Special Relativity and lightning

[friendbobbiny]

Homework Statement

This is a relativity problem:
[/B]
A tree and a pole are 3000 m apart. Each is hit by a bolt of lightning. Mark, who is standing at rest midway between the two, sees the two lightning bolts at the same instant of time. Nancy is flying her rocket at v = 0.5c in the direction from the tree toward the pole. The lightning hits the tree just as she passes by it. Define event 1 to be “lightning strikes tree” and event 2 to be “lightning strikes pole.” For Nancy, does event 1 occur before, after, or at the same time as event 2?

The actual answer: After

Homework Equations

1) The speed of light is constant in all reference frames[/B]

The Attempt at a Solution

I really can't figure this one out. The classic example, in which a train rider sees two lighting bolts, differs in that the rider is at the middle of the train. Thus, the example prepares us for a seemingly impossible result.

Here, Nancy literally passes the train as it begins to move right. Doesn't she see event 1 first then?[/B]

 

[friendbobbiny]

Homework Statement

This is a relativity problem:
[/B]
A tree and a pole are 3000 m apart. Each is hit by a bolt of lightning. Mark, who is standing at rest midway between the two, sees the two lightning bolts at the same instant of time. Nancy is flying her rocket at v = 0.5c in the direction from the tree toward the pole. The lightning hits the tree just as she passes by it. Define event 1 to be “lightning strikes tree” and event 2 to be “lightning strikes pole.” For Nancy, does event 1 occur before, after, or at the same time as event 2?

The actual answer: After

Homework Equations

1) The speed of light is constant in all reference frames[/B]

The Attempt at a Solution

I really can't figure this one out. The classic example, in which a train rider sees two lighting bolts, differs in that the rider is at the middle of the train. Thus, the example prepares us for a seemingly impossible result.

Here, Nancy literally passes the train as it begins to move right. Doesn't she see event 1 first then?

[/B]

 

[PeroK]

How are you going to work out the time the events 1 and 2 take place in Nancy's frame?

 

[TSny]

Yes, Nancy sees flash 1 before flash 2. But that doesn't mean that she assigns an earlier time to event 1. If an explosion occurs a mile away, I know that the explosion occurred at a time earlier than when I hear the explosion. Nancy knows that event 2 occurs before she sees flash 2.

 

[Chestermiller]

This is a perfect problem for getting some practice applying the Lorentz Transformation.

Event 1: x = 0, t = 0, x' = 0, t' = 0

Event 2: x = 3000, t = 0, x ' = ?, t' = ?

Chet

 

[friendbobbiny]

Yes, Nancy sees flash 1 before flash 2. But that doesn't mean that she assigns an earlier time to event 1. If an explosion occurs a mile away, I know that the explosion occurred at a time earlier than when I hear the explosion. Nancy knows that event 2 occurs before she sees flash 2.

Based on the suggestion you posted, I have the following. Is it sound?

Shortly after seeing flash 1, Nancy sees flash 2. How far ahead of flash 1 does she see flash 2? We can be exact about the distance by observing Nancy from an observer's frame of reference. Why determine this distance? If we know when flash 2's ray hits nancy, we know what distance the ray had to travel. If we know what distance ray 2 had to travel, we can determine the time it needed to travel that far -- according to Nancy.

_________A light ray from event 2 is traveling towards Nancy at a net speed of 1.5(c). The ray travels towards Nancy at that speed, because the light ray travels at c and nancy travels towards the ray at 0.5c; thus, both converge at the rate of 1c + 0.5(c).

let d = distance between points
We can represent the intersection of Nancy and light wave ii as if one object were moving from beginning to end at 1.5c
Then
$$d = 1.5c * t$$
$$t = d/1.5c $$
$$d_{travelled} = d_{initial} + t * v_{nancy}$$ where $$d_{initial} = 0$$
A little algebra tells us that the second event's ray hits Nancy at x = 1000.
_________

Nancy believes that ray 2 travels toward her at c (this is the second postulate). Yet, she also knows that ray 2 traveled 2000 m to reach her. Thus, while she observes event 1 at the moment it occurs, she knows that in the time since -- in the time it took her to travel 1000 m -- that ray 2 traveled 2000m. Ray 2 must have taken longer to reach her than it took her to reach x= 1000 from x = 0. This can only occur if, according to her, ray 2 occurred first

 

[TSny]

A light ray from event 2 is traveling towards Nancy at a net speed of 1.5(c). The ray travels towards Nancy at that speed, because the light ray travels at c and nancy travels towards the ray at 0.5c; thus, both converge at the rate of 1c + 0.5(c).

It is true that according to Mark, the light from the pole has a speed relative to Nancy of 1.5c.

let d = distance between points

When defining a distance, always state for which reference frame the distance is defined. I think you mean Mark's frame.

