I Choice of signature important for superluminal 4-velocity? (Minkowski)

[PeterDonis]

I think the new definition of 4-velocity should be

All you are doing is treating derivatives as fractions (bad) and taking an absolute value, which throws away information (bad), in order to reach the same conclusion you could reach by just using the standard definition of the tangent vector components, which are, as I've already said, just the derivatives of the coordinates with respect to the curve parameter.

it's the only way I know how to get the correct answer

You don't know how to take derivatives with respect to the curve parameter?

If so, learning how to do that would seem to be a much better idea than what you are doing now.

 

[PhDeezNutz]

All you are doing is treating derivatives as fractions (bad) and throwing away information about whether a curve is timelike or spacelike (bad) in order to reach the same conclusion you reach by just using the standard definition of the tangent vector components, which are, as I've already said, just the derivatives of the coordinates with respect to the curve parameter.
You don't know how to take derivatives with respect to the curve parameter?

If so, learning how to do that would seem to be a much better idea than what you are doing now.

I know in Multi-variable calculus

say we have a curve

$\vec{r} \left( t \right) = \left( x(t) , y(t), z(t) \right)$

that

$\vec{r}' \left( t \right) = \left( x'(t) , y'(t), z'(t) \right)$

should I just set $t = s$ for our problem?

 

[PeterDonis]

should I just set $t = s$ for our problem?

No, since $t$ is not the arc length along the curve.

In this case, you have a coordinate 4-vector $X(s) = (t(s), x(s), y(s), z(s))$. Just apply the same method you did for a 3-vector to that; the tangent vector $U$ is just $X'(s)$.

 

[vanhees71]

Well, take the approach with the Lagrangian as for usual subluminal massive particles for tachyons, which by definition travel along space-like world lines. Then you can define
$$L_0=-m c^2 \sqrt{-\dot{x}_{\mu} \dot{x}^{\mu}}.$$
with the same argument that this leads to a parameter-independent Lorentz-invariant action. Then the "four momentum" is
$$p_{\mu}=\frac{1}{c} \frac{\partial L_0}{\partial \dot{x}^{\mu}} = m c \dot{x}_{\mu} (-\cdot{x}_{\mu} \dot{x}^{\mu})^{-1/2}$$
and the on-shell condition is
$$p_{\mu} p^{\mu}=-m^2 c^2; \Rightarrow \; \vec{p}^2=(p^0)^2+m^2 c^2$$
This is then in accordance with the tachyonic feature that the particle moves in any inertial frame with a constant speed larger than the speed of light, i.e.,
$$\vec{v}=c \frac{\vec{p}}{p^0}=c \frac{\vec{p}}{\sqrt{\vec{p}^2-m^2 c^2}} \; \Rightarrow \; |\vec{v}| \geq c.$$
I think the real trouble with the tachyons starts when it comes to interacting ones and causality.

 

[PhDeezNutz]

No, since $t$ is not the arc length along the curve.

In this case, you have a coordinate 4-vector $X(s) = (t(s), x(s), y(s), z(s))$. Just apply the same method you did for a 3-vector to that; the tangent vector $U$ is just $X'(s)$.

Alright I've used your approach (or at least what I think is your approach) to come up with a more rigorous solution. However I am getting a factor of $\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}$ in front instead of $\frac{1}{\sqrt{\frac{u^2}{c^2} - 1}}$. I'm going to use signature (+,-,-,-) if that's okay.

The only questionable mathematical maneuver that might be troublesome is asserting $\frac{dt}{ds} = \frac{1}{\frac{ds}{dt}}$ but I think there's a version of the inverse function theorem that supports this.

My solution is as follows;

$\tilde{U} = c \frac{d \tilde{R}}{ds}$

$\frac{d \tilde{R}}{ds} = \left(c \frac{dt}{ds}, \frac{dx}{ds}, \frac{dy}{ds}, \frac{dz}{ds} \right)$

Using the chain rule we have

$\frac{d\tilde{R}}{ds} = \left( c \frac{dt}{ds}, \dot{x} \frac{dt}{ds}, \dot{y} \frac{dt}{ds}, \dot{z} \frac{dt}{ds} \right) = \frac{dt}{ds} \dot{\tilde{R}}$

