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About LinkBacksIf possible prove the equation E=mc^2/sqrt(1-v^2/c^2)
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If possible prove the equation E=mc^2/sqrt(1-v^2/c^2)
There are many ways to do this as George Gamow shows in his work The Great Physicists from Galileo to Einstein.The first person to write down the correct relation of e, m & c was Friedrich Hasenöhrl, Boltzmann's successor, who left in only the telltale 4/3 since he was considering a cylindrical rather than spherical rays in container; he published about 6 months before Einstein's article.Originally Posted by johnzxcv
Hence e=mc^2 is readily established without recourse to SR. Not so perhaps the equation you have written, but I'll let others deal with that.
TFOLZO
Posting lies again, Zoloft?
That derivation is a bit complicated. Let me see if I can find a text which you can download from the internet. If not I'll see if I can write up a derivation. In the mean time I strongly urge you to do a search using Google because I have a lot of work to do myself and it's better for you to do your own searching. It's good for the soul in a manner of speaking.
It's experimentally proven.
johnzxcv - I found the derivation that you were looking for. You need to first understand that the E that you're looking for is the total energy of a particle moving force free in an inertial frame. That has the value of kinetic energy + rest energy. Rest energy = E0. Therefore
E = Kinetic Energy + Rest Energy = K + E0
The kinetic energy of a relativistic particle is derived here Kinetic energy - Wikipedia, the free encyclopedia
Let m = rest mass and gamma = g = 1/sqrt(1 - v^2/c^2). The kinetic energy derived in the above link is found to be
K = mgc2 - mc2 = gE0 - E0
E = K + E0 = ( gE0 - E0 ) + E0
or
E = gE0 = mc2/sqrt(1 - v^2/c^2)
QED!
Last edited by Physicist; 08-30-2014 at 09:03 AM.
johnzxcv - I found the derivation that you were looking for. You need to first understand that the E that you're looking for is the total energy of a particle moving force free in an inertial frame. That has the value of kinetic energy + rest energy. Rest energy = E0. Therefore
E = Kinetic Energy + Rest Energy = K + E0
The kinetic energy of a relativistic particle is derived here Kinetic energy - Wikipedia, the free encyclopedia
Let m = rest mass and gamma = g = 1/sqrt(1 - v^2/c^2). The kinetic energy derived in the above link is found to be
K = mgc2 - mc2 = gE0 - E0
E = K + E0 = ( gE0 - E0 ) + E0
or
E = gE0 = mc2/sqrt(1 - v^2/c^2)
QED!
You obviously don't realize that your "proof" is invalid , since you are using the conclusion in your "proof".
There is no such thing as a "proof" that".
The total energy-momentum of a particle of rest massis DEFINED as :
The above reduces in the frame co-moving with the particle to:
So,is simply a CONSEQUENCE of the DEFINITION of the energy-momentum four-vector.
Last edited by x0x; 09-01-2014 at 01:53 PM.
Here is the derivation of energy-mass equivalence by Einstein, September 27, 1905 ...
http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf
Thank You,
SinceYouAsked
Einstein's paper is notoriously noted to be difficult to follow. There have been several articles on that paper claiming it's wrong. John Stachel and Roberto Torretti wrote a paper correcting the papers which claimed that Einstein was wrong. The best one of those is called Einstein's first derivation of mass-energy equivalence./ I recommend reading that paper first. You can download it at Einstein?s first derivation of mass?energy equivalence | Stachel, John | digital library booksc
However you need to keep in mind that Einstein's 1905 paper doesn't answer the original question which asks for the total energy of a moving object and Einstein's paper was only a paper deriving the relationship between rest mass and rest energy. What the person who started this thread is looking for is total energy which is the sum of rest energy and kinetic energy - assuming that he's setting potential energy to zero which is what it appears to be as most people do and it's absent from the question. I assume that he's ignoring potential since that's absent from his question.
