Pair of moving charges - confused

[Xezlec]

Pair of moving charges -- confused

Here was a subject I thought I kind of had a grasp on, but apparently my understanding of SR is not very good at all!

We have 2 charges, a positive and a negative one. I'm holding one in each hand, and I'm running forward, so their attraction is perpendicular to my direction of motion. In my frame, they attract electrostatically. In your frame, they attract electrostatically and repel magnetically, such that when I do the math, the total force is 2 factors of gamma (that is, gamma squared) less than in my frame.

Surely the relativistic force equals the relativistic mass times acceleration. But the relativistic mass is 1 factor of gamma greater than in my frame, and the acceleration is 2 factors of gamma less due to time dilation, since acceleration has 2 factors of time in it. Right? So the mass times acceleration is, in total, 1 factor of gamma less, but the force is 2 factors of gamma less? Clearly I've screwed something up pretty bad. Please help.

Thanks!

 

[ZikZak]

Surely the relativistic force equals the relativistic mass times acceleration.

Surely, it does not. Which is half of the reason you should never, ever use the concept of "relativistic mass." (the other half of the reason being that relativistic mass is not the gravitational charge either)

Force can be defined as $\vec F = d\vec p/dt$, with $\vec p = \gamma m \vec v$. If you work through that derivative, you will find that the coordinate acceleration is not even generally parallel to the force, and certainly not proportional through the so-called "relativistic mass," except in one particular limiting case, and then only by accident. You will even find that the "relativistic mass" has nothing to do with the speeding up of a particle towards c (force parallel with velocity) as it is so often misused. Anyway, give it a shot. It is a useful and enlightening calculation to go through.

 

[Xezlec]

Surely, it does not. Which is half of the reason you should never, ever use the concept of "relativistic mass." (the other half of the reason being that relativistic mass is not the gravitational charge either)

Oh? Are you sure it isn't? I was told that it is, the last couple times I asked a question about that.

Force can be defined as $\vec F = d\vec p/dt$, with $\vec p = \gamma m \vec v$. If you work through that derivative, you will find that the coordinate acceleration is not even generally parallel to the force, and certainly not proportional through the so-called "relativistic mass," except in one particular limiting case, and then only by accident.

Funny you should say that... I actually did work through that derivative, and even checked my answer against Wikipedia. The answer I came up with (and what appeared on Wikipedia) was that, for the perpendicular case I'm talking about (notice the geometry of my example), F= gamma m a, which is relativistic mass times acceleration.

You will even find that the "relativistic mass" has nothing to do with the speeding up of a particle towards c (force parallel with velocity) as it is so often misused. Anyway, give it a shot. It is a useful and enlightening calculation to go through.

Funny you should say that, too, because I did that calculation a long time ago and found that it does in fact explain the speeding up of a particle toward c, if we're talking about the same thing. Care to clarify?

 

[ZikZak]

Oh? Are you sure it isn't? I was told that it is, the last couple times I asked a question about that.

Yes, kinetic energy fits into different slots in the stress-energy tensor than the mass does. KE does not gravitate like mass. There are some common instances, specifically when the KE is heat confined to a box, where you can get away with calling it by a simple mass-equivalent name, but for a free particle, things are different.

Funny you should say that... I actually did work through that derivative, and even checked my answer against Wikipedia. The answer I came up with (and what appeared on Wikipedia) was that, for the perpendicular case I'm talking about (notice the geometry of my example), F= gamma m a, which is relativistic mass times acceleration.

In the case where the force is perpendicular to the velocity, this is the one degenerate case where, by accident, it does indeed equal $\gamma m \vec a$. Are you using the correct relativistic forms of the electric and magnetic fields? See, for instance, http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_22.pdf"

Funny you should say that, too, because I did that calculation a long time ago and found that it does in fact explain the speeding up of a particle toward c, if we're talking about the same thing. Care to clarify?

When force and velocity are parallel, $F = \gamma^3 m a,$ not $\gamma m a.$

 

[Jonathan Scott]

I think that the "powers of gamma" approach should work OK for the transverse case provided that you take care to keep track of the distinction between proper time and the time in the local frame, which also differ by a factor of gamma.

Note that the usual form of the Lorentz force equation gives you the rate of change of the (relativistic) momentum with respect to local time, not proper time.

