Contracting over indices chain rule

[black_hole]

Homework Statement

As part of a problem I am doing I am asked to show u^β∂_βu^α = a^α where u is 4 velocity and a refers to 4 acceleration. The way to do this is not immediately obvious to me, especially since the problem implies there should be a chain rule step involved which I am not seeing. I thought ∂ refers to ∂/∂t + ∂/∂x ect.

Homework Equations

The Attempt at a Solution

 

[Fredrik]

Hints: What is $\partial_\beta(u^\beta u^\sigma)$? What is the definition of $a^\sigma$, and is there a way to rewrite that in a way that involves partial derivatives?

Edit: Oops. You don't have to use the product rule here, so ignore my first hint. Just start with the definition of $a^\sigma$.

 

[black_hole]

Well if I do this (∂/∂t + ∂/∂x...)* each component of u^β there's still the u^α I have to worry about. a^α = du^α/dτ so this seems to imply that u^β∂_β = d/dτ which doesn't seem right...?

And I'm not seeing how I can get that?

Oh wait never mind...I think I got it! If say I transform into a rest frame, the only component of the four velocity that is nonzero is the time component which is γc but γ=dt/dτ so u^β∂_β = d/dτ which makes the whole thing work! (yay)

 

[black_hole]

yeah you really don't have to use chain rule

 

[Fredrik]

If say I transform into a rest frame, the only component of the four velocity that is nonzero is the time component which is γc but γ=dt/dτ so u^β∂_β = d/dτ which makes the whole thing work! (yay)

In the (momentarily) comoving inertial coordinate system, we have u=(c,0,0,0), not (γc,0,0,0), and this is only at one specific moment. The four-acceleration is the rate of change of the four-velocity as a function of proper time, and the four-velocity is just the normalized tangent vector to the world line. The four-acceleration is telling us how that tangent vector's direction in spacetime is changing with proper time, so you can't compute it from knowledge of that direction at only one point on the world line.

yeah you really don't have to use chain rule

You do, actually. Recall that the chain rule is about taking the derivative of the composition of two functions. Can you figure out what the two functions are in this case?

 

[andrien]

You do, actually. Recall that the chain rule is about taking the derivative of the composition of two functions. Can you figure out what the two functions are in this case?

I replied to this thread by using chain rule which gives answer in one line.someone deleted it,and send me an infraction for it.

 

[Fredrik]

I replied to this thread by using chain rule which gives answer in one line.someone deleted it,and send me an infraction for it.

Yes, I noticed. My reply to it was deleted as well. I just said that in the homework forum, we try to give hints, not complete solutions. It can be hard to do this when the calculation is this short.