# Julius Caesar Problem from a SpaceTime Physics book

• B
• NoahsArk
In summary: Enterprise frame and the firecracker explodes, then you've already missed the signal. If the Enterprise is closer to Earth, the signal might still reach Caesar in time.
NoahsArk
Gold Member
I've been trying to understand the following very interesting problem:
"Julius Ceasar was murdered on March 15 in the year 44 B.C. at the age of 55 approximately 2000 years ago. Is there some way we can use the laws of relativity to save his life?
Let Caesar's death be the reference event, labeled O: XO = 0, tO = 0. Event A is you reading this exercise. In the Earth frame the coordinates of Event A are XA = 0 light years, and tA = 2000 years. Simultaneous with event A in your frame, Starship Enterprise cruising the Andromeda galaxy sets off a firecracker: event B. The Enterprise moves along a straight line in space that connects it with Earth. Andromeda is 2 million light-years distant in our frame. Compared with this distance, you can neglect the orbit of the Earth around the Sun. Therefore, in our frame, event B has the coordinates XB = 2 x 106 light years, tB = 2000 years. Take Caesar's murder to be the reference event for the Enterprise too (X1O = 0, t1O).

a. How fast must the Enterprise be going in the Earth frame in order that Caesar's murder is happening NOW (that is, t1B) in the Enterprise rest frame? Under these circumstances is the Enterprise moving toward or away from Earth?

d. (I intentionally skipped parts b. and c. for now) Can the Enterprise firecracker explosion warn Caesar, thus changing the course of Earth history? Justify your answer."

For the answer to a. the book works out the answer this way:
"tB = vϒX1B
XB = ϒX1B

We do not yet know the value of X1B. Solve for v by diving the two sides of the first equation by the respective sides of the second equation. The unknown X1B drops out (along with ϒ), and we are left with v in terms of the known quantities tB and XB.
v = tB /XB = 2 x 103 years/ 2 x 106 years = 10-3 = .001

Since the velocity is a positive quantity, the Enterprise is moving away from Earth"

If first saw this problem about four months ago and have re-read it a few times since then, and I want to understand it as best as I can:

1) Is this problem saying that from the Enterprise frame, the exploding of the firecracker and Caesar's death have happened at the same time?
2) In the Earth frame, these events have happened 2,000 years apart, is the reason for the difference in time between these two events in the Enterprise frame the fact that, from the Enterprise frame, the location of Caesar being murdered compared with the firecracker's location is at the "front end" or "leading end" of the horizontal line between those two events? So Caesar's death happens much later (2000 years later) for someone in the Enterprise frame compared with someone in the Earth frame, and this is because of the leading clocks lag principle?
3) The leading clocks lag principle is built into the Lorentz transformations by the vϒX1 term. While qualitatively I think I understand the leading clocks lag principle with illustrations of a beam of light needing to catch up to the front end of a rocket, I still don't quite understand quantitatively why the term vϒX1 captures that principle.
4) With respect to question d., The book says it would be impossible to save Caesar's life because a signal connecting the two events would have to travel at infinite speed. Doesn't this depend on where the Enterprise is located at the moment the fire cracker explodes, and also how fast the Enterprise is moving? If we choose a little faster speed for the Enterprise, can't we create a situation where Caesar's death hasn't happened yet, and therefore the Enterprise can warn Caesar if the Enterprise is close enough to Earth?
5) Does it matter for this problem whether the firecracker exploded inside the Enterprise or outside of it?

NoahsArk said:
Is this problem saying that from the Enterprise frame, the exploding of the firecracker and Caesar's death have happened at the same time?

Yes.

NoahsArk said:
from the Enterprise frame, the location of Caesar being murdered compared with the firecracker's location is at the "front end" or "leading end" of the horizontal line between those two events?

This looks backwards. The Enterprise is moving away from Earth, so the location of Caesar being murdered, on Earth, is behind the Enterprise, and the location of the firecracker, in the Andromeda galaxy, is ahead of it.

