Light Speed Motor Scooter: Newsletter Challenge (07/10/07)

Hi TVP45

Very well stated!

So I take it that you also think the scooter will shed energy, while the streetcar will gain some?

Jorrie

Hi Jorrie,

Well, I'm quite certain the streetcar will gain energy.

The motor scooter will maintain constant total energy if I can really ignore everything else. Truth is, I'm not certain I can really do that, even in a gedanken experiment. If I can't ignore everything else, I'll likely measure some decrease in energy. I gather you're a real expert in relativity and can perhaps shed some light on this.

I think relativity theory is time-symmetric and so I could mentally turn off entropy, for example. Do I understand that correctly?

What effects do you see that absolutely have to be included?

Tom

Hi Tom.

My view is that you can't ignore the losses completely. Consider this: put a special rocket up in free space. This rocket 'consumes' its own mass and converts it at 100% efficiency into propulsion energy of some sort. How much of the rocket's mass will be remaining by the time it reaches 0.9999c relative to it's starting point?

The answer is that about one percent of its original mass will remain, so that: E = mc²/√(1-v²/c²) remains constant.

Now consider the tiniest inefficiency in the energy conversion/propulsion process...

Jorrie

PS: by the time this rocket 'reaches c', only one photon of its original mass will be remaining.

This is going to sound really obtuse Jorrie, but I have to ask ;

...by the time this rocket 'reaches c', only one photon of its original mass will be remaining..

How does a photon get the energy to travel anywhere ?

Is this a light-weight question ?

Hi Kris,

This 'rocket' will get white-hot as it travels faster and faster, while 'not losing energy' as per thought experiment. Just before it totally disappears, there must be at least one photon still to be emitted - and a photon is not a photon unless it moves at c.

Jorrie

Of course this wouldn't happen (entropy and all that). But if it did, in a lossless/gainless system the photon would have the same mass as the original scooter when measured from the reference frame.

Hi Fyz, you wrote: "in a lossless/gainless system the photon would have the same mass as the original scooter when measured from the reference frame."

Yep, the photon will have a frequency in the gazillion³ Hz range or so?

Anyway, I think the idea of the question was just to bring out the fact that an external power source can increase the total energy of the accelerating object while an internal power source cannot.

In practice, I still stand with the argument that the hyper-speed streetcar will gain some energy and the hyper-speed scooter will lose some, without trying to quantify anything accurately. I suppose this is a bit of an 'engineering attitude.

Jorrie

I wouldn't argue with that. My only "difference" is that, when there are "followers", I prefer to say what would happen in an ideal case first, before saying what I might expect to happen in practice - and as there is no way anything like this could be reduced to practice, perhaps I might just leave it at the ideal case with comment as to the different offsets one might see?

Regarding the frequency, I don't understand gazillions, but at a guess its apparent mass would be in the region of 10 Undecillion Hz. Is that particularly high by the standards of the early universe?

Hi Fyz.

You said "I don't understand gazillions, but at a guess its apparent mass would be in the region of 10 Undecillion Hz. Is that particularly high by the standards of the early universe?"

A gazillion is something like a "unspecified, large number", so it says I don't know.

I understand that the US value for an Undecillion is 10³⁶. It rings a bell in the sense that the our observable universe apparently had a radius of around 10³⁶ Planck lengths when inflation stopped at around 10^-32 seconds, with a temperature of around 10²²°K.

Now what frequency or energy would that represent?

Jorrie

Based on that, I think the radiation peak at the end of inflation would be around 10^33 Hz - a factor 10,000 lower. Was the universe even hotter before that?

Fyz

Hi Fyz, you asked: "Was the universe even hotter before that?"

Yep, I believe it was 10³² °K at the 'beginning of time' 10^-43 seconds, before inflation (George Smooth et. al, Wrinkles in Time, pp. 166-167).

Jorrie

Assuming the radiation then conformed to contemporary physics**, one photon near the frequency of the radiation peak would have had a mass equivalent to about a million motor scooters.

**Outside my field, so I can't comment. Nor whether, if such light were to hit the earth today there would usually be any significant interaction.

Fyz

that might be an alternative theory to explain the Tunguska Event--a photon with the energy of a million vespas--call it a megavespa--hitting siberia.

But my understanding of the big bang is that it was more of a big sigh--a cool sweep of almost energyless space spreading out to fill a void as objects coalesced out of it at random, the way a light snow will fall on a perfectly clear but very cold day--the moisture in the air becoming flakes at ground level.

Both would appear to "gain mass."

Everyone appears to agree on the streetcar. (The streetcar becomes clear when you remove confusion and assume DC power.)

Concerning the scooter, the rest mass of the batteries do not change as the batteries are discharged because this is a chemical reaction with no change in battery mass, i.e, no battery particles are shot out the rear. Thus, the rest mass of the scooter stays constant.

