Understanding Differential Notation

Hyperreality
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The four velocity component u^\mu with coordinate of x^\mu(\lambda) is given by

u^\mu = \frac{dx^\mu}{d\lambda}

where \lambda is the proper time. So, the component of acceleration a^\mu is

a^\mu = \frac{du^\mu}{d\lambda}

Using the chain rule we have

a^\mu = \frac{\partial u^\mu}{\partial x^\alpha} \frac{dx^\alpha}{d\lambda} = u^\alpha \partial_{\alpha}u^\mu

Everything was straight forward except the last part, I don't understand what the notation of \partial_{\alpha} meant.
 
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\partial_{\alpha} is the covariant 4-gradient. It means nothing other than the following:

\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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