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Sir Cumference

2:16 AM

Did Pythagoras or Euclid come up with the 2-norm?

Because it's called both the Pythagorean theorem and Euclidean norm

Nihar Karve

3:34 AM

Would you count $c = \sqrt(a^2 + b^2)$ as a Euclidean norm? How would you define 'coming up with it'?

If it's any help, Pythagoras came ~200 years before Euclid

Sir Cumference

3:57 AM

@NiharKarve more or less the idea that taking two lengths, squaring them, summing them and taking the square root would give you a new (useful) length

that's not exactly an intuitive thing for most people to come up with

Nihar Karve

4:15 AM

Well, I assume it would be Pythagoras then

That's not to say it wasn't known in the Early Babylonian era

Secret

4:41 AM

Quantum Field Theory visualized

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hmm...

123

Hi Morning...!

Secret

I am guessing trying to do the direct functional integration of the photon field with the electron field is too hard which is why it needs to be split up into feymann diagrams

123

Why Rocket motion can not be (or hard to) explain by newton's law??? What is the problem.

Pls help me out guys.

Nihar Karve

@123 @danielunderwood gave a brief answer to that - what part do you still not understand?

123

Sorry i did not see that i was busy. Is that yesterday??

JMac

4:47 AM

@123 It can be explained by Newton's laws quite well. The issue is that it's not as straightforward as most types of motion because the forces are due to some of the mass being ejected quickly from behind. There are different ways to look at it, but when you do it as force over time, there's a good amount to consider with the changing mass, and momentum transfer between the ship and the expelled fuel.

123

Thanks i see.. Excellent. great explanation

Azmuth

Most of the Rockets use channel profile, realtime SCADA kinda systems, they don't give a damn to any laws whatsoever....

JMac

Like the "rocket equation" is in terms of delta v

123

If mass is changing continuously in rocket. Does it follow law of conservation??

Azmuth

yes

JMac

4:50 AM

Well yeah, mass is only conserved in isolated systems, if you look at just the rocket when it's expelling mass, that's not an isolated system, mass is leaving.

Azmuth

That's the highest number of stars I've got till now!

Nov 3 at 13:26, by Azmuth

Programmers often confuse between Halloween and Christmas because Oct 31 = Dec 25.

Nihar Karve

@123 briefly, $F = m\ddot{x}$ won't hold because mass is also changing, so you have to make it $F = \frac{dp}{dt} = \dot{m}\dot{x} + m\ddot{x}$, using the product rule

123

Hi @Azmuth

How it follow law of conservation of momentum. If it changing mass continuously .

Sir Cumference

the total mass of the system is constant

just the mass of the rocket is changing

123

@NiharKarve Yes i used this so many times. But did not connect it with physical problem. Now it explain rocket problem. Thanks

danielunderwood

5:02 AM

Rockets work because of conservation of momentum. If you have no external forces, then $\frac{dp}{dt} = 0 \implies m \ddot{x} = - \dot{m} \dot{x}$, which is where the rocket equation mentioned above comes from.

If the mass is ejected one way, the rest of the rocket needs to speed up in the other direction to conserve momentum

123

@danielunderwood Superb.. :O

It means mass loss is taken by speed increase of rocket and momentum conserved?

Nihar Karve

Yes

123

Hmm... Thank you guys..

I have question about tension force of Atwood Machine. I have read SE thread of tension but did not explained me well.

3

A: Can two bodies of different mass attached at ends of string passed over a frictionless pulley produce equal tension on both sides?

That is a really good question that usually books fail to address in an intuitive way. Let me give you a step by step intuitive explanation by looking at the Atwood's machine: Start with the balance of forces around the cable: $$𝑭_𝟐−𝑭_𝟏=𝒎𝒂$$ F2 is greater than F1 since the cube has m...

4

Q: Pulley system: how can tensions be equal throughout a entire rope if the weights on opposite ends are different?

More weight means more pulling force, and thus more tension force. Like this answer says, however much force you pulling with is equal in magnitude to the tension force. If that is case, how then can tensions forces in this photo be equal if their sources are of different weight:

9

Q: When is tension constant in a rope?

