A dark part of special relativity(at least for me)

  • Thread starter ShayanJ
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In summary: E=pc for light.)In summary, the conversation discusses the reasoning behind the statement in relativity that the speed of light is the maximum speed. It is explained through equations that as an object approaches the speed of light, its velocity cannot change and it will continue to move at the speed of light forever. The conversation also touches on the use of LaTeX in mathematical equations and the properties of particles with zero and non-zero rest mass.
  • #1
ShayanJ
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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:

limv[tex]\rightarrow[/tex]c[tex]\frac{M_{0}}{\sqrt{1-\frac{v2}{c2}}}[/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?
 
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  • #2
And don't you have sth easier and faster than this latex?
It took me ten minutes to write this post!
 
  • #3
Shyan said:
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:

limv[tex]\rightarrow[/tex]c[tex]\frac{M_{0}}{\sqrt{1-\frac{v2}{c2}}}[/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?

Yes, well done. It a feature of relativity that anything which moves less than light speed will always move less than light speed; and anything which moves at light speed can never move at any other speed. Another point is... particles will non-zero rest mass move less than light speed. Particles with zero rest mass move at light speed.

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*}
& \lim_{v \to c} \frac{M_0}{\sqrt{1-\frac{v^2}{c^2}}} = \infty \\
& \lim_{m \to \infty} \frac{F}{m} = 0
\end{align*}[/tex]​
 
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  • #4
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 non-zero 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
does it have a mathematical reason that nothing with non zero rest mass can accelerate to c?
 
  • #6
Shyan said:
does it have a mathematical reason that nothing with non zero rest mass can accelerate to c?

Yes. You can't give something infinite energy.
 
  • #7
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
Shyan said:
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} \frac{F}{m}=\infty [/tex]

but Light has a constant speed.

The equation for energy of a particle is
[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
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
Shyan said:
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?

The momentum of a particle with non-zero rest mass m and velocity v is
[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 sub-relativistic 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
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
Shyan said:
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?
Are you talking about light? No one here has said that light has infinite energy.

If you are talking about a particle with non-zero mass, then you can't get to c! That's what we have been saying. So the energy of a non-zero mass particle does not "get infinite".
 
  • #13
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
Shyan said:
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!

You can't use that formula for a massless particle, because γ is undefined. (It diverges to infinite.) So you are multiplying a zero by an undefined infinite value.

For a massless particle, the velocity is c, and the momentum is just E/c, where E is its energy.
 
  • #15
Is it the Planck momentum?
 
  • #16
Shyan said:
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?
The equation

[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 non-zero 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.)
 
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  • #17
Shyan said:
Is it the Planck momentum?

I'm not familiar with that term... but the energy of a photon is hf and its momentum is hf/c, where f is the frequency.
 
  • #19
Shyan said:

Ah. Thank you. That is a unit of momentum in the Planck system of units. It's like an alternative to metric for ubernerds.

I'm not presuming any particular system of units; just giving the general laws which apply, regardless of what units you like.
 
  • #20
DrGreg said:
The equation

[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 non-zero 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.)
Slight correction:
[tex]
F=\frac{d}{dt}(\gamma m v)=\gamma m a +m v \frac{d \gamma}{dt}=\left(\frac{v^2}{c^2}\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)m a
[/tex]

Either way the conclusion remains the same.
 
  • #21
Cyosis said:
Slight correction:
[tex]
F=\frac{d}{dt}(\gamma m v)=\gamma m a +m v \frac{d \gamma}{dt}=\left(\frac{v^2}{c^2}\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)m a
[/tex]

Either way the conclusion remains the same.

Actually

[tex]
\frac{v^2/c^2}{1-v^2/c^2} + 1 = \frac{1}{1-v^2/c^2}
[/tex]​

so we are in agreement!:smile:

(l left out the intermediate steps in my calculation and just gave the final answer.)
 