We can represent the intersection of Nancy and light wave ii as if one object were moving from beginning to end at 1.5c
Then
$$d = 1.5c * t$$
$$t = d/1.5c $$
$$d_{travelled} = d_{initial} + t * v_{nancy}$$ where $$d_{initial} = 0$$
A little algebra tells us that the second event's ray hits Nancy at x = 1000.

Again, when stating the value of x, be sure to indicate which frame of reference. But I think your analysis is correct if x is measured in Mark's frame.

Nancy believes that ray 2 travels toward her at c (this is the second postulate).

Good.

Yet, she also knows that ray 2 traveled 2000 m to reach her. ]

Mark would say that ray 2 traveled 2000 m to reach Nancy, but Nancy would not agree.

Thus, while she observes event 1 at the moment it occurs, she knows that in the time since -- in the time it took her to travel 1000 m -- that ray 2 traveled 2000m.

From Nancy's point of view, she doesn't travel any distance. She is at rest in her own frame. Mark is the one who says Nancy travels 1000 m while ray 2 travels to Nancy from the pole.
------------------------------------------------------------
I don't think you need to do any calculations in order to answer this question. See if you can answer the following questions about Nancy's observations by drawing some pictures from the point of view of Nancy. For Nancy, the tree, pole and Mark are all traveling relative to Nancy at the same speed. Let D be the distance between the tree and the pole according to Nancy.

Draw a picture of the Tree, the pole, and Mark at the instant the light flash strikes the tree. Then draw a second picture that shows the flash 1 arriving at Mark. Remember, these pictures are to be drawn from the point of view of Nancy. Is the distance that flash 1 travels from the tree to Mark greater than D/2, less than D/2, or equal to D/2 according to Nancy?

Repeat by drawing two pictures for flash 2 and deduce whether the distance that flash 2 travels from the pole to Mark is greater than D/2, less than D/2, or equal to D/2 according to Nancy.

From your answers to these two questions you should be able to decide which lightning flash occurred first according to Nancy.

 

[Chestermiller]

You really should try the method I suggested in post #5. It only takes a minute or two, and there is no complicated reasoning involved.

Chet

 

[friendbobbiny]

Draw a picture of the Tree, the pole, and Mark at the instant the light flash strikes the tree. Then draw a second picture that shows the flash 1 arriving at Mark. Remember, these pictures are to be drawn from the point of view of Nancy. Is the distance that flash 1 travels from the tree to Mark greater than D/2, less than D/2, or equal to D/2 according to Nancy?

Repeat by drawing two pictures for flash 2 and deduce whether the distance that flash 2 travels from the pole to Mark is greater than D/2, less than D/2, or equal to D/2 according to Nancy.

From your answers to these two questions you should be able to decide which lightning flash occurred first according to Nancy.

1) The distance is = D/2. Before nancy can pass mark, the light ray overtakes her and reaches Mark.
2) She thinks that the light traveled > D/2. Right will only reach her after it first reaches Mark.

If flash 2 traveled further, she reasons that it must have started earlier.

 

[PeroK]

1) The distance is = D/2. Before nancy can pass mark, the light ray overtakes her and reaches Mark.
2) She thinks that the light traveled > D/2. Right will only reach her after it first reaches Mark.

If flash 2 traveled further, she reasons that it must have started earlier.

You should see Chet's comments above. When you're trying to learn SR, I think it's a bad idea to start worrying about the finite speed of light delaying observations. This makes the whole thing more complicated.

First, you need to understand the difference between an observer and an inertial reference frame. An inertial reference frame is essentially a set of observers, all at rest with respect to each other and with synchronised clocks - and each stationed at a critical position for your experiment. They can note down the time events happen at their position and "compare notes afterwards" to get a complete picture.

An observer has to take into account the finite speed of light when making observations any significant distance away. So, one observer in a reference frame is not enough to keep things simple. Wherever there is an event, you should imagine an observer (for each reference frame) at that point ready to measure the time of the event.

Another reason is that SR is not about the finiteness of the speed of light delaying observations. This happens in classical physics as well. SR is about the differences in observations between inertial reference frames.

 

[TSny]

1) The distance is = D/2. Before nancy can pass mark, the light ray overtakes her and reaches Mark.

If you visualize what's going on in Nancy's reference frame, then Nancy does not move. She's at rest as the tree, the pole, and Mark move at 0.5c.

2) She thinks that the light traveled > D/2. Right will only reach her after it first reaches Mark.

Which light flash are you referring to here?

If flash 2 traveled further, she reasons that it must have started earlier.

Yes.

The attachment shows flash 1 from the point of view of Nancy's reference frame. She remains at rest while the other objects move. Figure (a) shows flash 1 at the tree and figure (b) shows flash 1 at the instant it reaches Mark. The question is how far did flash 1 travel from the tree to Mark according to Nancy: Greater than, less than, or equal to D/2?

