# Showing Lorentz force is always spacelike: Using abstract index notation

So the Lorentz force on a massive particle is given by $$f^{\mu} = qg^{\mu\alpha}F_{\alpha\beta}\hat{v}^{\beta}$$, where $$\hat{v}^{\beta}$$ is the four vector of the particle and $$F_{\alpha\beta} = \partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}$$ is the EM field strength tensor, with EM potential $$A_{\alpha}$$.

I want to show that this force will always be spacelike. And since the four-velocity of a massive particle will always be time-like, then I can prove $$f^{\mu}$$ is always spacelike if: $$g_{\mu\nu}\hat{v}^{\mu}f^{\nu}=0$$

I feel like what I am doing is correct, but I am unable to show its zero. I think part of the problem is that I am quite new to abstract index notation, and am unsure how to simplify expressions. Thus far, I have worked out:

$$g_{\mu\nu}\hat{v}^{\mu}f^{\nu} = g_{\mu\nu}\hat{v}^{\mu}qg^{\nu\alpha}F_{\alpha\beta}\hat{v}^{\beta}=q\delta^{\alpha}_{\mu}F_{\alpha\beta}\hat v^{\beta}\hat v^{\mu}$$

Assuming this approach is valid, can anyone hint at where one can go from here? Also, what are some general guidelines or rules for manipulating/simplifying tensor expressions such as these?

• $\hat{v}^{\beta}$ is the four vector of the particle You mean the four-velocity. Why are you putting a hat on it? Oct 11 '20 at 18:59
• Contracted Kronecker deltas can be made to disappear. Oct 11 '20 at 19:01
• The hat just says its been normalized. So $\hat{v}^{\mu} = \frac{dx^{\mu}}{d\tau}$ Oct 11 '20 at 19:03
• @DescheleSchilder Does it mean that the force can only act orthogonal to the direction the particle is moving? Oct 11 '20 at 20:23
• @DescheleSchilder since $||u||^2 = c^2$, always, any 4-acceleration has to be orthogonal to the 4-velocity, and the 4-velocity is time-like.
– JEB
Oct 11 '20 at 20:45

Starting from your last equation, it should be clear that $$g_{\mu\nu} v^\mu f^\nu = q \delta^\alpha_\mu F_{\alpha \beta} v^\beta v^\mu = q F_{\alpha \beta} v^\beta v^\alpha,$$ since only the terms where $$\alpha = \mu$$ will be non-zero because of the Kronecker delta.

From here it's quite trivial, since $$F_{\alpha \beta}$$ is an antisymmetric tensor, and $$v^\alpha v^\beta$$ is a symmetric tensor, and so their contraction is zero. But if you want to show it explicitly:

$$F_{\alpha \beta} v^\beta v^\alpha = F_{\alpha \beta} v^\alpha v^\beta = F_{\beta\alpha} v^\beta v^\alpha = - F_{\alpha\beta} v^\beta v^\alpha,$$

where in the first step I've just changed the order of multiplication of $$v^\beta$$ and $$v^\alpha$$, in the second step I've flipped the dummy indices $$\alpha$$ and $$\beta$$, and in the third step I've used the fact that $$F_{\alpha\beta} = - F_{\beta \alpha}$$. From here it should be trivial that it's zero.

• Thank you!! I see why $F_{\alpha\beta}$ is antisymmetric from its definition. However, can you elaborate on $v^{\alpha}v^{\beta}$ being a symmetric tensor. I never really thought about the two four-vectors alone being a tensor. Oct 11 '20 at 19:13
• @ZacharyC The (tensor) product of two tensors is a tensor, and vectors are just tensors of rank 1. Oct 11 '20 at 19:30
• @ZacharyC Because $v^\alpha v^\beta=v^\beta v^\alpha$. These components are just commuting numbers. Oct 11 '20 at 19:34
• Well the symmetry isn't too hard to understand, since each individual term of $v^\alpha v^\beta$ is just a product. It shouldn't be too difficult to see that term-by-term, $v^\alpha v^\beta = v^\beta v^\alpha$. As to why it's a tensor, well it transforms like one! :) Oct 11 '20 at 19:35
• @ZacharyC This gets complicated. Tensor products are not commutative. But what you wrote is invalid. You can’t put a $\otimes$ between components of a tensor. The components of a tensor are not the same as the tensor itself. The components are just numbers that multiply normally and thus do commute. Discussing this further in comments is inappropriate because it could take way too long. Oct 11 '20 at 19:51