B Clarification on Length Contraction

[NoahsArk]

The rule for length contraction seems to be inconsistent with the lorentz transformations for distance.

The rule for length contraction is: $x = \gamma x ^\prime$ and $x ^\prime = \frac {1} {\gamma} x$

But the lorentz transformations for distance are $x = \gamma x ^\prime + \gamma vt ^\prime$ and $x ^\prime = \gamma x - \gamma vt$

Why in the lorentz transformations above, unlike in the rule for length contraction, aren't we multiplying $x$ by $\frac {1} {\gamma}$ in order to get $x ^\prime$ ? We are always multiplying by $\gamma$ whether or not we want to go from x to $x ^\prime$ or vice versa.

It's true that the rule for length contraction involves length, and the lorentz transfortions for distance involve finding the distance from an observer to an event. But isn't distance the same thing as length? When we say that a fire cracker happened at $x ^\prime = 10$ in Bob's frame, aren't we saying that the length between the fire cracker and Bob is 10 as measured by Bob?

Also, here's an example using the lorentz transformations for distance where the answers don't seem to make sense: take an event that happened at $(x ^\prime = 10, t ^\prime = 0)$ Say for example, Bob in a train passes Alice at 0,0 in both frames, and an explosion happens for Bob at $x ^\prime = 10, t ^\prime = 0$ as he passes Alice. Using the lorentz transformations to find x in Alice's frame, with a relative velocity between the frames at v = .6, we'd get $x = \gamma 10 + 0$, x =10.25. Now say we have another example where this time it is Alice who observes the explosion to happen at x = 10 and t= 0. If we try and find where the explosion happens for Bob we get $x ^\prime = \gamma 10 - 0$, $x ^\prime = 10.25$. Why are they both getting the same distance? Isn't distance frame dependent?

 

[Pencilvester]

The rule for length contraction is: $x = \gamma x ^\prime$ and $x ^\prime = \frac {1} {\gamma} x$

Where are you getting this from? And what is your interpretation of what those equations mean?

 

[NoahsArk]

By "x" and $"x ^\prime"$I meant the distance between any two points or the length. If someone standing on a road measures the distance, x, between to street lights, someone driving by in a car will measure the distance between the lights at $\frac {1} {\gamma} x$.

 

[Antarres]

From what I can understand in your question, you are mixing coordinate transformations with length contraction. Lorentz transformations are coordinate transformations for passing from one reference system to another that is moving at a constant velocity with respect to former one.

Distance between two events is defined as simultaneous distance between them. However, since we have relativity of simultaneity, what is simultaneous for one observer might not be simultaneous for another one. We will analyze the example you gave. So, we have Bob moving in a train at velocity $v$ with respect to Alice who is at rest. Alice's coordinates we denote with $x$ and $t$ and Bob's with $x'$ and $t'$. Now in Bob's frame, two events happen. One is passing by Alice at $t'_1 = 0$, $x'_1 = 0$ and the other is an explosion happening simultaneously for him at $t'_2 = 0$, $x'_2 = 10$. Now let's see what are the coordinates of these events for Alice.
We use Lorentz transformations for time and space which you're familiar with:
$$t = \gamma\left(t' + \frac{vx'}{c^2}\right) \qquad x = \gamma(x' + vt')$$
Transforming, we get the following:
$$t_1 = 0 \qquad x_1 = 0$$
$$t_2 = \gamma\frac{vx'_2}{c^2} \qquad x_2 = \gamma x'_2$$
So these are the coordinates in Alice reference frame, and we see that the explosion for her is happening after Bob passing by her. So in order for her to figure out the distance between these two in her frame, she will take into account that during this interval that has passed $t_2 - t_1$, Bob has moved a certain distance $v(t_2 - t_1)$ with respect to her. So she will subtract that distance in order to measure the simultaneous distance between two events in her own frame and will find that the distance between two events to her is:
$$x_2 - x_1 - v(t_2 - t_1) = \gamma \Delta x' - \gamma\frac{v^2\Delta x'}{c^2} = \gamma \Delta x' \left(1 - \frac{v^2}{c^2}\right) = \frac{\Delta x'}{\gamma}$$
Here $\Delta x' = x'_2 - x'_1$ is distance between events in Bob's frame.
So this is how length contraction happens, coordinate transformation is different from measuring lengths, because measuring lengths involves two simultaneous events, while coordinate transformation is just a transformation of coordinates of a single event.

In your second exercise, you're actually doing the same thing, you're just putting the two events happen simultaneously for Alice. Then they won't happen simultaneously for Bob, because now Alice measures proper length. Using the same treatment I did here, you'll find that in that case Bob will find the contraction happening. However this is not a paradox, because these two examples are different in a sense that the first example is about two events happening at the same time for Bob at rest, while the second is about two event happening at the same time for Alice at rest. Thus those two examples cannot be two pairs of same events. Hope that clarifies it somewhat.

 

[PeterDonis]

The rule for length contraction is...

...not what you wrote. $x$ and $x'$ aren't lengths, they're coordinates of events. That's why you have to use the Lorentz transformation to transform them.

If you want to write the rule for length contraction as $L' = L / \gamma$, you can, but then you have to recognize that $L'$ and $L$ are lengths, i.e., spacelike intervals between pairs of events, i.e., two different events with two different values for $x$ and $x'$--and then you have to recognize that the pair of events that defines $L'$ is different from the pair of events that defines $L$ (because of the relativity of simultaneity--the pair that defines $L'$ is along a line of constant $t'$, while the pair that defines $L$ is along a line of constant $t$, and those are two different lines in spacetime). And if you work out the coordinates of each pair of events, and transform those coordinates, you will see that they transform according to the Lorentz transformation.

 

[haushofer]

What do you know... confusion about relativity because of a sloppy application of physical definitions. :P

 

[Ibix]

As others have pointed out, $x$ and $x'$ are not lengths, they are coordinates. What you are implicitly doing when you treat them as lengths is assuming that the other end of the rod is at the spatial origin ($x=0$ or $x'=0$) and that the length of the rod is $L=x-0$ or $L'=x'-0$. Unfortunately, the rod is moving in at least one of the frames, so the end isn't at the origin most of the time.

So if the rod is stationary in the primed frame with one end at the origin and the other at $x'=L'$ and you transform $(x',t')=(0,L')$ you do indeed get $x=\gamma L'$. But this isn't the length, because at $t=\gamma vL/c^2$ (that you get from the time transform), the other end of the rod isn't at $x=0$.

You need to find two events at $x'=0$ and $x'=L'$ in the primed frame that are simultaneous in the unprimed frame and then find their $x$ coordinates.

 

[PeroK]

As others have pointed out, $x$ and $x'$ are not lengths, they are coordinates. What you are implicitly doing when you treat them as lengths is assuming that the other end of the rod is at the spatial origin ($x=0$ or $x'=0$) and that the length of the rod is $L=x-0$ or $L'=x'-0$. Unfortunately, the rod is moving in at least one of the frames, so the end isn't at the origin most of the time.