Suppose we have a massless rope with pulling forces applied at each end. In which scenarios is the tension in the rope constant throughout? For example if there is a knot in the rope the tension is not constant throughout (why?. Similarly if the rope is hung over a cylindrical pulley of non-negli...

homework-and-exercises newtonian-mechanics forces free-body-diagram string

These i have read. But how tension in rope segment by segment work still not clear.

What happened at left side of tension segment by segment where mass $m_1$ which is lower than right side $m_2$. Thus $m_2 > m_1$.

JMac

5:27 AM

@123 Imagine a very stiff massless vertical spring (i.e. an ideal rope) holding a mass still. When you pull on the spring with a force equal to it's weight, but in the opposite direction, it stays still, and tension in the spring equals the weight of the mass and the force opposing it.

Now imagine the upwards force is greater than the weight of the mass. The system accelerates upwards, but the tension in the spring is still just the weight of the mass downwards, and the rest of the upwards force accelerates the mass/spring system. The tension in the rope is very similar, except the upwards force is pointed downwards by the rope and pulley, and caused by another mass.

123

@JMac what happened to the tension segment by segment on the string by weight and acceleration on each side?

JMac

@123 Actually I think I worded that wrong, the tension in the spring is the force of the weight upwards along with the extra force accelerating it upwards. Anyways, the tension is the same all through the rope in a rope pulley system. On one side (with more mass) the rope is falling with the tension opposing the weight but not fully, on the other side, the tension is greater that the weight and so that mass accelerates upwards.

Since the rope doesn't stretch, the accelerations of both sides must be equal and opposite as well.

123

@JMac Thanks for explaining. I know Tension is same on each side due to massless string (or very low mass compared to hanging masses).

JMac

Yeah you need to assume the string is massless for it to make sense.. It's always important to pay attention to the assumptions like that too, since ideal cases aren't always exactly like real life.

123

I wanted to know the left side lower mass $m_1$ a bit of string going upward due to acceleration than lower bit of sting relaxes than it again stretched by weight.

This explanation i need to understand. What happened bit by bit

@JMac Yes you are right.

Secret

5:41 AM

$$\int D[\phi] \int_S d \mathbf{x} \Psi^* (\mathbf{x})[\phi]F_{\mu\nu}[\phi]F^{\mu\nu}[\phi] \frac{i\gamma^\mu\partial_{\mu}\psi (\mathbf{x})}{m_e} \Psi(\mathbf{x})[\phi]$$

pretty sure... that's not how it works...

JMac

@123 tension in an ideal massless rope is pretty boring bit by bit. At every point along the rope, tension is pointed in opposite lengthwise directions on each segment of the rope, and the tension in the opposite directions is always equal.

123

But SE thread what i shared. They emphasize it need to be understand at every point along the rope. That's why i confused.

@JMac Your explanation is pretty well to understand. But i cant figure out what happened at every point. That's why i asked question

JMac

Not sure what one you're talking about.

123

Above i shared SE thread.

JMac

You shared three though.

123

5:46 AM

Yes in those thread they said tension need to be understand segment by segment.

How make new line of text in LateX

I want to write line by line.

Kenzo Tenma

Some people just intentionally initiate fire (violence) so that they can later get credit for putting it off. That's what I learnt from this amazing song.

DJ Snake - Magenta Riddim

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123

What ideal conditions need to be taken in Atwood Machine \\For String : \(a) Massless (negligible small compared to hanging mass) \(b) Non-stretchable string \\\\For Pulley : \(a) Frictionless (because heat not loss and resist acceleration) \(b) Massless (Moment of interia not effect the acceleration)

What else ideal conditions need to be taken. And my conditions are Ok or not?

Nihar Karve

6:11 AM

@Secret where did that come from xD

Secret

just... smoshing the electromagnetic tensor with the spin 1/2 field equation and taking the intuition that the interaction dynamics has to be a functional integration across all possible field configurations and the spacetime region of interest

I am trying to figure out what the whole electron field and electromagnetic field interacting with each other look like without trying to break it down into pieces like feymann diagrams or asymptotics etc.

I just want to see what that functional integral look like without assuming the interacting field is decomposable into small steps

this also means, not assuming we are working on just one or a direct product of hilbert spaces

Azmuth

6:55 AM

hi

user400188

7:45 AM

How can I tell when an edited question (closed as a dupe) has left the review cue?

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