Relativistic Energy- well a basic algebraic simplification

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The discussion focuses on the algebraic simplification of relativistic energy, specifically solving for "u" in the equation E = mc^2 / √(1 - u^2/c^2). A mistake is identified in the transition between steps, emphasizing that the square root of a sum cannot be separated into the sum of square roots. The correct formulation is clarified as u = √(c^2(1 - (mc^2/E)^2)), leading to the final expression u = c√(1 - (mc^2/E)^2). Participants express gratitude for the clarification and assistance in correcting the simplification. The conversation highlights the importance of careful algebraic manipulation in physics equations.
Sneil
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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
 
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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)
 
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:
 
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)
 
great! thanks a lot for the help :smile:
-Neil
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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