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# Do R134 Air Conditioning Recharge Kits Work If Your Car Is Completely Out of Cool Air?

It must first be hooked up to a vacuum pump and tested for leaks. If no leaks then it must be charged from vacuum so no air is in syatem

1. Vacuum expectation value in presence of a source

My question is how does the presence of nonzero \$J(x)\$ results in a non-trivial spacetime dependent value of \$langle 0|phi(x)|0

angle\$?The equation \$phi(x)=mathrm e^-iPxphi(0)mathrm e^iPx\$ works both for \$J=0\$ and \$J

eq 0\$. Therefore, \$\$ langle phi(x)

angle_J= _Jlangle 0|mathrm e^-iPxphi(0)mathrm e^iPx|0

angle_J \$\$What it is not true when \$J

eq 0\$ is that \$P|0

angle_Jstackreltextno=0\$ (because the source breaks the invariance), and therefore we cannot conclude that \$\$ langle phi(x)

angle_Jstackreltextno= _Jlangle 0|phi(0)|0

angle_J \$\$Therefore, if \$J

eq 0\$ the vev depends on position \$x\$.To find the explicit dependence of \$langle phi(x)

angle_J\$ with \$x\$, instead of using operators, it is easier to work with path integrals: \$\$ langle phi(x)

angle_J=fracdeltadelta J(x)expleft[-iint mathrm dy,mathrm dz J^*(y)Delta(y-z)J(z)

ight] \$\$ which I believe you can calculate yourself (note that the result is proportional to \$J(x)\$ and so the vev goes to zero as \$Jto 0\$, as expected).The first thing we have to do is to differentiate from internal sources and external ones:An internal source is a term in the lagrangian that only includes dynamical fields, that is, fields that are part of the equations of motion. For example, you can have a KG theory, \$\$ mathcal Lsim (partialphi)^2-m^2phi^2gphi^3 \$\$ where the last term can be said to be an internal source (though the usual terminology is just interaction). This term is internal because it only depends on \$phi\$, which is itself a dynamical field (determined from the EoM's). Another (more illustrative) example is the lagrangian for QED, \$\$ mathcal Lsim barpsi(i

otpartial-m)psi-F^2eA_mu barpsigamma^mupsi \$\$ Again, the last term is an internal source, because it only depends on dynamical fields, \$psi\$ and \$A\$, which are determined from the EoM. I would like to stress that in general people do not say "internal source" but "interaction" instead.An exernal source is a function in the lagrangian that is externally determined (fixed), that is, a function that is not dynamical (there is not an equation of motion for that function). Typical examples are the \$J\$'s that are used in path integrals, \$\$ mathcal Lsim (partialphi)^2-m^2phi^2gphi^3phi(x)J(x) \$\$ and fixed (background) functions in effective theories, such as, for example, the electromagnetic field in a low energy treatment of the Hydrogen atom: \$\$ mathcal Lsim barpsi(i

otpartial-m)psieA_mu barpsigamma^mupsi \$\$ (here, \$A_mu\$ is an external source, because there is not a kinetic term \$F^2\$ for it, and so the value of \$A\$ has to be written by hand, say, a Coulomb potential \$A_0sim e/r\$.Note that external sources break the translational invariance of the theory (because of the obvious reason: an external source has a fixed dependence on position, and so the "physics do not look the same everywhere"). Therefore, if there are external sources, \$P_mu|0

angle

eq 0\$ and vev's depend on position, as discussed in the first part of this answer.On the other hand, internal sources do not break the translational invariance of the theory, because the sources themselves transform together with the fields. This might be easier to understand with an example. The action is not the same as before, and so the translation changed the theory. At this point, you might want to read this answer of mine. In the notation of that post, the \$(2)\$ derivative of a lagrangian with external sources is non-zero.To recapitulate,If there are only internal sources, then the theory is translationally invariant, and so all the vev's are position independent (as can be easily shown using \$P_mu|0

angle=0\$ and \$Q_alpha(x)=mathrm e^-iPxQ_alpha(x)mathrm e^iPx\$, where \$Q_alpha(x)\$ is any field). Most of the times we redefine every field \$Q_alpha(x)to Q_alpha(x)-langle Q

angle\$ so that all the vev's are zero (this is relevant for renormalisation). In some cases (e.g., in the case of the Higgs field) a non-zero vev is physically relevant (but only makes sense because of the form of the lagrangian for the Higgs field, and would not make sense for, say, a standard KG field). In any case, if the sources are internal then vev's are constant.If there are only external sources, then the theory is free. Therefore, the vev's depend on position, but in the limit \$Jto 0\$ we must have \$langlephi

angleto 0\$, as it must be for a free theory.If there are internal and external sources, the vev's are position-dependent and do not go to zero as the external sources go to zero (and therefore we must renormalise the fields).

2. What chores should a 16 year old do around the house?

If you do not get behind it's not hard. Pick up after yourself, do your own laundry (including ironing), wash dishes, vacuum, sweep & mop. "Spread out" your chores during the day, that way u do not get overwhelmed. :-)

3. How is vacuum created inside a vacuum toilet?

It's not a vacuum it's a pressure. In the tank is a pressurized tank that's full of water. When you flush it shoots water into the bowl to get a big flush with little water. Before I retired I worked for the NYCHA and had to remove them because some of them exploded and cracked the tank.

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