NoahsArk said:
The book says it would be impossible to save Caesar's life because a signal connecting the two events would have to travel at infinite speed. Doesn't this depend on where the Enterprise is located at the moment the fire cracker explodes, and also how fast the Enterprise is moving?

The "infinite speed" part does--that's only true in a frame in which the two events are simultaneous (which is the Enterprise frame in this case).

But the "impossible to save Caesar's life" part is not frame-dependent, or dependent on where the Enterprise is located.

NoahsArk said:
If we choose a little faster speed for the Enterprise, can't we create a situation where Caesar's death hasn't happened yet

Yes.

NoahsArk said:
and therefore the Enterprise can warn Caesar if the Enterprise is close enough to Earth?

No. "Hasn't happened yet" isn't enough. Remember the Enterprise is in the Andromeda galaxy, so the signal would have to travel from the Andromeda galaxy to Earth in time to reach Earth before Caesar is murdered. Try calculating the speed the signal would have to travel for various possible speeds for the Enterprise that make the firecracker explosion happen before Caesar's murder in the Enterprise frame. You will see that that speed, even if it is less than infinite, is always faster than the speed of light. (Note that we are assuming that the Enterprise itself travels slower than light; so the "Enterprise" name is perhaps not the best choice since the Star Trek series assumed FTL warp drives existed. )

NoahsArk said:
Does it matter for this problem whether the firecracker exploded inside the Enterprise or outside of it?

No, why would it?

PeterDonis said:
No, why would it?

It just seems to me like, for purposes of doing the Lorentz transformations, it matters whether or not the event that occurred, which we are measuring time for, occurred inside the rocket or inside. Something happening inside the rocket is stationary with respect to the rocket, whereas something happening outside is either moving towards or away from the rocket. It seems like that should affect the calculations. Or, do we just care how far the event happened from the rocket in the X direction, and we don't care if it happened inside the rocket, outside of it, or inside some other rocket moving at a different speed?

Also, why did we say that Caesar's death happened at t= 0 in the rocket frame? We didn't do any calculations to determine that and just randomly chose 0.

NoahsArk said:
Something happening inside the rocket is stationary with respect to the rocket, whereas something happening outside is either moving towards or away from the rocket.

Something happening outside the rocket could still be at rest relative to the rocket. The rocket is moving inertially, so something else could be moving inertially at the same speed.

However, since the firecracker explosion is instantaneous, i.e., a single point in spacetime, it doesn't even make sense to ask what "speed" it is moving at. "Speed" only makes sense for an object that persists over time, i.e., for something with a worldline (like the rocket or Earth), not for a single point in spacetime. The only thing that matters is where in spacetime that point is located, and what relationship that point in spacetime has to the point in spacetime where Caesar gets killed.

NoahsArk said:
Or, do we just care how far the event happened from the rocket in the X direction

What we care about is the spacetime separation between the two events--the firecracker explosion and Caesar's death. That spacetime separation (the spacetime interval) is frame-independent; you can calculate it in any frame. But in a frame where the two events are simultaneous, the spacetime interval is trivial: it's just the X separation between the two events. (And the critical point of the problem is to realize what kind of interval--timelike, null, or spacelike--this interval is, and what that implies about the question in part d.)

NoahsArk said:
why did we say that Caesar's death happened at t= 0 in the rocket frame?

Because the whole point of the problem is to find the inertial frame in which that is true and determine its properties. (More basically, it is to show that there is an inertial frame in which that is true, which tells you something very important about the spacetime relationship between the two points in question--the death of Caesar and the firecracker explosion.)

NoahsArk said:
Something happening inside the rocket is stationary with respect to the rocket, whereas something happening outside is either moving towards or away from the rocket.
Why can something outside the rocket not be at rest relative to it? Suppose a piece breaks off the rocket? Or we open a window, stick our hand out, and let go of whatever we’re holding? Or maybe the rocket doesn’t have an inside because it’s not a closed compartment? For that matter, you and I are on the outside of the Earth yet at rest relative to it.

