Question about the Minkowski diagram

[Leepappas]

TL;DR Summary

Simple case of two rulers moving past one another in constant relative motion both in inertial reference frames.

In the case of two rulers of equal rest length moving past one another ibex posted the minkowski diagram and I have a question about the one diagram that follows the following comment by him:

Note that, in the primed frame, "State 2" is after "State 3"! This is the relativity of simultaneity striking - the definitions of the states are anchored to spacelike separated events, and the ordering of spacelike separated events is frame-dependent.

Again we can transform this last diagram into the primed frame:

Now I wrote the formula for the equation of the line that goes from State two to State One in the primed frame and I got an interesting result that I want to share. Ibex can back me up on this.

The lower thick red line intersects the A' axis at
$\frac {L_0}{v}$
And it's slope is found from the lorentz transformation. For the unprimed frame we have
$t = \frac{t'}{\gamma}+\frac{vx}{c^2}$
So the slope is $\frac{v}{c^2}$
So for the primed frame the slope of a thick red line is
$\frac {-v}{c^2}$
So the equation of the lower thick red line is
$t' = \frac{-vx'}{c^2} + \frac{L_0}{v}$ (1)
Where x' is the distance from the B' axis to the lower thick red line. When $t'=\frac {L}{v}$ focus on the triple intersection point where the top thick red line intersects the B prime axis and also meets the thin Blue line representing state three.
Now I'm focused on the time in the prime frame corresponding to state three. For that state we have
$t' =\frac{L}{v}$ and $x'=L$

Now I checked and equation (1) can be solved explicitly for v. My question is what is v?

Thank you

Here's the attachment of his diagram

 

[PeterDonis]

ibex posted the minkowski diagram

Do you mean this post?

https://www.physicsforums.com/threa...ativity-self-contradicts.1055906/post-6935759

 

[PeterDonis]

I'll try to attach ibex's Minkowski diagram.

You don't have to, just give a link to the post that it's in.

 

[Leepappas]

Yes that's the one that I'm talking about. The fourth diagram down.

 

[Ibix]

The slope of the thick lines is $1/v$, as you can easily verify by looking at the stationary lines ($v=0$) which are vertical (gradient is infinite). Or by noting that the vertical axis is $t'$ and the horizontal is $x'$, so the gradient of a straight line is $\Delta t'/\Delta x'=1/v$. (Edit: this turns out to have been unclear. In general, the slope of a line is the inverse of its velocity, but that velocity is not necessarily the frame velocity $v$. It would have been better to write $\Delta t'/\Delta x'=1/u$ for a general line with velocity $u$, and then to note that $u=\pm v$ for the thick lines in this specific scenario. Note that not all lines represent physical objects and it is easy to draw lines for which $|u|>c$.)

Your problem seems to be that you are attempting to interpret the equation of a line of constant $t$ or $t'$, which describe (respectively) the red and blue fine lines in my diagram, as relevant to the thick lines, which are actually lines of constant $x$ (red) or $x'$ (blue). It isn't relevant, and your contradiction this time is in your mistaken assumption that it is.

 

[Ibix]

Given @Leepappas' propensity for applying special case maths to cases other than the special case and then (unsurprisingly) finding contradictions, I thought I'd set out a bit more maths.

All of the Minkowski diagrams drawn in the post linked in #2 are drawn using one of two frames. One frame (the "unprimed frame" because the coordinates have no primes) has its axes labelled $x$ and $t$ and the other (the "primed frame") has them labelled $x'$ and $t'$. If you find a point (called an event in relativity, because it's a place and a time) on a diagram using the unprimed frame you can read off its $x$ and $t$ coordinates. Then you can apply the Lorentz transforms to get the coordinates of the same event on a diagram that uses the primed frame:$$\begin{eqnarray}
x'&=&\gamma\left(x-vt\right)\\
t'&=&\gamma\left(t-\frac{v}{c^2}x\right)
\end{eqnarray}$$where $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}$ is the Lorentz gamma factor. To find where some event in a diagram using the primed frame will be in one using the unprimed frame you use the inverse Lorentz transforms:$$\begin{eqnarray}
x&=&\gamma\left(x'+vt'\right)\\
t&=&\gamma\left(t'+\frac{v}{c^2}x'\right)
\end{eqnarray}$$
You can use these equations to derive the equations of the thick lines in one frame given their equation in the other. The general approach is to pick two events on one line and read off their coordinates, use the forward ((1) and (2)) or inverse ((3) and (4)) Lorentz transforms to get the coordinates in the other frame, and then solve for a straight line through these new coordinates (the Lorentz transforms transform straight lines to straight lines, so you don't need to transform every event on the line explicitly, although you can do so).

