# Tag Info

### Odd notation $\stackrel{\leftarrow}{\nabla}$ for a gradient

$\stackrel{\leftarrow}{\nabla}$ means that the differential operator $\nabla$ differentiates to the left, i.e. it differentiates the fermion spinor $\hat{\bar{\psi}}$ in eq. (43).
• 208k

### When computing the Euler–Lagrange equations, why do we assume the coordinates do not depend on time?

Quoting from your question: But if we are differentiating with respect to x, which is really just a function of t, aren't we really differentiating with respect to t? In my opinion the problem here ...
• 21.8k

### When computing the Euler–Lagrange equations, why do we assume the coordinates do not depend on time?

This is something I recently had to grapple with, so I hope I have this right. The Euler-Lagrange formula is the general solution a path which extremizes an action (the integral of the Lagrangian). ...
• 50.3k
1 vote

### When computing the Euler–Lagrange equations, why do we assume the coordinates do not depend on time?

Consider a function $F$ of two variables $a$ and $b$: $$F(a,b)\;,$$ where the usual notation of parentheses and a comma indicates that $F$ is a function of two independent variables, $a$ and $b$. ...
• 22.1k
Accepted

• 74.8k

### Gondolo-Gelmini Change of Variables

Could someone maybe add how to arrive at $E_+\geq\sqrt{s}$? Weirdly enough this is the part I am having a hard time with getting...

• 2,385

### Field strength tensor written as commutator of covariant derivatives in QED

The derivative operators in the second commutator in the second line of Eq. 1 1/2 also act on the wave function via the chain rule. This cancels the third commutator.
• 25.7k
1 vote

### Why is $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$?

Having realized my error of thinking the multiplication was non-commutative, it becomes clear: (\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F^{\alpha\beta}(\partial_\mu F_{\alpha\beta})=\eta_{\...
1 vote
Accepted

### Why is $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$?

Here the solution asserts that this is equal to simply equal to twice the first part of the term, implying $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$. ...
• 22.1k

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