# B Julius Caesar Problem from a SpaceTime Physics book

#### PeterDonis

Mentor
I learned in the spacetime physics book that time and distance can both be measured in light years
Can you give a specific reference?

#### NoahsArk

Gold Member
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"

#### PeterDonis

Mentor
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".)

#### NoahsArk

Gold Member
I may have been mistaken.

I see I am botching this in more ways than one

#### NoahsArk

Gold Member
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.

#### PeterDonis

Mentor
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.

#### NoahsArk

Gold Member
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 wont do the math here for the Inverse LT solving for x^1, but I assume I'd get the same answer?

#### black hole 123

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

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

#### PeterDonis

Mentor
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.

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.

#### PeterDonis

Mentor
for me, no matter how fast i travel or in what direction, i will never see the murder right?
No. You can certainly see the murder in the sense of receiving light signals from it.

What you can't do, if you are not in the past light cone of the murder event, is to be present at the murder.

#### PeterDonis

Mentor
Since we know t^1 = zero since it's given that the firecracker happened at t^1 = 0
Only if Caesar's death happens at $t' = 0$ (and $t = 0$). But I thought you wanted to compute the different case where Caesar's death happens at $t = 2000$.

#### jbriggs444

Homework Helper
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 currently in the future light-cone of the murder then you will always remain so. You can never catch up to the light rays that carry the image of the murder. You will never be able to turn around, point a telescope in the right direction and see the murder directly.

Note that "seeing the murder" is not about being simultaneous. In order to see the murder (directly), you would need to be on the murder's future light-cone. The separation would need to become lightlike, not simultaneous.

#### NoahsArk

Gold Member
Peter, please let me know if 1,999,000 is the correct answer for the distance from Earth to the firecracker from the Enterprise perspective.

Thank you.

Mentor

#### PeterDonis

Mentor
And I don't mean what you showed in post #32. I already responded to that. I mean do the calculation I described in post #35.

#### NoahsArk

Gold Member
I posted four equations, in post 32, and I used the equation $x = \gamma x^1 + v \gamma t^1.$ Then I solved for x^1 by isolating it, and got 1,999,000. This value is less than the X distance, which is a sign that it's correct since according to the Enterprise frame, the Earth frame is moving and Enterprise should therefore see all distances in the Earth frame as shorter. (sorry for using the superscript 1 notation for now- I am still working on learning LaTex)

If I use the equation $x ^1 = \gamma x - v \gamma t$
$x^1 = \gamma x - 0$
$x^1 = \gamma 2,000,000 - 0$
$x^1= 1.0005(2,000,000)$
$x^1 = 2,001,000.00$

This answer must be wrong since it is a greater distance than the distance in the Earth frame. Please let me know if my first answer isn't right, and if not, what the correct answer is.

If you know of a good book or website where I can just drill myself on LT problems, where I can also look up the answers when I'm done, I would appreciate it.

#### Nugatory

Mentor
(I intentionally skipped parts b. and c. for now
I think you will find this problem much easier if you don’t skip these. They’re there for a reason.

#### PeterDonis

Mentor
If I use the equation x1=γx−vγt
Then you need to plug in the right value for $t$. You are trying to find the $x'$ (please start using a prime and stop using a superscript $1$, you need to learn standard notation) coordinate of the firecracker explosion; the firecracker explosion does not take place at $t = 0$ in the Earth frame.

This is why I keep telling you to stop guessing shorcuts and do the Lorentz transformation. The Lorentz transformation transforms coordinates of events. It doesn't transform lengths or times. If you would just go back and read what I posted in #35, and do the steps exactly as I described them, instead of trying to take shortcuts, you will (a) get the right answer, and (b) stop wasting other people's time by forcing them to repeat things over and over again because you didn't follow directions.

#### NoahsArk

Gold Member
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)(t,x)(t, x). Then, for each event, you use the LT equations for (t′,x′)(t′,x′) in terms of (t,x)(t,x) to obtain the coordinates in the primed (Enterprise) frame, (t′,x′)(t′,x′), for each of the events.

In this particular case, you are trying to find the relative velocity v (and its associated γγ\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 γ, 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.
Ok, I'll start by trying to find to find the $t^\prime and x^\prime$ coordinates for Caesar's death:

$t^\prime = \gamma t - v \gamma x$
$t^\prime = \gamma 0 - v \gamma 0$
$t^\prime = 0$

$x^\prime = \gamma x - \gamma vt$
$x^\prime = \gamma 0 - \gamma v0$
$x^\prime = 0$

Strange results so far because I am getting the prime coordinates for Caesar's death to be 0,0, which are the same coordinates for the unprimed frame. Now I will try and get the primed coordinates for the firecracker explosion.