So, although the total energy, counting the chemical potential energy in the batteries, cannot go up, when the kinetic energy increases, it appears that the total mass has also gone up. The scooter is not losing rest mass, it is converting chemical potential energy into kinetic energy, i.e., into "relativistic mass" at high speeds.

"It is converting chemical potential energy into kinetic energy". Precisely so: and as chemical potential energy has the same mass/energy relationship as any other form of energy, converting it to kinetic energy will not change its effective mass.

Fyz

The effective or apparent mass of the theoretical scooter has to increase as its speed increases. Otherwise, it would be possible to drive this theoretical scooter faster than the speed of light using this supper efficient and powerful battery postulated in the problem!

The REST mass of the scooter stays constant as the battery is discharged, propelling the scooter to relativistic speeds.

The conversion of chemical energy to kinetic energy has to have a mechanism to keep it from reaching the speed of light, which is precisely the relativistic mass effect.

Hi edykes, so you still reckon that the mass of the scooter increases, despite all that Fyz and myself have been preaching over and over?

Granted, we don't agree 100%, but that's only on the issue as to whether one should leave out losses at first, or make them part of the answer from the start. That's philosophy, not science, I guess. We do not disagree on the physics.

The situation with the scooter is the same as with your car. It uses at least as much mass-energy from the gas tank (actually more) than it is converting into kinetic energy of motion. So your car gets lighter the farther and faster you drive it. Electric cars driven form internal batteries do the same.

The rest mass of a discharged battery is less than the rest mass of a fully charged one. If the battery is huge enough, one can put it on a balance and see for oneself. But no need for that - it has been proven over and over that energy of every description is equivalent to mass.

Jorrie

Jorrie:

Well, the energy density of a battery is so low, I question whether we could measure such a difference. Having worked with batteries a lot, I definitely wish that there was a real world difference in weight when one of those things is discharged!!

Yes, if we had this magic high energy battery for the scooter, it would have a substantial decrease in mass as it discharges to propel the scooter to relativistic speeds.

Thank you. Again, it is important to put the normal work day aside when addressing these issues. It had not occurred to me that a charged chemical battery would weigh more when charged, but clearly it has to. It could also be a nuclear battery, and then the mass situation is very obvious.

Hi Jorrie

Thanks for translating!

Is it worth clarifying that internal losses convert to heat which carries the same KE (and therefore mass), albeit in a randomised form?

Fyz

Hi Jorrie,

Consider the following thought experiment:

Two identical battery powered electrical vehicles decide to have a race on a long straight track (drag race). Call one car the "Motor Scooter" and the other one the "Steet Car". Both vehicles line up at the starting line with batteries completely discharged. Before the race, the owner of the Street Car decides to do some modifications to his vehicle. He decides to install an electrical trolley over his vehicle's driving lane for the entire lenght of the track. He then removes the electrical line which connects his battery to the vehicle's electrical drive motor and reconnects it to the trolley (via varies slip ring devices etc.) An electrical line is then brought in from an outside source and first of all, run through a meter and into the batteries of the "Motor Scooter". The batteries are then charged with a recorded amount of energy 'E'. The meter is then connected in reverse and connected to the trolley which will now supply energy to the 'Street Car'. A contactor is designed to 'open' as soon as the meter reads zero. i.e.when the energy input to the Street Car equals the energy supplied to the batteries of the Motor Scooter. This ensures that at the finish of the race, both vehicles will have consumed equivelent amounts of energy 'E'.

The race is now set to commence. Both drivers and the guy with the flag starting the race all agree the "mass/energy" of the Scooter is its rest mass "m" plus its stored energy "E" and that the "mass/energy" of the Street Car is its rest mass "m" only. (remember that both vehicles are identical and thus have the same rest mass we also stipulate that both vehicle have been tweeked in such a manner to insure identical efficiencies).

The flag drops and the race is on. After a time, the batteries in the Scooter are emptied simultaneously with the meter zeroing out in the Street Car. The race ends in a dead heat-both vehicle attain the same velocity 'V'. Both drivers agree that the 'mass/energy' of the Scooter and of the Street Car are the same and are equal to m+E (assuming 100% conversion efficiency, the energy lost by the batteries is equal to the the kinetic energy gained by the 'Scooter' and the electrical energy gained by the 'Street Car' through the trolley is converted to an equal amount of kinetic energy). The guy flagging the finish of the race, however, observes the 'mass/energy' of both vehicles to be the relativistic mass 'M' plus 'E'.

The important point, I think, is that both vehicles recieve equivelent amounts of energy from outside sources. Relative mass/energy is dependent upon the ratio of the vehicle's relative velocity with the speed of light and NOT upon the ORDER in which a vehicle (or a particle) recieves energy.

What do ya think?

Hi Xeni.