So if the rod is stationary in the primed frame with one end at the origin and the other at $x'=L'$ and you transform $(x',t')=(0,L')$ you do indeed get $x=\gamma L'$. But this isn't the length, because at $t=\gamma vL/c^2$ (that you get from the time transform), the other end of the rod isn't at $x=0$.

You need to find two events at $x'=0$ and $x'=L'$ in the primed frame that are simultaneous in the unprimed frame and then find their $x$ coordinates.

... or, assuming you know that the object is moving with constant velocity $v$, you can construct a measurement of length from this data. Let's assume $v$ is positive: moving to the right in the unprimed frame.

1) At $t=0$, the rear of the rod was at $x = 0$.

2) At time $t=\gamma vL'/c^2$ (a later time), the front of the rod was at $x = \gamma L'$.

3) In that time difference, the rear of the rod moves to $\Delta x = \gamma v^2L'/c^2$. This is where the rear of the rod was (in frame S), when the front was at $x = \gamma L'$.

Now, we can calculate the length of the rod in S:

$L = \gamma L' - \gamma v^2L'/c^2 = \gamma L' (1 - v^2/c^2) = \gamma L' (1/\gamma^2) = L'/\gamma$

Hey, what do you know: its length is contracted!

 

[Pencilvester]

The rule for length contraction is: $x = \gamma x ^\prime$ and $x ^\prime = \frac {1} {\gamma} x$

Judging from this statement, I get the impression that you might think that length contraction is a one way phenomenon—i.e. only one frame observes lengths to be contracted, whereas the other frame sees them lengthened.

Just to make clear something that hasn’t been explicitly stated, but that you might have worked out from reading the previous responses or from carefully playing with the Lorentz transformation: length contraction is symmetrical. The primed frame measures unprimed’s meter stick to be short, but unprimed measures primed’s meter stick to be short also.

 

[Herman Trivilino]

When we say that a fire cracker happened at x′ = 10 in Bob's frame, aren't we saying that the length between the fire cracker and Bob is 10 as measured by Bob?

No. Bob need not be located at the origin. All we're saying is that the explosion occurred +10 units away from the origin of the primed system. Note that we usually include a reference event that occurs at x=x'=0. So you can say that the reference event and the explosion occurred 10 units of distance away from each other in the primed frame, but that 10 units of distance is not the length of any object in the primed frame unless the reference event and the explosion occurred at the same time in the primed frame. And then they will not occur simultaneously in the unprimed frame, and therefore the spatial separation of the reference event and the explosion in the unprimed frame is not the length of any object as measured in the unprimed frame.

 

[robphy]

Here's a spacetime diagram on rotated graph paper (so we can more easily view the tickmarks)
that could help clarify what is being calculated.

In the OP, the number should be 12.5 not 10.25.

Thinking spacetime-geometrically,
the length of an object is the distance between parallel worldlines.

• The proper-length of the object is the distance taken along the spacelike segment that is Minkowski-perpendicular to the object-worldlines (that is, in the rest space of the object where the endpoints are simultaneous according to the object).
• The apparent-length of the object is the distance taken along the spacelike segment that is Minkowski-perpendicular to the observer-worldlines (that is, in the rest space of the observer where the endpoints are simultaneous according to the observer). [The proper-length is a special case of the apparent-length, where the observer is the object itself.]

In Euclidean geometry terms,
the proper-length is measured along the perpendicular,
the apparent-length is the distance between the intercepts (the intersections with a transversal line).

In https://en.wikipedia.org/wiki/Distance_between_two_straight_lines,
it is written that these systems of equations yields the intersection points
y=m_1 x +b_1, y=-x/m
y=m_1 x +b_2, y=-x/m
and the distance is
d = \frac{|b_2-b_1|}{\sqrt{1+m^2}}.

In the spacetime version of this problem (using hyperbolic trigonometry),
the proper length is
L_{proper} = \frac{L_{apparent}}{\sqrt{1-(v/c)^2}},
where there is a minus-sign in the square-root for Minkowski spacetime,
and the slope is the velocity (v/c) in Minkowski spacetime.

Using angles and rapidities...
in Euclidean geometry we have \frac{1}{\sqrt{1+m^2}} = \frac{1}{\sqrt{1+\tan^2\phi}}=\cos\phi
(where \phi is the acute angle between the transversal and the line perpendicular to the two lines),
and
in Minkowski geometry we have \frac{1}{\sqrt{1-(v/c)^2}} = \frac{1}{\sqrt{1-\tanh^2\theta}}=\cosh\theta
(where \theta is the rapidity (shown in green) between the observer-worldline and
the object-worldline, which is numerically equal to the Minkowski-angle between their spacelike lines of simultaneity.

So, we have
d = \frac{|b_2-b_1|}{\sqrt{1+m^2}}=|b_2-b_1|\cos\phi
and
L_{proper} = \frac{L_{apparent}}{\sqrt{1-(v/c)^2}}= L_{apparent}\cosh\theta.

Thus,
observe that |b_2-b_1| is the length of the hypotenuse of a right triangle,
and d is the adjacent side...
so |b_2-b_1|=\frac{d}{\cos\phi}.
Analogously,
L_{apparent} is the length of the hypotenuse of a Minkowski-right triangle [the grey parallelograms indicate the Minkowski-right-angles],
and L_{proper} is the adjacent side...
so L_{apparent}=\frac{L_{proper}}{\cosh\theta}=\frac{L_{proper}}{\gamma}.

For \tanh\theta=(v/c)=3/5, we have \cosh\theta=\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=5/4.
So, for L_{proper}=10, we have
L_{apparent}=\frac{ L_{proper} } {\gamma }=\frac{10}{(5/4)}=10(4/5)=8.

Of course, this is all symmetrical between the two observers.
It might take some practice with visualizing Minkowski-spacetime geometry
to see that the corresponding triangles are Minkowski-congruent.
But counting light-clock diamonds helps you see this.Rapidity Cheat Sheet: (v/c)=\tanh\theta, \gamma=\cosh\theta, \gamma v/c=\sinh\theta, \gamma(1+(v/c))=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=\exp\theta=k.

 

[NoahsArk]

Thank you for the responses.

It's easier for me to do examples going from the unprimed to the primed frame, so I'll use the same example but reversed: In Alice's frame an explosion occurred at (x =10, t= 0), what are the coordinates in Bob's frame who is passing Alice at .6c? His coordinates, if I've done it right, would be $(x ^\prime =12.5, t ^\prime=-7.5)$. That means, I think, that for him, that if the reference event of him passing Alice at the origin is 0,0 in both frames, then when he passed Alice, the explosion had already occurred, but for Alice, the explosion and Bob's passing her happened at the same time. Is this correct? Robophy I see that's what you had on the graph except you had 7.5 and - 7.5. Does "12.5" mean that the explosion happened at a distance of 12.5 units from wherever he was when it happened? If so, is the reason why it is happening farther away from him than it is from Alice because he is behind the origin when it happens for him, but Alice is at the origin in her frame? If this is the reason, why does the fact that it happened 7.5 light seconds earlier for Bob mean that the explosion occurred 2.5 light seconds of distance further away from him?

With respect to the question "When we say that a fire cracker happened at x′=10 in Bob's frame, aren't we saying that the length between the fire cracker and Bob is 10 as measured by Bob?"

