# GRAVITATION by MTW: p.49, lambda or tau?

1. Jul 14, 2010

### Living_Dog

At the very bottom of the page they have that the straight line, using the definition of a vector as the derivative of the worldline is given as:

$$P(0) + \lambda(\frac{dP}{d\tau})_o$$

But since the worldline, $$P(\tau)$$ is paramtrized by $$\tau$$ and not $$\lambda$$, then shouldn't the equation be:

$$P(0) + \tau(\frac{dP}{d\tau})_o$$

?

2. Jul 14, 2010

### yossell

I'm not sure what you think the trouble is. It seems fine to me. At that point, the world line isn't straight, and they just seem to be looking for the vector equation of the straight line which is tangent to that point. The equation is just the vector pointing to the point, plus k x tangent vector, with k an arbitrary real. There's no reason why this k should be intimately connected with t is there

3. Jul 14, 2010

### Altabeh

Just a little typo. Your correction is completely apposite.

AB

4. Jul 15, 2010

### yossell

Altabeth

if you get a chance, could you explain that to me. For the equation of the straight line that is a tangent to the curve, why does tau have to appear as the parameter in the vector equation? I can't see any connection.

5. Jul 15, 2010

### Altabeh

The parameter of curve is apparently $$\tau$$ and when you differentiate the straight line $$P(0) + \tau(\frac{dP}{d\tau})_o$$ with respect to $$\tau$$ then you get the tangent to curve at $$\tau=0$$ which is $$(dP(\tau)/d\tau)_0.$$ I f we take $$\lambda$$ to be related to $$\tau$$ linearly, $$\lamda=a\tau+b$$ with both $$a,b$$ being constants (a=/=0) then it doesn't let the differentiation produce the tangent through including an extra factor $$a$$ multiplied by the true tangent above.

P.s. when you talk of straight line you're no longer allowed to asign a "vector equation" to this line. "tau" is just a parameter of curve and has the same role as $$t$$ in the Newtonian equation $$x=vt+x_0.$$

AB

Last edited: Jul 15, 2010
6. Jul 15, 2010

### yossell

Thanks Altabeth,

But in MTW they only say `the straight line' of the four velocity. For this, I'm struggling to see a connection between the parameter used on the curve and the parameter on the new straight line. It's surely irrelevant what it is, and provided to differentiate the function with respect the variable, whether it's tau or lambda, you get the right result.

And from MTW
P(0) and dP/d\tau are vectors, this is a vector equation, and they're calling it a straight line.

7. Jul 15, 2010

### Altabeh

Gotta be more careful here: They are saying that $$P(0)+\lambda (dP/d\tau)_0$$ is a straight line. You think incorrectly that $$P(0)$$ and $$dP/d\tau$$ are vectors whereas $$P(\tau)$$ and $$dP/d\tau$$ are respectively the straight world line of a free particle (a non-vector quantity) and the tangent on this world line so that both of $$P(0)$$ and $$dP/d\tau$$ due to being computed at some point ($$\tau=0$$) are to be considered as constants only. This means when taking the derivative of$$P(0)+\lambda (dP/d\tau)_0$$ with respect to the variable of world line, i.e. $$\tau$$, you get

$$\frac{d\lambda}{d\tau} (\frac{dP}{d\tau})_0$$

which was expected to be the tangent to the world line,

$$(\frac{dP}{d\tau})_0$$

$$\frac{d\lambda}{d\tau} =1$$
$$\lambda=\tau.$$