However, in the case of the thick red lines, it's actually easier to transform all of the events. You simply note that they are lines where $x=x_0$, where $x_0$ is a constant (0 for the left hand line, $L$ for the right) for all time. So from (3) we can immediately see that $x_0=\gamma(x'+vt')$ which we can rearrange to $t'=\frac{x_0}{v\gamma}-\frac 1vx'$. So we can see that the lines in the primed frame have slope $1/v$, as I said. You can find where they intersect the $x'$ axis by setting $t'=0$ and $x_0$ to 0 and $L$, and you will find that the locations differ by $L/\gamma$, which is length contraction. You can find where they cross the $t'$ axis by setting $x'=0$ and $x_0$ to 0 and $L$, and you will find they differ by $L/\gamma v$ because that's the time a point travelling at $v$ takes to travel the contracted length of the rod.

 

[Leepappas]

The slope of the thick lines is $1/v$, as you can easily verify by looking at the stationary lines ($v=0$) which are vertical (gradient is infinite). Or by noting that the vertical axis is $t'$ and the horizontal is $x'$, so the gradient of a straight line is $\Delta t'/\Delta x'=1/v$.

Your problem seems to be that you are attempting to interpret the equation of a line of constant $t$ or $t'$, which describe (respectively) the red and blue fine lines in my diagram, as relevant to the thick lines, which are actually lines of constant $x$ (red) or $x'$ (blue). It isn't relevant, and your contradiction this time is in your mistaken assumption that it is.

If ∆t'/∆x' = 1/v then the slope is positive because v is speed, however the slope is negative in the primed frame. Now I don't understand what the problem is with using the Lorentz transformation this time hopefully you can help me out. what is wrong with saying the following:
$t= \frac{t'}{\gamma} + \frac{vx}{c^2}$
So for constant t' the y-intercept is $\frac{t'}{\gamma}$ and the slope is $\frac{v}{c^2}$.
Then applying your argument we get
$\frac {1}{v}= \frac {v}{c^2}$ therefore
$v=c$ but this is impossible in special relativity.

Why can't you use the slope intercept form of the equation of a straight line and apply it to one of the diagonal lines in the primed frame and then use your argument to get an alternative expression for the slope then equate them and get a contradiction. I used the Lorentz transformation which everybody insisted that I do. So what's my problem?

 

[Ibix]

If ∆t'/∆x' = 1/v then the slope is positive because v is speed, however the slope is negative in the primed frame.

No, $v$ is velocity and is a signed quantity. That's why the slope is negative in this case where the sense of the motion is towards negative $x$.

Now I don't understand what the problem is with using the Lorentz transformation this time [snip snip]
So for constant t'

The problem is that $t'$ is not constant along the thick lines. It's constant along the fine lines that represent a "now" for the primed frame, but the thick lines are the positions of the ends of the rods. You can't have constant time along those - that would imply they exist everywhere but only at a specific time. That's a fine definition of "now", but it is not how objects behave.

Why can't you use the slope intercept form of the equation of a straight line and apply it to one of the diagonal lines in the primed frame and then use your argument to get an alternative expression for the slope then equate them

You can. You just need to use the equation of the line you're actually looking at.

 

[Vanadium 50]

So what's my problem?

Previously answered:

propensity for applying special case maths to cases other than the special case

 

[Ibix]

what is wrong with saying the following:
$t= \frac{t'}{\gamma} + \frac{vx}{c^2}$

The other issue is that $v$ is doing double duty here. It's being used as both the velocity of some worldline and as the velocity of one frame with respect to the other. Since we've mostly been talking about objects that are at rest in one frame or the other that mostly doesn't matter, but it may be what is confusing you.

Let's call the velocity of some object $u$ in some frame. Then the slope of its worldline in that frame will be $1/u$. In the case of objects that are at rest in one frame, their velocity in the other frame will be $u=\pm v$. That also resolves your issue above, since you'll end up with $u=c^2/v$ for the "now" line of the other frame. Yes, this is greater than $c$. That does not matter because "now" is not a physical object, it's just a line I drew on the graph.

