1

Technology / Re: Does cold weather damage the batteries used in an EV?

« on: 15/01/2025 07:43:54 »

Hi.

    Thank you to everyone who has replied.   I hope everyone has had at least one "thank you" mark put on their post.   I may be paying less attention to this thread from now on.

    I think I've found a partial answer to my earlier question about where the energy goes on a cold day.  @Halc was the only one who tried to address that question.   Specifically it seems you could charge the battery to full on a warm day but get much less than the full battery capacity displayed when you use it on a cold day.    The explanation seems to have two important points:
     1.    Most of the on-board sensors and computers are just trying to calculate the range and battery capacity that you have as best as they can.   Many will know what the ambient temperature is and some computers will adjust the figures that get shown on the drivers display according to some algorithm.   A very cold battery just won't deliver such a high voltage and this will usually appear to the sensors as if the battery capacity is lower than it actually is    etc.      Overall, it only appears as if some energy has just simply gone but actually it probably hasn't.   If it warms up again, the battery should still be at about full charge.
      2.     Many of these EV's monitor the battery temperature and they will heat it up, often called pre-conditioning of the battery.   The phrase "pre-conditioning" is usually applied to something that happens while the vehicle is plugged in and before the journey even begins.   For example, most of these EV's can be programmed to know when you start your morning commute to work and they will get everything pre-conditioned automatically.   However, it seems that this sort of conditioning for the battery doesn't completely stop when the vehicle is being driven.   I can only guess while this is still called "pre-conditioning" instead of just "active and current conditioning" but I suppose they would argue it's something that is done BEFORE more power is available from the batteries.   Most EV's will not allow the batteries to get too cold while you are driving.   Obviously if you do also put some heat into the cabin area, then the range is reduced even more.   While you could choose to let the driver freeze and wear winter gear inside the car, you simply cannot choose to let the batteries stay too cold.   So, inevitably, you will get less range on the cold days.

     NOTE:   I don't yet own an EV and claim no expertise.   I'm not really getting straight answers from most of the car salesmen.  I suspect that's because we just haven't had EV's for long.   Any comments I have made are just the best sense I can make out of it at the moment.

Best Wishes.

2

Technology / Re: Does cold weather damage the batteries used in an EV?

« on: 12/01/2025 00:35:02 »

Hi.

   

There is of course battery university, which says than lithium ion cannot be charged below 0C unless it is a specific chemistry...

   I have just checked this.  I think it's that Li-ion batteries should not be charged below freezing rather than that they cannot be charged.    According to your suggested Batteryuniversity site,  charging them below freezing temperature is always slow and would usually cause permanent damage resulting in lower battery performance.
    That's not something that Tesla ever seem to mention - and it does seem that they use Li-ion batteries of some variant or another.  Even with some clever technology it seems you would still be well advised to charge them at a small fraction of the usual rate when the temperature is below 0.    So, taking a long trip in the winter and putting your vehicle onto a fast charger along the way may be very bad news for the batteries.

Best Wishes.

3

Technology / Re: Does cold weather damage the batteries used in an EV?

« on: 11/01/2025 14:25:45 »

Hi.

    Thanks for your reply @alancalverd .    I know very little about batteries or real world engineering.

..... the effects being summarised as an increase in internal impedance.....

   Is that really a fair summary?
    I've found a few webpages suggesting the internal impedence of a Tesla battery is very low:   approx.   30 milli-Ohms     to   46.1 mΩ

[reference:    Tesla motors club,    https://teslamotorsclub.com/tmc/threads/model-x-100d-battery-internal-dc-resistance.108682/      and also a website from  circuitdigest dot com -  but the latter seems to be filled with adverts and strange links you must click,  I wouldn't want to list that actual website address and risk people causing problems to their computer ].

    Anyway, even if this internal impedance doubles in the cold weather, it hardly seems significant compared to the resistance of the motor and/or any other electrical equipment attached  (car radio,  or the heaters   etc.).

     I'm not a real-world engineer, as I keep saying, but this is a back-of-the envelope calculation that seems relevant:
According to one website,  a Tesla motor provides  220 KW power when driven at 320 V  and rotating at 5000 rpm.   I'm going to guess that's typical rpm.
   So  P = V² /  R         =>         R =  V² / P    =   320² / (220 x 10³)   =   0.47 Ohms   ≈   465  milli-Ohms.
    Let's drive just the batteries internal impedance and the motor.    So, let's assume   these are the only two things of relevance connected in series (so a current, I, passes through both).
The total power used =  I² R_total  =  0.51  I²
Power used to heat the battery =  I² R_{internal battery}  =  0.04 I²
Ratio of power lost as heat in the battery :  total power  =   0.04 : 0.51   =>    7.8% of the total battery capacity (in KWh) is wasted as heat in the battery.