John - Please note in my derivation that the Wiki did the derivation showing how to get the kinetic energy starting with the result of another derivation, i.e. the relationship between rest mass and momentum. You'll see that relation after the part which states Relativistic kinetic energy of rigid bodies.
If you need help finding or deriving relativistic momentum let me know and I'll either find one for you or derive it for you.
Last edited by Physicist; 09-01-2014 at 04:03 AM.
Maybe I am missing something, but this doesn't seem right. 4-momentum is invariant mass times 4-velocity, soThe total energy-momentum of a particle of rest massis DEFINED as :
The c should not be squared.
Thanks Markus. Since x0x is in my ignore list I didn't see this and as such I don't know the context in which it was used. Thanks for mentioning this!
There are several ways to arrive at the expression. Some ways are to postulate it while others derive it. The former is only done in hind sight because the author knows the answer in advance.
However contrary to x0x's claim, the choice of derivation I used is exactly correct. My experience in this forum with x0x is that he or she misses subtle facts like this very often. The derivation I outlined is the same as that found in many excellent special relativity textbooks such as the one by Ray D'Inverno and Hans Ohanian. In other textbooks the authors choose to define certain 4-vectors with the assumption that they're conserved.
However I do know that in the derivation I chose to use the derivation I chose was not what you assumed, i.e. that the energy and momentum are defined quantities. You're lack of knowledge an experience is special relativity and your tendency not to be able to see and/or admit to your own mistakes is what was responsible for that blunder. The total energy of a particle with proper massbut shown to be so by derivation and the momentum is not defined as
. Those are equalities and not identities (Also, it's bad netiquette to use all caps in a post since its netiquette for yelling. Shame on you!)
In Newtonian mechanics momentum is defined[ as p = mv. In special relativity momentum is defined aswhere
. Using relativity one can show, as I did, how to derive the relationship
Edit: Thanks for pointing that missing part of that sentance Jilan
Last edited by Physicist; 09-01-2014 at 04:31 PM.
Yes, it was a typo:Maybe I am missing something, but this doesn't seem right. 4-momentum is invariant mass times 4-velocity, so
The c should not be squared.
Thank you.
The point was that your "proof" is invalid, Peter.
Nope, it is circular. You may need to re-take the physics class.However contrary to x0x's claim, the choice of derivation I used is exactly correct.
Peter,The total energy of a particle with proper mass
I will have Ranch with your above word salad. You can't write coherent English, let alone physics.
The problem that x0x made here is that let out all the justification as to why P = mU should be equal to (Mc, p) whereMaybe I am missing something, but this doesn't seem right. 4-momentum is invariant mass times 4-velocity, so
The c should not be squared.is the relativistic mass of the particle and p = 3-momentum. It's inappropriate to simply defining both sides of a tensor equation and expect it to be correct unless you've already proved it in a derivation, like I always do.
It's true that 4-momentum is defined as P = mU where m = proper mass and U = 4-velocity of particle =where X = (ct, x, y, z) = (ct, r)
gamma is defined as
This means that the 4-momentum is equal to
That's what it means to define 4-momentum. What is lacking at this point is the justification of what the spatial and temporal components are. When that's done we end up with
M is called the time like component of this 4-vector because the time component of an event is X^0 = "ct" so the time component of P is P^0 = "Mc" and we have the symmetry that one would need to justify this as the name. This is also how Max Jammer justified calling it the time component in his last book on mass.
Note: I'm grateful to all the members in this forum for all the nice PMs they send me for helping them and letting me know how much they believe I understand the subject. That's very kind of all of you. I've gotten that from many forum members in many forums and by many physicists too. I wanted to take this time to thank you all for that and all the likes I get in my posts.
Last edited by Physicist; 09-02-2014 at 06:25 PM.
Good, you are learning. Now, going back to your "proof", do you understand why it is invalid?
