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HNQ
2:01 AM
1
Q: The fine structure constant

LSSI have a question about the really meaning of the fine structure constant. Actually, my doubt is why the physicist look to a meaning for this constant. I understand the difference between this constant and the ordinary constant, that is, it carry no units with it. But is not this just like pi or ...

 
 
16 hours later…
HNQ
5:43 PM
1
Q: Work done in assembling a point charge is infinite

Yasir SadiqGriffiths says The energy of a point charge is infinite$W=\frac{\epsilon_{0}}{2\left(4 \pi \epsilon_{0}\right)^{2}} \int\left(\frac{q^{2}}{r^{4}}\right)\left(r^{2} \sin \theta d r d \theta d \phi\right)=\frac{q^{2}}{8 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{1}{r^{2}} d r=\infty$ And then he s...

1
Q: Is Hawking radiation real for a far away observer?

AnOrAnIt is my understanding that Hawking radiation is observed by esternal observers and at the same time a necessary condition for having Hawking radiation is the formation of an event horizon during a gravitational collapse. Since the emergence of an event horizion takes infinite time for an observe...

 
HNQ
6:08 PM
1
Q: Why is a vector a rank-1 tensor?

TaeNyFanI learned that a tensor is a linear map from some number of vectors to the real numbers. For example, $t(\vec{u},\vec{v})=\vec{u}\cdot\vec{v}$ is a rank-2 tensor as it takes 2 vectors as arguments and combines them into a real number. I then read that a rank-1 tensor is a vector. How is that poss...

 
 
2 hours later…
HNQ
8:36 PM
1
Q: How do Rindler coordinates fit into special relativity?

AccidentalTaylorExpansionIt is often said that special relativity doesn't handle acceleration since you general relativity for that. If I understand correctly this is incorrect just with some caveats. Rindler coordinates are coordinates as experienced by a constantly accelerating observer such that the observer is at res...

 

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