No. Bob need not be located at the origin. All we're saying is that the explosion occurred +10 units away from the origin of the primed system.

What is meant exactly by primed "system"? Does this mean that Alice has her own coordinate plane, with its x and y axis, that is stationary with respect to her, and Bob has his own coordinate plane which is moving with respect to Alice? In a Galilean transformation, for example, if Bob has traveled five meters ahead of Alice, and an event occurred two meters ahead of him, that event would have occurred 7 meters ahead of Alice. Two would still be "length" between Bob and the event, even though he is not located at the origin.

 

[robphy]

Thank you for the responses.

It's easier for me to do examples going from the unprimed to the primed frame, so I'll use the same example but reversed: In Alice's frame an explosion occurred at (x =10, t= 0), what are the coordinates in Bob's frame who is passing Alice at .6c? His coordinates, if I've done it right, would be $(x ^\prime =12.5, t ^\prime=-7.5)$. That means, I think, that for him, that if the reference event of him passing Alice at the origin is 0,0 in both frames, then when he passed Alice, the explosion had already occurred, but for Alice, the explosion and Bob's passing her happened at the same time. Is this correct? Robophy I see that's what you had on the graph except you had 7.5 and - 7.5. Does "12.5" mean that the explosion happened at a distance of 12.5 units from wherever he was when it happened? If so, is the reason why it is happening farther away from him than it is from Alice because he is behind the origin when it happens for him, but Alice is at the origin in her frame? If this is the reason, why does the fact that it happened 7.5 light seconds earlier for Bob mean that the explosion occurred 2.5 light seconds of distance further away from him?

One of the reasons I moved to geometrical methods is that
it was always confusing which is primed and which is unprimed.
[I don't think there is a universal convention. "Alice and Bob" is better than "primed and unprimed".]
I need something else to reason with (like a picture).
[Similarly, I can't rely on the thin-lens formula because of sign conventions...
I need to draw a raytracing diagram.]
I still use the equations to calculate, but I use the geometry to understand what I am calculating!If Alice (RED) and Bob (BLUE) meet at O
(with their rulers and wristwatches zeroed there
and their space-axes [along which each observer faces] are pointing in the same direction ...
[otherwise, you have added layers of bookkeeping])
and
each mark an event (t=0,x=10) in their frames,
then each will determine that x=+12.5 is the spatial location of the "other observer's (0,10)-event".
However, Alice will determine Bob's-(0,10)-event occurs at t=+7.5
but Bob will determine Alice's-(0,10)-event occurs at t=-7.5.
Alice views Bob moving forward, along their common spatial-axis direction.
However, Bob views Alice moving backward, opposite their common spatial-axis direction.
(If they face each other and have their spatial-axes pointing opposite each other,
they would both get t=-7.5...
but then you'll have bookkeeping issues [e.g. deal with cases] with (say) velocity-composition.)

What is meant exactly by primed "system"? Does this mean that Alice has her own coordinate plane, with its x and y axis, that is stationary with respect to her, and Bob has his own coordinate plane which is moving with respect to Alice? In a Galilean transformation, for example, if Bob has traveled five meters ahead of Alice, and an event occurred two meters ahead of him, that event would have occurred 7 meters ahead of Alice. Two would still be "length" between Bob and the event, even though he is not located at the origin.

Primed-vs-unprimed, A-vs-B, whatever... each has a set of coordinate axes with which they resolve spacetime-displacements into their set of time-components and space-components.

In Galilean spacetime, it turns out that the Galilean-perpendiculars to their worldlines coincide.
So, their notions of length coincide since they will choose the same two events on the endpoint-worldlines that are simultaneous.
Geometrically, simultaneity is determined by perpendicularity, where perpendicular is defined by being tangent to their curve of equal-proper-time-displacements (equal-wristwatch-times along inertial worldlines) from O.
Play with https://www.desmos.com/calculator/wm9jmrqnw2 where the time-axis-runs-to-the-right
(Move the E-slider: E=1 in Minkowski, E=0 is Galilean, and E=-1 is Euclidean):
E=1:

E almost 0:

E=-1

.
(The focus there was time-dilation. I should add a something for length-contraction.)

With respect to the question "When we say that a fire cracker happened at x′=10 in Bob's frame, aren't we saying that the length between the fire cracker and Bob is 10 as measured by Bob?"

Yes, if implicitly, you are identifying the event on Bob's worldline that is Bob-simultaneous with the firecracker. "Length-of-an-object"-observed-by-an-observer is the distance between two events on the endpoint-worldlines that are simultaneous-to-that-observer.

With Bob-(0,0) and Bob-(0,10), Bob says the length is 10-0=10.
Assume that Bob has a 10meterstick with its distant endpoint where the firecracker went off. Otherwise, there is no object to have a length to speak of.

With only Alice-(12.5,7.5), all we can say is that Bob-(0,10) occurred at Alice(12.5,7.5).
Alice needs a second event, one on Bob's worldline, to describe the length of Bob's-10meterstick.
Alice needs to choose the event there that is Alice-simultaneous with Alice(12.5,7.5).
From the diagram, it looks like it's Alice(4.5,7.5).
So, Alice says that Bob's-10meterstick is 12.5-4.5=8.

With only Alice-(0,0), all we can say is that Bob-(0,0) occurred at Alice(0,0).
Alice needs a second event, one on Bob's-10meterstick-far-endpoint, to describe the length of Bob's-10meterstick.
Alice needs to choose the event there that is Alice-simultaneous with Alice(0,0).
From the diagram, it looks like it's Alice(0,8).
So, Alice says that Bob's-10meterstick is 8-0=8.

Now, maybe we start off this way.
Alice says I have two events on those Bob-and-endpoint-worldlines Alice-(0,0) and Alice-(0,8).
So, Alice measures a length 8-0=8.

What does Bob measure for the length?
So, we have to find two events on the Bob-and-endpoint-worldlines that are Bob-simultaneous.
We have Bob-(0,0). Where is the other one?
Thinking geometrically, since Alice's 8 is the hypotenuse, we have to multiply by \gamma=\cosh\theta=5/4 to get the adjacent side. So, 8(5/4)=10 is the adjacent side.

 

[NoahsArk]

What you are implicitly doing when you treat them as lengths is assuming that the other end of the rod is at the spatial origin (x=0 or x′=0) and that the length of the rod is L=x−0 or L′=x′−0.

Isn't that a correct assumption to make, though? When Bob measures an event at $x ^\prime = 10$, doesn't that mean by definition that the event is happening 10 units from his origin (even though it may be more than 10 units from Alice's origin)? I am assuming that Bob is always standing in his rocket at $x ^\prime = 0$, and Alice is always standing on the ground at x = 0. You always hear things in special relativity lessons things like "from Bob's perspective, he's not moving", so I took that to mean that Bob is always at his origin according to him. I am also assuming that when Bob measures something at $x ^\prime = 10$ he's measuring it at 10 units from where he is standing at $x ^\prime = 0$.

