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

[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 $u \gt c$

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

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

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

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

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

 

[PeterDonis]

I would think that parallel means "same sign".

Same sign of what?

A point that is stationary

I didn't describe a point, I described a curve. If you want to view it as a curve describing the "motion" of something, that something is certainly not staying at the same "point" (it goes from $x = - \infty$ to $x = \infty$), so "stationary" does not seem like a good word to describe it.

the x-axis is literally the position correct?

It's a coordinate line describing the range of possible positions in the $x$ direction.

time stays the same while position changes.

More precisely, coordinate time stays the same while position changes.

I don't know how to interpret that.

Well, we're talking about hypothetical objects that can "move" on spacelike trajectories. This particular spacelike trajectory covers every possible position on the $x$ axis in zero coordinate time. What does that suggest? (For example, what coordinate "speed" would you say such an object has?)

I don't see any other way how $\frac{dt}{ds}$ can change with frames without considering relative motion.

First, what is $dt / ds$ for the curve I described, where the only coordinate that changes as a function of $s$ is $x$?

Second, what does that same curve look like if we Lorentz transform to a different frame? I.e., what will the four functions $t'(s)$, $x'(s)$, $y'(s)$, and $z'(s)$ look like? Assume we are doing a Lorentz boost in the $x$ direction (which means $y' = y$ and $z' = z$, so only the $t'(s)$ and $x'(s)$ functions are of interest).

Third, given the answers above, will $dt / ds$ stay the same when we transform to the new frame, or will it change?

 

[PeterDonis]

Hopefully that definition is not limited to time like vectors.

No, it's valid in general since Rindler is taking the absolute value, which means this process discards the information about whether the vector is timelike or spacelike. (Note, however, that his definition as you give cannot possibly apply to null vectors, since their magnitude is zero.)

However, the fact that this definition throws away the information about whether the vector is timelike or spacelike means it can't possibly be used to determine a different formula for spacelike vectors than for timelike vectors. But that is what you are trying to use it to do.