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1

Maybe the answer is connected with the fact that propagator is the inverse of Lagrangian operator. An action of free theory may be written as $$ S[\psi ] = \int d^{4}x (\psi^{a})^{*}\Delta_{ab}\psi^{b}. $$ Here $()^{*}$ means conjugation which leaves $(\psi^{a})^{*}\Delta_{ab}\psi^{b}$-form lorentz-invariant. For example, $$ S_{KG}[\varphi ] = ...


3

If you do not define the direction of the force, you need to do the math explicitly. Usually, a letter like $g$ identifies a scalar quantity. If you want to show a vector, there are a number of typographical conventions. I have seen $\mathbf{g}$, $\vec{g}$, $\overline{g}$, $\underline{g}$ ... So when I have tension up and gravity down, I know that the two ...


3

$\boldsymbol g$ is a vector. You define it as a scalar only when you have mentioned a clear reference frame in which the sign of $\boldsymbol g$ makes sense. You could define a vector field from the law of universal gravitation for two bodies $A$ and $B$: $$\boldsymbol F_{AB}(\boldsymbol r) = -G\frac{m_Am_B}{|\boldsymbol r_{AB}|^2}\boldsymbol ...


2

$Q$ is not the internal energy, but the heat exchanged (absorbed or given away). We use it when we put something near something at a different temperature: the heat exchanged is $Q$. Instead,the internal energy is called $U$, and if we heat something up, then the $\Delta U$ is positive.


1

The version with the 'i' is correct. It is used in all references I checked, including printed books from pre-web days. The term without the 'i' may have come from some author using uncommon conventions or definitions, perhaps defining $\gamma^5 = \gamma^0 \gamma^1 \gamma^2 \gamma^3$ instead of the widely used standard definition $\gamma^5 = i\gamma^0 ...


3

Therefore it occurred to me that the definition in the book of Weinberg is not consistent with that in the book of Tung: in one of them the symbol ${\Lambda_\mu}^\nu$ is defined as the inverse of the Lorentz transformation of contravariant vectors, while in the other case, the same symbol is defined as the transpose of the original matrix. The symbol ...


2

In terms of $4\times 4$ matrices, elements of the representation of SO(3,1) group must obey the following relation: $$ g\cdot A^{T} \cdot g = A^{-1}, $$ where $g = \text{diag}(1, -1, -1, -1) = g^{-1}$. Does it answer your question?


-2

Rather, I pronounced it like $A$ bar dot $B$ bar. Or $A$ bar cross $B$ bar


5

The first bullet would be read "$A$ dot $B$" or "The dot product of $A$ and $B$" The second bullet would be read "$A$ cross $B$" or "The cross product of $A$ and $B$"



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