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I'm currently reading Schutz' first course in general relativity, and on the second page (already) I've encountered a problem:

We have the Galilean law of addition of velocities: $ v(t) = v'(t) = v(t) - V$ where $V$ is a constant velocity and $v'(t)$ is the velocity relative to the object moving with $V$ (relative to..).

He goes on to show that Newton's law are invariant under the above law, but I don't understand how he shows that the second law is invariant:

$$a'=\dfrac{dv'}{dt}=\dfrac{d(v-V)}{dt}=\dfrac{dv}{dt}=a$$

I think what mainly confuses me is the fact that it looks like he's using 2 different notations for derivatives at the same time, but even then this is what confuses me:

I'm not sure, but from what I know $a'$ should mean the relative acceleration, but $\dfrac{dv'}{dt}$ to me seems to mean the derivative of the relative velocity = relative acceleration. Of course those would cancel out, so my view is incorrect.

Can someone clear this up?

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Take the unprimed frame to be your and my rest frame. For some body we measure $v(t)$ and by differentiating it we get $a(t)$.

Now consider another observer in the primed frame moving at constant velocity $V$ relative to us. because of the law of addition of velocities we know that the other observer measures the velocity of body to be:

$$ v' = v(t) - V $$

That other observer works out $a'$ by differentiating $v'$:

$$ a' = \frac{dv'}{dt'} = \frac{d}{dt'}\left( v(t) - V \right) $$

In Galilean transformations time is absolute so $t = t'$ then:

$$ a' = \frac{d}{dt'}\left( v(t) - V \right) = \frac{d}{dt}\left( v(t) - V \right) $$

The relative velocity $V$ is constant so $dV/dt = 0$ and therefore:

$$ a' = \frac{d}{dt} v(t) = \frac{dv}{dt} = a $$

This shows that in a Galilean transformation the acceleration is invarient.

Incidentally, I commend your choice of Schutz's book. I think it's an excellent introduction to GR and it's the one I used. I still refer to it regularly.

Response to comment:

Let's take velocity to be positive if it's to the right (i.e. the usual coordinate convention in graphs) and suppose the observers in the unprimed frame see the primed frame moving to the right so $V$ is positive.

Now imagine you are in the primed frame. You see the unprimed frame moving to the left. That means a stationary object in the unprimed frame ($v = 0$) is moving to the left at velocity $-V$. That's why $v' = v - V$.

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  • $\begingroup$ Thanks, but what about the first minus term? That is what confused me? Should that have been an = sign? $\endgroup$ – user30117 Sep 25 '13 at 7:35
  • $\begingroup$ Oops, the minus term is correct. I made a mistake!! I'll add a footnote to my answer. $\endgroup$ – John Rennie Sep 25 '13 at 7:52

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