If I'm remembering the transformation law correctly for the field, I think that in the frame where the particles are moving with velocity $\mathbf{v}$ we have the following, where $\mathbf{E}$ is the electric field in the original frame:

$$\frac{d\mathbf{p}}{dt} = q \gamma (\mathbf{E} - \mathbf{v} \times (\mathbf{v} \times \mathbf{E})/c^2) = q \gamma \, \mathbf{E} ( 1 - v^2/c^2) = q \gamma \left ( \frac{\mathbf{E}}{\gamma^2} \right ) = \frac{q\mathbf{E}}{\gamma}$$

Making this relative to proper time:

$$\frac{d\mathbf{p}}{d\tau} = \frac{d\mathbf{p}}{dt} \frac{dt}{d\tau} = \frac{d\mathbf{p}}{dt} \gamma = q\mathbf{E}$$

If we substitute the usual expression for the relativistic momentum in terms of proper velocity, we get the following:

$$\frac{d}{d\tau} \left ( m \frac{d\mathbf{x}}{d\tau} \right ) = m \frac{d^2 \mathbf{x}}{d\tau^2} = q\mathbf{E}$$

That is, the proper acceleration is the same as in the original frame, so the transverse acceleration as seen from the moving frame and measured in terms of the local time will be time dilated by the square of the time dilation factor, as expected. (In other directions, the relationship between proper acceleration and local acceleration is also affected by space contraction, but not in the transverse direction).

Does that make sense? I hope I got that right.

 

[Xezlec]

Yes, kinetic energy fits into different slots in the stress-energy tensor than the mass does. KE does not gravitate like mass. There are some common instances, specifically when the KE is heat confined to a box, where you can get away with calling it by a simple mass-equivalent name, but for a free particle, things are different.

Wow. If you're right, then I am going to have to have a little talk with a certain lying physicist I know.

So how can the universe tell a free particle from a particle in a box? Also, why aren't mass and energy equivalent in this case? Kinetic energy doesn't obey E=mc^2? This seems to go against what my (possibly too low-level) SR text says.

As far as tensors, I don't know anything about GR. I thought I was asking a SR question.

In the case where the force is perpendicular to the velocity, this is the one degenerate case where, by accident, it does indeed equal $\gamma m \vec a$. Are you using the correct relativistic forms of the electric and magnetic fields? See, for instance, http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_22.pdf"

I was completely ignorant of the fact that the laws of electromagnetics changed in SR. So do Maxwell's equations get modified somehow? Is there somewhere you know of that I can find a derivation of that big complicated formula for electric field?

When force and velocity are parallel, $F = \gamma^3 m a,$ not $\gamma m a.$

Well, we can agree on that.

Johnathan Scott: I appreciate that, but I fear I may be a little behind.

 

[ZikZak]

Wow. If you're right, then I am going to have to have a little talk with a certain lying physicist I know.

Perhaps you should ;) Unfortunately, the important difference between energy density and energy flux is lost even on some physicists who do not work in Relativity.

So how can the universe tell a free particle from a particle in a box?

Take a look at the Wikipedia entry on the http://en.wikipedia.org/wiki/Stress-energy_tensor" . Note that "energy density", i.e. mass, is but one of 10 independent components of the source of the gravitational field. A stream of free particles in motion, for instance, constitutes a nonzero energy flux and momentum density. Its stress-energy is different from the same amount of energy density at rest, contributing only to $T_{00}$, and therefore so is its gravity.

A bunch of particles banging about in a box on average do not transport energy overall and look much different from a mass. A bunch of particles in a box DO have pressure, so you would think that even then you could not equate its solution to that of simple energy density, but the tension (negative pressure) of the container itself will cancel out the pressure terms to observers well outside the box.

Also, why aren't mass and energy equivalent in this case? Kinetic energy doesn't obey E=mc^2? This seems to go against what my (possibly too low-level) SR text says.

I am sure that somewhere your SR text gives the general equation, $E^2=(mc^2)^2 + (pc)^2$. The equation $E=mc^2$ is the energy extractable from a certain amount of rest mass, i.e. when p=0. It is an equation that is applicable to a mass at rest.

As far as tensors, I don't know anything about GR. I thought I was asking a SR question.

You probably were. I was making a point about how "relativistic mass" isn't "massy" in either sense of the word "mass."

I was completely ignorant of the fact that the laws of electromagnetics changed in SR. So do Maxwell's equations get modified somehow? Is there somewhere you know of that I can find a derivation of that big complicated formula for electric field?

I think you'll find it in those lecture notes I linked before if you read through them. They appear to be pretty good.

Well, we can agree on that.

Cool. :)

 

[PeterDonis]

Yes, kinetic energy fits into different slots in the stress-energy tensor than the mass does. KE does not gravitate like mass.