Complex set up! If I understand correctly, the salient points can be summarized as follows:

(1) At the start of the race the scooter has a fully charged battery and an otherwise identical 'streetcar' has a fully discharged battery, so that at the end of the race, both have fully discharged batteries.

(2) You know exactly how much electrical energy (E) the scooter's battery received and you are going the switch off the power to the 'streetcar' when it has received an equal amount of energy (E), where the race ends.

(3) You then measure the velocities and the relative positions of the two vehicles at this defined end of the race.

(4) You conclude that: "The race ends in a dead heat-both vehicle attain the same velocity 'V'. Both drivers agree that the 'mass/energy' of the Scooter and of the Street Car are the same and are equal to m+E… "

Not quite! When the scooter's battery runs out, the 'streetcar' will be way up the track and at a higher final speed. Why? Because it always had less mass-energy than the scooter, except right at the end. The easiest way to make this crystal clear is to look at the math.

Let the rest mass-energy (with empty batteries) of both vehicles be mc². Let the rest mass-energy of a fully charged scooter be mc²+E. Start by calculating the streetcar's velocity after external energy E has been applied to it, using the relativistic mass-energy equation:^[1]

where ε = mc²+E and mc² is the original rest-mass energy of the car. It is easy to extract v/c for the streetcar from it.

Then approximate the scooter's final velocity by assuming it consumes electrical energy at a constant rate, so that its average rest mass-energy over the race is ε = mc²+0.5E. You will find the scooter's final velocity significantly smaller than the streetcar's.

Jorrie

[1] Relativity 4 Engineers, eq. 2.5, pp 41. Chapter down-loadable free from => Inertial Movement.

Hi Jorrie

Assuming that the kinetic energies are applied at the same rate, I agree that the street-car will get there first. But I can't follow why the final velocity of the streetcar should be higher than the final velocity of the scooter. So far as I can see, towards the end of the race, the acceleration of the scooter will be greater than the acceleration of the streetcar. You can look at this as conservation of momentum and of energy, or in a non-rigorous way imagine that the removal of rest-mass during the acceleration means that the KE and momentum that has been imparted to the potential energy of the battery rest-mass is automatically transferred to the remainder of the vehicle as the potential energy is converted to KE

Fyz

Hi Fyz, you wrote: "So far as I can see, towards the end of the race, the acceleration of the scooter will be greater than the acceleration of the streetcar."

Why?

The scooter, with its almost empty battery towards the end, will still be 'heavier' than the streetcar, so given the assumption of electrical energy being supplied at the same rate, the scooter should always accelerate slower, not so?

The supply of energy to both is discontinued at the same time as measured in the static reference frame. So how can the scooter have the same velocity in the reference frame?

That said, there maybe something important in what you said about the conservation of energy – I just don't know what yet, but I will give it a long thought. Maybe my above argument is applying common sense to a relativistic situation an inappropriate way…

Jorrie

I'll try again:

When you convert rest mass to KE, what happens to the KE that was inherent in the rest mass you are converting? I say it must be added to the KE of the residual rest mass. Another way of looking at this is that, in the original frame, the amount of mass you are converting to KE is higher than the rest mass. However you look at it, if you start the streetcar with the same rest-mass as the scooter has at the end of the acceleration, and finish when both the moving masses and the rest masses are the same, the speeds must also be the same.

Fyz

Hi Fyz.

I gave it some deep thought and then I threw a spreadsheet with numerical integration, relativistic and all that, at it. Result: I agree with you.

More astonishingly, unless I have goofed with the spreadsheet, I have to agree with Xeni, the poster of this variant as well! The scooter and the streetcar in this ideal experiment reach the same point in space at the same time. The 'heavier' scooter starts slower and falls behind, but in a loss-less situation, it catches up with the streetcar just as the battery power 'runs out'.

It actually amazed me, but I should have known that one ignores the conservation of energy at your peril.

Thanks for your valuable inputs.

Jorrie

I think it must be something wrong in the spreadsheet. Assuming that you feed or convert power at the same rate for each vehicle (I think you can choose either the original frame, or separate frames relating to the individual vehicles), surely the scooter is going slower at all times until you cut the power.

Hi again Fyz.

Yep, I've been sloppy and paid the price (simple wrong spreadsheet reference). The scooter speed only catches up at the moment of cutting the power and hence the scooter ends up short of the streetcar.

Jorrie

That's a relief - it means that I don't yet have to add special relativity to my list of counter-intuitive physics (it's plenty long enough already)

Fyz

Hi Jorrie,

I think I agree with you, but I have some reservations about the question. In order to have such a race, we require simultaneity; yet, clearly, the streetcar and motor scooter are in different locations at different speeds. I don't think it possible to answer until it is reframed.

Tom

Yup, I believe you're right Jorrie!

I guess that's why they pay ya the Big Bucks!

