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age 53
visits member for 1 year, 10 months
seen 6 hours ago

BEE 1992, MEE 2001, PhD in progress

Starting building Heathkits in early 70s. Consumer electronics tech through 80s. EE since then. Sold company in 00s. Teach EE classes on occasion now.


10h
comment Is it really “time” that is dilating?
I believe that the "gravity clock" on the accelerated Earth will, by the equivalence principle, indeed be affected by the acceleration.
10h
answered Do I understand impedance correctly?
21h
awarded  Nice Answer
1d
answered What's the physical meaning of the imaginary component of impedance?
2d
awarded  Nice Answer
Apr
18
comment About accelerating particles
@user128932, to accelerate a particle arbitrarily close to c requires arbitrarily large energy. Assuming the final speed is ultra-relativistic, the required energy is approximately $\Delta E \approx mc^2 \frac{\beta}{\sqrt{1 - \beta^2}}$ which is unbounded as $\beta \rightarrow 1$ where $\beta = \frac{v}{c}$.
Apr
18
comment About accelerating particles
@user128932, a particle at rest in one frame of reference is travelling arbitrarily close to $c$ in another. Motion is relative, not absolute.
Apr
17
answered What happens in a circuit, when the wire and the battery are superconducting. And shorted
Apr
16
comment Why don't we substitute for $p$ in $E = pc$?
@rahulgarg12342, as others have mentioned, setting the invariant mass $m=0$ gives $p=0$ for any $v < c$. However, if $v = c$, $\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}$ is not defined thus, the equation $p = \gamma m v$ cannot be applied for the case $m=0, v=c$. However, one can rearrange the equation as I did and see that, in fact, $p$ can be non-zero for the case $m=0, v=c$
Apr
16
comment Why don't we substitute for $p$ in $E = pc$?
@rahulgarg12342, the momentum for a photon is $p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}$ so yes, a photon of red light will have less momentum than a photon of blue light.
Apr
16
comment Why don't we substitute for $p$ in $E = pc$?
@rahulgarg12342, I don't see a contradiction. If the invariant mass is non-zero, the momentum is proportional to the invariant mass. If the invariant mass is zero, the four-momentum is null, and the (three-) momentum is proportional to the energy.
Apr
16
answered Why don't we substitute for $p$ in $E = pc$?
Apr
16
answered About accelerating particles
Apr
15
comment How are magnetic and electric fields transmitted?
you must sharpen your thinking about fields and the terms you are using. Frankly, I don't have a clue as to what your final sentence is supposed to mean.
Apr
15
comment How are magnetic and electric fields transmitted?
A field isn't 'transmitted', a field is defined on all of space and time. In particular, a constant E field does not change with time. A uniform E field is the same over all space.
Apr
15
comment Is there any distinction between these products: scalar, dot, inner?
@garyp, this is the section I recall from Schutz: books.google.com/…
Apr
15
comment Is there any distinction between these products: scalar, dot, inner?
@garyp, without a metric, there is no mapping between $\vec A$ and $\tilde A$. So, while $\tilde A(\vec B)$ exists, $\vec A \cdot \vec B$ does not.
Apr
14
comment Is there any distinction between these products: scalar, dot, inner?
@jinawee, correct, $\mathbf g(\vec A,) = \tilde A$
Apr
14
comment Why can computer circuits recognise only two states?
To be sure, humans understand binary, computers don't have the capacity for understanding. But, as to your question, the early digital computers did not operate with binary representations but with decimal representation: en.wikipedia.org/wiki/Decimal_computer
Apr
14
comment Is there any distinction between these products: scalar, dot, inner?
I believe I read in Schutz that the contraction of a vector and co-vector (one-form), $\tilde p(\vec A)$ does not require the intervention of a metric whereas the dot product of two vectors $\vec A \cdot \vec B = \mathbf g(\vec A, \vec B) = \tilde A(\vec B) $ does