Solving Gravitation Question: Noam Struggling With Answers

[noamriemer]

Hello! I fail to understand this question... I don't even know how to approach it...

Given: $$m\frac{d^{2}x^{\mu}} {d\tau^{2}} =e{F^{\mu}_{nu}} \frac{dx^{\nu}} {d\tau}$$

I have to define $$u^{\mu}= \frac {dx^{\mu}} {d\tau^2}$$
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?
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

 

[cepheid]

I'm no expert in GR or tensors, but I can give you some help based on what I do know:

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:$$m\frac{d^{2}x^{\mu}} {d\tau^{2}} =e\sum_{\nu = 0}^3{F^{\mu}_{\nu}} \frac{dx^{\nu}} {d\tau}$$

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:$$m\frac{du^0} {d\tau} =e\sum_{\nu=0}^3{F^{0}_{\nu}} u^\nu$$$$m\frac{du^1} {d\tau} =e\sum_{\nu=0}^3{F^{1}_{\nu}} u^\nu$$$$m\frac{du^2} {d\tau} =e\sum_{\nu=0}^3{F^{2}_{\nu}} u^\nu$$$$m\frac{du^3} {d\tau} =e\sum_{\nu=0}^3{F^{3}_{\nu}} u^\nu$$ As you can see, tensor notation is really compact! So, you have some differential equations to solve

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).

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 = (x⁰, x¹, x², x³)

and,

x^ν = (x⁰, x¹, x², x³)

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.

 

[noamriemer]

Thank you so much for your reply...
Yap, all the typos you found are typos...

Actually, I did what you wrote (well, most of it) before, and still can't solve...
But you did clear some things for me, and thank you for that...
I know how the F tensor looks like (I just have no idea how to write it in latex- that is why I did not add this information... but ill try:)

F= 0 E 0 0
E 0 0 0
0 0 0 0
0 0 0 0
Meaning (if I understood right) that I have:
$$m \frac {du^{0}} {d\tau} = eE \frac {dx^{1}} {d\tau}$$
$$m \frac {du^{1}} {d\tau} = eE \frac {dx^{0}} {d\tau}$$

And how do I connect that to what I need? Have you got any clue for me?
Thank you for your help!
Noam

 

[cepheid]

Thank you so much for your reply...
Yap, all the typos you found are typos...

Actually, I did what you wrote (well, most of it) before, and still can't solve...
But you did clear some things for me, and thank you for that...
I know how the F tensor looks like (I just have no idea how to write it in latex- that is why I did not add this information... but ill try:)

F= 0 E 0 0
E 0 0 0
0 0 0 0
0 0 0 0

You can use code tags to preserve whitespace:

0  E  0  0
E  0  0  0
0  0  0  0
0  0  0  0

If this is the matrix you meant, then you can write in in LaTeX as follows:

[tex]
\begin{pmatrix}
0 & E & 0 & 0\\
E & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
[/tex]

which produces $$\begin{pmatrix}<br /> 0 & E & 0 & 0\\<br /> E & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0<br /> \end{pmatrix}$$

Meaning (if I understood right) that I have:
$$m \frac {du^{0}} {d\tau} = eE \frac {dx^{1}} {d\tau}$$
$$m \frac {du^{1}} {d\tau} = eE \frac {dx^{0}} {d\tau}$$

And how do I connect that to what I need? Have you got any clue for me?
Thank you for your help!
Noam

Yeah, that's right, but I would keep everything in terms of u⁰ and u¹, since what you really have is a coupled set of first-order ODEs for these two variables:$$\frac{du^0}{d\tau} = \frac{eE}{m} u^1$$$$\frac{du^1}{d\tau} = \frac{eE}{m} u^0$$One step that might help would be to decouple them. For instance, if you solve for u¹ using the first equation, and substitute the result into the second equation, you'll end up with a second-order ODE for u⁰ only. It should be one that you can solve as well. Here are two things that might help: remember that if you have two functions that are solutions to an ODE, then any linear combination of those functions will also be a solution to that ODE. Remember also to apply the initial conditions (at tau = 0). To get a unique solution for a second-order ODE, you need two initial conditions: the initial value of the function, and the initial value of its first derivative.

 

[noamriemer]

Got it- Thanks to you!

Thank you so much!