$t^\prime = \gamma t - v \gamma x$
$t^\prime = \gamma 2,000 - v \gamma 2,000,000$
$0 = 1.0005(2,000) - .001(1.0005 x 2,000,000)$
$0 = 2,001 - 2001$

Now I'll try and find the $x^\prime$ coordinate for the fire cracker explosion:

$x^\prime = \gamma x - \gamma vt$
$x^\prime = 1.005 x 2,000,000.00 - 1.005 x .001(2,000)$
$x^\prime = 2,001,000 - 2.01$
$x^\prime = 2,000,997.99$

This just feels like I've done something wrong so far The last step, which is to find the distance in the prime frame between earth and firecracker, is to subtract the x prime values. When I do that I get 2,000,997.99.

I think you will find this problem much easier if you don’t skip these. They’re there for a reason.
The work above attempts to solve part c. Part b asks us to draw a space time diagram which I am unable to do so far inside of a message in the forum. I also don't have the method down about how to draw a space time diagram. I was, though, able to follow the diagram given by the book which shows a horizontal line of simultaneity between me reading the exercise and the firecracker, and a diagonal line of simultaneity in the Enterprise frame between Caesar's death and the firecracker.

#### Nugatory

Mentor
Strange results so far because I am getting the prime coordinates for Caesar's death to be 0,0, which are the same coordinates for the unprimed frame. Now I will try and get the primed coordinates for the firecracker explosion.
That's not strange at all, it's a successful sanity check on your calculation. The origin of the primed coordinate system is the Caesar's death event - quoting from your very first post in this thread we have "Take Caesar's murder to be the reference event for the Enterprise too (X1O = 0, t1O)" so its coordinates in the primed frame are supposed to be (x'=0, t'=0).

#### PeterDonis

Mentor
Strange results so far because I am getting the prime coordinates for Caesar's death to be 0,0, which are the same coordinates for the unprimed frame.
Yes, that's because Caesar's death happens to be the spacetime origin $(0, 0)$, and the origin's coordinates by definition are the same in every frame. (More precisely, they are as long as we're only using Lorentz transformations; there is a broader class of transformations that allows the origin to be moved.)

Now I will try and get the primed coordinates for the firecracker explosion.
You're using the wrong value for $\gamma$. It doesn't make a difference in the $t'$ equation--in fact, what the $t'$ equation is actually for is to figure out that $v = .001$, since that's the value of $v$ that makes $t' = 0$, and the criterion for the Enterprise frame is that Caesar's death and the firecracker explosion should be simultaneous in this frame. But you'll notice that $\gamma$ drops out of the $t'$ equation when you have $t' = 0$, so using the wrong value for $\gamma$ doesn't affect that equation. But it does affect the $x'$ equation.

Try recomputing the value of $\gamma$ for $v = .001$, and then try the $x'$ equation again.

#### NoahsArk

Gold Member
Thank you for the help. I intend to keep at this but have been swamped for a while. I will try and rework this this weekend.

#### NoahsArk

Gold Member
Try recomputing the value of γ for v=.001, and then try the x′ equation again.
$\gamma = \frac {1} {\sqrt {1-v^2}}$
$\gamma = \frac {1} {\sqrt {1-.001^2}}$
$\gamma = \frac {1} {\sqrt {1 - .000001}}$
$\gamma = \frac {1} {\sqrt {.999999}}$
$\gamma = \frac {1} {.9999995}$
$\gamma = 1.0000005$

Now I'll try recomputing $x \prime$

$x \prime = \gamma x - \gamma vt$
$x \prime = 1.0000005(2,000,000) - 1.0000005(.001)(2,000)$
$x \prime = 1.0000005(2,000,000) - 1.0000005(.001)(2,000)$
$x \prime = 2,00,001 - 2.000001$
$x \prime = 1,999,999$

So the distance in the Enterprise frame from Earth to firecracker is 1,999,999. This makes more sense now since I am getting a value lower than in the Earth frame.

Regarding the same coordinates in both frames for Caesar's death:

That's not strange at all, it's a successful sanity check on your calculation.
Yes, that's because Caesar's death happens to be the spacetime origin (0,0)(0,0), and the origin's coordinates by definition are the same in every frame.
I had to think for a bit on why this is true. It would be easier for me to understand why this is true if the Enterprise were flying past earth during Caesar's death. Is the reason why it's 0,0 in both frames because both frames are using the same coordinate plane with 0,0 as the origin?

Thank you.

#### PeterDonis

Mentor
the distance in the Enterprise frame from Earth to firecracker is 1,999,999.
Yes.

I had to think for a bit on why this is true.
Just look at the Lorentz transformation equations. What are $t'$ and $x'$ if $t = 0$ and $x = 0$?

It would be easier for me to understand why this is true if the Enterprise were flying past earth during Caesar's death.
Yes, that is more like the usual convention in SR problems: if you have two observers, make the spacetime origin the point where their worldlines cross. But this particular problem doesn't do that.

"Julius Caesar Problem from a SpaceTime Physics book"

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