I think you meant to say that momentum, pressure, stress, etc. fit into different slots in the stress-energy tensor. Kinetic energy goes in the 00 (or tt) term, just like rest energy does. From the http://en.wikipedia.org/wiki/Stress-energy_tensor" :

In special relativity, the stress-energy of a non-interacting particle with mass m is
$$T^{\alpha \beta}[t,x,y,z] = \frac{m \, v^{\alpha}[t] v^{\beta}[t]}{\sqrt{1 - (v/c)^2}}\;\, \delta(x - x[t]) \delta(y - y[t]) \delta(z - z[t])$$.

If we align the x-axis of our coordinates along the direction of the particle's motion, the velocity 4-vector reduces to $$v^t = c$$, $$v^x = v$$, and we have (writing $$\gamma$$ for the relativistic time dilation/length contraction factor):

$$T^{tt} = \gamma m c^2$$,
$$T^{tx} = T^{xt} = \gamma m c v$$,
$$T^{xx} = \gamma m v^2$$.

Note that the $$T^{tt}$$ term is the total energy density, including rest energy and kinetic energy. The $$T^{tx}$$ and $$T^{xx}$$ terms do add additional effects, but they aren't kinetic energy terms.

 

[ZikZak]

I think you meant to say that momentum, pressure, stress, etc. fit into different slots in the stress-energy tensor.

You're right; I should have said that free bodies with KE will have off-(0,0) terms, but those terms are not equal to the KE.

 

[jtbell]

I was completely ignorant of the fact that the laws of electromagnetics changed in SR. So do Maxwell's equations get modified somehow?

No. Interestingly, Maxwell's equations are already relativistically correct, even though they pre-date Einstein's relativity theory by forty-odd years!

 

[Xezlec]

If I'm remembering the transformation law correctly for the field, I think that in the frame where the particles are moving with velocity $\mathbf{v}$ we have the following, where $\mathbf{E}$ is the electric field in the original frame:

$$\frac{d\mathbf{p}}{dt} = q \gamma (\mathbf{E} - \mathbf{v} \times (\mathbf{v} \times \mathbf{E})/c^2) = q \gamma \, \mathbf{E} ( 1 - v^2/c^2) = q \gamma \left ( \frac{\mathbf{E}}{\gamma^2} \right ) = \frac{q\mathbf{E}}{\gamma}$$

I should have read your post more closely. Rereading and pondering the above line until my brain started oozing out my nose finally showed me the mistake I made. I was thinking that length contraction didn't matter at all and I could entirely neglect it, not because I was truly thinking about the charges as points (I wasn't), but because I was thinking that if you (classically) take two parallel charged rods and then compress them both along their lengths, the net force between them does not change. Now, remembering Coulomb's Law, I realize that is not true, and therefore the way the electric field changes under SR should have been clear to me even without ZikZak showing me those cool slides. So my error rippled all the way down to my thinking about classical physics.

I am now absolutely 100% convinced that everything works out correctly. Yay. :)

No. Interestingly, Maxwell's equations are already relativistically correct, even though they pre-date Einstein's relativity theory by forty-odd years!

Thinking about it more, I can see how Gauss's Law still works out. Cool.

Thanks, everybody.

 

[Jonathan Scott]

Looking at it again, I see I wrote the wrong sign for the cross product term, but it didn't affect the conclusion. The corrected version is as follows:

$$\frac{d\mathbf{p}}{dt} = q \gamma (\mathbf{E} \, + \, \mathbf{v} \times (\mathbf{v} \times \mathbf{E})/c^2) = q \gamma \, \mathbf{E} ( 1 - v^2/c^2) = q \gamma \left ( \frac{\mathbf{E}}{\gamma^2} \right ) = \frac{q\mathbf{E}}{\gamma}$$

 

[PeterDonis]

If we align the x-axis of our coordinates along the direction of the particle's motion, the velocity 4-vector reduces to $$v^t = c$$, $$v^x = v$$, and we have (writing $$\gamma$$ for the relativistic time dilation/length contraction factor):

$$T^{tt} = \gamma m c^2$$,
$$T^{tx} = T^{xt} = \gamma m c v$$,
$$T^{xx} = \gamma m v^2$$.

Since I posted the above originally, I wanted to post again to correct it. In the course of working out the question in https://www.physicsforums.com/showthread.php?t=335864", I realized that the factor in front of all three of the tensor components above should be $$\gamma^2$$, not $$\gamma$$. That doesn't change the basic point, which is that a moving object's stress-energy tensor is more complicated than just the energy (t-t) term, but I wanted to post the correction for info.

 

[Bob S]

Here are the relativistic transformations for E and B fields; see bottom of first page.
http://pdg.lbl.gov/2002/elecrelarpp.pdf