The inertial frame trick.

by wotanub
Tags: frame, inertial, trick
wotanub is offline
Mar12-13, 11:42 PM
P: 209
Can anyone refer me to a discussion of applying the technique of changing reference frames to problem solving? Why it works, and what it means. I'm familiar with using it in some E&M problems, but I guess I don't really "get" it. For example a particle in an E&M field has

[itex] m\vec{a} = q(\vec{E}+\vec{v}\times\vec{B})[/itex]

It's common to set [itex]\vec{v} = \frac{\vec{E}\times\vec{B}}{B^{2}} + \vec{u}[/itex] to "cancel" the [itex]\vec{E}[/itex] field and just work out the circular motion for the magnetic field.

What does this say about the physics of the system? Is it right to say the particle really (in the original frame) is moving in a circle "plus" the velocity of the frame? That is, if I want the velocity, I find it in the intertial frame, then just tack on the [itex]\frac{\vec{E}\times\vec{B}}{B^{2}}[/itex] term?

That's right answer, but can someone point me to some formal proof for the general case of any moving particle? I've never seen it stated in general, it's always just the professor going "oh, let's wave our hands like this and make [itex]\vec{E}[/itex] go away..." during an example problem. I use it, but I've never really learned it, so I always have trouble explaining this. I usually say something like "if you add a constant to the velocity, it won't change the [itex]\frac{d\vec{v}}{dt}[/itex]"

Or maybe if you've taught a class before, how did you explain it to your students?
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Simon Bridge
Simon Bridge is offline
Mar13-13, 12:14 AM
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The idea is that physics does not care what reference frame you use: stuff still works the same way. Therefore it is a good idea to pick one which makes the math easy to do.

Another common transformation is to make B go away by changing to a reference frame that is stationary wrt to the moving charges. The general idea is to exploit something special about the relationships involved.

A very simple transform is in ballistics when someone throws something from a non-zero initial height and it turns out to be easier just to call the initial position zero or you rotate the coordinates so that "down" becomes "positive" so you don't risk misplacing a minus sign (you've transformed, in the first case, to the frame of someone sitting at that height, and, in the second case, to that for someone upside down).

Every day you do motion and distance calculations taking the Earth as non-rotating - which means you are using a reference frame that is rotating with the Earth (this is actually a non-inertial frame!)

In your example - you'd usually fix your coordinates to the lab frame: the apparatus.
However - you don't have to. Where would the zero of the coordinate system be to make the transformation shown? What is the coordinate system doing? (What effect would the E field normally have?)
A.T. is offline
Mar13-13, 08:05 AM
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Quote Quote by wotanub View Post
I use it, but I've never really learned it, so I always have trouble explaining this. I usually say something like "if you add a constant to the velocity, it won't change the [itex]\frac{d\vec{v}}{dt}[/itex]"

wotanub is offline
Mar14-13, 11:13 AM
P: 209

The inertial frame trick.

Yes thanks. This is what I was looking for.

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