Recent content by TheMercury79
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		Error propagation leads to different uncertainties, which do I choose?
What is up with your answer?. It is obvious I just forgot an extra '$' so it is not necessary to ask about my equation. Yes I said 'measure', you're right I should have said 'determine', but you still know what I meant. It depends on two quantitites, their fractional errors add in quadrature...- TheMercury79
 - Post #5
 - Forum: Introductory Physics Homework Help
 
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		Error propagation leads to different uncertainties, which do I choose?
I understand how to compute and propagate errors but have trouble with conceptualizing all things put together. I have performed an experiment to determine a value for some quantity. This quantity depend on two variables. The first one depend in turn on some other quantities as well but I think...- TheMercury79
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 - Error Error analysis Error propagation Propagation Uncertainties
 - Replies: 15
 - Forum: Introductory Physics Homework Help
 
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		Momentum transfer in electron-proton collision
anyway, thanks for the help- TheMercury79
 - Post #11
 - Forum: Advanced Physics Homework Help
 
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		Momentum transfer in electron-proton collision
Because I knew the answer is not zero. Why should I check and see if the answer i zero? That's just ridiculous, of course it's not zero.- TheMercury79
 - Post #10
 - Forum: Advanced Physics Homework Help
 
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		Momentum transfer in electron-proton collision
Now I see it! Pero said the same thing but he threw me off by insisting we should set ##\theta## to zero- TheMercury79
 - Post #8
 - Forum: Advanced Physics Homework Help
 
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		Momentum transfer in electron-proton collision
But the correct answer is supposed to be ##Q^2=-4\frac{EE'}{c^2}cos^2\frac{\theta}{2}## I just put in it the other form to illustrate the plus and minus difference. And I instead got ##Q^2=-4\frac{EE'}{c^2}sin^2\frac{\theta}{2}##- TheMercury79
 - Post #6
 - Forum: Advanced Physics Homework Help
 
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		Momentum transfer in electron-proton collision
maybe I formulated it badly but the original statement was"##\theta## is the angle between the outgoing electron and the incoming proton". I figured the collision was head on, so this is the same as what I said "##\theta## is the angle between the electron before and after the collision" I don't...- TheMercury79
 - Post #4
 - Forum: Advanced Physics Homework Help
 
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		Momentum transfer in electron-proton collision
In a head-on collision between the proton and electron, what is the squared 4-momentum transfer between the two particles. Starting with the difference in momentum of the electron with the 4-vectors before and after the event: $$(P-P')^2=P^2+P'^2-2P\cdot P'$$ The circumstances are such that the...- TheMercury79
 - Thread
 - Collision Momentum Momentum transfer Particle physics
 - Replies: 10
 - Forum: Advanced Physics Homework Help
 
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		Is the Cube of matrix associative?
Thanks. It turned out they are the same. I just ended up with different expressions on the transposes. e.g (3/2)*sqrt(1/2) was transposed with sqrt(1/2)*(1+sqrt(1/2)), but these are equal so there was no problem- TheMercury79
 - Post #3
 - Forum: Calculus and Beyond Homework Help
 
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		Is the Cube of matrix associative?
But I actually don't get the same matrix. What I get is the transpose of the other when I change the order i.e when I do [A]^2[A] I get the transpose of [A][A]^2 and vice versa What I'm trying to do is find the cube of the expectation value of x in the harmonic oscillator in matrix form. We're...- TheMercury79
 - Thread
 - associative Cube Matrix
 - Replies: 3
 - Forum: Calculus and Beyond Homework Help
 
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		I Scale factor from Friedmann's equations
No it didn't. The text we use showed how if you differentiate the first you get the second. I guess the intstructions threw me off a bit. "Use the second equation directly" it said, it sounded so definite, like that's what you have to work with, nothing else. Anyway thanks. Will try and see what...- TheMercury79
 - Post #5
 - Forum: Cosmology
 
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		I Scale factor from Friedmann's equations
If we take a flat universe dominated by radiation, the scale factor is ##a(t)=t^{1/2}## which can be derived from the first Friedmann Equation:$$(\dot a/a)^2 = \frac{8\pi G}{3c^2}\varepsilon(t)-\frac{kc^2}{R_0^2 a(t)^2}$$ But suppose I want to show this using the second Friedmann Equation (Also...- TheMercury79
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 - Cosmology Differential equation Scale Scale factor
 - Replies: 7
 - Forum: Cosmology
 
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		I Hubble relation to Scale Factor
I feel kinda stupid now, The Friedmann equations is in the next chapter. I was under the impression this should be solved with contents of the current chapter. But I see now I've read the curriculum wrong. I have a full time job and taking this course on the side is really stressful. I just knew...- TheMercury79
 - Post #4
 - Forum: Cosmology
 
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		I Hubble relation to Scale Factor
Imagine a Universe where the Hubble parameter is truly a constant, in both space and time. How much smaller would such a Universe be 14 billion years ago compared to today? Using the Hubble parameter in terms of scale factor: ##H(t) = \frac{\dot{a}}{a}## leads to the differential equation...- TheMercury79
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 - Cosmolgy Hubble Relation Scale Scale factor Space expansion
 - Replies: 3
 - Forum: Cosmology
 
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		I Can the Second Plane Autonomous System Have a Unique Critical Point?
After further inspection I still come up with nothing. I found that if y=a, then ##bx+x^2=1##, but since this is a quadratic, it yields two possible values for x in terms of b, so there can't be a unique point for the second system. And how to use it with ab>1 is a mystery to me since it seems...- TheMercury79
 - Post #2
 - Forum: Differential Equations