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5:22 AM
@JohnRennie Good morning sir :-)
@user8718165 morning :-)
5:42 AM
@JohnRennie, Hi sir :)
@M.GuruVishnu hi :-)
@JohnRennie May I ask a doubt regarding Rubens' tube?
@M.GuruVishnu yes, go ahead.
@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?
The flame is highest at the pressure nodes, though the reason for this is a little complicated.
5:55 AM
@JohnRennie Pressure nodes = Displacement antinodes; Particles here move violently. Shouldn't I expect flame to be highest at pressure antinodes = Displacement nodes? I'm slightly confused sir.
@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?
@JohnRennie Yes sir.
The height of the flames shows the gas flow rate through the holes in the tube. Where the flame is high the gas flow rate is large.
@JohnRennie Yes sir.
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.
OK so far?
6:01 AM
@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 Ok sir. Understood.
@JohnRennie Now yes :)
But anyhow, we still have to explain why there is any variation in height at all, given that the average pressure is the same everywhere.
@JohnRennie Yes sir.
And the reason is that the flow rate is not proportional to pressure ...
It is proportional to $\sqrt{P}$.
6:04 AM
@JohnRennie Is that a consequence of Bernoulli's Equation?
And the average value of $\sqrt{P}$ is not the same everywhere. It is highest at the nodes and lowest at the antinodes.
"It is highest at the nodes and lowest at the antinodes." - Why is it the opposite way sir? If possible, could you please explain this?
I'll have to have a think about the best way to explain this. It's probably to expand $\sqrt{P}$ using a binomial expansion.
@M.GuruVishnu do you know how to do a binomial expansion?
@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.
The equation is exactly the same for fractional powers. The expansion we need is:
$$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} + ... $$
6:16 AM
@JohnRennie Ok 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.
$$ (1 + x)^{1/2} = 1 + \tfrac12 x - \tfrac18 x^2 + ... $$
@JohnRennie Yes sir. I've seen the same in an approximation in the chapter gravitation (variation of $g$ with height).
@JohnRennie But isn't the $x^2$ term also neglected sir?
@M.GuruVishnu in this case the $x$ term is going to average to zero, and the first non-zero term is going to be the $x^2$ term.
@JohnRennie Fine sir. Here $x$ is the difference in pressure from the mean value. Am I right sir?
Correct. And that difference is going to be gven by the equation for the standing wave:
$$ x = P \cos(kx)\sin(\omega t) $$
Where $P$ is the amplitude of the sound wave, $x$ is the distance measured along the pipe and $\omega$ is the angular frequency of the sound wave.
6:22 AM
Ok so far sir.
So the nodes are where $kx = \pi/2$, $3\pi/2$, $5\pi/2$, etc.
OK. We'll write the pressure in atmospheres, so the total pressure is $P \cos(kx)\sin(\omega t)$
So the flow rate is proportional to $(1 + P \cos(kx)\sin(\omega t))^{1/2}$. Yes?
@JohnRennie Yes sir.
Now $P \ll 1$, so we can use the binomial expansion on this:
$$ (1 + P \cos(kx)\sin(\omega t))^{1/2} = 1 + \tfrac12 P \cos(kx)\sin(\omega t) - \tfrac18 P^2 \cos^2(kx)\sin^2(\omega t) $$
where I've considered only the first two terms.
This is the key step, so you need to be happy with the argument so far.
Ok sir.
Consider the three terms on the right hand side. The first term is $1$ and the average of $1$ is obviously just $1$.
The second term is $\tfrac12 P \cos(kx)\sin(\omega t)$ and the average of this term is zero. Can you see why it is zero?
6:30 AM
@JohnRennie Yes sir. Much like the mean current is zero but the mean squared current is non zero.
Yes, you've got it :-)
The second term is $\tfrac18 P^2 \cos^2(kx)\sin^2(\omega t)$ and the average of this is not zero.
The second term does not average to zero because it's squared so it is always positive.
@JohnRennie Yes sir.
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.
@JohnRennie So, as the average pressure is higher at the nodes, butane or any other fuel feels happy to come out of the holes near the nodes?
