Special Relativity

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shoelevy
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Post by shoelevy »

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.

i dare say u can't let m be a function of v. express m in terms of m knot(rest mass),rearrange and u find m knot and v are independent variables.

seriously, i do beleive u r very confused with the mathematics and variables involved.
relativity is a subject that is easy yet at the same time hard to understand. :cry: i suggest reading up from university texts and outside books and getting the variables and argument right. no offence chaosknight :oops: ...but my very strong opinion is based on previous study experience and experience with OTHERS, not u, who ask very silly questions easily explained if they had read up a little more. :!:
this thread is not getting anywhere, i do believe we are going in circles so i'm officially out of this thread :wink:
ChaosKnight
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Post by ChaosKnight »

Hahaa.......Shoelevy, if you understood my post, you would see relating m to v is only for the limiting purpose. Of course, everyone knows that m eventually is a fixed quantity, which will never be reached in the process. So is v.

But talking about being "very confused with the mathematics and variables involved", here is what you 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
You are suggesting i'm in error stating
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:
But i'm correct in saying
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.
So you are saying m is relativistic mass, and M is also relativistic mass, although there is a clearly defined mathematical relation between the two? This sounds to me like you are inconsistent. Care to clarify?

And oh yes, if you decide to come back and take me up on my offer, please utilize the quote/unquote tags.
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shoelevy
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Post by shoelevy »

just to clarify again...i am not going to clarify anything here anymore
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zong
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Post by zong »

The mathematics here is going crazy.. I don't feel like reading anymore.. :?

I guess no one here is good enough yet to talk about relativity huh. But ChaosKnight, don't give up tho, it's great you are asking these questions, and questions is what makes a topic go on :) It's just that all the formulae are giving me headache..

And shoelevy, discussing a topic with "others" doesn't mean you're absolutely correct, so try to accept others' words if they're correct, too..
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carlogambino
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Post by carlogambino »

hmm special relativity isn't that difficult a concept, its general relativity thats really really hard to grasp. I suggest reading up books for pple who want to know more, instead of getting confused by the convos....

Btw,
...but my very strong opinion is based on previous study experience and experience with OTHERS, not u, who ask very silly questions easily explained if they had read up a little more.
isnt this a little harsh? Well i can see that he did put in quite some effort in the math.
ChaosKnight
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Post by ChaosKnight »

Don't worry, zong. i can give you panadol. :D

For you, shoelevy,
just to clarify again...i am not going to clarify anything here anymore
Have it your way then. And i wish you plenty of good luck for your ph.d. You're going to need it. :)
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