Hi Xeni, I was only half right. As Fyz correctly pointed out, the distances will indeed differ, but the speeds will be identical when the power is simultaneously (in the reference frame) interrupted.

Jorrie

Hi again Xeni.

I an attempt to impress those who decide on the bucks, here's some interesting plots of the latest scenario, as I argued out with Fyz and was corrected by him. The sums are however solely my crimes and I must take the blame for any further errors.

The hypothetical amount of electrical energy loaded into the scooter was 50% of the rest mass of the scooter before charging the battery. The 'tram' of same rest mass as the uncharged scooter, received and equal amount of energy while running, all as kinetic energy.

To get it relativistic, yet stay 'realistic', years (y) and light-years (ly) have been used, e.g. acceleration is in ly/y². The race lasted for one year and the speed achieved was 0.745c.

Jorrie

Hi Jorrie

Starting from kinetic_energy=rest_mass, I have M=Mrest/sqrt(1-v²/c²), which gives a final velocity of exactly c*sqrt(0.75). Where did I go wrong?

Fyz

On that basis, the distances would be about 0.8 (streetcar) and 0.77 light-years (scooter)...

Hi again Fyz.

I think you started with electrical energy = rest energy, while I calculated for only 50% of rest energy of the scooter, as stated in my post.

When I plug your initial value in the spreadsheet, I indeed get your v=√(0.75)c, but I get distances: ~0.69 ly (car) and ~0.62 ly (scooter).

Don't know where the difference lies. I integrated numerically, since the acceleration is changing continuously. Did you integrate the velocity analytically?

Jorrie

Hi Jorrie

Ah - I misinterpreted this as the rest mass of the scooter at the beginning of the story (i.e. its total mass). I agree both your results (although my numerical integrations** gave just below 0.685 and just below 0.615, so I rounded down).

**I replied to myself with a correction. It was definitely there at one point. I tried to modify it, but the system timed out on me - it must disappear while you are editing it, and doesn't always replace after the time-out. I had no idea that could happen.

On the other hand, it means we should be able to remove posts if we've missed another contribution, rather than just voiding them. I shall try this now. If you see a void answer below mine, it means it didn't work

Regards

Fyz

Trial for removal

How strange - either it's not consistent, or something entirely different caused the other to vanish

Fyz, I didn't have such a problem before. Normally if something goes haywire in an attempt to either preview or submitting a post, I could back it until I had my original text again.

I'm not sure if one is supposed to reply to your own posts though - could result in funnies, I guess. I just post corrections on my last one as a reply to the one I originally replied to.

Jorrie

Thanks. Your approach is rational, but the system places answers to the first post before the correction, so the correction may become transparent (i.e. all but invisible).
I would be surprised if the system recognised self-replies as different from others, as clearly it allows them. In any event, either the cause was other than as I had imagined, or it is not repeatable. I suppose it comes back to 'if in doubt, check'.

BTW, I could in principle have recovered the reply if I had not closed the browser between times, as that is set to erase the previous session.

I just posted and deleted the word 'test with no problem. I didn't check the times on fyz/jorrie postings, but I wonder if your 15 min edit time evaporates if somebody replies to it straight away ? mgaulin would be the person to ask if this type of event is a problem. <sorry for the interruption guys, I was curious too>

Did you delete the word or the posting? I've (usually) no problem changing words in a post. The 'problem' was that the actual post disappeared while I was re-editing it, even though it had previously appeared in the thread - and I can't repeat that.

Fyz

My original #181 simply said 'test' before I went back to edit. I have previously posted comment then gone back to edit and deleted the whole thing (usually because of second thoughts about getting bogged down in another ling topic). The edit screen does 'hiccup' a bit - perhaps the number of people using it at the same time has some effect on it. Like other people, I've noticed that it occasionally freezes up. Overall it functions fine, and I've never encountered major problems with it.

the actual post disappeared while I was re-editing it,

Not sure what you mean Fyz. If you're saying it disappeared from the thread while editing then (a) how would you know whilst in the editor, and (b) I'd have expected that whilst amending the words (I think ?). Or do you mean that the text evaporated before your eyes whilst back in the edit screen ?

Sometimes when typing a reply, I copy it temporarily (ie select all text + press copy puts it on the clipboard) whilst I nip back to a thread to review other preceding posts.

******This line is added after I posted the next one (#144)****

just checking something. ignore this.

******In my previous post I should have said #184*********

I am now totally confused as to what the anomaly is !

What I mean is (long-winded version, so I can believe what I've written is clear:

I created a new entry, previewed it, and submitted it. Some time later I realised it could do with clarification, so I found it in the thread, pressed edit, and tried to resubmit, but I had run out of time. As it was only a minor clarification, I left didn't bother creating another new entry.

I only discovered it was no longer in the thread when I saw Jorrie's reply to a previous entry - and wondered why Jorrie hadn't seen the latest.