Average of $\sqrt{pressure}$ not average pressure.
6:37 AM
@JohnRennie Yes sir. My mistake.
The avreage pressure is 1 atm everywhere. The average root pressure is $\sqrt{1 atm}$ at the nodes and less than $\sqrt{1 atm}$ everywhere else.
And the gas flow rate is proportional to $\sqrt{P}$ so the gas flow rate is highest at the nodes.
@JohnRennie Thank you very much sir.
So again, thinking on the lines of displacement waves is incorrect?
It happens because the exponent is less than 1 and that makes the second term in the expansion negative.
If the exponent was greater than one the second term would be positive, and the flow rate would be greatest at the antinodes.
@M.GuruVishnu This probably seems like a lot of work to explain quite a simple experiment! :-)
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.
@M.GuruVishnu I've been doing this for 40 years. Over that time you pick up a lot of stuff.
The easy way to learn everything I know is to wait 40 years :-)
6:47 AM
@JohnRennie Thank you sir. That was a brief explanation of a long-term exploration of physics :-)
@JohnRennie hi
@psa hi :-)
I'm redoing/reviewing the homework sets for relativity for final exam but I'm getting pretty confused on one of the q's we worked on
they posted solutions but they used the inverse LT's and I prefer not to do that
they also drew a spacetime diagram, but they never showed how we could use it to actually find what we were looking for in the q
you might remember this Q
OK. I have to confess I'm not great on spacetime diagrams as I don't have a very graphical mind. I prefer to use equations.
I'm fine with working thru it with equations, I'm just also curious how the diagram would be useful in Q's like this.
I'll give you their solutions
6:56 AM
OK ...
notice that they use inverse LT's. I've been trying to do it without them but I don't understand how.
If the LT is the transformation from the Earth frame to the ship frame the inverse LT is just the transformation in the other direction.
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?
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 $\frac{3\sqrt{3}}{2}$
As I said, I'm not that good with the diagrams, but I think I see what they are doing.
@psa The line with slope $c/v$ is the trajectory of the ship in the Earth's frame. It is not the $x'$ axis.
no not that one, the one they labeled x' axis
7:05 AM
Ah sorry yes, the angle of the $x'$ axis to the $x$ axis is given by $\tan\theta = v/c$
So in this case the slope is $\sqrt{3}/2$
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
And lines parallel to the $x'$ axis are the lines of equal $t'$ i.e. the lines of simulaneity in the ship frame.
right yeah
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.
ah ah okay yes
I see how they get $3\sqrt{3}/2$ because we just moved 3 units to the right to get to Right (where the flare occurred)
7:09 AM
Now draw the trajectory of the ship on your diagram, and where the lines intersect is where the ship was in its frame when the flare happened.
or wait
the earth is just vertical*
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.
isn't the x' axis just the ship's trajectory?
cause the ship is moving at $\sqrt{3}/2$
I need to work I'm afraid. I'll be back in half an hour o so.
you're probably right, but we get multiple choice questions on these diagrams all the time so I'm just trying to understand the basics of it
OK sounds good
7:31 AM
@psa hi, I'm back! :-)
I'm still so confused how I would use the diagram for this purpose, even after knowing the answer with the equations : (
@psa the ship's trajectory has gradient $c/v$ not $v/c$. Remember time is on the vertical axis not the horizontal axis.
So on your spacetime diagram mark the position of the ship and the position of the flare.
Draw a line with gradient $c/v$ through the position of your ship and that's the ship trajectory.
Draw a line with gradient $v/c$ through the flare and that's the ship's $x'$ axis at the time of the flare in the ship's frame.
So where the two lines intersect is the position of the ship at the time of the flare in the ship's frame.
I think ... :-)
so then how do you find where the Earth is relative to the ship at the time of the flare in the ship's frame?
the original Q was just where is the Earth in the ship's frame when the flare occurs in the ship's frame
and it turned out to be that the flare had occurred in the past, and the Earth was ahead of the ship (to the right of it) when the flare occurred
7:38 AM
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.
haha OK no worries
I just feel like these questions would be so much easier if I understood these diagrams
Sorry :-(
it's totally fine
8:09 AM
@JohnRennie how do we find the simultaneous frame of reference for a spacelike separated event using the ST invariant?
