Comparing u-dot and Acceleration: What's the Difference?

[SpaceTrekkie]

Homework Statement

The problem says to show that the dot product u dot a = u*u-dot where u-dot is the differentiation with respect to time.

How is u-dot different from just the acceleration?

My teacher said that "a is the magnitude of the 3-acceleration.

u-dot is the time derivative of the speed.

In the first case, you first differentiate the velocity with time and then take the magnitude. In the second case, you first take the magnitude and then differentiate with time. These are not necessarily the same. For example, consider uniform circular motion. In that case, u-dot is always zero, but a is never zero."

but I am not sure I understand this...

Homework Equations

N/A

The Attempt at a Solution

No idea where to begin. I know that if u and a are perpendicular them the dot product is 0, but that is about it.

 

[SpaceTrekkie]

Okay, I figured out that u-dot is the change in SPEED over time, while a = the change in VELOCITY over time. So the u-dot has no direction...but where to go from there, is still a mystery to me...

 

[xboy]

Firstly, your topic name is a bit of a misnomer, isn't it? This doesn't have anything to do with either SR or 4-vectors, but perhaps you didn't know that.

Anyways here's a hint : we can write $$x \frac{dx}{dt} = \frac{1}{2} \frac {d}{dt} x^2$$

See if you can use it here.

 

[SpaceTrekkie]

Oh, okay, hmm...it was on our 4-vector HW for my relativity class and with no idea how to approach it I figured it was a 4-vector problem. Thanks for the tip, I will see if I can work it out from here.