@JohnRennie, Thank you sir. I saw some demonstrations of Rubens' tube. The flames show variation in height when a standing wave is formed in the tube. It is said that the height of the flames show the invisible nodes and antinodes in the tube. My doubt: Where is a pressure node/anti node formed - at the place where flame level is highest or lowest?
@M.GuruVishnu The average pressure is 1 atm everywhere in the tube. At the pressure nodes the pressure is a constant 1 atm while at the antinodes the pressure oscillates above and below 1 atm but with a time averaged value of 1 atm. OK so far?
The pressure inside the tube is actually slightly greater than 1 atm, so with no sound wave we'd get a steady flow rate through all the holes and all flames would be the same height.
And you might expect that since the average pressure is the same everywhere even when a sound wave is present, we'd also get flames with the same height.
@JohnRennie In addition to that I think the flames must show variation in height as a function of time. But we see something stable as time passes. Why don't we see flickers?
You would see flickers, but the flickering would be at the frequency of the sound, and that's far too fast for the eye to see. You could probably capture it with a high speed camera.
@JohnRennie I know only the basic form where we expand using the combinations of the exponent and the placement value. I don't know how to do that for square roots. I know only for integral exponents sir.
For a square root the expression never terminates, i.e. we get an infinite series, but if $x \ll 1$ the higher power terms rapidly become negligible so we'll consider only the $x$ and $x^2$ terms.
The second term is only zero at the nodes i.e. where $\cos(kx)=0$. Everywhere else it averages to a non-zero value.
That means at the nodes (where $\cos(kt)=0$) the average of $\sqrt{P}$ is just 1, but everywhere else the average is less than 1 because it is 1 minus a positive number.
Yes sir :) Thank you. May I know when did you learn these - in school, in college, or your own explanation? I do wish to understand/explain concepts like you.
that's the associated ST diagram, again, I dunno how it would be useful. Is there an obvious way you could read off "distance from ship to Earth in primed frame at time of flare" from that diagram?
yeah
no for sure, I just prefer to work with one set of eq's rather than dealing with inverse as well
it's worked for every other q it's just this one confuses me a lot
I also think their x' axis might be incorrect. I think the slope for x' is just v/c, I'm not sure why it says $\frac{3\sqrt{3}}{2}$
if I recall correctly, when you transform frames (i.e. t' = 0 is the x' axis) you get an x' axis with slope v/c, and similarly the ct' axis has slope c/v
So if you mark the point of the flare then draw a line through that point with gradient $\sqrt3/2$ you get the points in the ship frame at the time the flare happened.
Spacetime diagrams can make things like this very simple, but I find it's easy to make mistakes with them unless you're really experienced with using them.
To be honest I'd have to go away and think about this. As I said, I'm not good with spacetime diagrams and trying to think about this while doing bits and pieces of work isn't working.
The simple algebraic way to do this is to put the first event at the origin so the second point is at some $(T, X)$. Then write down the LT for $t'$ and set it to zero.
also, I was wondering, one part of the question just asked if there IS a frame where the events are simultaneous, and they actaully went through and did a calculation to show that $\Delta{x} > c\Delta{t}$
but is that really necessary?
I thought just the fact that they are some distance apart means they're spacelike separated
Suppose the two events in our frame $S$ are the green and red spots. We are free to choose our origin anywhere we want, so we can choose the origin to be at the green spot.
A manifold is a mathematical structure. It is just a set of points. A manifold has dimensionality, e.g. 2D, 3D, 4D, etc, but no notion of distance. So create a notion of distance we combine it with a metric.
Metrics can be classified by the sign of the time and space parts. If they are all the same this is a Riemannian metric. If the signs are different it is a non-Riemannian metric. If the time and space parts have opposite signs it is a Lorentzian metric.
So the Minkowski metric $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$ is a Lorentzian metric.
Though I wouldn't take a course in it from a maths dept unless it was specifically a course for physicists. Mathematicians and physicists tend to use different ways of formulating DG and I susect doing a maths course would just confuse you.
I believe you though, but it's such a big subject that I doubt I'd be able to learn it on my own. For instance, I plan to teach myself group theory over a summer with a group theory for physicists book because the only group theory course at my university is specifically focused for mathematicians.
DG I think would be ok though, I'd at least probably be able to translate its usefulness over time to physics.
@psa you need to start with the idea of parallel transport. What this means is you take some vector and you transport it along a curve while keeping the direction of the vector constant.
This is the sort of thing I mean. We transport the vector round some random loop and back to its starting point, and we keep it parallel to its previous position the whole time.
@psa so when you parallel transport round a loop on a flat surface the direction of the vector is always unchanged. When you parallel transport round a loop on a closed surface the direction of the vectr is in generl changed.
@psa Yes, the Riemann tensor describes the rotation of the vector. If the Riemann tensor is zero it means the direction never changes.
In real life it's a 4D loop so the Riemann tensor is a complicated object, but basically all it does is describe the rotation when you parallel transport round an infinitesimal loop of size $dt$, $dx$, $dy$, $dz$.
A geodesic is the equivalent of a straight line in a curved space. It is the path followed by a freely moving object with no external forces acting on it.
So if your vector is moving freely then it follows a geodesic and it is parallel transported along that geodesic.
i.e. if I fling you towards a black hole you will be following a geodesic and the vector from your feet to your head is parallel transported along that geodesic. So it's what happens in real life.
Yes. The pendulum is accelerating because the force is also inwards, so as the pendulum passes the centre point it has some speed $v > V_0$. As the pendulum moves inwards the acceleration decreases, and as it passes the central point the acceleration becomes zero.
So this is like your question i.e. the acceleration is decreasing with time but the velocity of the pendulum is increasing.