Just to make clear something that hasn’t been explicitly stated, but that you might have worked out from reading the previous responses or from carefully playing with the Lorentz transformation: length contraction is symmetrical. The primed frame measures unprimed’s meter stick to be short, but unprimed measures primed’s meter stick to be short also.

Right, and that's one of the things that makes me confused about the Lorentz transformation for distance. For an event happening in Bob's frame at $x ^\prime = 10, t ^\prime = 0$ Alice measures the x distance to be 12.5 which is longer than in Bob's frame, not shorter. Similarly if the event happened at x = 10, t = 0 in Alice's frame, Bob also measures $x ^\prime$ to be 12.5 as Robphy used in the most recent post. It seems like instead of each of them measuring the other's lengths to be shorter, which is what happens in length contraction, we are seeing length expansion. Again, I know I may be confusing transformations of coordinates with lengths, but, as I mentioned in the first paragraph above, it seems like x = 10 or $x ^\prime = 10$ is a length by definition.

 

[Ibix]

Isn't that a correct assumption to make, though?

Not if you are trying to measure the length of a rod whose other end isn't at the origin.

This is a case where a Minkowski diagram would help - will post one later (on my phone and still haven't checked if my program will work on the python install on this thing).

 

[PeroK]

Isn't that a correct assumption to make, though? When Bob measures an event at $x ^\prime = 10$, doesn't that mean by definition that the event is happening 10 units from his origin (even though it may be more than 10 units from Alice's origin)? I am assuming that Bob is always standing in his rocket at $x ^\prime = 0$, and Alice is always standing on the ground at x = 0. You always hear things in special relativity lessons things like "from Bob's perspective, he's not moving", so I took that to mean that Bob is always at his origin according to him. I am also assuming that when Bob measures something at $x ^\prime = 10$ he's measuring it at 10 units from where he is standing at $x ^\prime = 0$.
Right, and that's one of the things that makes me confused about the Lorentz transformation for distance. For an event happening in Bob's frame at $x ^\prime = 10, t ^\prime = 0$ Alice measures the x distance to be 12.5 which is longer than in Bob's frame, not shorter. Similarly if the event happened at x = 10, t = 0 in Alice's frame, Bob also measures $x ^\prime$ to be 12.5 as Robphy used in the most recent post. It seems like instead of each of them measuring the other's lengths to be shorter, which is what happens in length contraction, we are seeing length expansion. Again, I know I may be confusing transformations of coordinates with lengths, but, as I mentioned in the first paragraph above, it seems like x = 10 or $x ^\prime = 10$ is a length by definition.

This post simply restates your original false analysis. That's pointless.

I think you got several good answers to this already in posts #4, #5, #7. I suggest you choose one of those to understand and base your analysis on. Or, one of the later posts. Whichever you think is best for you.

There are too many different analyses here. You can't possibly digest them all. Pick one you like the look of and go with that.

I know it seems like the lengths are longer. If you do the analysis properly, you'll see that it's not what it seems.

 

[PeterDonis]

I am assuming that Bob is always standing in his rocket at $x ^\prime = 0$ ,

Yes, but that doesn't mean that either end of a rod that is moving relative to Bob will always be at $x^\prime = 0$. Nor does it mean that the pair of events that gives a distance $L'$ of 10.25, for example, is the same pair of events that gives a distance $L$ of 10.25. The two distances, primed and unprimed, are between two different pairs of events.

the Lorentz transformation for distance

You continue to fail to grasp the point we are all trying to get across to you: There is no such thing as a "Lorentz transformation for distance". Lorentz transformations transform coordinates of events, not distances or times.

 

[Herman Trivilino]

Again, I know I may be confusing transformations of coordinates with lengths, but, as I mentioned in the first paragraph above, it seems like x = 10 or x′=10 is a length by definition.

No, those are positions. Yes, the distance from x=0 to x=10 is indeed a length, but the length of what? If a golf ball has a position x=0 and then later on it has a position x=10 that doesn't mean the golf ball has a length of 10.

 

[NoahsArk]

So in order for her to figure out the distance between these two in her frame, she will take into account that during this interval that has passed t2−t1, Bob has moved a certain distance v(t2−t1)with respect to her.

Just to be clear, what are you calculating the distance of as measured by Alice: is it 1) the distance between Alice and the explosion, or 2) the distance between the first event (train passing her), and the second event (explosion), or 3) the distance between the train and the explosion as Alice measures it? It doesn't seem like the second one would be possible because, since you explained that the train is in front of her when the explosion occurs, there couldn't exist any distance between event one (train passing her) and event 2 (explosion) because when the train passed her, the explosion did not yet exist.

I also have some questions on the math (I read this and other posts over several times), but first I want to be clear on what it is we are calculating.

Thank you.

 

[NoahsArk]

After giving this some more thought, I explained my question poorly I think:

There is no such thing as a "Lorentz transformation for distance". Lorentz transformations transform coordinates of events, not distances or times.

Yes, however, in situations where time is zero, doesn't the Lorentz transformation essentially become the equation for length contraction? This seems to be exactly what professor Lagerstrom is saying here at around 14:30

https://www.coursera.org/lecture/ei...oring-the-lorentz-transformation-part-1-UPo37
With that clarification in mind, I think the following question better captures the problem:

Contrast these two situations:

1) Bob is in a rocket. Alice is standing on the road. Bob flies past her and, when they are side by (the reference event), an explosion occurs at some distance on the road which leaves a char mark on the road. Because Bob is moving with respect to the road, the distance between where Alice is standing and the char mark will be contracted from his point of view. Therefore, if Alice wants to transform Bob's measurement for that distance, she will need to multiply by gamma.

2) Same situation as above except now the explosion occurs in the rocket leaving a char mark in the rocket (assume its a long rocket and Bob is in the back of it when he and Alice meet, and the explosion happens further down the rocket (as opposed to further down the road in the first situation). We discussed in an earlier thread that neither Bob's nor Alice's measurements should be affected by the fact that the explosion occurred in the rocket vs. on the ground. However, in this case, where it occurred in the rocket which isn't moving with respect to Bob, he shouldn't be getting any length contraction effect. It would be Alice who would be seeing the length contraction effect. Therefore, she should not need to multiple by gamma to get the distance between herself and the char mark in the rocket, rather, she should be dividing by gamma.

I know I must be missing something here and sorry for it taking me a while to process this, and if the resolution to this problem has the same answer or similar answer to an above post. I did read through them, these things just take time to sync in for me. The only explanation I can think of why it doesn't matter where the explosion occurred (i.e. on the ground vs. in the rocket), is that when the explosion occurs we draw an imaginary horizontal line through space at the point of the explosion which is perpendicular to the ground/rocket, and we say that Alice thinks Bob is moving towards that imaginary line, but Bob thinks that that line is moving towards him, and the space between Alice and the line are moving away from him. This would be interesting because it means that space itself can be moving with respect to someone.