This means that the 4-momentum is equal to
Since Jilan and johnzxcv like this I'll do a bit better. Although the math isn't that difficult for those who know calculus its difficult to do in Latex and I'm not that good in Latex. I recall that I came across a set of lecture notes on SR one day and I know where they are. Fortunately they're still online. Merely download
http://physics.mq.edu.au/~jcresser/P...ivityNotes.pdf
and follow the derivation for kinetic energy (which the author labels T) and then find E as E = K + E0.
Note: Thanks for informing me of x0x lack of knowledge on this subject which led him to claim that I used circular logic. However anyone can verify that my derivation is correct as is all the derivations I've shown you. However as we all know x0x has a problem admitting to his mistakes and since he started with a mistake his unable to talk his way out of it. Since you've all confirmed that you understand this then it's not a problem. But I had to point it out so that you know for certain that his claim is false. Since x0x is unable to do anything other than get the last word in I'm certain that he'll post more ignorant claims and insults after this repeating and/or rephrasing his bogus claim. However if any member other than x0x please let me know and I'll prove it to you. It's only x0x who has these problems in this thread with ignorance and arrogance.
Nice find Physicist. I look forward to reading it all.
...because you did. The fact that you persist, shows that you don't know what you are doing.
You see, Peter, the wiki derivation use, i.e. it uses the momentum part of the four-vector
. In other words, your "proof" presupposes
.
QED had to define
You're most welcome Jilan. It's always a pleasure to be of help.Nice find Physicist. I look forward to reading it all.
Did you notice how I hit the nail right square on the head with respect to x0x's response! ROTFL!!!
Of course I'm assuming he posted his usual insults, claiming he's right, claiming I need to go back to school etc. (basically violating all the rules about being rude that the forum has written and which he ignores) as he usually does and how he was right all along. Nonsense like that is why he's in my ignore list.
Peter,You're most welcome Jilan. It's always a pleasure to be of help.
Did you notice how I hit the nail right square on the head with respect to x0x's response! ROTFL!!!
Of course I'm assuming he posted his usual insults, claiming he's right, claiming I need to go back to school etc. (basically violating all the rules about being rude that the forum has written and which he ignores) as he usually does and how he was right all along. Nonsense like that is why he's in my ignore list.
I know that you've been lying all along, you can see my responses, despite of your claims to the contrary.
Will someone ask this wacko x0x , if his posts are directed to me, why he posts to me when he knows I can't read them since he's on my ignore list? Thanks. Send me your reply in PM please. Thanks.You're most welcome Jilan. It's always a pleasure to be of help.
Did you notice how I hit the nail right square on the head with respect to x0x's response! ROTFL!!!
Of course I'm assuming he posted his usual insults, claiming he's right, claiming I need to go back to school etc. (basically violating all the rules about being rude that the forum has written and which he ignores) as he usually does and how he was right all along. Nonsense like that is why he's in my ignore list.
Interestingly enough, your own reference (paragraph 10.3) disproves your derivation. Here it is:
10.3 Total Relativistic Energy
We can now define a quantity E by
See, "physicist"? E is defined. Do you know why?
Although I do love it when he proves me right with every post that he makes about me.
You are severely delusional, I DISproved your claims.Although I do love it when he proves me right with every post that he makes about me.
Please lay off x0x. We came to this forum to escape from you. It was nice for a while and could be again if you would just STOP IT.
Please note that Jilan is asking x0x to "lay off" and not asking me to "lay off" x0x. x0x is famous for interpreting things to his benefit.
The problem with x0x is that he's one of those people who read a book or two on the introduction to relativity and now thinks he's an expert on the subject and perhaps on physics as well. That's the root cause of why he thinks P = mU is the only way to define energy and momentum, i.e. with so little experience and so little reading of different relativity and physics texts he's never seen other physicists do it other ways as in a great deal of the special relativity textbooks out there and the examples I posted here. So when he sees us do it correctly or differently he insults us. It's bad enough to think your an expert with so little training. Insulting others is the worst of all. Shame on him.