Do a similar calculation with the battery internal resistance doubled (for a very cold day) and we have: 
Ratio = 0.08 : 0.55     =>  14.5% of the total battery capacity wasted as heat in the battery.

    I'm going to assume that on typical driving, 1 KWh of energy used in the motor will move the car  X amount of miles.   I know you'd go further downhill etc. but let's just assume that on typical driving,  total energy delivered to the motor is linearly proportional to the number of miles the car will move.   Is that a reasonable assumption?  Probably.... it seems that car manufacturers are making an estimate like this when they show the increase in range  vs.  the choice of a larger capacity battery.  If it takes 100 Joules of energy to move 1 mile, it takes 200 J to move 2 miles etc.
    So, very roughly,   we get 92.2% of the battery capacity converted to range on a wam day  and  85.5% on a cold day.   Hence,  a drop in range of  6.7% of the battery capacity   (of which we were only getting  92.2% even on the warm day),  so a change of  6.7 / 92.2    ≈  a 7% reduction in range on a cold day compared to the warm day.
     Is that a reasonable back-of-the envelope calculation or have I made a mistake?  Are these estimates reasonable?

You can estimate the cold weather range loss from the cabin heater specification. You generally need about 5 kW to keep  the windows clear in a small car at 0 deg C, and maybe 20 kW on a really cold day.

     That does make sense.   The power demand to keep the window clear and the driver sufficiently alive and responsive may be the more significant power drain in cold weather.
      I've noticed that some top-of-the-line EV's are offering a heat pump as a means of heating the car.   This is the most efficient way of heating the car and will provide a bit more range on the cold days.

The advantage of an internal combustion vehicle is that  hot air is an inevitable waste product, so thanks to increased air density, it actually goes further on a cold day!

     I don't live right up in the highest hills.   The air is fairly dense most of the time  (I would have thought).  I would imagine a modern car with a turbo to ram the air in and suitable electronic clever bits would be able to get enough oxygen into the engine on every stroke.   I mean if it put too much oxygen in, then the ratio of fuel to oxygen is too weak and then you'd just be losing power again.
    Do you really get more  m.p.g. on a cold day?    Perhaps you were talking about the hot air waste product?   On a cold day, I suppose that will go further (further up) every second.   However, my vehicle wouldn't be a hot air balloon and that sort of vertical mileage wouldn't be something I want.

    ----------
    Anyway, I know very little about EV's or real world engineering in general.    If anyone has some data about the real world longevity of EV batteries after a few cold winters  and/or  the typical loss of range on a cold day, please let me know.   This is a science forum, not a help site just for me but I would argue that every government is putting all of us under pressure to go electric soon and so this sort of information is of value broadly.   However, any moderator is free to move the discussion somewhere else - like the "just Chat" section if they wish.

     Looking at the diagrams of modern EV's  the batteries seem to be really low down, often just under the floor that the passengers stand on.   I would be guessing that changing the batteries is an expensive piece of maintenance.   Do you have easy access from underneath or does everything have to be taken out to get to the batteries?   I don't know....
In terms of realistic cost, is the car dead as a whole once the batteries lose their ability to hold a sensible charge?

Best Wishes.

4

Technology / Does cold weather damage the batteries used in an EV?

« on: 11/01/2025 02:16:37 »

Hi.

    Supposedly, cold weather reduces the rate of chemical reactions inside a battery, so they don't perform as well either as something to provide power or to accept a charge.    Sometimes there is some more obvious permanent damage such as a cracked battery case or crystaline structures that have formed inside the cells and remain there for ever after, usually short-circuiting a cell.   
     This leads to two questions:

1.     Does it usually get back to normal when a warm day returns?
2.     If a rechargeable battery loses some capacity (in the sense of KWh of electrical energy it can store or deliver)  in the cold weather,  where does that energy go?    Suppose you charged it to full on a warm day, then a cold day arrives and you used it only to find it lasted just half the usual time  -  does the other half of the time just re-appear when you warm the device up again?

    Relevance / Background:   Like a lot of people, I'm thinking of getting an EV (Electric Vehicle) soon.   How many (British) winters will the batteries last?   Have we had EV's for enough years to know how well they last?  Also, how much range for the vehicle will usually be lost on the cold days?

Best Wishes.