 

[PeroK]

After giving this some more thought, I explained my question poorly I think:
Yes, however, in situations where time is zero, doesn't the Lorentz transformation essentially become the equation for length contraction? This seems to be exactly what professor Lagerstrom is saying here at around 14:30

https://www.coursera.org/lecture/ei...oring-the-lorentz-transformation-part-1-UPo37
With that clarification in mind, I think the following question better captures the problem:

Contrast these two situations:

1) Bob is in a rocket. Alice is standing on the road. Bob flies past her and, when they are side by (the reference event), an explosion occurs at some distance on the road which leaves a char mark on the road. Because Bob is moving with respect to the road, the distance between where Alice is standing and the char mark will be contracted from his point of view. Therefore, if Alice wants to transform Bob's measurement for that distance, she will need to multiply by gamma.

2) Same situation as above except now the explosion occurs in the rocket leaving a char mark in the rocket (assume its a long rocket and Bob is in the back of it when he and Alice meet, and the explosion happens further down the rocket (as opposed to further down the road in the first situation). We discussed in an earlier thread that neither Bob's nor Alice's measurements should be affected by the fact that the explosion occurred in the rocket vs. on the ground. However, in this case, where it occurred in the rocket which isn't moving with respect to Bob, he shouldn't be getting any length contraction effect. It would be Alice who would be seeing the length contraction effect. Therefore, she should not need to multiple by gamma to get the distance between herself and the char mark in the rocket, rather, she should be dividing by gamma.

I know I must be missing something here and sorry for it taking me a while to process this, and if the resolution to this problem has the same answer or similar answer to an above post. I did read through them, these things just take time to sync in for me. The only explanation I can think of why it doesn't matter where the explosion occurred (i.e. on the ground vs. in the rocket), is that when the explosion occurs we draw an imaginary horizontal line through space at the point of the explosion which is perpendicular to the ground/rocket, and we say that Alice thinks Bob is moving towards that imaginary line, but Bob thinks that that line is moving towards him, and the space between Alice and the line are moving away from him. This would be interesting because it means that space itself can be moving with respect to someone.

What's going on here? The last I knew we had a rod of length $L$ moving past the origin. Now we have a rocket and explosions going on up and down a road. When did that happen?

 

[NoahsArk]

What's going on here? The last I knew we had a rod of length LLL moving past the origin. Now we have a rocket and explosions going on up and down a road. When did that happen?

Yes, I caused confusion by explaning my question wrong initially.

 

[Pencilvester]

when they are side by [side] (the reference event), an explosion occurs at some distance on the road which leaves a char mark on the road.

Presumably, you mean that Alice determines that the explosion happened at the same time that Bob is passing her. Due to relativity of simultaneity, this means that Bob determines that explosion to have happened at some time other than when he passed Alice. Assuming the explosion happens in front of Bob, you’re implicitly saying that you are expecting Bob to measure the distance the explosion happened from him, then wait a little bit until he is passing Alice, then assign that length to the distance between him and the scorch mark, which has moved closer to him in that time? That doesn’t make much sense.

 

[NoahsArk]

Pencilvester, before I make the situation even more confusing, I want to ask does it necessarily have to be the case that the reference event is Bob passing Alice? Maybe that aspect throws an extra level of complexity unnecessarily.

 

[NoahsArk]

And here is another basic idea that I am not sure of which is likely making it confusing as well: When we say that the explosion happens at X distance from Alice, are we really saying that Alice could be standing anywhere in her coordinate system, but she observes the explosion to have happened X distance away from the origin? Or, does X = 10 mean 10 units from Alice and not ten units from what Alice has defined as her origin?

 

[Pencilvester]

Pencilvester, before I make the situation even more confusing, I want to ask does it necessarily have to be the case that the reference event is Bob passing Alice? Maybe that aspect throws an extra level of complexity unnecessarily.

All that that means is that both Bob and Alice label the event “Bob passes Alice” with the coordinates (0,0), i.e. the origins of their coordinates.

 

[PeterDonis]

n situations where time is zero, doesn't the Lorentz transformation essentially become the equation for length contraction?

No, because, as I have already explained and as you continue to fail to grasp, the "lengths" in the two different frames are spacelike intervals between two different pairs of events. Only one such pair of events occurs at the same time in the unprimed frame. The events in the pair that defines "length" in the primed frame do not happen at the same time in the unprimed frame.

 

[PeterDonis]

@NoahsArk I strongly suggest drawing a spacetime diagram and marking on it the various events of interest. It is a lot easier to see the relationships between events on a diagram than to try to grasp them from a description in words.

 

[PeroK]

And here is another basic idea that I am not sure of which is likely making it confusing as well: When we say that the explosion happens at X distance from Alice, are we really saying that Alice could be standing anywhere in her coordinate system, but she observes the explosion to have happened X distance away from the origin? Or, does X = 10 mean 10 units from Alice and not ten units from what Alice has defined as her origin?

Perhaps it's time to move away from the SR "baby talk" of "Bob and Alice" and into the grown up language of reference frames S and S', with by convention S' moving with velocity $v$ relative to $S$.

The baby talk gives you the impression that we are talking about the direct observations of two human observers - we are not. That the position of the observers matters - it does not. That the light travel time from an event to the observers is relevant - it is not. That the distance an event takes place from these human observers matters - it does not.

For example: length contraction applies equally to any object at rest in frame S', as measured in frame S. And, symmetrically, any object at rest in S, as measured in frame S'. With contraction in the direction of the relative motion.

Note also that it is length contraction; it's not distance contraction.

If I were you, I would forbid myself from making any statements such as: "explosion happens a distance X from Alice". That should be "an explosion occurs at coordinates $X, t$ in frame S".

The former statement is just setting yourself up for confusion. It's hiding the simultaneity two spatially separated events, for example.

From now on, everything is an event with coordinates in a specified frame. If that's not what you've written, then delete it and say what you mean using coordinates.

 

[robphy]

Perhaps it's time to move away from the SR "baby talk" of "Bob and Alice" and into the grown up language of reference frames S and S', with by convention S' moving with velocity $v$ relative to $S$.

The baby talk gives you the impression that we are talking about the direct observations of two human observers - we are not. That the position of the observers matters - it does not. That the light travel time from an event to the observers is relevant - it is not. That the distance an event takes place from these human observers matters - it does not.

In my opinion, the use of "Alice and Bob" isn't "baby talk"...
it's a direct replacement of "primed and unprimed".
If you don't like Alice and Bob, use Pete and Ulysses.
Prime and Unprimed Frames or Observers can carry the same baggage that Alice and Bob might.
But that baggage is really about my next point.

What is "baby talk" is
using common words that have technical meanings and definitions in this context
but without sticking to those definitions.
When definitions aren't adhered to,
that's when everyday Newtonian intuition creeps in... hence misconceptions.

What is also "baby talk" is
not thinking in terms of spacetime concepts and its associated spacetime geometry.

Anyone who studies relativity without understanding how to use simple space-time diagrams is as much inhibited as a student of functions of a complex variable who does not understand the Argand diagram.

...We have in the special theory of relativity the Minkowskian geometry of a flat 4- space with indefinite metric... Unfortunately, it has been customary to avoid this geometry, and to reason in terms of moving frames of reference, each with its own Euclidean geometry. As a result, intuition about Minkowskian space-time is weak and sometimes faulty...