So I guess we need to know $\frac{dt}{ds}$

$\frac{ds}{dt} = \sqrt{c^2 - \dot{x}^2 - \dot{y}^2 - \dot{z}^2} = \sqrt{c^2 - u^2}$

I believe from the inverse function theorem of single-variable calculus we can say

$\frac{dt}{ds} = \frac{1}{\sqrt{c^2 - u^2}}$

Giving us

$\tilde{U} = c \frac{d \tilde{R}}{ds}= c \frac{dt}{ds} \dot{\tilde{R}} = \frac{c}{\sqrt{c^2 - u^2}} \left( c, \dot{x},\dot{y},\dot{z} \right)$Your way (or at least what I perceive to be your way) is more satisfying than naively assuming derivatives are mere quotients of differentials. That said if I am wrong about the inverse function theorem then all of what I typed is pretty much bunk. I think if used (-,+,+,+) I would have gotten the correct answer, but I am still conflicted because the general impression I get from this thread is that signature should not matter.

 

[PeterDonis]

So I guess we need to know $\frac{dt}{ds}$

Yes, and you're almost there. See below.

I think if used (-,+,+,+) I would have gotten the correct answer

No, that's not the issue. The issue is that you don't need to flip around an equation for $ds / dt$ to get an equation for $dt / ds$. You just need to use the fact that the tangent vector you are trying to compute has a unit norm (since we are not considering the case of null curves, only timelike or spacelike ones). Unit norm means unit squared norm, but the sign of the squared norm depends on (a) whether the curve is timelike or spacelike, and (b) the signature you choose.

If we use your chosen signature, and $U$ is timelike, then we have

$$
U^2 = 1 = \left( \frac{dt}{ds} \right)^2 - \left( \frac{dx}{ds} \right)^2 - \left( \frac{dy}{ds} \right)^2 - \left( \frac{dz}{ds} \right)^2
$$

But if we use your chosen signature, and $U$ is spacelike, then we have

$$
U^2 = - 1 = \left( \frac{dt}{ds} \right)^2 - \left( \frac{dx}{ds} \right)^2 - \left( \frac{dy}{ds} \right)^2 - \left( \frac{dz}{ds} \right)^2
$$

Notice how, if we switch signature conventions, the signs of both the LHS and the RHS of each of the above equations are flipped, which is equivalent to changing neither sign; so the above equations are actually independent of the signature.

You already showed how to use the chain rule to express the last three terms in each equation above in terms of $dt / ds$, so the above equations give you two equations for $dt / ds$, one for each type of curve/vector (timelike and spacelike). What do those equations end up looking like?

 

[PhDeezNutz]

Yes, and you're almost there. See below.
No, that's not the issue. The issue is that you don't need to flip around an equation for $ds / dt$ to get an equation for $dt / ds$. You just need to use the fact that the tangent vector you are trying to compute has a unit norm (since we are not considering the case of null curves, only timelike or spacelike ones). Unit norm means unit squared norm, but the sign of the squared norm depends on (a) whether the curve is timelike or spacelike, and (b) the signature you choose.

If we use your chosen signature, and $U$ is timelike, then we have

$$
U^2 = 1 = \left( \frac{dt}{ds} \right)^2 - \left( \frac{dx}{ds} \right)^2 - \left( \frac{dy}{ds} \right)^2 - \left( \frac{dz}{ds} \right)^2
$$

But if we use your chosen signature, and $U$ is spacelike, then we have

$$
U^2 = - 1 = \left( \frac{dt}{ds} \right)^2 - \left( \frac{dx}{ds} \right)^2 - \left( \frac{dy}{ds} \right)^2 - \left( \frac{dz}{ds} \right)^2
$$

Notice how, if we switch signature conventions, the signs of both the LHS and the RHS of each of the above equations are flipped, which is equivalent to changing neither sign; so the above equations are actually independent of the signature.

You already showed how to use the chain rule to express the last three terms in each equation above in terms of $dt / ds$, so the above equations give you two equations for $dt / ds$, one for each type of curve/vector (timelike and spacelike). What do those equations end up looking like?

timelike

$\frac{dt}{ds} = \frac{1}{\sqrt{1-u^2}}$

spacelike

$\frac{dt}{ds} = - \frac{1}{\sqrt{1-u^2}}$

correct?

I would feel more comfortable if that $1$ was a $c^2$. It's probably just a matter of definitions and being off by a constant.

