greswd
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you don't need mass if you're only concerned about force and not acceleration.alba said:
Magnetic field at relativistic speed
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You have just found another of the many reasons why thinking in terms of mass increasing with speed just creates confusion. If you haven't already read this Insights article, do so now - the mass that you plug into Newton's ##F=Gm_1m_2/r^2## has nothing do with the speed of anything.alba said:
alba
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I'd just like to know what is the ratio Fm/FE after the SR corrections, and the final outcome, that is the rate at which A and B will diverge. After 32 posts , still no clue!greswd said:
greswd
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google transverse mass.alba said:
that will give you the transverse acceleration
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But it's the most simple way!greswd said:
greswd
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but he won't be able to understand what he's reading!vanhees71 said:
jartsa
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alba said:
I found some simple formulas, so now I will calculate the final speed of two electrons originally 1 cm apart and at rest, ignoring relativistic effects, then I will transform that result to another frame:
(k is Coulomb's constant, q is electrons charge, r is distance, m is electron's mass)
potential energy of the system:
$$ E_s = \frac {k*q*q}{r} $$
kinetic energy of one electron:
$$E_k=\frac{E_s}{2}$$
velocity of one electron:
$$ v=\sqrt{\frac{2E_k}{m}} = \sqrt{ \frac{E_s}{m}}= \sqrt{ \frac{\frac {k*q*q}{r}}{m}}=159 m/s $$So the electrons move apart at speed 318 m/s. Now we use this system as a clock: When the system has grown by 318 m, then one second has passed.
Now we observe that same clock from another frame, in which frame the clock moves at speed 0.99 c. The clock ticks like this: When system has grown by: $$ \frac{318 m}{\gamma} $$ then one second has passed. Because that's how time dilation of clocks works.
alba
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Are you sure speed is so low? I imagine it is over one kilometer in a secondgreswd said:
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Ok, let's analyze the problem in #1. I was a bit too quick with my posting on the force, because here you need not only the force but also its split into electric and magnetic components. So we have to do a bit more more work, but we should do it in a manifest covariant way. In the following I set ##c=1## and work in Heaviside-Lorentz units, which are most adequate for relativistic covariant formulations of electrodynamics ;-). In the following I use the four-velocity of the particle (proton in the example) ##u^{\mu} = \mathrm{d} x^{\mu}/\mathrm{d} \tau=p^{\mu}/m## and the west-coast convention for the metric, ##(\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)##.
Of course, the most simple frame is the rest frame of the two protons, which we call ##\Sigma'##. First we consider the electromagnetic field of one of the protons, which we put for simplicity in the origin of the reference frame. It's world line is given by ##(x^{\prime \mu})=(\tau,0,0,0)## and thus ##(u^{\prime \mu})=(1,0,0,0)##. The four-potential of its field is of course the static Coulomb potential (in Lorenz gauge):
$$(A^{\prime \mu})=\frac{q}{4 \pi r'} (1,0,0,0)=\frac{1}{4 \pi r'} (u^{\mu}).$$
Now in the frame ##\Sigma##, where this proton moves with velocity ##\vec{v}=v \vec{e}_1## along the common ##x^1## axes of ##\Sigma'## and ##\Sigma##, the four-potential is
$$A^{\mu} ={(\Lambda^{-1})^{\mu}}_{\nu} A^{\prime \nu}, \quad x^{\prime \mu}={\Lambda^{\mu}}_{\nu} x^{\nu}.$$
The Lorentz matrix ##{\Lambda^{\mu}}_{\nu}## is given in #27. The result of the transformation is
$$A^{\mu}=\frac{q}{4 \pi r'} u^{\mu}=\frac{\gamma q}{4 \pi r'} \begin{pmatrix}1 \\ v \\ 0 \\ 0 \end{pmatrix}, \quad r'=\sqrt{(x^1-v t)^2+(x^2)^2 + (x^3)^2}.$$
Now, written in conventional (1+3) form the electromagnetic field components in this frame are given by
$$\vec{E}=-\vec{\nabla} A^0-\partial_t \vec{A} = \frac{\gamma q}{4 \pi r^{\prime 3}} \begin{pmatrix} x^1-v t \\ x^2 \\ x^3\end{pmatrix}$$
and
$$\vec{B}=\vec{\nabla} \times \vec{A}=\vec{v} \times \vec{E}.$$
The force on the 2nd proton, whose world line is given by ##y^{\prime \mu}=(tau,0,L,0)## with ##L=1 \;\text{cm}##. It's four-velocity is ##\tilde{u}^{\prime \mu}=(1,0,0,0)## and thus in the frame ##\Sigma## it's ##\tilde{u}^{\mu}=\gamma (1,v,0,0)##. From this you get for the spatical part of the electric and magnetic forces at time ##t=0##
$$\vec{K}_E=q \tilde{u}^0 \vec{E}=q \gamma \vec{E}=\frac{\gamma^2 q^2}{4 \pi L^2} \vec{e}_2$$
and
$$\vec{K}_B=q \tilde{\vec{u}} \times \vec{B} = q \gamma \vec{v} \times (\vec{v} \times \vec{E})=-\frac{q^2 v^2 \gamma^2}{4 \pi L^2} \vec{e}_2.$$
Note that the total Minkowski force is
$$\vec{K}_{\text{tot}}=\vec{K}_E + \vec{K}_M=\frac{q^2}{4 \pi L^2} \vec{e}_2,$$
which immideately follows from the Lorentz transformation of the corresponding four vector, which stays unchanged in this case, because it has only a ##x^2## component in ##\Sigma'## and thus, via the Lorentz boost in ##x^1## direction also in ##\Sigma##.