Nugatory said:
Why can something outside the rocket not be at rest relative to it?

That's true. Maybe what I should have asked is does it matter whether or not the firecracker is at rest relative to the rocket for the calculations, or only how far in the x direction it is.

NoahsArk said:
does it matter whether or not the firecracker is at rest relative to the rocket for the calculations

The firecracker explosion only exists for an instant. It doesn't even make sense to ask whether it is "at rest" relative to anything. Please read my post #4.

I see sorry I initially missed your post #4. So each observer has a lattice of clocks extending infinitely in the positive and negative X directions, and those clocks are synchronized in the observer's own frame. To determine the time for an event outside of one's frame of reference that occurred at some X distance which does not = 0, we use the Lorentz transformations. All that we need to find the time of an event in another frame is the relative speed between the two frames and the X distance of the event in the other frame?

This might be silly and that I'm just getting stuck on how we chose our calendar system, but I'm still not sure why we decided that Caesar's death happened at t = 0. Say Caesar's death happened at t = 2,000. In that case, in the Earth frame, me reading the exercise and the firecracker explosion both would have occurred at t = 4,000, and in the Enterprise frame the explosion and Caesar's death both would have occurred at t= 2,000.

NoahsArk said:
To determine the time for an event outside of one's frame of reference that occurred at some X distance which does not = 0, we use the Lorentz transformations.

There is no such thing as "an event outside of one's frame of reference". All events are "in" all frames.

In general, if you have the ##T## and ##X## of an event in one frame, you can use the Lorentz transformations to get the ##T'## and ##X'## of that event in another frame.

NoahsArk said:
All that we need to find the time of an event in another frame is the relative speed between the two frames and the X distance of the event in the other frame?

In general, to do a Lorentz transformation, you need both the ##T## and ##X## of an event. Just ##X## alone isn't enough.

NoahsArk said:
I'm still not sure why we decided that Caesar's death happened at t = 0

It's just convenience; the math works out simpler that way. See below.

NoahsArk said:
Say Caesar's death happened at t = 2,000. In that case, in the Earth frame, me reading the exercise and the firecracker explosion both would have occurred at t = 4,000

Yes. And that means an extra 2,000 would be there in all the equations that didn't tell you anything useful.

NoahsArk said:
in the Enterprise frame the explosion and Caesar's death both would have occurred at t= 2,000

No. That logic only works if Caesar's death occurs at ##t = 0## in the Earth frame--in that one case, the time coordinate ##t'## in the other frame is also zero.

For any other choice of time coordinate ##t## for Caesar's death in the Earth frame, all you can conclude is that in the Enterprise frame, both the explosion and Caesar's death have the same time coordinate ##t'## (because that's how the Enterprise frame is defined). You cannot conclude that that ##t'## is numerically the same as the ##t## coordinate in the Earth frame. You have to do a full Lorentz transformation to find the actual numerical value of ##t'##.

Have you actually tried to write down the Lorentz transformation equations and evaluate them? If not, you should; it will help with understanding.

I did write down the Lorentz transformations for the case where Earth time for the firecracker = 2,000 years, but haven't done it experimenting with other times for that event in Earth's frame.

PeterDonis said:
No, why would it?

I realized that a source of my misconception might be the way that the leading clocks lag example is illustrated: Typically the example is given of a beam of light shooting from the middle of a rocket to the front. From Earth's perspective the light has to catch up with the nose of the rocket, but from the rocket frame, the nose isn't moving so the light gets there faster. The "event" in this case is the light hitting the nose.

If we are talking about an explosion that happened outside the rocket, the hypothetical light beam from the Earth's perspective doesn't need to "catch up" with anything like it did in the above example. So, that's why it seems like it should matter if the event is happening inside or outside the rocket.

