New answers tagged

2

Yes. It just means that the velocity is in a direction opposite the direction of your reference frame. If you make "down = positive" then $g$ would be positive, and so would the velocity.


2

Lucubration needs not light with insight. I fear you are expecting to make lemonade with apples. Here is why. The basic relation is the multiplication law of two Pauli vectors predicated on the abstract properties of the Pauli matrices, not their particular realization, $$(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) ...


1

The diagram above has a very important feature. It is the connection between the Earth and the outer conducting shell. Assume that the Earth is a conducting sphere and has some net positive charge on it. This will mean that the outer shell connected to it will also have some positive charge on it but the wire between the outer shell and the Earth means that ...


0

So, the potential is not necessarily zero both at the ground and at infinity. First off, there's nothing in physics which forces the potential to be zero at infinity; it just happens to be a nice way to think of the Coulomb potential. Potential energies are not absolute numbers; there is a constant of integration that enters into them! Second, there is ...


3

By convention, vectors are written as column vectors, whereas dual vectors are written as row vectors. This means that in principle, upper indices should index columns and lower indices should index rows. However, in practice, we normally translate rank-2 tensors to matrices by order of the indices, the first one indexing rows, the second one indexing ...


2

You are halfway correct. The trace of a combination of $\gamma$-matrices does not depend on the representation in which they are expressed. Sakurai primarily uses the Dirac-Pauli representation, while Peskin and Schroeder use the Weyl chiral representation. This difference in representation should no affect the traces of matrix combinations; the traces ...


3

$$\Gamma^\mu{}_{\alpha\beta} y^\alpha y^\beta$$ is defined to mean $$\sum_{\alpha = 0}^3 \sum_{\beta = 0}^3 \Gamma^\mu{}_{\alpha\beta} y^\alpha y^\beta $$ that is, each repeated index is summed over independently.


1

Yes, the charge conjugation operator is representation dependent. The time reversal operator depends on the representation as well, although the parity operator does not. A key reason for the representation dependence is that the $C$ and $T$ operators involve complex conjugation (although in different ways). Therefore, their form depends on whether ...


0

You can just use equation 3 as a definition. Yes, its dimension will differ from that of momentum, but this is not a big deal. That would not be a worse 4-vector than that of equation 1 - they differ just by a constant factor. And to use equation 2 you use a system of units where c=1, not 1 m/s.


0

The concept of electric charge is introduced to explain experiments (originally from static electricity). It is found that only two types of charges are necessary and to distinguish them and to distinguish between they are given labels. The most convenient label is positive and negative (that has some mathematical advantages). It is pure convention that ...


-1

All particles seems to be grouped under two distinct polarities based on the manner of attraction or repulsion. Those particles repelling one another are said to have like charges. Those that attract one another have different charges. Being a positive or negative charge, is a matter of convention already accepted by world scientific community. A Charge is ...


1

Force is a vector. Potential energy is a scaler. Forces which have associated potential energy functions as called conservative forces. Conservative forces act in such a direction that, if released from rest, the potential energy function associated with that force will decrease (and the kinetic energy will thus increase) with the velocity increasing, until ...


0

If you take down as positive then displacement $s = +30$ m, initial velocity $v_i = +8$ ms$^{-1}$ and acceleration $a = +10$ ms$^{-2}$. Using the constant acceleration kinematic equation $s = v_i t + \frac 1 2 a t^2$ where $t$ is the time gives $$(+30) = (+8)t+\frac 1 2 (+10)t^2$$ If you take up as positive then displacement $s = -30$ m, initial velocity ...


1

The question is rather incomplete and confusing. By the way, it is used to consider surfaces as vectors when needed for computing surface integrals, like flux integrals, where the scalar product between a vector field $\vec A$ and a infinitesimal surface $\mathrm d\vec S$ is considered: $\vec A\cdot\mathrm d\vec S$. To this aim, the differential surface is ...



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