Relativistic Energy- well a basic algebraic simplification

In summary, the conversation discusses the simplification and correction of a mathematical equation for solving for "u." The correct equation is u = c*root(1 - (mc^2/E)^2). The conversation also mentions the mistake of taking the root of two terms added together and the final equation is further simplified to u = c*root(1 - (mc^2/E)^2).
  • #1
Sneil
18
0
a little rough on this simplification, is this correct?
solving for "u"

E = mc^2 / (root)(1-u^2/c^2)

(root)(1-u^2/c^2) = mc^2/E

1 - u^2/c^2 = (mc^2/E)^2

1 = (mc^2/E)^2 + u^2/c^2

c^2 = (mc^2/E)^2 (c^2) + u^2 -> (sq root everything)


u = c - mc^3/E

pretty bad i don't know this basic simplification, but it'll all come back quickly enuf

thanks for the help :smile:

-Neil
 
Last edited:
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  • #2
c^2 = (mc^2/E)^2 (c^2) + u^2 -> (sq root everything)


u = c - mc^3/E

Your mistake occurs on the transition between these two lines

when you take the root of a function which has two terms added together, it is not equivalent to the root of each one added together

ie root (a^2 + b^2) is not equal to (a + b)

so you should really have
u = root ((mc^2/E)^2(c^2) + u^2)
 
  • #3
alright thank you :smile:


Warr said:
so you should really have
u = root ((mc^2/E)^2(c^2) + u^2)

but do you mean

u = root ((mc^2/E)^2(c^2) + c^2) but actually u = root (c^2 - (mc^2/E)^2(c^2) ) :confused:
 
  • #4
ah sorry, yes that's what I meant to type

also you could further make it so that

u = root (c^2(1 - (mc^2/E)^2))
u = c*root(1 - (mc^2/E)^2)
 
  • #5
great! thanks a lot for the help :smile:
-Neil
 

1. What is "relativistic energy"?

Relativistic energy refers to the energy of an object or system that is moving at a high speed, close to the speed of light. It takes into account the effects of special relativity, which states that as an object's velocity increases, its mass and energy also increase.

2. How is relativistic energy different from classical energy?

Classical energy, also known as Newtonian energy, only considers the kinetic energy of an object based on its mass and velocity. Relativistic energy takes into account the increase in energy as an object's velocity approaches the speed of light, as well as the increase in mass due to this high speed.

3. Can relativistic energy be simplified algebraically?

Yes, relativistic energy can be simplified using the equation E=mc^2, where E represents energy, m represents mass, and c represents the speed of light. This equation is a simplified version of the more complex equation for relativistic energy, but it is only accurate for objects moving at low speeds compared to the speed of light.

4. How is relativistic energy used in practical applications?

Relativistic energy is used in fields such as nuclear physics, where it helps explain the energy released in nuclear reactions. It is also used in particle accelerators, such as the Large Hadron Collider, to study the behavior of particles at high speeds.

5. Why is understanding relativistic energy important?

Understanding relativistic energy is crucial for accurately predicting and explaining the behavior of objects and systems moving at high speeds. It also plays a significant role in our understanding of the universe and is essential for advancements in fields such as nuclear physics and astrophysics.

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