NoahsArk said:
If we are talking about an explosion that happened outside the rocket, the hypothetical light beam from the Earth's perspective doesn't need to "catch up" with anything like it did in the above example.

But the explosion is a single event. When you talk about the light inside the rocket having to "catch up" with the nose from the Earth perspective, you're not talking about a single event; you're talking about a path through spacetime between two events (the light being emitted and the light hitting the nose). And if you want to talk about that for the explosion, you would have to talk about, for example, the path the light takes from the explosion to Earth, and then there is a difference between the frames--in the Earth frame, the Earth is at rest and the light travels to it, while in the rocket frame, the Earth is moving towards the light so they come together faster. And that's true regardless of whether the explosion takes place inside or outside the rocket.

PeterDonis said:
in the Earth frame, the Earth is at rest and the light travels to it, while in the rocket frame, the Earth is moving towards the light so they come together faster. And that's true regardless of whether the explosion takes place inside or outside the rocket.

Hmm, but in this example it means the event took longer in the rocket frame than in the Earth frame, and I am trying to understand this using examples where the opposite is true.

Does the shooting beam of light example even really explain relativity of simultaneity, or is it more fundamental than that?

NoahsArk said:
in this example it means the event took longer in the rocket frame than in the Earth frame

Single events don't take any time at all. They are single points in spacetime. You need at least two events (two points in spacetime) for the concept of "taking some time" to be meaningful.

NoahsArk said:
I am trying to understand this using examples where the opposite is true.

What set of events in the Julius Caesar problem are you referring to here?

NoahsArk said:
Does the shooting beam of light example even really explain relativity of simultaneity, or is it more fundamental than that?

Please start a separate thread if you want to discuss a different example from the one that we are discussing in this thread.

Just to practice solving Lorentz transformation problems, say the firecracker happened at t = 4,000 years in the Earth frame (I'll use the notation t and t1). If I want to know how fast the Enterprise is moving such that Caesar's murder happened simultaneously with it, I could write t = vϒX1 .
4,000 = vϒX1
v = 4,000/ϒX1
v = 4,000/2,000,000 = .002
?

NoahsArk said:
If I want to know how fast the Enterprise is moving such that Caesar's murder happened simultaneously with it, I could write t = vϒX1

I'm not sure what this means. Try writing out the complete Lorentz transformation instead of taking shortcuts. You have two events in the Earth frame, with ##(t, x)## coordinates given by:

Caesar's death: ##(2000, 0)##

Firecracker explosion: ##(4000, 2000000)##.

You are looking for a Lorentz transformation that makes the ##t'## coordinates of both events the same.

I will try it again. I just used the same form, I thought, that the book used. I wrote the Lorentz Transformation in the form where t1 = 0, and I eliminated the C2 term because C = 1.

NoahsArk said:
I wrote the Lorentz Transformation in the form where t1 = 0

Which is wrong. You can't assume ##t' = 0##. It isn't because you set ##t = 2000## instead of ##t = 0## for Caesar's death.

Hi Peter, what I should have said was that I want to find out how fast the Enterprise must be moving so that, in the Enterprise frame, the firecracker and Caeasar's death both happen simultaneously. So, because the problem specifies that the firecracker happened in the Enterprise frame at t′ = 0, t′ also = 0 for Caesar's murder if they both occur at the same time. In the Earth frame, Caesar's murder happened at t = 4,000 which is on the left side of the equation: 4,000 = vϒX1. I know ϒX1 is the same as X (2,000,000 light years), so then I just solved for v.

Also, and this brings up a question about the form of the Lorentz transformation for time: the book expresses the general form as t = vϒX1 + ϒt1. That's the form used for solving this problem with the ϒt1 part missing since t1 = 0. Most sources I've seen write this equation with the vϒX1 all over c2. The book form is much easier, but I'm not totally sure why we can leave out the C2.

Also, as a formatting question, is there a way on this site I can write fractions with the numerator vertically above the denominator instead of having to write the numbers sideways like 2/3?

