Special Relativity
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Destructos
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The math involved is too complexed when it comes to this subject, you either have to be God, a natural born mathematical genius or suffer from a special kind of autism (asperger's syndrom most likely), like what Einstein is said to suffer from, to be able to put these theories together to form the elusive grand-unified theory...
well...search no more!weixing wrote:Hi,
General Relativity and Special Relativity are subject that are very difficult to understand and I believe that not many people out there really understand it.... remember Einstein nearly spend his whole life study it. And there are so many books on this subject, so I was wondering how many author really understand the subject. Even the author understand the subject very well, but do you really understand what the author trying to said???
Anyway, it is good to have interest in this subject... hope someone here will do his PhD on this in the future.
Have a nice day.
here i am!
haha
i am considering doing that at phd level
wish me luck
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zong
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Geez shoelevy, you're abit overanxious about the PhD relativity... Take it one step at a time la.
And it looks asif you wanna totally thrash down my viewpoints on relativity. I'm voicing my opinions, and if you don't feel like supporting mine, just state yours, you don't have to denounce my point to support yours. You told me not to believe the internet, yet I have reason to believe you learnt your relativity from the internet too, because some information you give is not in books I find are already quite updated. As people say above, if truly less than 10 people fully understand relativity, you'll hafta prove to me you're one of the 10, or don't denounce me. I didn't denounce you either. And I'm not claiming the 10 positions. I'm still learning. And I believe, so are you.
And from this, I don't mean I'm admitting defeat or I'm a sore loser, I just don't want to keep on arguing about the same few points and bore everyone to death. I could argue on, in PM, if you wish, though.
And it looks asif you wanna totally thrash down my viewpoints on relativity. I'm voicing my opinions, and if you don't feel like supporting mine, just state yours, you don't have to denounce my point to support yours. You told me not to believe the internet, yet I have reason to believe you learnt your relativity from the internet too, because some information you give is not in books I find are already quite updated. As people say above, if truly less than 10 people fully understand relativity, you'll hafta prove to me you're one of the 10, or don't denounce me. I didn't denounce you either. And I'm not claiming the 10 positions. I'm still learning. And I believe, so are you.
And from this, I don't mean I'm admitting defeat or I'm a sore loser, I just don't want to keep on arguing about the same few points and bore everyone to death. I could argue on, in PM, if you wish, though.
Well, both of you have stated your points clearly. They all look fine to me. Maybe one day we'll see you presenting your theory to a bunch of famous physicist and cosmologist, but until then, this topic is still open for discussion. 
(Does the egg come first or the chickens?)
Regards,
Sam
(Does the egg come first or the chickens?) Regards,
Sam
We are the Borg, Resistance is Futile!
Yes, I'm bored to death with this subjectAnd from this, I don't mean I'm admitting defeat or I'm a sore loser, I just don't want to keep on arguing about the same few points and bore everyone to death.
Can we talk about observation (or lack of it) instead? i know it's been very cloudy for the past two months 
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ChaosKnight
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The equation relating kinetic energy to velocity is:
K = (y - 1) mc^2 --------------------(1)
where K is kinetic energy, m is rest mass and y represents the time dilation factor:
y = 1 / [ 1 - (v/c)^2 ] ^ 0.5 -------------------(2)
Lets use this equation to see if photons have any rest mass. Of course, you can't use the equations directly, because of the singular point caused by (2) when v = c.
Instead lets do a limiting approach, i.e. let v approach c. What happens is K increases to infinity.
We know that you could create photons, so therefore photons have finite energy. This is illustrated every time you turn on a torch. Therefore, for a photon, K must be a finite value. Since the term (y - 1) is going to infinity, the only way K can maintain a finite value is if m tends to zero.
Therefore, a photon does not have any rest mass.
The total mass of a particle is the sum of energy due to rest mass and kinetic energy. Therefore:
Total E = (y - 1) mc^2 + mc^2 ----------- (3)
where the sole mc^2 term on RHS represents rest energy. So total energy is:
Total E = y (mc^2) -------------- (4)
Now a new velocity-dependent mass is defined, which we call M.
Total E = y(mc^2) = Mc^2 ----------------(5)
It can be seen that M = ym, where y is a function of velocity.
For a photon, we know that total energy is not zero. Therefore y(mc^2) cannot be zero and M cannot be zero.
Hence, although a photon has zero rest mass, it has relativistic mass.
Anyone care to confirm??
K = (y - 1) mc^2 --------------------(1)
where K is kinetic energy, m is rest mass and y represents the time dilation factor:
y = 1 / [ 1 - (v/c)^2 ] ^ 0.5 -------------------(2)
Lets use this equation to see if photons have any rest mass. Of course, you can't use the equations directly, because of the singular point caused by (2) when v = c.
