Particle Accelerator

Guru Vishnu

6:00 AM

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

Are you free now sir? I've a doubt about drift speed.

@JohnRennie: Or please reply when you find time to the following question:

The drift speed is defined as $v_d=\Delta l/\Delta t$ where $\Delta l$ is the distance drifted in a long time $\Delta t$. Why don't we define the drift speed as the limit $\Delta l/\Delta t$ as $\Delta t\to 0$? Is that because the drift of electrons is significant only after a long period of time and in short time period the electron might not have drifted at all?

John Rennie

If you look at a single electron then the drift speed is the average speed of that electron.

Electrons actually have very high speeds, but they dance around in random directions as they scatter off the atoms in the metal.

It's only when you take the time average that you get the much slower average drift velocity.

Guru Vishnu

Ok sir. So, instantaneous drift speed has no meaning. Am I right? Further, don't we simply assume electrons move with uniform velocity (drift speed) neglecting the random motion for electricity analysis?

John Rennie

It's like the motion of gas molecules in air.

Air molecules move at several hundred metres per second, but wind speeds are tens of metres per second.

Guru Vishnu

6:14 AM

@JohnRennie: I understood that sir. But for simplicity of calculations don't we neglect the random motion and consider only the drift of electrons as a laminar flow?

John Rennie

Yes, just like we consider wind speeds without worrying what the air molecules are doing.

Guru Vishnu

@JohnRennie: If so why is the limit $\Delta t \to 0$ invalid? Or in other words, why is instantaneous drift speed meaningless? (at least on the theoretical basis)

John Rennie

Because the drift speed is the average speed.

If you just take one moment in time that doesn't give you an average.

Guru Vishnu

Ok sir. It seems I understood it now. Thank you sir :-)

Guru Vishnu

7:50 AM

@JohnRennie: Hi sir.

John Rennie

@GuruVishnu hi

Guru Vishnu

@JohnRennie: We know that two points in a circuit joined by a conducting wire will be of same potential. How does electric current move from one element to another through conducting wire in the absence of potential difference, sir? If possible could you explain this, sir or could you provide a link to a question on the main site? I think this must be on the main site but I'm unable to find the question I'm looking for. Thank you.

John Rennie

Suppose you have a ball that slides down a ramp, along a perfectly horizontal surface then up a ramp at the other end.

When the ball slides down the ramp it is being accelerated by the reduction in potential energy as it decends the ramp.

Guru Vishnu

Something like a U shaped ramp?

\____/ ?

John Rennie

But when the ball is on the horizontal section there is no change in potential energy. So why does the ball move on the horizontal surface?

@GuruVishnu yes

Guru Vishnu

7:55 AM

@JohnRennie Ah! Due to its inertia.

John Rennie

This analogy isn't quite right for electrons in a perfectly conducting wire, because it's more like we have a whole line of balls and the balls rolling down the slope push the other balls ahead of them.

Guru Vishnu

Ok sir. But when potential difference across two points is zero, the current should also be zero. But it isn't so. I'm trying to interpret using Ohm's law.

John Rennie

But basically electrons in the wire move, even though there is no PD, because of the potential gradients in other parts of the circuit.

I = V/R, but if you consider an ideal perfect conductor you get I = 0/0 and that is undefined.

Guru Vishnu

@JohnRennie Yes sir. But I think resistance of ordinary conductors is small but not zero. And so I think the denominator is non-zero unless and until we're using a superconductor (again I don't know whether Ohm's law is applicable for super conductors)

And so I=0/(small number)=0

John Rennie

If R is non-zero then V is also non-zero.

Guru Vishnu

8:03 AM

@JohnRennie Fine. So potential difference due to current across a resistor is also responsible for electron pumping much like battery. Am I right sir?

@JohnRennie Ok sir. I understood where I went wrong.

John Rennie

@GuruVishnu if you go back to my analogy of the ramp, the battery is like the downhill ramp and all resistances in the circuit are uphill ramps. When you go all the way round the circuit the up and down ramps cancel so your height is unchanged.

Guru Vishnu

@JohnRennie Wow!!! An interesting and intuitive explanation of Kirchhoff's voltage law! You've given me some more to think about, sir :-)

John Rennie

Electrical potential is actually just potential energy.

Guru Vishnu

@JohnRennie Shouldn't that be Electrical potential "energy", sir?

If not we'd be comparing two dimensionally inconsistent quantities.

John Rennie

It's the potential energy per unit charge, just like gravitational potential is the potential energy per unit mass.

Guru Vishnu

8:11 AM

Yes sir.
Downhills are batteries; up hills are resistors; what will be capacitors in this analogy?

John Rennie

A capacitor is like a rubber band stretched across the track.