5

Just Chat! / Re: What Order Should You Include People's Names When You List Them?

« on: 09/01/2025 17:37:14 »

Hi.

  .... if you got money on your birthday from your uncles A, B, and C. And A gave you the most, but C gave you the least, it would be in that order....

   What a terrible idea.  Money is less important than their shoe size.    List them in whatever order comes to mind and put up with the consequence that they would be able to analyse the situation based on your response.   For example, does uncle A only get added to the end and in a different colour pen which you had to hand when you finally remembered you should also include them?

Best Wishes.

6

Physics, Astronomy & Cosmology / Re: Am I safer on Mars?

« on: 08/01/2025 23:26:30 »

Hi.

...but you ( @alancalverd ) were the one who said....

so in the absence of an atmosphere the probability of any point being struck .....

    In the absence of atmosphere, collisions between objects broadly heading toward each other are extremely rare,  they just orbit in some way.    For example, when the entire Andromeda galaxy finally reaches us and starts occupying much the same space as the Milky Way galaxy,  head-on collisions of the stars, planets, asteroids and other stuff are still generally thought to be rare events.   Instead the "damage" will be done just by changing the orbits of stuff in our galaxy upto and including the possibility of having some astronomical bodies thrown out into deep space.

Best Wishes.

LATE ADDITION:
     The interesting and relevant point is that gravity isn't some problem or destroyer of worlds.    Actually, gravity is quite helpful , it's the great creator of solar systems, galaxies and most of the structure in the universe.   Everything about it, such as Newton's inverse square law, seems to be well crafted so that most bodies are kept safely away from each other and just orbit.    Your original inference, that the number of meteor impacts would be proportional to the gravitational attraction of the planet, just seems to be entirelly unfounded.   Your giving gravity a bad name which it doesn't deserve.
    Given a choice we'd probably be a lot safer on super-super massive planet.  I'd like to be the quivalent of the sun and just have everything spinning around me rather than being one of the little planets held in it's gravitational pull and made to move around in an orbit that could cross paths with all the other flotsam and jetsam that orbits the sun.

    However, I'm only being this critical because you ( @alancalverd ) aren't going to worry about it.   Most of the time you give sage advice.

7

Physics, Astronomy & Cosmology / Re: Am I safer on Mars?

« on: 07/01/2025 23:07:16 »

Hi.

Gravitational attraction increases with the cube of the radius of a planet, and surface area increases with r2, so in the absence of an atmosphere the probability of any point being struck by a meteorite increases with r3 /r2.

    I like the approach you've taken and can see what you're doing.  However, what reason do we have to assume the size of an objects gravitational attraction would be directly proportional to the number of meteors that strike it?    In the absence of any atmosphere, as you state, two objects generally heading toward each other will form an orbit almost all the time.  I agree that an object with a greater gravitational attraction may have more objects in orbit around it - but we don't really care about that too much, it's only the actual meteor impacts on the planet surface that matter.
    It's only when the approach of the meteor toward the planet was almost perfectly "head-on" that an impact would occurr instead of an orbit (either a closed elliptical or else a hyerbolic escape orbit) being the final result.   So, you'd have thought, the cross-section of the planet seen by the approaching meteor is a fair estimate of the critical solid angle for the approach that does result in an impact,   so that cross-sectional area is a circle, with area ~  r².

Best Wishes.

8

Just Chat! / Re: Free lectures

« on: 06/01/2025 04:18:26 »

Hi.

The mission statement for most universities never included free continuing access to lectures,.....

    In the days when you had to be in a lecture hall there were going to be problems with space and seats available.   That's not a problem any longer when these former students are just sat on their own chair in their own house with a computer screen.

    They now have the ability to do something they couldn't before.  There's nothing in any mission statement along the lines of ......   in the future if we can do more, then we just won't.  We were only pretending that we wanted to do these things......


Example:      The mission statement of Cambridge University is described here:   https://www.cam.ac.uk/about-the-university/how-the-university-and-colleges-work/the-universitys-mission-and-core-values 

These extracts are most relevant:
    *   education which enhances the ability of students to learn throughout life
    *   the contribution which the University can make to society through the pursuit, dissemination, and application of knowledge
   
Best Wishes.

9

Physics, Astronomy & Cosmology / Re: Am I safer on Mars?

« on: 06/01/2025 03:51:36 »

Hi.

    A brief survey of the evidence suggests that you'd be at higher risk of meteoroid impacts on Mars.

Why?
     The explanation usually given is that the atmosphere is too sparse.  Small meteoroids don't seem to break up or burn up in the atmosphere anything like as much as they do around Earth.