I am just pointing out your mistakes, Peter.
Want to help Peter (the "physicist") figure out his mistakes? In return, he could help you solve the electrostatics problem that you seem unable to get the correct answer.
Sheesh! What the hell is wrong with you x0x? Yet another post from x0x and with his track record it has to be a rude one at that. Don't you get it yet x0x? Everybody in this forum wants you to STOP IT. Just because they're not stating it in open forum it doesn't mean they don't want it. Just PM anybody and they'll tell you. I've talked to many people in the forum and that's the consensus. So STOP IT!!!!
Exposing your elementary mistakes is not rude, Peter. Just learn to accept that you produced a fake "proof". This is a science forum, posting fake proofs is not acceptable.
Thats is incorrect. Markus correct one of his mistake on x0x thanks him on post #12
On which post ? The discrepancies about your interpretations is far to be settled. And your are losing traction.
You are incorrect again. He has been very polite and helpful. Only you used the term "wacko" on post #23.
Put everybody you want on your ignore list, but stop complaining about them afterward.
This is a public place and every side of the story is important. That's how I process science anyway.
As far has I read, you are the only one doing that. You are barely special in that regards. Everybody has this tendency. Including me.
Unless you address his remarks, your post does not contain any clue on how your interpretation is useful.
You own link point to this exact phraseBold added.It is all well and good to define such a thing, but, apart from the neatness of the expression,
is there any real need to introduce such a quantity?
The layman book of Brian Cox & Jeff Forshaw, "why does E=mc2", which address precisely the question of this thread, do the exact same thing a page 133.
After having worked out the 4-momentum [Latex]mc=\gamma mv + \gamma mc[/Latext] they proceed to explain why it is relevant to multiply again by C (to square it).
Because in that case the good approximation for [Latex]\gamma \approx 1 + \frac{1}{2}(v^2/c^2)[/Latex] leads to a value for the time component (but C squared) [Latex]\gamma m c^2 \approx mc^2 + \frac{1}{2}mv^2[/Latex] where you right away recognize the old/newtonian kinetic energy + something else (especially correct at small/rest speed).
After that, it is really about considering we are still talking about a conserved 4 momentum quantity, it is just relabeling/defining what is beside [Latex] m c^2[/Latex] and calling it Energy, which is the correspondent conserved quantity.
Dang, Boing :-)
The layman book of Brian Cox & Jeff Forshaw, "why does E=mc2", which address precisely the question of this thread, do the exact same thing a page 133.
After having worked out the 4-momentumthey proceed to explain why it is relevant to multiply again by C (to square it).
Because in that case the good approximation forleads to a value for the time component (but C squared)
where you right away recognize the old/newtonian kinetic energy + something else (especially correct at small/rest speed).
After that, it is really about considering we are still talking about a conserved 4 momentum quantity, it is just relabeling/defining what is besideand calling it Energy, which is the correspondent conserved quantity.
I was hoping to lead the "physicist" through the steps of learning the above.
(I took the liberty of correcting your tex code).
No it isn’t. I said that x0x has a problem admitting to his mistakes and you misinterpreted that to mean that he has never admitted to any mistake he’s ever made. That’s an invalid intrepretation.
Markus correct one of his mistake on x0x thanks him on post #12
Losing traction? No way Jose. Far from it my friend. I know precisely what I'm talking about contrary to x0x who's lost. How could you possible convince yourself that such is true?
And your claim that The discrepancies about your interpretations is far to be settled. is laughable! Lol!!There are no interpretations there whatsoever. If you were a relativist then you'd know that implicitly. If you doubt it then there are thousands of relativists out there who'd be more than happy to confirm what I said. All you have to do is find one and e-mail them. Most are very happy to help. Are you up for the challenge?
