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age 53
visits member for 1 year, 10 months
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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.


22h
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 $\beta = \frac{v}{c}$.
1d
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.
1d
answered What happens in a circuit, when the wire and the battery are superconducting. And shorted
2d
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 $m=0, v=c$
2d
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.
2d
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.
2d
answered Why don't we substitute for $p$ in $E = pc$?
2d
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
Apr
11
answered What if photons are not the fastest particles?
Apr
11
comment Why is the black body radiation so important?
@CarlWitthoft, I guess I still don't understand what you want - maybe I haven't yet had enough coffee this morning? My comment was to stress that black body radiation is important because we needed new physics to properly model it. Is that what you want me to stress more?
Apr
11
comment Why is the black body radiation so important?
@CarlWitthoft, the opening question in my comment is a quote of the question title; they're not my words.
Apr
11
answered Does frequency f = $\frac1T$ if there is non-periodic signal
Apr
11
comment Why is the black body radiation so important?
Why is the black body radiation so important? In two words - Ultraviolet Catastrophe: en.wikipedia.org/wiki/Ultraviolet_catastrophe "The ultraviolet catastrophe, also called the Rayleigh–Jeans catastrophe, was a prediction of late 19th century/early 20th century classical physics that an ideal black body at thermal equilibrium will emit radiation with infinite power."