What evidence?

....  The Mars Global Surveyor, active from 1997 to 2006, was the first spacecraft able to image Mars in high enough resolution to detect new impacts, with a resolution of up to 1.5 meters (4.9 ft). The first detected impact, a 14.4-meter (47 ft)-diameter crater in southern Lucus Planum, happened between 27 January 2000, and 19 March 2001.[2] Since then, over 1,200 new impact craters have been found on Mars......

  .... Due to Mars's tenuous atmosphere, with just 0.6% the surface pressure of Earth's, incoming meteors are much less prone to breaking up..... 

[Extracts taken from Wikipedia]

More discussion and further references on this Wikipedia page:   https://en.wikipedia.org/wiki/Impact_events_on_Mars 

Best Wishes.

10

New Theories / Re: Origin of magnetic force

« on: 04/01/2025 13:29:55 »

Hi.

Drift velocity of electrons in copper wires, are small. Yet the force between two parallel current carrying wires can be significant.

     The force on each charge carrier is small  BUT  there are quite a lot of these you need to add together  (approximately 10²⁹  free electrons per cubic metre of copper wire).

According to relativistic explanation, positive charges on first wire see moving electrons in second wire length contracted,......

    This line of reasoning and many of the comments you copied-and-pasted from a YT video just a little earlier are all based on the notion that length contraction should apply to the free electrons just as it applies to the metal ions.   So that in a lab frame where the metal ions were stationary, the electrons are then drifting, so they would have higher density due to length contraction and thus the wire should be of net negative charge.    This was discussed in an earlier post  (post # 406):

.....it is just not possible for the (current carrying) wire to be of overall neutral charge in every frame.....    For some reason the frame of reference where the wire is overall neutrally charged is the usual one, the one where the wire (the positive metal atoms in it rather than the free electrons) is stationary....
NOTE:   text in blue italics added to the original quote for clarity

    I haven't read the original paper / discussion you cited by Jefimenko, who died in 2009.   At a guess, this work is quite old and from a time when we were still sorting out the details of electromagnetism and relativity.
    This sentence you quoted seems most relevant:

The true meaning of the calculations demonstrating the alleged relativistic nature of the magnetic field and of the calculations presented in this paper is, therefore, that the idea of a single force field, be it magnetic or electric, is incompatible with the relativity theory.

   Indeed it is helpful to consider that there probably is only a combined electro-magnetic interaction and we cannot sensibly declare that a thing like an electric field (or a magnetic field) exists in isolation.   For example, Maxwells equations where a separate E and B field appear, starts to look more like a "hack" or convenient way of performing calculations involving 2 fields described as ordinary 3-vector fields.   3-vectors are basically the ordinary vectors we might associate with Newtonian mechanics and 3D space.   However, these 2 separate fields may very well be just fictitious artifiacts, useful for calculations but not representing anything that really exists. 
    It may be that the underlying nature of the interaction requires description using 4-vectors such as the Electromagnetic four-potential   (Reference:  https://en.wikipedia.org/wiki/Electromagnetic_four-potential ).   Specifcally, it seems that the more important and presumably more physically real field that may exist is a single 4-vector valued field (rather than 2 seperate 3-vector valued fields).

     About a year ago, the Aharanov-Bohm effect was discussed in this forum.   Here it seems that something can be affected by an electromagnetic field even where the Electric and Magnetic field are both 0 valued.   (Reference:  https://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect  .   There's also an old forum post which had some diagrams and animation that still seem to be working  https://www.thenakedscientists.com/forum/index.php?topic=86694.msg722832#msg722832 ).    Anyway, this provides some evidence to support the idea that what physically exists and can exert physical effects may very well be a 4-vector valued field such as the  Elctromagnetic four-potential  rather than just 2 separate 3-vector valued fields like the E and B fields.

Best Wishes.

LATE EDITING:   I re-read the old forum post and it may cause some confusion when people read a sentence that said  ..... but back in the day it was important to imagine that magnetism (and possibly every other force) may demand that two fields existed throughout all of space..   That may seem to oppose the notion of needing to consider only ONE combined electro-magnetic object.
    The old forum thread emphasised that considering only a magnetic field (a B field) is not enough, we also need to consider a magnetic vector potential field.    That's why it was convenient to say we needed to consider 2 things (B field and a potential field).   There would be these two things to consider.   Note that both of these things (a B field and a magnetic vector potential field) are just ordinary 3-vector fields.   We get the next layer of complexity by considering 4-vector fields.
    In this post (this thread), you just need to be aware that we can recover ALL the information about EVERYTHING   (the E field, the scalar electric potential field, the B field and the magnetic vector potential field)  from just a single suitably defined 4-vector valued field we call the electromagnetic four-potential.   So if we're looking for the most fundamental description of electro-magnetism then we may need to be considering this object.  Specifically, the electromagnetic four-potential  seems to be a sufficiently integrated object to describe all the electro-magnetic effects that I am aware of.   Meanwhile a lesser set of objects such as:
  (i)  the B field on it's own.
   or (ii)  The B field plus a suitable potential for that B field,
 would NOT be sufficient.