As I have often said, a "spacetime diagram is worth a thousand words" .
https://www.google.com/search?&q="spacetime+diagram+is+worth+a+thousand+words"
Catchy buzzword phrases like "time dilation" and "length contraction"
should be backed up by a precise spacetime interpretation,
which usually has a trigonometric interpretation analogous to a problem in Euclidean geometry.
(Most textbook relativity problems, after learning to interpret the problem, boils down to find one feature of a Minkowski right-triangle given other features of that triangle. For example, the muon problem for time dilation is about being given two legs of a triangle, or a leg and an angle, and finding the hypotenuse.)

Rather than "prime and unprimed frames", or this "boxcar moving relative to the stationary boxcar in space",
we should really think in terms of
events and worldlines [or a family of worldlines] (which could be associated with local observers with 4-velocities) and proper-times (arc lengths, as measured by wristwatches) and light-cones (and causality) and invariants. If we use coordinates and components, we should be aware of how coordinates are operationally assigned with radar-measurements, or with tangents to hyperbolas [how perpendicularity is defined, a la Minkowski].

Rather than "Lorentz transformations",
we should think in terms of dot-products of tensors with 4-velocities
to decompose tensors into components with respect the observer with that 4-velocity.

The real problems aren't because of "Alice and Bob".

 

[Ibix]

@NoahsArk - I said I'd draw Minkowski diagrams. Here's one in the rest frame of a rod that's ten light nanoseconds long (about ten feet):

The two red lines represent the ends of the rod, not moving in this frame. I've also shown two crosses at time $t=0$, which represent a couple of firecrackers going off. Note that the rod is ten light nanoseconds long and the firecracker explosions are ten light nanoseconds apart.

Now here's the same thing in a frame where the rod is doing 0.6c.

The two red lines again represent the front and back of the rod. The crosses represent the firecrackers, and there's a thin diagonal line joining them for clarity. I've also added various dotted lines connecting events to axes, again for clarity.

You can immediately see that the rod has length contracted by a factor of $\gamma=0.8$ - just read off the horizontal distance between the red lines. You can also see that the firecrackers are further apart by a factor of $1/\gamma=1.25$. And you can see that this isn't contradictory, it's just that the firecrackers went off at different times when the rod was in a different place, so the distance between them isn't the same as the length of the rod.

This is the mistake you make when you treat the Lorentz transforms as transforming distance: your process measured the distance between the flash at the front of the rod 7.5ns after the back passed the origin, and assumed that the back was at the origin.

 

[PeroK]

If you don't like Alice and Bob, use Pete and Ulysses.

Penelope and Ulysses would be better. Penelope representing the stay-at-home "stationary" frame of reference, and Ulysses representing the wandering "moving" frame.

The geometric aspect is interesting. When I first learned SR (six years ago now), it took about six weeks to really get the basics nailed down - it might have been a bit longer. To be honest, the only thing I didn't understand were spacetime diagrams. I remember being hopelessly lost about what the geometry of these things really was. So, I passed on that first time round. In general, I am much stronger algebraically than geometrically, which might have something to do with it.

That said, I learned recently that Einstein was slow to warm to Minkowski's hyperbolic geometry interpretation of SR. Maybe it isn't quite so natural as JL Synge believes. It's possible that he (and you) have very strong geometric visualisation skills that not everyone shares?

The real issue, of course, is how best to teach SR and, in this case, to help the OP understand length contraction. I'm not sure anyone really knows. And, indeed, it might be different for different students.

 

[phinds]

And here is another basic idea that I am not sure of which is likely making it confusing as well: When we say that the explosion happens at X distance from Alice, are we really saying that Alice could be standing anywhere in her coordinate system, but she observes the explosion to have happened X distance away from the origin? Or, does X = 10 mean 10 units from Alice and not ten units from what Alice has defined as her origin?

Why would "distance from Alice" mean anything other than "distance from Alice" ? What does anything else have to do with it?

 

[robphy]

I think there have been papers on the history of Minkowski’s hyperbolic trigonometry interpretation ( which has hyperbolic geometry only in the space of velocities) that relativity attracted more attention after Minkowski’s 1907/1908 paper. Maybe it was just the time to digest Einstein’s ideas from 1905... or maybe it was Minkowski validating these new ideas by Einstein.

When we describe general relativity in terms of the geometry of Spacetime, it was Minkowski that formulated it for flat space... using ideas from Felix Klein and Arthur Cayley. Minkowski gave us the “Spacetime” idea, as well “proper time” and “light cone”. After initial resistance by Einstein, he embraced the idea and generalized the geometry to a curved one.

Byrecognizing Minkowski spacetime geometry, one can pull in one’s vast knowledge of Euclidean geometric thinking, suitably tweaked to handle the signature. Otherwise, one is just doing algebra with \gamma’s and v\gamma not realizing those are essentially cosines and sines, interpreted as ratios of sides of a right triangle. Just like algebra can be a crutch for physical understanding, geometry can as well.
(We tell students to draw position vs time graphs, free body diagrams, ray tracing diagrams, why not fancier position vs time graphs: Spacetime diagrams?)

(Without a Spacetime diagram and it’s geometry, how would one explain “light cone” if one only drew boxcars in relative motion? Or, later, explain the bending of light by the sun? Or the behavior near the event horizon of a black hole?)

 

[Ibix]

That said, I learned recently that Einstein was slow to warm to Minkowski's hyperbolic geometry interpretation of SR. Maybe it isn't quite so natural as JL Synge believes. It's possible that he (and you) have very strong geometric visualisation skills that not everyone shares?

I certainly learned the algebra first, but the realisation that you could just mentally sketch a displacement-time diagram and then sketch a boost by sliding events along hyperbolae was great for understanding it. Partly because it's impossible to forget the relativity of simultaneity like this...

 

[robphy]

I think spacetime diagrams are considered hard or too mathematical because
that was Einstein's first impression of them, and
many introductory physics textbooks take a historical/pseudo-historical development of relativity
[focusing on the ether, michelson morley, Einstein 1905, and then the classic effects (what I sometimes consider the classic traps and pitfalls) , and then appeal to experimental verification]
but very little on the more modern spacetime view of (that mathematician) Minkowski,
which might help develop intuition.

It's refreshing to see a new intro textbook (maybe not so new anymore) like Tom Moore's
Six Ideas that Shaped Physics, Unit R. http://www.physics.pomona.edu/sixideas/

(I like to point out that
the position-vs-time graphs of PHY 101
has an underlying non-euclidean geometry:
the Galilean geometry.
If you draw the clock effect/twin paradox scenario in Galilean physics, you get a triangle
where the kinked worldline of the traveler has the same length
(where arc-length is proper-time, measured with a wristwatch carried by that observer)
as the stay at home twin. That is, no time difference along the paths,
However, that triangle doesn't satisfy the triangle inequality.
In addition,
a t=const line is [Galilean-]perpendicular to all worldlines, regardless of their velocities (slopes)... which is the condition of "absolute time" and "absolute simultaneity".
That doesn't sound Euclidean.
However, we have been taught to read that ordinary position-vs-time graph a certain way.
So, we are generally unaware of it to complain about its nonEuclidean features
...but enter the spacetime diagram for special relativity... )

(Last quote, then I'll get off my soapbox.

http://aapt.scitation.org/doi/10.1119/1.17728
"Lapses in Relativistic Pedagogy" by Mermin
makes some good points

...Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it.