Btw, the "ϒ"symbol above is gamma as I'm sure shows from the context, but maybe I should be using a a symbol more clearly representing it:)

NoahsArk said:
what I should have said was that I want to find out how fast the Enterprise must be moving so that, in the Enterprise frame, the firecracker and Caeasar's death both happen simultaneously

Yes, we are already clear on that.

NoahsArk said:
because the problem specifies that the firecracker happened in the Enterprise frame at t′ = 0, t′ also = 0 for Caesar's murder if they both occur at the same time

Yes, but when you made the time ##t## in the Earth frame of Caesar's death equal to ##2000## instead of ##0##, you changed the conditions of the problem. The problem as stated in the OP says that the time ##t## in the Earth frame of Caesar's death is ##t = 0##, not ##t = 2000##. And that condition is what allows you to say that ##t' = 0## in the Enterprise frame for both the firecracker and Caesar's death. If you put ##t = 2000## for Caesar's death in the Earth frame, you can no longer put ##t' = 0## for Caesar's death in the Enterprise frame. You have to figure out what ##t'## is for Caesar's death in the Enterprise frame (which will also be ##t'## for the firecracker explosion in the Enterprise frame) by solving the Lorentz transformation equations. It won't end up being ##0##.

NoahsArk said:
In the Earth frame, Caesar's murder happened at t = 4,000

Not according to what you said earlier:

NoahsArk said:
Say Caesar's death happened at t = 2,000.

You need to make up your mind.

NoahsArk said:
the book expresses the general form as t = vϒX1 + ϒt1.

I assume that "Y" here is supposed to be ##\gamma##, correct? It will be really helpful if you start using the PF LaTeX feature to write equations the same way they appear in the books. You can find help on the LaTeX feature here:

https://www.physicsforums.com/help/latexhelp/
NoahsArk said:
I know ϒX1 is the same as X (2,000,000 light years)

No, you don't. You would if you could assume that ##t' = 0## for Caesar's death, but you can't with your changed conditions. See above.

NoahsArk said:
That's the form used for solving this problem with the ϒt1 part missing since t1 = 0.

NoahsArk said:
I'm not totally sure why we can leave out the C2.

Because we are using units in which ##c = 1##. We are measuring distance in years and time in light-years, so ##c = 1## light-year per year. Units like this are very convenient when solving relativity problems precisely because they free you from having to sprinkle factors of ##c## all over the equations.

NoahsArk said:
is there a way on this site I can write fractions with the numerator vertically above the denominator

Use the LaTeX feature. See above.

NoahsArk said:
the "ϒ"symbol above is gamma as I'm sure shows from the context, but maybe I should be using a a symbol more clearly representing it

Yes, using the LaTeX feature. See above.

Thank you for the helpful explanations. I will start learning LaTex. As I see it involves some learning code, excuse me for continuing the old notation for a bit until I get used to it.

Ok, so my main error in post #18, as you pointed out, was to say that Caesar's murder happened at t= 4,000 light years. What I meant to say, was that the firecracker happened in the Earth frame at t = 4,000 light years. Had I said that and based the equation on that, would my calculations have made sense? In this variation Caesar's murder still happened at t = 0, and we are just assuming the firecracker happened 4,000 years later instead of 2,000.

I'm still wondering about some of the fundamentals behind these equations:

The book gives the X distance from Earth to the firecracker as 2,000,000 light years. In order to find that same distance, X1, in the Enterprise frame, we'd multiply 2,000,000 by γ correct? Does it matter where the Enterprise is flying at the moment of the explosion in order to determine X1?

In the original problem, the Enterprise had to be traveling at .001c in order for the firecracker explosion to have happened at t1 = 0. The book specifically tells us that Caesar's death in the Enterprise frame is t1 = 0. The problem with that is, though, that in order for Caesar's death to be 0 both in the Enterprise frame and the Earth frame, the Enterprise would have to be traveling at a speed of v = 0 with respect to the Earth. That would contradict the finding above that v = .001 c. Please let me know what I'm missing.

