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Níckolas Alves
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I am reading the paper “alternatives to dark matter and dark energy”Alternatives to dark matter and dark energy, but cannot obtain one specific equation no matter how I tried. So I wrote an email to the author, the following is what he replies me.

In lines 4 and 5 when you integrated by parts you did not apply the derivatives to $g^{1/2}$. If you apply $\nabla_{\lambda}\nabla_{\mu}$ to $g^{1/2}$ you get a $g_{\lambda \mu}$ factor. This contracts with $R^{\lambda \mu}$ to give the Ricci scalar term. Philip.

In lines 4 and 5 when you integrated by parts you did not apply the derivatives to $g^{1/2}$. If you apply $\nabla_{\lambda}\nabla_{\mu}$ to $g^{1/2}$ you get a $g_{\lambda \mu}$ factor. This contracts with $R^{\lambda \mu}$ to give the Ricci scalar term. Philip.

I am wondering how the metric tensor factor is obtained. The square root of the determinant of the metric is a tensor density, not a true tensor, what does it mean by applying covariant derivatives to it?

I am reading the paper “alternatives to dark matter and dark energy”, but cannot obtain one specific equation no matter how I tried. So I wrote an email to the author, the following is what he replies me.

In lines 4 and 5 when you integrated by parts you did not apply the derivatives to $g^{1/2}$. If you apply $\nabla_{\lambda}\nabla_{\mu}$ to $g^{1/2}$ you get a $g_{\lambda \mu}$ factor. This contracts with $R^{\lambda \mu}$ to give the Ricci scalar term. Philip.

I am wondering how the metric tensor factor is obtained. The square root of the determinant of the metric is a tensor density, not a true tensor, what does it mean by applying covariant derivatives to it?

I am reading the paper Alternatives to dark matter and dark energy, but cannot obtain one specific equation no matter how I tried. So I wrote an email to the author, the following is what he replies me.

In lines 4 and 5 when you integrated by parts you did not apply the derivatives to $g^{1/2}$. If you apply $\nabla_{\lambda}\nabla_{\mu}$ to $g^{1/2}$ you get a $g_{\lambda \mu}$ factor. This contracts with $R^{\lambda \mu}$ to give the Ricci scalar term. Philip.

I am wondering how the metric tensor factor is obtained. The square root of the determinant of the metric is a tensor density, not a true tensor, what does it mean by applying covariant derivatives to it?

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Qmechanic
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Covariant derivative to the metric determinant?

I am reading the paper “alternatives to dark matter and dark energy”, but cannot obtain one specific equation no matter how I tried. So I wrote an email to the author, the following is what he replies me.

In lines 4 and 5 when you integrated by parts you did not apply the derivatives to $g^{1/2}$. If you apply $\nabla_{\lambda}\nabla_{\mu}$ to $g^{1/2}$ you get a $g_{\lambda \mu}$ factor. This contracts with $R^{\lambda \mu}$ to give the Ricci scalar term. Philip.

I am wondering how the metric tensor factor is obtained. The square root of the determinant of the metric is a tensor density, not a true tensor, what does it mean by applying covariant derivatives to it?