Quick calculation check please

[name123]

Why don't you work out the actual numbers, as I have repeatedly suggested? I have repeatedly given you the solution of the problem. All you have to do is work out the numbers.

Well, well I've laid out the numbers in the scenario, but I've not done the calculation with x and y before. And if others are reading it, maybe it'd be the same for them. Could you do it please?

 

[PeterDonis]

Could you do it please?

No, but I'll give the Lorentz transformation equations that apply. If the relative velocity between the frames is $v$ in the $x$ direction and $w$ in the $y$ direction, then the transformation from the unprimed (A-Team) frame to the primed (B-Team) frame is

$$
t' = \gamma \left( t - v x \right)
$$
$$
x' = \gamma \left( x - v t \right)
$$
$$
y' = y - w t
$$

where $\gamma = 1 / \sqrt{1 - v^2}$ (we assume that $w$ is too small to affect length contraction or time dilation), and we leave out the $z$ coordinate since there is no relative motion in that direction.

(Note that this transformation is approximate; it is not valid if $w$ is not small. That case is considerably more complicated mathematically, and you don't need that extra complication to solve this problem.)

So if you define coordinates for all the relevant events in the A-Team frame, you can use the above transformation to get the coordinates of those events in the B-Team frame. Answers to all the questions you want to ask can then be read off of those coordinate values.

 

[name123]

No, but I'll give the Lorentz transformation equations that apply. If the relative velocity between the frames is $v$ in the $x$ direction and $w$ in the $y$ direction, then the transformation from the unprimed (A-Team) frame to the primed (B-Team) frame is

$$
t' = \gamma \left( t - v x \right)
$$
$$
x' = \gamma \left( x - v t \right)
$$
$$
y' = y - w t
$$

where $\gamma = 1 / \sqrt{1 - v^2}$ (we assume that $w$ is too small to affect length contraction or time dilation), and we leave out the $z$ coordinate since there is no relative motion in that direction.

(Note that this transformation is approximate; it is not valid if $w$ is not small. That case is considerably more complicated mathematically, and you don't need that extra complication to solve this problem.)

So if you define coordinates for all the relevant events in the A-Team frame, you can use the above transformation to get the coordinates of those events in the B-Team frame. Answers to all the questions you want to ask can then be read off of those coordinate values.

I thought the gamma equation was $\gamma = 1 / \sqrt{1 - (v^2/c^2)}$ was that a mistake?

 

[PeterDonis]

I thought the gamma equation was $\gamma = 1 / \sqrt{1 - (v^2/c^2)}$ was that a mistake?

I am using units in which $c = 1$. So $v = 0.1$ in these units (given your specification of the problem). The easiest units of time and distance, given your numbers, are probably seconds and light-seconds, so 29,979,245.8 meters is a distance of 0.1 (i.e., 0.1 light-seconds).

 

[name123]

Ok, so I had a try at doing the calculations.

So with v = 29,979,245.8 and w = 0.00000001 gamma is roughly 1.0050378.

And at t = 0 the x coordinates are as follows:
A1: x = 0
A2: x = 29,979,245.8

B1: x = 50,000
B2: x = 29878973.1

According to the B-Team the time and coordinates are roughly as follows:
A1: t' = 0 x' = 0
A2: t' = -0.01005038 x' = 30,130,275.70

B1: t' = -0.00001676 x' = 50,251.89
B2: t' = -0.01001676 x' = 30,029,497.85

It seemed to me that from the B-Team's perspective at t' = -.006
A1: x' = 179,875.47
A2: x' = 30,008,848.42

B1: x' = 50,251.89
B2: x' = 30,029,497.85

So A-Team is within the B-Team bounds, and it seems to me that there is no time according to the B-Team when: B1.x' > A1.x' AND B2.x' < A2.x' is true from their perspective. I presume I've done it wrong. Taking into account the Y distance didn't seem to make any significant difference (as it was so small).

 

[PeterDonis]

I had a try at doing the calculations.

On re-checking the approximation I gave in my previous post, I realized that it's not good enough; there are additional terms that have to be taken into account. (Briefly, I was assuming that all of the corrections were second order in $w$, but I was wrong; there are correction terms that are first order in $w$ and therefore have to be included.)

