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

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Discussion Overview

The discussion revolves around understanding the concepts of Lorentz Contraction and Time Dilation, specifically through the use of a calculator that applies the formula E=mc². Participants are seeking clarification on the variables involved, particularly m₀ (rest mass) and E₀ (rest energy), as well as the application of these in calculations related to relativistic energy.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the meaning of m₀ and how to calculate E₀, despite understanding the square root component of the formula.
  • Another participant explains that E₀ is the rest energy calculated as E₀ = m₀c², and emphasizes that this does not involve division by the square root.
  • There is a discussion about the formula E = m₀c²/√(1 - v²/c²) and how to apply it with specific values, including an example calculation provided by one participant.
  • Some participants clarify that the square root factor is related to the Lorentz factor (γ), which indicates that moving objects have more energy than at rest, and they provide additional context about kinetic energy.
  • Another participant mentions that the square root factor is common in relativity and suggests verifying the relationship between kinetic energy and relativistic energy through trial numbers or Taylor expansion.
  • There is a reference to an alternative way to express energy in terms of momentum and rest mass, indicating a deeper exploration of the topic.

Areas of Agreement / Disagreement

Participants exhibit a mix of understanding and confusion regarding the calculations and definitions involved. While some explanations are provided, there is no consensus on the correct approach to the calculations, and multiple interpretations of the formulas exist.

Contextual Notes

Participants have not fully resolved the steps involved in the calculations, and there are indications of missing assumptions or definitions that could clarify the confusion. The discussion highlights the complexity of applying relativistic formulas in practical scenarios.

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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What specifically is confusing you about it?
 
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.
 
##m_o## is the mass. ##E_o## is the rest energy. ##E_o=m_oc^2##.
 
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.
 
Last edited:
Noah Englehart said:
The problem is,I do the square root part...
What square root?
 
jtbell said:
What square root?
The square root at the bottom part of the fraction.
 
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.
 
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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