    I hope that makes some sense.    Let's go back to your ( @hamdani yusuf ) original question and paraphrase all of the above comments.

What is the Origin of magnetic force?

I don't know but maybe the entire notion of a magnetic field existing as some sort of thing on its own is just a bit fictitious.
(i) We seem to need to have a E field with it.   We cannot adequately explain all effects with just a magnetic field.
(ii) We seem to need to consider potentials in addition to the forces.
(iii)  Overall we need to have an object like the electro-magnetic four-potential field in existance.   All of physics is just a model but considering the electromagnetic four-potential field as something that exists may be a good step closer to the truth rather than considering that we could attempt to identify and define a magnetic (or electric) field as something that could exist separately.

11

Just Chat! / Re: Free lectures

« on: 03/01/2025 14:23:59 »

Hi.

    Sorry to dig up this old forum thread but it seemed relevant.   It's a new year and I have a new grumble, suggestion, request or..... something.

    Students now usually have access to lectures and related material online.   They log in with their username and password  etc.    This is usefull because they can be ill or something and may miss a lecture.   A lot of universities had set up decent recording equipment for every lecture after the Covid crisis.   They just could NOT have all the students in one hall and they became aware that they cetainly had to make it possible for students to isolate themselves if they had a positive Covid test rather than feeling that they had to get to lectures.   So a lot of really good, really decent recording equipment and practices were established.

     Anyway, I have discovered that when the student finally finishes their degree, their online credentials are no longer valid.   They cannot log in and see the new lectures and new material for the next year.    That's wrong isn't it, or at least morally wrong?

     The student is reasonably expected to stay up-to-date and the University could make it a little easier for them to do this.    Once a student has paid their course fees for the degree, would it really hurt the universities to grant them continued access to future lecture material?    Surely it's even in the Universities own interests to do so?   They would presumably like the idea that their graduates are as good or better than graduates from other places.    It's a selling point -  "Our graduates are able to keep themselves as up-to-date".

    Morally, why wouldn't the universities want to help their graduates?    I mean these graduates are getting old and we can't always carry a ton of textbooks in our backpacks everywhere we go.   However, many of us have online access and could access all the lecture notes and related material while only carrying a pocket telephone.

    I guess I'll have to add to the problem of increasing population.   I'll have to keep having more children.  I need to keep one in University every 10 years or so,  such that I can "borrow" their active log in credentials.

But... more seriously.... it's just wrong....  Universities used to have mission statements about educating people and encouraging a policy of life-long learning  .... rather than making money.

Best Wishes.

12

General Science / Re: Can Human Consciousness Be Uploaded?

« on: 26/12/2024 10:26:51 »

Hi.

    Other than @evan_au 's link to a wikipedia article,  I can't see any mention of the term "connectome".    That's a keyword people will want if they're looking for more information.

Otherwise we will have recreated someone with "completely locked-in" syndrome: conscious, able to think, but in total isolation would quickly go mad.

     Even this much is not certain.    A real-life connectome is never without some input and would always have some syanpses within it that are firing, i.e. it is half-way through processing something.    It is never completely "off" and inactive.
     If you built a perfect copy of a connectome, it still isn't known if it would spontaneously start to work or ever get into some sort of normal firing pattern,  it may just stay basically in the "off" position.     Since so much of the brain structure involves feedback loops, arbitrarily firing a synapse here or there is as likely to set up something like an epileptic seizure rather than switch the connectome on and get it into a normal firing pattern.
    So, ideally, you don't just want to copy the connectome,  you'd aso want to copy the exact firing pattern that was happening at that time etc.


Para-phrasing:    It's not enough just to copy the hardware.   The whole thing is more like a syphon pump.  It must have the fluid in it and be working now in order for a copy of it to also be a syphon pump.    Otherwise all you have is just an empty tube rather than a syphon pump.   Put one end of this tube into some water with the other end lower down and it does nothing.