Construct similar figures for flash 2.

If you are familiar with the Lorentz transformation equations, then you can use them to easily answer the original question as Chet suggests. Sometimes these equations are not covered in introductory physics classes.

 

[friendbobbiny]

Yes, you guys are right. Using lorentz transformations makes the problem much easier. For anyone taking help like me, please see the image for a setup.

Let t' represent the time in Nancy's frame. To solve for the time of each event, substitute the appropriate spacetime coordinates (x, t) from Mark's frame.

Let me still try and understand this question conceptually:

The attachment shows flash 1 from the point of view of Nancy's reference frame. She remains at rest while the other objects move. Figure (a) shows flash 1 at the tree and figure (b) shows flash 1 at the instant it reaches Mark. The question is how far did flash 1 travel from the tree to Mark according to Nancy: Greater than, less than, or equal to D/2?

How far did flash 1 travel from the tree to Mark, according to Nancy?
< D/2

Why?
From Nancy's frame, Mark moves towards her. Thus, light hits him some distance before D/2.

How far did flash 2 travel from the tree to Mark, according to Nancy?
> D/2

Why?
From Nancy's frame, Mark moves away from the second light. Thus, light hits him at some distance past D/2.

 

[TSny]

Let me still try and understand this question conceptually:
How far did flash 1 travel from the tree to Mark, according to Nancy?
< D/2

Why?
From Nancy's frame, Mark moves towards her. Thus, light hits him some distance before D/2.

How far did flash 2 travel from the tree to Mark, according to Nancy?
> D/2

Why?
From Nancy's frame, Mark moves away from the second light. Thus, light hits him at some distance past D/2.

Sounds good!

 

[friendbobbiny]

How far did flash 1 travel from the tree to Mark, according to Nancy?
< D/2

How far did flash 2 travel from the tree to Mark, according to Nancy?
> D/2

I've learned how to understand what Nancy sees. It seems that I don't understand, however, why her observations lead her to conclude that event 2 took place first. This is what I originally thought:

If flash 2 traveled further, she reasons that it must have started earlier.

I find it problematic to claim that a greater distance implies an earlier start time. This claim makes sense only if we maintain that Nancy sees both rays hit Mark at the same time. In that case, event ii must take place earlier, if event ii's ray is to travel a greater distance.

Why is this true? Both rays might simultaneously hit Mark in his reference frame, but what justifies both rays hitting Mark in Nancy's frame?

 

[collinsmark]

Why is this true? Both rays might simultaneously hit Mark in his reference frame, but what justifies both rays hitting Mark in Nancy's frame?

Two events that share the same point in space and time in a given reference frame will appear simultaneous in all frames. [Edit: in a sense, these two events merge into a single event.] This is only true if the events share the same point in space (and time) though.

As we've encountered, absolute simultaneity does not hold true for events separated by a distance. It's only when that distance and time between events is zero does simultaneity hold for all frames for those two events.

If we treat Mark as a point particle, and he observes both rays striking him at the same time, then Nancy will observe Mark reacting to both rays simultaneously, regardless of her own relative velocity.

 

[friendbobbiny]

It's only when that distance and time between events is zero does simultaneity hold for all frames for those two events.

Why is this true? That simultaneity holds for all frames given those conditions (for distance and time) in one frame?

 

[collinsmark]

Why is this true? That simultaneity holds for all frames given those conditions (for distance and time) in one frame?

When two events happen at the exactly at the same place and at the same time, as measured in any given frame, those two events are really just one, single event. And it will be observed as a single event in all frames. [Edit: I must re-iterate though that this only applies to events that share the same point in both space and in time. It does not apply to events in the same place but at different times, nor does it apply to events measured to be "simultaneous" in a given frame, but at different points in space. It only applies when events share points in both space and time, in any given frame of reference.]

Transforming from one frame of reference to another only involves transforming spacetime coordinates, but does not alter the spacetime events themselves. If two events share the same spacetime coordinates in one frame of reference, they share the same coordinates in any frame of reference. (The numerical value of the shared coordinates will be different one one frame than the value of the shared coordinates of another frame; what I mean to say is that the events will still be shared, no matter what the frame. If they are shared in one frame, then they are shared in all frames.)

--- ---
Allow me to introduce a new tool that might be helpful.

If you continue to learn about special relativity, there is tool you will learn about called a "spacetime diagram," if you haven't learned about it already. Spacetime diagrams are very useful and insightful.

The general idea is that you plot time on the vertical axis and space on the horizontal one. This corresponds to some particular, inertial frame of reference.

Every point on the graph represents an "event." You can transition between frames, not by altering any of the events, but rather by altering the coordinates: You change the orientations of the axes, without altering the events at all.

