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The metric $\mathbf g: V\times V \rightarrow \mathbb R$ is, by definition, a multilinear map which eats two elements of the underlying vector space and spits out a real number. Because the metric provides an isomorphism$^\dagger$ between the vector space $V$ and the dual space $V^*$, we can map any vector $X\in V$ to a corresponding covector $X^\flat \...


6

Well, four-velocity $dx^\mu/d \tau$ lives on the tangent space $T \mathcal{M}$. This is actually how tangent space is defined, as the space of tangents to curves. In other words, this is where the whole story of tensors starts and all the other structure is built from its relation to $T \mathcal{M}$. So it is perhaps natural to define the metric first as ...


3

In some sub-fields people have conventions for what the "right" thing is, but the following rules generally should guide you to what makes sense: The zeroth rule: you are writing to communicate something accurately. This supersedes all other rules, whether of grammar or style. Think about what your reader will (mis)understand. The first rule: be ...


3

Keep in mind that electric field is negative gradient of potential.


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The conjugation formula you have in general dimensions is true only for scalar operators. In $d=2$, this corresponds to an operator with $h={\bar h} = \frac{\Delta}{2}$. We therefore need to prove $$ {\cal O}^\dagger(z,{\bar z}) = \frac{1}{(z {\bar z})^{\Delta} } {\cal O} \left( \frac{1}{ {\bar z} } , \frac{1}{z} \right) \qquad \qquad (1) $$ To derive this, ...


2

The correct relation is $$\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\right)= \partial_0 \left(\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}\right)+\partial_i \left(\frac{\partial \mathcal{L}}{\partial(\partial_i \phi)}\right).$$ Essentially you're asking if $$A_\mu B^\mu \overset{?}{=} A_0 B^0 + A_i B^i,$$ or $$A_\...


2

There should be no special "minus" sign for transposing a single spinor, yet here the minus sign must appear when you transpose the quantity $\psi_L^\dagger \psi_R$ because when you transpose it an ordinary way, the order of components is changed ($\psi_R$ becomes first and $\psi_L$ second). Spinors consist of anticommuting Grassmann components. In ...


2

Placement of the metric varies, as it's simply a convention. Different authors will use different conventions and sometimes not stick to the same one, depending on what makes an equation more readable. There is no difference in that $g_{ab}A^aB^b = A^ag_{ab}B^B = A^aB^bg_{ab}$. The only time there would be a difference in an index expression like this is if ...


1

By the Einstein convention there is an implicit sum $$ \sum_{a,b=1}^d g_{ab}A^aB^b\,. $$ Clearly you will agree that for any term of the sum $g_{ab}$, $A^a$ and $A^b$ are just real numbers. Therefore they commute as always.


1

This makes sense to me if I pick the y axis to be upward and the x axis to point to the right in the sketch given in the book. Then, the torque would point in the negative z direction. In the diagram $x$ points down, $y$ points right (conventionally). (See: polar coordinates) But what if I picked my y axis to point downwards? Then the torque would ...


1

I'm assuming you mean that the real part of the Green function will change its sign. The imaginary part of the Green function is related to the spectral density and therefore it is always negative (from the realtion $-\rm{Im}\{G^r(\omega)\} = \pi A(\omega)$). To see what you are looking for you can examine it in the following manner: $$ \rm{Re}\{G^r(\omega)\...


1

Whenever you are calculating fields/potentials in a region (of interest) you are always considering the fields/potentials created inside your region. Any variation in field that you’ve calculated is a result of an effect/charge outside your region. If you want to incorporate those variations then you must extend your region to include the source of the ...


1

To understand the reason for the difference in signs, it may be helpful to consider the following definition of potential difference, or voltage $V$. The potential difference, $V$, between two points is the work required per unit charge to move the charge between the two points. By convention, the direction of the electric field is the direction of the ...


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Also what does it mean if the value is negative? If $V_{ab}$ is negative, it means that $V_a-V_b$ is negative. In other words, it means that point b is at a higher potential than point a. What is the difference between $V_{ab}$ and $V_{ba}$ ? $V_{ab}$ is $V_a-V_b$. $V_b$ is $V_b-V_a$. So $V_{ab}=-V_{ba}$. So if $V_{ab}$ is positive 4 for example , ...


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If you have a circuit and enough information about it then you may be able to deduce the voltage between $A$ and $B$ using Kirchoff's laws. There is no certain simple answer to this as it depends on the other components and what you know. E.g. if you had a battery connected to three resistors in series and you knew the battery's voltage and all of the ...


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