I found it using the Einstein v addition formula, but was curious on how you would do it this way.
You mean the frame in which the two events happen simultaneously?
well, I'd like to find the speed at which the observer would have to be moving
to see those events as simultaneous
just in one dimensional motion
if the events are simultaneous, the $c\Delta{t}$ term goes to zero, but then we're just left with the proper length aren't we?
Isn't it just given by the gradient of the line connecting the points on a spacetime diagram?
I can give you the actual question I'm wondering about to be concrete about it
The line connecting the two points is the $x'$ axis of the frame in which both events happen at time zero ...
I don't think there is an easy way to do this just using the proper time/length.
8:17 AM
If you consider the four vector connecting the two points then what Lorentz transformations do is rotate this vector.
Specifically they rotate it by an angle $\tan\theta = v/c$
So you're just looking for the velocity that rotates the vector to be horizontal.
alternatively we could just set $\Delta t' = 0$ where the primed frame is the one we're interested in I guess? (using LT's)
I don't know why, but I find ST diagrams so damn confusing
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.
Solve for $v$ and you're done.
so you're saying I'd be solving for the A = that matrix such that Av = 0?
That seems like making a meal of it.
8:23 AM
haha yeah
I'm just thinking about the rotation though
I guess I'm just associating "rotating stuff" with linear transformations
Well it is a linear transformation, but you can short cut it.
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.
The dashed lines are the $x'$ and $t'$ axes of the frame $S'$ moving at some velocity $v$.
Since the $x'$ axis passes through both events that means both events happen at time $t' = 0$ in the $S'$ frame i.e. they are simultaneous in $S'$.
8:31 AM
If the coordinates in $S$ of the red event are $(T,X)$ the gradient of the line is $T/X$, so the velocity of $S'$ is $v = T/X$ (using $c=1$).
If $T > X$ ($cT > X$) that would mean $v > c$
Algebraically write:
$$ t' = \gamma\left(T - \frac{vX}{c^2}\right) = 0 $$
And solve for $v$
9:00 AM
@JohnRennie what the hell is a Lorentzian manifold?
@psa Do you know what a metric is?
I learned about a metric as like a "distance" function, if that's what you're referring to.
Yes, exactly.
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.
why call it a manifold? does it have special properties?
it just sounds like a space with some basis to give it dimension
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.
9:05 AM
that's the usual thing we work with
in SR
If you combine a manifold with a Lorentzian metric the result is a Lorentzian manifold.
so is spacetime a manifold with a metric?
In SR spacetime is just a Lorentzian manifold.
what do we mean by spacetime is "flat" then? that doesn't look very flat to me
This is a grossly simplified description and no doubt any mathematicians reading this would be screaming at me, but it covers the basics.
9:07 AM
looks more like a spherical term with some other stuff
Terms like flat in SR have a precise mathematical meaning that is not always in accordance with what you think of as the everyday meanings.
what does flat mean in SR?
i just thought it meant no curvature
The curvature of a spacetime is described by an object called the Riemann tensor.
(You can Google all these terms, but be warned this is a deeeeeep rabbit hole :-)
i've been exposed to tensors but only on a really basic level
The term flat means that the Riemann tensor is zero.
9:09 AM
how does a tensor give you curvature?
nonzero tensor*
How long have you got? :-)
Seriously, what you're asking about is a subject called differential geometry, which is fascinating but complicated.
DG is the foundation of general relativity.
yeah, my advisor told me to take both classes (classical and modern)
sounds pretty nuts
I would have thought differential geometry was a bit advanced for a first year undergrad ...
You'd normally only do it in your final year if you opt for the theoretical physics courses.
9:12 AM
i'm in 2nd, and he meant take it in 3rd and 4th
Ah, yes, that makes sense.
DG is such an elegant subject. Everyone I know who has studied it loves it!
that's pretty awesome
I'd definitely love to know more about it so I can know more about gravity!