As I said before, I've tried this since, and didn't see a repetition.

P.S. I still haven't discovered how to remove an entry completely should I wish to do so.

I had run out of time

Well I guess that if you have it back in the editor and run over your 15 minutes, the system is not yet advanced enough to guess what you would have liked the post to say. Perhaps at 15 mins it went from edit screen straight to the circular filing cabinet. That way, you could re-do the post from scratch anyway.

If your post is on the thread at t+15 mins, it's there for eternity. The only real way round this is to include (eg) slightly disguised vulgarity and report your own post to get it deleted by admin for breaching rules on posting. Within the 15 mins you just take it to the edit box and backspace it clear or 'cut' , then exit ( I usually use the backarrow at top left of screen) to make the post disappear.

"Well I guess that if you have it back in the editor and run over your 15 minutes, the system is not yet advanced enough to guess what you would have liked the post to say."
No, I tried that with an experimental post, and it didn't remove it.

Maybe the time it vanished I had used the back-arrow to look at something, and removed it that way; I didn't know that this removed posts that had already been submitted - I shall try it now.

This is the trial, Kris. If it's visible five minutes after the initial posting, either your method doesn't work with my browser, or I've failed to follow your meaning.

OK, I tried pressing the edit button, clearing the text and then using the back arrow - then I pressed the "refresh" arrow and it was still there. Then I tried going back from the preview screen - same result. What have I misunderstood?

Fyz

.

*** Yeah , you're right. The best I could do was leave a '.' in the title and message field. Must have confused myself with cancelling a post befor submitting it.***

(I won't try the alternate method for getting a post deleted. f*6* it.)

Hello Master,

I started out so darn sure you were off the mark and ended up seeing this as the best question on relativity that I've seen in years! Your "race" clarifies the issue of transferring relativistic energy. WOW! The two stay side by each the whole blasted way. I'm going to go back and review Michael Fowler's notes on relativity to see if I missed anything, but I think you've absolutely nailed it.

Grasshopper aka Tom

Hi Tom.

You wrote: "WOW! The two stay side by each the whole blasted way."

Are you darn sure of this?

Jorrie

Hi Jorrie,

No. That's the part I have to check because I see two different ways of defining the race and I have to think through that. I am sure that I hadn't really thought about energy momentum in this problem until Xeni raised this most excellent point.

Tom

Hi Jorrie and Xeni,

Sorry to be so late in replying. I have a verrry non-relativistic brain. Anyway, I had to think through all this before answering, and I had to download some of my old college notes to see if my memory was correct.

I still think the problem needs to be reframed to make it clearer, but I believe I understand the gist. This is a race and, like most races (apart from the Tour de Steroids), should be fair, i.e., spectators are not allowed to assist or impede either racer.

That said, this race is about a vehicle already carrying its energy vs a vehicle which will be acquiring energy during the race. The great beauty of Xeni's question is that it makes us think about what we have to do to get that "down the road" energy. That energy will have to be velocity matched to the streetcar, else we add or subtract momentum (both unfair, I submit). Further, in an acceleration field, that energy will have positional energy. If it is to be fair race, the streetcar must be responsible for supplying both energy terms.

The only question is whether the streetcar has to supply that energy on a "pay as you go" method or supply it instantaneously when it is needed. These are the two instances I spoke of. If we are to be able to approach c, the energy must be supplied from the very start, i.e., the streetcar must know the motor scooter has 10 units of energy, e.g., and begin the race prepositioning all 10 of its units. Thus, there is no distinction between the two and the race reduces to the trivial. But, that may indeed be a great point.

I await "holes in my baloon".

Tom

Hi Tom. I think Xeni agreed that what Physicist? and myself have worked out is the correct interpretation for his question. The idea was that the the streetcar gets energy from the outside as it accelerates.

It seems to me as if you are worried about the slow moving electrons in the overhead wires that have to be accelerated to relativistic speeds when they enter the streetcar's wires. This could have an effect on the efficiency of the streetcar, but I think Xeni agreed that one can idealize that away, because the mass of the electrons involved is very, very small in comparison with a streetcar.

Jorrie

Jorrie,

Yes. Initially I too saw the question that way. Then, I realized that the question invited an examination of whether the energies are equal. Certainly in the original gedanken I concur that the external energy source creates added "mass" while the internal energy does not. My point of view (at least at the moment) is that I have not fully understood the energy of an energy source and that I have been sloppy in applying STR to accelerating systems. I am trying to reason my way past these limitations.

Tom

Hi Jorrie

As I see it, the wires are only a waveguide*. If we therefore consider the energy being extracted from the guided wave, all the street-car 'sees' is a "standard" red shift.

Returning from the mass question to the race, as we have loosely defined the "final" energies as being identical, it shouldn't be an issue that the street-car receives less power in its reference frame than is seen to be extracted by the stationary frame.