The math is a little different from Euclidean math though. The length between two points on angled line is not $\sqrt{ \left( \Delta x \right)^2 + \left( c \Delta t \right)^2}$ as you might expect, but rather it's $\sqrt{ \left( \Delta x \right)^2 - \left( c \Delta t \right)^2}$ for space-like separations and $\sqrt{ -\left( \Delta x \right)^2 + \left( c \Delta t \right)^2}$ for time-like separations. Also, hyperbolic functions replace trigonometric ones. But that's all in the details. One neat thing about them is that you can use spacetime diagrams to do Lorentz transformations geometrically (among other uses).

The thing I'm trying to point out with this is that as you change from one frame to another, none of the events on the paper are altered. The only thing that changes is the coordinate system that indicates relative distance and time intervals.

If two events share the same spacetime coordinates (both time and space) in a given frame, they correspond to a single point on the spacetime diagram. When changing frames, the points (i.e., events) are not modified. Only the diagram's axes are modified. But since the points are not modified, a single point in one frame will remain a single point in all frames.

 

[kumusta]

Given two events $\dots~\rm E_1~,$ lightning strikes tree $\dots~\rm E_2~,$ lightning strikes pole $\dots$
Given two observers $\dots$ Mark (in fixed frame) $\dots$ Nancy (in frame moving with speed $v = (0.5)c~$ from the tree on the left to the pole on the right) $\dots$
Mark, standing at rest midway between tree and pole separated by a distance of 3000 m, sees the two lightning bolts struck and hit at the same time $\dots$
The lightning hits the tree as Nancy (in moving frame using primed variables) just passes by it$~\dots$
$~\Rightarrow~(x_{1\rm N}' ,t_{1\rm N}') = (0~,0)~\Leftarrow~\begin{cases}\begin{align} &\text {origin of fixed and moving frame} \nonumber \\ &\text {set at the tree at that instant}\nonumber~\dots\end{align}\end{cases}$
$\rm E_1~$according to Mark (in fixed frame using unprimed variables):$~(x_{1\rm M} ,t_{1\rm M}~) = (0~,0)$ ...
$\rm E_2~$according to Mark: $~(x_{2\rm M} ,t_{2\rm M}) = (3000~\rm m~,0)$ ...
$\rm E_2~$according to Nancy: $~(x_{2\rm N}' ,t_{2\rm N}') = (? , ?)$ ...
To know if Nancy sees $\rm E_1~$occur before, after, or at the same time as $\rm E_2~$, the time$~t_{2\rm N}'~$above must be found. For this purpose, we use the Lorentz-Einstein transformation equation for time from the fixed to the moving frame directed toward the $(+x)$ axis: $t' = γ( t - vx/c^2 )~~\Leftarrow~~\begin{cases}\begin{align} &γ = (1 - v^2/c^2)^{-1/2}~\Rightarrow~v = (0.5)c~\Rightarrow~v/c = 0.5~\dots\nonumber \\ &\Rightarrow~1 - (v/c)^2 = 1 - 0.25 = 0.75~\dots\nonumber \\ &\Rightarrow~[1 - (v/c)^2]^{1/2} = 0.866025~\dots\nonumber \\ &\Rightarrow~γ = [1 - (v/c)^2]^{-1/2} = (0.866025)^{-1} = 1.1547~\dots\nonumber\end{align}\end{cases}$
$\Rightarrow ~t_{2\rm N}' = γ( t_{2\rm M} - vx_{2\rm M}/c^2 ) =\begin{cases}\begin{align} &(1.1547)\{ 0 - [(0.5)]c(3000~\rm m)/c^2\}~\dots\nonumber \\ & -(1.1547)(0.5)(3000~\rm m)/c\nonumber \\ &-(1.1547)(0.5)(3000~\rm m)/[3×10^8~\rm m ⋅ {\rm s}^{-1}]~\dots\nonumber \\ &\rm -5.7735×10^{-6}~\rm s ~\dots\nonumber\end{align}\end{cases}$
$t_{2\rm N}'~\lt~0 = t_{1\rm N}'~\Rightarrow~\rm E_1~$occurs after$~\rm E_2~$.
The same result comes out by using the suggestion made by Chestermiller in post #5:

This is a perfect problem for getting some practice applying the Lorentz Transformation.
Event 1: x = 0, t = 0, x' = 0, t' = 0
Event 2: x = 3000, t = 0, x ' = ?, t' = ?

Exactly how it should be done he did not say, but I suppose it was by using the Lorentz-Einstein transformation equations for the space coordinates:
$\dots~x' = γ(x - vt)~\dots~x = γ(x' + vt')~\dots~γ = (1 - v^2/c^2)^{-1/2}~\dots~$

 

[PeroK]

@kumusta the necromancer strikes again!