In post #10 Markus quoted him where he noted that the total energy-momentum of a particle of rest mass m_0 is defined as (E, p) = gamma m_0(c^2, v). However as Griffith points out in his text on electrodynamics in his EM section one defines E and p as such. Griffiths, on the other hand, stated that Einstein “identified” E as total energy. Whether it was a definition or derivation he does say.
From there I was told that x0x claimed I was wrong because I used circular logic because I derived E = gamma m_0 c^2. That claims is very bogus indeed because he was approaching if from a different perspective and as I said above he has very limited perspectives in physics.
The way I derived it was from a different perspective. It’s very similar to the way that it’s derives in Classical Electrodynamics – Third Edition by J.D. Jackson, (1999) on page 536. The section its derived in is 11.5 Relativistic Momentum and Energy of a Particle. You can easily see how it’s derived and in the same way I did it if you download and read Special Relativity: A Modern Introduction by Hans C. Ohanian, page 137.
It’s also in the post where I derived it myself, correctly too.
Perhaps to you but you don’t read how he treats others. Jilan tells me that since I put him in my ignore list he turned to him/her to start tormenting/insulting and I’ve been told that she’s not the only one.
He is a wacko and still causes problems. I saw how he reasons and behaves and that’s why he’s in my ignore list. However that does solve the entire problem since, like this thread, pumps out misinformation like wildfire. I had to let that go because I have better ways to spend my time than dealing with the likes of him.
Not when you start insulting people and ignoring what’s being said. He was doing that constantly before I kill filed him
And you’re in the minority on that. I don’t do that contrary to this claim. If I ever get pissed off at someone then it was someone being rude and so blind to their mistake they couldn’t see it. x0x is one of those. I pay heed not to what I see people saying in public but to what I hear them say in private, i.e. PM and e-mail. That’s where people are truthful, otherwise they’ll be subject to attack. In PM I’ve gotten many messages from members thanking me for my help. There’s even one who asked me to start tutoring them.
I did address it. He didn’t understand it. Apparently you don’t either. No matter since those who are really looking for the answer have what they were looking for.
As far as your comment post does not contain any clue on how your interpretation is useful. goes, that’s just silly. Something is useful it’s the answer to the question asked. That was the answer, period. x0x spoiled the thread by claiming I made aa mistake by using circular logic, which is nonsense. There are various perspectives to learning relativity and as such the various perspetives require different paths to results. E.g. Rindler’s text on relativity merely defines E as E = mc^2/sqrt(1 – v2/ c2) (where m = proper mass). However Ohanian’s and Jackson’s texts derive it. x0x doesn’t understand that so he turned to insults.
I read that page and I don’t see it. Please quote what you’re referring to.
By the way, to use Latex here you don’t type in “Latex”, you type in “tex”.
I see no derivation of that. In fact I don’t really see any equations in that book at all for the most part.
You used the fact that
...and Jackson uses the same exact approach as Feynman, "Lectures on Physics" vol I. He starts with Einstein's re-DEFINITION of momentum asIt’s very similar to the way that it’s derives in Classical Electrodynamics – Third Edition by J.D. Jackson, (1999) on page 536. The section its derived in is 11.5 Relativistic Momentum and Energy of a Particle.. (paragraph 15-8) and follows with the DEFINITION of total energy as
Why did Einstein had to re-DEFINE momentum as
Yep, so does W.Pauli on page 87of his "Theory of Relativity". Why do you think is that?Rindler’s text on relativity merely defines E as E = mc^2/sqrt(1 – v2/ c2) (where m = proper mass).
Grow up, children. No more name calling, or bans will ensue. Yes, I am saying that people in this forum act like children, because some people here do act like children, but that does not mean you can say anything similar. People have been acting up in this forum for too long. If you don't like what I am saying, tough luck.
You can claim people are wrong, if you are prepared to back up your argument in a scientific manner.
You cannot call people names, troll, threaten, abuse or insult people.