)

Now... back to the ideas for this thread.
As others have said, and as some of us have done,
draw a spacetime diagram.

(Since you don't usually solve a geometry problem with rotation matrices,
it might be good to try to solve special relativity problems using methods other than just the formulas for Lorentz transformations, length-contraction, or time-dilation.
Try trigonometry.)

 

[PeroK]

(Since you don't usually solve a geometry problem with rotation matrices,
it might be good to try to solve special relativity problems using methods other than just the formulas for Lorentz transformations, length-contraction, or time-dilation.
Try trigonometry.)

In other words, rely on the student's intuitive understanding of hyperbolic trigonometry.

 

[robphy]

In other words, rely on the student's intuitive understanding of hyperbolic trigonometry.

Um... no...

develop the understanding of hyperbolic trigonometry using circular trigonometry, which they probably developed by drawing free body diagrams and breaking vectors into components. Why not use what they know and build on it (even though it wasn’t Einstein’s method of calculation or intuition)?

[From asking folks at the Einstein Papers project if Einstein reasoned with Spacetime diagrams, no one could point to an example that wasn’t a simple sketch, and such sketches didn’t solve a problem].I think that’s better than using transformation equations expressed in terms of slopes... (we don’t even teach rotations with slopes!) at least use the analogue of the angle (scary word: rapidity).

Let me just say that as a physics professor whose been a regular at AAPT meetings, I’ve spoken to all sorts of teachers and students who struggle with relativity. Since the standard way doesn’t seem to help, I’m looking for alternative ways to teach relativity.
Some of these methods have been in the literature but are obscure or forgotten or not developed enough. I visit here at PF and at PSE to look for problems (some simple and some more challenging) that could be clarified and solved with a Spacetime diagram... using my diamonds or my Spacetime trig methods...
in some sense, I’m doing research.

 

[NoahsArk]

PeroK: You cleared up some misconceptions I was having! Getting some rest over the weekend is letting me see things more clearly:) The fact that the position of the observers doesn't matter takes away a lot of confusion. Although using human names in the problem still can be a helpful crutch, I think, I see that it's necessary to know, when doing so, that what really matters is what's happening in the primed and unprimed frames, S and $S ^\prime$, and that when we say something happened at x, it doesn't mean that something happened x units away from an observer like Alice, it means that something happened x units away from the origin of the observer's frame: the S origin.

That still leaves an important question, and I think I answered the question for myself today: Should we assume, when doing Lorentz transformations, that the origins of the S and $S ^\prime$ frames are aligned at the reference event? I think the answer must be yes, and the same is true for Galilean transformations? If that's true, I didn't need, in my example, to say that Alice and Bob were aligned at (0,0). The origins of their S and $S ^\prime$ frames were aligned at some reference event, and it doesn't really matter what the reference event was.

Ibix: thank you for the space-time diagram. Can we illustrate both frame's perspective on a single diagram, and is that a good idea to do? I will follow up with some additional questions on the diagram you posted.

PeterDonis: you told me to try and draw a space time diagram which I have done. It was helpful, but you will see I ran into a problem:

The explosion (marked with a circle) happened in the S frame at (0, 10) (side note- the left number when I indicate a pair of coordinates should be t, and the right number should be x, right?). In the $S ^\prime$ frame, the event happened at (-7.5, 12.5). The red line here is the world line of the $S ^\prime$ frame. The dotted line is the $S ^\prime$ frame's line of simultaneity at the moment the explosion happened. The first thing I want to confirm is that on this diagram I should only be drawing one circle for the explosion, and I should not be drawing two circles - one for where it happened in each frame?

What I am not sure how to do is how to illustrate that the explosion happened at x = 12.5 in the $S ^\prime$ frame without drawing a new diagram? Another problem you see in this picture is that when you draw the line of simultaneity with the right slope (which is .6 as Pencilvester showed in another thread), the line of simultaneity intercepts the time axis at the wrong time. Rather than intersecting at -7.5, it intersects at around - 6. The only reason why I can think of of this happening is because the gaps between the lines of simultaneity in the $S ^\prime$ frame are smaller than they are in the S frame, and therefore you can draw 7.5 of them from the origin until the point where the line of simultaneity of the explosion intercepts with the t axis?

 

[Pencilvester]

the line of simultaneity intercepts the time axis at the wrong time. Rather than intersecting at -7.5, it intersects at around - 6.

A time axis can be thought of as the worldline of the spacial origin for a specific frame, i.e. the line given by equation $x=0$. There is a unique one for each frame. The worldline of the S’ spatial origin is the red line you drew. That is the time axis for S’. But because Euclidean geometry isn’t applicable to spacetime diagrams, you could measure the distance of the point of intersection with that line to the origin in a naively Euclidean manner, and it still won’t be 7.5. You’d have to use the Minkowski metric to get the correct “distance” ($ds^2 = -dt^2 + dx^2$).

 

[PeroK]

PeroK: You cleared up some misconceptions I was having! Getting some rest over the weekend is letting me see things more clearly:) The fact that the position of the observers doesn't matter takes away a lot of confusion. Although using human names in the problem still can be a helpful crutch, I think, I see that it's necessary to know, when doing so, that what really matters is what's happening in the primed and unprimed frames, S and $S ^\prime$, and that when we say something happened at x, it doesn't mean that something happened x units away from an observer like Alice, it means that something happened x units away from the origin of the observer's frame: the S origin.

Yes!

That still leaves an important question, and I think I answered the question for myself today: Should we assume, when doing Lorentz transformations, that the origins of the S and $S ^\prime$ frames are aligned at the reference event? I think the answer must be yes, and the same is true for Galilean transformations? If that's true, I didn't need, in my example, to say that Alice and Bob were aligned at (0,0). The origins of their S and $S ^\prime$ frames were aligned at some reference event, and it doesn't really matter what the reference event was.

The often unstated rule is that to use the Lorentz Transformation the origins (which means $x= 0, t=0$ and $x'=0, t'=0$) must coincide. That's the reference event.

 

[jbriggs444]

Why would "distance from Alice" mean anything other than "distance from Alice" ? What does anything else have to do with it?

Distance from where Alice is, distance from where Alice was or distance from where Alice will be? Presumably the intent is "distance from here to where Alice is at this moment". Using coordinates makes it somewhat more obvious that a choice of simultaneity convention may become important.

 

[NoahsArk]

@Ibix the diagram helps me understand the difference between length contraction and the Lorentz transformation for x. If a rod, 10 light nanoseconds long, is in a rocket flying with it, then the rod is stationary with respect to the rocket observer. I am assuming that is what your first diagram represents? (It still helps me to visualize by using terms like "rocket observer", but I'm doing so now keeping in mind the clarifications that have been made above.) Your second diagram could be the from the frame of reference of someone on Earth watching the rocket go by correct?

I've also shown two crosses at time t=0, which represent a couple of firecrackers going off.

