# Horndeski: denominator null implies an infinite quantity

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• fab13
fab13
TL;DR Summary
Summary of My Issue with Tensor Calculus Calculations :
In my exploration of tensor calculus within a specific theoretical framework-potentially related to Horndeski theory or another complex field theory-I have encountered a significant computational issue involving the higher-order derivatives of the metric tensor, ##g_{\rho \sigma}##.
1. Core Equation:I am examining the following equation:

## \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi \lambda}} \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}} = \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi}} \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}}##
2. Assumptions and Context:The assumption that third-order derivatives of ##g_{\rho \sigma}## are zero is crucial in this framework. This assumption has direct implications for how the derivatives in my calculations are evaluated.
3. Problem Identified:The term:

## \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}}##

acts like a delta function, being 1 when the indices match. However, the term:

## \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi \lambda}}##

is zero because all third-order derivatives of ##g_{\rho \sigma}## are assumed to be zero.

4. Mathematical Concern:Typically, having a zero numerator wouldn't be problematic, but in this case, since the calculations require non-zero values for correct progression, having a zero value in these derivatives leads to a division by zero in subsequent calculations, effectively rendering the result as undefined or infinite. This represents a significant challenge to the consistency and applicability of the theoretical model.
5. Conclusion:The essence of the issue lies in the presence of a zero denominator, leading to an infinite or undefined quantity. This issue may necessitate a reevaluation of the assumptions or the mathematical techniques employed in my study.

Hoping you will help me to understand why the denominator concerned equal to zero doens
6. 't imply the inverse to be infinite.

Regards

fab13 said:
Core Equation:I am examining the following equation:

## \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi \lambda}} \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}} = \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi}} \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}}##
1) The left side of your equation has two free unprimed indices ##\mu \nu## but the right has three ##\mu \nu \lambda##. Can you explain?
2) I am unfamiliar with tensor-derivative expressions like yours. Can you display a simple example of the tensor ##\mathcal{A}_{\mu \nu}## to illustrate how it explicitly depends on the metric ##g## and its derivatives?

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PeroK
1) I don't understand why you mean : this is all the issue that I have , how the author can remove the denominator ##x^{\lambda}## of ##\frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi \lambda}} by \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi}}##, i.e by removing the partial derivative ##\dfrac{\partial}{\partial x^{\lambda}}##.

2) for this point, I am just using thre chain rule in tensor calculus format ( summation of repeated indices ).

fab13 said:
1) I don't understand why you mean : this is all the issue that I have , how the author can remove the denominator ##x^{\lambda}## of ##\frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi \lambda}} by \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi}}##, i.e by removing the partial derivative ##\dfrac{\partial}{\partial x^{\lambda}}##.
Since I don't understand the notation I can't address your issue. But I can tell you that what you wrote is not a valid tensor equation because the free indices are unequal on the two sides. Can you please cite the technical reference that discusses this "equation"?

Sorry the notation is following : ##g_{\rho\sigma,\pi\lambda}=\dfrac{\partial g_{\rho\sigma}}{\partial x^{\pi} \partial x^{\lambda}}## . I can't provide the technical reference, this is a report. Regards

fab13 said:
Sorry the notation is following : ##g_{\rho\sigma,\pi\lambda}=\dfrac{\partial g_{\rho\sigma}}{\partial x^{\pi} \partial x^{\lambda}}## . I can't provide the technical reference, this is a report. Regards
Thanks, but I am familiar with the comma notation for partial derivatives with respect to coordinates ##x##. What I'm trying to understand is the meaning of differentiating one tensor object (like ##A_{\mu\nu}##) with respect to derivatives of another (like ##g_{\rho\sigma,\tau\pi\lambda}##). So let me focus my question. Suppose I define the quantity:$$\Gamma{}^{\alpha}{}_{\mu\nu} \equiv\tfrac{1}{2}g^{\alpha\lambda}\left(g_{\lambda\mu,\nu}+g_{\lambda\nu,\mu}-g_{\mu\nu,\lambda}\right)$$My question is: what is the explicit tensor expression that results from evaluating the following derivative:$$\frac{\partial\Gamma{}^{\alpha}{}_{\mu\nu}}{\partial g_{\sigma\tau,\beta}}$$Thanks!​

renormalize said:
what is the explicit tensor expression that results from evaluating the following derivative:$$\frac{\partial\Gamma{}^{\alpha}{}_{\mu\nu}}{\partial g_{\sigma\tau,\beta}}$$​
That expression isn't even well-defined. ##g_{\sigma \tau, \beta}## means ##\partial / \partial \beta ( g_{\sigma \tau} )##. You can't take a derivative with respect to that. If the OP is trying to do that, the OP is trying to do something that isn't well-defined.

fab13 said:
I am examining the following equation:

## \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi \lambda}} \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}} = \frac{\partial \mathcal{A}_{\mu \nu}}{\partial g_{\rho \sigma, \tau \pi}} \frac{\partial g_{\rho \sigma, \tau \pi \lambda}}{\partial g_{\rho^{\prime} \sigma^{\prime}, \tau^{\prime} \pi^{\prime} \lambda^{\prime}}}##
Per my post #7 just now, this equation isn't even well-defined. Where are you getting it from?

my issue is on (4.2.6) where indice ##\lambda^{\prime}## disappears on the denominator.

fab13 said:
my issue is on (4.2.6) where indice ##\lambda^{\prime}## disappears on the denominator.
Where did you get that from? You need to give a reference.