Also here's the question for completeness. 5.4.

 

[PeterDonis]

correct?

Timelike is correct, spacelike is wrong (expression under the square root is negative). Look again at spacelike and be careful about signs.

I would feel more comfortable if that $1$ was a $c^2$.

I used units in which $c = 1$, since that way you don't have to quibble about whether $U$ is a unit vector. In conventional units, you would have $c^2$ in place of $1$ on the LHS of the two equations for the squared norm of $U$ (but the signs would still be the same, as given). But then you have to explain why your tangent vector is called a "unit" vector when its magnitude is $c$ instead of $1$.

 

[PeterDonis]

In conventional units, you would have $c^2$ in place of 11 on the LHS of the two equations for the squared norm of $U$ (but the signs would still be the same, as given).

And you would also have to put a factor of $c^2$ in front of the $dt / ds$ terms on the RHS of those equations. After a while you get very tired of having to insert $c$'s all over the place and you begin to see the advantages of using units where $c = 1$.

 

[PhDeezNutz]

for spacelike

$\left(\frac{dt}{ds}\right)^2 = \frac{-1}{1 - u^2}$

$1-u^2$ is a negative quantity so overall the expression above is positive (as it must be for a quantity squared). We need to reflect that somehow so we factor out a $-1$ in the denominator and cancel.

$\left(\frac{dt}{ds}\right)^2 = \frac{1}{u^2-1} \Rightarrow \frac{dt}{ds} = \frac{1}{\sqrt{u^2 - 1}}$

 

[PeterDonis]

$1-u^2$ is a negative quantity so overall the expression above is positive

Yes.

We need to reflect that somehow so we factor out a $-1$ in the denominator and cancel.

Yes, you've got it.

 

[PeterDonis]

$\left(\frac{dt}{ds}\right)^2 = \frac{1}{u^2-1} \Rightarrow \frac{dt}{ds} = \frac{1}{\sqrt{u^2 - 1}}$

Btw, there is one other implicit assumption that is being made here when going from the squared expression to the expression with the square root. It is also made in the timelike case, but there it doesn't cause any issue while in the spacelike case it can. Can you see what that assumption is?

 

[PhDeezNutz]

Btw, there is one other implicit assumption that is being made here when going from the squared expression to the expression with the square root. It is also made in the timelike case, but there it doesn't cause any issue while in the spacelike case it can. Can you see what that assumption is?

Sorry I cannot. I thought the only assumption we were making was that $u$ is real and that $u^2 > 1$ (or in other units $u^2 > c^2$).

 

[PeterDonis]

How many square roots does a squared expression have?

 

[PhDeezNutz]

How many square roots does a squared expression have?

By that do you mean in order to square a quantity you have to raise its square root to the 4th power?

 

[PeterDonis]

By that do you mean in order to square a quantity you have to raise its square root to the 4th power?

No. I mean if I have a quantity $a^2$, how many square roots does that quantity have? (Hint: the answer is not "one".)

 

[PhDeezNutz]

No. I mean if I have a quantity $a^2$, how many square roots does that quantity have? (Hint: the answer is not "one".)

2 so I arbitrarily took the positive root.

 

[PeterDonis]

I arbitrarily took the positive root.

Exactly. This is the implicit assumption I referred to. Can you see any problem with it?

More precisely: can you see any problem that there might be with it when $U$ is spacelike, that wouldn't be a problem if $U$ is timelike? (Hint: your implicit assumption amounts to assuming that $dt / ds$ is positive. What does that actually mean? Think about $s$ as the parameter describing arc length along the curve from some chosen "zero" point.)

 

[PeterDonis]

there is one other implicit assumption

Just to be even more irritating , there is still another implicit assumption in the derivation, this one being made in the step where $dt / ds$ is factored out of the other terms by applying the chain rule. This assumption is guaranteed to be true for a timelike vector, which is why it's never talked about, but it is not guaranteed to be true for a spacelike vector. What is it?

 

[PhDeezNutz]

@PeterDonis you are anything but irritating, you have helped me a lot. I also glossed over some of the articles on your website and I enjoyed them.

would it be alright if we rewinded a little?

The constraint that $U^2 = - 1$ made everything fall into place to the point where I think the problem statement may be flawed.