Fyz

*What guides the wave is the difference between the movements of the "stationary" matrix of nuclei-minus-inner orbitals and of the free electrons in the conductors. If you want to go into detail, you will find that relativistic transformations of these (to produce net currents) corresponds exactly with what is required to guide the red shifted wave.

Hi Edykes,

I think you've hit on the crux of the question: does a battery lose mass as it discharges? Any ideas how to settle that question?

Tom

Hi Physicist,

Interesting point and certainly a different way to see this. I'd have to ponder this some, but I think I will always measure a photon as traveling at c in vacuuo no matter what the frame (well, OK, a heavily accelerated frame makes my measurement difficult). I think I can never measure its proper mass since the little dickens won't sit still, but I believe the proper mass to be 0. I think the photon will have a relativistic momentum proportional to frequency and that I will measure that frequency as different depending on relative motion. Is that correct?

I enjoy these theoretical discussions, but have a nagging worry that somewhere the ghost of Rutherford is screaming, "For goodness sakes, get some experimental data!"

Tom

Hi TPV45

This was Jorrie's concept, I think. Yes, the rest-mass is zero - but the relativistic mass is substantial. You can think of this as zero times infinity, with the resultant mass defined by the frequency. Both frequency and mass will, of course, be different when observed from different frames; but as you say, the velocity will be 'always' C. (There's a slight issue about the velocity in the frame of another photon, but it's beyond me and of no practical consequence, so I like to say it's philosophy, not physics)

Fyz

Thanks Jorrie,

I can't wait to see if we wave goodbye to that photon to next week !

Kris

The reason for choosing a streetcar and scooter would presumably be to allow initial sums to push against an infinite mass object as the "reference". I think the first port of call should be to describe what happens in the "ideal" case - otherwise all you will do is spread confusion (perhaps Kris got to you?, or maybe someone already did the above - and I failed to find it).

After that, maybe (just maybe) it makes sense to add modifiers. The problem is that the situation is so far from any known reality that no sensible set can be made to fit.

I'll selecting a few examples of modifiers that I didn't yet see mentioned specifically (again, they may well be there somewhere): internal losses in the scooter will not modify its mass, whereas those in the streetcar will increase the mass. External losses for the scooter will reduce the mass, but those for the streetcar will leave things unchanged.

But then we have wear that will certainly lighten both vehicles. But perhaps the battery in the scooter is a fuel cell, or a nuclear battery that uses any material the scooter collides with as fuel.

Regards

Fyz

Thanks. I used to think I understood this part of relativity.

we will have billions & billions of rpm on motors and wheels , at light speed they will seem to be at stand still ,

both scooter and car will be miniturised to sub photonic level

As Jorrie says, a rather open and thought-provoking challenge.

I propose the following constraints for initial calculations:

1) Both wheeled vehicles gain their momentum by traction against a stationary object of effectively infinite mass in the original (= "stationary") frame of reference.

2) The traction is lossless, and there are no other sources of Kinetic loss.

3) The battery is 100% efficient, and there is no energy interchange (other than that already) defined electrical between either the scooter or the streetcar and the environment.

I think that gives enough information so that the problem can be solved numerically. In many cases the effects of different assumptions will be clear by using this as a reference.

Fyz

I've stayed out of this one because it is waaaaay over my head. but I do have a couple of questions.

Assuming this will be along trip, what if a passenger on the streetcar gets hungry for a hotdog? How would you cook it? Will cooking the hot dog affect the energy going into the car's motor?

I can answer that one. You can cook a hotdog by hooking up each end of the dog to a high voltage electrical source. The resistance of the hot dog will result in enough heat not only to cook it, but to make it explode like the barrel of a plugged blunderbus on Thanksgiving morning. You need a fair amount of voltage, though. House current would be enough, probably, but not a 12 volt battery. You could also use electricity to create hydrogen by electrolyzing water, then burn the hydrogen to cook the dog. Most hotdogs are precooked, so anything you do to warm it up would probably be fine. In that case, just jam it into your armpit and walk around with it in there for a few hours. Rinse off and eat. There are other places you could jam it, but they'll all be within a degree or two of the same temperature, so use your armpit. Is cooking a hotdog in your armpit more disgusting than the stuff they make a hotdog out of?

I don't buy the answer. Both systems are seeing an increase in kinetic energy and therefore should be subject to relativistic mass effects. By the argument posed by the answer, a chemical fuel rocket would not be subject to relativistic mass increase while a solar sail would. Again, I don't buy it. The apparent mass increase is the heart of the light speed barrier - as mass increases to infinity at c, the energy required to accelerate the mass also increases to infinity (basic pre-calculus limit work here). So we trick relativity simply by taking our fuel source with us? No dice.

Hi Guest, you said you don't buy the 'official answer'.