Instead lets do a limiting approach, i.e. let v approach c. What happens is K increases to infinity.
We know that you could create photons, so therefore photons have finite energy. This is illustrated every time you turn on a torch. Therefore, for a photon, K must be a finite value. Since the term (y - 1) is going to infinity, the only way K can maintain a finite value is if m tends to zero.
Therefore, a photon does not have any rest mass.
The total mass of a particle is the sum of energy due to rest mass and kinetic energy. Therefore:
Total E = (y - 1) mc^2 + mc^2 ----------- (3)
where the sole mc^2 term on RHS represents rest energy. So total energy is:
Total E = y (mc^2) -------------- (4)
Now a new velocity-dependent mass is defined, which we call M.
Total E = y(mc^2) = Mc^2 ----------------(5)
It can be seen that M = ym, where y is a function of velocity.
For a photon, we know that total energy is not zero. Therefore y(mc^2) cannot be zero and M cannot be zero.
Hence, although a photon has zero rest mass, it has relativistic mass.
Anyone care to confirm??
Instead lets do a limiting approach, i.e. let v approach c. What happens is K increases to infinity.
We know that you could create photons, so therefore photons have finite energy. This is illustrated every time you turn on a torch. Therefore, for a photon, K must be a finite value. Since the term (y - 1) is going to infinity, the only way K can maintain a finite value is if m tends to zero.
not right because the m in this equation is relativistic m.
not rest m so when u use this method, you haven't shown anything about rest m. and besides, we know photons do travel at c....so K is either infinity if m not =0 or it is 0 if m=0. neither of these 2 cases are true as we all know photons haf K not equivalent to these 2 values.
your second part is correct
We know that you could create photons, so therefore photons have finite energy. This is illustrated every time you turn on a torch. Therefore, for a photon, K must be a finite value. Since the term (y - 1) is going to infinity, the only way K can maintain a finite value is if m tends to zero.
not right because the m in this equation is relativistic m.
not rest m so when u use this method, you haven't shown anything about rest m. and besides, we know photons do travel at c....so K is either infinity if m not =0 or it is 0 if m=0. neither of these 2 cases are true as we all know photons haf K not equivalent to these 2 values.
your second part is correct
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weixing
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Hi,
I would like to point out something on mass. Below is taken from some website:
Today, when physicists talk about the mass of an object they always mean the rest mass. They use other terms, like energy, to refer to the relativistic mass.
Relativistic Mass = E/c^2. Physicists usually work in units where c = 1 so that energy and relativistic mass become identical concepts. Instead of talking about relativistic mass they simply talk about energy.
If you see "relativistic mass" in your first-year physics textbook, complain! There is no reason for books to teach obsolete terminology.
Have a nice day.
I would like to point out something on mass. Below is taken from some website:
Today, when physicists talk about the mass of an object they always mean the rest mass. They use other terms, like energy, to refer to the relativistic mass.
Relativistic Mass = E/c^2. Physicists usually work in units where c = 1 so that energy and relativistic mass become identical concepts. Instead of talking about relativistic mass they simply talk about energy.
If you see "relativistic mass" in your first-year physics textbook, complain! There is no reason for books to teach obsolete terminology.
Have a nice day.
Yang Weixing
"The universe is composed mainly of hydrogen and ignorance."
"The universe is composed mainly of hydrogen and ignorance." -
ChaosKnight
- Posts: 293
- Joined: Sun Oct 05, 2003 6:54 pm
If the first part is wrong, how can the second be correct?shoelevy wrote:.
not right because the m in this equation is relativistic m.
not rest m so when u use this method, you haven't shown anything about rest m. and besides, we know photons do travel at c....so K is either infinity if m not =0 or it is 0 if m=0. neither of these 2 cases are true as we all know photons haf K not equivalent to these 2 values.
your second part is correct
The limiting method is more than just zeros or infinity. What is infinity multiplied by zero? Undefined. Let me illustrate with an example.
Let A = a/x --------(1)
Let B = bx ---------(2)
Let C = c/(x^2) --------(3)
A, B and C are functions of x and a constant.
Consider the function BA. Let B tend to infinity, and let A tend to zero, so we get infinity multiplied by zero. To do this is simple: we simply let x tend to infinity.
Therefore,
[(lim A -> 0)A] * [(lim B -> infinity)B] = (lim x -> infinity) AB = (lim x -> infinity) ab ---------(4)
You can see that in this case, a multiplication of an infinity term and a zero term gives a finite constant (ab).
Similarly, if you take B*C, and let B tend to infinity and C tend to zero, you get zero.
Putting things in perspective, during the limiting process, if you let m be a suitable function of v, you could get a constant for K. However, this doesn't change the fact that m (in a case similar to A) has to tend to zero.