Guru Vishnu

Ok sir. Is that something like the following?:
Top view:
                                  \
                                   \
                                  o|
                                   /
                                 /

John Rennie

I'm not sure what that shows ...

Ah yes, with your latest edit I think I agree.

Guru Vishnu

@JohnRennie Ok sir. Then I understood your analogy. Except...

John Rennie

Suppose you put the rubber band across the track on the ramp. Then the weight of the ball stretches the rubber band.

Guru Vishnu

8:17 AM

20 mins ago, by John Rennie

This analogy isn't quite right for electrons in a perfectly conducting wire, because it's more like we have a whole line of balls and the balls rolling down the slope push the other balls ahead of them.

Shouldn't that be positrons instead of electrons?

John Rennie

If the ramp is steep it will stretch the rubber band a lot. If the ramp is shallw it will stretch it a little.

Guru Vishnu

@JohnRennie And steepness or shallowness depends upon the capacitance of the capacitor. Am I right?

John Rennie

No. The elasticity of the rubber band depends on the capacitance. A high capacitance is very flexible i.e. it will stretch a long way. A low capacitance is very stiff i.e. it will only stretch a short way.

Guru Vishnu

@JohnRennie Ok sir. I thought potential gradient depends upon the slope and since capacitance affects the charge as well as the potential difference across its terminals, the slope would be affected by capacitance. However, I find your reason more understandable.

Sir, shall we continue after some time? I'm going to have my lunch. And I also see you're busy in another room :-)

John Rennie

OK. I'll be around for another four hours or so.

Guru Vishnu

9:45 AM

@JohnRennie: Hi sir :)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

@JohnRennie: Is our previous discussion on potential difference across a conducting wire over? I think I got some understanding to proceed based on your analogy.

John Rennie

@GuruVishnu yes unless there is something more you want to ask about it.

Guru Vishnu

@JohnRennie: Ok sir. I think I don't have anything more to discuss on that. I have a different doubt:
This is a question from my book: Can the potential difference across a battery be greater that its emf?
My answer: Yes. This can happen when the battery of small emf is connected in parallel to a battery of large emf.
Book's answer: No. The potential difference across the terminals of the battery can only be lesser than or equal to the emf. It's equal to the emf when the internal resistance is zero. It's less than the emf in all other cases.

John Rennie

Suppose the internal resistance of the battery is R and it's voltage is E. Then the potential we measure across the terminals of the battery is $V = E - IR$. Yes?

Guru Vishnu

9:54 AM

@JohnRennie Yes if and only if current flows from positive to negative terminal. Else it must be E+IR.

John Rennie

The current $I$ can be positive if it flows into the anode and out of the cathode, or negative if it flows the other way.

Guru Vishnu

@JohnRennie Ok sir.

John Rennie

And if the current is negative then we get $V > E$. So you are correct.

Guru Vishnu

@JohnRennie Thank you sir. When potential difference across the terminals is less than emf, the battery is discharging. When it's more than the emf, battery is charging. When it's equal does it mean the battery is neither charging nor discharging? I don't think so. Because this is what we observe in a circuit which has a single ideal battery, sir.

John Rennie

> When potential difference across the terminals is less than emf, the battery is discharging

I don't think that's true.

Guru Vishnu

9:58 AM

@JohnRennie May I explain my reasoning behind it, sir? It's based on the direction of current.

John Rennie

@GuruVishnu it is correct to say if the current is positive the battery is discharging and if the current is negative the battery is charging.

Hmm ...

Guru Vishnu

@JohnRennie But is it incorrect to say it on the lines of potential difference and it's relation to the emf, sir?

I think both are same.

John Rennie

Yes, OK, since $V = E - IR$ is always true then yes a positive current, i.e. discharging, must mean $V < E$. And the opposite for charghing.

Guru Vishnu

Ok sir. What if V=E? Was I correct?

John Rennie

10:14 AM

$V=E$ means $I=0$ so the battery is neither charging nor discharging.

Guru Vishnu

Ok sir. I think while considering ideal batteries this condition will always be satisfied.

@JohnRennie: Why is it wrong to connect positive and negative terminals of a battery sir? Is that because due to the low resistance of the conducting wire, a very large current flows through the battery and the internal resistance liberates energy as a large amount of heat?

John Rennie

@GuruVishnu yes

Guru Vishnu

@JohnRennie If the battery is ideal will it not be a big issue?

John Rennie

If you do that with batteries that have a very low internal resistance, e.g. lithum batteries, they will get so hot they catch fire.

@GuruVishnu there's always a problem when you start talking about zero resistances because then you start finding you get division by zero and undefined results.

Guru Vishnu

@JohnRennie "...very low internal resistance" : Are you sure sir? Very low internal resistance means close to ideal battery and so - more ideal a battery is, more easily it can catch fire. Isn't this counter-intuitive?