Also, and sorry if this seems very basic, but what is really meant by "reference event"? Does that mean an event at which a particular observer starts their stop watch? I am assuming that two different frames of reference can have the same reference event, even if they both use different calendar systems? For example, in the Earth frame, it's the year 2,000, and if the Enterprise frame uses the Hebrew calendar where it is now the year 5,779, that wouldn't change the problem correct?

NoahsArk said:
What I meant to say, was that the firecracker happened in the Earth frame at t = 4,000 light years. Had I said that and based the equation on that, would my calculations have made sense?

Not if Caesar's death happens at ##t = 2000##.

NoahsArk said:
In this variation Caesar's murder still happened at t = 0, and we are just assuming the firecracker happened 4,000 years later instead of 2,000.

If Caesar's death still happens at ##t = 0##, then yes, you can still assume that ##t' = 0## for both events (Caesar's death and firecracker explosion) in the Enterprise frame. Then the only change from the original scenario will be a different relative speed for the two frames.

NoahsArk said:
The book gives the X distance from Earth to the firecracker as 2,000,000 light years. In order to find that same distance, X1, in the Enterprise frame, we'd multiply 2,000,000 by γ correct?

What do the Lorentz transformation equations say?

NoahsArk said:
Does it matter where the Enterprise is flying at the moment of the explosion in order to determine X1?

Does changing where the Enterprise is flying change where the explosion is?

NoahsArk said:
in order for Caesar's death to be 0 both in the Enterprise frame and the Earth frame, the Enterprise would have to be traveling at a speed of v = 0 with respect to the Earth

Why?

NoahsArk said:
what is really meant by "reference event"?

As the book is using the term, it appears to mean the event that is chosen as the origin of the coordinates, i.e., ##(0, 0)##.

NoahsArk said:
in the Earth frame, it's the year 2,000, and if the Enterprise frame uses the Hebrew calendar where it is now the year 5,779, that wouldn't change the problem correct?

It wouldn't change the physics, but if Hebrew years aren't the same length as Earth years, it would change the units.

PeterDonis said:
What do the Lorentz transformation equations say?

If I'm not mistaken we would use the inverse Lorentz transformation for distance here:
X1 = γ X - γvt1
So, in the original problem, at a speed of .001, it would be:
X1 = 1.0005(2,000,000) - 1.0005(.001 x 0)
X1 = 2,001,000 - 0
X1 = 2,001,000

Is this correct?

PeterDonis said:
Does changing where the Enterprise is flying change where the explosion is?

If the Enterprise were passing someone on Earth and going to their left, then I think we'd have to use the regular and not the inverse Lorentz transformations to get from X1 to X.

PeterDonis said:
Why?

Because if Caesar's death were zero in both frames, then the equation would be:
t = γt1 + vX1γ
o = o + vγX1
Since γ can't be zero and 1 isn't zero, v must be zero.

PeterDonis said:
It wouldn't change the physics, but if Hebrew years aren't the same length as Earth years, it would change the units.

Ok that makes sense.

NoahsArk said:
If I'm not mistaken we would use the inverse Lorentz transformation for distance here

You're mistaken. Do you know what the Lorentz transformation equations are? Just write them down. You're making this way harder than it needs to be. The equations you're looking for will have ##t'## and ##x'## on the left and only unprimed quantities (plus ##\gamma## and ##v##) on the right. You need to write down the transformations for two events: Caesar's death and the firecracker explosion.

Also, the transformation you're looking for is from the Earth (unprimed) frame to the Enterprise (primed) frame, which is normally not called the "inverse" transformation--the inverse transformation would be from the primed to the unprimed frame.

NoahsArk said:
If the Enterprise were passing someone on Earth and going to their left, then I think we'd have to use the regular and not the inverse Lorentz transformations to get from X1 to X.

