The link Relativity 4 Engineers somehow disappeared from the blog article. It happens if one is forced to copy in and out of CR4 for spell-checking - so a spell checker must be the highest priority on the wish-list for CR4!

Hi Jorrie,

A spellchecker is indeed high up on our long "short list" of improvements.

- Chris

Interesting explanation. As you know I've been under the impression that it was relativistic mass that caused the decline in acceleration. This gives me a fresh perspective I need to think about, thanks.

Hi Roger; modern thinking steers away from relativistic mass and only considers rest mass M₀, since it's independent of choice of coordinate system. The old-fashioned relativistic mass: M₀/sqrt(1-(v/c)^2) alone does in any case not explain the relativistic acceleration. You need that plus velocity time dilation to get to the correct answer.

is it true that maximum velocity attainable in any system is "c" ? Or is that just an assumption of relativity theory? Is that velocity c merely the fastest thing we are able to measure at this stage of human investigational development, rather than the fastest thing possible?

What is "energy" ?

How can light, which has both a particulate and a wave nature travel at the speed of light without having its photonic mass increase to infinity?

Guest wrote: "is it true that maximum velocity attainable in any system is "c"? Or is that just an assumption of relativity theory?"

In a way it is an assumption: we use light to synchronize clocks and we use light to determine distance (the meter is defined in terms of a number of wavelengths of light of a specific type). If we then use these measurements to determine the speed of anything, we are virtually assured not to measure speeds exceeding "c". If we used, e.g., a different method of clock synchronization, we could measure speeds exceeding "c".

You also asked about energy. If you Wiki it, you'll find: "In physics, energy is the ability to do work and has many different forms (potential, kinetic, electromagnetic, etc.) No matter what its form, physical energy has the same units as work; a force applied through a distance. The SI unit of energy, the joule, equals one newton applied through one meter, for example."

Finally, the energy of light is not calculated by "E=mc^2", but by "E=pc", where p is the finite, measurable momentum of light.

Hope it helps - Regards, Jorrie

Doesn't a photon have a non-zero yet infintesimal rest mass, which means it is merely a nominal mass even as actually reaches the speed of light? It would be interesting to find out if the photons in these slow-light expeririments with Einstein-Bose condensates decrease the apparent mass.

Scott

Guest wrote: "Doesn't a photon have a non-zero yet infinitesimal rest mass, which means it is merely a nominal mass even as actually reaches the speed of light?"

All observations of photons show that they must have a precisely zero rest mass - actually, it is the reason why they always move at the speed of light. A stationary photon is 'nothing at all'! The energy and momentum of a photon depend only on it's frequency and not on it's speed.

hi i am a astudent from egypt and i i have 16 years and i had reached for a low for the force done by photon and it bring the same results brought by the old law but the proplem that i use the acceleration in my law and i dont know if if the photon has acceleration or no so please i am waiting for ur answer and here is the law :f=4m.a and i need ur help

Hi guest, a photon has no real mass and no acceleration by itself - it always moves at the speed of light. So you cannot use force F= m.a. The energy of a photon is not expressed as kinetic energy (1/2 m v²), but as E_photon = h/λ, where λ is the wavelength of the photon and h is Planck's constant.

When a photon is absorbed or reflected from a body, it imparts its energy to the body, which actually may cause a tiny acceleration of that body. But the photon is not decelerated in the process - it simple ceases to exist!

Regards, Jorrie

dear mr u didnt understand me nice i dont use f=m.a i am using a law states that force done by a photon =4.m.a and λ=h2a/pw and 2a=rate of number of photn per time * c and they brings the same result and i proved them with units and with another things and the bring typical results..and i am in diffrent with u that photon doesnt have acceleration coz compton effect states that when a photon collide with somethinh its frequncy decrease so its speed will decrease since mass is constatnt and as the photon is particle and we can apply on it law of momentuem and enisten has proved that.........please answer

Dear Guest, you are writing nonsense when you declare: "..and i am in diffrent with u that photon doesnt have acceleration coz compton effect states that when a photon collide with somethinh its frequncy decrease so its speed will decrease since mass is constatnt.. "

A photon's speed remains c in every inertial reference frame, irrespective of energy gained or lost!