A better approximation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = \gamma \left( x - v t \right) + \left( \gamma - 1 \right) \frac{w}{v} y
$$

$$
y' = y - \gamma w t + \left( \gamma - 1 \right) \frac{w}{v} x
$$

Note that even this might not be good enough, depending on how small you pick $w$ and how many significant figures you look at. The fully correct transformation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = - \gamma v t + \left[ 1 + \left( \gamma - 1 \right) \frac{v^2}{v^2 + w^2} \right] x + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} y
$$

$$
y' = - \gamma w t + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} x + \left[ 1 + \left( \gamma - 1 \right) \frac{w^2}{v^2 + w^2} \right] y
$$

where now we are using $\gamma = 1 / \sqrt{1 - v^2 - w^2}$ so we take both speeds into account (and remember units are always such that $c = 1$).

Sorry for the mixup on my part.

Taking into account the Y distance didn't seem to make any significant difference (as it was so small).

If you leave out the motion in the y direction, you leave out the critical feature of the problem, because if you only look at the "fit" in the x direction, the A-Team will not fit. You have to look at the y coordinates. What you should be checking for is the path in the x'-y' plane of the two A-Team poles, and where each pole passes relative to the two B-Team members (who are at rest in the B-Team frame). You can't figure that out just from looking at x coordinates.

I also suggest picking a value for w that doesn't require you to do calculations accurate to 30 or more significant figures in order to see actual variation in the y coordinates. Try, for example, a value of w that is 1/100 or 1/1000 the value of v, so you only need 7 or 8 figures at most.

 

[name123]

On re-checking the approximation I gave in my previous post, I realized that it's not good enough; there are additional terms that have to be taken into account. (Briefly, I was assuming that all of the corrections were second order in $w$, but I was wrong; there are correction terms that are first order in $w$ and therefore have to be included.)

A better approximation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = \gamma \left( x - v t \right) + \left( \gamma - 1 \right) \frac{w}{v} y
$$

$$
y' = y - \gamma w t + \left( \gamma - 1 \right) \frac{w}{v} x
$$

Note that even this might not be good enough, depending on how small you pick $w$ and how many significant figures you look at. The fully correct transformation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = - \gamma v t + \left[ 1 + \left( \gamma - 1 \right) \frac{v^2}{v^2 + w^2} \right] x + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} y
$$

$$
y' = - \gamma w t + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} x + \left[ 1 + \left( \gamma - 1 \right) \frac{w^2}{v^2 + w^2} \right] y
$$

where now we are using $\gamma = 1 / \sqrt{1 - v^2 - w^2}$ so we take both speeds into account (and remember units are always such that $c = 1$).

Sorry for the mixup on my part.
If you leave out the motion in the y direction, you leave out the critical feature of the problem, because if you only look at the "fit" in the x direction, the A-Team will not fit. You have to look at the y coordinates. What you should be checking for is the path in the x'-y' plane of the two A-Team poles, and where each pole passes relative to the two B-Team members (who are at rest in the B-Team frame). You can't figure that out just from looking at x coordinates.

I also suggest picking a value for w that doesn't require you to do calculations accurate to 30 or more significant figures in order to see actual variation in the y coordinates. Try, for example, a value of w that is 1/100 or 1/1000 the value of v, so you only need 7 or 8 figures at most.

Have you the equations in the normal form where the unit of c isn't 1? (a link maybe, I'm having trouble finding it)

 

[PeterDonis]

Have you the equations in the normal form where the unit of c isn't 1?

It's easy to modify them. The general rule is to put $ct$, $ct'$ in place of $t$, $t'$ and $v/c$, $w/c$ in place of $v$, $w$. If you work it out, though, you will see that the only equation that actually changes is the $t'$ equation; if you put $v/c^2$, $w/c^2$ in place of $v$, $w$ in that equation, that will do it.

 

[name123]

It's easy to modify them. The general rule is to put $ct$, $ct'$ in place of $t$, $t'$ and $v/c$, $w/c$ in place of $v$, $w$. If you work it out, though, you will see that the only equation that actually changes is the $t'$ equation; if you put $v/c^2$, $w/c^2$ in place of $v$, $w$ in that equation, that will do it.

Not clear if that is $v/c$, $w/c$ in place of $v$, $w$ or $v/c^2$, $w/c^2$ in place of $v$, $w$. Have you a link to the equations in normal form? Also regarding the calculations I'd done they'd be correct for x right? So could the gap between the A-Team members if there was a y-direction also be worked out using gap = SQRT(x-length ^2 + y-length ^2) so that I could work out the minimum y-length that would be required to allow the B-Team members through, and just use the normal equation to see if that y-length is big enough?

