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A dark part of special relativity(at least for me) 
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#1
Jun709, 09:28 AM

P: 774

Hi
For some days I was looking for the reason that relativity said the speed of light is the maximum speed.At last after a lot of thinking I have noticed the equations below: lim_{v[tex]\rightarrow[/tex]c}[tex]\frac{M_{0}}{\sqrt{1\frac{v^{2}}{c^{2}}}}[/tex]=[tex]\infty[/tex] lim _{m[tex]\rightarrow[/tex][tex]\infty[/tex]} [tex]\frac{F}{m}[/tex]=0 So if sth moved with the speed of light,It couldn't change its speed.It means that not only the speed of light is the maximum speed but also When sth reaches this speed if will move with the speed of light for ever.because acceleration that means the change in velocity is 0 so that thing can't increase nor decrease its speed. Is my reasoning right and if yes,have you noticed about it? 


#2
Jun709, 09:33 AM

P: 774

And don't you have sth easier and faster than this latex?
It took me ten minutes to write this post! 


#3
Jun709, 09:50 AM

Sci Advisor
P: 1,750

LaTeX is designed to allow writing production quality typesetting of mathematics. It takes a bit of getting used to, but once you know it, it's the best there is for this task. You should put the whole formulae in LaTeX; not just bits of it. If you don't know LaTeX, you can try just writing ascii. But in the meantime, just to help, here are your formulae rewritten a bit. (Note that a superscript in LaTeX is written x^2.) Click on the formula to see the LaTeX code. [tex]\begin{align*} 


#4
Jun709, 09:53 AM

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PF Gold
P: 39,361

A dark part of special relativity(at least for me)
Well, that happens when you don't know Latex! Practice, practice, practice!
[tex]\lim_{v\to c}\frac{M_{0}}{\sqrt{1\frac{v^2}{c^2}}}[/tex] [tex]\lim_{m\to\infty} \frac{F}{m}= 0[/tex] That took me only a few seconds. Your only error is in thinking that it is possible to an object with nonzero mass to accelerate to c! Since nothing with mass can ever get to c so "what happens at c" is not relevant. It is true that light itself, which has mass 0, can neither accelerate nor decelerate and always moves at "c". Darn! Sylas got in while I was typing! 


#5
Jun709, 11:31 AM

P: 774

does it have a mathematical reason that nothing with non zero rest mass can accelerate to c?



#7
Jun709, 11:35 AM

P: 774

And I have a problem with photons too.I read that they're rest mass is 0.So we have:
[tex] \lim_{m \rightarrow 0} a \ = \ \lim_{m \rightarrow 0} \frac{F}{m} \ = \ \infty [/tex] but Light has a constant speed. any way.I downloaded the Latex reference.With having that read,It got easier. 


#8
Jun709, 11:41 AM

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P: 1,750

[tex]E^2 = (pc)^2 + (mc^2)^2[/tex]The E is total energy, and p is momentum, and m is rest mass. A photon has no rest mass, so it reduces to [tex]E = pc[/tex]The fundamental quantities are energy and momentum. You are much better to think in terms of these than in terms of mass. Cheers  sylas 


#9
Jun709, 11:43 AM

P: 774

I still don't understand.because for getting to c you don't need an impossible acceleration.the same goes for force.so how does the energy gets infinite?



#10
Jun709, 11:47 AM

Sci Advisor
P: 1,750

[tex]\frac{mv}{\sqrt{1\frac{v^2}{c^2}}}[/tex]The total energy is [tex]\frac{mc^2}{\sqrt{1\frac{v^2}{c^2}}}[/tex]As you show in your first post, this diverges to infinite values as v approaches c. Note that F = ma is only true at subrelativistic velocity. But [itex]F = \frac{dp}{dt}[/itex] is always true. As you apply arbitrarily large forces, you can get momentum as large as you like... but never infinite. 


#11
Jun709, 11:50 AM

Math
Emeritus
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PF Gold
P: 39,361

In saying
[tex]\frac{F}{0}= \infty[/tex] you are assuming that F is not 0. What force are you assuming is acting on light? 


#12
Jun709, 11:53 AM

Math
Emeritus
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PF Gold
P: 39,361

If you are talking about a particle with nonzero mass, then you can't get to c! That's what we have been saying. So the energy of a nonzero mass particle does not "get infinite". 


#13
Jun709, 11:54 AM

P: 774

I may seem stupid but as I know [tex]p=\gamma m_{0} v [/tex] and [tex]m_{0}[/tex] is the rest mass.if it is,so pc must become 0,too.Hey its really confusing!



#14
Jun709, 11:57 AM

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P: 1,750

For a massless particle, the velocity is c, and the momentum is just E/c, where E is its energy. 


#16
Jun709, 12:09 PM

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PF Gold
P: 1,843

[tex]F = ma[/tex]isn't valid in relativity. The correct version is [tex]F = \frac{dp}{dt} = \frac{d}{dt}\left( \frac{mv}{\sqrt{1  v^2/c^2}} \right)[/tex]where m is nonzero rest mass. Unless I've made a mistake, this works out to be [tex]F = \frac {ma} {(1  v^2/c^2)^{3/2}}[/tex]So, if m and a are constant, the force really does diverge to infinity. (In practice, it's more likely F and m would be constant, so a would converge to zero.) 


#17
Jun709, 12:09 PM

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#18
Jun709, 12:11 PM

P: 774

so take a look at http://en.wikipedia.org/wiki/Planck_momentum



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