My advice to you: first study and understand science before you develop your own theory...

Jorrie

Mr. JOrrie,

The individual was kind enough to request for you to write back with your answer. His indifference was not a personal one against you, but with the misunderstanding you had initially pertinent to his original question. You simply didnt understand the question. Hey, we all make mistakes, but should not go about belittling others over those mistakes! Stop the arrogance, keep it simple stupid, and leave the educational advice for the University counselours.

and u made a great mistake how E_photon = h/λ its supposed that λ=h/pl where pl is momentuem of photon

sorry mister i replyed for you a lot but here is my laws and they work nice and check them by units and sorry i canot tell u thier proves :

f=4.m.a

2a=Φc

λ=2ha/pw or λ=hcΦ/p

where a is acceleration and λ is wavelength and Φ no of photns per time

Hi Guest, when you say: "... here is my laws and they work nice and check them by units and sorry i canot tell u thier proves", you better provide references from reputable scientific publications that show the proofs. Otherwise, you are just wasting your own and everyone else's time.

Regards, Jorrie

Dear mister

first thanks for your advice and i will tell you the prove.... Einstein said that :

1) e=mc^2 and u know that power is the differentiation of the energy therefore p=2mca

2) and there is a law states that f=2p/c so by replacing the power with 2mca we deduced that equation becomes f=4m.a

3)you know that hvΦ=p so by replacing p by 2mca we reach to hvΦ=2mca after that multiply by c in both sides we reach to hvcΦ=2mc^2a so since hv=mc^2 so the equation became 2a=Φc

where a is acceleration and Φ is number of photons per time so please check them and tell me out what is the wrong in this conclustion....and if they are wrong as u said how they bring the typical answer of the old laws .....

and i will give you sir an example: calculate the force applied by a photon if its power is 100kw and its mass is 10 kg :

answer :

1) according to old law f=2x100000/3x10^8=6.6x10^-4

2) according to my conclusion p=2mca therefor a=100000/2x10x3x10^8

therefore a= 1.666666x10^-5

there fore f=4.m.a = 4x10x1.666666x10^-5=6.6x10^-4

so as u saw the answer are typical ...waiting for Ur answer Mr Jorrie and please i want Ur comments and corrects and Ur advice....and another request please: the old law and my conclusion are same how we can develop my conclusion in other things

Hi Guest, I would be more interested if you register to CR4 (for free). So do that and log in and we can talk!

Regards, Jorrie

Hi Guest (and future member of CR4!), here is one tip which may solve your puzzle:

It is true that power is the time differential of energy transfer (the rate at which energy is transferred), so P = dE/dt. For this to be non-zero, E must change over time, i.e., energy transfer!

What you did when you differentiated E=mc² was to find dE/dc. Now, c is a constant and you cannot differentiate with respect to it!

Think it over...

Regards, Jorrie

hi my mister iam the guest my name is sherif an ambitious studient have 16 years try to prove my self and i would like to know my mister why you invited me to join your site that its really a very good sitee...second about my conclusion i asked a physics doctor in nuclar physics he said to me that the law is theoratically is right and photon has acceleration which proved by compton and he said also that the scientists have different opinons in this subjuct and he said to me 2 say 2 u even conclusion is rong its very enought that one has 16 yeras reach to it...waiting for u mister and for ur answer and i would like you to write me ur mail.....

Hi chavo, welcome as member of CR4!

You wrote: "...second about my conclusion i asked a physics doctor in nuclar physics he said to me that the law is theoratically is right and photon has acceleration which proved by compton and he said also that the scientists have different opinons... "

The Compton scattering has shown that the frequency and hence the momentum of a photon changes, NOT its speed! Hence no acceleration of photons!