 

[PeterDonis]

Not clear if that is $v/c$, $w/c$ in place of $v$, $w$ or $v/c^2$, $w/c^2$ in place of $v$, $w$.

Modify just the $t'$ equation using the latter.

Have you a link to the equations in normal form?

Have you tried googling "Lorentz transformation"?

regarding the calculations I'd done they'd be correct for x right?

Not if you used the formulas I gave before. Note that the new formulas I gave add terms in all the equations ($t'$, $x'$, and $y'$).

could the gap between the A-Team members if there was a y-direction also be worked out using gap = SQRT(x-length ^2 + y-length ^2) so that I could work out the minimum y-length that would be required to allow the B-Team members through, and just use the normal equation to see if that y-length is big enough?

No. You are still missing a key aspect of the situation: in the B-Team frame, the line connecting the two A-Team members is at an angle from the line connecting the two B-Team members. You need to look at the x-y plane; there is no shortcut.

Here is an outline of the steps I recommend:

(1) We have four observers, two A-Team (call them A1 and A2) and two B-Team (call them B1 and B2). The key condition of the problem is that there is an instant of time in the A-Team frame (call it $t = 0$) at which these four observers are lined up along the $x$ axis (i.e., they all have $y = 0$) in the following order (going from smaller to larger $x$ coordinates): A1, B1, B2, A2. This is what it means to say that the two B-Team members pass between the two A-Team members.

(2) The above condition gives us $( t, x, y )$ coordinates for four events, in the A-Team frame.

(3) Use the Lorentz transformation to obtain the $( t', x', y' )$ coordinates for these four events in the B-Team frame.

(4) In the B-Team frame, B1 and B2 are at rest, so the $( x', y' )$ coordinates obtained for them are valid at any time $t'$. So to find out whether they still pass between A1 and A2 in the B-Team frame, compute the worldlines of A1 and A2 in that frame, using the coordinates obtained above and the fact that A1 and A2 both move with speed $( v_x, v_y ) = ( - v, - w )$ in this frame, and check to see where they are in relation to the fixed coordinates of B1 and B2.

Note that you can actually do all of this using general formulas; you don't need to plug in numbers. However, picking specific numbers and then graphing the results in the $x', y'$ plane may help to visualize what is going on.

 

[name123]

Modify just the $t'$ equation using the latter.

With the gamma equation you gave wouldn't it be wrong if $w$ = 0, and I divided $v^2$ by $c$ instead of $c^2$? By wrong I mean different to the normal equation for a single boost.

 

[PeterDonis]

With the gamma equation you gave wouldn't it be wrong if $w = 0$, and I divided $v^2$ by $c$ instead of $c^2$?

I don't think I ever said to divide $v^2$ by $c$ anywhere, but you are correct that you do have to include $c$ in the $\gamma$ equation, and I should have mentioned that. The correct equation with $c$ included is

$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2} - \frac{w^2}{c^2}}}
$$

 

[name123]

I don't think I ever said to divide $v^2$ by $c$ anywhere, but you are correct that you do have to include $c$ in the $\gamma$ equation, and I should have mentioned that. The correct equation with $c$ included is

$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2} - \frac{w^2}{c^2}}}
$$

Could you just include c with the others, I'm sure I'm not the only one that would find it useful?

 

[PeterDonis]

Could you just include c with the others

The $x'$ and $y'$ equations are unchanged. The $t'$ equation becomes

$$
t' = \gamma \left( t - \frac{v}{c^2} x - \frac{w}{c^2} y \right)
$$

 

[name123]

Thanks, and when looking at it from the B-Team's perspective are they considering the A-Team to be going at $-v$,$-w$?

 

[PeterDonis]

when looking at it from the B-Team's perspective are they considering the A-Team to be going at $-v$,$-w$?

Yes.

 

[name123]

That was great thanks, I the figures were pretty much the same, but I could see the gap (one A-Team member at a lower y and one at a higher y than the B-Team members).

Regarding a point earlier, regarding "now" I had said:

Now does have an absolute meaning for you personally though doesn't it? And would have for the next team A observer down. So can you see how it could true that simultaneously to you experiencing, a team A member was experiencing simultaneously to the team B member opposite to it who was experiencing simultaneously with another team B member experiencing who was simultaneously experiencing with you in the future?

If by "now" you mean "what is happening to me at a given reading on my clock", then yes, of course. But I don't see what that has to do with the rest of what you said.