For the ratio of the magnetic to the electric force parts you get ##v^2##, i.e., for this simple example the non-relativistic approximation is exact (because the additional ##\gamma## factor of the exact transformation cancels in the ratio).
Of course, the most simple frame is the rest frame of the two protons, which we call ##\Sigma'##. First we consider the electromagnetic field of one of the protons, which we put for simplicity in the origin of the reference frame. It's world line is given by ##(x^{\prime \mu})=(\tau,0,0,0)## and thus ##(u^{\prime \mu})=(1,0,0,0)##. The four-potential of its field is of course the static Coulomb potential (in Lorenz gauge):
$$(A^{\prime \mu})=\frac{q}{4 \pi r'} (1,0,0,0)=\frac{1}{4 \pi r'} (u^{\mu}).$$
Now in the frame ##\Sigma##, where this proton moves with velocity ##\vec{v}=v \vec{e}_1## along the common ##x^1## axes of ##\Sigma'## and ##\Sigma##, the four-potential is
$$A^{\mu} ={(\Lambda^{-1})^{\mu}}_{\nu} A^{\prime \nu}, \quad x^{\prime \mu}={\Lambda^{\mu}}_{\nu} x^{\nu}.$$
The Lorentz matrix ##{\Lambda^{\mu}}_{\nu}## is given in #27. The result of the transformation is
$$A^{\mu}=\frac{q}{4 \pi r'} u^{\mu}=\frac{\gamma q}{4 \pi r'} \begin{pmatrix}1 \\ v \\ 0 \\ 0 \end{pmatrix}, \quad r'=\sqrt{(x^1-v t)^2+(x^2)^2 + (x^3)^2}.$$
Now, written in conventional (1+3) form the electromagnetic field components in this frame are given by
$$\vec{E}=-\vec{\nabla} A^0-\partial_t \vec{A} = \frac{\gamma q}{4 \pi r^{\prime 3}} \begin{pmatrix} x^1-v t \\ x^2 \\ x^3\end{pmatrix}$$
and
$$\vec{B}=\vec{\nabla} \times \vec{A}=\vec{v} \times \vec{E}.$$
The force on the 2nd proton, whose world line is given by ##y^{\prime \mu}=(tau,0,L,0)## with ##L=1 \;\text{cm}##. It's four-velocity is ##\tilde{u}^{\prime \mu}=(1,0,0,0)## and thus in the frame ##\Sigma## it's ##\tilde{u}^{\mu}=\gamma (1,v,0,0)##. From this you get for the spatical part of the electric and magnetic forces at time ##t=0##
$$\vec{K}_E=q \tilde{u}^0 \vec{E}=q \gamma \vec{E}=\frac{\gamma^2 q^2}{4 \pi L^2} \vec{e}_2$$
and
$$\vec{K}_B=q \tilde{\vec{u}} \times \vec{B} = q \gamma \vec{v} \times (\vec{v} \times \vec{E})=-\frac{q^2 v^2 \gamma^2}{4 \pi L^2} \vec{e}_2.$$
Note that the total Minkowski force is
$$\vec{K}_{\text{tot}}=\vec{K}_E + \vec{K}_M=\frac{q^2}{4 \pi L^2} \vec{e}_2,$$
which immideately follows from the Lorentz transformation of the corresponding four vector, which stays unchanged in this case, because it has only a ##x^2## component in ##\Sigma'## and thus, via the Lorentz boost in ##x^1## direction also in ##\Sigma##.
For the ratio of the magnetic to the electric force parts you get ##v^2##, i.e., for this simple example the non-relativistic approximation is exact (because the additional ##\gamma## factor of the exact transformation cancels in the ratio).
alba
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...so, what is the distance between A and B in the two frames after one millisecond? The net force in the CM mass is 2.3*10^-24/2 N, what is the net force acting on B in the rest frame? That is the questionvanhees71 said:
alba
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Is the electric force affected by motion? does A affect B in the same way it affects a charge D at rest at 1 cm from A while is it passing?vanhees71 said:
jartsa
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alba said:
I have a strong feeling that this is one of the rare cases when I managed to calculate something correctly.
An electron was accelerated by an electric field created by the smallest charge that exists, at macroscopic distance. Still the electron reached speed 159 m/s.
alba
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1/1000 s in the rest/lab frame means 1/7000 s in the moving frame. Consider LHC, for example:Nugatory said:
http:// http://i.stack.imgur.com/c1ggS.png
In the AB frame Fe = 2.3*10^-24N and time is 1/7000 s. In the C (lab) frame time is 1/ 7 = 1/1000 but mass is 7 time bigger, so they compensate each other in the impulse J = Ft.. On the other hand, besides the Fe we have a contrasting Fm = .98 Fe, which reduces the latter to 10^-26N, I cannot find a way to make the distance between A and B the same in both frames, be it 159*2 after a second or a few centimeters after 1/1000 s
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alba
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I was talking about distance, jartsa,jartsa said:
Can you show me how to find the distance between A and B after a period of time? While you are at it , try 1/7000 s