Talk science, not trash.
Thank you
In reply to SpeedFreek, regarding your #38 post, as well as the "closed thread".
First, Thank You! I apologize for being a participant in the "trolling category" of meaningless insults. (No bs, I AM sorry)
Second, thanks for stepping-in with moderation...I know no one cares overmuch for my own "stuff", but still...it's a lot of years on my part to "think it up", and don't mind people
being critical of what I write, what I don't get is the constant character assassination!
Thank You
And that includes the use of the term "wacko", of course.
I used that because I lacked sufficient flow of speech to think of a more accurate term. Just like your use of saying that I've been acting like a child which is not only rude but inaccurate. People get angry and they argue about it. Just because adults do that too does not make them children. Had you and the moderators taken action when reports were filed against the people doing all the insulting then this would never have gotten to this point. What is it you expect when you make rules, have moderators to enforce the rules but never take action when the rules are violated and at that time and not let it accumulate to a boiling point?
I am all for liberating code
First I am not again going to forget \vect before vector quantity. Second what is the correct [tags] for latex ?
Here is the derivation of energy-mass equivalence by Einstein, September 27, 1905 ...
http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf
OK, so the final eqn in Einstein's famous paper is wrt the relation between mass and radiant energy per an observer at rest with the mass (v=0) ...
E = mc2 ... whereby his paper referred to energy E instead as L ...
K0 - K1 = ½(L/c2)v2
where m = L/c2
However, the eqn derived just preceding the final eqn, which is also a derived eqn, was ...
K0 - K1 = L(1/√(1-v²/c²)-1)
or ...
K0 - K1 = L(γ-1)
Which means the kinetic energy of motion as seen by the observer who moves relatively at v, is the total energy of the moving body (L*γ) less its rest energy L*1. So the total energy of the moving system per a stationary observer is L*γ, yes?
Thank You,
SinceYouAsked
Thanks SinceYouAsked, but that's not what the OP was looking for. The purpose of Einstein's derivation was to show that the rest mass m (aka proper mass) of a body is related to its rest energy E0 (i.e. the amount of energy a body has when it's at rest) is.
The OP wanted someone to show how the relationship between total relativistic energy, usually defined as E = Kinetic energy + rest energy is related to m and speed byis derived.
Last edited by Physicist; 09-04-2014 at 01:45 PM.
I don't see where there is interpretation here. We have different perspectives, they are both valid. Mine is based on the fact that when *I see him wrong*, he admitted its mistake gracefully. I give you the link to that post, just because you apparently don't see his posts.
I have no problem to understand that you don't observe the same reality. For once, you have a biased version (called an ignore list). Second you see him as being wrong. That's not an invalid interpretation either.
You are, in regards to my point of view. Even though my point of view is that BOTH of you know a lots in physics.
Easy, by understanding that there is an interpretation here. I am doing it. And it is not easy to reconcile your dispute.
I know you don't meant to be rude here. But there is nothing laughable about the fact the I interpret that you don't agree.
I am not (not in that sense**), and I generally distrust "implicit" things, they sound too much like dogma to me.
I don't doubt that "thousands of relativists" would confirm what you said. What I doubt is that there would not be another "thousands" that would disagree.
I am not going to go private, nor to start another stochastic sampling of physics idea, to see what are the ones more likely to be correct.
Instead I will take up the challenge to reply to your questions, and those of anyone here, in the hope they would do the same. On this site, here and now. This forum is the space, and arguments are the objects.
(**)I am a relativist in the sense that I do believe different coherent perspective may seem different, but there is a may to "coordinate transform" them so they can be absolutely reconciled (if not, one of them or both is false). I am speaking here of idea also called "point of view", and that last expression is for me is not an accident (we use the same in french).