I'm slightly confused, for me, while I can imagine all the clock times that I will be alive for, there will only ever be one "now". Are you saying that you think the way that "now" should be understood given relativity, is that any clock time in your life has equal right for the claim of being now?

Also with your understanding of relativity would it be true to say that everything the human you experience being has ever done exists in spacetime, so in theory if you could travel to those spacetime coordinates you could see yourself?

 

[PeterDonis]

while I can imagine all the clock times that I will be alive for, there will only ever be one "now"

But which clock time is "now" to you, on this definition of "now", is continually changing.

Are you saying that you think the way that "now" should be understood given relativity, is that any clock time in your life has equal right for the claim of being now?

No. It's true that, since each of them is your "now" when it's your "now", each of them has an equal claim to the term "now". This concept of "now" is just a label that means "whichever instant of my clock time I'm experiencing". But none of this has anything to do with relativity. Your entire worldline is represented in the spacetime model used in relativity; which instant you want to label "now" has nothing to do with the model.

with your understanding of relativity would it be true to say that everything the human you experience being has ever done exists in spacetime

Relativity models spacetime this way, yes. But relativity does not talk about what "exists" except in the sense that it's all there in the model.

in theory if you could travel to those spacetime coordinates you could see yourself?

In theory, yes. But the "if" there is a big "if". Traveling to spacetime coordinates that you've already visited once requires what are called "closed timelike curves", and while they are possible mathematically, pretty much all physicists believe that they aren't physically possible. Spacetimes which include them have a number of properties that don't appear to be physically reasonable.

 

[name123]

But which clock time is "now" to you, on this definition of "now", is continually changing.

The way I mean it if I were to always wear a watch (which had the date and always kept time) only one of its times would ever be "now".

Are you saying that you think the way that "now" should be understood given relativity, is that any clock time in your life has equal right for the claim of being now?

No. It's true that, since each of them is your "now" when it's your "now", each of them has an equal claim to the term "now". This concept of "now" is just a label that means "whichever instant of my clock time I'm experiencing". But none of this has anything to do with relativity. Your entire worldline is represented in the spacetime model used in relativity; which instant you want to label "now" has nothing to do with the model.

Reality has to do with what instant I label "now". Do you think that an interpretation of a model should take into account the reality of the situation?

What is moving along your worldline in the spacetime model used in relativity?

Also with your understanding of relativity would it be true to say that everything the human you experience being has ever done exists in spacetime, so in theory if you could travel to those spacetime coordinates you could see yourself?

In theory, yes. But the "if" there is a big "if". Traveling to spacetime coordinates that you've already visited once requires what are called "closed timelike curves", and while they are possible mathematically, pretty much all physicists believe that they aren't physically possible. Spacetimes which include them have a number of properties that don't appear to be physically reasonable.

Can my history change? (without me visiting it again)

 

[PeterDonis]

The way I mean it if I were to always wear a watch (which had the date and always kept time) only one of its times would ever be "now".

The way you are saying this makes me think that either you are confused or I'm misunderstanding you. The time you call "now" is whatever time your watch says it is when you look at it. But that time isn't always the same, right? Look at your watch today, it tells you one time and date. Look at your watch tomorrow, it tells you a different time and date. Both of those times and dates are "now" when you look at them, but they're not the same. Is this what you're trying to say? Or is it something else?

Reality has to do with what instant I label "now".

Why? This does not seem to me to be a statement about physics; it seems to be a statement about philosophy or metaphysics, both of which are off topic for this forum.

What is moving along your worldline in the spacetime model used in relativity?

Taking the model by itself, nothing. Nothing "moves" in spacetime; spacetime is just a 4-dimensional geometric object that is what it is. Your particular history is just a particular curve in this 4-dimensional geometry. Saying that you and I are "moving" relative to each other is just an informal way of saying that our two histories are described by different curves.

Can my history change? (without me visiting it again)

No, not even if you do visit it again--at least, not according to the way relativity models the universe. As above, your "history" is just a particular curve in the 4-dimensional geometry of the universe; it can't "change", because the 4-dimensional geometry doesn't "change", it just is what it is.

 

[Dale]

if I were to always wear a watch (which had the date and always kept time) only one of its times would ever be "now".

Reality has to do with what instant I label "now".

The first statement is obviously wrong, and the second is just philosophy. Since you have resolved your physics questions, it is a good time to close the thread.