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bio website martin-ueding.de
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visits member for 2 years, 9 months
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22h
awarded  Enlightened
22h
awarded  Nice Answer
23h
answered Could two identical stars revolve around each other in a common orbit if we only account for Newtonian physics?
2d
comment Given eigenvalues of $\vec l^2$ and $\vec s^2$, calculate the eigenvalue for $\vec j^2$
Thanks, I think this is going to become interesting in the review of the exam, then.
2d
accepted Given eigenvalues of $\vec l^2$ and $\vec s^2$, calculate the eigenvalue for $\vec j^2$
Jul
22
asked Given eigenvalues of $\vec l^2$ and $\vec s^2$, calculate the eigenvalue for $\vec j^2$
Jul
21
comment Twins Paradox Paradox
What you can see in the drawing not really the length, since the metric has this minus sign. The perceived time is $\Delta t_\text T$, which is related to $\Delta s_\text T$ with $\Delta t_\text T = \Delta s_\text T \gamma$. Therefore, from the diagram with $\beta = 4/5$, you can read off that $\Delta s_\text T \approx 1/3 \Delta s_\text E$. With $\gamma = 5/3$ you have a duration ratio of about 0.55, which matches your 3/5 pretty close. I think that you can only see the stretching qualitatively, but I am not sure how to construct this directly in a drawing.
Jul
19
comment Definition of Electromagnetic field?
You need the retarded potential for that. Calculate the charge and current density of your moving charge (with a $\delta$-distibution) and that will give you $\phi$ and $\vec A$. From that, you can apply the gradient and curl to get $E$ and $B$.
Jul
19
comment Twins Paradox Paradox
Your move through spacetime can be written as a parametric curve $t(\tau), \vec r(\tau)$ where $\tau$ is the time the traveler itself will experience. Then $|u|=1$ means $\dot t^2 - \dot{\vec r}^2 = 1$. So when you move faster through space ($\dot r \gg 0$), then you also have to move faster through time ($\dot t \gg 0$), to experience the same amount of time. Since you moved through more $t$ with the same $\tau$, others (where $\tau = t$) will see you as younger, saying that you moved through time at a slower rate.
Jul
18
comment Twins Paradox Paradox
Time is a special dimension, yes. The straight line distance $s$ between two points is given by $s^2 = t^2 - x^2 - y^2 - z^2$. This distance $s$ is the time that one would experience when traveling along this straight line. This $s$ is the same in every coordinate system. For anything massive, that $s$ has to be positive, since you cannot go backwards in time. This defines a forward light-cone in which you have to travel.
Jul
18
revised Twins Paradox Paradox
Add a sketch
Jul
18
comment Twins Paradox Paradox
One key point is that minus sign in the metric. All possible four-velocites do not lie on a sphere, but on a hyperboloid. That makes the whole geometry really unintuitive.
Jul
18
comment Twins Paradox Paradox
You are right, his comment on the question clarified it. I will draw a picture and write a bit more.
Jul
18
comment Twins Paradox Paradox
They meet up again. However, the whole concept of perceived time depends on the path!
Jul
18
answered Twins Paradox Paradox
Jul
17
comment Gauss' Law, charge distribution on conductor
I can give you the answer. Would you like to trade the answer for a little effort on your side?
Jul
17
comment Relation between component and algebraic definition of covariant vectors
Let the vector space have basis vectors $\vec e_i$. Then the a vector $\vec v$ with components $v^i$ can be written as $\vec v = v^i \vec e_i$ (summation convention). The dual space has a basis $\vec e^i$ such that $\langle e^i, e_j \rangle = \delta^i_j$. Try Applications of Differential Geometry to Physics which I started to read yesterday.
Jul
17
comment Why is a rainbow curved in shape?
By the way, the Wikipedia article you linked has some explanations on it.
Jul
15
answered Rotation of Taylor expansion of a scalar
Jul
14
comment Ball thrown from a moving train
You can use LaTeX with $V_g$ to make the formula prettier. If the person floating next to the train observes $v_g$, it is at rest. However, it sounds like the person is floating (and moving) with the train.