My magic formula to do that is to observe that any idea as many dimensions, and each of those can be again split (recursively add infinitude). Nothing will ever be black and white for me. Just box of more or less grey box inside more or less grey box inside ... you get the picture. That is a fractal tree, incidentally my brain also is
Back to physicts. I don't think I would read Griffith in this life time. So you view on his opinion is what matters to me. I've added a not in your quote. Is is right ?
In that phrase I understand that some physicists do define E as...(not derive)
Silly ? No. But surely way too exclusive. I am sorry. Let me rephrase. (I have added the rephrasing in underscore, I always assume that I am talking in my stead, and this is apparently a mistake)
That was an answer. I give it a great interest and start processing it. This thread exist, and those books exist. You post #6 exist. I also find it circular. And x0x explain that better then me.
I'll try anyway:
Your own wiki article of post#10contains:When I read that, I see definition of rest EnergyThe constant of integrationis found by observing that, when
,
and
, giving
and giving the usual formula:
I also quoted another linked article where o page 30 is writtenThe book I mention is for Layman, and contains the bare minimum of math. I am at work now (lunch time boss, don't worry), I'll retype what I meant tonight.This quantity E is known as the total relativistic energy of the particle of rest mass. It
is all well and good to define such a thing, but, apart from the neatness of the expression,is there any real need to introduce such a quantity?
I am sure of it. I want to learn more about them. And I somewhat count on you for that. Thanks in advance (double tanks for the {tex} tip)
"tex" and "/tex" (within square brackets).I am all for liberating code
First I am not again going to forget \vect before vector quantity. Second what is the correct [tags] for latex ?
OK, so that's what the OP asked for then, thanx.Thanks SinceYouAsked, but that's not what the OP was looking for. The purpose of Einstein's derivation was to show that the rest mass m (aka proper mass) of a body is related to its rest energy E0 (i.e. the amount of energy a body has when it's at rest) is
The OP wanted someone to show how the relationship between total relativistic energy, usually defined as E = Kinetic energy + rest energy is related to m and speed by
Here is a good fundamental visual aid (E=mc2 is Incomplete) to help in understanding the derivation, conceptually ...
..... https://www.youtube.com/watch?v=NnMIhxWRGNw
Another related visual aid (Einstein's Proof of E=mc2) ...
..... https://www.youtube.com/watch?v=hW7DW9NIO9M
Thank You,
SinceYouAsked
Last edited by SinceYouAsked; 09-05-2014 at 05:58 PM.
I those videos were shown in secondary schools, science would definitely be much less intimidating for peoples.
They are sooo good !
Since you liked those, you should watch some of the other ones, this one inparticular (Special Relativity) in the context of spacetime geometry (Minkowski spacetime diagram teaser). Einstein's mass energy equivalence paper, used the special theory as foundation ...I those videos were shown in secondary schools, science would definitely be much less intimidating for peoples.
They are sooo good !
..... https://www.youtube.com/watch?v=ajhFNcUTJI0
Thank You,
SinveYouAsked
This one is a very good book too: Special Relativity by A.P. French which you can download at - Special relativity | A.P. French | digital library BookOS - Turn to pages 20 to 21. It was written by an MIT physicists and used to teach special relativity at MIT for many years.Nice find Physicist. I look forward to reading it all.
Here's another one from a very well known undergraduate textbook in classical dynamics. It's called Classical Dynamics of Particles and Systems - 5th Ed. by Marion and Thornton (2003). It to can be downloaded at Classical Dynamics of Particles and Systems | Stephen T.(Stephen T. Thornton) Thornton, Jerry B. Marion | digital library BookOS
Momentum is derived on pages 562 -> 564, Eq. (14.45). Rest mass is defined on page 567, Eq. (14.61). For a derivation ofsee
Mass Energy Equivalence
The derivation of(m = proper mass) is found on pages 562 through 567.
I highly recommend downloading this text and carefully read and absorb it cover to cover with great care. Another good text is Classical Mechanics by John R. Taylor which you can download from Classical Mechanics | John R. Taylor | digital library BookOS
The benefit of reading this one is to see a different kind of derivation so that you don't get the false impression that there's only one way to do this.
