Understanding E=mc^2: Lorentz Contraction and Time Dilation Calculator

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In summary, the website says that:-The square root at the bottom part of the fraction is what you need to do-The problem is, you don't understand what the m0 means and you don't know how to get the E0.
  • #1
Noah Englehart
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So, I've been viewing this calculator website, and from it i figured out how to do the Lorentz Contraction and Time Dilation formula. However, When i stumbled upon this one... No matter how i tried, I couldn't figure it out. So, Please help.

http://keisan.casio.com/exec/system/1224060366
 
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  • #2
What specifically is confusing you about it?
 
  • #3
jtbell said:
What specifically is confusing you about it?
Well.. I understand the square root part, because that is in all the formulas, but i don't exactly understand what the m0 means, and i have no clue as to how to get the E0. Oh, And sorry, I forgot to post the website. Posting now.
 
  • #4
##m_o## is the mass. ##E_o## is the rest energy. ##E_o=m_oc^2##.
 
  • #5
Mister T said:
##m_o## is the mass. ##E_o## is the rest energy. ##E_o=m_oc^2##.
I know all that, The problem is,I do the square root part... And then I am not sure how to divide that. Because, I multiply the mass with speed of light, divide the result with the square root part, And then run it through the calculator to find its inconsistent with my results.
 
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  • #6
Noah Englehart said:
The problem is,I do the square root part...
What square root?
 
  • #7
jtbell said:
What square root?
The square root at the bottom part of the fraction.
 
  • #8
What fraction? I don't see one in ##E_0 = m_0 c^2##. Isn't that the equation we're talking about?

Argh. :headbang: I finally clicked your link. Sorry about that. Now I think I see what you're asking about.

How about showing us exactly what numbers you're trying to plug in, and the intermediate steps of your arithmetic? Then someone can probably tell you where you're going wrong.
 
  • #9
jtbell said:
What fraction? I don't see one in ##E_0 = m_0 c^2##. Isn't that the equation we're talking about?

Argh. :headbang: I finally clicked your link. Sorry about that. Now I think I see what you're asking about.

How about showing us exactly what numbers you're trying to plug in, and the intermediate steps of your arithmetic? Then someone can probably tell you where you're going wrong.
Well, I know that when i do the square root, its right, because the calculator confirms. However, I then tried to divide m0 x c2, which worked. So then i declared that must be the value of E0, But then i get confused because i realize that the E0 is a fraction and that messes up my thought processes.
 
  • #10
The formula you link to says that
[tex]E=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

If you take, say, [itex]m_0= 1 kg[/itex] and [itex]v= c/2= 299792.5/2=149896.25 km/s.[/itex] then that formula says that
[tex]E= \frac{c^2}{\sqrt{1- \frac{1}{4}}}= \frac{c^2}{\frac{\sqrt{3}}{2}}= \frac{2c^2}{\sqrt{3}}[/tex] which would be 103779337954.2 Joules.

That is, much to my surprise, what I get when I enter the values into that website.
 
  • #11
[itex]E_0 = m_0c^2 [/itex] is the "rest energy" of m0. It is the energy equivalence of the rest mass m0

[itex] E = \frac{E_o}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{m_oc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] is the energy of m0 when moving at velocity v. (If v=0, then E=E0= m0c2)
 
  • #12
Noah Englehart said:
Well.. I understand the square root part, because that is in all the formulas, but i don't exactly understand what the m0 means, and i have no clue as to how to get the E0. Oh, And sorry, I forgot to post the website. Posting now.
I don't know if anyone will be able to help until you post your calculations step by step.

Regardless, E0 is not going to include you dividing by the square root. It's just rest mass times the square of the speed of light. You don't even need the square root for rest energy. If you want to get the relativistic mass energy, then you divide by the square root.
 
  • #13
That square root factor is common in relativity so there's a special name for it https://en.wikipedia.org/wiki/Lorentz_factor:
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
so
$$E = \gamma m c^2$$
It just means that an object moving at a faster velocity has more energy than an object at rest. How much more? Well that's what the formula tells you. The kinetic energy T is equal to the total energy minus the rest energy.
$$T = E - E_0 = (\gamma-1) m c^2$$
Note that for small ##v##, ##\gamma-1 \approx \frac{1}{2} \frac{v^2}{c^2}## (why don't you verify this with some trial numbers, or Taylor expansion if you know how?) so ##T \approx \frac{1}{2} m v^2##
 
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  • #14
##\gamma## can only be defined for massive particles, so there's a better way to write the energy:
$$E^2 = m^2 c^4 + p^2 c^2$$
 
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What is E=mc^2 and why is it important?

E=mc^2 is a famous equation proposed by Albert Einstein in his theory of relativity. It states that energy (E) is equal to mass (m) multiplied by the speed of light (c) squared. This equation is important because it helps to explain the relationship between mass and energy, and has many practical applications in modern science and technology.

How did Albert Einstein come up with E=mc^2?

Albert Einstein developed E=mc^2 as part of his theory of special relativity, which he published in 1905. He based his theory on the idea that the laws of physics should be the same for all observers, regardless of their relative motion. Through mathematical calculations, he was able to show that mass and energy are two forms of the same thing and can be converted into each other.

Can you explain the meaning of each component in E=mc^2?

E represents energy, which is measured in joules. m stands for mass, which is measured in kilograms. And c is the speed of light, which is approximately 299,792,458 meters per second. When you square the speed of light, you get a very large number, which is why even a small amount of mass can produce a huge amount of energy according to this equation.

What are some real-life examples of E=mc^2?

One of the most well-known examples of E=mc^2 is the atomic bomb. In this case, a small amount of mass is converted into a massive amount of energy, resulting in a powerful explosion. This equation is also used in nuclear power plants, where the energy produced by the fission of atoms is converted into electricity. Additionally, the equation is used in medical imaging techniques such as PET scans, which utilize the conversion of mass into energy to create images of the body.

Is E=mc^2 still relevant in modern science?

Yes, E=mc^2 is still a fundamental concept in modern science and is used in many areas of research and technology. It has been tested and confirmed through numerous experiments and continues to play a crucial role in our understanding of the universe. In fact, it is considered one of the most famous and significant equations in physics.

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