  • #22
Cyosis said:
Slight correction:
[tex]
F=\frac{d}{dt}(\gamma m v)=\gamma m a +m v \frac{d \gamma}{dt}=\left(\frac{v^2}{c^2}\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)m a
[/tex]

Either way the conclusion remains the same.

Keep going.
[tex]\begin{align*}
\frac{v^2}{c^2}\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &=
\frac{\frac{v^2}{c^2}}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1-\frac{v^2}{c^2}}{(1-\frac{v^2}{c^2})^{3/2}} \\
&= \frac{1}{(1-\frac{v^2}{c^2})^{3/2}}
\end{align*}[/tex]​

PS. By the way, I believe the quantity
[tex]\frac{a}{(1-\frac{v^2}{c^2})^{3/2}} = \gamma^3 a[/tex]​
is called proper acceleration, and is an invariant... corresponding the acceleration experienced by the moving body in its accelerated reference frame.
 
  • #23
Ugh, it looks like I am particular sharp tonight!

DrGreg said:
l left out the intermediate steps in my calculation and just gave the final answer.

I understood that, but I thought you had forgotten to apply the product rule, silly me. Anyway it's good we agree in the end!
 
  • #24
You know,I just read it in wikipedia and I haven't noticed that its a unit.
 
  • #25
Cyosis said:
I understood that, but I thought you had forgotten to apply the product rule, silly me. Anyway it's good we agree in the end!
(As a minor aside, I actually worked it out by doing the substitution [itex]v = c \tanh \phi[/itex], which might ring a bell with you from another thread!)
 
  • #26
Please note that you guys are treating a vector equation like a scalar, and in general the coordinate acceleration isn't even in the same direction as the force. Let's see if I can get this right...
[tex]\begin{align*}
\vec F &=& \frac{d\gamma m \vec v}{dt} \\
&=& \frac{d\gamma}{dt}m\vec v + \gamma m \vec a \\
&=& \frac{v}{(1-v^2)^{(3/2)}}\dot v m \vec v + \gamma m \vec a \\
&=& \frac{\vec v\cdot \vec a}{(1-v^2)^{(3/2)}} m \vec v+ \gamma m \vec a \\
\vec F &=& \gamma^3 (\vec v \cdot \vec a) m\vec v+ \gamma m \vec a \\
\end{align*}[/tex]
 
  • #27
Yes we calculated the x component of the force.
 
  • #28
Cyosis said:
Yes we calculated the x component of the force.

No, you calculated the v component. ;) But ok.
 
  • #29
ZikZak said:
Please note that you guys are treating a vector equation like a scalar, and in general the coordinate acceleration isn't even in the same direction as the force.
You are right. We are considering the special case of motion in a straight line, where the force is parallel to the velocity. I think that's sufficient for the questioner.
 

FAQ: A dark part of special relativity(at least for me)

1. What is the dark part of special relativity?

The dark part of special relativity refers to the concept of time dilation and length contraction, which can be difficult to understand and often leads to confusion and misconceptions.

2. How does time dilation work in special relativity?

Time dilation is the phenomenon where time appears to pass slower for objects moving at high speeds relative to an observer. This is due to the fact that as an object's speed increases, its mass also increases, causing time to slow down for that object.

3. What is length contraction in special relativity?

Length contraction is the effect where objects in motion appear shorter in the direction of motion when measured by an observer at rest. This is due to the fact that as an object's speed increases, its length appears to decrease in the direction of motion.

4. How does special relativity affect our daily lives?

Special relativity plays a crucial role in modern technology, such as GPS systems, which rely on precise timing measurements that take into account the effects of time dilation. However, these effects are only noticeable at extremely high speeds or in extreme environments, so they do not have a significant impact on our daily lives.

5. Can special relativity be proven?

Special relativity has been extensively tested and validated through numerous experiments and observations. It has consistently been found to accurately predict the behavior of objects at high speeds and has been a fundamental part of modern physics for over a century.

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