bigdogsmhs306
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Theory of Relativity help
- Thread starter bigdogsmhs306
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JuicyBurger
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Well, as v approaches c...
\(\displaystyle \frac{v^2}{c^2} = 1\) correct?
So then the value of
\(\displaystyle \frac{v^2}{c^2} = \frac{1}{1} = 1\)
Now you have...
\(\displaystyle m = \frac{m_0}{\sqrt{1-1}} = \frac{m_0}{0}\)
What is this equal to?
Hmmm this doesn't seem right.
\(\displaystyle \frac{v^2}{c^2} = 1\) correct?
So then the value of
\(\displaystyle \frac{v^2}{c^2} = \frac{1}{1} = 1\)
Now you have...
\(\displaystyle m = \frac{m_0}{\sqrt{1-1}} = \frac{m_0}{0}\)
What is this equal to?
Hmmm this doesn't seem right.
You're on the right track. This is basic relativity. The rest mass of a particle as is the mass of the particle as measured in the instantaneous rest frame of the particle. The rest mass is the \(\displaystyle m_{0}\).
As our velocity approaches the speed of light, the mass becomes infinite. This says that no one can reach the speed of light because the mass becomes infinitely large. \(\displaystyle \sqrt{1-\frac{v^{2}}{c^{2}}}\) is called the Lorentz factor.
Here is a cool way to look at it.
\(\displaystyle m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}={\gamma}m_{0}\)........[1]
Now, consider the binomial expansion, which is valid for |x|<1,
\(\displaystyle (1+x)^{n}\sim 1+nx\)
If n=-1/2:
\(\displaystyle (1-x)^{\frac{-1}{2}}\sim 1+\frac{x}{2}\)
Let \(\displaystyle x=\frac{v^{2}}{c^{2}}\) and sub into [1]:
\(\displaystyle m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\sim m_{0}\left(1+\frac{1}{2}\frac{v^{2}}{c^{2}}\right)\sim m_{0}+\frac{1}{2}m_{0}\frac{v^{2}}{c^{2}}\)
multiply through by c^2 and we can relate the relativistic energy to the rest mass energy plus the Newtonian energy of the particle:
\(\displaystyle mc^{2}=m_{0}c^{2}+\frac{1}{2}m_{0}v^{2}\).
Cool, huh?.
Imagine a space ship traveling faster and faster and approaching the speed of light. The closer it gets, its mass becomes infinite. A basic tenet of relativity. Hence \(\displaystyle E=mc^{2}\). That is why the Enterprise has to be in a Warp Bubble
As our velocity approaches the speed of light, the mass becomes infinite. This says that no one can reach the speed of light because the mass becomes infinitely large. \(\displaystyle \sqrt{1-\frac{v^{2}}{c^{2}}}\) is called the Lorentz factor.
Here is a cool way to look at it.
\(\displaystyle m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}={\gamma}m_{0}\)........[1]
Now, consider the binomial expansion, which is valid for |x|<1,
\(\displaystyle (1+x)^{n}\sim 1+nx\)
If n=-1/2:
\(\displaystyle (1-x)^{\frac{-1}{2}}\sim 1+\frac{x}{2}\)
Let \(\displaystyle x=\frac{v^{2}}{c^{2}}\) and sub into [1]:
\(\displaystyle m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\sim m_{0}\left(1+\frac{1}{2}\frac{v^{2}}{c^{2}}\right)\sim m_{0}+\frac{1}{2}m_{0}\frac{v^{2}}{c^{2}}\)
multiply through by c^2 and we can relate the relativistic energy to the rest mass energy plus the Newtonian energy of the particle:
\(\displaystyle mc^{2}=m_{0}c^{2}+\frac{1}{2}m_{0}v^{2}\).
Cool, huh?.
Imagine a space ship traveling faster and faster and approaching the speed of light. The closer it gets, its mass becomes infinite. A basic tenet of relativity. Hence \(\displaystyle E=mc^{2}\). That is why the Enterprise has to be in a Warp Bubble
JuicyBurger
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So I was correct when I arrived with
\(\displaystyle m = \frac{m_0}{0}\)
I just did not think it right for something traveling at the speed of light to have infinite mass, however I have not taken a relativity class =D
\(\displaystyle m = \frac{m_0}{0}\)
I just did not think it right for something traveling at the speed of light to have infinite mass, however I have not taken a relativity class =D
bigdogsmhs306
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Thanks this now makes a lot more sense