If we take a look at the problem statement it asks us to derive $\tilde{U}$ before proving that $\tilde{U} \cdot \tilde{U} = - c^2$. With your approach (at least it seems to me) we imposed that $\tilde{U} \cdot \tilde{U} = - c^2$ on the way to deriving $\tilde{U}$ instead of the other way around. But I really do not see any other way to do it, therefore I think the problem statement may be flawed.
Here it is again,

 

[PeterDonis]

you are anything but irritating, you have helped me a lot

I'm glad I've helped. The "irritating" part was tongue in cheek, that's why the wink emoji was there.

If we take a look at the problem statement it asks us to derive $\tilde{U}$ before proving that ~$\tilde{U} \cdot \tilde{U} = - c^2$. With your approach (at least it seems to me) we imposed that $\tilde{U} \cdot \tilde{U} = - c^2$ on the way to deriving $\tilde{U}$ instead of the other way around.

Hm, yes, that's an interesting point. I'm not sure if the intent of the problem statement was to derive those things in exactly that order (textbooks vary on things like this). I'll have to think about it some more to see if I think there's a way to do it in exactly the order given in the problem.

 

[PhDeezNutz]

I'm glad I've helped. The "irritating" part was tongue in cheek, that's why the wink emoji was there.
Hm, yes, that's an interesting point. I'm not sure if the intent of the problem statement was to derive those things in exactly that order (textbooks vary on things like this). I'll have to think about it some more to see if I think there's a way to do it in exactly the order given in the problem.

I very much look forward to hearing what you come up with. Thank you again.

 

[PeterDonis]

we imposed that $\tilde{U} \cdot \tilde{U} = - c^2$

We did just state that outright without proving it, but it's straightforward to prove it from the problem statement if we don't care in what order we prove the things the problem asks us to prove. If we have $\mathbf{U} = c \ \text{d}\mathbf{R} / \text{d}s$, then, assuming the signature convention we have been using, we will have $\mathbf{U}^2 = - c^2 ( \text{d}\mathbf{R} / \text{d}s )^2$. But $\text{d}\mathbf{R} / \text{d}s$ is a unit vector (physically, this is just saying that the rate of change of arc length along the curve, with respect to the curve parameter, is $1$--which it must be because we chose arc length to be the curve parameter), so we have $( \text{d}\mathbf{R} / \text{d}s )^2 = 1$ and hence $\mathbf{U}^2 = - c^2$.

The question is whether (a) you can prove $\mathbf{U} = ( u^2 / c^2 - 1 )^{- 1/2} ( \mathbf{u}, c )$ without using $\mathbf{U}^2 = - c^2$, and (b) you can prove $\mathbf{U}^2 = - c^2$ from $\mathbf{U} = ( u^2 / c^2 - 1 )^{- 1/2} ( \mathbf{u}, c )$, without using any of the above. I'm not sure that (a) is possible, but I want to think about it some more. It looks to me like (b) is straightforward, but without (a) it's not enough by itself if we insist on proving what the problem asks us to prove in the exact order given.

 

[PhDeezNutz]

@PeterDonis I think your argument that $\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \pm 1$ makes a lot of sense. I think that is a perfectly reasonable way to get around imposing $U^2 = \pm c^2$ from the get go. Although the two conclusions go hand in hand so one could argue that I'm starting from the conclusion. Then again we didn't use $U^2 = \pm c^2$ to come to the conclusion $\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \pm 1$.

I'm going to attempt to solve the problem with each signature. My only problem seems to be that when I use $(-,+,+,+)$ I get $U^2 = c^2$ (instead of $U^2 = - c^2$). If I recall correctly this may not actually be a problem; we are considering a space-like vector with signature $(-,+,+,+)$ so it makes sense that $U^2$ should have the same signs as the space components.

