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Apr
27
comment What are the anomalies with General Relativity?
I guess you could call dark matter and dark energy an anomaly.
Apr
27
answered Is there a handwavy way to explain what quantum correlation means?
Apr
27
comment While Space-man lives for 1 day, then how long does Earth-man live ? 1000 years or 1 second?
@C.TowneSpringer: Of course no true contradiction (otherwise it would not be a veridical paradox). But an apparent contradiction (because at first view the situation looks symmetric, until you think about it). If there were no apparent contradiction, it would not be asked so often.
Apr
27
comment Velocity measurement in relativistic perspectives
Since $|u|<c$ and $|v|<c$, $1+uv/c^2 \ne 0$, and thus we can just multiply the equation with it, giving the simplified equation $u+v=0$ and thus $u=-v$. That's simple algebra (and anyway, if it would not give that result, relativity would clearly be inconsistent). And your instinct is misled by that word "addition" (and I don't really get that "no transform needed" anyway, to be honest; A's velocity is transformed from $0$ to $-v$, and B's velocity from $v$ to $0$).
Apr
27
comment Velocity measurement in relativistic perspectives
Well, the relevant case here is not $u=0$ or $v=0$, but $s=0$ (using your symbols).
Apr
27
comment Velocity measurement in relativistic perspectives
… the "addition theorem" is the transformation formula for velocities (which therefore exists). That, and only that, is what I claimed. Also note that to derive the Lorentz transformations, you use that the speed of light is constant; the "sign change property" is already true for Galilean transformations.
Apr
27
comment Velocity measurement in relativistic perspectives
It of course doesn't matter whether you imagine one observer being at rest relative to earth and the other relative to a train; all that matters is the velocity of one observer relative to another. The question which leads to the sign reversal is: What is the velocity $u'$ in B's frame so that the velocity in A's frame is $u=0$ (that is, the velocity of A in A's frame). Inserting into the "addition formula" and solving for $u'$ gives $u'=-v$. And yes, this doesn't invalidate your more basic argument (that's why I upvoted your answer). But that doesn't change the fact that …
Apr
27
answered What is the difference between “kinematics” and “dynamics”?
Apr
27
comment Velocity measurement in relativistic perspectives
Basically, the "addition theorem" tells you the answer to the question "if observer B moves with velocity $v$ relative to observer $A$, and an object moved with velocity $u'$ relative to observer $B$, then what is the velocity $u$ of that object relative to observer $A$? That is, how does $u'$ in inertial system $B$ transform to inertial system $A$? The answer being, of course, $u = (v + u')/(1+ vu'/c^2)$. Note that the fact that the transformation is in the unusual direction (B to A, rather than A to B) is also due to the unfortunate "addition" interpretation.
Apr
27
comment Velocity measurement in relativistic perspectives
Yes, but there's $v=0$ relative to an observer. Which is your "missing" second velocity, as each observer is at rest (i.e. $v=0$) relative to himself.
Apr
27
comment Velocity measurement in relativistic perspectives
"Not moving at all" makes no sense in relativity, since there's no such thing as absolute rest. The "stationary frame" is just the frame of some inertial observer which you have chosen to call "stationary" (another unfortunate convention, as your comments clearly demonstrate). Especially, given that observer A is not accelerated, it is a perfect choice for the "stationary" observer.
Apr
27
comment Velocity measurement in relativistic perspectives
Being stationary just means moving at velocity $0$.
Apr
27
comment Velocity measurement in relativistic perspectives
"The Special Theory of Relativity gives us no transform equation for velocity (at least I did not find any)." — It does. It's known under the (IMHO unfortunate) name "relativistic addition of velocities" (the term is because under Galilean transformations velocities transform by addition).
Apr
27
comment While Space-man lives for 1 day, then how long does Earth-man live ? 1000 years or 1 second?
@C.TowneSpringer: From Wikipedia: "A paradox is a statement that apparently contradicts itself and yet might be true." The twin paradox fits that description quite nicely. More exactly, the twin paradox is a veridical paradox, a claim that actually is true, despite the apparent contradiction.
Apr
27
comment While Space-man lives for 1 day, then how long does Earth-man live ? 1000 years or 1 second?
As long as the two never meet again, that question has no answer. Only if they meet again, their age can be meaningfully compared.
Apr
26
comment Is the universe a quantum computer - is light speed barrier a computational constraint
"But yes, a uranium nucleus in a 700MeV excited state is performing an impossibly complex quantum computation that we can't compute even with a computer the size of the universe." — for all we know, the universe might be infinite. That should be more than enough space to calculate an uranium atom, shouldn't it? (The calculation might take a lot of time, though).
Apr
26
revised Mass-Energy Equivalence theory energy or momentum is not conserved?
added 435 characters in body
Apr
26
answered Mass-Energy Equivalence theory energy or momentum is not conserved?
Apr
26
comment Probability of measuring two observables in a mixed state
The state you're measuring is given by $\rho$. There may, or may not, be a $|v_1\rangle$ or $|v_2\rangle$ "underlying" that mixed state $\rho$. Quantum mechanics simply doesn't care how you produced it. But yes, if $\rho$ is created by randomly choosing between $|v_1\rangle$ and $|v_2\rangle$, then two measurements done at the same time necessarily are done also on the same of the two states. But again, it doesn't really matter. You can get the very same mixed state in infinitely many ways, and all of them are indistinguishable by quantum measurements. What you measure is just $\rho$.
Apr
26
comment Probability of measuring two observables in a mixed state
… $\operatorname{tr}(B\rho)=53/14$ which, not coincidentally equals $3 p(b=3) + 4 p(b=4) =$ $3 \left(p(a=1 \land b=3) + p(a=2 \land b=3)\right) + 4 \left(p(a=1 \land b=4) + p(a=2 \land b=4)\right)$.)