PeterDonis said:
That expression isn't even well-defined. ##g_{\sigma \tau, \beta}## means ##\partial / \partial \beta ( g_{\sigma \tau} )##. You can't take a derivative with respect to that.
I disagree. Consider the Lagrangian ##\mathcal{L}\left(A_{\alpha},\partial_{\beta}A_{\alpha}\right)## for the electromagnetic potential ##A##. It's well known that the Euler-Lagrange equations for that field are:$$\partial_{\beta}\left[\frac{\mathcal{\partial L}}{\partial\left(\partial_{\beta}A_{\alpha}\right)}\right]-\frac{\partial\mathcal{L}}{\partial A_{\alpha}}=0$$which manifestly involves the derivative operator ##\mathcal{\partial}/\partial\left(A_{\alpha,\beta}\right)##. Why should the analogous derivative ##\mathcal{\partial}/\partial\left(g_{\sigma\tau,\beta}\right)## be problematic?

renormalize said:
which manifestly involves the derivative operator ##\mathcal{\partial}/\partial\left(A_{\alpha,\beta}\right)##.
It involves a functional derivative with respect to the function that is the derivative with respect to ##x^\beta## of ##A_\alpha##. The Lagrangian is a functional of that function (as well as of the function ##A_\alpha## itself, considered as a function on spacetime).

It's possible that whatever reference the OP is using is also dealing with functional derivatives. But we won't know unless the OP gives us the reference.

Sorry, I didn't give you the definition of the tenseur ##A_{\mu\nu}## wiht all its dependances :

fab13 said:
I didn't give you the definition of the tenseur ##A_{\mu\nu}## wiht all its dependances :
That's true, but more importantly, you still haven't told us what reference this is from or given a link to it. We can't comment on quotes from a reference we don't know.

PeterDonis said:
It's possible that whatever reference the OP is using is also dealing with functional derivatives.
From post #14 it looks like it is.

fab13 said:
my issue is on (4.2.6) where indice ##\lambda^{\prime}## disappears on the denominator.
Actually, the contravariant index ##\lambda^{\prime}## appears on both sides of eq.(4.2.6), so it doesn't "disappear". I think what you're really asking is: how is that equation derived? Here is my take. To minimize the clutter of indices, I consider here the second equation in (4.2.6) that involves ##\phi## and its derivatives, but the first equation follows the same logic. First write:$$\frac{\partial\nabla^{\nu}A_{\mu\nu}}{\partial\phi_{,\tau^{\prime}\pi^{\prime}\lambda^{\prime}}}=g^{\nu\lambda}\frac{\partial\left(\partial_{\lambda}A_{\mu\nu}\right)}{\partial\phi_{,\tau^{\prime}\pi^{\prime}\lambda^{\prime}}}\tag{1}$$where the covariant- is replaced by the partial-derivative since the Christoffel connection has no dependence on ##\phi##. Now since ##A_{\mu\nu}\equiv A_{\mu\nu}\left(\phi,\partial_{\tau}\phi,\partial_{\pi}\partial_{\tau}\phi\right)##, we have:$$dA_{\mu\nu}=\frac{\partial A_{\mu\nu}}{\partial\phi}d\phi+\frac{\partial A_{\mu\nu}}{\partial\phi_{,\tau}}d\left(\phi_{,\tau}\right)+\frac{\partial A_{\mu\nu}}{\partial\phi_{,\tau\pi}}d\left(\phi_{,\tau\pi}\right)$$which implies:$$\partial_{\lambda}A_{\mu\nu}=\frac{\partial A_{\mu\nu}}{\partial\phi}\phi_{,\lambda}+\frac{\partial A_{\mu\nu}}{\partial\phi_{,\tau}}\phi_{,\tau\lambda}+\frac{\partial A_{\mu\nu}}{\partial\phi_{,\tau\pi}}\phi_{,\tau\pi\lambda}\tag{2}$$Only the last term on the right survives from eq.(2) when that equation is inserted into (1), resulting in the equation in your reference:$$\frac{\partial\nabla^{\nu}A_{\mu\nu}}{\partial\phi_{,\tau^{\prime}\pi^{\prime}\lambda^{\prime}}}=g^{\nu\lambda}\frac{\partial}{\partial\phi_{,\tau^{\prime}\pi^{\prime}\lambda^{\prime}}}\left(\frac{\partial A_{\mu\nu}}{\partial\phi_{,\tau\pi}}\phi_{,\tau\pi\lambda}\right)=g^{\nu\lambda}\frac{\partial A_{\mu\nu}}{\partial\phi_{,\tau\pi}}\frac{\partial\phi_{,\tau\pi\lambda}}{\partial\phi_{,\tau^{\prime}\pi^{\prime}\lambda^{\prime}}}\tag{4.2.6}$$

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