Without further ado

Using signature (-,+,+,+)

$\tilde{R} = \left( ct, x, y, z \right)$

$\tilde{U} = c \frac{d \tilde{R}}{ds}$

$\frac{d \tilde{R}}{ds} = \left( \frac{dt}{ds} \right) \left( c , \dot{x} , \dot{y} , \dot{z} \right)$

So what is $\frac{dt}{ds}$ equal to?

consider

$\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \left(\frac{dt}{ds}\right)^2 \left(u^2 - c^2\right)$

We know that $u^2 \gt c^2$ so the quantity directly above is positive. We also know that the derivative of a curve with respect to its own arc-length has to have magnitude of 1. Therefore,

$\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \left(\frac{dt}{ds}\right)^2 \left(u^2 - c^2\right) = 1 \Rightarrow \frac{dt}{ds} = \frac{1}{\sqrt{u^2 - c^2}}$

$\tilde{U} = c \frac{d \tilde{R}}{ds} = c \left( \frac{dt}{ds} \right) \dot{\tilde{R}} = \frac{c}{\sqrt{u^2 - c^2}} \left( c , \dot{x}, \dot{y}, \dot{z} \right)$

So

$\tilde{U} = \frac{1}{\sqrt{\frac{u^2}{c^2} - 1}} \left( c , \tilde{u} \right)$

As for $\tilde{U} \cdot \tilde{U}$ we have

$\tilde{U} \cdot \tilde{U} = \frac{1}{\left(\frac{u^2}{c^2} -1\right)} \left(u^2 - c^2\right) = c^2$

$\tilde{U} \cdot \tilde{U} = c^2$

Using signature (+,-,-,-)

$\tilde{R} = \left( ct, x, y, z \right)$

$\tilde{U} = c \frac{d \tilde{R}}{ds}$

$\frac{d \tilde{R}}{ds} = \left( \frac{dt}{ds} \right) \left( c , \dot{x} , \dot{y} , \dot{z} \right)$

So what is $\frac{dt}{ds}$ equal to?

consider

$\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \left(\frac{dt}{ds}\right)^2 \left(c^2 - u^2\right)$

We know that $u^2 \gt c^2$ so the quantity directly above is negative. We also know that the derivative of a curve with respect to its own arc-length has to have magnitude of 1. Therefore,

$\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \left(\frac{dt}{ds}\right)^2 \left(c^2 - u^2\right) = -1$

$\left( \frac{dt}{ds} \right)^2 = \frac{(-1)}{(-1)\left(u^2 - c^2 \right)} = \frac{1}{u^2 - c^2} \Rightarrow \frac{dt}{ds} = \frac{1}{\sqrt{u^2 - c^2}}$

$\tilde{U} = c \frac{d \tilde{R}}{ds} = c \left( \frac{dt}{ds} \right) \dot{\tilde{R}} = \frac{c}{\sqrt{u^2 - c^2}} \left( c , \dot{x}, \dot{y}, \dot{z} \right)$

So

$\tilde{U} = \frac{1}{\sqrt{\frac{u^2}{c^2} - 1}} \left( c , \tilde{u} \right)$

As for $\tilde{U} \cdot \tilde{U}$ we have

$\tilde{U} \cdot \tilde{U} = \frac{1}{\left(\frac{u^2}{c^2} -1\right)} \left(c^2 - u^2\right) = -c^2$

$\tilde{U} \cdot \tilde{U} = - c^2$I think proving that $\tilde{U}$ is tangent to the worldline is as simple as noting that $\tilde{U}$ is just a scalar (function) multiple of $\dot{\tilde{R}}$.

 

[PeterDonis]

when I use $(-,+,+,+)$ I get $U^2 = c^2$

That's what you should get, since with that signature a spacelike vector has positive squared norm.

proving that $\tilde{U}$ is tangent to the worldline is as simple as noting that $\tilde{U}$ is just a scalar (function) multiple of $\dot{\tilde{R}}$.

Of course, but the original question didn't ask us to prove that $U$ was tangent to the worldline; in fact it already told us that, by telling us that $U = c \ dR / ds$. The question asked us to prove that $U$ had a specific form in terms of components, and what its squared norm was.

 

[PeterDonis]

I think your argument that $\frac{d \tilde{R}}{ds} \cdot \frac{d \tilde{R}}{ds} = \pm 1$ makes a lot of sense.

I actually was arguing that $dR / ds = 1$, without the possibility of the minus sign. Your putting in the possibility of the minus sign brings me back to the question I asked in post #49, which nobody has responded to.

Focus on this step:

$$
\frac{d \tilde{R}}{ds} = \left( \frac{dt}{ds} \right) \left( c , \dot{x} , \dot{y} , \dot{z} \right)
$$

If you're saying that $dR / ds = \pm 1$ (i.e., either sign is possible), then you are saying that the RHS of the equation just above can have either sign. What does that mean?

