Question about electromagnetic force and mass

[tiny-tim]

Perhaps you'll see the difficulties I keep encountering …

honestly, no …

you just measure everything in the same frame as you're measuring the E field in.

(you must always measure everything in the same frame, anyway)

 

[h0dgey84bc]

show me the money, this is a simple situation so shouldn't be a problem if you got the physics down...

and of course I'm measuring everything in the same frame

 

[h0dgey84bc]

$$F^{\mu} = \frac{dp^{\mu}}{d\tau}$$

firstly what is the form of $$F^{\mu}$$ in the scenario I've described above (i.e. external electric field in y direction)?

secondly, say we end up with an equation "like" (obviously it will depend on the form of F) :

$$qE_y = \frac{dp^{y}}{d\tau}$$ implying that $$p^y -> qEt$$
and another equation $$0=\frac{dp^{x}}{d\tau}$$ implying that $$p^x -> constant= p^x{}_0$$, and similarly $$0=\frac{dp^{z}}{d\tau}$$ implying that $$p^z -> constant= 0$$, now where do we go from here?

Obviously for a normal situation where the particle in question has mass, once you have the momentum you got the velocity too just by division by $$\gamma m_0$$, and then with knowledge of velocity getting position as a function of time (trajectory) is trivial. How do we proceed to get these things from knowledge of the momentum for a rest massless particle.

If you feel otherwise please post the few lines proceeding from the above, where you would take it next.

 

[Wiemster]

And the Newtonian formula, technically, isn't F=ma, it's F = d(mv)/dt, which works fine when translated into relativity.

Technically and nontechnically Newton's second law written as F=ma works as fine as F = d(mv)/dt. Actually F = d(mv)/dt = vdm/dt + mdv/dt =ma as it only applies to systems not changing their mass. Both equations need revising with a gamma factor in relativity, although your version a derivative less.

 

[h0dgey84bc]

Technically and nontechnically Newton's second law written as F=ma works as fine as F = d(mv)/dt. Actually F = d(mv)/dt = vdm/dt + mdv/dt =ma as it only applies to systems not changing their mass. Both equations need revising with a gamma factor in relativity, although your version a derivative less.

yeah adding in the gamma factor:

F = d(mv)/dt

$$F=\frac{ d\gamma m_0 v}{dt}$$
$$F= m_0 \frac{d\gamma v}{dt} +\gamma v \frac{ d m_0}{dt}$$
$$F=m_0 \frac{d\gamma v}{dt}$$
$$F=m_0 v \frac{d\gamma }{dt} +m_0 \gamma \frac{dv }{dt}$$
$$F=m_0 \gamma^3 v^2 \frac{dv}{dt} +m_0 \gamma \frac{dv }{dt}$$

but what if $$m_0 =0$$ ?

so seems we are left with the generic $$F^{\mu}=\frac{d p^{\mu} }{dt}$$ to use. But as my previous post adressed, seems this allows us to get the four momentum as a function of time quite easily. But how does one then infer the particles trajectory velocity etc from this quantity, since $$p^{\mu}=m_0 U^{\mu}$$ does not apply (for one we have the same problem of $$m_0 =0$$, and two the four velocity is undefined for a photon as I discussed above (if you don't believe me on this one please see Schutz or another intro Gen Rel book, he states it clear as day))

sigh, can someone just kill this thread and put me out of my misery

If someone could just show me the equations for my simple example of a "charged photon" moving in a uniform E-field in the y direction, I'd be supremeley grateful .

 

[Mårten]

Mm, what will happen with a charged photon, was just one of the thing I wondered, when starting this thread. Hopefully someone can help you in your efforts h0dgey84bc! :-) Would be interesting to know!