# Lorentz Force Law - Tensorial Form

1. Sep 27, 2016

### Dewgale

1. The problem statement, all variables and given/known data
Show that the Lorentz Force Law, $\frac{dp^{\nu}}{d \tau} = -q U_{\mu} F^{\mu \nu}$, is consistent with $P^\mu P_\mu= -m^2.$ Here U is the 4-velocity, F is the Electromagnetic Tensor, and p is the 4-momentum. (Minkowski Space)

2. Relevant equations
As stated above.

3. The attempt at a solution
I initially tried differentiating $p^\nu p_\nu = -m^2$ w.r.t. tau. This gave me, through the product rule,
$$p_\nu \frac{d p^\nu}{d \tau} + p^\nu \frac{d p_\nu}{d \tau} = 0$$. Since the only difference between the covariant and contravariant forms of the Lorentz Force Law is whether the $\nu$ on the F is raised of lowered. Substituting the Lorentz Force Law into the above, we get:
$$-q ( U_\mu F^{\mu \nu} p_\nu + U_\mu F^\mu_\nu p^\nu) = 0$$
-q cancels, leaving
$$( U_\mu F^{\mu \nu} p_\nu + U_\mu F^\mu_\nu p^\nu) = 0$$

This is where I'm stuck. Both the left hand term and the right hand term are constants, and if I showed that one was the negative of the other I'd probably be set, but I'm not sure where to go from here.

Thanks!

2. Sep 28, 2016

### Orodruin

Staff Emeritus
First off, the metric is constant. What does this tell you about the derivative $dp^2/d\tau$?

Hint: use the fact that you can raise and lower indices using the metric.

Second, what is the 4-momentum expressed using the mass and 4-velocity?

3. Sep 28, 2016

### Dewgale

Well, first off, considering that $p^2 = p^\mu p_\mu = \eta_{\mu \nu} p^\mu p^\nu = m^2$, the constancy of the metric just tells me that the derivative is zero.

I suppose what I could do is take advantage of the fact that $p_\nu = \eta_{\nu \alpha} p^{\alpha} = m \eta_{\nu \alpha} U^\alpha$, and also of the fact that $$U^{\mu}U_{\mu} = -1$$, but I'm not sure of exactly how that would work with the indices.

4. Sep 28, 2016

### Dewgale

Well, first off, considering that $p^2 = p^\mu p_\mu = \eta_{\mu \nu} p^\mu p^\nu = m^2$, the constancy of the metric just tells me that the derivative is zero.

I suppose what I could do is take advantage of the fact that $p_\nu = \eta_{\nu \alpha} p^{\alpha} = m \eta_{\nu \alpha} U^\alpha$, and also of the fact that $$U^{\mu}U_{\mu} = -1$$, but I'm not sure of exactly that would work with the indices.

EDIT: oops! Accidental double post. I can't seem to delete it, so if someone with that ability could do that, that would be nice. Thanks.

5. Sep 28, 2016

### Orodruin

Staff Emeritus
No it doesnt. It also tells you something about the form of the derivative regardless of the fact that p^2 = m^2.

6. Sep 28, 2016

### Dewgale

I've got to admit, I'm very confused. If I add the metric in, the form of $\frac{\partial p^2}{\partial \tau}$ is $$\frac{\partial p^2}{\partial \tau} = \frac{\partial \eta_{\mu \nu} p^\mu p^\nu}{\partial \tau} = \eta_{\mu \nu} p^\mu \frac{\partial p^\nu}{\partial \tau} + \eta_{\mu \nu} p^\nu \frac{\partial p^\mu}{\partial \tau}.$$
This can be re-written as $$\eta_{\mu \nu} p^\mu f^\nu + \eta_{\mu \nu} p^\nu f^\mu...$$

Or as $2 p_\mu f^\mu = -q U_\nu F^{ \nu \mu} p_{\mu} = 0.$

I sort of just figured it out while typing. I feel like I'm probably (big if) on the right path now. Thank you!

7. Sep 28, 2016

### Orodruin

Staff Emeritus
Right (apart from a factor of two in the middle step). Now you can use the relation $p = mU$. What does the result tell you?

Edit: Note that what you want to show is that left-hand side expression is equal to zero given the Lorentz force (otherwise it is incompatible with a fixed mass). Do not go around assuming that this is the case for the Lorentz force, you have to show it!

8. Sep 28, 2016

### Dewgale

I get $m U_\mu f^\mu = -q m U_\nu F^{\nu \mu} U_\mu$. Looking just at the left hand side, we can see that $$m U_\mu f^\mu = m (f^0 U_0 + f^i U_i) = m(\frac{\partial E}{\partial \tau} \gamma + \frac{\partial E}{\partial \tau} \gamma^{-1}) = m \frac{\partial E}{\partial \tau} (\frac{\gamma^2+1}{\gamma})$$

However, in an object's rest frame, $\frac{\partial E}{\partial \tau} = 0$ I believe, due to conservation of energy. Therefore, the left hand side of the equation is equal to zero, showing that the law is consistent.

9. Sep 28, 2016

### Orodruin

Staff Emeritus
You are seriously overthinking this problem. Just look at the symmetry properties of the right-hand side.

Also, what you want to show is that the right-hand side is equal to zero. That the left-hand side is follows from $p^2$ being a constant.