It can't mean the individual components flip sign; their signs are fixed by the nature of the curve, i.e., the sign of $d(c dt) / dt = c$ is fixed at positive, and the signs of $\dot{x}$, $\dot{y}$, and $\dot{z}$ are fixed by the curve. So the only way to have $dR / ds$ be negative in the above is to have $dt / ds$ be negative.

Is this just another sign convention choice we can make? Not quite. What is $dt / ds$? It's the rate of change of the $t$ coordinate, with respect to the curve parameter. For timelike curves, the usual conventions make this always positive: $t$ increases towards the future, and the curve parameter $s$, which in this case is just proper time $\tau$, also increases towards the future, so $dt / ds = dt / d\tau$ is always positive.

But for a spacelike curve, this is not always true. It's perfectly possible to have a spacelike curve whose curve parameter $s$ increases with decreasing $t$. In fact, it's even possible to have a spacelike curve whose curve parameter $s$ goes from $- \infty$ to $\infty$ without $t$ changing at all! (Can you see how?) For this case, we would have $dt / ds = 0$ and the entire derivation breaks down.

Furthermore, even if $dt / ds$ is positive for a spacelike curve in one frame, it won't be in all frames if we hold the parameterization constant. We can always find a frame in which $dt / ds$ is negative (and indeed we can always find a frame in which it is zero).

In other words, the sign of $dt / ds$, and therefore the sign of $dR / ds$, is not an invariant; it changes from frame to frame. As far as I can tell, this fact is ignored in the problem statement, which, as far as I can tell, expects you to assume that $dt / ds$ and hence $dR / ds$ is positive, just as you would for a timelike curve. But in general, as we've shown above, the sign of $dt / ds$ and hence of $dR / ds$ depends on the choice of frame and the direction of parameterization of the curve (i.e., in which direction along the curve $s$ increases). In order for the derivation to hold in any frame, we have to assume we are changing the parameterization of the curve as we change frames, in order to keep $dt / ds$ and hence $dR / ds$ positive. (And even that won't help in a frame where $dt / ds = 0$.)

 

[vanhees71]

That's finally the point! The temporal ordering of space-like separated events is not an invariant under the part of the Poincare group that is connected continuously with the identity transformation, which is the proper orthochronous Poincare group, and which is the group that must be a symmetry of any (fundamental) relativistic model. For time-like curves, if you choose $\mathrm{d}t/\mathrm{d} s>0$ in one inertial frame it's also fulfilled in any other inertial frame connected by this one by a proper orthochronous Poincare transformation, but that's not the case for space-like curves.

That makes classical tachyons at least very complicated, if not impossible to implement at all in any theory. It's also the reason for the tension with the microcausality condition in any QFT with tachyonic fields/irreps. of the proper orthochronous Poincare group. I've quoted one paper above, where it could however be shown that there's at least one example for a model of tachyonic Majorana fermions, where this problem is circumvented.

 

[PhDeezNutz]

I actually was arguing that $dR / ds = 1$, without the possibility of the minus sign. Your putting in the possibility of the minus sign brings me back to the question I asked in post #49, which nobody has responded to.

Focus on this step:
If you're saying that $dR / ds = \pm 1$ (i.e., either sign is possible), then you are saying that the RHS of the equation just above can have either sign. What does that mean?

It can't mean the individual components flip sign; their signs are fixed by the nature of the curve, i.e., the sign of $d(c dt) / dt = c$ is fixed at positive, and the signs of $\dot{x}$, $\dot{y}$, and $\dot{z}$ are fixed by the curve. So the only way to have $dR / ds$ be negative in the above is to have $dt / ds$ be negative.

As for a

Is this just another sign convention choice we can make? Not quite. What is $dt / ds$? It's the rate of change of the $t$ coordinate, with respect to the curve parameter. For timelike curves, the usual conventions make this always positive: $t$ increases towards the future, and the curve parameter $s$, which in this case is just proper time $\tau$, also increases towards the future, so $dt / ds = dt / d\tau$ is always positive.

But for a spacelike curve, this is not always true. It's perfectly possible to have a spacelike curve whose curve parameter $s$ increases with decreasing $t$. In fact, it's even possible to have a spacelike curve whose curve parameter $s$ goes from $- \infty$ to $\infty$ without $t$ changing at all! (Can you see how?) For this case, we would have $dt / ds = 0$ and the entire derivation breaks down.