One cannot view a photon that is scattered or reflected as being accelerated. It is absorbed and re-emitted in a different direction and perhaps at a different energy level. A photon either moves as a constant speed C, or it ceases to exist (i.e., it is absorbed as energy).

Regards, Jorrie

and please mister also i want you to say ur opinon about me and advice me and what u think about me and about my science future ... i want to be a mechincal enginering

Hi chavo, my best advice to you is to study physics and, if you are inclined towards design, study engineering. Do not waste your time by trying to "rewrite physics" until you know quite a lot in your chosen branch of the subject!

People proposing a new theory or attacking an established theory without showing a deep understanding of the existing theories, are not taken seriously.

Regards, Jorrie

my respectful sir its not over there a math doctor and physics doctor i disccused the matter with them and they said that all is right may be u are right or wrong and we can develop the law to another law

Hi again chavo, you wrote: 1) e=mc^2 and u know that power is the differentiation of the energy therefore p=2mca

Do you agree that your "differentiation" of e = mc² to get p=2mca is wrong, as I pointed out previously? Even if c was a variable and one could differentiate against c, where is the a coming from?

You see, you cannot start with a wrong step and hope to produce anything useful from that!

Regards, Jorrie

hi mister i love physics too much and i didnt study any subject exept physics and i want to be if god wishes a design mechanical enginering and i want to ask you what is your branch in physics ...and i want you to say your opinon about me and why u invited me to join your site sir....waiting for u

and now my sir about the conclusion i will throw it in the trash because i must study to enter the best enginering in egypt coz in egypt i must bring 96% to enter it and i brought 95.5% in the last year ...until i enter enginering and study alot of physics and in that case i will be able to look on the conclusion again and try to develop it as u said that it possible.......... and thanks for you very much..sheriff..and i would like u to say to me whats ur nationalty

mister the last question about conclusion if they are wrong and there doctors said that its right how it bring the same result typically of old law...send me please ur mail

Hi chavo, you can mail me through CR4's private mail if you want. Just click on my avatar (Jorrie) on the left and look for "Send Jorrie a Message" top right.

Jorrie

hi sir why u didnt replyed to me untill now please chek the conclusions and if they are wrong its just a try from a person has 16 years

Give Newton his due. He lived in a real world where there is no perfect vacuum, thus there is always friction, thus there is always a terminal velocity. But forget that, assume a perfect vacuum is possible. Where does this constant force come from? If it is external to the object, any force I can think of will decrease as the object moves away from it. If the object is drawn toward the force such as with gravity or magnetism it will A:impact before achieving lightspeed B:enter orbit, or C: bypass the source of the force causing deceleration. If the constant force is somehow generated by the object, what does it react against in a vacuum? Answer, it needs to eject mass to achieve thrust. "No Problem" you think. Less mass at the same force = greater acceleration. True, but before you have time to achieve the speed of light you will have ejected ALL of your mass. Not to mention you've contaminated your vacuum. No relativity required, just common sense. slo

Guest #2 wrote: "If the constant force is somehow generated by the object, what does it react against in a vacuum? Answer, it needs to eject mass to achieve thrust. "No Problem" you think. Less mass at the same force = greater acceleration. True, but before you have time to achieve the speed of light you will have ejected ALL of your mass. Not to mention you've contaminated your vacuum. No relativity required, just common sense."

I'm afraid Newton might not quite have agreed with you! If you could 'instantaneously' eject 70% of the mass of your spaceship backwards at half the speed of light, then even from a standing start, the remaining 30% of your ship would exceed the speed of light (1.167c) in the original frame of reference. It is simply Newtonian conservation of momentum at work. The limiting speed "c" is only true in a relativistic calculation.

Another way to look at it: if you can make a drive that will push a ship at 1g for more than a year, the Newtonian prediction is a speed increase of more than "c".

