# Mass spectometer and relativity

1. Sep 6, 2004

### evelyncanarvon

Hi, I had a homework question in my physics class that I'm not totally sure about. We're supposed to design (not actually build, just explain on paper) a mass spectrometer that can measure the speed of electrons going at .998c, so taking into account special relativity. Here are my questions:

Would it basically be the same as a normal spectrometer, same design?

How would you account for relativity? I think you would assume that the mass would get bigger in your reference frame, so you would have to divide by that lorentz factor to get the object's rest mass. Is this right? What about length contraction? Do you just multiply r by the lorentz factor?

Also, how would you actually set up the magnetic field and the electric field?

Any help would be greatly appreciated. Thanks!

2. Sep 6, 2004

### Tide

Your mass spectrometer would basically still have the same design. I think the problem has to do with where the electrons will land after their partial orbit of the magnetic field so the only change necessary is in the equation of motion for the electron. So instead of "mass times acceleration" you will use
$$\frac{d\vec p}{dt}$$.

The Lorentz force (due to the magnetic field) should remain unchanged. You will also have to consider how large to make the device and what the magnetic field strength will be to accomodate that size.

3. Sep 6, 2004

### evelyncanarvon

Thanks for your prompt reply! So the distance they travel would change? Would it just be multiplied by the lorentz factor? I'm still a little confused.

4. Sep 6, 2004

### Tide

Remember

$$\vec p = \gamma m_0 \vec v$$

5. Sep 6, 2004

### evelyncanarvon

Hmmm... I still don't totally understand. Here's what I have so far.

Not taking into account relativity, the mass of the electron measured by the spectrometer will equal

B^2*e*R/E

where B is the magnetic field, e is the charge of the electron, R is the radius of the circle it travels in (one-half the distance from the end of the velocity filter to the plate that catches the electrons at the end), and E is the electric field.

How do I adjust this for relativity?

There would be length contraction for the electron, so would I have to divide R by the lorentz factor? Or would I just assume that the M measured would be too large and divide the entire thing by the lorentz factor?

Sorry, but I don't understand what I can do with your previous statement.

6. Sep 6, 2004

### evelyncanarvon

Okay, Tide, I kinda get what you're saying.

dp/dt = qvB

p= mv/(1-v^2/c^2)^1/2

but how do you translate these to make them equal eachother?

How do you transform the second equation to become a variation of mv^2/r ?

If you have to take its derivative, it'll get really ugly... unless you don't need to take the derivative of the lorentz part. Is that just a constant?

Last edited: Sep 6, 2004
7. Sep 6, 2004

### Tide

Make sure you recognize that the differential equation is a VECTOR equation! Namely,
$$\frac {d \vec p}{dt} = q \vec v \times \vec B$$

You should have no trouble deducing that $p^2$ is a constant which means that $v^2$ is a constant and, therefore, so is $\gamma$.

You end up with (after a little rearranging)
$$\frac {d \vec v}{dt} = \frac {q}{\gamma m} \vec v \times \vec B$$

Now, remember $\gamma$ is a constant so you can solve the differential equations it just like you did for the nonrelativistic case.

8. Sep 6, 2004