John Rennie

10:20 AM

Any time you want to understand what happens with a zero resistance you need to start with a small but non-zero resistance and see what happens in the limit of $R \to 0$.

Guru Vishnu

@JohnRennie Fine sir. I remember you telling this to me on a previous occasion while discussing about charging and discharging of capacitors.

John Rennie

@GuruVishnu you're welcome to try shorting a lithium battery to see what happens, but make your will first.

Guru Vishnu

@JohnRennie Let me use an online simulator to be on a safer side or see someone else doing it on YouTube :-)

5 mins ago, by Guru Vishnu

@JohnRennie "...very low internal resistance" : Are you sure sir? Very low internal resistance means close to ideal battery and so - more ideal a battery is, more easily it can catch fire. Isn't this counter-intuitive?

Is that because heat generated is given by $H=i^2Rt$? Current has a power 2 whereas resistance only has power 1? So current has higher priority to cause fire than resistance.

@JohnRennie: Am I right sir?

John Rennie

Yes

Guru Vishnu

Ok sir. Thank you.

Guru Vishnu

11:13 AM

@JohnRennie: Kindly reply when you find time.

**Question:** A current passes through a resistor. Let $K_1$ and $K_2$ represent the average kinetic energy of the conduction electrons and the metal ions respectively.
(a) $K_1<K_2$
(b) $K_1=K_2$
(c) $K_1>K_2$
(d) Any of these three cases may occur.
**My approach:**
Mass of an electron is much less than the metal atoms of the lattice. Energy of small mass, large velocity electrons is almost comparable to the energy of large mass, small velocity metallic atoms of the lattice. So kinetic energies of electrons and the metal atoms may either be equal or energy of electrons may be higher. Or in…

John Rennie

I'm busy for a few moments.

@GuruVishnu the metal ions don't move. They are held in place by all the metal ions around them, so the KE of the metal ions is zero.

Guru Vishnu

11:36 AM

@JohnRennie Don't they vibrate?

John Rennie

Lower the temperature to absolute zero and they don't vibrate.

Guru Vishnu

@JohnRennie Ok sir. When temperatures are significant - say close to room temperature, I think they'd be significant. Even though their mean position remains the same, due to the large mass, even a small velocity would contribute to a large energy. I don't know whether it would be comparable to the large velocity, less mass of electrons.

John Rennie

The vibrational energy is going to be of order kT and won't depend on the mass.

Guru Vishnu

@JohnRennie Is kT kilo joules, sir?

John Rennie

The electrons are oscillating around as well. Remember that the drift velocity is just their average velocity. So even if the drift velocity is zero the electrons still have a KE just like the ions.

k is Boltzmann's constant and T is the temperature.

Anything in thermal equilibrium at temperature T has an energy of order kT.

Guru Vishnu

11:43 AM

@JohnRennie Yes sir. I remember that. So when there is no electric current in the resistor, can we say both kinetic energies are equal or still is it uncertain?

@JohnRennie Ok sir.

John Rennie

Offhand I'm not sure what the electron energies would be, but I think they will be a lot higher than kT.

At room temperature kT is about 1/40 of an electron volt, and I'd guess the electron energies would around one electron volt.

Guru Vishnu

@JohnRennie Ok sir. I just did a qualitative analysis. I don't have any values. However, which option do you feel, the most appropriate one? I think (d) is good one.

John Rennie

It's not a very good question though as it fails to make clear what it means. Are you really supposed to take into account the random motion of the electrons and nuclei?

More likely the equation means the values of $\tfrac12 m v_{av}^2$ i.e. the KE associated with the average velocity.

Guru Vishnu

@JohnRennie There was no specification about that sir. Further this is an introductory book and so we're not required to worry about complicated motion. But I thought it would be relevant to this question.

@JohnRennie Yes. Is there any other energy an electron could have? I think besides electrostatic potential energy, kinetic energy is the only choice and it's been mentioned in the question "average kinetic energy", sir.

John Rennie

If you're trying to calculate the energy associated with the random motion of the electrons that's really complicated. Far too complicated for an introductory text. That's why I think the book means the KE associated with the verage velocity.

Guru Vishnu

11:49 AM

@JohnRennie Yes sir. I get it. Is that complicated because we can't determine the velocity or position of a quantum particle due to the Uncertainty principle?

I guess I know the name by which Physicists calculate the energy - Statistical Mechanics?

John Rennie

It's complicated because calculating the wavefunctions of the electrons in the conduction band is complicated, and you need the wavefunction to calculate the KE.

Guru Vishnu

@JohnRennie Ok sir. Thank you for your help and time.

Good bye. Let's see tomorrow :-)

John Rennie

@GuruVishnu Bye :-)

Transcript for

♦ Particle Accelerator