You have it backwards. The transformation from the primed (Enterprise) to unprimed (Earth) frame is the inverse transformation. See above.

NoahsArk said:
then the equation would be

No, it wouldn't. You are very confused.

Also, please start using the LaTeX feature and the correct symbols like ##\gamma##. I suspect that your trying to cobble together your notation from ordinary characters is not helping. Help on the LaTeX feature is here:

https://www.physicsforums.com/help/latexhelp/

What I meant to say, was that the firecracker happened in the Earth frame at t = 4,000 light years.

That's even worse! "Light years" is a measure of distance, not time. I presume you meant to say "4000 years" but where did you get the 4000? There is no "4000 years" in your first post.

HallsofIvy said:
That's even worse! "Light years" is a measure of distance, not time.

Because I learned in the spacetime physics book that time and distance can both be measured in light years and by keeping them in the same units we can do geometric calculations. 4,000 light years of time is the same as saying 4000 years in my e.g.

The 4,000 wasn't in the first post because in the first post I used the e.g. from the book. I put the 4,000 in there just to play with the numbers and see if I could solve a different example.

NoahsArk said:
I learned in the spacetime physics book that time and distance can both be measured in light years

Can you give a specific reference?

PeterDonis said:
Can you give a specific reference?

When I looked near the beginning I saw these two examples:

Solution to Sample Problem 1-2:
"The space separation measured in the lab equals two meters, as given in the problem. A flash of light would take 2 meters of light-travel time to travel between the two detectors..."

Part 1.5 "Unity of Spacetime"

"time and space: equal floating but distinct nature

When time and space are measured in the same unit - whether meter or second or year - the expression for the square of the spacetime interval between two events takes on a particularly simple form:

(time separation)2 - (space separation)2 = (interval)2"

NoahsArk said:
When I looked near the beginning I saw these two examples:

Those are examples using meters. Can you find a place where the book specifically uses "light years" for time instead of "years"? (Note that in the second example you quote, "year" is given as a possible unit, not "light year".)

I may have been mistaken.

I see I am botching this in more ways than one

After thinking about it a bit more, no, I did not botch this part of the problem.

Not only are we allowed to express time in terms of a distance, we must express it in terms of a distance when we express the X term in terms of a distance. Alternatively, if we express the X term in terms of a time, we must express the t term in terms of a time. For example, if I say 5 meters of time happened on Earth when you traveled a distance of 4 meters in your rocket, I mean that 5 meters of light travel time occurred (or about 16 nanoseconds). When I express both my time and your distance of travel in terms of a distance, I can figure out, by taking the square of the difference of the squares, that 3 meters of time occurred for you in your rocket. I couldn't do that if I used 16 nanoseconds for me and 4 meters for you. I could use 16 nanoseconds only if for you I converted 4 meters into a time which is about 13.2 nanosecond. If I take the square of the difference of those times I get around 9.9 nanoseconds of time having passed for you which is the same as the meters of light travel time.

Same thing in the example where t = 4,000 years. It means that in that time light traveled a distance of 4,000 light years.

NoahsArk said:
Not only are we allowed to express time in terms of a distance, we must express it in terms of a distance when we express the X term in terms of a distance.

You are confusing yourself even more, not to mention going off on a tangent that has nothing to do with the actual substance of your original errors.

We don't "express time in terms of a distance". We choose units of time and distance, and the most convenient choices of units are those for which ##c## comes out to be numerically equal to ##1##, which means we can just leave ##c## out of all the formulas altogether, making everything simpler.

Whether you want to call units of time and distance chosen this way by the same names (like "meters" of both distance and time), or related names (like "years" and "light-years", or for that matter "seconds" and "light-seconds", another fairly common choice), or even completely different names, is a matter of words, not physics. It doesn't really matter what choice you make, as long as it's consistent. But if you're going to use one of the common choices from the literature, you need to just use it, not make up your own version of it. Still less should you bother trying to argue that your made-up version is somehow more "right" than another version. That's pointless; you're arguing over choice of words, not physics.