 

[Leepappas]

The slope of the thick lines is $1/v$, as you can easily verify by looking at the stationary lines ($v=0$) which are vertical (gradient is infinite). Or by noting that the vertical axis is $t'$ and the horizontal is $x'$, so the gradient of a straight line is $\Delta t'/\Delta x'=1/v$.

Your problem seems to be that you are attempting to interpret the equation of a line of constant $t$ or $t'$, which describe (respectively) the red and blue fine lines in my diagram, as relevant to the thick lines, which are actually lines of constant $x$ (red) or $x'$ (blue). It isn't relevant, and your contradiction this time is in your mistaken assumption that it is.

They are relevant or rather related.
For a constant t' the horizontal line through that t prime coordinate intersects the lower thick red line at the point x'. As t' varies so does x' according to the Lorentz coordinate transformations. The formula for this equation of the lower thick red line is:
$t'= \frac{L_0}{v} - \frac{v(L_0+x')}{c^2 }$
That's the right equation of the straight line. as you can see as t prime varies so does x prime. Check to see that's the right equation.

 

[Dale]

If ∆t'/∆x' = 1/v then the slope is positive because v is speed, however the slope is negative in the primed frame.

$v$ is velocity. The relative velocity is positive in one frame and negative in the other, in the standard configuration. The negative slope is correct.

what is wrong with saying the following:
t=t′γ+vxc2

You can say it but it isn't really relevant to that diagram. The diagram is in the primed frame, so you need to use equations with both $x'$ and $t'$. This equation contains $x$ and $t$ so it is an equation describing things from the unprimed frame. There is nothing wrong with that, but it corresponds to a different diagram.

I am not sure what the point is of this thread. You are just throwing symbols around with no thought and unsurprisingly getting problems. That is not an issue with the formulas nor with @Ibix diagrams.

 

[Leepappas]

$v$ is velocity. The relative velocity is positive in one frame and negative in the other, in the standard configuration. The negative slope is correct.You can say it but it isn't really relevant to that diagram. The diagram is in the primed frame, so you need to use equations with both $x'$ and $t'$. This equation contains $x$ and $t$ so it is an equation describing things from the unprimed frame. There is nothing wrong with that, but it corresponds to a different diagram.

I am not sure what the point is of this thread. You are just throwing symbols around with no thought and unsurprisingly getting problems. That is not an issue with the formulas nor with @Ibix diagrams.

The point is to solve a quadratic equation in ∆t, to throw out the negative root because amount of time must be strictly positive, and solve for the speed v.

 

[Dale]

The point is to solve a quadratic equation in ∆t, to throw out the negative root because amount of time must be strictly positive, and solve for the speed v.

There is no quadratic equation, the Lorentz transform is linear.

The $v$ in the Lorentz transform is arbitrary. The Lorentz transforms work for any $-c<v<c$. It is not even necessary that the $v$ in the Lorentz transform corresponds to the relative velocity between the rods, there is no requirement to use a frame where either rod is at rest.

You don't solve the Lorentz transform for $v$, you take some unprimed coordinates $(t,x)$ and use $v$ to transform the unprimed coordinates into primed coordinates $(t',x')$

Edit: I should clarify. with the Lorentz transform we have two equations in five variable. If you set any three as known then you can solve for the remaining two unknowns. Usually the three knowns are $t$, $x$, and $v$, and you solve for $t'$ and $x'$. But you could pick any three and solve for the other two. So if you want to solve for $v$ then you need any three of $x$, $y$, $x'$, and $y'$.

 

[Ibix]

The point is to solve a quadratic equation in ∆t, to throw out the negative root because amount of time must be strictly positive, and solve for the speed v.

You need to re-read #10. The problem is that I re-used $v$ as the speed of the worldline as well as the speed of the frame. Up to sign, that works out for the physical objects in this scenario because they're stationary in one frame and moving at speed $\pm v$ in the other, but you aren't using the equation of motion of any physical object. Again, you are using maths that works in a specific case in a more general setting.

Let the speed of the frame be $v$ and we will define the "speed" of a line (worldline or otherwise) as $u$, where $1/u=\Delta t/\Delta x$ is the gradient of the line. That makes your "quadratic equation" turn into $u=c^2/v$. That is not quadratic, but does give $|u|>c$. This is not surprising because the line whose gradient you are studying is not a physical object. It's just the line a different frame defines as "now". You can call $u$ its velocity if you like, but nothing is travelling with that speed.