamma/du + g[beta,gamma]*d/dx_alpha{dx_beta/du*dx_gamma/du} f is a summation over all indices (ALPHA INCLUDED), so do we not need extra terms to determine these if we are differentiating with respect to alpha?? Why can we jump to just df/dx_alpha=dg[beta,gamma]/dx_xlpha*dx_beta/du*dx_gamma/du ??? Please help. I feel I'm missing a point here. 2. Relevant equations 3. The attempt at a solution 


#2
Oct1308, 09:29 AM

HW Helper
P: 5,003

Hi Corals, welcome to PF.
Your post is really hard to read. This forum supports LaTeX, so you could for example write your first equation like
[tex] \frac{\partial L}{\partial x^{\alpha}}  \frac{d}{du} \left( \frac{\partial L}{\partial \dot{x}^{\alpha}} \right)=0 [/tex] Anyways...I will try to decipher your post and see if I can help... 


#3
Oct1308, 09:44 AM

P: 5

Cheers for that. I'll change the layout tonight. Any light on my question in the meantime would be appreciated. I'm a banker by day, reminiscing my background of physics, contemplating a move back...



#4
Oct1308, 09:45 AM

HW Helper
P: 5,003

Variational method for geodesics  I'm stuck!
[tex]\frac{\partial}{\partial x^{\alpha}} (g_{\beta \gamma}\dot{x}^{\beta}\dot{x}^{\gamma})[/tex] If so, the product rule gives: [tex]\frac{\partial}{\partial x^{\alpha}} (g_{\beta \gamma}\dot{x}^{\beta}\dot{x}^{\gamma})=\frac{\partial g_{\beta \gamma}}{\partial x^{\alpha}} (\dot{x}^{\beta}\dot{x}^{\gamma})+\frac{\partial \dot{x}^{\beta}}{\partial x^{\alpha}} (g_{\beta \gamma}\dot{x}^{\gamma})+\frac{\partial \dot{x}^{\gamma}}{\partial x^{\alpha}} (g_{\beta \gamma}\dot{x}^{\beta})[/tex] But(!) [itex] \frac{\partial \dot{x}^{\beta}}{\partial x^{\alpha}}[/itex] and [itex]\frac{\partial \dot{x}^{\gamma}}{\partial x^{\alpha}}[/itex] are both zero so it simplifies to [tex]\frac{\partial g_{\beta \gamma}}{\partial x^{\alpha}} (\dot{x}^{\beta}\dot{x}^{\gamma})[/tex] 


#5
Oct1308, 10:40 AM

P: 5

Thanks for that, that helps.
But just to explore this a bit. The first term for the Lagrangian is the below. [tex] \frac{\partial L}{\partial x^{\alpha}} \frac{d}{du} \left( \frac{\partial L}{\partial \dot{x}^{\alpha}} \right)=0[/tex] [tex] (g_{\beta \gamma}\dot{x}^{\beta}\dot{x}^{\gamma})[/tex] I can imagine writing this out explicity as a big sum in which case there may well be instances of x^alpha in the sum, in which case they would need to be differentiated to. I'm explicity trying to solve a problem in Ray D'inverno's book, one in which you use the euler lagrange equation you showed me to calculate the christoffel symbols for spherical coordinates. There the Lagrangian is [tex]L=e^{\nu}\dot{t}^{2}[/tex] And now differentiating with respect to t gives a term [tex]\frac{\partial \dot{t}}{\partial \t}[/tex] i.e the derivative of dt/du with respect to t which may or may not be zero. (Ps this may well be an embarassing question on my part) ps i'm trying to latex the equations, but they're previewing as text  how can I activate them> 


#6
Oct1308, 10:57 AM

HW Helper
P: 5,003

Get rid of the & signs in your [&tex] and [&/tex] and make sure that you use a forward slash in the closing [&/tex].



#7
Oct1308, 02:10 PM

P: 5

Thanks for that, that helps.
But just to explore this a bit. The first term for the Lagrangian is the below. [tex]g_{\beta\gamma}\dot{x}^{\beta}\dot{x}^{\gamma}[/tex] I can imagine writing this out explicity as a big sum in which case there may well be instances of x^alpha in the sum, in which case they would need to be differentiated too. I'm explicity trying to solve a problem in Ray D'inverno's book, one in which you use the euler lagrange equation you showed me to calculate the christoffel symbols for spherical coordinates. There the Lagrangian is [tex]2L=e^{\nu}\dot{t}^{2}[/tex] And now differentiating with respect to t gives a term [tex]e^{\nu}\dot{t}\frac{\partial \dot{t} }{\partial t}[/tex] and the derivative of dt/du with respect to t which may or may not be zero, surely. I suspect I am being naive somewhere and making a basic error. Can you help? 


#8
Oct1308, 02:38 PM

HW Helper
P: 5,003

The derivatives are not ordinary derivatives, but partial derivatives. This is very important, because if a function has no explicit dependence on x, then its partial derivative with respect to x is zero. For ordinary derivatives, this is not always the case; if f had an implicit dependence on x, then the ordinary derivative would not necessarily be zero.
The quantities [itex]\dot{x}^{\beta}[/itex] are the time derivatives of the coordinates [itex]x^{\beta}[/itex] and they have no explicit dependence on the coordinates and so the partial derivatives I posted above are zero. The same thing applies to [itex]\frac{\partial \dot{t} }{\partial t}[/itex], If [itex]\dot{t}[/itex] has no explicit dependence on [itex]t[/itex] then [itex]\frac{\partial \dot{t} }{\partial t}=0[/itex]....Out of curiosity, what exactly do [itex]t[/itex] and [itex]\dot{t}[/itex] represent here? (surely [itex]t[/itex] isn't time because then [itex]\dot{t}=\frac{dt}{dt}=1[/itex]) 


#9
Oct1408, 05:15 PM

P: 5

Well t is in fact time and [tex]\dot{t}[/tex] is the derivative of t with respect to u (the curve along which the integration takes place).
Thanks for your answer  it's all clear now... 


Register to reply 
Related Discussions  
Geodesics on R^2  General Math  0  
Newton's Method..stuck on a simple problem  Calculus & Beyond Homework  7  
Evaluate the ground state energy using the variational method  Advanced Physics Homework  3  
Why is it that in general geodesics are paths of stationary character  Introductory Physics Homework  2  
Light geodesic path  Special & General Relativity  8 