Reputation
47,867
Next tag badge:
120/100 score
19/20 answers
Badges
4 70 151
Impact
~1.8m people reached

Apr
18
comment Stanford: “Objects in spacetime all move at constant speed $c$.” Are they right?
I've deleted some comments that seemed to be trying to pick a fight. That's not what we do here. But allow me to expand a little: physicist can and do disagree on theory, interpretation, and pedagogy and that is not a problem. And sometimes they air those disagreements in public and that too is not a problem.
Apr
18
comment Is the LHC more dangerous if it works with more luminosity?
Here's the thing: high energy cosmic rays have been running more energetic experiments that we can for billions of years. The LHC simply isn't dangerous, full stop.
Apr
18
answered Is a spring shaped like a sine wave?
Apr
17
comment Can $E=mc^2$ be derived from the Minkowski spacetime metric? $s^2=x^2+y^2+z^2-(ct)^2$?
@Ilja Most professionals in relativity have long since discarded the notion of "relativistic mass", and use only the invariant mass so they don't write $m_0$ because there is only $m$ which is not a function of velocity. There are a lot of questions around the site that touch on it, but physics.stackexchange.com/questions/133376/… is probably the best. On top of which the form you suggest doesn't deal with massless particles.
Apr
17
comment Can $E=mc^2$ be derived from the Minkowski spacetime metric? $s^2=x^2+y^2+z^2-(ct)^2$?
Look at our formatting page for some minimal information about using the MathJax markup tool for mathematics.
Apr
17
comment What information am I losing out when I assume that the displacement in S.H.M. is small?
This. If the oscillator is truly harmonic (as per the title) then nothing is lost but the approximation is unnecessary.
Apr
17
comment Can $E=mc^2$ be derived from the Minkowski spacetime metric? $s^2=x^2+y^2+z^2-(ct)^2$?
Though it is quite orthogonal to your question I would encourage you to not imagine that $E = mc^2$ is in anyway emblematic of relativity. If you really must chose an equation for such a purpose you should be using $(mc^2)^2 = E^2 - (\vec{p}c)^2$. This reduces to the simpler form when $m \ne 0$ and $\vec{p} = 0$ but covers cases the other does not and gives mass it's proper definition as a Lorentz invariant.
Apr
17
comment Can 2 photons make up the same colour as another photon?
Color is a sensation; a feature of the human visual system. It is linked to the physical properties of the incident light (frequency in wave terms or energy in photon terms), but there is not a one-to-one correspondence between them. Or rather there is a nice one-to-one correspondence between the perceived color and frequency for a monochromatic source, but poly-chromatic sources have surprises up their sleeves.
Apr
16
comment How much ionizing (carcinogenic) radiation is one exposed to on a commercial flight, what are the sources, and how could exposure be minimized?
The word you want is not "carcinogenic", but "ionizing", not withstanding the ionizing radiation is a cancer risk. And most of the dose is in charged particles, not photons so there is basically nothing you can do about it.
Apr
16
comment Why positron emission is unlikely to occur for nuclei with an excess of neutrons?
Follow the logic. "Positron emission can only occur when a ___ is converted into a ___ inside the nucleus, but in a neutron rich nucleus adding a ___ takes more energy than you get from removing a ___ so the event results in a net energy gain to the nucleus."
Apr
16
comment Symbol $p^{0}$ of particle
I'm a bit confused because you both asked for a particle and tagged the question with particle physics, but accepted an answer about something quite different. Did you actually meant the zeroth component of momentum $p^0$ which is the conventional meaning of the symbol you supplied and the answer given below, or did you mean $\rho^0$ the symbol for a particular uncharged, vector meson which is the particle whose symbol most closely resembles the one you supplied and is often misidentified as a roman p?
Apr
15
comment Why only light nuclei are able to undergo nuclear fusion not heavy nuclei?
Related (in effect duplicates): physics.stackexchange.com/q/80256 physics.stackexchange.com/q/215769 plus physics.stackexchange.com/q/168237 and several others where the meaning of binding energy is explored in some detail.
Apr
13
comment How did we realize that light travels at a finite speed?
And there were already quite reasonable purely terrestrial measurements by the beginning of the 20th century.
Apr
13
comment Antique X-Ray Tube Safety
The dosimetry is the really stinker and the reason a E&HS rep would look askance at using this thing for a classroom demo. If you really want to run it for the cool factor, take video of yourself demonstrating it, then show the video. Our new display case is going to have a video-display on the opposite wall, so if we run our we'll put the video in the loop the screen shows. The place these things hold in the history of physics is important and worth explaining, but is it worth a unmetered (if probably small) dose delivered to a lot of students?
Apr
13
comment Antique X-Ray Tube Safety
We found something similar here, while cleaning out the prep room for a move to temporary quarters. It is destined for the display case when the re-model of our space done. You could operate it behind a bunch of shielding, but what would you gain by doing so?
Apr
13
comment Why do we use Bequerel to measure Radioactivity in food?
I used to approximate the activity of a banana as part of party trick we did for people visiting a place I worked at that had a low background Ge-detector system. But "decays" isn't a good measure of health effects.
Apr
11
comment What is the desest material on earth?
It always amuses me that people use lead as an exemplar for density when it is only middling dense compared to gold, mercury, and tungsten much less exotics like uranium metal, osmium, and heavymet.
Apr
11
comment Can special relativity be derived from the invariance of the interval?
@David Getting the general invariance of the interval takes some work, but the invariance of light-like intervals is the invariance of $c$. Like this: $c = (\Delta x)/(\Delta t)$ so $c \Delta t = \Delta x$ so $(c\Delta t)^2 = (\Delta x)^2$ so $(c\Delta t)^2 - (\Delta x)^2 = 0$. All those games Einstein plays with light clocks and so on rely on the invariance of light-like intervals.
Apr
10
comment Dalton's law of additive pressures
We have the MathJax math rendering engine active on the site, and it is preferred to posting images of math because it is editable and agrees in size and typeface with the rest of the post. I've done this one for you. You can fine some basic help on our formating page, and much more on on various latex site (MathJax's input language is approximate the same as LaTeX mathmode).
Apr
10
revised Dalton's law of additive pressures
remove image of math in favor of MathJax