If one ignores losses, the anwser is correct. See this earlier reply and replies to it.

If losses are taken into account, the scooter actually lose 'relativistic mass'.

Jorrie

Hi Guest,

It seems like a trick only because the question poses an extraordinarily implausible situation. No real object with any degree of size and complexity can ever be accelerated to near c in this sort of set-up. Besides all the various losses we've kicked around, there is the problem that the rider needs an infinitely large battery at the start and, if there are any limits on energy drain from that battery, he needs an infinitely long time (unless we are willing to wind up with a single photon at some point). That's too many infinites to relate to real, large, objects. Einstein used an ideal when he talked about accelerating a gainless, lossless, object.

Tom

The glitch isn't in the trick, it's in the base principle behind the trick. On-board power vs. external power. Applying the same base principle to a chemical rocket negates relativistic mass. No dice. This would seem to invalidate the base principle used to arrive to the answer - regardless of how improbable the situation may be. The thought experiment fails to prove itself.

Hi Guest.

You wrote: "Applying the same base principle to a chemical rocket negates relativistic mass. No dice."

It looks like you are forgetting that a super chemical rocket on the launch pad will consist of about 90% expendable fuel and the rest is 'payload'. Say this super-rocket can boost that 10% payload to a velocity increment (Δv) of 0.2c.

Let the starting mass be 1 unit (0.9 fuel and 0.1 payload). The initial energy is then Eo = mc² = c² units.

The final energy, with all that fuel blown out of the tail nozzles, is:

E = 0.1c²/√[1-0.2^2] = 0.104c² units.

If you could push your rocket's efficiency to near 100%, you will at best end up with a Δv of ~0.94c and a final energy of E = ~0.86c² units, which is way less than the c² units you started with.

To exceed c² units as the final energy of the payload, your rocket engine needs to be 'over-unity' (efficiency > 100%)

Jorrie

Okay - so you're saying that in the case of the rocket, the mass expelled as propellant more than offsets the Lorentz factor for the remaining payload. I get that. So let's go back to the origional question. The electric scooter is powered by an electrochemical resction in its batteries. By reading some of the previous posts, presumably some mass is lost (not sure I agree, but I will accept for the sake of argument) and mass/energy total of the system is not increased (although I am hard pressed to see enough mass loss to offset Lorentz). Official answer stands. What happens if we replace the battery with a capacitor? On-board energy source that should loose no mass while increasing velocity. Now how do we offset Lorentz and validate the principle of the official answer?

There is energy stored in the electric field in the capacitor. That has mass. When you discharge the capacitor, the capacitor becomes lighter by the same amount that the energy applied to the remainder of the scooter increases (neglecting losses, of course).

Fyz

P.S. We are not trying to gang up on you - just the first to see your post is answering. (Jorrie in S.A will doubtless be taking some well-earned rest).

No worries, I'm not feeling ganged up on - I appreciate the dialog!

Welcome! Which I mean in two ways - many of us would appreciate it if you would create a username by joining CR4. Among other things, this makes it easier for people like me to respond to others who share their interests.

Fyz

I tried to suggest at #4 that the battery scooter would disappear to nothing. Whether that implies I would have seen an increasingly heavy streetcar is just as puzzling to me. Keeping my head stuck up my own rear-end is undoubtedly my safest bet. This is a very interesting theoretical area , but I much prefer things I can try to simulate with assorted junk in my garden shed. I hope CR4 can 'Make it so'.

My question to the creater of this question is.......

How can the street car take on electrical energy when it is traveling at the same speed as the electrons in the wire?

Seems to me your answer is flawed in that the power from the powerhouse will never catch up to the streetcar going the speed of light therefore the mass can not increase because the electrons would appear "still" when viewed from the streetcar and they would not have enough speed to "jump on board".....

Dwight

The question only says "approaching the speed of light". And perhaps the streetcar is going towards the power source, which would increase the available power.

The more serious point is that it's not the speed of the electrons that is important - it has been mentioned elsewhere that they generally move quite slowly relative to their wires. What conveys the energy is electric and magnetic fields - the wires and electrons merely serve to guide those fields

Fyz

Don't forget also, that Jorrie solved the problem (which I posed) of near light-speed current trying to catch the near light-speed streetcar moving in the same direction by suggesting that the power line be fed in parallel, from both ends! That way, power could always reach the streetcar almost instantly, no matter which direction it was travelling.

I know that Jorrie is tops on this - sorry that I overlooked that post.

Fyz

If your answer is true, then the relativistic mass equation given by Guest #70 is false and a spaceship carrying enough of its own fuel would be able to attain a velocity grater than the speed of light.