PeterDonis said:
You're mistaken. Do you know what the Lorentz transformation equations are? Just write them down.

Lorentz transformation for t: $$t = \gamma t^1 + v\gamma x^1$$
for x: $$x = \gamma x^1 + v \gamma t^1$$
Inverse Lorentz transformation for t^1: $$t^1 = \gamma t - v \gamma x$$
Inverse Lorentz transformation for x^1: $$x^1 = \gamma x - v \gamma t$$

Regarding my question about how to find the x^1 distance from Earth to the firecracker, can't I choose either the LT for x or the Inverse LT for x^1 and solve for the unknown x^1 in both cases? E.g.

x = ## \gamma x^1 + v \gamma t^1 ##
## 2,000,000 = \gamma x^1 + .001 \gamma t^1 ##
Since we know t^1 = zero since it's given that the firecracker happened at t^1 = 0:
## 2,000,000 = \gamma x^1 ##
2,000,000 = 1.0005x^1
x^1 = 2,000,000/1.0005
x^1 = 1,999,000

I won't do the math here for the Inverse LT solving for x^1, but I assume I'd get the same answer?

hi i just read this question, and have a question of my own. for me, no matter how fast i travel or in what direction, i will never see the murder right? because the murder has a time like interval with me, and simultaneous events have spacelike interval, and interval length is the same to all observers and no matter how i travel i will never reach an event with spacelike interval to the murder. is this correct reasoning?

kent davidge
black hole 123 said:
hi i just read this question, and have a question of my own. for me, no matter how fast i travel or in what direction, i will never see the murder right? because the murder has a time like interval with me, and simultaneous events have spacelike interval, and interval length is the same to all observers and no matter how i travel i will never reach an event with spacelike interval to the murder. is this correct reasoning?
if you are far far away, the interesting information "Caesar was murdered" might not have reached you yet, so you would see the murder in a sense

NoahsArk said:
Lorentz transformation for t:

$$t = \gamma t^1 + v\gamma x^1$$

for x:

$$x = \gamma x^1 + v \gamma t^1$$

Inverse Lorentz transformation for t^1:

$$t^1 = \gamma t - v \gamma x$$

Inverse Lorentz transformation for x^1:

$$x^1 = \gamma x - v \gamma t$$

You have these backwards. The ones with the plus signs are the inverse transformations. The ones with the minus signs are the ordinary transformations.

Also, standard notation has primes: ##t'##, ##x'##; not superscript ##1##s.

NoahsArk said:
Regarding my question about how to find the ##x^1## distance from Earth to the firecracker, can't I choose either the LT for ##x## or the Inverse LT for ##x^1## and solve for the unknown ##x^1## in both cases?

No. I don't understand why you keep trying to guess shortcuts instead of using the transformation equations as they are meant to be used. Perhaps it's because you don't understand how they are meant to be used. Here's how:

You have two events, Caesar's death and the firecracker explosion, whose coordinates in the Earth (unprimed) frame you know. In other words, for each event, you have a pair of coordinates ##(t, x)##. Then, for each event, you use the LT equations for ##(t', x')## in terms of ##(t, x)## to obtain the coordinates in the primed (Enterprise) frame, ##(t', x')##, for each of the events.

In this particular case, you are trying to find the relative velocity ##v## (and its associated ##\gamma## factor) that will make the value of ##t'## the same for both events. So you need to write out the transformation equations for both events and find the value of ##v## that makes ##t'## numerically the same for both. Then, once you know ##v## (and therefore ##\gamma##), you compute ##x'## for both events using the transformation equation for ##x'## and take the difference of the ##x'## values to get the distance in the Enterprise frame.

Replies
27
Views
2K
Replies
98
Views
4K
Replies
24
Views
2K
Replies
124
Views
15K
Replies
4
Views
3K
Replies
29
Views
9K
Replies
2
Views
7K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
10
Views
5K