# Lorentz Contraction of Moving Line of Charge

1. Sep 13, 2013

### leonardthecow

1. The problem statement, all variables and given/known data

A point charge +q rests halfway between two steady streams of positive charge of equal charge per unit length λ, moving opposite directions and each at c/3 relative to the point charge. With equal electric forces on the point charge, it would remain at rest. Consider the situation from a frame moving right at c/3.

a) Find the charge per unit length of each stream in this frame.

b) Calculate the electric force and the magnetic force on the point charge in this frame, and explain why they must be related the way they are.

2. The attempt at a solution

I know that the charge density of each moving line of charge will change due to relativistic length contraction. I also managed to show that the new charge density is simply the original charge density multiplied by the Lorentz factor:

L=L0/γ, where L is the contracted length
λ0=Q/L0, λ=Q/L, where lamda is the new charge density

Solve each for Q, set them equal, and rearrange to obtain

λ/λ0=L0/L=γ, ∴ λ=γλ0

But here's where I'm stuck. I know I now need to find the Lorentz factor γ for each moving line of charge. I also know that γ=(1-v2/c2)-1/2. In this formula, I need to know v, which I thought would be computed using a Lorentz transformation for velocity, namely u'=(u-v)/(1-uv/c2).

I do have the final answer to part a, which is given as λ(√8)/3 and 5λ(√8)/12 in the textbook. However, I can't produce these values, because when I insert the values that I thought would be correct into the velocity transformation equation for u and v (namely c/3 for each), one u' would be equal to 0, yielding a Lorentz factor of 1, and the other u' would be incorrect as well.

I think I understand the method behind solving the problem (although I could be wrong), but it seems like I'm having trouble conceptually trying to figure out the velocities relative to one another.

Any help would be greatly appreciated, thanks!

2. Sep 13, 2013

### TSny

I think you have the right idea. In order to see where you might be making a mistake, we need to see more details of your calculation. What did you get for the speed of each line charged line in the new frame? EDIT: You are correct that you can use the relativistic formula for addition of velocities.

Last edited: Sep 13, 2013
3. Sep 13, 2013

### leonardthecow

I did the following:

For the line of charge moving in the same direction as the frame:

Let u1=velocity of stream of particles moving in same direction as moving frame, relative to stationary point charge
Let v=velocity of frame relative to stationary point charge
Let u1'=velocity of stream of particles moving in same direction as moving frame, relative to moving frame

u1'=(u1-v)/(1-u1v/c2)
u1'=(c/3-c/3)/(1-(c/3)(c/3)/c2)=0

γu1=(1-0/c2)-1/2=1

∴ λ10. (But I know this is incorrect)

Similarly, for the line of charge moving in the opposite direction as the frame:

Let u2=velocity of stream of particles moving in opposite direction as moving frame, relative to stationary point charge
Let v=velocity of frame relative to stationary point charge
Let u2'=velocity of stream of particles moving in opposite direction as moving frame, relative to moving frame

u2'=(-u2-v)/(1-u2v/c2)
u2'=(-c/3-c/3)/(1-(c/3)(c/3)/c2)
u2'=(-2c/3)/(1-(c2/9)/c2)
u2'=(-2c/3)/(1-1/9)
u2'=(-2c/3)/(8/9)=-18c/24=-3c/4

γu2=(1-(-3c/4)2/c2)-1/2
γu2=(1-(9/16))-1/2
γu2=(7/16)-1/2

∴ λ2=(7/16)-1/2λ0. (But I know this is incorrect)

Is it a sign error in setting up the relative velocity transformations?

4. Sep 13, 2013

### TSny

This is correct. But you want to express λ1 in terms of λ rather than λ0. In the statement of the problem, the symbol λ stands for the charge density of each line in the original frame of reference (where each line is moving at c/3).

I think there is a sign error in the second line. See if you can spot it.

5. Sep 13, 2013

### leonardthecow

Ah I see it now!

So, because of length contraction,

λ=Q/L, L=L0
λ=Qγ/L0
λ/γ=Q/L=λ0.

For the moving line of charge,

γ=(1-(c/3)2/c2)-1/2
γ=(1-1/9)-1/2
γ=(8/9)-1/2

So, because λ10,

λ1=λ/γ
∴ λ=(8/9)-1/2λ1
λ=√(8)λ1/3.

I'm not sure what the sign error would be in the second line, my logic is that v is still in the same direction, so its sign wouldn't change, but now the stream of charge I'm considering is flowing in the opposite direction as the first. Should it instead be a positive u plus a positive v in the numerator?

6. Sep 13, 2013

### TSny

Good.

The numerator is fine. u1 is negative. v is positive. The sign error occurs when you substitute values in the denominator of
u1'=(u1-v)/(1-u1v/c2).

7. Sep 13, 2013

### yands

I would like to jump in and help. However I really don't get what you mean by steady stream of positive charge of equal linear charge density. :(

8. Sep 13, 2013

### yands

Do you mean
<<<<<<rod moving left point charge rod moving right >>>>>>

9. Sep 13, 2013

### leonardthecow

So, assuming I'm not making another dumb mistake, here's where I end up:

u2'=(-c/3-c/3)/(1-(-c/3)(c/3)/c2)
u2'=(-2c/3)/(1+(c2/9)/c2)
u2'=(-2c/3)/(1+1/9)
u2'=(-2c/3)/(10/9)=-18c/30=-3c/5

γu2=(1-(-3c/5)2/c2)-1/2
γu2=(1-(9/25))-1/2
γu2=(16/25)-1/2
γ2=5/4

∴ λ=4λ2/5

.....but that still doesn't work out. Sorry to be a bother here, I just can't see the error

10. Sep 13, 2013

### leonardthecow

And hey yands, yeah I believe the set up is meant to be a line of charge moving left, a line of charge moving right, a stationary charge in between, and a moving frame. The way I set it up was one line of positive charge moving in the -x direction, one in the +x direction, and the moving frame also moving in the +x direction

11. Sep 13, 2013

### TSny

Looks good.

Oops. Can you spot the error with the left hand side of this equation? Did you really want to use the charge density λ of the lines in the original frame where they are moving at c/3?

12. Sep 13, 2013

### yands

The rod moving with the frame will have the same value of its linear charge density since in this frame its speed is zero.
The other rod would have a negative speed according to Lorenz transformation
U'=(U-v)/(1-(Uv/c^2))

Last edited: Sep 13, 2013
13. Sep 13, 2013

### yands

Calculating Lorentz fraction µ for the second rod in the frame of reference of the first rod to find L' = Lµ^-1
Then since charge is invariant Q=λL=λ'L'

14. Sep 13, 2013

### leonardthecow

Ohhhhhh, okay, so let me think this through. Really, it should be something like

λ22=λ/γ1,

because not only is the line of charge lorentz contracted because it is moving at c/3, but also because it is moving relative to the moving frame, hence why both lorentz factors are involved.

I think solving for λ2 then yields the correct answer, because then you would have

λ2=λγ21
λ2=(λ5√(8))/12.

Phew!! I think all the subscripts got me lost, thanks a bunch for helping me out though, would've been lost without the guidance

15. Sep 13, 2013

Good work!