I'm no expert in GR or tensors, but I can give you some help based on what I do know:
noamriemer said:
Hello! I fail to understand this question... I don't even know how to approach it...
Given: m\frac{d^{2}x^{\mu}} {d\tau^{2}} =e{F^{\mu}_{nu}} \frac{dx^{\nu}} {d\tau}
It looks like there's a slight typo above, and it should be m\frac{d^{2}x^{\mu}} {d\tau^{2}} =e{F^{\mu}_{\nu}} \frac{dx^{\nu}} {d\tau} Now remember that the Einstein summation convention is in effect: whenever you see repeated indices, one on the top and one on the bottom, you sum over them (and so that index disappears). So the expression above really is a shorthand for:m\frac{d^{2}x^{\mu}} {d\tau^{2}} =e\sum_{\nu = 0}^3{F^{\mu}_{\nu}} \frac{dx^{\nu}} {d\tau}
noamriemer said:
I have to define u^{\mu}= \frac {dx^{\mu}} {d\tau^2}
I'm going to assume there was a slight typo here as well, and that the definition should have been u^\mu = \frac{dx^\mu}{d\tau}. I'm assuming that because it sounds like u^\mu is meant to be a 4-velocity. If so, then the expression above becomes:m\frac{du^{\mu}} {d\tau} =e{F^{\mu}_{\nu}} u^\nu At this point, I would say that in order to solve the problem, you'd have to know what the components of the tensor F_\nu^\mu are.
Since each index can take on any of four different values from 0 to 3, the above equation is really a shorthand for four different equations, one for each component of the "4-acceleration." You get the four equations by considering each possible value for \mu in turn:m\frac{du^0} {d\tau} =e\sum_{\nu=0}^3{F^{0}_{\nu}} u^\num\frac{du^1} {d\tau} =e\sum_{\nu=0}^3{F^{1}_{\nu}} u^\num\frac{du^2} {d\tau} =e\sum_{\nu=0}^3{F^{2}_{\nu}} u^\num\frac{du^3} {d\tau} =e\sum_{\nu=0}^3{F^{3}_{\nu}} u^\nu As you can see, tensor notation is really compact! So, you have some differential equations to solve
noamriemer said:
and obtain:
u^0= cosh(\frac {eE\tau} {m}) u^0(0)+ sinh(\frac {eE\tau} {m})u^1(0)
u^1= sinh(\frac {eE\tau} {m}) u^0(0)+ cosh(\frac {eE\tau} {m})u^1(0)
I don't understand what I should do. First, what does u(0) mean? u(x=0)? how do I obtain these equations?
Well, the parameter tau is the independent variable that u
μ depends on, so u
μ(0) = u
μ(τ = 0).
noamriemer said:
What is the relation between x^{\mu} and x^{\nu}?
If I define this four vector, u, and u refers to mu, than what am I to do with the other index, nu?
Thank you! I have been struggling for two days...
Noam
They are both the same -- symbols for the position coordinates. Either μ or ν is an index that can range from 0 to 3, so:
x
μ = (x
0, x
1, x
2, x
3)
and,
x
ν = (x
0, x
1, x
2, x
3)
So the only difference is what symbol you're using for the index. Hopefully the preceding discussion shows why its important to choose different symbols for different indices once you start doing tensor operations.
