# 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?

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#### 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.

• NoahsArk and Orodruin

#### PeterDonis

Mentor
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.

• NoahsArk, Orodruin and Antarres

#### haushofer

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

• vanhees71

#### 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.

• NoahsArk and PeroK

#### PeroK

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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!

• NoahsArk and Ibix

#### 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.

• NoahsArk

#### Mister T

Gold Member
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.

• NoahsArk

#### robphy

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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
$$L_{proper} = \frac{L_{apparent}}{\sqrt{1-(v/c)^2}},$$
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

#### 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

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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.

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#### 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

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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

Mentor
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.

#### Mister T

Gold Member
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

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

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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

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?