Higgs field at the event horizon

Antiphon
Messages
1,685
Reaction score
4
If the postulated Higgs field has a non-zero vacuum expectation, shouldn't the typical black hole be ingesting Higgs particles at an alarming rate?
 
Physics news on Phys.org
The Higgs field is a background field and does not contain any Higgs bosons. It is the 'vacuum state' for Higgs bosons, i.e. the state with occupation number zero. A black hole has no effect on it, and vice versa.
 
That makes sense of course. So then if there are no Higgs particles in empty space, how do They impart mass to the typical particleslike the electron or proton? Is it via virtual Higgs particles in the immediate vicinity of say the electron?
 
A Higgs vev is not the same thing as a Higgs particle, which is an excitation of the Higgs field of definite momentum.
 
Yes, that makes sense too. (What's a vev?)
 
Antiphon said:
Yes, that makes sense too. (What's a vev?)

vacuum expectation value
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

Similar threads

Replies
21
Views
3K
Replies
7
Views
2K
Replies
46
Views
7K
Replies
20
Views
2K
Replies
40
Views
3K
Replies
44
Views
3K
Replies
12
Views
1K
Back
Top