Furthermore, even if $dt / ds$ is positive for a spacelike curve in one frame, it won't be in all frames if we hold the parameterization constant. We can always find a frame in which $dt / ds$ is negative (and indeed we can always find a frame in which it is zero).

In other words, the sign of $dt / ds$, and therefore the sign of $dR / ds$, is not an invariant; it changes from frame to frame. As far as I can tell, this fact is ignored in the problem statement, which, as far as I can tell, expects you to assume that $dt / ds$ and hence $dR / ds$ is positive, just as you would for a timelike curve. But in general, as we've shown above, the sign of $dt / ds$ and hence of $dR / ds$ depends on the choice of frame and the direction of parameterization of the curve (i.e., in which direction along the curve $s$ increases). In order for the derivation to hold in any frame, we have to assume we are changing the parameterization of the curve as we change frames, in order to keep $dt / ds$ and hence $dR / ds$ positive. (And even that won't help in a frame where $dt / ds = 0$.)

I am unable to figure out how to break up your post and respond to individual parts.

I believe if I am saying $\frac{d\tilde{R}}{ds} = \pm 1$ in the equation $\frac{d\tilde{R}}{ds} = \left( \frac{dt}{ds} \right) \left(c , \dot{x} , \dot{y}, \dot{z} \right)$ that $\frac{d\tilde{R}}{ds}$ is either parallel or anti-parallel to the trajectory. It should be parallel to the trajectory so we have to restrict ourselves $\frac{d\tilde{R}}{ds}$ = 1.

As for a space-like curve whose curve parameter $s$ goes from $-\infty$ to $+\infty$ without $t$ changing at all I don't know. It would correspond to a horizontal line in a Minkowski diagram but I don't know how to interpret that. I know how to interpret a vertical line in a Minkowski diagram; just a particle that stays in the same place for all time.

So I take it there are a lot of implicit assumptions in the problem statement. I never thought about that; $\frac{dt}{ds}$ changes depending on which frame we are in and can be either positive or negative. I think if the particle is moving towards us $\frac{dt}{ds}$ is negative and away from us $\frac{dt}{ds}$ is positive.

That said I think I found a way around using $\frac{dt}{ds}$. We've already established that

1) In order for $U^2 = - c^2$ for a space-like curve we must use signature $(+,-,-,-)$

2) $\frac{d\tilde{R}}{ds}$ must be a unit vector parallel to $\dot{\tilde{R}}$

A key formula is equation $(5.13)$ from Rindler's book; the magnitude of a 4-vector is $A = \left|A^2 \right|^{\frac{1}{2}} \geq 0$. And as @robphy pointed out.

$\tilde{U} = c \frac{d \tilde{R}}{ds}$

$\frac{d \tilde{R}}{ds}$ is a unit vector. A unit vector is a unit vector regardless of which parameter we use for the curve so we might as well use $t$.

$\frac{d\tilde{R}}{ds} = \frac{1}{\left|\dot{\tilde{R}} \right|} \dot{\tilde{R}} = \frac{1}{\left| \dot{\tilde{R}} \right|} \left( c , \dot{x}, \dot{y} , \dot{z} \right)$

$\left| \dot{\tilde{R}} \right| = \left| \dot{\tilde{R}} \cdot \dot{\tilde{R}} \right| = \left| c^2 - u^2 \right|^{\frac{1}{2}} = \sqrt{u^2 - c^2}$
$u \gt c$

$\frac{d\tilde{R}}{ds} = \frac{1}{\sqrt{u^2 - c^2}} \left( c , \dot{x} , \dot{y} , \dot{z} \right)$

$\tilde{U} = c \frac{d\tilde{R}}{ds} = \frac{c}{\sqrt{u^2 - c^2}} \left( c , \tilde{u} \right) = \frac{1}{\sqrt{\frac{u^2}{c^2} - 1}} \left( c , \tilde{u} \right)$
$\tilde{U} \cdot \tilde{U} = \frac{c^2}{\frac{u^2}{c^2} - 1} = - c^2$

 

[PeterDonis]

I am unable to figure out how to break up your post and respond to individual parts.