So, does this mean that according to the rocket observer, two fire crackers are going off at the same time, one on each end of the rod? For the rocket observer, one fire cracker goes off at x = 10 and the other at x = 0 which is where the ends of the rods are sitting in the rocket frame? If this is the case, it makes more sense that for the Earth observer, because of the leading clocks lag rule, the explosion at the front of the ship would happen after it happened for the rocket observer. As we've been discussing, it happened 7.5 seconds later according to the Earth observer. When I look at your second diagram, I see that the front of the rod begins at x = 8 in the Earth frame, and ends up at x = 12.5. That makes sense because it took 7.5 seconds of travel time in the Earth frame, and 7.5 x .6v = 4.5 nanoseconds of distance that it traveled to get from x = 8 to x = 12.5. Have I worked out what's going on correctly? I assume the back of the rod starts out at 0,0 in both frames?

@robphy I agree that visualizing things is helpful. In my case, it helped me a lot to understand the term $\gamma vX ^\prime$ as the leading clocks lag rule by seeing v as the slope of the line and realizing that the bigger $x ^\prime$ is, the higher on the time access you will go due to the fact that $x ^\prime$ is slopping upward.. Some of your comments regarding light cones and using right triangles as visual aides gave me some food for thought. Some of the concepts you brought up though, while interesting, are advanced for me at this point, and I will need time to be able to get the background. Once the examples go beyond the use of basic algebra and geometry as tools, I would need to learn the higher level math to be able to follow (which I intend to do in the future, but am not there yet).

Thank you.

 

[Ibix]

I am assuming that is what your first diagram represents? (It still helps me to visualize by using terms like "rocket observer", but I'm doing so now keeping in mind the clarifications that have been made above.) Your second diagram could be the from the frame of reference of someone on Earth watching the rocket go by correct?

Yes.

So, does this mean that according to the rocket observer, two fire crackers are going off at the same time, one on each end of the rod? For the rocket observer, one fire cracker goes off at x = 10 and the other at x = 0 which is where the ends of the rods are sitting in the rocket frame? If this is the case, it makes more sense that for the Earth observer, because of the leading clocks lag rule, the explosion at the front of the ship would happen after it happened for the rocket observer. As we've been discussing, it happened 7.5 seconds later according to the Earth observer. When I look at your second diagram, I see that the front of the rod begins at x = 8 in the Earth frame, and ends up at x = 12.5. That makes sense because it took 7.5 seconds of travel time in the Earth frame, and 7.5 x .6v = 4.5 nanoseconds of distance that it traveled to get from x = 8 to x = 12.5. Have I worked out what's going on correctly? I assume the back of the rod starts out at 0,0 in both frames?

Yes, assuming you meant 7.5 × 0.6$c$ and the v was a typo.

 

[NoahsArk]

To try and further understand what's happening, I thought about it and made this diagram:

The bottom is the rocket, $x ^\prime$ coordinate system. The top is the earth, x, coordinate system- say it's a road. At each place where a number is drawn on the Earth system there is a post on the road. In the rocket, which is flying next to the road, one person stands at $x ^\prime = 0$ and the other at $x ^\prime = 10$. Instead of firecrackers going off at those points, as in the previous example, each person, simultaneously in the rocket frame, waves a paint brush out of the window and puts paint on Earth observer's posts. The event representing the second paintbrush being waved at $x ^\prime = 10$ in the rocket frame, similar to the firecracker example, will be seen as happening at x = 12.5 in the Earth frame.

I think the above diagram, correct me if I'm wrong, illustrates another point: As we discussed, the Earth observer will explain why this event happened at 12.5 in their frame as being due to the relativity of simultaneity- i.e. the second paint brush wasn't waved until a later time and therefore it had to travel further down the road before making it's mark. However, the rocket observer won't say that ROS is the reason for the second paint brush getting paint on post number 12.5. They will say that Earth's posts have been cramped together by a factor of .8, and that's why the person standing 10 units from the origin of the rocket frame was able to reach out to a post 12.5 units from the origin in the Earth frame. Does this describe things correctly from the rocket perspective?

 

[PeterDonis]

The bottom is the rocket, $x ^\prime$ coordinate system. The top is the earth, $x$, coordinate system

No, they aren't, because you've left out time. You need to draw a spacetime diagram. Just space alone is wrong.

 

[PeterDonis]

The first thing I want to confirm is that on this diagram I should only be drawing one circle for the explosion

That's correct. The explosion is one event. That means it is one point in spacetime, so it's one point on the diagram.

What I am not sure how to do is how to illustrate that the explosion happened at x = 12.5 in the $S^\prime$ frame

That is the spacetime interval between the explosion event and the event where your $x^\prime = - 7.5$ line of simultaneity (the dotted line) crosses the worldline of the $S^\prime$ origin (the red line). So you should find that

$$
\sqrt{\left( \Delta x \right)^2 - \left( \Delta t \right)^2} = 12.5
$$

where $\Delta x$ is the difference in $x$ coordinates between the two events, and $\Delta t$ is the difference in $t$ coordinates between the two events. We know the explosion is at $t = 0$, $x = 10$. The other event, as you can calculate from the inverse Lorentz transform and the fact that in the primed frame its coordinates are $t^\prime = - 7.5$, $x^\prime = 0$, will have coordinates $t = - 9.375$,. $x = - 15.625$. So we have $\Delta x = 21.25$ and $\Delta t = 18.25$, and thus

$$
\sqrt{\left( \Delta x \right)^2 - \left( \Delta t \right)^2} = = \sqrt{15.625^2 - 9.375^2} = 12.5
$$

 

[NoahsArk]

No, they aren't, because you've left out time. You need to draw a spacetime diagram. Just space alone is wrong.

What I meant to say is, these are their x axis. In the question I am only trying to understand what is happening to their respective x coordinates, and not what's happening to their t coordinates.

 

[PeterDonis]

In the question I am only trying to understand what is happening to their respective x coordinates, and not what's happening to their t coordinates.

And what we are all trying to tell you is that you can't do that, because what happens to their x coordinates and what happens to their t coordinates are connected. You simply can't look at them separately. If you keep trying to all you are going to do is continue to get wrong answers and have problems understanding what's going on and frustrate all of us who are trying to help you.

 

[robphy]

To try and further understand what's happening, I thought about it and made this diagram:
[snip]

While the diagram is correct only if appropriately interpreted, it is limited.

A spacetime diagram below suggest the following interpretation:
It describes how the Earth frame
compares its x-axis tickmarks with the rocket frame's x'-axis tickmarks.
The correspondence follows the Earth frame lines.
Is that how you interpret it?By the principle of relativity, there is a similar diagram with the Earth's x-axis at the bottom and the rocket x-axis at the top... and it will be description of how the rocket frame compares...

But your diagram, as is... with the bottom axis as the rocket frame and the top axis as the Earth frame, can't be used to explain this.

As @PeterDonis says, you really need a spacetime diagram.

Based on my spacetime diagram back in post #11,
you can make a comparison of x-axes by extending lines
parallel to the worldlines (and the timelike diagonal of the diamonds).

In fact, every calculation that you might want to make about this situtation
can be read off by doing some counting on this diagram.