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.
unfortunately we only have it offered in the math department, and the only GR course we have is quite hand-wavy
the only undergrad GR course I mean
There's a simple way of describing what the Riemann tensor tells you. I can try and explain if you want ...
9:16 AM
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.
That would be a great idea if you want to do theoretical physics. I never formally studied group theory and I do find myself wishing I had.
@JohnRennie hi.alesha here
yeah, he recommended group theory, DG, calculus of variations, and complex analysis (for QFT calculations, poles and the like)
@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.
@yuvrajsingh hi :-)
today i have exam on kinematics ,i came across some question swhich i do not get.
9:18 AM
oh, they have an exam
@yuvrajsingh give us a few minutes to finish talking about general relativity. I'll ping you when we are finished.
should we continue this conversation some other time?
so I'd imagine its basis vectors and its components stay the same through the movement
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.
9:22 AM
So obviously when we get back to the starting point the direction of the vector hasn't changed.
@psa OK so far?
@psa OK. Now I'll do the same thing but on the surface of a sphere. It will take me a moment to draw this ...
@psa there!
I start on the equator with a vector pointing North, and I parallel transport it up to the North Pole keeping its direction constant the whole time.
Step 2 is to parallel transport it back to the equator along a line at 90° to the first line. Again keeping its direction constant.
and now they're orthogonal
And the final step is to transport it along the equator back to the starting point, and as you say we find the direction of the vector has changed!
9:31 AM
sir can i meet you 1 hour later @JohnRennie till,then i can do some questions
@yuvrajsingh yes I'll be here in an hour, though psa and I have almost finished now.
is this what we mean by curvature? when parallel transport can result in the vectors not being parallel?
@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.
And that's what we mean by a flat spacetime.
the rotation of the vector under some transformation like parallel transport?
@psa Yes
9:35 AM
the components of this vector changed
like, they're now completely switched
it's weird
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 $dt$, $dx$, $dy$, $dz$.
why do you want to parallel transport? is this what objects do in space?
Do you know what a geodesic is?
the shortest distance between two points on a curved surface, I believe?
like a generalization of a straight line in cartesian coords.
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.
9:38 AM
so by parallel transporting, you're following the geodesic?
wow that is so cool
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.
that's awesome
I told you DG was cool :-)
it's actually extremely cool!
it's cooler than I could have imagined.
Shall I see if @yuvrajsingh is still here. I probably ought to answer her question now.
9:40 AM
yeah for sure
I'm gonna go to bed
thanks for the talk!
@yuvrajsingh are you still there?
@psa you're welcome. Sleep well :-)
10:40 AM
@JohnRennie hi
@yuvrajsingh hi :-)
tjhe acceleration time graph of a particle
Yes ... ?
moving along a straight line.
actually i can,t post diagram ,but it is line ihaving negative slope
where intial acceleration intially is 10ms
and on the x axis graph cut at t=4 sec
at what time particle acquire it intial velocity.
Like that?
10:45 AM
yes sir.
So $a = 10 - 2.5t$. Yes?
sorry got it.
And $a = dv/dt$ so we have the differential equation:
$$ \frac{dv}{dt} = 10 - 2.5 t $$
If we integrate this we get:
$$ v(t) = 10t - 1.25 t^2 + V_0 $$
where $V_0$ is the initial velocity i.e. the velocity at time $t = 0$. Yes?
10:49 AM
And the question is asking when is the velocity equal to $V_0$ again.
i.e. we need to set $v(t) = V_0$ in our equation and solve for $t$.
ok sir ,but i have doubt
sir i can see from the diagram that acceleration is decreasing,
with respect to time
OK ... ?
10:52 AM
how can the velocity be same again.
although magnitude of velocity can be same.
but direction would be different.
Let me give you an example of where this sort of behaviour happens, though the details of the motion will be different.
@yuvrajsingh Consider this pendulum. We start at some displacement $x$ and the pendulum moving inwards with an initial velocity $V_0$. OK so far?
this velocity is inward direction,
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.
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