Hi Xeni,

Many folks don't use the relativistic mass equation because it isn't always so clear; instead, they use relativistic energy given by

E = sqr(p^2*c^2+m'^2*c^4)

where p is the relativistic momentum and m' is the proper or rest mass (I can't figure out to type a subscript). The "increased mass" is now hidden in p and you don't have to think about it. The calculation of p will approach infinity as v approaches c and, thus, the energy required will approach infinity. Since energy/mass is conserved in this instance, the only place to get the added energy for the momentum is by stealing from the second term, m'^2*c^4, thus decreasing the proper mass since c is a constant. Since approaching c requires infinite energy, there is nothing more beyond that.

But, to believe these results requires us to accept Minkowski space-time, a constant velocity of c in that space-time, and relativity. If those are not the case, none of this holds water.

Tom

Hi Tom, Xeni.

Tom is right, provided one remembers that p is not the good old mv, but rather the relativistic momentum: p = mv/√[1-v²/c²].

If you substitute that into Tom's equation and simplify, you get the more general form:

E = mc²/√[1-v²/c²].

Jorrie

PS: Tom, the subscripts, superscripts, etc. are in the toolbar at the top of the CR4 editor window, shown as x₂, x², etc. Just highlight what you want to change and select the appropriate icon.

Thanks, Jorrie

I didn't know the toolbar was there. I even found γ. Now I can be really long-winded.

Tom

I have enjoyed reading this form very much. Jorrie your answers have been great from a Physics stand point. they are clear and to the point. Some of the readers seem to be having a hard time seeing that all forms of energy have to follow the rules "Laws of Physics" so lets see if this fly's or hits the ground like a rock.

if the scooter has a lead acid battery the size of the earth (neatly tucked under the seat of course) because of the formulae E=MC² the battery has to convert mass to energy and the energy required to approach the speed of light is on the scale of holy cow batman.

Perhaps a up graded battery would be in order. Say a magnetic bottle filled with antimatter this could be mixed with an equal amount of matter the math would be much easer this way. a billion tons of lead are not needed. Bottom line, Jorrie is still right you get up to speed you get an empty bottle and an equal amount of matter lighter and this missing mass is still greater than the relativistic effects

Hi Cforce, you're welcome.

I like your "magnetic bottle filled with antimatter this could be mixed with an equal amount of matter..."

If the 'magnetic bottle' was created by a 'force field', 'shield' or whatever, then by the time you have used up all the matter-antimatter at 100% efficiency, you will have 'nothing', but that nothing will be going at the speed of light. Or, maybe there will be one photon left?

Jorrie

Hi Xeni, also read my reply #130 above.

Jorrie

Hello Jorrie.

I'm using this to reply to a question from you that I have lost in these threads and wormholes of CR4.

Let us say that I used to visit Slangpark at halfway house in the mid60's which according to GEarth is nearby and worked at 26.10S 28.03E and 26.07 28.03 etc before heading into the dark heart.You seem to have an awful lot of new roads since then - reminiscent of the start of the Hitch-hikers guide.IW

I wonder if the Streetcar is called Cemetery or Desire.

Actually the mass of both should increase as both are getting energy from different sources. One carries the source with it whereas the other gets the energy supplied through cables.

Thought the scooter carries the batteries, they generate power through chemical reactions and I assume that some sort of recharging facility would also be there based on a dynamo or otherwise

Are you trying to invent perpetual motion, or to modify the question? So long as the scooter is powered by the batteries, recharging the batteries will slow the scooter down at least proportionately.

Fyz

The answer implies that the motor scooter already has infinite mass since it contains infinite energy to enable it to accelerate sufficiently to approach the speed of light. Is that correct?

Er... no. What I think it is trying to say is that, as the energy in the battery is converted to kinetic energy, the rest mass gets smaller, so that the mass viewed from the original ("stationary") frame remains constant - and because the rest mass reduces towards zero, it does not need infinite energy to increase its velocity towards the speed of light.

So, if the scooter starts out with some finite mass m, its mass viewed from the original frame of reference will remain at m whatever speed it gets to. But its mass in a frame of reference that is moving with the scooter will get smaller and smaller. If it were to reach the speed of light (!!!), its rest mass would reduce to zero - but the mass of the resultant photon (as viewed from the original frame of reference) would be the same as the scooters original mass. This was first mentioned by Jorrie.

The position for the streetcar is quite different - the power is provided from the outside, and the rest mass (measured in a frame of reference that is moving with the streetcar) remains constant. It would of course take infinite energy supplied from the outside to accelerate it to the speed of light, and its apparent mass would then become infinite.

N.B. I'm not certain whether the universe is finite, but even if it's infinite we would have a bit of a job concentrating inifinite energy into one streetcar.

Fyz

It would likewise be a bit of a job to accelerate a streetcar up to speeds approaching the speed of light, don't you think?

I thought I said that on purely theoretical grounds. The practical difficulty of achieving even the speed of sound would defeat me, of course.

Fyz

I was correct at #147, the streetcar is called Cemetary. Board here all ye who dare..