I must confess. If we make it practical and eject the mass regularly over time and at a practical ejection speed, we cannot achieve a 'delta-v' of even near "c" in Newton dynamics. The simulation is relatively easy to do, but even with an exhaust speed of "c" relative to the ship, one reaches only 63% of "c" when you have ejected 99% of your ship's mass - a sobering thought!

I'm glad I read your 2nd reply to my post. It saved alot of self dout and alot of recalculation. When I read your first reply I didn't pick up on the subtle importance of "instantaneously" eject 70% of the ships mass. In other words, if you want to achieve the speed of light, throw something twice your own mass behind you at several times the speed of light. Kind of a chicken/egg problem. slo

Oops. Foot in mouth. No edit as a guest. You said half of C, not several times C but I was still focused on the practical implications of 'instantaneous'. slo

I suppose if you do not mind the pollution, find a suitable moon without an atmosphere. Detonate a nuclear device under a probe with a suitable protective plate. Then the moon serves as the 99.999....% of the reaction mass and maybe the probe will speed away at near the speed of light!

If you can eject your reaction mass at near lite speed, a molecule can hav the mass uv a planet, so it bekumz very efficient.

I've alwayz wondered about the speed uv lite particlez.

If they arent moving in a strate line, but a side to side or spiral path with a width (amplitude) determined by their wavelength, arent the photonz actually going faster than c ?

Hi Zman, you wrote: "If they arent moving in a strate line, but a side to side or spiral path with a width (amplitude) determined by their wavelength, arent the photonz actually going faster than c?"

You must not think of photons as a moving along a path that is 'waving' at a certain frequency and wavelength. The best description for a photon is a 'packet of energy' or perhaps a 'wave packet' that has frequency, wavelength and thus propagation speed. Quantum physics has other descriptions for photons that I'm not very knowledgeable about.

This still leavez the idea open. If a foton iz not a single particle, but a packet uv energy, then that packet must be comprized uv something. So if a foton iz a cloud uv much smaller particlez that are vibrating, theze little particlez must be exceeding c at least haf the time depending on the direction uv vibration.

Herez a related Q for you:

A car iz going 100 mph. How fast are the tops uv the tirez going?

Quoting Zman "So if a foton iz a cloud uv much smaller particlez that are vibrating, theze little particlez must be exceeding c at least haf the time depending on the direction uv vibration. . . . A car iz going 100 mph. How fast are the tops uv the tirez going?"

This is a bit off-topic for this thread, but I will make my concluding comments here. If you want to, you can start another thread on this question.

As far as anyone knows, photons are not "clouds of much smaller particles that are vibrating", so the car tire analogy is of no use. I think the only photon characterisitics that we can really observe and measure are the speed, momentum/energy and spin, which is for the 'whole packet'. In the case of a single photon, even wavelength and frequency can only be calculated from energy/momentum. IMO, the "size of a photon" has no meaning at all.

What happens 'inside the packet' is not currently known and it is irrelevant to relativity theory. All that matters is that no information can be transferred faster than the speed of light. The 'quantum mechanics' may have different ideas though - you should ask them!

In kinematics, you analyse the tire by assuming the bottom is fixed to the road, ie. zero speed, therfore the top is going at 200!

Guest gets it!

But now Im trying to recall why I thot this iz related to acceleration by constant force.

Is'nt it normal that if delta T tend to 0 (formula A.26) that accelaration tend also to 0?

I still think that it is the growth of the mass that make it impossible to reach the speed of light

Hi Guest, you asked: "Isn't it normal that if delta T tend to 0 (formula A.26) that acceleration tend also to 0?"

No, but I think you associated -> 0 with v -> c. This is not what the derivation says. It says as delta T -> 0, meaning "in the limit, when the time increment tends to zero", irrespective of velocity. Eq. A.26 tells us that when the velocity factor v/c -> 1, the acceleration also tends to zero, as you suggested.

Regards, Jorrie