There are other ways too such as using Lagrangian mechanics.
In reply to Physicist, re: your post "links".
In your years of reading, have you ever come across any explanation(s) of why "velocity of a particle serves as causation to alter it's "rest state" to an "excited state?" This is presuming
that the velocity-particle is not inter-acting w/others, rather the particle is moving in a "space frame" only.
(Thanks for reading!)
I don't understand this question. It sounds like you're asking me if I've ever read about why velocity causes a particle to go from a state of rest to an excited state. To me that seems circular since velocity is merely a description of a state of motion and not related to causation. However its common knowledge that whatever causes a force causes a particle to go from a state of rest to a state of motion. Is that what you mean?
In reply to Physicist, re: your #54 post.
I meant "why does velocity serve as a pro-forma mandate of a given atoms ability to radiate energy". <(is this better?)
(Cheerio!)
Are you saying that you believe that an atom has to be moving at a certain speed or above to emit radiation? If so then where did you get such an idea from?
In reply to Physicist, re: your #56 post.
No...I don't think much of the idea of "velocity=energy" of a SINGLE particle. It seems the "idea" is popping-up here and there on threads in other forums, and also an article I read in
"Science" about 2yrs. ago seemed to advocate this idea, that a single particle in a velocity frame could emit energy.
(I do not share in this idea...it just that I'm seeing more references to it, and I wonder if it has any "official' standing anywhere?)
(Thanks for reading!)
Then what do you mean by "why does velocity serve as a pro-forma mandate of a given atoms ability to radiate energy"?
You do know that a single particle that is moving has energy, right? I.e. kinetic energy?
Your derivation of.
All the gyrations that you are going through are totally unnecessary, since for,
and the rest energy is
. For systems with variable mass, the consequence is
. As an aside, it is bad form to link in personal webpages, even more so when they contain errors.
I tried surfing the page when my antivirus software inform me that the download link will direct me to a harmful malware page. I tried SinceYouAsked link and it seemes normal, and i am able to download the pdf without malware warning.
Could the equivalence be derived based on fundamental concepts in Newtonian mechanics, in absolute space and time?
This is a relativistic effect, so your question makes no sense.
I heard of something called Einstein's box which could yield the equivalence without working with the gamma factor. Is it true?
no
Nice findings. Good to read all the facts. E=mc^2 - Deriving the Equation - Easy. When Einstein first proposed his Special Theory of Relativity in 1905 few people understood it and even fewer believed it. It wasn't until 1919 that the Special Theory was "proved by inference" from an experiment carried out on his General Theory of Relativity.
Theories cannot be "proven". They can only be falsified. Via experiment. In the 113 years that passed since 1905, no such experiment has falsified the special theory of relativity. No experiment has falsified GR, either.
I just found the box here Mass and Energy
They got Einstein equation E = mc2 without dealing with gamma factor or relative space-time.
The so-called "proof" is invalid. It starts with. But this is just a CONSEQUENCE of the general relationship:
.....which is a CONSEQUENCE of:
...which are the DEFINITIONS of RELATIVISTIC energy and momentum in QED.
By putting(mass of photon is zero) one gets
In the frame comoving with an object,, so, one gets
Besides, the relationship comes from QED (Quantum Electro Dynamics) the RELATIVISTIC theory of quantum mechanics and electrodynamics. So, you are doubly wrong.
I think E = pc can also come from quantum mechanics or Maxwell. After we have E = pc, we can get E = mc2 with the box.
No, "we" can't. You understood nothing. What is your agenda here? What are you trying to prove?
I am still not clear what is wrong with Einstein's box.
If it is wrong, why some educational links and textbooks use the box to derive the equivalence?
The "proof" uses the conclusion. This is the third time I explain this to you.
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