Highlight just the part you want to respond to, then click the "Reply" button that pops up.

It should be parallel to the trajectory

What does "parallel" mean? I think if you try to answer that, you will find that it is not obvious for a spacelike vector that $dR / ds = 1$ must always be true.

As for a space-like curve whose curve parameter $s$ goes from $-\infty$ to $+\infty$ without $t$ changing at all I don't know.

Consider this curve: $t(s) = 0$, $x(s) = s$, $y(s) = 0$, $z(s) = 0$.

It would correspond to a horizontal line in a Minkowski diagram but I don't know how to interpret that.

Think harder. See above. Or just ask yourself: what is the $x$ axis in a Minkowski diagram?

I know how to interpret a vertical line in a Minkowski diagram; just a particle that stays in the same place for all time.

Ok, just exchange "time" and "place" (and start with "for all times" in the quote above). What do you get?

Remember, we're talking about spacelike vectors now, so you have to be very careful not to use intuitive ideas that are only valid for timelike vectors.

I think if the particle is moving towards us $\frac{dt}{ds}$ is negative and away from us $\frac{dt}{ds}$ is positive.

No. See above.

A unit vector is a unit vector regardless of which parameter we use for the curve

This is not a valid statement. $dR / ds$ is the derivative with respect to the curve parameter so its magnitude cannot possibly be independent of your choice of curve parameter.

so we might as well use $t$.

No, you can't, because $t$ is not arc length along the curve. (Also this would break down anyway in the case of a curve for which $t$ is constant; I gave an example of one above.)

 

[PhDeezNutz]

What does "parallel" mean? I think if you try to answer that, you will find that it is not obvious for a spacelike vector that dR/ds=1dR/ds=1dR / ds = 1 must always be true.

That's a tough question. I would think that parallel means "same sign". Hard to know because this Minkowski space just seems so different from Euclidean space.

Consider this curve: t(s)=0t(s)=0t(s) = 0, x(s)=sx(s)=sx(s) = s, y(s)=0y(s)=0y(s) = 0, z(s)=0z(s)=0z(s) = 0.

A point that is stationary, and time is always 0. Would that be the origin of a particle's rest frame?

Think harder. See above. Or just ask yourself: what is the xxx axis in a Minkowski diagram?

the x-axis is literally the position correct? So time stays the same while position changes. I don't know how to interpret that. Again maybe the origin of a rest frame of a particle?

No. See above.

I'm not trying to be obtuse. I don't see any other way how $\frac{dt}{ds}$ can change with frames without considering relative motion. Thanks for being patient with me, this subject is very new to me.

No, you can't, because ttt is not arc length along the curve. (Also this would break down anyway in the case of a curve for which ttt is constant; I gave an example of one above.)

This is not a valid statement. dR/dsdR/dsdR / ds is the derivative with respect to the curve parameter so its magnitude cannot possibly be independent of your choice of curve parameter.

Would the following be valid?

$\left|\frac{d\tilde{R}}{ds} \cdot \frac{d\tilde{R}}{ds}\right|= 1$

According to Rindler for a general 4-vector $\left| \tilde{A} \right| = \left| \tilde{A} \cdot \tilde{A} \right|^{\frac{1}{2}} = \left| A^2 \right| ^ {\frac{1}{2}} \gt 0$. Hopefully that definition is not limited to time like vectors.

We know $\frac{d\tilde{R}}{ds} = \left(\frac{dt}{ds}\right) \dot{\tilde{R}}$

So using the first equation

$\left|\left(\frac{dt}{ds}\right)^2 \left( c^2 - u^2 \right) \right|= 1$

We know $u \gt c$

$\left|-\left(\frac{dt}{ds}\right)^2 \left(u^2 - c^2\right) \right| = 1$

$\left(\frac{dt}{ds}\right)^2 \left( u^2 - c^2 \right) = 1$

$\frac{dt}{ds} = \frac{1}{\sqrt{u^2 - c^2}}$

$\tilde{U} = c \left(\frac{dt}{ds} \right) \left( c , \dot{x} , \dot{y} , \dot{z} \right)$

$\tilde{U} = \frac{c}{\sqrt{u^2 - c^2}} \left( c , \tilde{u} \right) = \frac{1}{\sqrt{\frac{u^2}{c^2} - 1}} \left( c, \tilde{u} \right)$