Best Wishes.

13

New Theories / Re: How does Hamdani explain the twins paradox?

« on: 24/12/2024 17:15:46 »

Hi.

If we are willing to accept this explanation, we are forced to also accept reverse time jump or rewind when the direction of the acceleration is away from the observed twin.

    "Reverse Time Jump" --->  It sounds like you're describing something amazing, maybe time travel into the past.  Of course, that's a feature that SR does not exhibit.   So it either isn't that or else it will (probably) be wrong.   Quite a few people have looked for ways to travel back into the past and SR doesn't seem to offer that ability.   However, you're welcome to try again and it's certainly quite educational to give it a go.

Looking at the next post I can now see what this thing you're describing will end up being.

Look at the lines of simultaneity. The video only shows one side of them, i.e. the side where they leave gaps. The other side is where they overlap. This is where the reverse time jump appears in the analysis.

     Although it may sound like you'd be looking at something new, I suspect it's just going to be some re-wording of the Andromeda paradox that you're going to end up with.

Best Wishes.

14

New Theories / Re: what is temperature?

« on: 22/12/2024 22:46:05 »

Hi.

I reckon it would disintegrate before adequate cooking

   Apparently you adjust the slap depth, so that the chicken isn't turned to pulp and rely on having a very high rate of slapping.   If it's the video I think it is, they built a slapping machine that managed about 10 slaps per second.
    Of course, there's a limited market for chicken slapping but it it may have more potential as a steak slapping machine because then your steak should come out both cooked and very tender.

The inverse principle is important in the manufacture of sausages.

    Sausages need a bigger slap.  They're building a special machine to deliver the bangers.

Best Wishes.

15

Just Chat! / Re: A holiday wish.

« on: 22/12/2024 12:55:05 »

Hi.

     Thank you, merry christmas and/or best wishes to everyone.


(that's a copyright free animated GIF as far far as I could tell).

16

New Theories / Re: Origin of magnetic force

« on: 21/12/2024 06:09:12 »

Hi.

    I think I can see what you ( @paul cotter ) are saying and yes, you do seem to have the right idea.

I can only assume a relativistic effect of the movement of + and- charges in the conductor produces an electric field with force on the charge

Yes.
In post #406 I said the following:

....the same force and overall effect should appear.   However, it doesn't have to appear as a magnetic force and in this case it will be partly a force due to an electric field....

  In this situation described by @hamdani yusuf ,  the velocity of the test particle became 0 exactly and so we have the most extreme case.    Using the Lorentz force law and putting v=0 we would have that   F = q(E + v x B)  =  qE  =  a force entirely due to an Electric field.
This was mentioned in reply #408 by Hamdani:

In the reference frame where the test particle is stationary, v is 0. Thus special theory of relativity interpretes that the force is purely electric.

     In this frame of reference, the distance between the -ve charges (the free electrons) and between the +ve charges (the metal atoms in the lattice) has been affected (by length contraction under SR) such that the wire had an overall +ve charge of exactly the right magnitude to repel the positive test charge as required.   Although a B field would still be there circling the wire, it plays no part in exerting a force on the particle (in this frame).

---------
Minor notes:    You said the following (with red text added by me):

and as the (test)  charge is moving in this frame it also produces a magnetic field...

.
    Yes that's actually completely true  BUT  you are over-thinking this. 
     We were just considering it as a test charge.   They do not create or alter the field(s) around them, they just experience the field(s) created by other things    https://en.wikipedia.org/wiki/Test_particle .       
     In the real world, any moving charge would create a magnetic field in the space around it and this would combine with the magnetic field from the wire  -  but  we're talking about 1 small charge  (maybe just a proton) compared to billions  (~10²⁴ atoms per cm³)  of free electrons and metal ions in the wire.   The alteration in the magnetic (or electric) field caused by our little postive charge is utterly insignificant and we can consider it as an idealised test charge.
   
Best Wishes.

17

New Theories / Re: Origin of magnetic force

« on: 20/12/2024 02:03:39 »

Hi.

Sorry to hear you're poorly @paul cotter .

As we know a static B field will have no effect on a static charge.

   Yes.   Stationary test charge implies no force.

Yet if we change frames of reference as you have done we get a force - what am I missing?

    Well, the "static" test charge won't be static (stationary) in both frames.   It's the motion of the test charge that undergoes an important change when you switch frames.   The B field doesn't need to change, the test particle now has motion.

