Wikipedia:Reference desk/Archives/Mathematics/2013 June 4

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June 4

What topics in mathematics should I read to understand the article Einstein field equations?

I read the article Einstein field equations and understood everything except the mathematical equations. So, my question is - What topics in mathematics should I read to understand these equations? Once I was suggested a nice link by someone. Suggest me some useful links (like the previous one) related to those mathematical topics. Thank you! Concepts of Physics (talk) 12:29, 4 June 2013 (UTC)

There is mathematics of general relativity for a start. Sławomir Biały (talk) 12:35, 4 June 2013 (UTC)

In addition to Sławomir's recommendation, there is a wikibook on general relativity that while incomplete, has some math sections that may be worth checking out. Outside of the wikiverse, Bernard F. Schutz' book A First Course in General Relativity is an approachable book for self study and this site has a number of links to online resources for learning GR. Good luck, it is a beautiful subject. --Mark viking (talk) 22:17, 4 June 2013 (UTC)

To solve a simple mathematical equation, can we take derivative or integral of both sides and then solve the equation?

I have an equation like this one: 2x²*31 = 5x + 50 . I can add, subtract, multiply and divide any number on both sides of the equation. I can also square (or take log of) both sides and then solve the problem. My question is - Is it correct to take derivative or integral (as in the process of squaring) of both sides and then solve the problem? Concepts of Physics (talk) 13:23, 4 June 2013 (UTC)

No, because the equality is only valid of specific values of x. If you integrate for differentiate, you would use that the equation is an identity that is valid for all values of x (or at least for x in some interval). Count Iblis (talk) 13:30, 4 June 2013 (UTC)

The original question is equivalent to asking for the value of x where the graph of y = 2x²*31 meets the graph of y = 5x + 50. Taking the derivatives and comparing is equivalent to asking for the value of x where the two graphs have the same slope, which is unlikely to give the same answer (in fact only at points where the graphs are tangents to each other). AndrewWTaylor (talk) 13:37, 4 June 2013 (UTC)

And sometimes it may not happen at all. For a simple equation x = 1, which is solved by the only real number, namely by x = 1, differentiating with respect to x gives 1 = 0, which is not satisfied by any real number. On the other hand x = x+1 has no solution, while (x)' = (x+1)' is equivalent to 1 = 1, which is solved by every real x. --CiaPan (talk) 05:22, 5 June 2013 (UTC)

There are lots of things you can't do to both sides of an equation and preserve equality. For instance, you can't add one to the x variable on both sides. It's not something special about calculus. Sławomir Biały (talk) 15:42, 4 June 2013 (UTC)

When you "do an operation" to both sides of an equation, you are applying a function. One of the defining properties of a function, say f, is that if a and b are equal, then f(a) and f(b) are equal. The reason you can't differentiate the equation you have above is an issue of domains, things like "multiplying by 2" are functions of the real numbers, things like "take the derivative" are functions of differentiable functions. Your equation works with numbers, not differentiable functions. Best case scenario, you could argue that you are working with constant functions, but in that case both sides have a derivative of 0, which while true, is not helpful.Phoenixia1177 (talk) 07:26, 5 June 2013 (UTC)

Indeed. You can, however, differentiate or integrate both sides of a functional equation, in which the two sides of the equation are equal as functions throughout the whole of a domain, not just at a single point. For example, if you are told that

$f'(x)=f(x)$

for all real x and that f is infinitely differentiable, you can conclude that

$f''(x)=f'(x)$

and by induction, that

$f^{(n)}(x)=f(x)\quad \forall \,x\in \mathbb {N}$

and thus

$f(x)=f(0)\left(1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots \right)=f(0)e^{x}$

Gandalf61 (talk) 08:08, 5 June 2013 (UTC)