    Just to be clear,  @hamdani yusuf   was discussing a situation with this sort of diagram:



      ------------------------------------------------------------ this is the wire


                                   *                this is the test charge

The test charge is something outside the wire,  we can move the wire while leaving the test charge static in the lab frame.    Of course, the moment you switch frames, so that the wire is stationary, then you see the test charge will be moving.       

Well that's a short answer anyway.
----------

I've read what you've written a couple of times now.   It seems that you think it would be important to know if a B field is moving or not.   However, it is NOT important to know.

You said:

Looked at from the frame of the moving wire the test charge and the B field will move in unison.

It's not meaningful to talk about the B field as if it can have motion and therefore move in unison with the test particle.   All that matters is that the test charge has motion in this frame.
 
The Magnetic force law (Lorentz force law where E=0 ) is often mis-interpreted in some fashion similar to the following:
   There's a force when a charge moves through a magnetic field   OR  when the magnetic field moves past the charge.

But that's just pop-sci and sloppy speech.   The Lorentz force law (or just Magnetic force law) actually says nothing at all about a magnetic field moving past a static charge because it's not even clear if a magnetic field is some sort of "stuff" that can move through space.   All we need is that the magnetic field, B, is something that has a value (this one has magnitude and direction) at every point in space and at every time.

     Much as you seem to suggest, quite often we just couldn't tell if the B field was moving or staying still.   For convenience we'll align our straight wire with the x-axis, then the current carrying wire has a B field that depends only on y and z co-ordinates of a point in space.   At any time, the B field at x=a is no different to the B field at x=b.   We can't tell if the B field at x=b,t=now   has just come from position x=a,  a second ago, or if the B field just wasn't moving.
   Fortunately, the Lorentz force law does not ask about how the test charge was moving relative to the B field.  It just asks how the test charge is moving   (the implication being this is relative to the frame you have chosen).

    Read perhaps just the first section of the Wikipedia article about the Lorentz force law to yourself:   https://en.wikipedia.org/wiki/Lorentz_force
    This time pay carefull attention to the LACK of something:   The lack of any need to know about the movement of the B field.

      The Lorentz force law states that a particle of charge q moving with a velocity v in an electric field E and a magnetic field B experiences a force (in SI units) of    F = q ( E + v x B ) .

    The velocity, v, that appears in these formulae is never stated as being the velocity of the test charge relative to the B field.   It is just the velocity of the test charge  (in whatever frame you have chosen to use).   Specifically, we start by picking a frame of reference,  then the test particle has a velocity v in that frame.... that's the only v we're interested in.... we don't care if the B field is something that might also be moving, it doesn't matter, we don't ever need its velocity. 

Best Wishes.

18

New Theories / Re: Origin of magnetic force

« on: 19/12/2024 03:16:05 »

Hi.

Let me simplify the question. A long straight metal wire moves to the right at speed 1 m/s relative to the lab. Electrons in the wire move to the left 1 m/s relative to the lab. A positively charged test particle is stationary relative to the lab, 1 cm below the moving wire. Will it experience a force by the wire? Which way?

   Yes, it should do.  It will be a force pushing the charged test particle directly away from the wire.

   Why?   Translate to the more typical frame of reference.    So we'll have the wire stationary, so we can assume the positively charged atoms of it are stationary.   Adjust the velocities of everything else appropriately.
    The free electrons of the metal now move with velocity 2 m/s to the left   ( my left direction is    < ---   ).   Actually slightly less if we use the the velocity addition of special relativity but  2 m/s will be OK.
     By conventional electromagnetism stuff, we have a magnetic field around the wire.   Conventional current is going  to the right  ( ---> that way ),   right hand rule shows the B field circles around the wire.
    The test charge, in this frame, is now moving at  1 m/s   (  <-- that way), or more importantly in the oppositie direction to the current.   A moving charge does feel force in a magnetic field,  we use the other right hand rule   for  F  = q  v x B.    We see the test charge is accelerated downwards,  i.e. directly away from the wire   (if I twisted my hands around in all the right directions).

    In the atypical frame where your situation applies   ( the wire moves ---> that way),   the same force and overall effect should appear.   However, it doesn't have to appear as a magnetic force and in this case it will be partly a force due to an electric field.
    We could draw the diagrams and do the maths - to show that due to relativisitic effects,  the average distance between charges has altered and the wire is no longer of overall neutral charge as far your test particle is concerned.

    The smart person would then ask a follow up question:   Aren't the electrons and positively charged ions affected equally?   Aren't the distances between +ve charges and between -ve charges equally contracted and the charge does remain overall neutral?    Indeed, in your original situation where the test paticle was at rest, it does seem that both the positively charged ions and the electrons are moving with equal speed, albeit in opposing directions.
    That's a good question.   The answer tends to be that the free electrons are not like the positively charged atoms in the metal wire.   The free electrons are, as in their name, "free",  while the metal atoms are much more locked in place in the crytsal lattice of the overall structure.
      An approximation is made here.   Here's the step-by-step reasoning:
1.   It does seem that when a current is passed along a wire, the free electrons do start drifting. 
2.   So there would be a corresponding length contraction for the distance between them, which would make the density of negative charges exceed the density of positive charges in the lab frame.
3.    However, the electrons are free to move.   This is the important approximation step.    By electrostatics, they will re-distribute themselves and the wire remains overall neutral in the lab frame, to a good approximation.    Note that the metal atoms could NOT do this, they are not like the free electrons.   We know that a static test charge placed next to this wire which has a current in it, doesn't seem to experience any force, that's the experimental evidence that we've got.  So it seems that the wire has remained overall neutral in charge, even when a current is made to pass through it.     

    One consequence of this is that the wire cannot be of overall neutral charge in other frames of reference.   In other frames, there must be length contraction as usual.   So, the easiest one to consider is where we move to a frame where the electrons are stationary and the metal atoms were the things moving with (the negative or direction reversal of)  the electron drift velocity.   In this frame, the electrons are not moving, while they were in the first frame, so the average distance between electrons is now bigger,   -ve charge density is therefore lower.   Meanwhile, the metal atoms are now moving whereas they were stationary, so the average distance between them has decreased,  hence positive charge density has increased.   So overall,  the positive charge density now exceeds the -ve charge density and the wire has a net positive charge.

    So the irritating and relentless person may then ask - but the electrons are still "free" in this new frame, why don't they just re-distribute themselves again?   Why isn't a current carrying wire of overall neutral charge in any arbitrary frame I wish to use?
One half of that we can answer well:   Since the electrons do drift when there's a current in the wire, it is just not possible for the wire to be of overall neutral charge in every frame.   Do the maths carefully, there is only one frame where that can apply.    The second half... well I don't have a satisfactory answer for it.... 
    For some reason the frame of reference where the wire is overall neutrally charged is the usual one, the one where the wire (the positive metal atoms in it rather than the free electrons) is stationary.  You can try and dig up a few reasons, for example a current just does not flow unless there's a cell (a battery) or something that is maintaining a potential difference across the wire.  You can attempt to argue that as part of that setting up procedure,  just the right number of electrons are initially pulled out of the wire at the +ve terminal of this battery,  so the slight over-density of -ve charges that should appear in the wire when the electrons do start drifting along does not emerge.  However, we can't make too much speculation.   Let's just leave it as  "for some reason" the frame where the current carrying wire is overall neutrally charged is the one where the wire is stationary.

Best Wishes.

19

Just Chat! / Re: What does it cost? What does it do?

« on: 17/12/2024 14:37:45 »

Hi.

Not sure about the forum cost but the sponsors obviously think it is money well spent.

   Thanks @alancalverd, that's a partial answer.   By "the sponsors" do you mean these people? 

     
            

[Taken from  https://www.thenakedscientists.com/our-sponsors ]
   Is that where all the money is coming from?    Also, it's not clear if they have ear-marked some money for a forum or if they just support  TNS.    3 out of 5 of those sponsors seem to have expertise strongly related to the production and distribution of podcasts.   Is a forum expected or required?

[ Also I've got to say that "Reithian" was a good word I hadn't seen before.  https://www.bbc.co.uk/historyofthebbc/100-voices/inventingthefuture/reithianism/ .   There's nothing wrong with learning a bit of vocabulary and history, thank you for that ].

Best Wishes.

20

Just Chat! / What does it cost? What does it do?

« on: 16/12/2024 21:58:34 »

Hi.

      How much does it cost to run this forum?   A typical cost per forum post made would be useful.

      Where does that money come from?

      What does the forum do?

      Is it doing what the funder wanted it to do?

- - - - - - - - - - - - -
    Why ask?   I'd like to know what I'm using, that's only sensible and responsible.
Specific example:   There are some days when I may have answered a question somone asked and it's easy to assume that's a good thing.   However, is the cost per forum post such that more peope would have had more benefit if we all just left everything with short replies or no replies at all?

    For example, the Naked Scientists radio show and podcast sometimes go out (sometimes overseas) and enagage with school children that are trying to learn some science in disadvantaged circumstances.   Would it be better if they just had more money to get on with that?

Best Wishes.